Theorems and Proofs about Pisot sequences with initial terms a<, 20, a+2<=b<=, 40, that satisfy Non-Trivial Linear Recurrences With Constant Coefficients of Order at most, 6 By Shalosh B. Ekhad ------------------------------------------------------------------------- Theorem , 1, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 5, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) with initial conditions, a(1) = 2, a(2) = 5, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) with initial conditions, b(1) = 2, b(2) = 5, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t + 1 = 0 whose roots are 1/2 1/2 5 5 [3/2 + ----, 3/2 - ----] 2 2 In floating-point [2.618033988, 0.381966012] The largest root is, 2.618033988 and the remaining roots are [0.381966012] whose absolute values are [0.381966012] so the largest absolute value is, 0.381966012 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.381966012 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2000000000, -0.07692307692, -0.02941176471, -0.01123595506, -0.004291845494, -0.001639344262, -0.0006261740764, -0.0002391772303, -5 -0.00009135757354, -0.00003489548801, -0.00001332889037, -0.5091183089 10 , -5 -6 -6 -0.1944658897 10 , -0.7427936022 10 , -0.2837219094 10 , -6 -7 -7 -0.1083721260 10 , -0.4139446871 10 , -0.1581128010 10 , -8 -8 -9 -0.6039371593 10 , -0.2306834678 10 , -0.8811324406 10 , -9 -9 -10 -0.3365626437 10 , -0.1285554906 10 , -0.4910382795 10 , -10 -11 -11 -0.1875599330 10 , -0.7164151948 10 , -0.2736462543 10 , -11 -12 -12 -0.1045235683 10 , -0.3992445045 10 , -0.1524978309 10 , -13 -13 -14 -0.5824898820 10 , -0.2224913368 10 , -0.8498412846 10 , -14 -14 -15 -0.3246104857 10 , -0.1239901724 10 , -0.4736003159 10 , -15 -16 -16 -0.1808992236 10 , -0.6909735488 10 , -0.2639284103 10 , -16 -17 -17 -0.1008116821 10 , -0.3850663611 10 , -0.1470822620 10 , -18 -18 -19 -0.5618042495 10 , -0.2145901283 10 , -0.8196613536 10 , -19 -19 -20 -0.3130827778 10 , -0.1195869798 10 , -0.4567816168 10 , -20 -21 -21 -0.1744750522 10 , -0.6664353975 10 , -0.2545556705 10 , -22 -22 -22 -0.9723161412 10 , -0.3713917181 10 , -0.1418590132 10 , -23 -23 -24 -0.5418532142 10 , -0.2069695109 10 , -0.7905531854 10 ] The largest is -24 -0.7905531854 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 2, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 7, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 7, 25, 89, 317, 1129, 4021, 14321, 51005, 181657, 646981, 2304257, 8206733, 29228713, 104099605, 370756241, 1320467933, 4702916281, 16749684709, 59654886689] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 2, a(2) = 7, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 2, b(2) = 7, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 2 = 0 whose roots are 1/2 1/2 17 17 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.561552813, -0.561552813] The largest root is, 3.561552813 and the remaining roots are [-0.561552813] whose absolute values are [0.561552813] so the largest absolute value is, 0.561552813 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.561552813 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2857142857, -0.1600000000, 0.08988764045, -0.05047318612, 0.02834366696, -0.01591643870, 0.008937923329, -0.005019115773, 0.002818498599, -0.001582735814, 0.0008887897487, -0.0004991023834, 0.0002802723473, -0.0001573877250, 0.00008838151965, -0.00004963089096, -5 -5 0.00002787036642, -0.00001565068266, 0.8788684869 10 , -0.4935310709 10 , -5 -5 -6 0.2771437611 10 , -0.1556308586 10 , 0.8739494640 10 , -6 -6 -6 -0.4907687797 10 , 0.2755925887 10 , -0.1547597934 10 , -7 -7 -7 0.8690579728 10 , -0.4880219491 10 , 0.2740500982 10 , -7 -8 -8 -0.1538936035 10 , 0.8641938593 10 , -0.4852904925 10 , -8 -8 -9 0.2725162411 10 , -0.1530322617 10 , 0.8593569702 10 , -9 -9 -9 -0.4825743238 10 , 0.2709909689 10 , -0.1521757408 10 , -10 -10 -10 0.8545471532 10 , -0.4798733575 10 , 0.2694742337 10 , -10 -11 -11 -0.1513240139 10 , 0.8497642566 10 , -0.4771875085 10 , -11 -11 -12 0.2679659877 10 , -0.1504770541 10 , 0.8450081300 10 , -12 -12 -12 -0.4745166922 10 , 0.2664661832 10 , -0.1496348347 10 , -13 -13 -13 0.8402786233 10 , -0.4718608245 10 , 0.2649747732 10 , -13 -14 -14 -0.1487973292 10 , 0.8355755877 10 , -0.4692198216 10 , -14 0.2634917106 10 ] The largest is 0.2857142857 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 3, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 9, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 9, 41, 187, 853, 3891, 17749, 80963, 369317, 1684659, 7684661, 35053987, 159900613, 729395091, 3327174229, 15177080963, 69231056357, 315801119859, 1440543486581, 6571115193187] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 2 a(n - 2) with initial conditions, a(1) = 2, a(2) = 9, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 2 b(n - 2) with initial conditions, b(1) = 2, b(2) = 9, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 5 t + 2 = 0 whose roots are 1/2 1/2 17 17 [5/2 + -----, 5/2 - -----] 2 2 In floating-point [4.561552813, 0.438447187] The largest root is, 4.561552813 and the remaining roots are [0.438447187] whose absolute values are [0.438447187] so the largest absolute value is, 0.438447187 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.438447187 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2222222222, -0.09756097561, -0.04278074866, -0.01875732708, -0.008224106913, -0.003605836949, -0.001580969085, -0.0006931714489, -0.0003039190720, -0.0001332524623, -0.00005842416727, -0.00002561591180, -5 -5 -0.00001123122448, -0.4924298781 10 , -0.2159044949 10 , -6 -6 -6 -0.9466271851 10 , -0.4150460266 10 , -0.1819757629 10 , -7 -7 -7 -0.7978676139 10 , -0.3498228111 10 , -0.1533788275 10 , -8 -8 -8 -0.6724851551 10 , -0.2948492247 10 , -0.1292758132 10 , -9 -9 -9 -0.5668061667 10 , -0.2485145695 10 , -0.1089605140 10 , -10 -10 -11 -0.4777343086 10 , -0.2094612638 10 , -0.9183770195 10 , -11 -11 -12 -0.4026598210 10 , -0.1765450659 10 , -0.7740568756 10 , -12 -12 -13 -0.3393830598 10 , -0.1488015480 10 , -0.6524162015 10 , -13 -13 -14 -0.2860500484 10 , -0.1254178391 10 , -0.5498909879 10 , -14 -14 -15 -0.2410981569 10 , -0.1057088087 10 , -0.4634772985 10 , -15 -16 -16 -0.2032103179 10 , -0.8909699228 10 , -0.3906432565 10 , -16 -17 -17 -0.1712764370 10 , -0.7509567204 10 , -0.3292548618 10 , -17 -18 -18 -0.1443608680 10 , -0.6329461652 10 , -0.2775134658 10 , -18 -19 -19 -0.1216749985 10 , -0.5334806083 10 , -0.2339030721 10 , -19 -20 -20 -0.1025541441 10 , -0.4496457600 10 , -0.1971459187 10 ] The largest is -20 -0.1971459187 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 4, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 11, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 11, 61, 338, 1873, 10379, 57514, 318707, 1766077, 9786506, 54230761, 300513323, 1665258898, 9227834459, 51134948989, 283358248322, 1570196088577, 8701055187851, 48215864204986, 267182486588483] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) + 3 a(n - 2) with initial conditions, a(1) = 2, a(2) = 11, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) + 3 b(n - 2) with initial conditions, b(1) = 2, b(2) = 11, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 5 t - 3 = 0 whose roots are 1/2 1/2 37 37 [5/2 + -----, 5/2 - -----] 2 2 In floating-point [5.541381265, -0.541381265] The largest root is, 5.541381265 and the remaining roots are [-0.541381265] whose absolute values are [0.541381265] so the largest absolute value is, 0.541381265 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.541381265 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2727272727, -0.1475409836, 0.07988165680, -0.04324612920, 0.02341266018, -0.01267517474, 0.006862102182, -0.003715013558, 0.002011238740, -0.001088846974, 0.0005894813522, -0.0003191341602, 0.0001727732554, -0.00009353620361, 0.00005063874825, -0.00002741486959, -5 -5 -5 0.00001484189678, -0.8035124858 10 , 0.4350066061 10 , -0.2355044268 10 , -5 -6 -6 0.1274976845 10 , -0.6902485774 10 , 0.3736876481 10 , -6 -6 -7 -0.2023074917 10 , 0.1095254858 10 , -0.5929504607 10 , -7 -7 -8 0.3210122706 10 , -0.1737900292 10 , 0.9408666587 10 , -8 -8 -8 -0.5093675820 10 , 0.2757620660 10 , -0.1492924162 10 , -9 -9 -9 0.8082411714 10 , -0.4375666279 10 , 0.2368903746 10 , -9 -10 -10 -0.1282480107 10 , 0.6943107029 10 , -0.3758868067 10 , -10 -10 -11 0.2034980750 10 , -0.1101700453 10 , 0.5964399850 10 , -11 -11 -12 -0.3229014337 10 , 0.1748127867 10 , -0.9464036762 10 , -12 -12 -12 0.5123652195 10 , -0.2773849308 10 , 0.1501710048 10 , -13 -13 -13 -0.8129976854 10 , 0.4401417155 10 , -0.2382844788 10 , -13 -14 -14 0.1290027526 10 , -0.6983967340 10 , 0.3780989075 10 , -14 -14 -15 -0.2046956649 10 , 0.1108183980 10 , -0.5999500452 10 , -15 0.3248017145 10 ] The largest is 0.2727272727 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 5, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 13, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 13, 85, 556, 3637, 23791, 155626, 1018009, 6659185, 43560268, 284944321, 1863929443, 12192673138, 79756923637, 521720446045, 3412772351404, 22324245121693, 146031398797639, 955247056218394, 6248635197135841] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) - 3 a(n - 2) with initial conditions, a(1) = 2, a(2) = 13, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) - 3 b(n - 2) with initial conditions, b(1) = 2, b(2) = 13, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 7 t + 3 = 0 whose roots are 1/2 1/2 37 37 [7/2 + -----, 7/2 - -----] 2 2 In floating-point [6.541381265, 0.458618735] The largest root is, 6.541381265 and the remaining roots are [0.458618735] whose absolute values are [0.458618735] so the largest absolute value is, 0.458618735 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.458618735 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2307692308, -0.1058823529, -0.04856115108, -0.02227110256, -0.01021394645, -0.004684307249, -0.002148311066, -0.0009852557032, -0.0004518567241, -0.0002072299591, -0.00009503954169, -0.00004358691437, -5 -5 -0.00001998977552, -0.9167685561 10 , -0.4204472353 10 , -5 -6 -6 -0.1928249791 10 , -0.8843314798 10 , -0.4055709845 10 , -6 -7 -7 -0.1860024518 10 , -0.8530420912 10 , -0.3912210846 10 , -7 -8 -8 -0.1794213189 10 , -0.8228597827 10 , -0.3773789125 10 , -8 -9 -9 -0.1730730394 10 , -0.7937453837 10 , -0.3640265037 10 , -9 -10 -10 -0.1669493746 10 , -0.7656611095 10 , -0.3511465294 10 , -10 -11 -11 -0.1610423770 10 , -0.7385705122 10 , -0.3387222739 10 , -11 -12 -12 -0.1553443807 10 , -0.7124384335 10 , -0.3267376130 10 , -12 -13 -13 -0.1498479907 10 , -0.6872309592 10 , -0.3151769931 10 , -13 -14 -14 -0.1445460738 10 , -0.6629153750 10 , -0.3040254106 10 , -14 -15 -15 -0.1394317492 10 , -0.6394601240 10 , -0.2932683931 10 , -15 -16 -16 -0.1344983794 10 , -0.6168347660 10 , -0.2828919800 10 , -16 -17 -17 -0.1297395620 10 , -0.5950099377 10 , -0.2728827048 10 , -17 -18 -18 -0.1251491209 10 , -0.5739573148 10 , -0.2632275776 10 , -18 -19 -19 -0.1207210986 10 , -0.5536495751 10 , -0.2539140677 10 ] The largest is -19 -0.2539140677 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 6, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 15, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 15, 113, 851, 6409, 48267, 363505, 2737603, 20617241, 155271099, 1169366657, 8806650995, 66324023593, 499494769131, 3761759478289, 28330295424547, 213359105884985, 1606834922893083, 12101280883791521, 91136305878112979] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) + 4 a(n - 2) with initial conditions, a(1) = 2, a(2) = 15, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) + 4 b(n - 2) with initial conditions, b(1) = 2, b(2) = 15, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 7 t - 4 = 0 whose roots are 1/2 1/2 65 65 [7/2 + -----, 7/2 - -----] 2 2 In floating-point [7.531128874, -0.531128874] The largest root is, 7.531128874 and the remaining roots are [-0.531128874] whose absolute values are [0.531128874] so the largest absolute value is, 0.531128874 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.531128874 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2666666667, -0.1415929204, 0.07520564042, -0.03994382899, 0.02121532310, -0.01126807059, 0.005984797650, -0.003178698838, 0.001688298735, -0.0008967042063, 0.0004762654955, -0.0002529583564, 0.0001343534871, -0.00007135901632, 0.00003790083400, -0.00002013022729, -5 -5 -5 0.00001069174496, -0.5678694462 10 , 0.3016118596 10 , -0.1601947674 10 , -6 -6 -6 0.8508406647 10 , -0.4519060443 10 , 0.2400203485 10 , -6 -7 -7 -0.1274817375 10 , 0.6770923171 10 , -0.3596232801 10 , -7 -7 -8 0.1910063079 10 , -0.1014489652 10 , 0.5388247470 10 , -8 -8 -9 -0.2861853812 10 , 0.1520013193 10 , -0.8073228960 10 , -9 -9 -9 0.4287925008 10 , -0.2277440782 10 , 0.1209614559 10 , -10 -10 -10 -0.6424612186 10 , 0.3412297037 10 , -0.1812369484 10 , -11 -11 -11 0.9626017635 10 , -0.5112655909 10 , 0.2715479177 10 , -11 -12 -12 -0.1442269398 10 , 0.7660309215 10 , -0.4068611409 10 , -12 -12 -13 0.2160956997 10 , -0.1147746657 10 , 0.6096013897 10 , -13 -13 -14 -0.3237768998 10 , 0.1719672603 10 , -0.9133677734 10 , -14 -14 -14 0.4851159971 10 , -0.2576591134 10 , 0.1368501948 10 , -15 -15 -15 -0.7268508990 10 , 0.3860514997 10 , -0.2050430984 10 , -15 0.1089043100 10 ] The largest is 0.2666666667 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 7, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 17, 145, 1237, 10553, 90029, 768049, 6552325, 55898729, 476879261, 4068318433, 34707348853, 296092865945, 2526006398093, 21549686119057, 183843149479141, 1568389600836041, 13380133809607805, 114147645883126081, 973808277709703509] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 9 a(n - 1) - 4 a(n - 2) with initial conditions, a(1) = 2, a(2) = 17, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 9 b(n - 1) - 4 b(n - 2) with initial conditions, b(1) = 2, b(2) = 17, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 9 t + 4 = 0 whose roots are 1/2 1/2 65 65 [9/2 + -----, 9/2 - -----] 2 2 In floating-point [8.531128874, 0.468871126] The largest root is, 8.531128874 and the remaining roots are [0.468871126] whose absolute values are [0.468871126] so the largest absolute value is, 0.468871126 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.468871126 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2352941176, -0.1103448276, -0.05173807599, -0.02425850469, -0.01137411279, -0.005332993077, -0.002500486469, -0.001172405906, -0.0005497072770, -0.0002577418698, -0.0001208477207, -0.00005666200686, -5 -0.00002656717895, -0.00001245658310, -0.5840532144 10 , -5 -5 -6 -0.2738456882 10 , -0.1283983361 10 , -0.6020227242 10 , -6 -6 -7 -0.2822710725 10 , -0.1323487556 10 , -0.6205451002 10 , -7 -7 -8 -0.2909556798 10 , -0.1364207171 10 , -0.6396373524 10 , -8 -8 -9 -0.2999074855 10 , -0.1406179604 10 , -0.6593170141 10 , -9 -9 -10 -0.3091347107 10 , -0.1449443398 10 , -0.6796021581 10 , -10 -10 -11 -0.3186458290 10 , -0.1494038286 10 , -0.7005114131 10 , -11 -11 -12 -0.3284495749 10 , -0.1540005220 10 , -0.7220639813 10 , -12 -12 -13 -0.3385549518 10 , -0.1587386414 10 , -0.7442796552 10 , -13 -13 -14 -0.3489712399 10 , -0.1636225381 10 , -0.7671788367 10 , -14 -14 -15 -0.3597080049 10 , -0.1686566972 10 , -0.7907825551 10 , -15 -15 -16 -0.3707751069 10 , -0.1738457418 10 , -0.8151124869 10 , -16 -16 -17 -0.3821827094 10 , -0.1791944373 10 , -0.8401909754 10 , -17 -17 -18 -0.3939412886 10 , -0.1847076955 10 , -0.8660410514 10 , -18 -18 -19 -0.4060616428 10 , -0.1903905796 10 , -0.8926864542 10 ] The largest is -19 -0.8926864542 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 8, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 19, 181, 1724, 16421, 156409, 1489786, 14190119, 135160001, 1287390604, 12262315441, 116797791989, 1112491705106, 10596414305899, 100930187278621, 961353757037084, 9156834749726861, 87218281532727169, 830748707543178826, 7912829775552245279] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 9 a(n - 1) + 5 a(n - 2) with initial conditions, a(1) = 2, a(2) = 19, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 9 b(n - 1) + 5 b(n - 2) with initial conditions, b(1) = 2, b(2) = 19, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 9 t - 5 = 0 whose roots are 1/2 1/2 101 101 [9/2 + ------, 9/2 - ------] 2 2 In floating-point [9.524937810, -0.524937810] The largest root is, 9.524937810 and the remaining roots are [-0.524937810] whose absolute values are [0.524937810] so the largest absolute value is, 0.524937810 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.524937810 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2631578947, -0.1381215470, 0.07250580046, -0.03806101943, 0.01997966869, -0.01048808352, 0.005505591602, -0.002890093201, 0.001517119197, -0.0007963932299, 0.0004180569184, -0.0002194538835, 0.0001151996411, -0.00006047264738, 0.00003174437911, -0.00001666382487, -5 -5 -5 0.8747471742 10 , -0.4591878664 10 , 0.2410450732 10 , -5 -6 -6 -0.1265336730 10 , 0.6642230926 10 , -0.3486758160 10 , -6 -7 -7 0.1830331194 10 , -0.9608100498 10 , 0.5043655239 10 , -7 -7 -8 -0.2647605338 10 , 0.1389828150 10 , -0.7295733459 10 , -8 -8 -8 0.3829806348 10 , -0.2010410159 10 , 0.1055340307 10 , -9 -9 -9 -0.5539880303 10 , 0.2908092637 10 , -0.1526567782 10 , -10 -10 -10 0.8013531491 10 , -0.4206605676 10 , 0.2208206373 10 , -10 -11 -11 -0.1159171019 10 , 0.6084926967 10 , -0.3194208240 10 , -11 -12 -12 0.1676760680 10 , -0.8801950800 10 , 0.4620476782 10 , -12 -12 -13 -0.2425462966 10 , 0.1273217219 10 , -0.6683598592 10 , -13 -13 -14 0.3508473611 10 , -0.1841730456 10 , 0.9667939532 10 , -14 -14 -14 -0.5075067011 10 , 0.2664094565 10 , -0.1398483968 10 , -15 -15 -15 0.7341171123 10 , -0.3853658296 10 , 0.2022930949 10 , -15 -16 -0.1061912943 10 , 0.5574382554 10 ] The largest is 0.2631578947 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 9, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 21, 221, 2326, 24481, 257661, 2711866, 28542221, 300405101, 3161745006, 33277169561, 350240140141, 3686255693746, 38797611930501, 408342452766781, 4297778920782086, 45233855864769041, 476083519908549021, 5010749439670194026, 52737826236829389181] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 11 a(n - 1) - 5 a(n - 2) with initial conditions, a(1) = 2, a(2) = 21, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 11 b(n - 1) - 5 b(n - 2) with initial conditions, b(1) = 2, b(2) = 21, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 11 t + 5 = 0 whose roots are 1/2 1/2 101 101 [11/2 + ------, 11/2 - ------] 2 2 In floating-point [10.52493781, 0.475062190] The largest root is, 10.52493781 and the remaining roots are [0.475062190] whose absolute values are [0.475062190] so the largest absolute value is, 0.475062190 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.475062190 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2380952381, -0.1131221719, -0.05374032674, -0.02553000286, -0.01212833917, -0.005761715365, -0.002737173116, -0.001300327453, -0.0006177364070, -0.0002934632100, -0.0001394132751, -0.00006622997569, -5 -0.00003146335726, -0.00001494705139, -0.7100778958 10 , -5 -5 -6 -0.3373311598 10 , -0.1602532794 10 , -0.7613027376 10 , -6 -6 -7 -0.3616661453 10 , -0.1718139108 10 , -0.8162229266 10 , -7 -7 -8 -0.3877566506 10 , -0.1842085234 10 , -0.8751050444 10 , -8 -8 -9 -0.4157293184 10 , -0.1974972802 10 , -0.9382349034 10 , -9 -9 -9 -0.4457199274 10 , -0.2117446846 10 , -0.1005918935 10 , -10 -10 -10 -0.4778740515 10 , -0.2270198932 10 , -0.1078485675 10 , -11 -11 -11 -0.5123477661 10 , -0.2433970515 10 , -0.1156287362 10 , -12 -12 -12 -0.5493084058 10 , -0.2609556539 10 , -0.1239701643 10 , -13 -13 -13 -0.5889353768 10 , -0.2797809295 10 , -0.1329133409 10 , -14 -14 -14 -0.6314210275 10 , -0.2999642558 10 , -0.1425016761 10 , -15 -15 -15 -0.6769715825 10 , -0.3216036022 10 , -0.1527817114 10 , -16 -16 -16 -0.7258081432 10 , -0.3448040056 10 , -0.1638033458 10 , -17 -17 -17 -0.7781677611 10 , -0.3696780803 10 , -0.1756200782 10 , -18 -18 -18 -0.8343045888 10 , -0.3963465646 10 , -0.1882892668 10 ] The largest is -18 -0.1882892668 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 10, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 23, 265, 3053, 35173, 405221, 4668469, 53784485, 619640149, 7138748549, 82244074933, 947517315557, 10916154920725, 125762808021317, 1448887817758837, 16692342843475109, 192309098184779221, 2215554137093422085, 25524950097136318261, 294067775891060033381] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 11 a(n - 1) + 6 a(n - 2) with initial conditions, a(1) = 2, a(2) = 23, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 11 b(n - 1) + 6 b(n - 2) with initial conditions, b(1) = 2, b(2) = 23, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 11 t - 6 = 0 whose roots are 1/2 1/2 145 145 [11/2 + ------, 11/2 - ------] 2 2 In floating-point [11.52079729, -0.520797290] The largest root is, 11.52079729 and the remaining roots are [-0.520797290] whose absolute values are [0.520797290] so the largest absolute value is, 0.520797290 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.520797290 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2608695652, -0.1358490566, 0.07075008189, -0.03684644472, 0.01918952868, -0.009993854516, 0.005204772343, -0.002710631328, 0.001411689448, -0.0007352040381, 0.0003828922702, -0.0001994092564, 0.0001038518002, -0.00005408573606, 0.00002816770474, -0.00001466966428, -5 -5 -5 0.7639921391 10 , -0.3978850352 10 , 0.2072174478 10 , -5 -6 -6 -0.1079182851 10 , 0.5620355037 10 , -0.2927065669 10 , -6 -7 -7 0.1524407866 10 , -0.7939074847 10 , 0.4134648660 10 , -7 -7 -8 -0.2153313815 10 , 0.1121439998 10 , -0.5840429112 10 , -8 -8 -9 0.3041679650 10 , -0.1584098517 10 , 0.8249942139 10 , -9 -9 -9 -0.4296547504 10 , 0.2237630294 10 , -0.1165351792 10 , -10 -10 -10 0.6069120542 10 , -0.3160781528 10 , 0.1646126452 10 , -11 -11 -11 -0.8572981942 10 , 0.4464785757 10 , -0.2325248320 10 , -11 -12 -12 0.1210983022 10 , -0.6306766755 10 , 0.3284547031 10 , -12 -13 -13 -0.1710583191 10 , 0.8908670890 10 , -0.4639611651 10 , -13 -13 -14 0.2416297172 10 , -0.1258401017 10 , 0.6553718389 10 , -14 -14 -15 -0.3413158772 10 , 0.1777563837 10 , -0.9257504280 10 , -15 -15 -15 0.4821283136 10 , -0.2510911188 10 , 0.1307675741 10 , -16 -16 -0.6810339812 10 , 0.3546806514 10 ] The largest is 0.2608695652 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 11, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 25, 313, 3919, 49069, 614383, 7692565, 96317047, 1205966221, 15099658591, 189059764357, 2367178985095, 29638968220093, 371103512950639, 4646511859037749, 58178033089786903, 728435359013003245, 9120591468630320767, 114197076938116150501, 1429838451383728031911] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 13 a(n - 1) - 6 a(n - 2) with initial conditions, a(1) = 2, a(2) = 25, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 13 b(n - 1) - 6 b(n - 2) with initial conditions, b(1) = 2, b(2) = 25, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 13 t + 6 = 0 whose roots are 1/2 1/2 145 145 [13/2 + ------, 13/2 - ------] 2 2 In floating-point [12.52079729, 0.479202710] The largest root is, 12.52079729 and the remaining roots are [0.479202710] whose absolute values are [0.479202710] so the largest absolute value is, 0.479202710 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.479202710 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2400000000, -0.1150159744, -0.05511610105, -0.02641178748, -0.01265660020, -0.006065077123, -0.002906401397, -0.001392755428, -0.0006674121762, -0.0003198257239, -0.0001532613538, -0.00007344325618, -5 -0.00003519420744, -0.00001686515960, -0.8081830196 10 , -5 -5 -6 -0.3872834937 10 , -0.1855872999 10 , -0.8893393718 10 , -6 -6 -7 -0.4261738376 10 , -0.2042236582 10 , -0.9786453057 10 , -7 -7 -7 -0.4689694832 10 , -0.2247314475 10 , -0.1076919188 10 , -8 -8 -8 -0.5160625941 10 , -0.2472985939 10 , -0.1185061565 10 , -9 -9 -9 -0.5678847144 10 , -0.2721318944 10 , -0.1304063415 10 , -10 -10 -10 -0.6249107230 10 , -0.2994589124 10 , -0.1435015225 10 , -11 -11 -11 -0.6876631857 10 , -0.3295300626 10 , -0.1579116992 10 , -12 -12 -12 -0.7567171430 10 , -0.3626209061 10 , -0.1737689211 10 , -13 -13 -13 -0.8327053801 10 , -0.3990346753 10 , -0.1912184980 10 , -14 -14 -14 -0.9163242257 10 , -0.4391050528 10 , -0.2104203315 10 , -14 -15 -15 -0.1008339932 10 , -0.4831992288 10 , -0.2315503802 10 , -15 -16 -16 -0.1109595698 10 , -0.5317212663 10 , -0.2548022721 10 , -16 -17 -17 -0.1221019395 10 , -0.5851158036 10 , -0.2803890791 10 , -17 -18 -18 -0.1343632067 10 , -0.6438721287 10 , -0.3085452693 10 ] The largest is -18 -0.3085452693 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 12, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 27, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 27, 365, 4934, 66697, 901599, 12187666, 164750851, 2227074725, 30105227382, 406957479041, 5501183819207, 74364092002978, 1005241482773163, 13588687920071965, 183689633340347686, 2483086048865023673, 33565946068627741551, 453738901234215805874, 6133567338525199667219] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 13 a(n - 1) + 7 a(n - 2) with initial conditions, a(1) = 2, a(2) = 27, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 13 b(n - 1) + 7 b(n - 2) with initial conditions, b(1) = 2, b(2) = 27, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 13 t - 7 = 0 whose roots are 1/2 1/2 197 197 [13/2 + ------, 13/2 - ------] 2 2 In floating-point [13.51783442, -0.517834425] The largest root is, 13.51783442 and the remaining roots are [-0.517834425] whose absolute values are [0.517834425] so the largest absolute value is, 0.517834425 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.517834425 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2592592593, -0.1342465753, 0.06951763275, -0.03599862063, 0.01864132502, -0.009653119802, 0.004998717730, -0.002588508116, 0.001340418609, -0.0006941148979, 0.0003594365882, -0.0001861286385, 0.00009638381629, -0.00004991085798, 0.00002584556038, -0.00001338372087, -5 -5 -5 0.6930551384 10 , -0.3588878083 10 , 0.1858444614 10 , -6 -6 -6 -0.9623665959 10 , 0.4983465517 10 , -0.2580609995 10 , -6 -7 -7 0.1336328690 10 , -0.6919969970 10 , 0.3583398662 10 , -7 -8 -8 -0.1855607181 10 , 0.9608972756 10 , -0.4975856871 10 , -8 -8 -9 0.2576669976 10 , -0.1334288412 10 , 0.6909404711 10 , -9 -9 -10 -0.3577927607 10 , 0.1852774081 10 , -0.9594301987 10 , -10 -10 -10 0.4968259841 10 , -0.2572735972 10 , 0.1332251250 10 , -11 -11 -11 -0.6898855583 10 , 0.3572464906 10 , -0.1849945306 10 , -12 -12 -12 0.9579653616 10 , -0.4960674411 10 , 0.2568807975 10 , -12 -13 -13 -0.1330217198 10 , 0.6888322561 10 , -0.3567010544 10 , -13 -14 -14 0.1847120850 10 , -0.9565027611 10 , 0.4953100562 10 , -14 -14 -15 -0.2564885975 10 , 0.1328186251 10 , -0.6877805621 10 , -15 -15 -16 0.3561564511 10 , -0.1844300706 10 , 0.9550423936 10 , -16 -16 -0.4945538276 10 , 0.2560969964 10 ] The largest is 0.2592592593 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 13, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 29, 421, 6112, 88733, 1288211, 18702034, 271513033, 3941781257, 57226127624, 830799445561, 12061408790047, 175105535731778, 2542153174446341, 36906558866572669, 535803310777465648, 7778703749595976037, 112929933068497381019, 1639498069780288883026, 23801961515224851578257] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 15 a(n - 1) - 7 a(n - 2) with initial conditions, a(1) = 2, a(2) = 29, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 15 b(n - 1) - 7 b(n - 2) with initial conditions, b(1) = 2, b(2) = 29, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 15 t + 7 = 0 whose roots are 1/2 1/2 197 197 [15/2 + ------, 15/2 - ------] 2 2 In floating-point [14.51783442, 0.482165575] The largest root is, 14.51783442 and the remaining roots are [0.482165575] whose absolute values are [0.482165575] so the largest absolute value is, 0.482165575 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.482165575 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2413793103, -0.1163895487, -0.05611910995, -0.02705870420, -0.01304677572, -0.006290706134, -0.003033161948, -0.001462486278, -0.0007051605390, -0.0003400041376, -0.0001639382909, -0.00007904540050, -5 -0.00003811297108, -0.00001837676266, -0.8860642356 10 , -5 -5 -6 -0.4272296727 10 , -0.2059954413 10 , -0.9932391065 10 , -6 -6 -6 -0.4789057061 10 , -0.2309118457 10 , -0.1113377431 10 , -7 -7 -7 -0.5368322707 10 , -0.2588420411 10 , -0.1248047219 10 , -8 -8 -8 -0.6017654065 10 , -0.2901505639 10 , -0.1399006138 10 , -9 -9 -9 -0.6745526008 10 , -0.3252460435 10 , -0.1568224459 10 , -10 -10 -10 -0.7561438501 10 , -0.3645865352 10 , -0.1757910768 10 , -11 -11 -11 -0.8476040583 10 , -0.4086854992 10 , -0.1970540792 10 , -12 -12 -12 -0.9501269363 10 , -0.4581185017 10 , -0.2208889713 10 , -12 -13 -13 -0.1065050581 10 , -0.5135307273 10 , -0.2476068390 10 , -13 -14 -14 -0.1193874942 10 , -0.5756453993 10 , -0.2775563956 10 , -14 -15 -15 -0.1338281394 10 , -0.6452732196 10 , -0.3111285337 10 , -15 -16 -16 -0.1500154687 10 , -0.7233229492 10 , -0.3487614265 10 , -16 -17 -17 -0.1681607542 10 , -0.8108132693 10 , -0.3909462472 10 , -17 -18 -18 -0.1885008225 10 , -0.9088860771 10 , -0.4382335791 10 ] The largest is -18 -0.4382335791 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 14, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 31, 481, 7463, 115793, 1796599, 27875329, 432502727, 6710543537, 104118174871, 1615456971361, 25064799969383, 388895655311633, 6033953229429559, 93620463683936449, 1452578581094483207, 22537642425888739697, 349685265037086961111, 5425580114963414334241, 84181183844747910702503] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 15 a(n - 1) + 8 a(n - 2) with initial conditions, a(1) = 2, a(2) = 31, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 15 b(n - 1) + 8 b(n - 2) with initial conditions, b(1) = 2, b(2) = 31, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 15 t - 8 = 0 whose roots are 1/2 1/2 257 257 [15/2 + ------, 15/2 - ------] 2 2 In floating-point [15.51560977, -0.515609770] The largest root is, 15.51560977 and the remaining roots are [-0.515609770] whose absolute values are [0.515609770] so the largest absolute value is, 0.515609770 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.515609770 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2580645161, -0.1330561331, 0.06860511858, -0.03537346817, 0.01823890584, -0.009404158064, 0.004848875785, -0.002500127733, 0.001289090288, -0.0006646675480, 0.0003427090822, -0.0001767041514, 0.00009111038700, -0.00004697740577, 0.00002422200943, -0.00001248910473, -5 -5 -5 0.6439504431 10 , -0.3320271405 10 , 0.1711964378 10 , -6 -6 -6 -0.8827055610 10 , 0.4551316121 10 , -0.2346703063 10 , -6 -7 -7 0.1209983029 10 , -0.6238790722 10 , 0.3216781455 10 , -7 -8 -8 -0.1658603949 10 , 0.8551924024 10 , -0.4409455587 10 , -8 -8 -9 0.2273558385 10 , -0.1172268918 10 , 0.6044333084 10 , -9 -9 -10 -0.3116517197 10 , 0.1606906718 10 , -0.8285368048 10 , -10 -10 -10 0.4272016722 10 , -0.2202693563 10 , 0.1135730324 10 , -11 -11 -11 -0.5855936520 10 , 0.3019378088 10 , -0.1556820844 10 , -12 -12 -12 0.8027120389 10 , -0.4138861705 10 , 0.2134037536 10 , -12 -13 -13 -0.1100330605 10 , 0.5673412112 10 , -0.2925266719 10 , -13 -14 -14 0.1508296103 10 , -0.7776922082 10 , 0.4009857014 10 , -14 -14 -15 -0.2067521456 10 , 0.1066034264 10 , -0.5496576829 10 , -15 -15 -16 0.2834088720 10 , -0.1461283836 10 , 0.7534522238 10 , -16 -16 -0.3884873285 10 , 0.2003078625 10 ] The largest is 0.2580645161 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 15, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 33, 545, 9001, 148657, 2455161, 40548481, 669682889, 11060221265, 182666298393, 3016845302561, 49825039756393, 822890913438193, 13590545210398137, 224456141269262785, 3707030039894282249, 61223861548048695953, 1011149405997673573209, 16699749009576061176929, 275806537914811651422121] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 17 a(n - 1) - 8 a(n - 2) with initial conditions, a(1) = 2, a(2) = 33, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 17 b(n - 1) - 8 b(n - 2) with initial conditions, b(1) = 2, b(2) = 33, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 17 t + 8 = 0 whose roots are 1/2 1/2 257 257 [17/2 + ------, 17/2 - ------] 2 2 In floating-point [16.51560977, 0.484390230] The largest root is, 16.51560977 and the remaining roots are [0.484390230] whose absolute values are [0.484390230] so the largest absolute value is, 0.484390230 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.484390230 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2424242424, -0.1174311927, -0.05688256860, -0.02755336109, -0.01334657890, -0.006464952411, -0.003131559779, -0.001516896959, -0.0007347700653, -0.0003559154402, -0.0001724019616, -0.00008350982568, -5 -0.00004045134359, -0.00001959423559, -0.9491256264 10 , -5 -5 -5 -0.4597471796 10 , -0.2226970416 10 , -0.1078722710 10 , -6 -6 -6 -0.5225227406 10 , -0.2531049100 10 , -0.1226015453 10 , -7 -7 -7 -0.5938699063 10 , -0.2876647800 10 , -0.1393420087 10 , -8 -8 -8 -0.6749590749 10 , -0.3269435809 10 , -0.1583682760 10 , -9 -9 -9 -0.7671204551 10 , -0.3715856529 10 , -0.1799924595 10 , -10 -10 -10 -0.8718658871 10 , -0.4223233167 10 , -0.2045692881 10 , -11 -11 -11 -0.9909136434 10 , -0.4799888867 10 , -0.2325019268 10 , -11 -12 -12 -0.1126216616 10 , -0.5455283244 10 , -0.2642485900 10 , -12 -13 -13 -0.1279994351 10 , -0.6200167566 10 , -0.3003300588 10 , -13 -14 -14 -0.1454769460 10 , -0.7046761118 10 , -0.3413382232 10 , -14 -15 -15 -0.1653409001 10 , -0.8008951648 10 , -0.3879457923 10 , -15 -16 -16 -0.1879171512 10 , -0.9102523192 10 , -0.4409173294 10 , -16 -16 -17 -0.2135760462 10 , -0.1034541499 10 , -0.5011217938 10 , -17 -17 -18 -0.2427385005 10 , -0.1175801579 10 , -0.5695467960 10 ] The largest is -18 -0.5695467960 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 16, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 35, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 35, 613, 10736, 188029, 3293117, 57675250, 1010117303, 17691071401, 309839269544, 5426487224857, 95038836248465, 1664498601247618, 29151825747445691, 510561525117805309, 8941912358729701472, 156607563824465172805, 2742805796244475250933, 48037166610576265821106, 841317084545996796217199] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 17 a(n - 1) + 9 a(n - 2) with initial conditions, a(1) = 2, a(2) = 35, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 17 b(n - 1) + 9 b(n - 2) with initial conditions, b(1) = 2, b(2) = 35, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 17 t - 9 = 0 whose roots are 1/2 1/2 5 13 5 13 [17/2 + -------, 17/2 - -------] 2 2 In floating-point [17.51387819, -0.513878188] The largest root is, 17.51387819 and the remaining roots are [-0.513878188] whose absolute values are [0.513878188] so the largest absolute value is, 0.513878188 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.513878188 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2571428571, -0.1321370310, 0.06790238450, -0.03489355365, 0.01793103616, -0.009214368382, 0.004735062934, -0.002433245563, 0.001250391823, -0.0006425490850, 0.0003301919599, -0.0001696784463, 0.00008719405262, -0.00004480712182, 0.00002302540260, -0.00001183225218, -5 -5 -5 0.6080336319 10 , -0.3124552214 10 , 0.1605639232 10 , -6 -6 -6 -0.8251029803 10 , 0.4240024250 10 , -0.2178855981 10 , -6 -7 -7 0.1119666565 10 , -0.5753722263 10 , 0.2956712375 10 , -7 -8 -8 -0.1519390000 10 , 0.7807813808 10 , -0.4012265217 10 , -8 -8 -9 0.2061815582 10 , -0.1059522057 10 , 0.5444652754 10 , -9 -9 -10 -0.2797888295 10 , 0.1437773769 10 , -0.7388405802 10 , -10 -10 -10 0.3796740590 10 , -0.1951062177 10 , 0.1002608298 10 , -11 -11 -11 -0.5152185359 10 , 0.2647595680 10 , -0.1360541672 10 , -12 -12 -12 0.6991526902 10 , -0.3592793180 10 , 0.1846258052 10 , -13 -13 -13 -0.9487517435 10 , 0.4875428274 10 , -0.2505376250 10 , -13 -14 -14 0.1287458210 10 , -0.6615966927 10 , 0.3399801101 10 , -14 -15 -15 -0.1747083631 10 , 0.8977881719 10 , -0.4613537596 10 , -15 -15 -16 0.2370796343 10 , -0.1218300530 10 , 0.6260580698 10 , -16 -16 -0.3217175869 10 , 0.1653236508 10 ] The largest is 0.2571428571 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 17, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 37, 685, 12682, 234793, 4346929, 80478514, 1489969405, 27585112069, 510707404666, 9455174680033, 175051952278633, 3240890521173730, 60001452331793173, 1110859579613506717, 20566318941670489066, 380762323675217731801, 7049387279354102502625, 130511497394650987963666, 2416273964984181848786029] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 19 a(n - 1) - 9 a(n - 2) with initial conditions, a(1) = 2, a(2) = 37, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 19 b(n - 1) - 9 b(n - 2) with initial conditions, b(1) = 2, b(2) = 37, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 19 t + 9 = 0 whose roots are 1/2 1/2 5 13 5 13 [19/2 + -------, 19/2 - -------] 2 2 In floating-point [18.51387819, 0.486121812] The largest root is, 18.51387819 and the remaining roots are [0.486121812] whose absolute values are [0.486121812] so the largest absolute value is, 0.486121812 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.486121812 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2432432432, -0.1182481752, -0.05748304684, -0.02794376323, -0.01358407280, -0.006603514076, -0.003210112224, -0.001560505569, -0.0007585957937, -0.0003687699613, -0.0001792671216, -0.00008714565785, -5 -0.00004236340504, -0.00002059377519, -0.00001001108330, -0.4866605947 10 , -5 -5 -6 -0.2365763298 10 , -0.1150049140 10 , -0.5590639709 10 , -6 -6 -7 -0.2717731902 10 , -0.1321148755 10 , -0.6422392258 10 , -7 -7 -8 -0.3122064958 10 , -0.1517703872 10 , -0.7377889555 10 , -8 -8 -9 -0.3586553034 10 , -0.1743501657 10 , -0.8475541838 10 , -9 -9 -10 -0.4120145750 10 , -0.2002892715 10 , -0.9736498346 10 , -10 -10 -10 -0.4733124212 10 , -0.2300874915 10 , -0.1118505481 10 , -11 -11 -11 -0.5437299106 10 , -0.2643189690 10 , -0.1284912160 10 , -12 -12 -12 -0.6246238266 10 , -0.3036432660 10 , -0.1476076145 10 , -13 -13 -13 -0.7175528092 10 , -0.3488180713 10 , -0.1695680727 10 , -14 -14 -14 -0.8243073863 10 , -0.4007137997 10 , -0.1947957181 10 , -15 -15 -15 -0.9469444735 10 , -0.4603303627 10 , -0.2237766297 10 , -15 -16 -16 -0.1087827006 10 , -0.5288164345 10 , -0.2570692030 10 , -16 -17 -17 -0.1249669466 10 , -0.6074915844 10 , -0.2953149094 10 , -17 -18 -0.1435590187 10 , -0.6978717018 10 ] The largest is -18 -0.6978717018 10 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 18, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 2, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [2, 39, 761, 14849, 289741, 5653569, 110315221, 2152524889, 42001125101, 819546625809, 15991397141381, 312032011944329, 6088522198356061, 118802241888208449, 2318127817859521141, 45232450958212986169, 882597846384641948621, 17221683590890326885489, 336037966690762630310501, 6556938203033393244754409] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 19 a(n - 1) + 10 a(n - 2) with initial conditions, a(1) = 2, a(2) = 39, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 19 b(n - 1) + 10 b(n - 2) with initial conditions, b(1) = 2, b(2) = 39, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 19 t - 10 = 0 whose roots are 1/2 1/2 401 401 [19/2 + ------, 19/2 - ------] 2 2 In floating-point [19.51249220, -0.512492200] The largest root is, 19.51249220 and the remaining roots are [-0.512492200] whose absolute values are [0.512492200] so the largest absolute value is, 0.512492200 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.512492200 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2564102564, -0.1314060447, 0.06734460233, -0.03451358282, 0.01768794190, -0.009064932209, 0.004645707026, -0.002380888601, 0.001220186831, -0.0006253362299, 0.0003204799385, -0.0001642434679, 0.00008417349573, -0.00004313825978, 0.00002210802154, -0.00001133018853, -5 -5 -5 0.5806633217 10 , -0.2975854216 10 , 0.1525102066 10 , -6 -6 -6 -0.7816029088 10 , 0.4005653921 10 , -0.2052866380 10 , -6 -7 -7 0.1052078001 10 , -0.5391817667 10 , 0.2763264483 10 , -7 -8 -8 -0.1416151487 10 , 0.7257665870 10 , -0.3719497129 10 , -8 -9 -9 0.1906213256 10 , -0.9769194200 10 , 0.5006635801 10 , -9 -9 -10 -0.2565861783 10 , 0.1314984143 10 , -0.6739191127 10 , -10 -10 -11 0.3453782868 10 , -0.1770036771 10 , 0.9071300340 10 , -11 -11 -11 -0.4648970643 10 , 0.2382561180 10 , -0.1221044014 10 , -12 -12 -12 0.6257755297 10 , -0.3207050762 10 , 0.1643588492 10 , -13 -13 -13 -0.8423262775 10 , 0.4316856448 10 , -0.2212355246 10 , -13 -14 -14 0.1133814801 10 , -0.5810712387 10 , 0.2977944759 10 , -14 -15 -15 -0.1526173453 10 , 0.7821519862 10 , -0.4008467900 10 , -15 -15 -16 0.2054308522 10 , -0.1052817088 10 , 0.5395605428 10 , -16 -16 -0.2765205681 10 , 0.1417146335 10 ] The largest is 0.2564102564 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 19, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 5, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 3, a(2) = 5, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 3, b(2) = 5, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.2000000000, 0.1250000000, -0.07692307692, 0.04761904762, -0.02941176471, 0.01818181818, -0.01123595506, 0.006944444444, -0.004291845494, 0.002652519894, -0.001639344262, 0.001013171226, -0.0006261740764, 0.0003869969040, -0.0002391772303, 0.0001478196600, -0.00009135757354, 0.00005646208571, -0.00003489548801, 0.00002156659765, -5 -5 -5 -0.00001332889037, 0.8237707281 10 , -0.5091183089 10 , 0.3146524192 10 , -5 -5 -6 -0.1944658897 10 , 0.1201865295 10 , -0.7427936022 10 , -6 -6 -6 0.4590716928 10 , -0.2837219094 10 , 0.1753497834 10 , -6 -7 -7 -0.1083721260 10 , 0.6697765733 10 , -0.4139446871 10 , -7 -7 -8 0.2558318861 10 , -0.1581128010 10 , 0.9771908509 10 , -8 -8 -8 -0.6039371593 10 , 0.3732536915 10 , -0.2306834678 10 , -8 -9 -9 0.1425702237 10 , -0.8811324406 10 , 0.5445697969 10 , -9 -9 -9 -0.3365626437 10 , 0.2080071532 10 , -0.1285554906 10 , -10 -10 -10 0.7945166260 10 , -0.4910382795 10 , 0.3034783465 10 , -10 -10 -11 -0.1875599330 10 , 0.1159184135 10 , -0.7164151948 10 , -11 -11 -11 0.4427689404 10 , -0.2736462543 10 , 0.1691226861 10 , -11 -12 -12 -0.1045235683 10 , 0.6459911781 10 , -0.3992445045 10 ] The largest is 0.3333333333 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 20, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 7, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 7, 16, 37, 86, 200, 465, 1081, 2513, 5842, 13581, 31572, 73396, 170625, 396655, 922111, 2143648, 4983377, 11584946, 26931732] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 3, a(2) = 7, a(3) = 16, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 3, b(2) = 7, b(3) = 16, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + ------------------- + 1, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- + 1 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- + 1 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [2.324717958, 0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] The largest root is, 2.324717958 and the remaining roots are [0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] whose absolute values are [0.6558656185, 0.6558656185] so the largest absolute value is, 0.6558656185 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6558656185 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.4285714286, -0.4375000000, -0.1081081081, 0.1162790698, 0.1250000000, 0.03440860215, -0.03052728955, -0.03541583764, -0.01078397809, 0.007952286282, 0.01000886862, 0.003338056570, -0.002051282051, -0.002821091377, -0.001022653455, 0.0005229403335, 0.0007930365292, 0.0003105754658, -0.0001314063277, -0.0002223333854, -0.00009361203497, -5 0.00003242433815, 0.00006216369902, 0.00002803038580, -0.7811902501 10 , -5 -5 -5 -0.00001733278008, -0.8344149430 10 , 0.1821209362 10 , 0.4819146869 10 , -5 -6 -5 0.2470872453 10 , -0.4044670174 10 , -0.1335999089 10 , -6 -7 -6 -6 -0.7281907795 10 , 0.8295882229 10 , 0.3692589368 10 , 0.2136683863 10 , -7 -6 -7 -0.1455389243 10 , -0.1017395131 10 , -0.6244236808 10 , -8 -7 -7 -9 0.1598029464 10 , 0.2793931149 10 , 0.1817950744 10 , 0.2579288246 10 , -8 -8 -9 -0.7645916928 10 , -0.5274100990 10 , -0.2725402894 10 , -8 -8 -9 -9 0.2084664184 10 , 0.1524972140 10 , 0.1330477632 10 , -0.5661368067 10 , -9 -10 -9 -0.4395338064 10 , -0.5328004264 10 , 0.1530906782 10 , -9 -10 -10 0.1262983136 10 , 0.1943354158 10 , -0.4120532415 10 , -10 -0.3618474206 10 ] The largest is 0.3333333333 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 21, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 8, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368, 121393, 317811, 832040, 2178309, 5702887, 14930352, 39088169, 102334155, 267914296] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) with initial conditions, a(1) = 3, a(2) = 8, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) with initial conditions, b(1) = 3, b(2) = 8, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t + 1 = 0 whose roots are 1/2 1/2 5 5 [3/2 + ----, 3/2 - ----] 2 2 In floating-point [2.618033988, 0.381966012] The largest root is, 2.618033988 and the remaining roots are [0.381966012] whose absolute values are [0.381966012] so the largest absolute value is, 0.381966012 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.381966012 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.1250000000, 0.04761904762, 0.01818181818, 0.006944444444, 0.002652519894, 0.001013171226, 0.0003869969040, 0.0001478196600, -5 -5 0.00005646208571, 0.00002156659765, 0.8237707281 10 , 0.3146524192 10 , -5 -6 -6 -7 0.1201865295 10 , 0.4590716928 10 , 0.1753497834 10 , 0.6697765733 10 , -7 -8 -8 -8 0.2558318861 10 , 0.9771908509 10 , 0.3732536915 10 , 0.1425702237 10 , -9 -9 -10 0.5445697969 10 , 0.2080071532 10 , 0.7945166260 10 , -10 -10 -11 0.3034783465 10 , 0.1159184135 10 , 0.4427689404 10 , -11 -12 -12 0.1691226861 10 , 0.6459911781 10 , 0.2467466736 10 , -13 -13 -13 0.9424884271 10 , 0.3599985451 10 , 0.1375072083 10 , -14 -14 -15 0.5252307989 10 , 0.2006203132 10 , 0.7663014082 10 , -15 -15 -16 0.2927010923 10 , 0.1118018687 10 , 0.4270451385 10 , -16 -17 -17 0.1631167282 10 , 0.6230504602 10 , 0.2379840991 10 , -18 -18 -18 0.9090183708 10 , 0.3472141212 10 , 0.1326239929 10 , -19 -19 -20 0.5065785758 10 , 0.1934957980 10 , 0.7390881815 10 , -20 -20 -21 0.2823065646 10 , 0.1078315124 10 , 0.4118797270 10 , -21 -22 -22 0.1573240564 10 , 0.6009244230 10 , 0.2295327049 10 , -23 -23 -23 0.8767369176 10 , 0.3348837033 10 , 0.1279141924 10 , -24 0.4885887385 10 ] The largest is 0.3333333333 The smallest is -24 0.4885887385 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 22, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 10, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 2) with initial conditions, a(1) = 3, a(2) = 10, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 2) with initial conditions, b(1) = 3, b(2) = 10, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 1 = 0 whose roots are 1/2 1/2 13 13 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.302775638, -0.302775638] The largest root is, 3.302775638 and the remaining roots are [-0.302775638] whose absolute values are [0.302775638] so the largest absolute value is, 0.302775638 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.302775638 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.1000000000, 0.03030303030, -0.009174311927, 0.002777777778, -0.0008410428932, 0.0002546473135, -0.00007710100231, 0.00002334430516, -5 -5 -6 -0.7068086881 10 , 0.2140044513 10 , -0.6479533422 10 , -6 -7 -7 0.1961844864 10 , -0.5939988298 10 , 0.1798483745 10 , -8 -8 -9 -0.5445370629 10 , 0.1648725565 10 , -0.4991939343 10 , -9 -10 -10 0.1511437618 10 , -0.4576264887 10 , 0.1385581520 10 , -11 -11 -12 -0.4195203283 10 , 0.1270205349 10 , -0.3845872347 10 , -12 -13 -13 0.1164436452 10 , -0.3525629895 10 , 0.1067474840 10 , -14 -15 -15 -0.3232053754 10 , 0.9785871365 10 , -0.2962923443 10 , -16 -16 -17 0.8971010351 10 , -0.2716203380 10 , 0.8224002107 10 , -17 -18 -18 -0.2490027483 10 , 0.7539196590 10 , -0.2282685055 10 , -19 -19 -20 0.6911414234 10 , -0.2092607852 10 , 0.6335906770 10 , -20 -21 -21 -0.1918358213 10 , 0.5808321313 10 , -0.1758618190 10 , -22 -22 -23 0.5324667439 10 , -0.1612179580 10 , 0.4881287004 10 , -23 -24 -24 -0.1477934786 10 , 0.4474826472 10 , -0.1354868439 10 , -25 -25 -26 0.4102211556 10 , -0.1242049720 10 , 0.3760623961 10 , -26 -27 -27 -0.1138625318 10 , 0.3447480068 10 , -0.1043812976 10 , -28 -29 -29 0.3160411395 10 , -0.9568955757 10 , 0.2897246682 10 , -30 -0.8772157117 10 ] The largest is 0.3333333333 The smallest is -0.1000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 23, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 11, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 11, 40, 145, 526, 1908, 6921, 25105, 91065, 330326, 1198213, 4346356, 15765820, 57188385, 207443151, 752472043, 2729490816, 9900859685, 35914032730, 130273308376] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 3, a(2) = 11, a(3) = 40, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 3, b(2) = 11, b(3) = 40, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (540 + 60 21 ) 10 (540 + 60 21 ) [------------------- + ------------------- + 1, - ------------------- 6 1/2 1/3 12 (540 + 60 21 ) 5 - ------------------- + 1 1/2 1/3 (540 + 60 21 ) / 1/2 1/3 \ 1/2 |(540 + 60 21 ) 10 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (540 + 60 21 ) / 1/2 1/3 (540 + 60 21 ) 5 - ------------------- - ------------------- + 1 12 1/2 1/3 (540 + 60 21 ) / 1/2 1/3 \ 1/2 |(540 + 60 21 ) 10 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (540 + 60 21 ) / In floating-point [3.627365085, -0.313682542 + 0.4210528072 I, -0.313682542 - 0.4210528072 I] The largest root is, 3.627365085 and the remaining roots are [-0.313682542 + 0.4210528072 I, -0.313682542 - 0.4210528072 I] whose absolute values are [0.5250544768, 0.5250544768] so the largest absolute value is, 0.5250544768 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5250544768 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.4545454545, -0.3750000000, 0.1103448276, 0.03422053232, -0.05188679245, 0.02311804652, -0.0001991635132, -0.006248284193, 0.003974861198, -0.0007711483684, -0.0006120069318, 0.0005965436622, -0.0002055312455, -0.00003551334409, 0.00007894113881, -0.00003973451728, -5 -5 -5 -5 0.3165381694 10 , 0.8968249331 10 , -0.6499005902 10 , 0.1604862650 10 , -6 -6 -6 -7 0.7848254764 10 , -0.9348041733 10 , 0.3701010826 10 , 0.2552037746 10 , -6 -7 -8 -0.1180408758 10 , 0.6701921014 10 , -0.9503743690 10 , -7 -7 -8 -0.1251368657 10 , 0.1047066304 10 , -0.3119127715 10 , -9 -8 -9 -0.9297436375 10 , 0.1443176698 10 , -0.6490848965 10 , -11 -9 -9 0.9355068744 10 , 0.1730721111 10 , -0.1111584256 10 , -10 -10 -10 0.2202401412 10 , 0.1682730224 10 , -0.1662849064 10 , -11 -12 -11 0.5793146678 10 , 0.9497609882 10 , -0.2192914325 10 , -11 -13 -12 0.1113925679 10 , -0.9429062450 10 , -0.2479348403 10 , -12 -13 -13 0.1815399092 10 , -0.4554057740 10 , -0.2147675403 10 , -13 -13 -15 0.2602849234 10 , -0.1040860843 10 , -0.6455946201 10 , -14 -14 -15 0.3274491631 10 , -0.1876322772 10 , 0.2744203261 10 , -15 -15 -16 0.3451070654 10 , -0.2921609236 10 , 0.8815168627 10 ] The largest is 0.4545454545 The smallest is -0.3750000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 24, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 13, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 13, 56, 241, 1037, 4462, 19199, 82609, 355448, 1529413, 6580721, 28315366, 121834667, 524227237, 2255632184, 9705479209, 41760499493, 179686059838, 773148800711, 3326685824041] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 3 a(n - 2) with initial conditions, a(1) = 3, a(2) = 13, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 3 b(n - 2) with initial conditions, b(1) = 3, b(2) = 13, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 5 t + 3 = 0 whose roots are 1/2 1/2 13 13 [5/2 + -----, 5/2 - -----] 2 2 In floating-point [4.302775638, 0.697224362] The largest root is, 4.302775638 and the remaining roots are [0.697224362] whose absolute values are [0.697224362] so the largest absolute value is, 0.697224362 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.697224362 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.2307692308, 0.1607142857, 0.1120331950, 0.07810993250, 0.05445988346, 0.03797072764, 0.02647411299, 0.01845839616, 0.01286964345, 0.008973028943, 0.006256214382, 0.004361985083, 0.003041282267, 0.002120456089, 0.001478433645, 0.001030799955, 0.0007186988413, 0.0005010943413, 0.0003493751825, 0.0002435928888, 0.0001698388966, 0.0001184158163, 0.00008256239204, 0.00005756451113, 0.00004013537957, -5 0.00002798336442, 0.00001951068341, 0.00001360332380, 0.9484568761 10 , -5 -5 -5 -5 0.6612872406 10 , 0.4610655746 10 , 0.3214661512 10 , 0.2241340323 10 , -5 -5 -6 -6 0.1562717077 10 , 0.1089564418 10 , 0.7596708561 10 , 0.5296610282 10 , -6 -6 -6 -6 0.3692925726 10 , 0.2574797784 10 , 0.1795211743 10 , 0.1251665363 10 , -7 -7 -7 -7 0.8726915843 10 , 0.6084618333 10 , 0.4242344137 10 , 0.2957865685 10 , -7 -7 -7 -8 0.2062296016 10 , 0.1437883025 10 , 0.1002527075 10 , 0.6989863005 10 , -8 -8 -8 -8 0.4873502776 10 , 0.3397924865 10 , 0.2369115997 10 , 0.1651805390 10 , -8 -9 -9 -9 0.1151678960 10 , 0.8029786282 10 , 0.5598562620 10 , 0.3903454252 10 ] The largest is 0.3333333333 The smallest is -9 0.3903454252 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 25, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 14, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 14, 65, 302, 1403, 6518, 30281, 140678, 653555, 3036254, 14105681, 65531486, 304442987, 1414366406, 6570794585, 30526277558, 141817493987, 658848808622, 3060847716449, 14219937291662] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 3 a(n - 2) with initial conditions, a(1) = 3, a(2) = 14, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 3 b(n - 2) with initial conditions, b(1) = 3, b(2) = 14, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t - 3 = 0 whose roots are 1/2 1/2 [2 + 7 , 2 - 7 ] In floating-point [4.645751311, -0.645751311] The largest root is, 4.645751311 and the remaining roots are [-0.645751311] whose absolute values are [0.645751311] so the largest absolute value is, 0.645751311 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.645751311 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.2142857143, 0.1384615385, -0.08940397351, 0.05773342837, -0.03728137465, 0.02407450216, -0.01554614083, 0.01003894087, -0.006482659224, 0.004186185694, -0.002703234900, 0.001745617481, -0.001127234777, 0.0007279133350, -0.0004700509904, 0.0003035360433, -0.0001960087979, 0.0001265729382, -0.00008173464082, 0.00005278025147, -5 -0.00003408291658, 0.00002200908807, -0.00001421239748, 0.9177674303 10 , -5 -5 -5 -0.5926495214 10 , 0.3827042054 10 , -0.2471317424 10 , -5 -5 -6 0.1595856467 10 , -0.1030526406 10 , 0.6654637775 10 , -6 -6 -6 -0.4297241068 10 , 0.2774949054 10 , -0.1791926990 10 , -6 -7 -7 0.1157139203 10 , -0.7472241573 10 , 0.4825209792 10 , -7 -7 -7 -0.3115885550 10 , 0.2012087179 10 , -0.1299307934 10 , -8 -8 -8 0.8390298016 10 , -0.5418045944 10 , 0.3498710272 10 , -8 -8 -9 -0.2259296745 10 , 0.1458943835 10 , -0.9421148944 10 , -9 -9 -9 0.6083719282 10 , -0.3928569703 10 , 0.2536879036 10 , -9 -9 -10 -0.1638192964 10 , 0.1057865254 10 , -0.6831178747 10 , -10 -10 -10 0.4411242632 10 , -0.2848565713 10 , 0.1839465044 10 , -10 -11 -11 -0.1187836964 10 , 0.7670472767 10 , -0.4953217846 10 ] The largest is 0.3333333333 The smallest is -0.2142857143 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 26, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 16, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 16, 85, 452, 2404, 12786, 68004, 361688, 1923684, 10231360, 54416800, 289422728, 1539331888, 8187134016, 43544322000, 231596059712, 1231773338304, 6551344434976, 34844165376064, 185323161193088] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - 4 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 3, a(2) = 16, a(3) = 85, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - 4 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 3, b(2) = 16, b(3) = 85, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 6 t + 4 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (135 + 3 489 ) 8 (135 + 3 489 ) [------------------- + ------------------- + 2, - ------------------- 3 1/2 1/3 6 (135 + 3 489 ) 4 - ------------------- + 2 1/2 1/3 (135 + 3 489 ) / 1/2 1/3 \ 1/2 |(135 + 3 489 ) 8 | + 1/2 I 3 |------------------- - -------------------|, | 3 1/2 1/3| \ (135 + 3 489 ) / 1/2 1/3 (135 + 3 489 ) 4 - ------------------- - ------------------- + 2 6 1/2 1/3 (135 + 3 489 ) / 1/2 1/3 \ 1/2 |(135 + 3 489 ) 8 | - 1/2 I 3 |------------------- - -------------------|] | 3 1/2 1/3| \ (135 + 3 489 ) / In floating-point [5.318628218, 0.340685891 + 0.5098724705 I, 0.340685891 - 0.5098724705 I] The largest root is, 5.318628218 and the remaining roots are [0.340685891 + 0.5098724705 I, 0.340685891 - 0.5098724705 I] whose absolute values are [0.6132184052, 0.6132184052] so the largest absolute value is, 0.6132184052 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6132184052 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.4375000000, -0.4235294118, -0.1238938053, 0.07487520799, 0.09760675739, 0.03835068525, -0.01057264825, -0.02162517337, -0.01075907797, 0.0008009291248, 0.004591539888, 0.002827366882, 0.0001998999890, -0.0009269878171, -0.0007067930957, -0.0001330073277, 0.0001751527827, 0.0001693598152, 0.00004953310507, -0.00002993506492, -5 -5 -0.00003902317946, -0.00001533260696, 0.4226946277 10 , 0.8645746560 10 , -5 -6 -5 0.4301480339 10 , -0.3202116499 10 , -0.1835698137 10 , -5 -7 -6 -0.1130381543 10 , -0.7992001123 10 , 0.3706098315 10 , -6 -7 -7 0.2825759477 10 , 0.5317633749 10 , -0.7002610270 10 , -7 -7 -7 -0.6771007083 10 , -0.1980333921 10 , 0.1196804265 10 , -7 -8 -8 0.1560147110 10 , 0.6129977562 10 , -0.1689933720 10 , -8 -8 -9 -0.3456570372 10 , -0.1719732226 10 , 0.1280206919 10 , -9 -9 -10 0.7339123110 10 , 0.4519266465 10 , 0.3195201874 10 , -9 -9 -10 -0.1481698515 10 , -0.1129738911 10 , -0.2125990289 10 , -10 -10 -11 0.2799644393 10 , 0.2707049301 10 , 0.7917376539 10 , -11 -11 -11 -0.4784824931 10 , -0.6237469725 10 , -0.2450765550 10 , -12 -11 -12 0.6756357409 10 , 0.1381937193 10 , 0.6875490983 10 ] The largest is 0.3333333333 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 27, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 17, 96, 542, 3060, 17276, 97536, 550664, 3108912, 17552144, 99095040, 559465952, 3158605632, 17832701888, 100679000064, 568408596608, 3209093579520, 18117744283904, 102288278544384, 577494182698496] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - 2 a(n - 2) with initial conditions, a(1) = 3, a(2) = 17, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - 2 b(n - 2) with initial conditions, b(1) = 3, b(2) = 17, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 6 t + 2 = 0 whose roots are 1/2 1/2 [3 + 7 , 3 - 7 ] In floating-point [5.645751311, 0.354248689] The largest root is, 5.645751311 and the remaining roots are [0.354248689] whose absolute values are [0.354248689] so the largest absolute value is, 0.354248689 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.354248689 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.1176470588, 0.04166666667, 0.01476014760, 0.005228758170, 0.001852280621, 0.0006561679790, 0.0002324466462, 0.00008234391967, -5 -5 0.00002917022559, 0.00001033351417, 0.3660633847 10 , 0.1296774741 10 , -6 -6 -7 -7 0.4593807518 10 , 0.1627350290 10 , 0.5764867068 10 , 0.2042196601 10 , -8 -8 -9 -9 0.7234454684 10 , 0.2562796087 10 , 0.9078671538 10 , 0.3216107490 10 , -9 -10 -10 0.1139301862 10 , 0.4035961908 10 , 0.1429734215 10 , -11 -11 -12 0.5064814710 10 , 0.1794203971 10 , 0.6355944043 10 , -12 -13 -13 0.2251584844 10 , 0.7976209791 10 , 0.2825561861 10 , -13 -14 -14 0.1000951585 10 , 0.3545857866 10 , 0.1256115500 10 , -15 -15 -16 0.4449772691 10 , 0.1576326142 10 , 0.5584114691 10 , -16 -17 -17 0.1978165308 10 , 0.7007624669 10 , 0.2482441852 10 , -18 -18 -18 0.8794017713 10 , 0.3115269245 10 , 0.1103580046 10 , -19 -19 -20 0.3909417844 10 , 0.1384906146 10 , 0.4906011864 10 , -20 -21 -21 0.1737948271 10 , 0.6156658963 10 , 0.2180988366 10 , -22 -22 -23 0.7726122692 10 , 0.2736968834 10 , 0.9695676212 10 , -23 -23 -24 0.3434680586 10 , 0.1216731095 10 , 0.4310253951 10 , -24 -25 -25 0.1526901811 10 , 0.5409029647 10 , 0.1916141661 10 , -26 0.6787906712 10 ] The largest is 0.3333333333 The smallest is -26 0.6787906712 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 28, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 19, 120, 758, 4788, 30244, 191040, 1206728, 7622448, 48148144, 304133760, 1921098848, 12134860608, 76651361344, 484177889280, 3058370058368, 19318576128768, 122028196889344, 770806333593600, 4868894395340288] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 3, a(2) = 19, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 3, b(2) = 19, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 6 t - 2 = 0 whose roots are 1/2 1/2 [3 + 11 , 3 - 11 ] In floating-point [6.316624790, -0.316624790] The largest root is, 6.316624790 and the remaining roots are [-0.316624790] whose absolute values are [0.316624790] so the largest absolute value is, 0.316624790 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.316624790 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.1052631579, 0.03333333333, -0.01055408971, 0.003341687552, -0.001058061103, 0.0003350083752, -0.0001060719566, 0.00003358501101, -5 -5 -6 -0.00001063384707, 0.3366939599 10 , -0.1066056545 10 , 0.3375399300 10 , -6 -7 -7 -0.1068735096 10 , 0.3383880256 10 , -0.1071420377 10 , -8 -8 -9 0.3392382522 10 , -0.1074112405 10 , 0.3400906150 10 , -9 -10 -10 -0.1076811197 10 , 0.3409451194 10 , -0.1079516769 10 , -11 -11 -12 0.3418017708 10 , -0.1082229140 10 , 0.3426605747 10 , -12 -13 -13 -0.1084948326 10 , 0.3435215363 10 , -0.1087674344 10 , -14 -14 -15 0.3443846612 10 , -0.1090407212 10 , 0.3452499548 10 , -15 -16 -16 -0.1093146945 10 , 0.3461174224 10 , -0.1095893563 10 , -17 -17 -18 0.3469870697 10 , -0.1098647082 10 , 0.3478589020 10 , -18 -19 -19 -0.1101407519 10 , 0.3487329249 10 , -0.1104174892 10 , -20 -20 -21 0.3496091438 10 , -0.1106949219 10 , 0.3504875642 10 , -21 -22 -22 -0.1109730516 10 , 0.3513681918 10 , -0.1112518801 10 , -23 -23 -24 0.3522510320 10 , -0.1115314092 10 , 0.3531360905 10 , -24 -25 -25 -0.1118116406 10 , 0.3540233727 10 , -0.1120925762 10 , -26 -26 -27 0.3549128843 10 , -0.1123742176 10 , 0.3558046308 10 , -27 -28 -28 -0.1126565666 10 , 0.3566986179 10 , -0.1129396251 10 ] The largest is 0.3333333333 The smallest is -0.1052631579 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 29, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 20, 133, 884, 5876, 39058, 259620, 1725704, 11470820, 76246976, 506816544, 3368828808, 22392732976, 148845346176, 989380666576, 6576450850112, 43713918459328, 290568075489568, 1931417028474944, 12838202309726592] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + 4 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 3, a(2) = 20, a(3) = 133, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + 4 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 3, b(2) = 20, b(3) = 133, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 6 t - 4 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (351 + 3 1401 ) 16 (351 + 3 1401 ) [-------------------- + -------------------- + 2, - -------------------- 3 1/2 1/3 6 (351 + 3 1401 ) 8 - -------------------- + 2 1/2 1/3 (351 + 3 1401 ) / 1/2 1/3 \ 1/2 |(351 + 3 1401 ) 16 | + 1/2 I 3 |-------------------- - --------------------|, | 3 1/2 1/3| \ (351 + 3 1401 ) / 1/2 1/3 (351 + 3 1401 ) 8 - -------------------- - -------------------- + 2 6 1/2 1/3 (351 + 3 1401 ) / 1/2 1/3 \ 1/2 |(351 + 3 1401 ) 16 | - 1/2 I 3 |-------------------- - --------------------|] | 3 1/2 1/3| \ (351 + 3 1401 ) / In floating-point [6.647037962, -0.323518981 + 0.4429688124 I, -0.323518981 - 0.4429688124 I] The largest root is, 6.647037962 and the remaining roots are [-0.323518981 + 0.4429688124 I, -0.323518981 - 0.4429688124 I] whose absolute values are [0.5485306736, 0.5485306736] so the largest absolute value is, 0.5485306736 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5485306736 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.4500000000, -0.3909774436, 0.1176470588, 0.04152484683, -0.06226637309, 0.02779446884, 0.0007509978536, -0.008848887874, 0.005499601715, -0.0008959454962, -0.001075041864, 0.0009651702641, -0.0003010368624, -0.00009562384550, 0.0001524500055, -0.00006986907410, -6 -5 -0.6621137566 10 , 0.00002145103198, -0.00001368041133, 0.2397432411 10 , -5 -5 -6 -6 0.2565013095 10 , -0.2381014455 10 , 0.7688304683 10 , 0.2189511777 10 , -6 -6 -8 -0.3729999714 10 , 0.1754658193 10 , -0.1302614125 10 , -7 -7 -8 -0.5195235018 10 , 0.3400708103 10 , -0.6372142779 10 , -8 -8 -8 -0.6109232904 10 , 0.5870193526 10 , -0.1960056018 10 , -9 -9 -9 -0.4980278104 10 , 0.9119961188 10 , -0.4402465645 10 , -10 -9 -10 0.1044946708 10 , 0.1257027819 10 , -0.8447856921 10 , -10 -10 -10 0.1683864662 10 , 0.1452316673 10 , -0.1446355155 10 , -11 -11 -11 0.4988650835 10 , 0.1124032255 10 , -0.2228306237 10 , -11 -13 -12 0.1103593270 10 , -0.4360081356 10 , -0.3038442729 10 , -12 -13 -13 0.2097176493 10 , -0.4427282298 10 , -0.3445488649 10 , -13 -13 -14 0.3561468774 10 , -0.1267706546 10 , -0.2513414770 10 , -14 -14 -15 0.5440625023 10 , -0.2764039864 10 , 0.1514313653 10 ] The largest is 0.4500000000 The smallest is -0.3909774436 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 30, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 22, 161, 1178, 8619, 63062, 461401, 3375898, 24700179, 180721942, 1322274641, 9674587418, 70785326139, 517909672022, 3789350745481, 27725257603738, 202855307102499, 1484216168801302, 10859452814897921, 79454541675176858] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 8 a(n - 1) - 5 a(n - 2) with initial conditions, a(1) = 3, a(2) = 22, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 8 b(n - 1) - 5 b(n - 2) with initial conditions, b(1) = 3, b(2) = 22, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 8 t + 5 = 0 whose roots are 1/2 1/2 [4 + 11 , 4 - 11 ] In floating-point [7.316624790, 0.683375210] The largest root is, 7.316624790 and the remaining roots are [0.683375210] whose absolute values are [0.683375210] so the largest absolute value is, 0.683375210 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.683375210 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.2272727273, 0.1552795031, 0.1061120543, 0.07251421279, 0.04955440677, 0.03386425257, 0.02314199066, 0.01581466272, 0.01080734845, 0.007385474014, 0.005047049852, 0.003449028751, 0.002356980746, 0.001610702211, 0.001100713961, 0.0007522006341, 0.0005140352660, 0.0003512789577, 0.0002400553313, 0.0001640478624, 0.0001121062423, 0.00007661062686, 0.00005235380319, 0.00003577729123, 0.00002444931390, -5 -5 0.00001670805501, 0.00001141787060, 0.7802689712 10 , 0.5332164718 10 , -5 -5 -5 -5 0.3643869182 10 , 0.2490129866 10 , 0.1701693019 10 , 0.1162894824 10 , -6 -6 -6 -6 0.7946934940 10 , 0.5430738331 10 , 0.3711231945 10 , 0.2536163909 10 , -6 -6 -7 -7 0.1733151543 10 , 0.1184392799 10 , 0.8093846772 10 , 0.5531134235 10 , -7 -7 -7 -7 0.3779840017 10 , 0.2583048964 10 , 0.1765191627 10 , 0.1206288198 10 , -8 -8 -8 -8 0.8243474505 10 , 0.5633386118 10 , 0.3849716420 10 , 0.2630800765 10 , -8 -8 -9 -9 0.1797824024 10 , 0.1228588370 10 , 0.8395868347 10 , 0.5737528292 10 , -9 -9 -9 -9 0.3920884599 10 , 0.2679435335 10 , 0.1831059684 10 , 0.1251300795 10 ] The largest is 0.3333333333 The smallest is -9 0.1251300795 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 31, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 23, 176, 1347, 10309, 78898, 603831, 4621307, 35368304, 270684663, 2071634161, 15854862442, 121342207899, 928669767503, 7107399412016, 54395144721627, 416303010111469, 3186096794388418, 24384192611276271, 186619832250875987] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) + 5 a(n - 2) with initial conditions, a(1) = 3, a(2) = 23, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) + 5 b(n - 2) with initial conditions, b(1) = 3, b(2) = 23, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 7 t - 5 = 0 whose roots are 1/2 1/2 69 69 [7/2 + -----, 7/2 - -----] 2 2 In floating-point [7.653311932, -0.653311932] The largest root is, 7.653311932 and the remaining roots are [-0.653311932] whose absolute values are [0.653311932] so the largest absolute value is, 0.653311932 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.653311932 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.2173913043, 0.1420454545, -0.09279881218, 0.06062663692, -0.03960810160, 0.02587644556, -0.01690539062, 0.01104449340, -0.007215499313, 0.004713971793, -0.003079694017, 0.002012000846, -0.001314464159, 0.0008587551186, -0.0005610349652, 0.0003665308367, -0.0002394589689, 0.0001564414015, -0.0001022050341, 0.00006677176826, -0.00004362279289, 0.00002849929108, -0.00001861892690, 0.00001216396710, -5 -5 -5 -0.7946864837 10 , 0.5191781616 10 , -0.3391852875 10 , -5 -5 -6 0.2215937953 10 , -0.1447698704 10 , 0.9457988366 10 , -6 -6 -6 -0.6179016647 10 , 0.4036825300 10 , -0.2637306134 10 , -6 -6 -7 0.1722983564 10 , -0.1125645720 10 , 0.7353977796 10 , -7 -7 -7 -0.4804441438 10 , 0.3138798915 10 , -0.2050614782 10 , -7 -8 -8 0.1339691104 10 , -0.8752361825 10 , 0.5718022409 10 , -8 -8 -8 -0.3735652264 10 , 0.2440546196 10 , -0.1594437949 10 , -8 -9 -9 0.1041665336 10 , -0.6805323927 10 , 0.4445999319 10 , -9 -9 -9 -0.2904624402 10 , 0.1897625778 10 , -0.1239741562 10 , -10 -10 -10 0.8099379547 10 , -0.5291421295 10 , 0.3456948667 10 , -10 -10 -11 -0.2258465810 10 , 0.1475482661 10 , -0.9639504269 10 ] The largest is 0.3333333333 The smallest is -0.2173913043 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 32, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 25, 208, 1731, 14406, 119892, 997785, 8303931, 69108345, 575144874, 4786565589, 39835546092, 331525955916, 2759080023459, 22962071113911, 191098737752193, 1590393453156648, 13235834865238407, 110153449281462354, 916737214701200688] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 9 a(n - 1) - 6 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 3, a(2) = 25, a(3) = 208, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 9 b(n - 1) - 6 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 3, b(2) = 25, b(3) = 208, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 9 t + 6 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (156 + 4 149 ) 14 (156 + 4 149 ) [------------------- + ------------------- + 3, - ------------------- 2 1/2 1/3 4 (156 + 4 149 ) 7 - ------------------- + 3 1/2 1/3 (156 + 4 149 ) / 1/2 1/3 \ 1/2 |(156 + 4 149 ) 14 | + 1/2 I 3 |------------------- - -------------------|, | 2 1/2 1/3| \ (156 + 4 149 ) / 1/2 1/3 (156 + 4 149 ) 7 - ------------------- - ------------------- + 3 4 1/2 1/3 (156 + 4 149 ) / 1/2 1/3 \ 1/2 |(156 + 4 149 ) 14 | - 1/2 I 3 |------------------- - -------------------|] | 2 1/2 1/3| \ (156 + 4 149 ) / In floating-point [8.322365034, 0.338817483 + 0.4956583176 I, 0.338817483 - 0.4956583176 I] The largest root is, 8.322365034 and the remaining roots are [0.338817483 + 0.4956583176 I, 0.338817483 - 0.4956583176 I] whose absolute values are [0.6003952486, 0.6003952486] so the largest absolute value is, 0.6003952486 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6003952486 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.4400000000, -0.4182692308, -0.1247833622, 0.06622240733, 0.08985587028, 0.03701799486, -0.007306057818, -0.01829488175, -0.009763604361, -0.00002132217727, 0.003505081308, 0.002382851752, 0.0003512113914, -0.0006209640676, -0.0005473896996, -0.0001470887165, 0.00009764754626, 0.0001191911166, 0.00004556862218, -0.00001208646106, -6 -5 -0.00002461653287, -0.00001232416295, 0.5223475343 10 , 0.4796506875 10 , -5 -6 -6 0.3061987829 10 , 0.3458918162 10 , -0.8693800049 10 , -6 -6 -6 -0.7138074538 10 , -0.1703116060 10 , 0.1419002535 10 , -6 -7 -7 0.1575495568 10 , 0.5560967198 10 , -0.1910953246 10 , -7 -7 -8 -0.3299515358 10 , -0.1547017146 10 , 0.1410780892 10 , -8 -8 -9 -8 0.6532596089 10 , 0.3918165054 10 , 0.3002516272 10 , -0.1208937411 10 , -9 -9 -9 -0.9274513029 10 , -0.1926823775 10 , 0.2037541867 10 , -9 -10 -10 0.2075280363 10 , 0.6718007439 10 , -0.2928498839 10 , -10 -10 -11 -0.4406123280 10 , -0.1930094175 10 , 0.2803955877 10 , -11 -11 -12 0.8857555004 10 , 0.4991434516 10 , 0.1894482443 10 , -11 -11 -12 -0.1670907882 10 , -0.1200556858 10 , -0.2112196973 10 , -12 -12 0.2896402268 10 , 0.2724096504 10 ] The largest is 0.3333333333 The smallest is -0.4400000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 33, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 26, 225, 1947, 16848, 145791, 1261575, 10916802, 94466493, 817448031, 7073632800, 61210351107, 529672261563, 4583419300746, 39661756922025, 343205554395987, 2969864718797808, 25699165805992311, 222382898097537375, 1924348585459859442] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 9 a(n - 1) - 3 a(n - 2) with initial conditions, a(1) = 3, a(2) = 26, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 9 b(n - 1) - 3 b(n - 2) with initial conditions, b(1) = 3, b(2) = 26, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 9 t + 3 = 0 whose roots are 1/2 1/2 69 69 [9/2 + -----, 9/2 - -----] 2 2 In floating-point [8.653311932, 0.346688068] The largest root is, 8.653311932 and the remaining roots are [0.346688068] whose absolute values are [0.346688068] so the largest absolute value is, 0.346688068 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.346688068 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.1153846154, 0.04000000000, 0.01386748844, 0.004807692308, 0.001666769554, 0.0005778491172, 0.0002003333943, 0.00006945319755, -5 -5 -5 0.00002407859491, 0.8347761563 10 , 0.2894069333 10 , 0.1003339307 10 , -6 -6 -7 -7 0.3478457665 10 , 0.1205939769 10 , 0.4180849295 10 , 0.1449450567 10 , -8 -8 -9 -9 0.5025072175 10 , 0.1742132566 10 , 0.6039765746 10 , 0.2093914721 10 , -10 -10 -11 0.7259352503 10 , 0.2516730898 10 , 0.8725205741 10 , -11 -11 -12 0.3024924726 10 , 0.1048705311 10 , 0.3635736186 10 , -12 -13 -13 0.1260466356 10 , 0.4369886465 10 , 0.1514987498 10 , -14 -14 -15 0.5252280896 10 , 0.1820903119 10 , 0.6312853855 10 , -15 -16 -16 0.2188591110 10 , 0.7587584247 10 , 0.2630524928 10 , -17 -17 -17 0.9119716064 10 , 0.3161696748 10 , 0.1096122539 10 , -18 -18 -19 0.3800126059 10 , 0.1317458363 10 , 0.4567470954 10 , -19 -20 -20 0.1583487683 10 , 0.5489762864 10 , 0.1903235284 10 , -21 -21 -22 0.6598289647 10 , 0.2287548293 10 , 0.7930656995 10 , -22 -23 -23 0.2749464156 10 , 0.9532064177 10 , 0.3304652919 10 , -23 -24 -24 0.1145683738 10 , 0.3971948821 10 , 0.1377027265 10 , -25 -25 -26 0.4773989229 10 , 0.1655085105 10 , 0.5737982584 10 , -26 0.1989290099 10 ] The largest is 0.3333333333 The smallest is -26 0.1989290099 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 34, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 28, 261, 2433, 22680, 211419, 1970811, 18371556, 171256437, 1596422601, 14881572720, 138723422283, 1293155518707, 12054569935212, 112370595973029, 1047499073562897, 9764603449985160, 91023928270555131, 848509164784951659, 7909654267876230324] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 9 a(n - 1) + 3 a(n - 2) with initial conditions, a(1) = 3, a(2) = 28, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 9 b(n - 1) + 3 b(n - 2) with initial conditions, b(1) = 3, b(2) = 28, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 9 t - 3 = 0 whose roots are 1/2 1/2 93 93 [9/2 + -----, 9/2 - -----] 2 2 In floating-point [9.321825380, -0.321825380] The largest root is, 9.321825380 and the remaining roots are [-0.321825380] whose absolute values are [0.321825380] so the largest absolute value is, 0.321825380 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.321825380 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.1071428571, 0.03448275862, -0.01109741060, 0.003571428571, -0.001149376357, 0.0003698984834, -0.0001190427202, 0.00003831096871, -5 -5 -6 -0.00001232944208, 0.3967927390 10 , -0.1276979742 10 , 0.4109644914 10 , -6 -7 -7 -0.1322588038 10 , 0.4256423986 10 , -0.1369825269 10 , -8 -8 -9 0.4408445383 10 , -0.1418749613 10 , 0.4565896340 10 , -9 -10 -10 -0.1469421327 10 , 0.4728970777 10 , -0.1521902820 10 , -11 -11 -12 0.4897869540 10 , -0.1576258728 10 , 0.5072800650 10 , -12 -13 -13 -0.1632555999 10 , 0.5253979557 10 , -0.1690863970 10 , -14 -14 -15 0.5441629405 10 , -0.1751254454 10 , 0.5635981309 10 , -15 -16 -16 -0.1813801829 10 , 0.5837274639 10 , -0.1878583132 10 , -17 -17 -18 0.6045757311 10 , -0.1945678147 10 , 0.6261686100 10 , -18 -19 -19 -0.2015169512 10 , 0.6485326949 10 , -0.2087142813 10 , -20 -20 -21 0.6716955299 10 , -0.2161686695 10 , 0.6956856431 10 , -21 -22 -22 -0.2238892968 10 , 0.7205325813 10 , -0.2318856721 10 , -23 -23 -24 0.7462669467 10 , -0.2401676441 10 , 0.7729204344 10 , -24 -25 -25 -0.2487454129 10 , 0.8005258715 10 , -0.2576295432 10 , -26 -26 -27 0.8291172576 10 , -0.2668309769 10 , 0.8587298067 10 , -27 -28 -28 -0.2763610468 10 , 0.8893999904 10 , -0.2862314903 10 ] The largest is 0.3333333333 The smallest is -0.1071428571 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 35, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 29, 280, 2703, 26094, 251904, 2431809, 23475987, 226630449, 2187825390, 21120639165, 203892596172, 1968320676708, 19001603584899, 183436034112855, 1770838890555213, 17095171030428744, 165031880719528539, 1593170469329994954, 15380011021378412052] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 9 a(n - 1) + 6 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 3, a(2) = 29, a(3) = 280, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 9 b(n - 1) + 6 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 3, b(2) = 29, b(3) = 280, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 9 t - 6 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (300 + 4 301 ) 22 (300 + 4 301 ) [------------------- + ------------------- + 3, - ------------------- 2 1/2 1/3 4 (300 + 4 301 ) 11 - ------------------- + 3 1/2 1/3 (300 + 4 301 ) / 1/2 1/3 \ 1/2 |(300 + 4 301 ) 22 | + 1/2 I 3 |------------------- - -------------------|, | 2 1/2 1/3| \ (300 + 4 301 ) / 1/2 1/3 (300 + 4 301 ) 11 - ------------------- - ------------------- + 3 4 1/2 1/3 (300 + 4 301 ) / 1/2 1/3 \ 1/2 |(300 + 4 301 ) 22 | - 1/2 I 3 |------------------- - -------------------|] | 2 1/2 1/3| \ (300 + 4 301 ) / In floating-point [9.653713345, -0.326856673 + 0.4515816230 I, -0.326856673 - 0.4515816230 I] The largest root is, 9.653713345 and the remaining roots are [-0.326856673 + 0.4515816230 I, -0.326856673 - 0.4515816230 I] whose absolute values are [0.5574596370, 0.5574596370] so the largest absolute value is, 0.5574596370 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5574596370 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.4482758621, -0.3964285714, 0.1198668146, 0.04483789377, -0.06656107088, 0.02957798084, 0.001349080658, -0.01007360225, 0.006166006237, -0.0009003153669, -0.001327607618, 0.001147657944, -0.0003376703147, -0.0001359080230, 0.0001937797366, -0.00008444145313, -5 -5 -0.5018727814 10 , 0.00002952194064, -0.00001773926047, 0.2422116210 10 , -5 -5 -6 -6 0.3929305018 10 , -0.3321338983 10 , 0.9501278898 10 , 0.4110321635 10 , -6 -6 -7 -0.5639601388 10 , 0.2409354013 10 , 0.1775426894 10 , -7 -7 -8 -0.8647958835 10 , 0.5101552225 10 , -0.6475022985 10 , -7 -8 -8 -0.1162083839 10 , 0.9608883359 10 , -0.2670149055 10 , -8 -8 -9 -0.1240556506 10 , 0.1640747194 10 , -0.6870614504 10 , -10 -9 -9 -0.6073940527 10 , 0.2532182331 10 , -0.1466566850 10 , -10 -10 -10 0.1718101821 10 , 0.3434375345 10 , -0.2779016461 10 , -11 -11 -11 0.7494093821 10 , 0.3737117073 10 , -0.4771877247 10 , -11 -12 -12 0.1958088677 10 , 0.2028858256 10 , -0.7411272512 10 , -12 -13 -12 0.4214357226 10 , -0.4518452715 10 , -0.1014281624 10 , -13 -13 -13 0.8034654337 10 , -0.2100366541 10 , -0.1123821565 10 , -13 -14 -15 0.1387369681 10 , -0.5577018843 10 , -0.6656356788 10 ] The largest is 0.4482758621 The smallest is -0.3964285714 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 36, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 31, 320, 3303, 34093, 351902, 3632271, 37491667, 386982440, 3994365171, 41229139801, 425559981614, 4392555819147, 45339194139319, 467983244798480, 4830441333808047, 49858971958299157, 514635602204634398, 5311978820542884279, 54829317810539286283] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 11 a(n - 1) - 7 a(n - 2) with initial conditions, a(1) = 3, a(2) = 31, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 11 b(n - 1) - 7 b(n - 2) with initial conditions, b(1) = 3, b(2) = 31, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 11 t + 7 = 0 whose roots are 1/2 1/2 93 93 [11/2 + -----, 11/2 - -----] 2 2 In floating-point [10.32182538, 0.678174620] The largest root is, 10.32182538 and the remaining roots are [0.678174620] whose absolute values are [0.678174620] so the largest absolute value is, 0.678174620 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.678174620 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.2258064516, 0.1531250000, 0.1038449894, 0.07042501393, 0.04776045604, 0.03238992906, 0.02196602781, 0.01489680255, 0.01010263340, 0.006851349564, 0.004646411384, 0.003151078272, 0.002136981308, 0.001449246486, 0.0009828421839, 0.0006665386241, 0.0004520295778, 0.0003065549869, 0.0002078978116, 0.0001409910193, 0.00009561653086, 0.00006484470443, 0.00004397603276, 0.00002982342928, 0.00002022549281, -5 -5 -5 0.00001371641589, 0.9302125126 10 , 0.6308465168 10 , 0.4278240965 10 , -5 -5 -5 -6 0.2901394438 10 , 0.1967652069 10 , 0.1334411693 10 , 0.9049641424 10 , -6 -6 -6 -6 0.6137237130 10 , 0.4162118455 10 , 0.2822643100 10 , 0.1914244910 10 , -6 -7 -7 -7 0.1298192314 10 , 0.8804010783 10 , 0.5970656663 10 , 0.4049147810 10 , -7 -7 -7 -8 0.2746029276 10 , 0.1862287359 10 , 0.1262956021 10 , 0.8565047191 10 , -8 -8 -8 -8 0.5808597620 10 , 0.3939243481 10 , 0.2671494949 10 , 0.1811740070 10 , -8 -9 -9 -9 0.1228676133 10 , 0.8332569689 10 , 0.5650937278 10 , 0.3832322239 10 , -9 -9 -9 -10 0.2598983676 10 , 0.1762564766 10 , 0.1195326689 10 , 0.8106402226 10 ] The largest is 0.3333333333 The smallest is -10 0.8106402226 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 37, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 32, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 32, 341, 3634, 38727, 412708, 4398169, 46870646, 499493643, 5323030952, 56726765021, 604528866874, 6442376023887, 68655462306988, 731651255237089, 7797100788519806, 83092566671857683, 885505372238215472, 9436701689085158501, 100565554496519093314] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 10 a(n - 1) + 7 a(n - 2) with initial conditions, a(1) = 3, a(2) = 32, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 10 b(n - 1) + 7 b(n - 2) with initial conditions, b(1) = 3, b(2) = 32, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 10 t - 7 = 0 whose roots are 1/2 1/2 [5 + 4 2 , 5 - 4 2 ] In floating-point [10.65685425, -0.656854248] The largest root is, 10.65685425 and the remaining roots are [-0.656854248] whose absolute values are [0.656854248] so the largest absolute value is, 0.656854248 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.656854248 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.2187500000, 0.1436950147, -0.09438635113, 0.06199808919, -0.04072370780, 0.02674954055, -0.01757054938, 0.01154129003, -0.007580945398, 0.004979576200, -0.003270855788, 0.002148475524, -0.001411235278, 0.0009269758891, -0.0006088880519, 0.0003999507044, -0.0002627093198, 0.0001725617331, -0.0001133479077, 0.00007445305482, -0.00004890480545, 0.00003212332928, -0.00002110034534, 0.00001385985151, -5 -5 -5 -0.9103902359 10 , 0.5979936951 10 , -0.3927946998 10 , -5 -5 -5 0.2580088677 10 , -0.1694742212 10 , 0.1113198624 10 , -6 -6 -6 -0.7312092465 10 , 0.4802979008 10 , -0.3154857172 10 , -6 -6 -7 0.2072281340 10 , -0.1361186804 10 , 0.8941013367 10 , -7 -7 -7 -0.5872942625 10 , 0.3857667320 10 , -0.2533925172 10 , -7 -7 -8 0.1664419517 10 , -0.1093281033 10 , 0.7181262924 10 , -8 -8 -8 -0.4717043068 10 , 0.3098409784 10 , -0.2035203633 10 , -8 -9 -9 0.1336832155 10 , -0.8781038820 10 , 0.5767862664 10 , -9 -9 -9 -0.3788645101 10 , 0.2488587635 10 , -0.1634639363 10 , -9 -10 -10 0.1073719812 10 , -0.7052774213 10 , 0.4632644712 10 , -10 -10 -10 -0.3042972366 10 , 0.1998789330 10 , -0.1312913265 10 ] The largest is 0.3333333333 The smallest is -0.2187500000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 38, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 34, 385, 4360, 49376, 559172, 6332496, 71714080, 812145680, 9197365504, 104158076928, 1179566581824, 13358323828480, 151279985594880, 1713211502838016, 19401731444611072, 219720205255008256, 2488281457514562560, 28179222773913128960, 319123302447861080064] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 12 a(n - 1) - 8 a(n - 2) + 4 a(n - 3) with initial conditions, a(1) = 3, a(2) = 34, a(3) = 385, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 12 b(n - 1) - 8 b(n - 2) + 4 b(n - 3) with initial conditions, b(1) = 3, b(2) = 34, b(3) = 385, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 12 t + 8 t - 4 = 0 whose roots are 1/2 1/3 1/2 1/3 (1350 + 30 105 ) 40 (1350 + 30 105 ) [--------------------- + --------------------- + 4, - --------------------- 3 1/2 1/3 6 (1350 + 30 105 ) 20 - --------------------- + 4 1/2 1/3 (1350 + 30 105 ) / 1/2 1/3 \ 1/2 |(1350 + 30 105 ) 40 | + 1/2 I 3 |--------------------- - ---------------------|, | 3 1/2 1/3| \ (1350 + 30 105 ) / 1/2 1/3 (1350 + 30 105 ) 20 - --------------------- - --------------------- + 4 6 1/2 1/3 (1350 + 30 105 ) / 1/2 1/3 \ 1/2 |(1350 + 30 105 ) 40 | - 1/2 I 3 |--------------------- - ---------------------|] | 3 1/2 1/3| \ (1350 + 30 105 ) / In floating-point [11.32477304, 0.337613479 + 0.4891064301 I, 0.337613479 - 0.4891064301 I] The largest root is, 11.32477304 and the remaining roots are [0.337613479 + 0.4891064301 I, 0.337613479 - 0.4891064301 I] whose absolute values are [0.5943130162, 0.5943130162] so the largest absolute value is, 0.5943130162 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5943130162 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.4411764706, -0.4155844156, -0.1247706422, 0.06254050551, 0.08629902785, 0.03618162570, -0.006050694647, -0.01686523039, -0.009250704668, -0.0002893915181, 0.003072017583, 0.002176524472, 0.0003845869278, -0.0005090823109, -0.0004795852643, -0.0001440169733, -5 0.00007214919089, 0.00009958502016, 0.00004175882148, -0.6977540013 10 , -6 -5 -0.00001946097131, -0.00001067604975, -0.3349865809 10 , 0.3544673800 10 , -5 -6 -6 0.2511779230 10 , 0.4440140390 10 , -0.5873701730 10 , -6 -6 -7 -0.5534374680 10 , -0.1662320759 10 , 0.8323414109 10 , -6 -7 -8 0.1149164284 10 , 0.4819570897 10 , -0.8046355523 10 , -7 -7 -9 -0.2245622427 10 , -0.1232101114 10 , -0.3877616644 10 , -8 -8 -9 -9 0.4090052100 10 , 0.2898673939 10 , 0.5126238188 10 , -0.6776972918 10 , -9 -9 -10 -0.6386622940 10 , -0.1918739183 10 , 0.9602216493 10 , -9 -10 -11 0.1326081497 10 , 0.5562480428 10 , -0.9278886923 10 , -10 -10 -12 -0.2591247830 10 , -0.1421942713 10 , -0.4488468748 10 , -11 -11 -12 0.4719341361 10 , 0.3345162795 10 , 0.5918351528 10 , -12 -12 -12 -0.7819150822 10 , -0.7370110293 10 , -0.2214710822 10 , -12 -12 0.1107749183 10 , 0.1530235604 10 ] The largest is 0.3333333333 The smallest is -0.4411764706 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 39, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 35, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 35, 408, 4756, 55440, 646256, 7533312, 87814720, 1023643392, 11932461824, 139094968320, 1621409772544, 18900537397248, 220320809676800, 2568247566532608, 29937687559684096, 348979260450078720, 4068000375162208256, 47420087460146184192, 552769048021105377280] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 12 a(n - 1) - 4 a(n - 2) with initial conditions, a(1) = 3, a(2) = 35, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 12 b(n - 1) - 4 b(n - 2) with initial conditions, b(1) = 3, b(2) = 35, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 12 t + 4 = 0 whose roots are 1/2 1/2 [6 + 4 2 , 6 - 4 2 ] In floating-point [11.65685425, 0.343145752] The largest root is, 11.65685425 and the remaining roots are [0.343145752] whose absolute values are [0.343145752] so the largest absolute value is, 0.343145752 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.343145752 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.1142857143, 0.03921568627, 0.01345668629, 0.004617604618, 0.001584511401, 0.0005437183539, 0.0001865746426, 0.00006402229576, -5 -5 -6 0.00002196897873, 0.7538561694 10 , 0.2586825410 10 , 0.8876581468 10 , -6 -6 -7 -7 0.3045961210 10 , 0.1045208645 10 , 0.3586589051 10 , 0.1230722792 10 , -8 -8 -9 -9 0.4223172960 10 , 0.1449163855 10 , 0.4972744185 10 , 0.1706376036 10 , -10 -10 -11 0.5855356854 10 , 0.2009240822 10 , 0.6894624498 10 , -11 -12 -12 0.2365861098 10 , 0.8118351821 10 , 0.2785777928 10 , -13 -13 -13 0.9559278580 10 , 0.3280225823 10 , 0.1125595552 10 , -14 -14 -15 0.3862433304 10 , 0.1325377575 10 , 0.4547976826 10 , -15 -16 -16 0.1560618921 10 , 0.5355197510 10 , 0.1837613269 10 , -17 -17 -18 0.6305691842 10 , 0.2163771360 10 , 0.7424889471 10 , -18 -19 -19 0.2547819270 10 , 0.8742733556 10 , 0.3000031868 10 , -19 -20 -20 0.1029448187 10 , 0.3532507706 10 , 0.1212165008 10 , -21 -21 -22 0.4159492714 10 , 0.1427312249 10 , 0.4897761330 10 , -22 -23 -23 0.1680645987 10 , 0.5767065286 10 , 0.1978943946 10 , -24 -24 -25 0.6790662055 10 , 0.2330186827 10 , 0.7995937077 10 , -25 -26 -26 0.2743771829 10 , 0.9415136436 10 , 0.3230764059 10 , -26 0.1108622958 10 ] The largest is 0.3333333333 The smallest is -26 0.1108622958 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 40, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 37, 456, 5620, 69264, 853648, 10520832, 129664576, 1598058240, 19695357184, 242736519168, 2991619658752, 36870381981696, 454411062415360, 5600414276911104, 69022615572594688, 850673043978780672, 10484166990035746816, 129212696056344084480, 1592489020636272001024] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 12 a(n - 1) + 4 a(n - 2) with initial conditions, a(1) = 3, a(2) = 37, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 12 b(n - 1) + 4 b(n - 2) with initial conditions, b(1) = 3, b(2) = 37, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 12 t - 4 = 0 whose roots are 1/2 1/2 [6 + 2 10 , 6 - 2 10 ] In floating-point [12.32455532, -0.324555320] The largest root is, 12.32455532 and the remaining roots are [-0.324555320] whose absolute values are [0.324555320] so the largest absolute value is, 0.324555320 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.324555320 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.1081081081, 0.03508771930, -0.01138790036, 0.003696003696, -0.001199557663, 0.0003893228216, -0.0001263567931, 0.00004100976946, -5 -5 -6 -0.00001330993886, 0.4319811471 10 , -0.1402017796 10 , 0.4550323349 10 , -6 -7 -7 -0.1476831652 10 , 0.4793135699 10 , -0.1555637692 10 , -8 -8 -9 0.5048904895 10 , -0.1638648946 10 , 0.5318322335 10 , -9 -10 -10 -0.1726089809 10 , 0.5602116309 10 , -0.1818196653 10 , -11 -11 -12 0.5901053972 10 , -0.1915218462 10 , 0.6215943415 10 , -12 -13 -13 -0.2017417506 10 , 0.6547635850 10 , -0.2125070051 10 , -14 -14 -15 0.6897027911 10 , -0.2238467103 10 , 0.7265064077 10 , -15 -16 -16 -0.2357915199 10 , 0.7652739226 10 , -0.2483737231 10 , -17 -17 -18 0.8061101327 10 , -0.2616273323 10 , 0.8491254265 10 , -18 -19 -19 -0.2755881748 10 , 0.8944360836 10 , -0.2902939896 10 , -20 -20 -21 0.9421645880 10 , -0.3057845297 10 , 0.9924399598 10 , -21 -21 -22 -0.3221016691 10 , 0.1045398104 10 , -0.3392895165 10 , -22 -23 -23 0.1101182177 10 , -0.3573945342 10 , 0.1159942975 10 , -24 -24 -25 -0.3764656639 10 , 0.1221839342 10 , -0.3965544589 10 , -25 -26 -26 0.1287038594 10 , -0.4177152233 10 , 0.1355716981 10 , -27 -27 -28 -0.4400051591 10 , 0.1428060154 10 , -0.4634845206 10 ] The largest is 0.3333333333 The smallest is -0.1081081081 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 41, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 38, 481, 6088, 77056, 975300, 12344400, 156243424, 1977577488, 25030254848, 316808651776, 4009856170048, 50752864274176, 642380455257600, 8130627801964800, 102909588722735104, 1302529608909570048, 16486154527904580608, 208665729561022468096, 2641088109391144542208] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 12 a(n - 1) + 8 a(n - 2) + 4 a(n - 3) with initial conditions, a(1) = 3, a(2) = 38, a(3) = 481, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 12 b(n - 1) + 8 b(n - 2) + 4 b(n - 3) with initial conditions, b(1) = 3, b(2) = 38, b(3) = 481, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 12 t - 8 t - 4 = 0 whose roots are 1/2 1/3 1/2 1/3 (2214 + 6 4449 ) 56 (2214 + 6 4449 ) [--------------------- + --------------------- + 4, - --------------------- 3 1/2 1/3 6 (2214 + 6 4449 ) 28 - --------------------- + 4 1/2 1/3 (2214 + 6 4449 ) / 1/2 1/3 \ 1/2 |(2214 + 6 4449 ) 56 | + 1/2 I 3 |--------------------- - ---------------------|, | 3 1/2 1/3| \ (2214 + 6 4449 ) / 1/2 1/3 (2214 + 6 4449 ) 28 - --------------------- - --------------------- + 4 6 1/2 1/3 (2214 + 6 4449 ) / 1/2 1/3 \ 1/2 |(2214 + 6 4449 ) 56 | - 1/2 I 3 |--------------------- - ---------------------|] | 3 1/2 1/3| \ (2214 + 6 4449 ) / In floating-point [12.65702861, -0.328514307 + 0.4561888764 I, -0.328514307 - 0.4561888764 I] The largest root is, 12.65702861 and the remaining roots are [-0.328514307 + 0.4561888764 I, -0.328514307 - 0.4561888764 I] whose absolute values are [0.5621654035, 0.5621654035] so the largest absolute value is, 0.5621654035 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5621654035 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.4473684211, -0.3991683992, 0.1208935611, 0.04671926910, -0.06890187635, 0.03050581640, 0.001731861688, -0.01077863403, 0.006534550806, -0.0008870157757, -0.001482318959, 0.001254249511, -0.0003556206428, -0.0001627274611, 0.0002193033685, -0.00009266183794, -5 -5 -0.8424951750 10 , 0.00003481934949, -0.00002021477187, 0.2277726550 10 , -5 -5 -5 -6 0.4891941638 10 , -0.3933975410 10 , 0.1038734378 10 , 0.5607758084 10 , -6 -6 -7 -0.6967169148 10 , 0.2805410021 10 , 0.3585993998 10 , -6 -7 -8 -0.1122203628 10 , 0.6239917451 10 , -0.5533048428 10 , -7 -7 -8 -0.1608463628 10 , 0.1231667522 10 , -0.3009181373 10 , -8 -8 -9 -0.1915319876 10 , 0.2209411370 10 , -0.8463480592 10 , -9 -9 -9 -0.1421652534 10 , 0.3608779660 10 , -0.1921786725 10 , -10 -10 -10 0.1221864430 10 , 0.5270621554 10 , -0.3849094907 10 , -11 -11 -11 0.8632912676 10 , 0.6492221715 10 , -0.6993834302 10 , -11 -12 -11 0.2543412801 10 , 0.5391660564 10 , -0.1158042122 10 , -12 -13 -12 0.5904741912 10 , -0.2198245529 10 , -0.1721644215 10 , -12 -13 -13 0.1200640649 10 , -0.2447641403 10 , -0.2186213496 10 , -13 -14 -14 0.2209932786 10 , -0.7610801496 10 , -0.1983534927 10 ] The largest is 0.4473684211 The smallest is -0.3991683992 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 42, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 3, a(2) = 40, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [3, 40, 533, 7102, 94631, 1260916, 16801145, 223867786, 2982938699, 39746331712, 529602195677, 7056713754070, 94027572795887, 1252875595355788, 16694010179818049, 222440262159250594, 2963917578611145875, 39492883741122786904, 526225114168218703781, 7011715644684956770798] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 14 a(n - 1) - 9 a(n - 2) with initial conditions, a(1) = 3, a(2) = 40, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 14 b(n - 1) - 9 b(n - 2) with initial conditions, b(1) = 3, b(2) = 40, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 14 t + 9 = 0 whose roots are 1/2 1/2 [7 + 2 10 , 7 - 2 10 ] In floating-point [13.32455532, 0.675444680] The largest root is, 13.32455532 and the remaining roots are [0.675444680] whose absolute values are [0.675444680] so the largest absolute value is, 0.675444680 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.675444680 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.2250000000, 0.1519699812, 0.1026471417, 0.06933245976, 0.04683024087, 0.03163123704, 0.02136515077, 0.01443097742, 0.009747326918, 0.006583780108, 0.004446979246, 0.003003688472, 0.002028825398, 0.001370359321, 0.0009256019126, 0.0006251928874, 0.0004222832095, 0.0002852289472, 0.0001926563749, 0.0001301287234, 0.00008789475389, 0.00005936804389, 0.00004009982939, 0.00002708521641, 0.00001829456532, -5 -5 -5 0.00001235696682, 0.8346447492 10 , 0.5637563553 10 , 0.3807862308 10 , -5 -5 -5 -6 0.2572000337 10 , 0.1737243944 10 , 0.1173412179 10 , 0.7925750133 10 , -6 -6 -6 -6 0.5353405760 10 , 0.3615929439 10 , 0.2442360301 10 , 0.1649679271 10 , -6 -7 -7 -7 0.1114267087 10 , 0.7526257756 10 , 0.5083570759 10 , 0.3433670823 10 , -7 -7 -7 -8 0.2319254689 10 , 0.1566528241 10 , 0.1058103166 10 , 0.7146901538 10 , -8 -8 -8 -8 0.4827336620 10 , 0.3260598837 10 , 0.2202354137 10 , 0.1487568384 10 , -8 -9 -9 -9 0.1004770151 10 , 0.6786666527 10 , 0.4584017798 10 , 0.3096250433 10 , -9 -9 -10 0.2091345882 10 , 0.1412588449 10 , 0.9541253527 10 , -10 0.6444588932 10 ] The largest is 0.3333333333 The smallest is -10 0.6444588932 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 43, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 6, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 6, 9, 14, 22, 35, 56, 90, 145, 234, 378, 611, 988, 1598, 2585, 4182, 6766, 10947, 17712, 28658] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) with initial conditions, a(1) = 4, a(2) = 6, a(3) = 9, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) with initial conditions, b(1) = 4, b(2) = 6, b(3) = 9, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + 1 = 0 whose roots are 1/2 1/2 5 5 [1, 1/2 - ----, ---- + 1/2] 2 2 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [-0.6180339880, 1.618033988] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, -0.2222222222, -0.4285714286, -0.3181818182, -0.4000000000, -0.3571428571, -0.3888888889, -0.3724137931, -0.3846153846, -0.3783068783, -0.3829787234, -0.3805668016, -0.3823529412, -0.3814313346, -0.3821138211, -0.3817617499, -0.3820224719, -0.3818879855, -0.3819875776, -0.3819362074, -0.3819742489, -0.3819546271, -0.3819691578, -0.3819616629, -0.3819672131, -0.3819643503, -0.3819664703, -0.3819653768, -0.3819661866, -0.3819657689, -0.3819660782, -0.3819659187, -0.3819660368, -0.3819659759, -0.3819660210, -0.3819659977, -0.3819660150, -0.3819660061, -0.3819660127, -0.3819660093, -0.3819660118, -0.3819660105, -0.3819660115, -0.3819660110, -0.3819660113, -0.3819660111, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113] The largest is 0. The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 44, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 7, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 7, 12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 4, a(2) = 7, a(3) = 12, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 4, b(2) = 7, b(3) = 12, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (100 + 12 69 ) 2 (100 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (100 + 12 69 ) 1 - ----------------------- + 2/3 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (100 + 12 69 ) / 1/2 1/3 (100 + 12 69 ) 1 - ------------------- - ----------------------- + 2/3 12 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (100 + 12 69 ) / In floating-point [1.754877667, 0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] The largest root is, 1.754877667 and the remaining roots are [0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] whose absolute values are [0.7548776666, 0.7548776666] so the largest absolute value is, 0.7548776666 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7548776666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.4285714286, -0.2500000000, 0.1904761905, 0.1891891892, -0.06153846154, -0.1228070175, 0.005000000000, 0.07122507122, 0.01461038961, -0.03700277521, -0.01739588824, 0.01682186843, 0.01403628894, -0.006145142411, -0.009504752376, 0.001171924490, 0.005703456367, 0.0007302348065, -0.003071062271, -0.001168903168, 0.001463490759, 0.001024822391, -0.0005827491423, -0.0007268299193, 0.0001539116948, 0.0004519041666, 0.00002306671899, -0.0002518590338, -0.00007488062000, 0.0001251645128, 0.00007335061177, -0.00005334390925, -0.00005487391748, -6 0.00001694668605, 0.00003542338034, -0.9738428538 10 , -0.00002042437999, -5 -5 -5 -0.4451536794 10 , 0.00001054746355, 0.5122083903 10 , -0.4754832539 10 , -5 -5 -5 -0.4084285431 10 , 0.1708345581 10 , 0.2746144054 10 , -6 -5 -6 -0.3003429039 10 , -0.1638484280 10 , -0.2304816029 10 , -6 -6 -6 0.8771781708 10 , 0.3463536640 10 , -0.4149524457 10 , -6 -6 -6 -0.2990803846 10 , 0.1631453405 10 , 0.2104186200 10 , -7 -6 -8 -0.4138848514 10 , -0.1300502497 10 , -0.8293394300 10 , -7 0.7207497597 10 ] The largest is 0.2500000000 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 45, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 9, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 9, 20, 44, 97, 214, 472, 1041, 2296, 5064, 11169, 24634, 54332, 119833, 264300, 582932, 1285697, 2835694, 6254320, 13794337] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 3) with initial conditions, a(1) = 4, a(2) = 9, a(3) = 20, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 3) with initial conditions, b(1) = 4, b(2) = 9, b(3) = 20, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (172 + 12 177 ) 8 (172 + 12 177 ) [-------------------- + ---------------------- + 2/3, - -------------------- 6 1/2 1/3 12 3 (172 + 12 177 ) 4 - ---------------------- + 2/3 1/2 1/3 3 (172 + 12 177 ) / 1/2 1/3 \ 1/2 |(172 + 12 177 ) 8 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (172 + 12 177 ) / 1/2 1/3 (172 + 12 177 ) 4 - -------------------- - ---------------------- + 2/3 12 1/2 1/3 3 (172 + 12 177 ) / 1/2 1/3 \ 1/2 |(172 + 12 177 ) 8 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (172 + 12 177 ) / In floating-point [2.205569431, -0.1027847152 + 0.6654569515 I, -0.1027847152 - 0.6654569515 I] The largest root is, 2.205569431 and the remaining roots are [-0.1027847152 + 0.6654569515 I, -0.1027847152 - 0.6654569515 I] whose absolute values are [0.6733480912, 0.6733480912] so the largest absolute value is, 0.6733480912 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6733480912 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, 0.4444444444, -0.2000000000, -0.1590909091, 0.1237113402, 0.04672897196, -0.06567796610, -0.007684918348, 0.03135888502, -0.002962085308, -0.01360909661, 0.004140618657, 0.005319148936, -0.002970801031, -0.001800983731, 0.001717181421, 0.0004635617879, -0.0008738601556, -0.00003053889152, 0.0004024840048, -0.00006889214593, -0.0001683231834, 0.00006583763810, 0.00006278313027, -0.00004275692283, -5 -0.00001967620755, 0.00002343071517, 0.4104507522 10 , -0.00001146719251, -6 -5 -5 0.4963301608 10 , 0.5097167843 10 , -0.1272856820 10 , -5 -6 -6 -0.2049383478 10 , 0.9984008870 10 , 0.7239449544 10 , -6 -6 -6 -0.6014935694 10 , -0.2045862517 10 , 0.3147724510 10 , -7 -6 -7 -7 0.2805133261 10 , -0.1484835865 10 , 0.1780527797 10 , 0.6366188856 10 , -7 -7 -7 -0.2115980939 10 , -0.2451434080 10 , 0.1463320696 10 , -8 -8 -8 0.8106604528 10 , -0.8301131744 10 , -0.1969056531 10 , -8 -10 -8 0.4168491467 10 , 0.3585119014 10 , -0.1897354150 10 , -9 -9 -9 0.3737831665 10 , 0.7834175232 10 , -0.3305191039 10 , -9 -9 -10 -0.2872550413 10 , 0.2089074407 10 , 0.8729577750 10 , -9 -0.1126634863 10 ] The largest is 0.4444444444 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 46, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 10, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 10, 25, 63, 159, 401, 1011, 2549, 6427, 16205, 40859, 103021, 259755, 654941, 1651355, 4163693, 10498251, 26470077, 66741115, 168279693] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 4, a(2) = 10, a(3) = 25, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 4, b(2) = 10, b(3) = 25, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 2 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (27 + 3 78 ) 1 (27 + 3 78 ) [----------------- + ------------------- + 1, - ----------------- 3 1/2 (1/3) 6 (27 + 3 78 ) 1 - --------------------- + 1 1/2 (1/3) 2 (27 + 3 78 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(27 + 3 78 ) 1 | (27 + 3 78 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 (1/3)| 6 \ (27 + 3 78 ) / 1 - --------------------- + 1 1/2 (1/3) 2 (27 + 3 78 ) / 1/2 1/3 \ 1/2 |(27 + 3 78 ) 1 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 (1/3)| \ (27 + 3 78 ) / In floating-point [2.521379707, 0.2393101464 + 0.8578736270 I, 0.2393101464 - 0.8578736270 I] The largest root is, 2.521379707 and the remaining roots are [0.2393101464 + 0.8578736270 I, 0.2393101464 - 0.8578736270 I] whose absolute values are [0.8906270297, 0.8906270297] so the largest absolute value is, 0.8906270297 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8906270297 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, -0.2400000000, 0.2857142857, 0.3270440252, -0.06982543641, -0.2927794263, -0.08473911338, 0.1916913023, 0.1589632829, -0.07596857485, -0.1624523155, -0.01749340725, 0.1204871889, 0.07154367171, -0.06133017012, -0.08610348524, 0.007437228082, 0.07185831402, 0.02849351526, -0.04336162610, -0.04335528080, 0.01364444032, 0.04092063036, 0.008762448829, -0.02826503359, -0.02047873770, 0.01261875173, 0.02228366341, 0.0006560113662, -0.01736178926, -0.008830063711, 0.009545410126, 0.01157277927, -0.002032609855, -0.01015256786, -0.003246925319, 0.006499140050, 0.005686135071, -0.002433725524, -0.005675166615, -0.0007857786537, 0.004125546221, 0.002597862739, -0.002029061530, -0.003031817628, 0.0001583956542, 0.002480699159, 0.001061670911, -0.001459594275, -0.001540726330, 0.0004203513823, 0.001423318257, 0.0003477993471, -0.0009625357088, -0.0007365693059, 0.0004109621942, 0.0007809537767] The largest is 0.3270440252 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 47, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 11, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 11, 30, 82, 224, 612, 1672, 4568, 12480, 34096, 93152, 254496, 695296, 1899584, 5189760, 14178688, 38736896, 105831168, 289136128, 789934592] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 4, a(2) = 11, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 4, b(2) = 11, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 2 t - 2 = 0 whose roots are 1/2 1/2 [1 + 3 , 1 - 3 ] In floating-point [2.732050808, -0.732050808] The largest root is, 2.732050808 and the remaining roots are [-0.732050808] whose absolute values are [0.732050808] so the largest absolute value is, 0.732050808 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.732050808 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.1818181818, 0.1333333333, -0.09756097561, 0.07142857143, -0.05228758170, 0.03827751196, -0.02802101576, 0.02051282051, -0.01501642421, 0.01099278598, -0.008047277757, 0.005891016200, -0.004312523163, 0.003156986065, -0.002311074198, 0.001691823733, -0.001238500930, 0.0009066456060, -0.0006637106481, 0.0004858699159, -0.0003556814643, 0.0002603769032, -0.0001906091223, 0.0001395355619, -0.0001021471208, 0.00007477688224, -0.00005474047703, 0.00004007281042, -0.00002933533323, 0.00002147495438, -0.00001572075770, 0.00001150839337, -5 -5 -5 -0.8424728658 10 , 0.6167329418 10 , -0.4514798481 10 , -5 -5 -5 0.3305061874 10 , -0.2419473214 10 , 0.1771177320 10 , -5 -6 -6 -0.1296591788 10 , 0.9491710652 10 , -0.6948414448 10 , -6 -6 -6 0.5086592408 10 , -0.3723644080 10 , 0.2725896656 10 , -6 -6 -6 -0.1995494848 10 , 0.1460803615 10 , -0.1069382466 10 , -7 -7 -7 0.7828422979 10 , -0.5730803364 10 , 0.4195239231 10 , -7 -7 -7 -0.3071128267 10 , 0.2248221928 10 , -0.1645812678 10 , -7 -8 -8 0.1204818500 10 , -0.8819883559 10 , 0.6456602882 10 , -8 -0.4726561354 10 ] The largest is 0.2500000000 The smallest is -0.1818181818 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 1, : Consider the Pisot Sequence a(n), defined by, a(1) = 4, a(2) = 13, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 4, 13, 42, 136, 440, 1424, 4609, 14918, 48285, 156284, 505844, 1637264, 5299328, 17152321, 55516872, 179691313, 581606398, 1882483892, 6093030640, 19721296176, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 3 a(n - 1) + a(n - 2) - a(n - 3) + a(n - 4) - a(n - 5) + a(n - 6), . Alas, it breaks down at the, 25, -th term. a(25), equals , 7005609443657, while the corresponding term for the solution of the recurrence is , 7005609443656 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.07253346951707 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 48, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 14, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 14, 49, 172, 604, 2121, 7448, 26154, 91841, 322504, 1132488, 3976785, 13964668, 49037590, 172197809, 604680724, 2123364868, 7456295833, 26183134320, 91943310482] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 4, a(2) = 14, a(3) = 49, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 4, b(2) = 14, b(3) = 49, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (332 + 12 321 ) 20 (332 + 12 321 ) [-------------------- + ---------------------- + 4/3, - -------------------- 6 1/2 1/3 12 3 (332 + 12 321 ) 10 - ---------------------- + 4/3 1/2 1/3 3 (332 + 12 321 ) / 1/2 1/3 \ 1/2 |(332 + 12 321 ) 20 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (332 + 12 321 ) / 1/2 1/3 (332 + 12 321 ) 10 - -------------------- - ---------------------- + 4/3 12 1/2 1/3 3 (332 + 12 321 ) / 1/2 1/3 \ 1/2 |(332 + 12 321 ) 20 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (332 + 12 321 ) / In floating-point [3.511547142, 0.244226429 + 0.4744767784 I, 0.244226429 - 0.4744767784 I] The largest root is, 3.511547142 and the remaining roots are [0.244226429 + 0.4744767784 I, 0.244226429 - 0.4744767784 I] whose absolute values are [0.5336429161, 0.5336429161] so the largest absolute value is, 0.5336429161 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5336429161 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, -0.2448979592, 0.02325581395, 0.08112582781, 0.03300330033, -0.006981740064, -0.01280874818, -0.004268246208, 0.001562771314, 0.001978828915, 0.0005215268112, -0.0003087792707, -0.0002993417907, -5 -0.00005828181008, 0.00005677707034, 0.00004433011086, 0.5484492691 10 , -5 -5 -6 -0.9945180620 10 , -0.6419596998 10 , -0.3035340634 10 , -5 -6 -7 0.1679877124 10 , 0.9069796232 10 , -0.3536981798 10 , -6 -6 -7 -0.2755613946 10 , -0.1245263194 10 , 0.1764769383 10 , -7 -7 -8 0.4408201940 10 , 0.1650637057 10 , -0.4490862676 10 , -8 -8 -9 -0.6894172455 10 , -0.2088593893 10 , 0.9431066600 10 , -8 -9 -9 0.1055441972 10 , 0.2469606736 10 , -0.1799345889 10 , -9 -10 -10 -0.1582177312 10 , -0.2604107339 10 , 0.3233657995 10 , -10 -11 -11 0.2321073537 10 , 0.2128708190 10 , -0.5570058031 10 , -11 -13 -12 -0.3326913133 10 , -0.3882828165 10 , 0.9284551092 10 , -12 -13 -12 0.4645638665 10 , -0.3748303387 10 , -0.1506047594 10 , -13 -13 -13 -0.6288910315 10 , 0.1217007225 10 , 0.2385373593 10 , -14 -14 -14 0.8185696074 10 , -0.2794615317 10 , -0.3696117487 10 , -14 -15 -15 -0.1009543241 10 , 0.5594466947 10 , 0.5607557727 10 , -15 0.1145864609 10 ] The largest is 0.08112582781 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 49, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 15, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 15, 56, 209, 780, 2911, 10864, 40545, 151316, 564719, 2107560, 7865521, 29354524, 109552575, 408855776, 1525870529, 5694626340, 21252634831, 79315912984, 296011017105] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - a(n - 2) with initial conditions, a(1) = 4, a(2) = 15, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - b(n - 2) with initial conditions, b(1) = 4, b(2) = 15, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t + 1 = 0 whose roots are 1/2 1/2 [2 + 3 , 2 - 3 ] In floating-point [3.732050808, 0.267949192] The largest root is, 3.732050808 and the remaining roots are [0.267949192] whose absolute values are [0.267949192] so the largest absolute value is, 0.267949192 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.267949192 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, 0.06666666667, 0.01785714286, 0.004784688995, 0.001282051282, -5 0.0003435245620, 0.00009204712813, 0.00002466395363, 0.6608686457 10 , -5 -6 -6 -7 0.1770792199 10 , 0.4744823398 10 , 0.1271371598 10 , 0.3406629929 10 , -8 -8 -9 -9 0.9128037383 10 , 0.2445850245 10 , 0.6553635980 10 , 0.1756041468 10 , -10 -10 -11 0.4705298933 10 , 0.1260781049 10 , 0.3378252640 10 , -12 -12 -13 0.9052000667 10 , 0.2425476269 10 , 0.6499044074 10 , -13 -14 -14 0.1741413611 10 , 0.4666103708 10 , 0.1250278720 10 , -15 -16 -16 0.3350111735 10 , 0.8976597338 10 , 0.2405272008 10 , -17 -17 -18 0.6444906920 10 , 0.1726907605 10 , 0.4627234980 10 , -18 -19 -20 0.1239863876 10 , 0.3322205244 10 , 0.8901822121 10 , -20 -21 -21 0.2385236048 10 , 0.6391220729 10 , 0.1712522433 10 , -22 -22 -23 0.4588690030 10 , 0.1229535788 10 , 0.3294531214 10 , -24 -24 -25 0.8827669782 10 , 0.2365366989 10 , 0.6337981746 10 , -25 -26 -26 0.1698257090 10 , 0.4550466159 10 , 0.1219293733 10 , -27 -28 -28 0.3267087710 10 , 0.8754135134 10 , 0.2345663440 10 , -29 -29 -30 0.6285186244 10 , 0.1684110578 10 , 0.4512560694 10 , -30 -31 -32 0.1209136994 10 , 0.3239872810 10 , 0.8681213031 10 , -32 -33 0.2326124021 10 , 0.6232830529 10 ] The largest is 0.2500000000 The smallest is -33 0.6232830529 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 50, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 17, 72, 305, 1292, 5473, 23184, 98209, 416020, 1762289, 7465176, 31622993, 133957148, 567451585, 2403763488, 10182505537, 43133785636, 182717648081, 774004377960, 3278735159921] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + a(n - 2) with initial conditions, a(1) = 4, a(2) = 17, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + b(n - 2) with initial conditions, b(1) = 4, b(2) = 17, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t - 1 = 0 whose roots are 1/2 1/2 [2 + 5 , 2 - 5 ] In floating-point [4.236067977, -0.236067977] The largest root is, 4.236067977 and the remaining roots are [-0.236067977] whose absolute values are [0.236067977] so the largest absolute value is, 0.236067977 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.236067977 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.05882352941, 0.01388888889, -0.003278688525, 0.0007739938080, -5 -0.0001827151471, 0.00004313319531, -0.00001018236618, 0.2403730590 10 , -6 -6 -7 -0.5674438188 10 , 0.1339553146 10 , -0.3162256020 10 , -8 -8 -9 0.7465073831 10 , -0.1762264881 10 , 0.4160143063 10 , -10 -10 -11 -0.9820765590 10 , 0.2318368270 10 , -0.5472925087 10 , -11 -12 -13 0.1291982356 10 , -0.3049956618 10 , 0.7199970903 10 , -13 -14 -15 -0.1699682569 10 , 0.4012406265 10 , -0.9472006318 10 , -15 -16 -16 0.2236037374 10 , -0.5278568206 10 , 0.1246100920 10 , -17 -18 -18 -0.2941645241 10 , 0.6944282425 10 , -0.1639322707 10 , -19 -20 -20 0.3869915960 10 , -0.9135632337 10 , 0.2156630249 10 , -21 -21 -22 -0.5091113411 10 , 0.1201848846 10 , -0.2837180264 10 , -23 -23 -24 0.6697674066 10 , -0.1581106371 10 , 0.3732485832 10 , -25 -25 -26 -0.8811203813 10 , 0.2080043064 10 , -0.4910315591 10 , -26 -27 -28 0.1159168270 10 , -0.2736425092 10 , 0.6459823370 10 , -28 -29 -30 -0.1524957438 10 , 0.3599936182 10 , -0.8498296536 10 , -30 -31 -31 0.2006175675 10 , -0.4735938342 10 , 0.1118003386 10 , -32 -33 -33 -0.2639247982 10 , 0.6230419331 10 , -0.1470802490 10 , -34 -35 -35 0.3472093692 10 , -0.8196501356 10 , 0.1934931498 10 , -36 -0.4567753653 10 ] The largest is 0.2500000000 The smallest is -0.05882352941 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 51, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 18, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 18, 81, 365, 1645, 7414, 33415, 150602, 678766, 3059211, 13787921, 62142417, 280077032, 1262312405, 5689265544, 25641625878, 115567285897, 520863912185, 2347543363253, 10580421705998] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 2 a(n - 2) - a(n - 3) with initial conditions, a(1) = 4, a(2) = 18, a(3) = 81, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 2 b(n - 2) - b(n - 3) with initial conditions, b(1) = 4, b(2) = 18, b(3) = 81, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 5 t + 2 t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (532 + 228 I 3 ) 38 (532 + 228 I 3 ) [--------------------- + ----------------------- + 5/3, - --------------------- 6 1/2 1/3 12 3 (532 + 228 I 3 ) 19 - ----------------------- + 5/3 1/2 1/3 3 (532 + 228 I 3 ) / 1/2 1/3 \ 1/2 |(532 + 228 I 3 ) 38 | + 1/2 I 3 |--------------------- - -----------------------|, | 6 1/2 1/3| \ 3 (532 + 228 I 3 ) / 1/2 1/3 (532 + 228 I 3 ) 19 - --------------------- - ----------------------- + 5/3 12 1/2 1/3 3 (532 + 228 I 3 ) / 1/2 1/3 \ 1/2 |(532 + 228 I 3 ) 38 | - 1/2 I 3 |--------------------- - -----------------------|] | 6 1/2 1/3| \ 3 (532 + 228 I 3 ) / In floating-point -9 -9 [4.507018645 - 0.2 10 I, -0.2851424815 + 0.1 10 I, -9 0.7781238385 + 0.1 10 I] -9 The largest root is, 4.507018645 - 0.2 10 I and the remaining roots are -9 -9 [-0.2851424815 + 0.1 10 I, 0.7781238385 + 0.1 10 I] whose absolute values are [0.2851424815, 0.7781238385] so the largest absolute value is, 0.7781238385 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7781238385 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, -0.2469135802, -0.2328767123, -0.1696048632, -0.1352845967, -0.1043244052, -0.08144646153, -0.06329869204, -0.04927610420, -0.03833667164, -0.02983245727, -0.02321283882, -0.01806260789, -0.01405490452, -0.01093646800, -0.008509923101, -0.006621774979, -0.005152560686, -0.004009330373, -0.003119755515, -0.002427556140, -0.001888939298, -0.001469828697, -0.001143708746, -0.0008899470387, -0.0006924890052, -0.0005388422023, -0.0004192859624, -0.0003262564022, -0.0002538678837, -0.0001975406520, -0.0001537110902, -0.0001196062634, -0.00009306848471, -0.00007241880650, -0.00005635079964, -0.00004384790047, -0.00003411909659, -0.00002654888238, -0.00002065831825, -0.00001607472987, -5 -5 -0.00001250813050, -0.9732874508 10 , -0.7573381664 10 , -5 -5 -5 -0.5893028805 10 , -0.4585506190 10 , -0.3568091674 10 , -5 -5 -5 -0.2776417187 10 , -0.2160396397 10 , -0.1681055935 10 , -5 -5 -6 -0.1308069696 10 , -0.1017840212 10 , -0.7920057318 10 , -6 -6 -6 -0.6162785395 10 , -0.4795410223 10 , -0.3731423006 10 , -6 -0.2903509190 10 ] The largest is 0. The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 52, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 19, 90, 426, 2016, 9541, 45154, 213697, 1011348, 4786332, 22651920, 107203069, 507352048, 2401107571, 11363544486, 53779407822, 254517831936, 1204537747753, 5700626846950, 26978935702753] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 3 a(n - 2) + 2 a(n - 3) + a(n - 4) with initial conditions, a(1) = 4, a(2) = 19, a(3) = 90, a(4) = 426, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 3 b(n - 2) + 2 b(n - 3) + b(n - 4) with initial conditions, b(1) = 4, b(2) = 19, b(3) = 90, b(4) = 426, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 4 t - 3 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 4 _Z - 3 _Z - 2 _Z - 1 In floating-point [4.73262615271643, -0.0831313726785751 + 0.605119600086641 I, -0.566363407359285, -0.0831313726785751 - 0.605119600086641 I] The largest root is, 4.73262615271643 and the remaining roots are [-0.0831313726785751 + 0.605119600086641 I, -0.566363407359285, -0.0831313726785751 - 0.605119600086641 I] whose absolute values are [0.610803205240805, 0.566363407359285, 0.610803205240805] so the largest absolute value is, 0.610803205240805 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.610803205240805 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, 0.3157894737, 0.4000000000, -0.4929577465, 0.1076388889, 0.06697411173, 0.004805775790, -0.05755813137, 0.02577154451, 0.006997425168, -0.005006154004, -0.005047383485, 0.004558398865, 0.00007656216748, -0.001119475707, -0.0001788020804, 0.0006378877573, -0.0001472444590, -0.0001523944321, 0.00004566232877, 0.00006886485806, -5 -0.00003958690463, -0.00001282281890, 0.00001334005540, 0.4582813685 10 , -5 -7 -5 -0.6881121504 10 , 0.8124693364 10 , 0.2187305991 10 , -6 -6 -6 -0.1864645581 10 , -0.9025678962 10 , 0.2861936560 10 , -6 -6 -7 0.2514478100 10 , -0.1272281425 10 , -0.8474972434 10 , -7 -7 -7 0.6840595107 10 , 0.1636615622 10 , -0.2604511311 10 , -8 -7 -8 -0.3019805955 10 , 0.1092370037 10 , -0.1088686358 10 , -8 -9 -8 -0.3668369321 10 , 0.8880584380 10 , 0.1293453448 10 , -9 -9 -9 -0.5874358919 10 , -0.3616356672 10 , 0.2661149904 10 , -10 -9 -10 0.9813462457 10 , -0.1198237569 10 , -0.1429684025 10 , -10 -11 -10 0.4572560791 10 , -0.1500978279 10 , -0.1724452676 10 , -11 -11 -11 0.3673333713 10 , 0.5683405937 10 , -0.2236406903 10 , -11 -11 -12 -0.1793269128 10 , 0.1157848366 10 , 0.4621782110 10 ] The largest is 0.4000000000 The smallest is -0.4929577465 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 53, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 21, 110, 576, 3016, 15792, 82688, 432960, 2267008, 11870208, 62153216, 325438464, 1704017920, 8922353664, 46718050304, 244618887168, 1280841121792, 6706571182080, 35116062605312, 183870090903552] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - 4 a(n - 2) with initial conditions, a(1) = 4, a(2) = 21, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - 4 b(n - 2) with initial conditions, b(1) = 4, b(2) = 21, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 6 t + 4 = 0 whose roots are 1/2 1/2 [3 + 5 , 3 - 5 ] In floating-point [5.236067977, 0.763932023] The largest root is, 5.236067977 and the remaining roots are [0.763932023] whose absolute values are [0.763932023] so the largest absolute value is, 0.763932023 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.763932023 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, 0.1904761905, 0.1454545455, 0.1111111111, 0.08488063660, 0.06484295846, 0.04953560372, 0.03784183296, 0.02890858788, 0.02208419600, 0.01687082451, 0.01288816309, 0.009845680496, 0.007521430614, 0.005745861701, 0.004389447750, 0.003353239698, 0.002561647184, 0.001956924314, 0.001494957149, 0.001142045639, 0.0008724452345, 0.0006664888525, 0.0005091521771, 0.0003889576524, 0.0002971372061, 0.0002269926268, 0.0001734069365, 0.0001324711117, 0.0001011989243, 0.00007730909890, 0.00005905889628, 0.00004511698208, 0.00003446630737, 0.00002632991590, 0.00002011426590, 0.00001536593183, 0.00001173852738, -5 -5 -5 -5 0.8967436965 10 , 0.6850512257 10 , 0.5233325684 10 , 0.3997905074 10 , -5 -5 -5 -5 0.3054127709 10 , 0.2333145958 10 , 0.1782364910 10 , 0.1361605631 10 , -5 -6 -6 -6 0.1040174143 10 , 0.7946223371 10 , 0.6070374491 10 , 0.4637353462 10 , -6 -6 -6 -6 0.3542622809 10 , 0.2706323008 10 , 0.2067446809 10 , 0.1579388822 10 , -6 -7 -7 -7 0.1206545697 10 , 0.9217188947 10 , 0.7041305794 10 , 0.5379078976 10 ] The largest is 0.2500000000 The smallest is -7 0.5379078976 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 2, :The Pisot Sequence a(n), defined by, a(1) = 4, a(2) = 22, and for n>1, 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [4, 22, 121, 666, 3666, 20180, 111084, 611479, 3365980, 18528553, 101993261, 561437544, 3090519047, 17012235968, 93646461397, 515491305709, 2837600933313, 15620009430934, 85982736951530, 473305159408853] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 54, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 23, 132, 758, 4353, 24998, 143556, 824399, 4734276, 27187526, 156129801, 896606582, 5148942468, 29568831047, 169804921092, 975138691862, 5599928802129, 32158710192326, 184677819625284, 1060549283773439] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - 2 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 4, a(2) = 23, a(3) = 132, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - 2 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 4, b(2) = 23, b(3) = 132, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 6 t + 2 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (1620 + 60 249 ) 20 (1620 + 60 249 ) [--------------------- + --------------------- + 2, - --------------------- 6 1/2 1/3 12 (1620 + 60 249 ) 10 - --------------------- + 2 1/2 1/3 (1620 + 60 249 ) / 1/2 1/3 \ 1/2 |(1620 + 60 249 ) 20 | + 1/2 I 3 |--------------------- - ---------------------|, | 6 1/2 1/3| \ (1620 + 60 249 ) / 1/2 1/3 (1620 + 60 249 ) 10 - --------------------- - --------------------- + 2 12 1/2 1/3 (1620 + 60 249 ) / 1/2 1/3 \ 1/2 |(1620 + 60 249 ) 20 | - 1/2 I 3 |--------------------- - ---------------------|] | 6 1/2 1/3| \ (1620 + 60 249 ) / In floating-point [5.742699833, 0.128650083 + 0.7112324000 I, 0.128650083 - 0.7112324000 I] The largest root is, 5.742699833 and the remaining roots are [0.128650083 + 0.7112324000 I, 0.128650083 - 0.7112324000 I] whose absolute values are [0.7227740800, 0.7227740800] so the largest absolute value is, 0.7227740800 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7227740800 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.4347826087, -0.2424242424, 0.1649076517, 0.1690787962, -0.04264341147, -0.09929922818, -0.003272687133, 0.05103209023, 0.01484022489, -0.02284089250, -0.01362953412, 0.008425254947, 0.009287920431, -0.002011589663, -0.005369613999, -0.0003307433765, 0.002719998751, 0.0008726372585, -0.001196404079, -0.0007637027420, 0.0004285034824, 0.0005092141400, -0.00009283035090, -0.0002898999382, -0.00002609650747, 0.0001447297789, 0.00005087187373, -0.00006251783779, -5 -0.00004266143754, 0.00002168267157, 0.00002786539109, -0.4157309201 10 , -5 -5 -5 -0.00001562662269, -0.1848944462 10 , 0.7687651001 10 , 0.2943926861 10 , -5 -5 -5 -0.3258574224 10 , -0.2376346062 10 , 0.1090852655 10 , -5 -6 -6 0.1522085385 10 , -0.1782311870 10 , -0.8409999262 10 , -6 -6 -6 -0.1232810277 10 , 0.4076201251 10 , 0.1692830276 10 , -6 -6 -7 -0.1693851678 10 , -0.1320166869 10 , 0.5451929737 10 , -7 -8 -7 0.8299365439 10 , -0.7126728929 10 , -0.4518979026 10 , -8 -7 -8 -0.7904320538 10 , 0.2157347051 10 , 0.9680093361 10 , -8 -8 -8 -0.8779342469 10 , -0.7315830000 10 , 0.2703985022 10 ] The largest is 0.2500000000 The smallest is -0.4347826087 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 55, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 25, 156, 973, 6069, 37855, 236118, 1472770, 9186303, 57298942, 357398265, 2229247441, 13904779737, 86730120658, 540973246008, 3374283935917, 21046867223484, 131278407014845, 818840161120305, 5107459975407127] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 4, a(2) = 25, a(3) = 156, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 4, b(2) = 25, b(3) = 156, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 6 t - t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (2268 + 12 9357 ) 26 (2268 + 12 9357 ) [---------------------- + ---------------------- + 2, - ---------------------- 6 1/2 1/3 12 (2268 + 12 9357 ) 13 - ---------------------- + 2 1/2 1/3 (2268 + 12 9357 ) / 1/2 1/3 \ 1/2 |(2268 + 12 9357 ) 26 | + 1/2 I 3 |---------------------- - ----------------------|, | 6 1/2 1/3| \ (2268 + 12 9357 ) / 1/2 1/3 (2268 + 12 9357 ) 13 - ---------------------- - ---------------------- + 2 12 1/2 1/3 (2268 + 12 9357 ) / 1/2 1/3 \ 1/2 |(2268 + 12 9357 ) 26 | - 1/2 I 3 |---------------------- - ----------------------|] | 6 1/2 1/3| \ (2268 + 12 9357 ) / In floating-point [6.237432186, -0.118716092 + 0.6832815260 I, -0.118716092 - 0.6832815260 I] The largest root is, 6.237432186 and the remaining roots are [-0.118716092 + 0.6832815260 I, -0.118716092 - 0.6832815260 I] whose absolute values are [0.6935179553, 0.6935179553] so the largest absolute value is, 0.6935179553 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6935179553 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, 0.4400000000, -0.2243589744, -0.1582733813, 0.1454934915, 0.04157971206, -0.07984990556, -0.001039537742, 0.03865200179, -0.008677245733, -0.01653008584, 0.008098244577, 0.006027644420, -0.005326146424, -0.001634500393, 0.002949784482, 0.00008576722539, -0.001439113344, 0.0003004406076, 0.0006208319779, -0.0002919075567, -0.0002292915398, 0.0001948391378, 0.00006402061689, -0.0001089117804, -5 -0.4932651843 10 , 0.00005355415923, -0.00001034303753, -0.00002330202150, -5 -5 -5 0.00001050731120, 0.8712733081 10 , -0.7122354805 10 , -0.2499462160 10 , -5 -6 -5 -6 0.4019071475 10 , 0.2479022769 10 , -0.1991901344 10 , 0.3537086363 10 , -6 -6 -6 0.8740573043 10 , -0.3776515714 10 , -0.3307262154 10 , -6 -7 -6 0.2601630491 10 , 0.9729736500 10 , -0.1482314070 10 , -7 -7 -7 -0.1160192977 10 , 0.7404910935 10 , -0.1200149471 10 , -7 -7 -7 -0.3276564819 10 , 0.1355194419 10 , 0.1254153281 10 , -8 -8 -8 -0.9495803552 10 , -0.3777455942 10 , 0.5464059218 10 , -9 -8 -9 -8 0.5194887105 10 , -0.2751376345 10 , 0.4034082916 10 , 0.1227539536 10 , -9 -9 -0.4854835317 10 , -0.4751367798 10 ] The largest is 0.4400000000 The smallest is -0.2243589744 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 56, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 26, 169, 1099, 7147, 46478, 302253, 1965594, 12782536, 83126641, 540584313, 3515496307, 22861770102, 148673327051, 966843690484, 6287521375694, 40888641503177, 265904623471883, 1729215405168943, 11245332549810954] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) - 4 a(n - 2) + 5 a(n - 3) with initial conditions, a(1) = 4, a(2) = 26, a(3) = 169, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) - 4 b(n - 2) + 5 b(n - 3) with initial conditions, b(1) = 4, b(2) = 26, b(3) = 169, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 7 t + 4 t - 5 = 0 whose roots are 1/3 1/3 / 1/3 \ %1 74 %1 37 1/2 |%1 74 | [----- + ------- + 7/3, - ----- - ------- + 7/3 + 1/2 I 3 |----- - -------|, 6 1/3 12 1/3 | 6 1/3| 3 %1 3 %1 \ 3 %1 / 1/3 / 1/3 \ %1 37 1/2 |%1 74 | - ----- - ------- + 7/3 - 1/2 I 3 |----- - -------|] 12 1/3 | 6 1/3| 3 %1 \ 3 %1 / 1/2 %1 := 2276 + 12 13461 In floating-point [6.503141550, 0.248429225 + 0.8409174095 I, 0.248429225 - 0.8409174095 I] The largest root is, 6.503141550 and the remaining roots are [0.248429225 + 0.8409174095 I, 0.248429225 - 0.8409174095 I] whose absolute values are [0.8768461492, 0.8768461492] so the largest absolute value is, 0.8768461492 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8768461492 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, -0.2485207101, 0.2611464968, 0.3208339163, -0.04137441370, -0.2672330796, -0.1009659167, 0.1552988390, 0.1547901112, -0.04249416131, -0.1401253792, -0.03695045323, 0.08937753749, 0.07281767949, -0.03253865971, -0.07215364849, -0.01083250314, 0.05009377341, 0.03321818397, -0.02201032152, -0.03647611953, -0.001200630730, 0.02744845538, 0.01456111295, -0.01386918455, -0.01808646672, 0.001677035864, 0.01473919519, 0.006033889283, -0.008334376472, -0.008780216471, 0.002045437010, 0.007767042592, 0.002286467748, -0.004835711081, -0.004160635604, 0.001650733840, 0.004019123885, 0.0007277538188, -0.002728549611, -0.001915243127, 0.001146265651, 0.002042084012, 0.0001333098425, -0.001503838892, -0.0008496915559, 0.0007340638894, 0.001018018989, -0.00005858041390, -0.0008118194063, -0.0003583192438, 0.0004461408490, 0.0004971658864, -0.00009599841000, -0.0004299481709, -0.0001398141243, 0.0002611017638] The largest is 0.3208339163 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 57, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 27, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 27, 182, 1227, 8272, 55767, 375962, 2534607, 17087452, 115197747, 776623742, 5235731187, 35297505832, 237963690927, 1604269674722, 10815436502967, 72913967391412, 491560986863307, 3313935758136902, 22341419483137947] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + 5 a(n - 2) with initial conditions, a(1) = 4, a(2) = 27, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + 5 b(n - 2) with initial conditions, b(1) = 4, b(2) = 27, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 6 t - 5 = 0 whose roots are 1/2 1/2 [3 + 14 , 3 - 14 ] In floating-point [6.741657387, -0.741657387] The largest root is, 6.741657387 and the remaining roots are [-0.741657387] whose absolute values are [0.741657387] so the largest absolute value is, 0.741657387 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.741657387 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.1851851852, 0.1373626374, -0.1018744906, 0.07555609284, -0.05603672423, 0.04156005128, -0.03082331896, 0.02286034220, -0.01695454165, 0.01257446106, -0.009325941928, 0.006916653719, -0.005129787323, 0.003804544661, -0.002821668651, 0.002092711398, -0.001552074867, 0.001151107790, -0.0008537275952, 0.0006331733773, -0.0004695977123, 0.0003482806122, -0.0002583048887, 0.0001915737287, -0.0001420820710, 0.0001053762175, -0.00007815305010, 0.00005796278691, -0.00004298852907, 0.00003188276013, -0.00002364608456, 0.00001753729328, -5 -5 -5 -0.00001300666311, 0.9646487771 10 , -0.7154388912 10 , 0.5306105384 10 , -5 -5 -5 -0.3935312253 10 , 0.2918653402 10 , -0.2164640855 10 , -5 -5 -6 0.1605421880 10 , -0.1190672996 10 , 0.8830714227 10 , -6 -6 -6 -0.6549364437 10 , 0.4857384514 10 , -0.3602515105 10 , -6 -6 -6 0.2671831938 10 , -0.1981583893 10 , 0.1469656332 10 , -6 -7 -7 -0.1089981475 10 , 0.8083928122 10 , -0.5995505005 10 , -7 -7 -7 0.4446610575 10 , -0.3297861579 10 , 0.2445883401 10 , -7 -7 -8 -0.1814007491 10 , 0.1345372056 10 , -0.9978051229 10 ] The largest is 0.2500000000 The smallest is -0.1851851852 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 3, : Consider the Pisot Sequence a(n), defined by, a(1) = 4, a(2) = 29, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 4, 29, 210, 1521, 11016, 79785, 577855, 4185203, 30311971, 219539073, 1590045219, 11516145003, 83407436559, 604091080091, 4375221779983, 31688211024659, 229506701245947, 1662237286788051, 12039007064231835, 87194344781356587, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 7 a(n - 1) + 2 a(n - 2) - 2 a(n - 3) + 2 a(n - 4) - 2 a(n - 5) + 2 a(n - 6) , . Alas, it breaks down at the, 23, -th term. a(23), equals , 33126935817454044668, while the corresponding term for the solution of the recurrence is , 33126935817454044667 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.07574660319045 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 58, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 30, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 30, 225, 1688, 12664, 95010, 712800, 5347688, 40120324, 300997440, 2258193600, 16941799688, 127103617984, 953578132320, 7154094186000, 53672648194688, 402671965078144, 3020993316218400, 22664603965824000, 170038202391874688] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 8 a(n - 1) - 4 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 4, a(2) = 30, a(3) = 225, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 8 b(n - 1) - 4 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 4, b(2) = 30, b(3) = 225, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 8 t + 4 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (395 + 3 1713 ) 52 (395 + 3 1713 ) [-------------------- + ---------------------- + 8/3, - -------------------- 3 1/2 1/3 6 3 (395 + 3 1713 ) 26 - ---------------------- + 8/3 1/2 1/3 3 (395 + 3 1713 ) / 1/2 1/3 \ 1/2 |(395 + 3 1713 ) 52 | + 1/2 I 3 |-------------------- - ----------------------|, | 3 1/2 1/3| \ 3 (395 + 3 1713 ) / 1/2 1/3 (395 + 3 1713 ) 26 - -------------------- - ---------------------- + 8/3 6 1/2 1/3 3 (395 + 3 1713 ) / 1/2 1/3 \ 1/2 |(395 + 3 1713 ) 52 | - 1/2 I 3 |-------------------- - ----------------------|] | 3 1/2 1/3| \ 3 (395 + 3 1713 ) / In floating-point [7.502368126, 0.248815937 + 0.4524081368 I, 0.248815937 - 0.4524081368 I] The largest root is, 7.502368126 and the remaining roots are [0.248815937 + 0.4524081368 I, 0.248815937 - 0.4524081368 I] whose absolute values are [0.5163162720, 0.5163162720] so the largest absolute value is, 0.5163162720 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5163162720 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, -0.2488888889, 0.009478672986, 0.07106759318, 0.03283864856, -0.002603815937, -0.01004995056, -0.004307043981, 0.0005358185106, 0.001414822892, 0.0005612211321, -0.00009788549057, -0.0001983226687, -5 -0.00007259712306, 0.00001674270911, 0.00002768482777, 0.9313539573 10 , -5 -5 -5 -0.2745576263 10 , -0.3849112859 10 , -0.1183518678 10 , -6 -6 -6 -7 0.4371494854 10 , 0.5330448774 10 , 0.1487237213 10 , -0.6809076832 10 , -7 -7 -7 -0.7353127704 10 , -0.1843970038 10 , 0.1042596852 10 , -7 -8 -8 0.1010399555 10 , 0.2248689553 10 , -0.1574528729 10 , -8 -9 -9 -0.1382996949 10 , -0.2684815686 10 , 0.2350777887 10 , -9 -10 -10 0.1885546863 10 , 0.3116319883 10 , -0.3475757736 10 , -10 -11 -11 -0.2560404158 10 , -0.3475625552 10 , 0.5096007186 10 , -11 -12 -12 0.3462476529 10 , 0.3645323811 10 , -0.7416326938 10 , -12 -13 -12 -0.4662380179 10 , -0.3430860582 10 , 0.1072178374 10 , -13 -14 -13 0.6250108700 10 , 0.2520134604 10 , -0.1540759630 10 , -14 -16 -14 -0.8339134778 10 , -0.4242382838 10 , 0.2201955891 10 , -14 -16 -15 0.1107072885 10 , -0.3608813718 10 , -0.3130848573 10 , -15 -16 -16 -0.1461805385 10 , 0.1071884657 10 , 0.4430321211 10 ] The largest is 0.07106759318 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 59, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 31, 240, 1858, 14384, 111356, 862080, 6673928, 51667264, 399990256, 3096587520, 23972719648, 185588582144, 1436763217856, 11122928578560, 86109902192768, 666633360385024, 5160847078694656, 39953509908787200, 309306385112908288] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 8 a(n - 1) - 2 a(n - 2) with initial conditions, a(1) = 4, a(2) = 31, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 8 b(n - 1) - 2 b(n - 2) with initial conditions, b(1) = 4, b(2) = 31, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 8 t + 2 = 0 whose roots are 1/2 1/2 [4 + 14 , 4 - 14 ] In floating-point [7.741657387, 0.258342613] The largest root is, 7.741657387 and the remaining roots are [0.258342613] whose absolute values are [0.258342613] so the largest absolute value is, 0.258342613 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.258342613 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, 0.06451612903, 0.01666666667, 0.004305705059, 0.001112347052, -5 0.0002873666439, 0.00007423904974, 0.00001917911011, 0.4954781426 10 , -5 -6 -7 -7 0.1280031182 10 , 0.3306866005 10 , 0.8543044052 10 , 0.2207032325 10 , -8 -8 -9 -10 0.5701704984 10 , 0.1472993365 10 , 0.3805369553 10 , 0.9830891146 10 , -10 -11 -11 0.2539738109 10 , 0.6561225800 10 , 0.1695044219 10 , -12 -12 -13 0.4379021531 10 , 0.1131287866 10 , 0.2922598635 10 , -14 -14 -15 0.7550317688 10 , 0.1950568802 10 , 0.5039150417 10 , -15 -16 -17 0.1301827287 10 , 0.3363174633 10 , 0.8688513234 10 , -17 -18 -18 0.2244613214 10 , 0.5798792434 10 , 0.1498075191 10 , -19 -20 -20 0.3870166596 10 , 0.9998289521 10 , 0.2582984243 10 , -21 -21 -22 0.6672948992 10 , 0.1723907080 10 , 0.4453586601 10 , -22 -23 -24 0.1150551201 10 , 0.2972364039 10 , 0.7678882932 10 , -24 -25 -25 0.1983782683 10 , 0.5124956025 10 , 0.1323994532 10 , -26 -27 -27 0.3420442073 10 , 0.8836459436 10 , 0.2282834022 10 , -28 -28 -29 0.5897533069 10 , 0.1523584105 10 , 0.3936066991 10 , -29 -30 -31 0.1016853832 10 , 0.2626966763 10 , 0.6786574583 10 , -31 -32 -32 0.1753261413 10 , 0.4529421350 10 , 0.1170142548 10 , -33 -34 0.3022976837 10 , 0.7809637358 10 ] The largest is 0.2500000000 The smallest is -34 0.7809637358 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 60, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 33, 272, 2242, 18480, 152324, 1255552, 10349064, 85303616, 703127056, 5795623680, 47771243552, 393761195776, 3245632053312, 26752578818048, 220511894651008, 1817600314844160, 14981826308055296, 123489811094130688, 1017882141369156096] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 8 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 4, a(2) = 33, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 8 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 4, b(2) = 33, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 8 t - 2 = 0 whose roots are 1/2 1/2 [4 + 3 2 , 4 - 3 2 ] In floating-point [8.242640686, -0.242640686] The largest root is, 8.242640686 and the remaining roots are [-0.242640686] whose absolute values are [0.242640686] so the largest absolute value is, 0.242640686 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.242640686 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.06060606061, 0.01470588235, -0.003568242640, 0.0008658008658, -5 -0.0002100785168, 0.00005097359568, -0.00001236826828, 0.3001045114 10 , -6 -6 -7 -0.7281756485 10 , 0.1766850397 10 , -0.4287097944 10 , -7 -8 -9 0.1040224391 10 , -0.2524007609 10 , 0.6124269406 10 , -9 -10 -11 -0.1485996937 10 , 0.3605633178 10 , -0.8748733119 10 , -11 -12 -12 0.2122798615 10 , -0.5150773146 10 , 0.1249787135 10 , -13 -14 -14 -0.3032492093 10 , 0.7358059651 10 , -0.1785364650 10 , -15 -15 -16 0.4332021053 10 , -0.1051124565 10 , 0.2550455867 10 , -17 -17 -18 -0.6188443641 10 , 0.1501568217 10 , -0.3643415440 10 , -19 -19 -20 0.8840408257 10 , -0.2145042734 10 , 0.5204746429 10 , -20 -21 -22 -0.1262883250 10 , 0.3064268595 10 , -0.7435162373 10 , -22 -23 -23 0.1804072907 10 , -0.4377414898 10 , 0.1062138959 10 , -24 -25 -25 -0.2577181267 10 , 0.6253290336 10 , -0.1517302664 10 , -26 -27 -27 0.3681593609 10 , -0.8933044030 10 , 0.2167519942 10 , -28 -28 -29 -0.5259285280 10 , 0.1276116594 10 , -0.3096378072 10 , -30 -30 -31 0.7513073030 10 , -0.1822977202 10 , 0.4423284410 10 , -31 -32 -33 -0.1073268769 10 , 0.2604186715 10 , -0.6318816538 10 , -33 -34 -35 0.1533201987 10 , -0.3720171835 10 , 0.9026650503 10 , -35 -0.2190232680 10 ] The largest is 0.2500000000 The smallest is -0.06060606061 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 61, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 34, 289, 2457, 20889, 177595, 1509885, 12836807, 109136533, 927861799, 7888536445, 67067107743, 570194100309, 4847701398919, 41214401973549, 350398423965647, 2979032804998789, 25327272745179415, 215328526638688285, 1830689583157479327] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 9 a(n - 1) - 4 a(n - 2) - 2 a(n - 3) with initial conditions, a(1) = 4, a(2) = 34, a(3) = 289, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 9 b(n - 1) - 4 b(n - 2) - 2 b(n - 3) with initial conditions, b(1) = 4, b(2) = 34, b(3) = 289, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 9 t + 4 t + 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (540 + 3 I 4101 ) 23 (540 + 3 I 4101 ) [---------------------- + ---------------------- + 3, - ---------------------- 3 1/2 1/3 6 (540 + 3 I 4101 ) 23 - ------------------------ + 3 1/2 1/3 2 (540 + 3 I 4101 ) / 1/2 1/3 \ 1/2 |(540 + 3 I 4101 ) 23 | + 1/2 I 3 |---------------------- - ----------------------|, | 3 1/2 1/3| \ (540 + 3 I 4101 ) / 1/2 1/3 (540 + 3 I 4101 ) 23 - ---------------------- - ------------------------ + 3 6 1/2 1/3 2 (540 + 3 I 4101 ) / 1/2 1/3 \ 1/2 |(540 + 3 I 4101 ) 23 | - 1/2 I 3 |---------------------- - ----------------------|] | 3 1/2 1/3| \ (540 + 3 I 4101 ) / In floating-point -9 -8 [8.501844190 - 0.2 10 I, -0.2961586320 - 0.1732050808 10 I, -8 0.7943144420 + 0.1732050808 10 I] -9 The largest root is, 8.501844190 - 0.2 10 I and the remaining roots are -8 -8 [-0.2961586320 - 0.1732050808 10 I, 0.7943144420 + 0.1732050808 10 I] whose absolute values are [0.2961586320, 0.7943144420] so the largest absolute value is, 0.7943144420 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7943144420 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, -0.2491349481, -0.2417582418, -0.1790416008, -0.1460626707, -0.1148802723, -0.09158850795, -0.07265013632, -0.05773665007, -0.04585228940, -0.03642373167, -0.02893112726, -0.02298063990, -0.01825378672, -0.01449926639, -0.01151697077, -0.009148097963, -0.007266465809, -0.005771858888, -0.004584670826, -0.003641670262, -0.002892631278, -0.002297658800, -0.001825063567, -0.001449674349, -0.001151497271, -0.0009146509125, -0.0007265204291, -0.0005770856692, -0.0004583874813, -0.0003641037964, -0.0002892129038, -0.0002297259863, -0.0001824746686, -0.0001449422645, -0.0001151297340, -0.00009144921039, -0.00007263942851, -0.00005769854712, -0.00004583078925, -0.00003640405779, -0.00002891626885, -0.00002296860995, -0.00001824429859, -0.00001449170986, -5 -5 -0.00001151097443, -0.9143333228 10 , -0.7262681630 10 , -5 -5 -5 -0.5768852906 10 , -0.4582283176 10 , -0.3639773704 10 , -5 -5 -5 -0.2891124818 10 , -0.2296462197 10 , -0.1824113088 10 , -5 -5 -6 -0.1448919370 10 , -0.1150897580 10 , -0.9141745693 10 ] The largest is 0. The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 4, :The Pisot Sequence a(n), defined by, a(1) = 4, a(2) = 35, and for n>1, 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [4, 35, 306, 2675, 23384, 204415, 1786927, 15620713, 136551003, 1193682799, 10434772307, 91217259049, 797390503943, 6970518763745, 60933923335503, 532663800055841, 4656367231233475, 40704391381265113, 355824056703601003, 3110493856623101673] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 62, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 37, 342, 3161, 29216, 270033, 2495818, 23067949, 213208764, 1970611997, 18213658622, 168342302241, 1555927412056, 14380878004873, 132917288164338, 1228506735609269, 11354646338942324, 104946916240158357, 969986638028987302, 8965237966608764521] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 10 a(n - 1) - 7 a(n - 2) with initial conditions, a(1) = 4, a(2) = 37, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 10 b(n - 1) - 7 b(n - 2) with initial conditions, b(1) = 4, b(2) = 37, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 10 t + 7 = 0 whose roots are 1/2 1/2 [5 + 3 2 , 5 - 3 2 ] In floating-point [9.242640686, 0.757359314] The largest root is, 9.242640686 and the remaining roots are [0.757359314] whose absolute values are [0.757359314] so the largest absolute value is, 0.757359314 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.757359314 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, 0.1891891892, 0.1432748538, 0.1085099652, 0.08218099671, 0.06224054097, 0.04713845320, 0.03570074652, 0.02703829285, 0.02047770290, 0.01550897900, 0.01174586968, 0.008895843787, 0.006737350138, 0.005102594871, 0.003864497745, 0.002926813357, 0.002216649353, 0.001678800031, 0.001271454838, 0.0009629481624, 0.0007292977586, 0.0005523404493, 0.0004183201832, 0.0003168186865, 0.0002399455827, 0.0001817250217, 0.0001376311375, 0.0001042362238, 0.00007894427480, 0.00005978918172, 0.00004528189358, 0.00003429466381, 0.00002597338302, -5 0.00001967118352, 0.00001489815403, 0.00001128325570, 0.8545478784 10 , -5 -5 -5 -5 0.6471997940 10 , 0.4901627913 10 , 0.3712293548 10 , 0.2811540091 10 , -5 -5 -5 -6 0.2129346071 10 , 0.1612680078 10 , 0.1221378275 10 , 0.9250222114 10 , -6 -6 -6 -6 0.7005741864 10 , 0.5305863845 10 , 0.4018445396 10 , 0.3043407044 10 , -6 -6 -6 -6 0.2304952667 10 , 0.1745677368 10 , 0.1322105012 10 , 0.1001308544 10 , -7 -7 -7 -7 0.7583503506 10 , 0.5743437004 10 , 0.4349845503 10 , 0.3294396002 10 ] The largest is 0.2500000000 The smallest is -7 0.3294396002 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 5, :The Pisot Sequence a(n), defined by, a(1) = 4, a(2) = 38, and for n>1, 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [4, 38, 361, 3430, 32590, 309653, 2942160, 27954857, 265612349, 2523708847, 23978954173, 227835411329, 2164772253241, 20568527434197, 195431330098546, 1856885715629157, 17643151480210182, 167635946323262769, 1592788597389578202, 15133839558981299651] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 63, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 4, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [4, 39, 380, 3703, 36085, 351641, 3426670, 33392202, 325400215, 3170958894, 30900349305, 301117617443, 2934329920985, 28594448104046, 278647079364720, 2715359098939987, 26460621991825940, 257853378018263039, 2512728709701852505, 24486030072922865633] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 10 a(n - 1) - 3 a(n - 2) + 5 a(n - 3) with initial conditions, a(1) = 4, a(2) = 39, a(3) = 380, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 10 b(n - 1) - 3 b(n - 2) + 5 b(n - 3) with initial conditions, b(1) = 4, b(2) = 39, b(3) = 380, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 10 t + 3 t - 5 = 0 whose roots are 1/3 %1 182 [----- + ------- + 10/3, 6 1/3 3 %1 1/3 / 1/3 \ %1 91 1/2 |%1 182 | - ----- - ------- + 10/3 + 1/2 I 3 |----- - -------|, 12 1/3 | 6 1/3| 3 %1 \ 3 %1 / 1/3 / 1/3 \ %1 91 1/2 |%1 182 | - ----- - ------- + 10/3 - 1/2 I 3 |----- - -------|] 12 1/3 | 6 1/3| 3 %1 \ 3 %1 / 1/2 %1 := 7460 + 12 51549 In floating-point [9.744796555, 0.127601722 + 0.7048490230 I, 0.127601722 - 0.7048490230 I] The largest root is, 9.744796555 and the remaining roots are [0.127601722 + 0.7048490230 I, 0.127601722 - 0.7048490230 I] whose absolute values are [0.7163060412, 0.7163060412] so the largest absolute value is, 0.7163060412 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7163060412 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.4358974359, -0.2394736842, 0.1625708885, 0.1643619232, -0.04146842945, -0.09491605553, -0.002945687739, 0.04794914164, 0.01374820187, -0.02109384491, -0.01243734653, 0.007649078843, 0.008333603437, -0.001797934785, -0.004734763944, -0.0002858179020, 0.002356438889, 0.0007480228724, -0.001018177453, -0.0006436486993, 0.0003581597269, 0.0004216561037, -0.00007616163982, -0.0002357860750, -0.00002109531206, 0.0001155969053, 0.00004032461382, -0.00004902113795, -5 -0.00003320069459, 0.00001667953699, 0.00002129176394, -0.3124444582 10 , -5 -5 -5 -0.00001172205268, -0.1388373387 10 , 0.5660201264 10 , 0.2156869397 10 , -5 -5 -6 -0.2353776756 10 , -0.1707369427 10 , 0.7719829810 10 , -5 -6 -6 0.1073054314 10 , -0.1222529437 10 , -0.5817774720 10 , -7 -6 -6 -0.8574432133 10 , 0.2766244844 10 , 0.1145904478 10 , -6 -7 -7 -0.1126905814 10 , -0.8755473535 10 , 0.3547662985 10 , -7 -8 -7 0.5397759771 10 , -0.4427589228 10 , -0.2882553614 10 , -8 -7 -8 -0.5084605145 10 , 0.1349261082 10 , 0.6052242912 10 , -8 -8 -8 -0.5378429058 10 , -0.4477965236 10 , 0.1616849375 10 ] The largest is 0.2500000000 The smallest is -0.4358974359 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 64, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 7, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 7, 10, 14, 20, 29, 42, 61, 89, 130, 190, 278, 407, 596, 873, 1279, 1874, 2746, 4024, 5897] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 5, a(2) = 7, a(3) = 10, a(4) = 14, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 5, b(2) = 7, b(3) = 10, b(4) = 14, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t + t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [1, ------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.465571232, -0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.2857142857, -0.4000000000, -0.4285714286, 0.05000000000, -0.1724137931, -0.4047619048, -0.1475409836, -0.1123595506, -0.3076923077, -0.2421052632, -0.1402877698, -0.2334152334, -0.2600671141, -0.1844215349, -0.2017200938, -0.2454642476, -0.2134013110, -0.1985586481, -0.2274037646, -0.2241379310, -0.2060007896, -0.2166909110, -0.2241011690, -0.2133647720, -0.2133125663, -0.2206663786, -0.2172807828, -0.2138410261, -0.2177537536, -0.2182799455, -0.2153657483, -0.2163638549, -0.2178878591, -0.2164974644, -0.2161050407, -0.2172365282, -0.2169775571, -0.2163261190, -0.2168061388, -0.2170271673, -0.2165967439, -0.2166463309, -0.2169169401, -0.2167571215, -0.2166468869, -0.2168072594, -0.2168078119, -0.2166981289, -0.2167488177, -0.2168000586, -0.2167416162, -0.2167338624, -0.2167773493, -0.2167623938, -0.2167396844, -0.2167604618, -0.2167662837] The largest is 0.2857142857 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 65, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 8, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 5, a(2) = 8, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 5, b(2) = 8, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.1250000000, -0.07692307692, 0.04761904762, -0.02941176471, 0.01818181818, -0.01123595506, 0.006944444444, -0.004291845494, 0.002652519894, -0.001639344262, 0.001013171226, -0.0006261740764, 0.0003869969040, -0.0002391772303, 0.0001478196600, -0.00009135757354, 0.00005646208571, -0.00003489548801, 0.00002156659765, -0.00001332889037, -5 -5 -5 0.8237707281 10 , -0.5091183089 10 , 0.3146524192 10 , -5 -5 -6 -0.1944658897 10 , 0.1201865295 10 , -0.7427936022 10 , -6 -6 -6 0.4590716928 10 , -0.2837219094 10 , 0.1753497834 10 , -6 -7 -7 -0.1083721260 10 , 0.6697765733 10 , -0.4139446871 10 , -7 -7 -8 0.2558318861 10 , -0.1581128010 10 , 0.9771908509 10 , -8 -8 -8 -0.6039371593 10 , 0.3732536915 10 , -0.2306834678 10 , -8 -9 -9 0.1425702237 10 , -0.8811324406 10 , 0.5445697969 10 , -9 -9 -9 -0.3365626437 10 , 0.2080071532 10 , -0.1285554906 10 , -10 -10 -10 0.7945166260 10 , -0.4910382795 10 , 0.3034783465 10 , -10 -10 -11 -0.1875599330 10 , 0.1159184135 10 , -0.7164151948 10 , -11 -11 -11 0.4427689404 10 , -0.2736462543 10 , 0.1691226861 10 , -11 -12 -12 -0.1045235683 10 , 0.6459911781 10 , -0.3992445045 10 , -12 0.2467466736 10 ] The largest is 0.1250000000 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 66, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 9, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 9, 16, 28, 49, 86, 151, 265, 465, 816, 1432, 2513, 4410, 7739, 13581, 23833, 41824, 73396, 128801, 226030] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 5, a(2) = 9, a(3) = 16, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 5, b(2) = 9, b(3) = 16, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (100 + 12 69 ) 2 (100 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (100 + 12 69 ) 1 - ----------------------- + 2/3 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (100 + 12 69 ) / 1/2 1/3 (100 + 12 69 ) 1 - ------------------- - ----------------------- + 2/3 12 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (100 + 12 69 ) / In floating-point [1.754877667, 0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] The largest root is, 1.754877667 and the remaining roots are [0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] whose absolute values are [0.7548776666, 0.7548776666] so the largest absolute value is, 0.7548776666 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7548776666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, 0.4444444444, 0., -0.2500000000, -0.06122448980, 0.1279069767, 0.06622516556, -0.05660377358, -0.05161290323, 0.01960784314, 0.03421787709, -0.002785515320, -0.02018140590, -0.003359607184, 0.01067668066, 0.004531531910, -0.004973221117, -0.003801297073, 0.001902159145, 0.002632393930, -0.0004386683642, -0.001607571533, -0.0001440807769, 0.0008807416143, 0.0002979924718, -0.0004288374476, -0.0002749257528, 0.0001769784138, 0.0002000451328, -0.00005181390101, -0.0001266945210, -5 -0.1530008233 10 , 0.00007182060353, 0.00001847669429, -0.00003639722319, -5 -0.00001945053714, 0.00001597284320, 0.00001499900035, -0.5425379648 10 , -5 -6 -5 -5 -0.9876916443 10 , 0.6705471088 10 , 0.5792631012 10 , 0.1037798473 10 , -5 -5 -5 -0.3046486958 10 , -0.1338141377 10 , 0.1408002678 10 , -5 -6 -6 0.1107659774 10 , -0.5308245068 10 , -0.7613061097 10 , -6 -6 -7 -6 0.1158720611 10 , 0.4622257251 10 , 0.4727327944 10 , -0.2518071051 10 , -7 -6 -7 -0.8866176457 10 , 0.1217568554 10 , 0.8036837027 10 , -7 -7 -0.4968187944 10 , -0.5797527374 10 ] The largest is 0.4444444444 The smallest is -0.2500000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 67, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 11, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 11, 24, 52, 113, 246, 536, 1168, 2545, 5545, 12081, 26321, 57346, 124941, 272212, 593075, 1292147, 2815232, 6133614, 13363453] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 4) with initial conditions, a(1) = 5, a(2) = 11, a(3) = 24, a(4) = 52, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 4) with initial conditions, b(1) = 5, b(2) = 11, b(3) = 24, b(4) = 52, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 3 t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 3 _Z + 2 _Z - 1 In floating-point [2.17872417610522, 0.667076110379437 + 0.670769076539608 I, -0.512876396864097, 0.667076110379437 - 0.670769076539608 I] The largest root is, 2.17872417610522 and the remaining roots are [0.667076110379437 + 0.670769076539608 I, -0.512876396864097, 0.667076110379437 - 0.670769076539608 I] whose absolute values are [0.946003007966020, 0.512876396864097, 0.946003007966020] so the largest absolute value is, 0.946003007966020 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.946003007966020 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, 0.3636363636, -0.3333333333, -0.4423076923, -0.4601769912, -0.1300813008, 0.1940298507, 0.3981164384, 0.3457760314, 0.1110910730, -0.1643075904, -0.3170472246, -0.2767586231, -0.08508816161, 0.1339433971, 0.2549576360, 0.2202272652, 0.06567842366, -0.1094759142, -0.2048270009, -0.1753019154, -0.05057331886, 0.08940795846, 0.1645435108, 0.1395127000, 0.03887775956, -0.07298416291, -0.1321644971, -0.1110124654, -0.02983064247, 0.05954884047, 0.1061433093, 0.08831978147, 0.02284208343, -0.04856447220, -0.08523427419, -0.07025409670, -0.01745165828, 0.03958874634, 0.06843528141, 0.05587425484, 0.01330054341, -0.03225813310, -0.05494020471, -0.04443009310, -0.01010932647, 0.02627407369, 0.04410066931, 0.03532376743, 0.007660637210, -0.02139154954, -0.03539525373, -0.02807889468, -0.005785539369, 0.01740962171, 0.02840469015, 0.02231593233, 0.004352877342] The largest is 0.3981164384 The smallest is -0.4601769912 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 68, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 12, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) with initial conditions, a(1) = 5, a(2) = 12, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) with initial conditions, b(1) = 5, b(2) = 12, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 2 t - 1 = 0 whose roots are 1/2 1/2 [1 + 2 , 1 - 2 ] In floating-point [2.414213562, -0.414213562] The largest root is, 2.414213562 and the remaining roots are [-0.414213562] whose absolute values are [0.414213562] so the largest absolute value is, 0.414213562 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.414213562 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.08333333333, -0.03448275862, 0.01428571429, -0.005917159763, 0.002450980392, -0.001015228426, 0.0004205214466, -0.0001741856819, -5 0.00007215007215, -0.00002988553839, 0.00001237899532, -0.5127547750 10 , -5 -6 -6 0.2123899820 10 , -0.8797481105 10 , 0.3644035988 10 , -6 -7 -7 -0.1509409128 10 , 0.6252177321 10 , -0.2589736641 10 , -7 -8 -8 0.1072704040 10 , -0.4443285616 10 , 0.1840469164 10 , -9 -9 -9 -0.7623472887 10 , 0.3157745862 10 , -0.1307981163 10 , -10 -10 -11 0.5417835369 10 , -0.2244140888 10 , 0.9295535919 10 , -11 -11 -12 -0.3850337047 10 , 0.1594861825 10 , -0.6606133979 10 , -12 -12 -13 0.2736350289 10 , -0.1133433401 10 , 0.4694834868 10 , -13 -14 -14 -0.1944664275 10 , 0.8055063171 10 , -0.3336516411 10 , -14 -15 -15 0.1382030349 10 , -0.5724557140 10 , 0.2371189206 10 , -16 -16 -16 -0.9821787280 10 , 0.4068317498 10 , -0.1685152284 10 , -17 -17 -17 0.6980129306 10 , -0.2891264226 10 , 0.1197600855 10 , -18 -18 -19 -0.4960625163 10 , 0.2054758220 10 , -0.8511087223 10 , -19 -19 -20 0.3525407758 10 , -0.1460271706 10 , 0.6048643455 10 , -20 -20 -21 -0.2505430153 10 , 0.1037783149 10 , -0.4298638551 10 , -21 -22 -22 0.1780554388 10 , -0.7375297759 10 , 0.3054948358 10 ] The largest is 0.08333333333 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 69, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 13, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) with initial conditions, a(1) = 5, a(2) = 13, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) with initial conditions, b(1) = 5, b(2) = 13, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t + 1 = 0 whose roots are 1/2 1/2 5 5 [3/2 + ----, 3/2 - ----] 2 2 In floating-point [2.618033988, 0.381966012] The largest root is, 2.618033988 and the remaining roots are [0.381966012] whose absolute values are [0.381966012] so the largest absolute value is, 0.381966012 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.381966012 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, -0.07692307692, -0.02941176471, -0.01123595506, -0.004291845494, -0.001639344262, -0.0006261740764, -0.0002391772303, -0.00009135757354, -5 -0.00003489548801, -0.00001332889037, -0.5091183089 10 , -5 -6 -6 -0.1944658897 10 , -0.7427936022 10 , -0.2837219094 10 , -6 -7 -7 -0.1083721260 10 , -0.4139446871 10 , -0.1581128010 10 , -8 -8 -9 -0.6039371593 10 , -0.2306834678 10 , -0.8811324406 10 , -9 -9 -10 -0.3365626437 10 , -0.1285554906 10 , -0.4910382795 10 , -10 -11 -11 -0.1875599330 10 , -0.7164151948 10 , -0.2736462543 10 , -11 -12 -12 -0.1045235683 10 , -0.3992445045 10 , -0.1524978309 10 , -13 -13 -14 -0.5824898820 10 , -0.2224913368 10 , -0.8498412846 10 , -14 -14 -15 -0.3246104857 10 , -0.1239901724 10 , -0.4736003159 10 , -15 -16 -16 -0.1808992236 10 , -0.6909735488 10 , -0.2639284103 10 , -16 -17 -17 -0.1008116821 10 , -0.3850663611 10 , -0.1470822620 10 , -18 -18 -19 -0.5618042495 10 , -0.2145901283 10 , -0.8196613536 10 , -19 -19 -20 -0.3130827778 10 , -0.1195869798 10 , -0.4567816168 10 , -20 -21 -21 -0.1744750522 10 , -0.6664353975 10 , -0.2545556705 10 , -22 -22 -22 -0.9723161412 10 , -0.3713917181 10 , -0.1418590132 10 , -23 -23 -24 -0.5418532142 10 , -0.2069695109 10 , -0.7905531854 10 , -24 -0.3019644469 10 ] The largest is -24 -0.3019644469 10 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 70, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 14, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 14, 39, 109, 305, 853, 2386, 6674, 18668, 52217, 146058, 408544, 1142753, 3196435, 8940862, 25008803, 69953012, 195668057, 547310079, 1530900481] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 5, a(2) = 14, a(3) = 39, a(4) = 109, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 5, b(2) = 14, b(3) = 39, b(4) = 109, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t - 2 t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 2 _Z - 2 _Z - _Z + 1 In floating-point [0.451027959780797, 2.79713555395954, -0.624081756870166 + 0.634960473669366 I, -0.624081756870166 - 0.634960473669366 I] The largest root is, 2.79713555395954 and the remaining roots are [0.451027959780797, -0.624081756870166 + 0.634960473669366 I, -0.624081756870166 - 0.634960473669366 I] whose absolute values are [0.451027959780797, 0.890310531433038, 0.890310531433038] so the largest absolute value is, 0.890310531433038 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.890310531433038 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, -0.3571428571, -0.3589743590, 0.4403669725, -0.3967213115, 0.08675263775, 0.1793797150, -0.3047647588, 0.2327512321, -0.05140088477, -0.1214449054, 0.1918251155, -0.1433914415, 0.02682332036, 0.08013377234, -0.1213023670, 0.08787757416, -0.01353913378, -0.05275925861, 0.07658315642, -0.05376891232, 0.006408363364, 0.03462131712, -0.04829270777, 0.03283449438, -0.002703473022, -0.02265198217, 0.03041629177, -0.02000934821, 0.0008653779533, 0.01478033341, -0.01913421724, 0.01216695851, -0.00001956201036, -0.009619757663, 0.01202253641, -0.007380963032, -0.0003170489032, 0.006246270200, -0.007545056846, 0.004466340836, 0.0004058870832, -0.004046871207, 0.004729429435, -0.002695337297, -0.0003845740135, 0.002616478021, -0.002960958716, 0.001621801894, 0.0003227383915, -0.001688356166, 0.001851525061, -0.0009727257122, -0.0002534958596, 0.001087438083, -0.001156366326, 0.0005813733684, 0.0001909480286] The largest is 0.4403669725 The smallest is -0.3967213115 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 71, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 16, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 16, 51, 163, 521, 1665, 5321, 17005, 54345, 173677, 555041, 1773813, 5668793, 18116461, 57897009, 185028613, 591318761, 1889750301, 6039308129, 19300561909] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 3) with initial conditions, a(1) = 5, a(2) = 16, a(3) = 51, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 3) with initial conditions, b(1) = 5, b(2) = 16, b(3) = 51, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t - 2 = 0 whose roots are 1/2 1/3 1/2 (1/3) 1 (2 + 3 ) 1 [(2 + 3 ) + --------------- + 1, - ------------- - ----------------- + 1 1/2 (1/3) 2 1/2 (1/3) (2 + 3 ) 2 (2 + 3 ) 1/2 1/3 1/2 / 1/2 (1/3) 1 \ (2 + 3 ) + 1/2 I 3 |(2 + 3 ) - ---------------|, - ------------- | 1/2 (1/3)| 2 \ (2 + 3 ) / 1 1/2 / 1/2 (1/3) 1 \ - ----------------- + 1 - 1/2 I 3 |(2 + 3 ) - ---------------|] 1/2 (1/3) | 1/2 (1/3)| 2 (2 + 3 ) \ (2 + 3 ) / In floating-point [3.195823345, -0.097911673 + 0.7850032635 I, -0.097911673 - 0.7850032635 I] The largest root is, 3.195823345 and the remaining roots are [-0.097911673 + 0.7850032635 I, -0.097911673 - 0.7850032635 I] whose absolute values are [0.7910858483, 0.7910858483] so the largest absolute value is, 0.7910858483 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7910858483 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, -0.4375000000, -0.03921568627, 0.2822085890, -0.03071017274, -0.1705705706, 0.05262168765, 0.09644222288, -0.05181709449, -0.05020814500, 0.04225994116, 0.02314561907, -0.03097943425, -0.008418421236, 0.02103597441, 0.001149054714, -0.01338967833, 0.001902913839, 0.008006850945, -0.002758803824, -0.004470583794, 0.002601950508, 0.002288243876, -0.002076435961, -0.001025406867, 0.001500267151, 0.0003479295310, -0.001007025141, -0.00002054112030, 0.0006342357011, -0.0001113431782, -0.0003751117753, 0.0001431360764, 0.0002067218726, -0.0001300579327, -0.0001039016453, 0.0001017388095, 0.00004510056309, -6 -0.00007250160125, -0.00001402718482, 0.00004811957172, -0.6444873339 10 , -5 -5 -0.00002998783164, 0.6275648527 10 , 0.00001753797091, -0.7361750535 10 , -5 -5 -5 -0.9533954551 10 , 0.6474078177 10 , 0.4698733462 10 , -5 -5 -5 -0.4971708716 10 , -0.1966969795 10 , 0.3496557539 10 , -6 -5 -6 -5 0.5462551850 10 , -0.2295174034 10 , 0.1075929756 10 , 0.1415289297 10 , -6 -6 -0.3444801786 10 , -0.8182545846 10 ] The largest is 0.2822085890 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 72, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 17, 58, 198, 676, 2308, 7880, 26904, 91856, 313616, 1070752, 3655776, 12481600, 42614848, 145496192, 496755072, 1696027904, 5790601472, 19770350080, 67500197376] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 2 a(n - 2) with initial conditions, a(1) = 5, a(2) = 17, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 2 b(n - 2) with initial conditions, b(1) = 5, b(2) = 17, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t + 2 = 0 whose roots are 1/2 1/2 [2 + 2 , 2 - 2 ] In floating-point [3.414213562, 0.585786438] The largest root is, 3.414213562 and the remaining roots are [0.585786438] whose absolute values are [0.585786438] so the largest absolute value is, 0.585786438 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.585786438 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, -0.1176470588, -0.06896551724, -0.04040404040, -0.02366863905, -0.01386481802, -0.008121827411, -0.004757656854, -0.002786970911, -0.001632569767, -0.0009563372284, -0.0005602093783, -0.0003281630560, -0.0001922334675, -0.0001126077581, -0.00006596409749, -0.00003864087368, -5 -0.00002263529974, -0.00001325945160, -0.7767206918 10 , -5 -5 -5 -0.4549924471 10 , -0.2665284047 10 , -0.1561287247 10 , -6 -6 -6 -0.9145808947 10 , -0.5357490842 10 , -0.3138345475 10 , -6 -6 -7 -0.1838400216 10 , -0.1076909913 10 , -0.6308392218 10 , -7 -7 -7 -0.3695370604 10 , -0.2164697982 10 , -0.1268050719 10 , -8 -8 -8 -0.7428069137 10 , -0.4351262158 10 , -0.2548910359 10 , -8 -9 -9 -0.1493117119 10 , -0.8746477581 10 , -0.5123567944 10 , -9 -9 -9 -0.3001316614 10 , -0.1758130567 10 , -0.1029889042 10 , -10 -10 -10 -0.6032950330 10 , -0.3534020482 10 , -0.2070181269 10 , -10 -11 -11 -0.1212684111 10 , -0.7103739052 10 , -0.4161273993 10 , -11 -11 -12 -0.2437617868 10 , -0.1427923487 10 , -0.8364582129 10 , -12 -12 -12 -0.4899858767 10 , -0.2870270812 10 , -0.1681365714 10 , -13 -13 -13 -0.9849212320 10 , -0.5769534999 10 , -0.3379715354 10 , -13 -13 -0.1979791417 10 , -0.1159734961 10 ] The largest is -13 -0.1159734961 10 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 73, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 18, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 18, 65, 235, 850, 3074, 11117, 40204, 145395, 525811, 1901559, 6876856, 24869672, 89939441, 325259740, 1176279253, 4253932199, 15384050265, 55635349010, 201201374550] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - a(n - 2) - a(n - 3) - 2 a(n - 4) + 2 a(n - 5) with initial conditions, a(1) = 5, a(2) = 18, a(3) = 65, a(4) = 235, a(5) = 850, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - b(n - 2) - b(n - 3) - 2 b(n - 4) + 2 b(n - 5) with initial conditions, b(1) = 5, b(2) = 18, b(3) = 65, b(4) = 235, b(5) = 850, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 4 t + t + t + 2 t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 4 _Z + _Z + _Z + 2 _Z - 2 In floating-point [3.61643052717899, 0.787548685432515 + 0.288476563352837 I, -0.595763949022011 + 0.656683916532892 I, -0.595763949022011 - 0.656683916532892 I, 0.787548685432515 - 0.288476563352837 I] The largest root is, 3.61643052717899 and the remaining roots are [0.787548685432515 + 0.288476563352837 I, -0.595763949022011 + 0.656683916532892 I, -0.595763949022011 - 0.656683916532892 I, 0.787548685432515 - 0.288476563352837 I] whose absolute values are [0.838720251055348, 0.886661406167698, 0.886661406167698, 0.838720251055348] so the largest absolute value is, 0.886661406167698 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.886661406167698 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, -0.2777777778, -0.3846153846, 0.4680851064, 0.03058823529, 0.1929082628, 0.4858325088, 0.01445129838, 0.2539014409, 0.1906863873, -0.09145758822, 0.1323407092, -0.04876618397, -0.1095178366, 0.04264207123, -0.1187442894, -0.04588760607, 0.01405509903, -0.08346752402, 0.02073513214, 0.006639586965, -0.03059467047, 0.04529184513, -0.003282848320, 0.0003625224187, 0.03390960773, -0.01321427435, 0.01002015933, 0.01209456244, -0.01552180582, 0.01004581912, -0.002858347513, -0.01010620959, 0.007620426587, -0.007688986431, -0.002461829465, 0.004716966129, -0.006434591934, 0.002625321637, 0.001764598423, -0.003489927202, 0.003953487259, -0.0005805493263, -0.001064310933, 0.002878869583, -0.001726490329, 0.0003475532061, 0.001205356783, -0.001685996779, 0.0009138227205, -0.00001215619186, -0.0009920578629, 0.0009128091432, -0.0005441883708, -0.0002455469388, 0.0006089948143, -0.0005840194451, 0.0002144693718] The largest is 0.4858325088 The smallest is -0.3846153846 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 74, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 19, 72, 273, 1035, 3924, 14877, 56403, 213840, 810729, 3073707, 11653308, 44181045, 167503059, 635052312, 2407666113, 9128155275, 34607464164, 131206858317, 497442967443] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 3 a(n - 2) with initial conditions, a(1) = 5, a(2) = 19, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 3 b(n - 2) with initial conditions, b(1) = 5, b(2) = 19, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 3 = 0 whose roots are 1/2 1/2 21 21 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.791287848, -0.791287848] The largest root is, 3.791287848 and the remaining roots are [-0.791287848] whose absolute values are [0.791287848] so the largest absolute value is, 0.791287848 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.791287848 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, -0.1578947368, 0.1250000000, -0.09890109890, 0.07826086957, -0.06192660550, 0.04900181488, -0.03877453327, 0.03068181818, -0.02427814967, 0.01921100482, -0.01520143465, 0.01202871050, -0.009518172441, 0.007531614183, -0.005959674775, 0.004715818224, -0.003731569652, 0.002952745717, -0.002336471803, 0.001848821744, -0.001462950178, 0.001157614697, -0.0009160064420, 0.0007248247658, -0.0005735450287, 0.0004538392112, -0.0003591174525, 0.0002841652760, -0.0002248565296, 0.0001779262393, -0.0001407908709, 0.0001114061052, -0.00008815429715, 0.00006975542404, -0.00005519661934, 0.00004367641411, -0.00003456061570, 0.00002734739521, -0.00002163966149, 0.00001712320116, -5 -5 -0.00001354938099, 0.00001072146052, -0.8483761413 10 , 0.6713097307 10 , -5 -5 -5 -0.5311992318 10 , 0.4203314967 10 , -0.3326032053 10 , -5 -5 -5 0.2631848744 10 , -0.2082549927 10 , 0.1647896449 10 , -5 -5 -6 -0.1303960434 10 , 0.1031808045 10 , -0.8164571670 10 , -6 -6 -6 0.6460526343 10 , -0.5112135983 10 , 0.4045171078 10 , -6 -0.3200894715 10 ] The largest is 0.2000000000 The smallest is -0.1578947368 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 6, :The Pisot Sequence a(n), defined by, a(1) = 5, a(2) = 21, and for n>1, 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [5, 21, 88, 369, 1547, 6486, 27193, 114009, 477993, 2004029, 8402073, 35226452, 147690090, 619204077, 2596069167, 10884255079, 45633225081, 191321428631, 802132415328, 3363012791217] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 75, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 22, 97, 428, 1888, 8328, 36735, 162039, 714758, 3152815, 13907144, 61344752, 270593200, 1193593217, 5264968845, 23223906222, 102441217809, 451870714852, 1993212764432, 8792110207000] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 3 a(n - 2) + 2 a(n - 3) - a(n - 4) with initial conditions, a(1) = 5, a(2) = 22, a(3) = 97, a(4) = 428, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 3 b(n - 2) + 2 b(n - 3) - b(n - 4) with initial conditions, b(1) = 5, b(2) = 22, b(3) = 97, b(4) = 428, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t + 3 t - 2 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z + 3 _Z - 2 _Z + 1 In floating-point [0.575233114691156, 4.41102443446646, 0.00687122542119004 + 0.627743628074015 I, 0.00687122542119004 - 0.627743628074015 I] The largest root is, 4.41102443446646 and the remaining roots are [0.575233114691156, 0.00687122542119004 + 0.627743628074015 I, 0.00687122542119004 - 0.627743628074015 I] whose absolute values are [0.575233114691156, 0.627781232856093, 0.627781232856093] so the largest absolute value is, 0.627781232856093 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.627781232856093 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, -0.3181818182, 0.4948453608, 0.3738317757, -0.05084745763, -0.06808357349, 0.06508779094, 0.05417831510, -0.009691392052, -0.01273274835, 0.008679280232, 0.008033547841, -0.001644206137, -0.002230365389, 0.001168606915, 0.001212170628, -0.0002614922448, -0.0003763938897, 0.0001582416276, 0.0001852346891, -0.00003984697178, -5 -0.00006206178138, 0.00002145975907, 0.00002855550682, -0.5878334090 10 , -5 -5 -6 -0.00001007689139, 0.2901799875 10 , 0.4427498556 10 , -0.8433555449 10 , -5 -6 -6 -0.1618782247 10 , 0.3893526356 10 , 0.6889002736 10 , -6 -6 -7 -0.1177654880 10 , -0.2580407426 10 , 0.5154066281 10 , -6 -7 -7 0.1073942921 10 , -0.1596652496 10 , -0.4089343297 10 , -8 -7 -8 0.6680331447 10 , 0.1675461411 10 , -0.2088264766 10 , -8 -9 -8 -0.6451070298 10 , 0.8383395825 10 , 0.2613765164 10 , -9 -8 -10 -0.2600687584 10 , -0.1013889821 10 , 0.9994791697 10 , -9 -10 -9 0.4075063663 10 , -0.3002280253 10 , -0.1588474568 10 , -10 -10 -11 0.1089593909 10 , 0.6347009457 10 , -0.3009455538 10 , -10 -12 -11 -0.2481822640 10 , 0.9814846646 10 , 0.9873096876 10 , -12 -11 -0.2059668748 10 , -0.3867929273 10 ] The largest is 0.4948453608 The smallest is -0.3181818182 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 76, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 23, 106, 489, 2256, 10408, 48017, 221525, 1021999, 4714962, 21752337, 100353760, 462979088, 2135940257, 9854096869, 45461582919, 209735661114, 967609236601, 4464036443696, 20594699406392] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 5, a(2) = 23, a(3) = 106, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 5, b(2) = 23, b(3) = 106, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 5 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (748 + 36 93 ) 38 (748 + 36 93 ) [------------------- + --------------------- + 5/3, - ------------------- 6 1/2 1/3 12 3 (748 + 36 93 ) 19 - --------------------- + 5/3 1/2 1/3 3 (748 + 36 93 ) / 1/2 1/3 \ 1/2 |(748 + 36 93 ) 38 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (748 + 36 93 ) / 1/2 1/3 (748 + 36 93 ) 19 - ------------------- - --------------------- + 5/3 12 1/2 1/3 3 (748 + 36 93 ) / 1/2 1/3 \ 1/2 |(748 + 36 93 ) 38 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (748 + 36 93 ) / In floating-point [4.613470269, 0.193264866 + 0.4235625860 I, 0.193264866 - 0.4235625860 I] The largest root is, 4.613470269 and the remaining roots are [0.193264866 + 0.4235625860 I, 0.193264866 - 0.4235625860 I] whose absolute values are [0.4655712327, 0.4655712327] so the largest absolute value is, 0.4655712327 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.4655712327 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, -0.4782608696, -0.1415094340, 0.04907975460, 0.04964539007, 0.008551114527, -0.007455692775, -0.004735357183, -0.0002142859240, 0.0009435919102, 0.0004111742108, -0.00004559869007, -0.0001067499619, -5 -0.00003137821846, 0.00001101014141, 0.00001105718208, 0.1887409122 10 , -5 -5 -7 -0.1667177140 10 , -0.1053521865 10 , -0.4584592284 10 , -6 -7 -7 0.2106369757 10 , 0.9135485891 10 , -0.1034557960 10 , -7 -8 -8 -0.2380064016 10 , -0.6957182705 10 , 0.2469787203 10 , -8 -9 -9 0.2462661260 10 , 0.4165491904 10 , -0.3727893655 10 , -9 -11 -10 -0.2343839483 10 , -0.9791819876 10 , 0.4701943163 10 , -10 -11 -11 0.2029684964 10 , -0.2346434957 10 , -0.5306442424 10 , -11 -12 -12 -0.1542492573 10 , 0.5539870277 10 , 0.5484778600 10 , -13 -13 -13 0.9192267141 10 , -0.8335533513 10 , -0.5214415852 10 , -14 -13 -14 -0.2087450945 10 , 0.1049572719 10 , 0.4509379335 10 , -15 -14 -15 -0.5320086591 10 , -0.1183074771 10 , -0.3419772014 10 , -15 -15 -16 0.1242548755 10 , 0.1221540097 10 , 0.2028309580 10 , -16 -16 -18 -0.1863766479 10 , -0.1160050590 10 , -0.4441041114 10 , -17 -17 -18 0.2342826448 10 , 0.1001834565 10 , -0.1205841833 10 , -18 -19 -0.2637635980 10 , -0.7581505877 10 ] The largest is 0.04964539007 The smallest is -0.4782608696 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 77, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 24, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 24, 115, 551, 2640, 12649, 60605, 290376, 1391275, 6665999, 31938720, 153027601, 733199285, 3512968824, 16831644835, 80645255351, 386394631920, 1851327904249, 8870244889325, 42499896542376] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - a(n - 2) with initial conditions, a(1) = 5, a(2) = 24, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - b(n - 2) with initial conditions, b(1) = 5, b(2) = 24, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 5 t + 1 = 0 whose roots are 1/2 1/2 21 21 [5/2 + -----, 5/2 - -----] 2 2 In floating-point [4.791287848, 0.208712152] The largest root is, 4.791287848 and the remaining roots are [0.208712152] whose absolute values are [0.208712152] so the largest absolute value is, 0.208712152 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.208712152 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, 0.04166666667, 0.008695652174, 0.001814882033, 0.0003787878788, -5 -6 0.00007905763301, 0.00001650028876, 0.3443810783 10 , 0.7187651615 10 , -6 -7 -8 -8 0.1500150240 10 , 0.3130995857 10 , 0.6534768849 10 , 0.1363885673 10 , -9 -10 -10 0.2846595145 10 , 0.5941190001 10 , 0.1239998554 10 , -11 -12 -12 0.2588027673 10 , 0.5401528264 10 , 0.1127364591 10 , -13 -14 -14 0.2352946904 10 , 0.4910886132 10 , 0.1024961615 10 , -15 -16 -17 0.2139219450 10 , 0.4464810961 10 , 0.9318603063 10 , -17 -18 -19 0.1944905704 10 , 0.4059254559 10 , 0.8472157566 10 , -19 -20 -21 0.1768242242 10 , 0.3690536445 10 , 0.7702598054 10 , -21 -22 -23 0.1607625820 10 , 0.3355310453 10 , 0.7002940671 10 , -23 -24 -25 0.1461598821 10 , 0.3050534361 10 , 0.6366835929 10 , -25 -26 -27 0.1328836032 10 , 0.2773442285 10 , 0.5788511092 10 , -27 -28 -29 0.1208132610 10 , 0.2521519575 10 , 0.5262717782 10 , -29 -30 -31 0.1098393156 10 , 0.2292480000 10 , 0.4784684354 10 , -32 -32 -33 0.9986217706 10 , 0.2084244993 10 , 0.4350072589 10 , -34 -34 -35 0.9079130136 10 , 0.1894924794 10 , 0.3954938326 10 , -36 -36 -37 0.8254436911 10 , 0.1722801295 10 , 0.3595695667 10 , -38 -38 -39 0.7504653826 10 , 0.1566312454 10 , 0.3269084438 10 ] The largest is 0.2000000000 The smallest is -39 0.3269084438 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 78, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) + a(n - 2) with initial conditions, a(1) = 5, a(2) = 26, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) + b(n - 2) with initial conditions, b(1) = 5, b(2) = 26, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 5 t - 1 = 0 whose roots are 1/2 1/2 29 29 [5/2 + -----, 5/2 - -----] 2 2 In floating-point [5.192582404, -0.192582404] The largest root is, 5.192582404 and the remaining roots are [-0.192582404] whose absolute values are [0.192582404] so the largest absolute value is, 0.192582404 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.192582404 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, -0.03846153846, 0.007407407407, -0.001426533524, 0.0002747252747, -5 -6 -0.00005290725358, 0.00001018900606, -0.1962223277 10 , 0.3778896751 10 , -7 -7 -8 -0.7277490192 10 , 0.1401516553 10 , -0.2699074264 10 , -9 -9 -10 0.5197942092 10 , -0.1001032182 10 , 0.1927811836 10 , -11 -12 -12 -0.3712626370 10 , 0.7149865099 10 , -0.1376938206 10 , -13 -14 -15 0.2651740693 10 , -0.5106785962 10 , 0.9834771151 10 , -15 -16 -17 -0.1894003867 10 , 0.3647518170 10 , -0.7024478163 10 , -17 -18 -19 0.1352790888 10 , -0.2605237208 10 , 0.5017228434 10 , -20 -20 -21 -0.9662299111 10 , 0.1860788787 10 , -0.3583551771 10 , -22 -22 -23 0.6901290133 10 , -0.1329067042 10 , 0.2559549254 10 , -24 -25 -25 -0.4929241473 10 , 0.9492851707 10 , -0.1828156198 10 , -26 -27 -27 0.3520707148 10 , -0.6780262448 10 , 0.1305759239 10 , -28 -29 -30 -0.2514662527 10 , 0.4842797537 10 , -0.9326375896 10 , -30 -31 -32 0.1796095887 10 , -0.3458964629 10 , 0.6661357221 10 , -32 -33 -34 -0.1282860185 10 , 0.2470562978 10 , -0.4757869565 10 , -35 -35 -36 0.9162819566 10 , -0.1764597816 10 , 0.3398304886 10 , -37 -37 -38 -0.6544537231 10 , 0.1260362710 10 , -0.2427236801 10 , -39 -40 -40 0.4674430971 10 , -0.9002131518 10 , 0.1733652125 10 , -41 -0.3338708932 10 ] The largest is 0.2000000000 The smallest is -0.03846153846 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 79, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 27, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 27, 146, 789, 4264, 23044, 124537, 673037, 3637303, 19657126, 106233273, 574117920, 3102713272, 16768035473, 90619721829, 489737393363, 2646694445946, 14303566738285, 77300959976680, 417758627805916] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) + 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 5, a(2) = 27, a(3) = 146, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) + 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 5, b(2) = 27, b(3) = 146, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 5 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (1468 + 60 69 ) 62 (1468 + 60 69 ) [-------------------- + ---------------------- + 5/3, - -------------------- 6 1/2 1/3 12 3 (1468 + 60 69 ) 31 - ---------------------- + 5/3 1/2 1/3 3 (1468 + 60 69 ) / 1/2 1/3 \ 1/2 |(1468 + 60 69 ) 62 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (1468 + 60 69 ) / 1/2 1/3 (1468 + 60 69 ) 31 - -------------------- - ---------------------- + 5/3 12 1/2 1/3 3 (1468 + 60 69 ) / 1/2 1/3 \ 1/2 |(1468 + 60 69 ) 62 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (1468 + 60 69 ) / In floating-point [5.404313582, -0.202156790 + 0.3796972588 I, -0.202156790 - 0.3796972588 I] The largest root is, 5.404313582 and the remaining roots are [-0.202156790 + 0.3796972588 I, -0.202156790 - 0.3796972588 I] whose absolute values are [0.4301597100, 0.4301597100] so the largest absolute value is, 0.4301597100 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.4301597100 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.4814814815, -0.1575342466, -0.02534854246, 0.03939962477, -0.01123936817, -0.002746171820, 0.003190017785, -0.0007816230872, -0.0002742516887, 0.0002555131668, -0.00005256063075, -0.00002602850890, -5 -5 -5 0.00002024936079, -0.3370844600 10 , -0.2384010320 10 , 0.1587619986 10 , -6 -6 -6 -0.2007653093 10 , -0.2125968941 10 , 0.1231048974 10 , -7 -7 -8 -0.1043461059 10 , -0.1856015232 10 , 0.9434914598 10 , -9 -8 -9 -0.3803422363 10 , -0.1592034301 10 , 0.7140586182 10 , -11 -9 -10 0.5882251673 10 , -0.1345058067 10 , 0.5329408793 10 , -11 -10 -11 0.3341077906 10 , -0.1121224132 10 , 0.3915037150 10 , -12 -12 -12 0.4917810157 10 , -0.9232619416 10 , 0.2822894732 10 , -13 -13 -13 0.5670449862 10 , -0.7516050206 10 , 0.1989596013 10 , -14 -14 -14 0.5863295142 10 , -0.6052106094 10 , 0.1362019943 10 , -15 -15 -16 0.5691826664 10 , -0.4821528771 10 , 0.8962088960 10 , -16 -16 -17 0.5298136007 10 , -0.3800429758 10 , 0.5562121834 10 , -17 -17 -18 0.4783374078 10 , -0.2963183526 10 , 0.3129523577 10 , -18 -18 -19 0.4217688137 10 , -0.2284347422 10 , 0.1431627432 10 , -19 -19 -21 0.3648070097 10 , -0.1739868866 10 , 0.2842329575 10 , -20 -20 0.3104488438 10 , -0.1307780557 10 ] The largest is 0.4814814815 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 80, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 28, 157, 880, 4932, 27642, 154923, 868285, 4866410, 27274393, 152862688, 856737724, 4801691880, 26911672341, 150829775481, 845344015912, 4737834442565, 26553775483864, 148823898554320, 834101832124234] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) + 3 a(n - 2) + 2 a(n - 3) + a(n - 4) with initial conditions, a(1) = 5, a(2) = 28, a(3) = 157, a(4) = 880, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) + 3 b(n - 2) + 2 b(n - 3) + b(n - 4) with initial conditions, b(1) = 5, b(2) = 28, b(3) = 157, b(4) = 880, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t - 3 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z - 3 _Z - 2 _Z - 1 In floating-point [5.60462291491303, -0.0430216465845746 + 0.584989130927212 I, -0.518579621743881, -0.0430216465845746 - 0.584989130927212 I] The largest root is, 5.60462291491303 and the remaining roots are [-0.0430216465845746 + 0.584989130927212 I, -0.518579621743881, -0.0430216465845746 - 0.584989130927212 I] whose absolute values are [0.586568960462300, 0.518579621743881, 0.586568960462300] so the largest absolute value is, 0.586568960462300 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.586568960462300 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.3214285714, 0.4840764331, -0.3818181818, -0.01459854015, 0.07087041459, 0.03095085946, -0.04365502110, 0.001719748233, 0.01040580445, 0.0008290839423, -0.004852691651, 0.0007551507033, 0.001281650897, -0.0002025927633, -0.0005104013690, 0.0001586673627, 0.0001385980772, -0.00005440302722, -0.00004928754812, 0.00002621669484, 0.00001301285260, -5 -5 -5 -0.9263775929 10 , -0.4134480277 10 , 0.3778670868 10 , -6 -5 -6 0.9752142519 10 , -0.1320652619 10 , -0.2547588817 10 , -6 -7 -6 0.4933471057 10 , 0.3636789664 10 , -0.1682895824 10 , -9 -7 -8 -0.4088924214 10 , 0.5916968964 10 , -0.5589497210 10 , -7 -8 -8 -0.1954578437 10 , 0.3433073403 10 , 0.6518709141 10 , -8 -8 -9 -0.1788300031 10 , -0.2065010295 10 , 0.7805401140 10 , -9 -9 -9 0.6497787618 10 , -0.3278064712 10 , -0.1936261380 10 , -9 -10 -10 0.1285475340 10 , 0.5602507570 10 , -0.4929076647 10 , -10 -10 -11 -0.1490967513 10 , 0.1817701038 10 , 0.3599569284 10 , -11 -12 -11 -0.6581239166 10 , -0.6631423456 10 , 0.2316719723 10 , -13 -12 -13 0.3126252929 10 , -0.8010520420 10 , 0.5882447861 10 , -12 -13 -13 0.2702110488 10 , -0.4331287461 10 , -0.8933431128 10 ] The largest is 0.4840764331 The smallest is -0.3818181818 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 7, : Consider the Pisot Sequence a(n), defined by, a(1) = 5, a(2) = 29, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 5, 29, 168, 973, 5635, 32634, 188993, 1094514, 6338652, 36708995, 212592569, 1231185991, 7130159589, 41292847820, 239139006610, 1384924206044, 8020502734690, 46449086409517, 269000298316358, 1557859714533878, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 6 a(n - 1) - a(n - 2) - a(n - 3) - a(n - 4) - a(n - 5) - a(n - 6) - a(n - 7), . Alas, it breaks down at the, 34, -th term. a(34), equals , 74368074923307896332150649, while the corresponding term for the solution of the recurrence is , 74368074923307896332150648 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.05843714354031 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 81, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 31, 192, 1189, 7363, 45596, 282357, 1748519, 10827848, 67052341, 415227147, 2571328324, 15923162533, 98605496111, 610622660112, 3781331140229, 23416204681043, 145006777066156, 897966416057877, 5560731027074359] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) - 5 a(n - 2) with initial conditions, a(1) = 5, a(2) = 31, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) - 5 b(n - 2) with initial conditions, b(1) = 5, b(2) = 31, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 7 t + 5 = 0 whose roots are 1/2 1/2 29 29 [7/2 + -----, 7/2 - -----] 2 2 In floating-point [6.192582404, 0.807417596] The largest root is, 6.192582404 and the remaining roots are [0.807417596] whose absolute values are [0.807417596] so the largest absolute value is, 0.807417596 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.807417596 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, 0.1612903226, 0.1302083333, 0.1051303616, 0.08488387885, 0.06853671375, 0.05533774619, 0.04468066976, 0.03607595895, 0.02912836406, 0.02351875370, 0.01898945558, 0.01533242059, 0.01237966618, 0.009995560309, 0.008070591279, 0.006516337413, 0.005261405491, 0.004248151376, 0.003430032173, 0.002769468333, 0.002236117465, 0.001805480589, 0.001457776797, 0.001177034638, 0.0009503584782, 0.0007673361582, 0.0006195607165, 0.0005002442246, 0.0004039059894, 0.0003261208032, 0.0002633156750, 0.0002126057094, 0.0001716615909, 0.0001386025891, 0.0001119101694, 0.00009035823997, 0.00007295683294, 0.00005890663069, 0.00004756225017, 0.00003840259771, 0.00003100693314, 0.00002503554343, 0.00002021413830, 0.00001632125096, 0.00001317806522, 0.00001064020175, -5 -5 -5 -5 0.8591086120 10 , 0.6936594106 10 , 0.5600728140 10 , 0.4522126453 10 , -5 -5 -5 -5 0.3651244472 10 , 0.2948079035 10 , 0.2380330889 10 , 0.1921921045 10 , -5 -5 -5 0.1551792871 10 , 0.1252944870 10 , 0.1011649735 10 ] The largest is 0.2000000000 The smallest is -5 0.1011649735 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 8, :The Pisot Sequence a(n), defined by, a(1) = 5, a(2) = 32, and for n>1, 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [5, 32, 205, 1313, 8410, 53868, 345037, 2210042, 14155832, 90671390, 580771301, 3719975001, 23827303423, 152619409609, 977562747068, 6261516322880, 40106465573985, 256891222172427, 1645447911821220, 10539475843590807] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 82, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 33, 218, 1440, 9512, 62832, 415040, 2741568, 18109568, 119623680, 790180352, 5219576832, 34478182400, 227747401728, 1504397139968, 9937372446720, 65641823240192, 433600429228032, 2864169868328960, 18919420926885888] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + 4 a(n - 2) with initial conditions, a(1) = 5, a(2) = 33, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + 4 b(n - 2) with initial conditions, b(1) = 5, b(2) = 33, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 6 t - 4 = 0 whose roots are 1/2 1/2 [3 + 13 , 3 - 13 ] In floating-point [6.605551275, -0.605551275] The largest root is, 6.605551275 and the remaining roots are [-0.605551275] whose absolute values are [0.605551275] so the largest absolute value is, 0.605551275 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.605551275 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.1212121212, -0.07339449541, 0.04444444444, -0.02691337258, 0.01629742806, -0.009868928296, 0.005976142120, -0.003618860483, 0.002191405581, -0.001327008445, 0.0008035716563, -0.0004866038414, 0.0002946635768, -0.0001784339048, 0.0001080508786, -0.00006543034736, -5 0.00003962143030, -0.00002399280765, 0.00001452887528, -0.8797978955 10 , -5 -5 -5 0.5327627377 10 , -0.3226151554 10 , 0.1953600188 10 , -5 -6 -6 -0.1183005086 10 , 0.7163702385 10 , -0.4337989116 10 , -6 -6 -7 0.2626874842 10 , -0.1590707411 10 , 0.9632549018 10 , -7 -7 -7 -0.5833002344 10 , 0.3532182009 10 , -0.2138917321 10 , -7 -8 -8 0.1295224112 10 , -0.7843246128 10 , 0.4749487697 10 , -8 -8 -8 -0.2876058333 10 , 0.1741600792 10 , -0.1054628581 10 , -9 -9 -9 0.6386316822 10 , -0.3867242297 10 , 0.2341813505 10 , -9 -10 -10 -0.1418088155 10 , 0.8587250911 10 , -0.5200020742 10 , -10 -10 -10 0.3148879193 10 , -0.1906807811 10 , 0.1154669902 10 , -11 -11 -11 -0.6992118320 10 , 0.4234086167 10 , -0.2563956279 10 , -11 -12 -12 0.1552606995 10 , -0.9401831461 10 , 0.5693291033 10 , -12 -12 -12 -0.3447579646 10 , 0.2087686252 10 , -0.1264201073 10 , -13 0.7655385721 10 ] The largest is 0.1212121212 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 83, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 34, 231, 1569, 10657, 72385, 491657, 3339457, 22682425, 154064689, 1046445801, 7107720929, 48277413657, 327912236945, 2227261715017, 15128117185857, 102753945818745, 697930433219569, 4740517609642921, 32198778184336289] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) - 2 a(n - 2) + 4 a(n - 3) with initial conditions, a(1) = 5, a(2) = 34, a(3) = 231, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) - 2 b(n - 2) + 4 b(n - 3) with initial conditions, b(1) = 5, b(2) = 34, b(3) = 231, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 7 t + 2 t - 4 = 0 whose roots are 1/2 1/3 1/2 1/3 (334 + 3 3561 ) 43 (334 + 3 3561 ) [-------------------- + ---------------------- + 7/3, - -------------------- 3 1/2 1/3 6 3 (334 + 3 3561 ) 43 - ---------------------- + 7/3 1/2 1/3 6 (334 + 3 3561 ) / 1/2 1/3 \ 1/2 |(334 + 3 3561 ) 43 | + 1/2 I 3 |-------------------- - ----------------------|, | 3 1/2 1/3| \ 3 (334 + 3 3561 ) / 1/2 1/3 (334 + 3 3561 ) 43 - -------------------- - ---------------------- + 7/3 6 1/2 1/3 6 (334 + 3 3561 ) / 1/2 1/3 \ 1/2 |(334 + 3 3561 ) 43 | - 1/2 I 3 |-------------------- - ----------------------|] | 3 1/2 1/3| \ 3 (334 + 3 3561 ) / In floating-point [6.792249461, 0.103875268 + 0.7603396940 I, 0.103875268 - 0.7603396940 I] The largest root is, 6.792249461 and the remaining roots are [0.103875268 + 0.7603396940 I, 0.103875268 - 0.7603396940 I] whose absolute values are [0.7674024509, 0.7674024509] so the largest absolute value is, 0.7674024509 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7674024509 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, 0.4411764706, -0.02597402597, -0.2651370300, -0.03978605611, 0.1478759411, 0.05415157315, -0.07583508337, -0.04764499387, 0.03476150203, 0.03528016832, -0.01314180128, -0.02350693747, 0.002855713568, 0.01443666479, 0.001317476534, -0.008228139578, -0.002485270945, 0.004329288677, 0.002363004317, -0.002058630915, -0.001819270331, 0.0008343867802, 0.001244724464, -0.0002327836390, -0.0007813872796, -0.00002524582463, 0.0004549192307, 0.0001093771456, -0.0002451817406, -0.0001153495528, 0.0001204251942, 0.00009294850269, -0.00005160908074, -5 -0.00006545979363, 0.00001679361685, 0.00004203858226, -0.1156332398 10 , -5 -5 -0.00002499702390, -0.4512173471 10 , 0.00001378350391, 0.5520778732 10 , -5 -5 -5 -0.6970250585 10 , -0.4699295904 10 , 0.3128544770 10 , -5 -5 -5 0.3417402860 10 , -0.1132453134 10 , -0.2247798577 10 , -6 -5 -6 -6 0.1999276718 10 , 0.1365278321 10 , 0.1658985959 10 , -0.7695557835 10 , -6 -6 -6 -0.2575743929 10 , 0.3996852005 10 , 0.2347220550 10 , -6 -6 -7 -0.1866135877 10 , -0.1769984220 10 , 0.7312644133 10 ] The largest is 0.4411764706 The smallest is -0.2651370300 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 84, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 36, 259, 1863, 13401, 96397, 693410, 4987888, 35879244, 258089225, 1856506454, 13354359190, 96061561753, 690997113013, 4970531412138, 35754393258349, 257191139392236, 1850046278338971, 13307889377853115, 95727291671958377] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 8 a(n - 1) - 6 a(n - 2) + a(n - 3) + 3 a(n - 4) with initial conditions, a(1) = 5, a(2) = 36, a(3) = 259, a(4) = 1863, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 8 b(n - 1) - 6 b(n - 2) + b(n - 3) + 3 b(n - 4) with initial conditions, b(1) = 5, b(2) = 36, b(3) = 259, b(4) = 1863, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 8 t + 6 t - t - 3 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 8 _Z + 6 _Z - _Z - 3 In floating-point [7.19327377572487, 0.651032645029501 + 0.646620217906537 I, -0.495339065783873, 0.651032645029501 - 0.646620217906537 I] The largest root is, 7.19327377572487 and the remaining roots are [0.651032645029501 + 0.646620217906537 I, -0.495339065783873, 0.651032645029501 - 0.646620217906537 I] whose absolute values are [0.917584443579776, 0.495339065783873, 0.917584443579776] so the largest absolute value is, 0.917584443579776 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.917584443579776 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, 0.3611111111, -0.3474903475, -0.4347826087, -0.4328781434, -0.1186343973, 0.1709003331, 0.3417751160, 0.2915292474, 0.09658042098, -0.1220560055, -0.2390759763, -0.2091036142, -0.07068779822, 0.08387530641, 0.1687976976, 0.1491311015, 0.05207453852, -0.05776668421, -0.1190565105, -0.1063841355, -0.03827708973, 0.03973153208, 0.08396112805, 0.07587033567, 0.02809617998, -0.02729684989, -0.05919815918, -0.05409698709, -0.02059525164, 0.01873120061, 0.04172965009, 0.03856378414, 0.01507781828, -0.01283690668, -0.02940942870, -0.02748481885, -0.01102543040, 0.008785321121, 0.02072204643, 0.01958455779, 0.008053213605, -0.006003628085, -0.01459760922, -0.01395221831, -0.005876078446, 0.004096188844, 0.01028093544, 0.009937617066, 0.004283277393, -0.002789981279, -0.007239091205, -0.007076713372, -0.003119308852, 0.001896774376, 0.005096061137, 0.005038393872, 0.002269631971] The largest is 0.3611111111 The smallest is -0.4347826087 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 85, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 37, 274, 2029, 15025, 111262, 823909, 6101149, 45179770, 334561837, 2477472169, 18345990694, 135854351365, 1006018431637, 7449692075554, 55165899823789, 408510374993185, 3025070324423662, 22401023395945189, 165882374744887309] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) + 3 a(n - 2) with initial conditions, a(1) = 5, a(2) = 37, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) + 3 b(n - 2) with initial conditions, b(1) = 5, b(2) = 37, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 7 t - 3 = 0 whose roots are 1/2 1/2 61 61 [7/2 + -----, 7/2 - -----] 2 2 In floating-point [7.405124838, -0.405124838] The largest root is, 7.405124838 and the remaining roots are [-0.405124838] whose absolute values are [0.405124838] so the largest absolute value is, 0.405124838 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.405124838 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.08108108108, -0.03284671533, 0.01330704781, -0.005391014975, 0.002184034082, -0.0008848064531, 0.0003584570710, -0.0001452198628, -5 -5 0.00005883217338, -0.00002383437471, 0.9655897191 10 , -0.3911843785 10 , -5 -6 -6 0.1584785079 10 , -0.6420357985 10 , 0.2601046488 10 , -6 -7 -7 -0.1053748537 10 , 0.4268997053 10 , -0.1729476739 10 , -8 -8 -8 0.7006539838 10 , -0.2838523316 10 , 0.1149956299 10 , -9 -9 -10 -0.4658758591 10 , 0.1887378819 10 , -0.7646240383 10 , -10 -10 -11 0.3097681896 10 , -0.1254947876 10 , 0.5084105550 10 , -11 -12 -12 -0.2059697437 10 , 0.8344345904 10 , -0.3380501782 10 , -12 -13 -13 0.1369525237 10 , -0.5548286896 10 , 0.2247748830 10 , -14 -14 -14 -0.9106188804 10 , 0.3689143263 10 , -0.1494563567 10 , -15 -15 -16 0.6054848228 10 , -0.2452969407 10 , 0.9937588336 10 , -16 -16 -17 -0.4025963864 10 , 0.1631017958 10 , -0.6607658860 10 , -17 -17 -18 0.2676926725 10 , -0.1084489506 10 , 0.4393536352 10 , -18 -19 -19 -0.1779930703 10 , 0.7210941375 10 , -0.2921331456 10 , -19 -20 -20 0.1183503933 10 , -0.4794668390 10 , 0.1942439254 10 , -21 -21 -21 -0.7869303882 10 , 0.3188050460 10 , -0.1291558426 10 , -22 -22 -23 0.5232423980 10 , -0.2119784917 10 , 0.8587775211 10 ] The largest is 0.08108108108 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 86, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 38, 289, 2198, 16717, 127142, 966985, 7354454, 55934677, 425414054, 3235508401, 24607825046, 187156075165, 1423425126182, 10825932783961, 82337186893142, 626219696793253, 4762746013666598, 36223309018953025, 275498234110624406] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 8 a(n - 1) - 3 a(n - 2) with initial conditions, a(1) = 5, a(2) = 38, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 8 b(n - 1) - 3 b(n - 2) with initial conditions, b(1) = 5, b(2) = 38, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 8 t + 3 = 0 whose roots are 1/2 1/2 [4 + 13 , 4 - 13 ] In floating-point [7.605551275, 0.394448725] The largest root is, 7.605551275 and the remaining roots are [0.394448725] whose absolute values are [0.394448725] so the largest absolute value is, 0.394448725 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.394448725 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, -0.07894736842, -0.03114186851, -0.01228389445, -0.004845366992, -0.001911248840, -0.0007538896674, -0.0002973708177, -0.0001172975398, -5 -0.00004626786495, -0.00001825030032, -0.7198807683 10 , -5 -5 -6 -0.2839560509 10 , -0.1120061021 10 , -0.4418066411 10 , -6 -7 -7 -0.1742700661 10 , -0.6874060529 10 , -0.2711464408 10 , -7 -8 -8 -0.1069533677 10 , -0.4218761949 10 , -0.1664085270 10 , -9 -9 -9 -0.6563963122 10 , -0.2589146881 10 , -0.1021285685 10 , -10 -10 -11 -0.4028448358 10 , -0.1589016317 10 , -0.6267854594 10 , -11 -12 -12 -0.2472347250 10 , -0.9752142195 10 , -0.3846720050 10 , -12 -13 -13 -0.1517333817 10 , -0.5985103890 10 , -0.2360816596 10 , -14 -14 -14 -0.9312210950 10 , -0.3673189732 10 , -0.1448885005 10 , -15 -15 -16 -0.5715108421 10 , -0.2254317227 10 , -0.8892125550 10 , -16 -16 -17 -0.3507487582 10 , -0.1383524003 10 , -0.5457292783 10 , -17 -18 -18 -0.2152622178 10 , -0.8490990724 10 , -0.3349260461 10 , -18 -19 -19 -0.1321111517 10 , -0.5211107529 10 , -0.2055514718 10 , -20 -20 -20 -0.8107951588 10 , -0.3198171162 10 , -0.1261514536 10 , -21 -21 -22 -0.4976027997 10 , -0.1962787897 10 , -0.7742191823 10 , -22 -22 -23 -0.3053897690 10 , -0.1204606049 10 , -0.4751553194 10 , -23 -0.1874244097 10 ] The largest is -23 -0.1874244097 10 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 87, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 5, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [5, 39, 304, 2370, 18477, 144050, 1123040, 8755424, 68258877, 532158613, 4148805281, 32344840127, 252166253170, 1965933947675, 15326778417156, 119490350592101, 931568493783483, 7262677315028274, 56621152533831550, 441428797562781157] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) + 6 a(n - 2) + 2 a(n - 3) - 3 a(n - 4) with initial conditions, a(1) = 5, a(2) = 39, a(3) = 304, a(4) = 2370, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) + 6 b(n - 2) + 2 b(n - 3) - 3 b(n - 4) with initial conditions, b(1) = 5, b(2) = 39, b(3) = 304, b(4) = 2370, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 7 t - 6 t - 2 t + 3 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 7 _Z - 6 _Z - 2 _Z + 3 In floating-point [0.474746247419215, 7.79618177674897, -0.635464012084091 + 0.637755171229264 I, -0.635464012084091 - 0.637755171229264 I] The largest root is, 7.79618177674897 and the remaining roots are [0.474746247419215, -0.635464012084091 + 0.637755171229264 I, -0.635464012084091 - 0.637755171229264 I] whose absolute values are [0.474746247419215, 0.900303376136999, 0.900303376136999] so the largest absolute value is, 0.900303376136999 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.900303376136999 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, -0.3589743590, -0.3552631579, 0.4341772152, -0.4102397575, 0.09996528983, 0.1724747115, -0.3158936677, 0.2542428584, -0.07060844095, -0.1180134057, 0.2064222345, -0.1670702497, 0.04484017038, 0.07834438048, -0.1359555175, 0.1092687503, -0.02868360317, -0.05211689686, 0.08948415613, -0.07148574544, 0.01832173449, 0.03465667160, -0.05889685116, 0.04676277676, -0.01169352990, -0.02304176587, 0.03876256653, -0.03058801956, 0.007456320260, 0.01531655513, -0.02550993124, 0.02000651130, -0.004749858892, -0.01017947234, 0.01678735658, -0.01308458963, 0.003022644076, 0.006764100934, -0.01104667801, 0.008556916560, -0.001921382513, -0.004493837056, 0.007268712686, -0.005595548242, 0.001219911849, 0.002985030035, -0.004782553201, 0.003658776230, -0.0007735610700, -0.001982466615, 0.003146579340, -0.002392195143, 0.0004898600171, 0.001316407785, -0.002070113706, 0.001563956235, -0.0003097530733] The largest is 0.4341772152 The smallest is -0.4102397575 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 88, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 8, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 8, 11, 15, 20, 27, 36, 48, 64, 85, 113, 150, 199, 264, 350, 464, 615, 815, 1080, 1431] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) - a(n - 4) with initial conditions, a(1) = 6, a(2) = 8, a(3) = 11, a(4) = 15, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) - b(n - 4) with initial conditions, b(1) = 6, b(2) = 8, b(3) = 11, b(4) = 15, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [1, ------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.324717958, -0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, 0.1250000000, 0.4545454545, -0.3333333333, 0.4500000000, 0., 0., 0.3333333333, -0.1093750000, 0.2235294118, 0.1150442478, 0.006666666667, 0.2311557789, 0.01515151515, 0.1314285714, 0.1400862069, 0.04065040650, 0.1656441718, 0.07500000000, 0.1006289308, 0.1350210970, 0.07006369427, 0.1301081731, 0.09956906328, 0.09467556925, 0.1241922978, 0.08877182714, 0.1134020619, 0.1075041690, 0.09671869755, 0.1154540559, 0.09877340219, 0.1067251462, 0.1087812521, 0.1000534825, 0.1100621118, 0.1033910821, 0.1046724093, 0.1080103532, 0.1026209240, 0.1072403917, 0.1051890587, 0.1044192125, 0.1069874322, 0.1041663191, 0.1059647414, 0.1057118851, 0.1046892225, 0.1062348095, 0.1049593066, 0.1054822431, 0.1057523363, 0.1049997768, 0.1057928117, 0.1053103493, 0.1053508277, 0.1056614024, 0.1052194201] The largest is 0.4545454545 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 89, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 9, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 9, 14, 22, 35, 56, 90, 145, 234, 378, 611, 988, 1598, 2585, 4182, 6766, 10947, 17712, 28658, 46369] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) with initial conditions, a(1) = 6, a(2) = 9, a(3) = 14, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) with initial conditions, b(1) = 6, b(2) = 9, b(3) = 14, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + 1 = 0 whose roots are 1/2 1/2 5 5 [1, 1/2 - ----, ---- + 1/2] 2 2 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [-0.6180339880, 1.618033988] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2222222222, -0.4285714286, -0.3181818182, -0.4000000000, -0.3571428571, -0.3888888889, -0.3724137931, -0.3846153846, -0.3783068783, -0.3829787234, -0.3805668016, -0.3823529412, -0.3814313346, -0.3821138211, -0.3817617499, -0.3820224719, -0.3818879855, -0.3819875776, -0.3819362074, -0.3819742489, -0.3819546271, -0.3819691578, -0.3819616629, -0.3819672131, -0.3819643503, -0.3819664703, -0.3819653768, -0.3819661866, -0.3819657689, -0.3819660782, -0.3819659187, -0.3819660368, -0.3819659759, -0.3819660210, -0.3819659977, -0.3819660150, -0.3819660061, -0.3819660127, -0.3819660093, -0.3819660118, -0.3819660105, -0.3819660115, -0.3819660110, -0.3819660113, -0.3819660111, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112] The largest is -0.2222222222 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 90, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 10, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 10, 17, 29, 49, 83, 141, 240, 409, 697, 1188, 2025, 3452, 5885, 10033, 17105, 29162, 49718, 84764, 144514] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) + a(n - 5) - a(n - 6) with initial conditions, a(1) = 6, a(2) = 10, a(3) = 17, a(4) = 29, a(5) = 49, a(6) = 83, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) + b(n - 5) - b(n - 6) with initial conditions, b(1) = 6, b(2) = 10, b(3) = 17, b(4) = 29, b(5) = 49, b(6) = 83, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 3 t - 2 t + t - t + 1 = 0 whose roots are [1, RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 %1 := _Z - _Z - _Z - 1 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.70490277604165, 0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] The largest root is, 1.70490277604165 and the remaining roots are [0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] whose absolute values are [0.829475823582982, 0.923307702773949, 0.923307702773949, 0.829475823582982] so the largest absolute value is, 0.923307702773949 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.923307702773949 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, -0.1000000000, 0.4705882353, -0.2068965517, -0.4081632653, -0.4698795181, -0.4893617021, 0.004166666667, -0.2029339853, -0.1162123386, -0.2954545455, -0.4054320988, -0.2001738123, -0.3119796092, -0.1316655038, -0.2423852675, -0.2826966600, -0.2284082224, -0.3262233967, -0.1894210942, -0.2611513828, -0.2363869665, -0.2290618245, -0.2947863021, -0.2163822362, -0.2754326428, -0.2313142787, -0.2389210400, -0.2681338343, -0.2265492575, -0.2732278565, -0.2342034935, -0.2494644815, -0.2549138947, -0.2340397150, -0.2652935457, -0.2366488323, -0.2545189369, -0.2491937408, -0.2408644694, -0.2584638325, -0.2390892107, -0.2551840565, -0.2465790844, -0.2457396866, -0.2538946798, -0.2418356534, -0.2540264661, -0.2455532802, -0.2484315092, -0.2509915455, -0.2443707845, -0.2525008724, -0.2455370134, -0.2495814714, -0.2492221067, -0.2462864389, -0.2511214943] The largest is 0.4705882353 The smallest is -0.4893617021 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 91, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 11, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 11, 20, 36, 65, 117, 211, 381, 688, 1242, 2242, 4047, 7305, 13186, 23802, 42965, 77556, 139996, 252706, 456158] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) + a(n - 6) with initial conditions, a(1) = 6, a(2) = 11, a(3) = 20, a(4) = 36, a(5) = 65, a(6) = 117, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) + b(n - 6) with initial conditions, b(1) = 6, b(2) = 11, b(3) = 20, b(4) = 36, b(5) = 65, b(6) = 117, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 t - 2 t + t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 3 %1 := _Z - 2 _Z + _Z - _Z - 1 In floating-point [1.80509425159131, 0.647170796637930 + 0.726361507447275 I, -0.202712494937727 + 0.895728484592812 I, -0.694010854991715, -0.202712494937727 - 0.895728484592812 I, 0.647170796637930 - 0.726361507447275 I] The largest root is, 1.80509425159131 and the remaining roots are [0.647170796637930 + 0.726361507447275 I, -0.202712494937727 + 0.895728484592812 I, -0.694010854991715, -0.202712494937727 - 0.895728484592812 I, 0.647170796637930 - 0.726361507447275 I] whose absolute values are [0.972846894183278, 0.918380026848806, 0.694010854991715, 0.918380026848806, 0.972846894183278] so the largest absolute value is, 0.972846894183278 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.972846894183278 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, 0.3636363636, -0.2000000000, 0.3611111111, -0.4000000000, -0.4786324786, -0.03317535545, 0.3727034121, 0.09883720930, 0.1529790660, 0.1779661017, -0.1771682728, -0.4125941136, -0.09752768087, 0.1391899840, 0.1162806936, 0.1738098922, 0.1933483814, -0.08343292205, -0.2839323217, -0.1518943221, 0.01299106725, 0.06775379767, 0.1639705778, 0.1897454118, -0.0006583536292, -0.1789858690, -0.1545769096, -0.06307251040, 0.01341659383, 0.1250741996, 0.1730009409, 0.05535840629, -0.09178683851, -0.1290036528, -0.09744546703, -0.03259992018, 0.07624291475, 0.1429986889, 0.08536770444, -0.02502401813, -0.08986251879, -0.1019332352, -0.06278505496, 0.02949929541, 0.1052181150, 0.09312786159, 0.02067438476, -0.04849421221, -0.08732000255, -0.07597211272, -0.007900320041, 0.06597933168, 0.08456125544, 0.04674864695, -0.01240463241, -0.06296876904, -0.07468457877] The largest is 0.3727034121 The smallest is -0.4786324786 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 92, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 13, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 13, 28, 60, 129, 277, 595, 1278, 2745, 5896, 12664, 27201, 58425, 125491, 269542, 578949, 1243524, 2670964, 5736961, 12322413] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 6, a(2) = 13, a(3) = 28, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 6, b(2) = 13, b(3) = 28, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (188 + 12 93 ) 14 (188 + 12 93 ) [------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (188 + 12 93 ) 7 - --------------------- + 1/3 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 14 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (188 + 12 93 ) / 1/2 1/3 (188 + 12 93 ) 7 - ------------------- - --------------------- + 1/3 12 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 14 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (188 + 12 93 ) / In floating-point [2.147899035, -0.5739495177 + 0.3689894078 I, -0.5739495177 - 0.3689894078 I] The largest root is, 2.147899035 and the remaining roots are [-0.5739495177 + 0.3689894078 I, -0.5739495177 - 0.3689894078 I] whose absolute values are [0.6823278039, 0.6823278039] so the largest absolute value is, 0.6823278039 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6823278039 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, 0.3076923077, -0.4285714286, 0.3500000000, -0.2015503876, 0.06859205776, 0.01512605042, -0.04929577465, 0.04954462659, -0.03392130258, 0.01587176248, -0.002426381383, -0.004604193410, 0.006414802655, -0.005219965720, 0.003005446076, -0.001019682773, -0.0002287563591, 0.0007373241687, -0.0007398713223, 0.0005060206559, -0.0002363978200, 0.00003577216951, 0.00006899718539, -0.00009585629560, 0.00007791024469, -5 -0.00004480516112, 0.00001515903265, 0.3458955094 10 , -0.00001102814072, -5 -5 -6 0.00001104880212, -0.7548524235 10 , 0.3520939277 10 , -0.5273070744 10 , -5 -5 -5 -0.1033952755 10 , 0.1432372374 10 , -0.1162840209 10 , -6 -6 -7 0.6679517843 10 , -0.2253562603 10 , -0.5229290111 10 , -6 -6 -6 0.1649463625 10 , -0.1649957001 10 , 0.1126041238 10 , -7 -8 -7 -0.5244091381 10 , 0.7771633758 10 , 0.1549392995 10 , -7 -7 -8 -0.2140371635 10 , 0.1735577731 10 , -0.9957725445 10 , -8 -9 -8 -8 0.3350112816 10 , 0.7904392311 10 , -0.2467060582 10 , 0.2463930696 10 , -8 -9 -9 -0.1679751236 10 , 0.7810495747 10 , -0.1145222013 10 , -9 -9 -0.2321742882 10 , 0.3198308839 10 ] The largest is 0.3500000000 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 93, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 14, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 14, 33, 78, 184, 434, 1024, 2416, 5700, 13448, 31728, 74856, 176608, 416672, 983056, 2319328, 5472000, 12910112, 30458880, 71861760] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 3) with initial conditions, a(1) = 6, a(2) = 14, a(3) = 33, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 3) with initial conditions, b(1) = 6, b(2) = 14, b(3) = 33, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (35 + 3 129 ) 4 (35 + 3 129 ) [------------------ + -------------------- + 2/3, - ------------------ 3 1/2 1/3 6 3 (35 + 3 129 ) 2 - -------------------- + 2/3 1/2 1/3 3 (35 + 3 129 ) / 1/2 1/3 \ 1/2 |(35 + 3 129 ) 4 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (35 + 3 129 ) / 1/2 1/3 (35 + 3 129 ) 2 - ------------------ - -------------------- + 2/3 6 1/2 1/3 3 (35 + 3 129 ) / 1/2 1/3 \ 1/2 |(35 + 3 129 ) 4 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (35 + 3 129 ) / In floating-point [2.359304086, -0.1796520430 + 0.9030131455 I, -0.1796520430 - 0.9030131455 I] The largest root is, 2.359304086 and the remaining roots are [-0.1796520430 + 0.9030131455 I, -0.1796520430 - 0.9030131455 I] whose absolute values are [0.9207103766, 0.9207103766] so the largest absolute value is, 0.9207103766 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.9207103766 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, -0.2142857143, 0.3636363636, 0.05128205128, -0.3260869565, 0.07373271889, 0.2500000000, -0.1523178808, -0.1571929825, 0.1856038073, 0.06656580938, -0.1812546756, 0.008697227759, 0.1505260733, -0.06145733305, -0.1055202196, 0.09001169591, 0.05710872222, -0.09682299546, -0.01362259984, 0.08697224476, -0.01970150152, -0.06664820271, 0.04064808409, 0.04189316514, -0.04951007514, -0.01772398210, 0.04833836607, -0.002343418139, -0.04013480049, 0.01640713117, 0.02812742605, -0.02401474886, -0.01521523539, 0.02582438132, 0.003619264923, -0.02319194094, 0.005264880767, 0.01776829138, -0.01084729912, -0.01116483671, 0.01320690934, 0.004719220425, -0.01289123257, 0.0006313535250, 0.01070114790, -0.004380169345, -0.007497631641, 0.006407032519, 0.004053726347, -0.006887810588, -0.0009615561376, 0.006184340419, -0.001406940338, -0.004736992950, 0.002894694937, 0.002975509198, -0.003522967505] The largest is 0.3636363636 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 94, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 15, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 15, 38, 96, 243, 615, 1556, 3937, 9961, 25202, 63763, 161325, 408164, 1032685, 2612769, 6610498, 16725047, 42315601, 107061588, 270873705] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) - 2 a(n - 4) + 2 a(n - 5) with initial conditions, a(1) = 6, a(2) = 15, a(3) = 38, a(4) = 96, a(5) = 243, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) - 2 b(n - 4) + 2 b(n - 5) with initial conditions, b(1) = 6, b(2) = 15, b(3) = 38, b(4) = 96, b(5) = 243, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 t - 3 t + t + 2 t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 %1 := _Z - 3 _Z + _Z + 2 _Z - 2 In floating-point [2.53007367037607, 0.828616128606492 + 0.426151215131703 I, -0.593652963794526 + 0.747034176337147 I, -0.593652963794526 - 0.747034176337147 I, 0.828616128606492 - 0.426151215131703 I] The largest root is, 2.53007367037607 and the remaining roots are [0.828616128606492 + 0.426151215131703 I, -0.593652963794526 + 0.747034176337147 I, -0.593652963794526 - 0.747034176337147 I, 0.828616128606492 - 0.426151215131703 I] whose absolute values are [0.931777627304411, 0.954192801292142, 0.954192801292142, 0.931777627304411] so the largest absolute value is, 0.954192801292142 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.954192801292142 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2666666667, -0.4736842105, 0.09375000000, 0.4814814815, -0.1934959350, 0.4196658098, 0.3167386335, -0.2448549342, 0.2983493374, -0.08636670169, -0.3516441965, 0.1546167717, -0.2709141703, -0.1979275627, 0.2076864708, -0.1915351867, 0.06876988466, 0.2518715956, -0.1243831659, 0.1734222208, 0.1240396856, -0.1675065861, 0.1259500790, -0.05025395035, -0.1779468595, 0.09950591518, -0.1104487253, -0.07844403225, 0.1305024468, -0.08495417659, 0.03454430428, 0.1245777034, -0.07870415232, 0.07022308654, 0.05037645020, -0.09916053416, 0.05870565874, -0.02257696733, -0.08674328806, 0.06142107187, -0.04472588212, -0.03303346609, 0.07395812531, -0.04142087784, 0.01407314915, 0.06025549323, -0.04728985225, 0.02863295631, 0.02220066721, -0.05439564285, 0.02970309521, -0.008340688648, -0.04186058295, 0.03595155993, -0.01848221339, -0.01531063239, 0.03959010482] The largest is 0.4814814815 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 9, : Consider the Pisot Sequence a(n), defined by, a(1) = 6, a(2) = 16, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 6, 16, 43, 116, 313, 845, 2281, 6157, 16619, 44858, 121081, 326823, 882164, 2381146, 6427213, 17348397, 46826965, 126395808, 341168818, 920886256, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 3 a(n - 1) - a(n - 2) + a(n - 4) + a(n - 5), . Alas, it breaks down at the, 47, -th term. a(47), equals , 405131322121380785922, while the corresponding term for the solution of the recurrence is , 405131322121380785923 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.00714432316835 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 95, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 17, 48, 136, 385, 1090, 3086, 8737, 24736, 70032, 198273, 561346, 1589270, 4499505, 12738896, 36066072, 102109441, 289089922, 818464798, 2317218881] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 6, a(2) = 17, a(3) = 48, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 6, b(2) = 17, b(3) = 48, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (316 + 12 249 ) 20 (316 + 12 249 ) [-------------------- + ---------------------- + 2/3, - -------------------- 6 1/2 1/3 12 3 (316 + 12 249 ) 10 - ---------------------- + 2/3 1/2 1/3 3 (316 + 12 249 ) / 1/2 1/3 \ 1/2 |(316 + 12 249 ) 20 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (316 + 12 249 ) / 1/2 1/3 (316 + 12 249 ) 10 - -------------------- - ---------------------- + 2/3 12 1/2 1/3 3 (316 + 12 249 ) / 1/2 1/3 \ 1/2 |(316 + 12 249 ) 20 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (316 + 12 249 ) / In floating-point [2.831177208, -0.4155886033 + 0.4248482988 I, -0.4155886033 - 0.4248482988 I] The largest root is, 2.831177208 and the remaining roots are [-0.4155886033 + 0.4248482988 I, -0.4155886033 - 0.4248482988 I] whose absolute values are [0.5943147013, 0.5943147013] so the largest absolute value is, 0.5943147013 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5943147013 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, -0.4705882353, 0.3333333333, -0.1102941176, -0.02597402597, 0.06055045872, -0.04115359689, 0.01281904544, 0.003880983182, -0.007753598355, 0.005073812370, -0.001478588963, -0.0005631516357, 0.0009903311586, -0.0006242299176, 0.0001690508465, 0.00007997301640, -0.0001261821919, 0.00007663249556, -0.00001912637618, -0.00001116995308, -5 -5 -5 0.00001603983705, -0.9386608236 10 , 0.2136504548 10 , 0.1539629675 10 , -5 -5 -6 -0.2034339789 10 , 0.1147084322 10 , -0.2348812589 10 , -6 -6 -6 -0.2099336633 10 , 0.2574544770 10 , -0.1398396315 10 , -7 -7 -7 -7 0.2529602775 10 , 0.2836726954 10 , -0.3251303693 10 , 0.1700449297 10 , -8 -8 -8 -0.2649818370 10 , -0.3803687722 10 , 0.4097480789 10 , -8 -9 -9 -0.2062232235 10 , 0.2668093868 10 , 0.5066350929 10 , -9 -9 -10 -0.5153432756 10 , 0.2493930215 10 , -0.2526541520 10 , -10 -10 -10 -0.6708806295 10 , 0.6468606520 10 , -0.3006941070 10 , -11 -11 -11 0.2145246039 10 , 0.8837735867 10 , -0.8103446893 10 , -11 -12 -11 0.3613823988 10 , -0.1415099426 10 , -0.1158818802 10 , -11 -12 -14 0.1013166498 10 , -0.4328145522 10 , 0.1885088395 10 , -12 -12 0.1513075700 10 , -0.1264292355 10 ] The largest is 0.3333333333 The smallest is -0.4705882353 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 96, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 19, 60, 189, 595, 1873, 5896, 18560, 58425, 183916, 578949, 1822473, 5736961, 18059374, 56849086, 178955183, 563332848, 1773314929, 5582216355, 17572253481] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 3 a(n - 2) + a(n - 3) with initial conditions, a(1) = 6, a(2) = 19, a(3) = 60, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 3 b(n - 2) + b(n - 3) with initial conditions, b(1) = 6, b(2) = 19, b(3) = 60, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 3 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (188 + 12 93 ) 14 (188 + 12 93 ) [------------------- + --------------------- + 4/3, - ------------------- 6 1/2 1/3 12 3 (188 + 12 93 ) 7 - --------------------- + 4/3 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 14 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (188 + 12 93 ) / 1/2 1/3 (188 + 12 93 ) 7 - ------------------- - --------------------- + 4/3 12 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 14 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (188 + 12 93 ) / In floating-point [3.147899035, 0.4260504820 + 0.3689894078 I, 0.4260504820 - 0.3689894078 I] The largest root is, 3.147899035 and the remaining roots are [0.4260504820 + 0.3689894078 I, 0.4260504820 - 0.3689894078 I] whose absolute values are [0.5636241623, 0.5636241623] so the largest absolute value is, 0.5636241623 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5636241623 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, 0.4736842105, 0.3500000000, 0.1481481481, 0.01512605042, -0.03416978110, -0.03392130258, -0.01804956897, -0.004604193410, 0.001810609191, 0.003005446076, 0.001985763301, 0.0007373241687, -5 -0.2547153628 10 , -0.0002363978200, -0.0002006256505, -0.00009585629560, -0.00001794605091, 0.00001515903265, 0.00001861798775, 0.00001104880212, -5 -6 -5 0.3500277883 10 , -0.5273070744 10 , -0.1561259829 10 , -5 -6 -7 -0.1162840209 10 , -0.4948884251 10 , -0.5229290111 10 , -6 -6 -7 -7 0.1126534614 10 , 0.1126041238 10 , 0.6016321000 10 , 0.1549392995 10 , -8 -8 -8 -0.5909786401 10 , -0.9957725445 10 , -0.6607612629 10 , -8 -11 -9 -0.2467060582 10 , -0.3129885472 10 , 0.7810495747 10 , -9 -9 -10 0.6665273734 10 , 0.3198308839 10 , 0.6079099004 10 , -10 -10 -10 -0.4980131805 10 , -0.6174735844 10 , -0.3679448960 10 , -10 -11 -11 -0.1173720110 10 , 0.1687305941 10 , 0.5166337470 10 , -11 -11 -12 0.3866230956 10 , 0.1653217356 10 , 0.1805140243 10 , -12 -12 -12 -0.3713650140 10 , -0.3737847731 10 , -0.2005300260 10 , -13 -13 -13 -0.5213079876 10 , 0.1928210986 10 , 0.3299080974 10 , -13 -14 -16 0.2198611060 10 , 0.8254123063 10 , 0.4897018109 10 ] The largest is 0.4736842105 The smallest is -0.03416978110 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 97, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 20, 67, 224, 749, 2504, 8371, 27985, 93556, 312765, 1045598, 3495516, 11685784, 39066492, 130602345, 436613877, 1459634416, 4879672270, 16313126905, 54536061992] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 2) + a(n - 3) - 2 a(n - 4) + a(n - 5) with initial conditions, a(1) = 6, a(2) = 20, a(3) = 67, a(4) = 224, a(5) = 749, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 2) + b(n - 3) - 2 b(n - 4) + b(n - 5) with initial conditions, b(1) = 6, b(2) = 20, b(3) = 67, b(4) = 224, b(5) = 749, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 3 t - t - t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 3 _Z - _Z - _Z + 2 _Z - 1 In floating-point [3.34307838770613, 0.454543710297994 + 0.314012459972290 I, -0.626082904151057 + 0.766858862893121 I, -0.626082904151057 - 0.766858862893121 I, 0.454543710297994 - 0.314012459972290 I] The largest root is, 3.34307838770613 and the remaining roots are [0.454543710297994 + 0.314012459972290 I, -0.626082904151057 + 0.766858862893121 I, -0.626082904151057 - 0.766858862893121 I, 0.454543710297994 - 0.314012459972290 I] whose absolute values are [0.552461591053456, 0.989975918125260, 0.989975918125260, 0.552461591053456] so the largest absolute value is, 0.989975918125260 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.989975918125260 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, 0.4500000000, -0.1044776119, 0.4687500000, 0.1829105474, -0.3190894569, 0.3522876598, -0.1211005896, -0.2272756424, 0.3704506578, -0.2606890985, -0.04440431684, 0.3099992264, -0.3432726696, 0.1276057409, 0.1776633087, -0.3470797718, 0.2605742998, 0.01382228467, -0.2727594944, 0.3279409535, -0.1433427206, -0.1419169723, 0.3181885894, -0.2593353262, 0.01289203318, 0.2380205868, -0.3106756838, 0.1557448102, 0.1094599409, -0.2897001912, 0.2554761318, -0.03597715903, -0.2053306081, 0.2923674718, -0.1648578066, -0.08010610610, 0.2618754042, -0.2494032519, 0.05564262725, 0.1747544396, -0.2733542202, 0.1710163143, 0.05376065585, -0.2349221904, 0.2414732790, -0.07212854625, -0.1463395475, 0.2539311268, -0.1745434618, -0.03030843462, 0.2090129100, -0.2320149677, 0.08567762260, 0.1201042175, -0.2343589472, 0.1757478438, 0.009618588703] The largest is 0.4687500000 The smallest is -0.3470797718 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 98, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 21, 74, 261, 921, 3250, 11469, 40473, 142825, 504015, 1778618, 6276563, 22149356, 78162837, 275828746, 973371746, 3434930440, 12121522097, 42775625450, 150950855676] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 2) + 3 a(n - 3) + a(n - 5) - a(n - 6) with initial conditions, a(1) = 6, a(2) = 21, a(3) = 74, a(4) = 261, a(5) = 921, a(6) = 3250, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 2) + 3 b(n - 3) + b(n - 5) - b(n - 6) with initial conditions, b(1) = 6, b(2) = 21, b(3) = 74, b(4) = 261, b(5) = 921, b(6) = 3250, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 t - 3 t - t - 3 t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 3 %1 := _Z - 3 _Z - _Z - 3 _Z - _Z + 1 In floating-point [0.496503936931801, 3.52889885508405, 0.0595834696248379 + 0.851463869804557 I, -0.572284865632765 + 0.675197067605786 I, -0.572284865632765 - 0.675197067605786 I, 0.0595834696248379 - 0.851463869804557 I] The largest root is, 3.52889885508405 and the remaining roots are [0.496503936931801, 0.0595834696248379 + 0.851463869804557 I, -0.572284865632765 + 0.675197067605786 I, -0.572284865632765 - 0.675197067605786 I, 0.0595834696248379 - 0.851463869804557 I] whose absolute values are [0.496503936931801, 0.853546080440351, 0.885099456296164, 0.885099456296164, 0.853546080440351] so the largest absolute value is, 0.885099456296164 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.885099456296164 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2380952381, -0.4459459459, -0.03448275862, -0.4875135722, 0.2187692308, 0.3316766937, -0.4563536185, 0.03063189218, 0.1775324147, -0.09953739364, 0.08372496221, -0.1037956589, -0.03928831294, 0.1764148107, -0.09850066266, -0.05368976002, 0.08215386814, -0.03822279760, 0.02211931886, -0.0003187099248, -0.04869430108, 0.05579997157, -0.002627181879, -0.03782236084, 0.02886762154, -0.007476633016, -0.002535087371, 0.01309381603, -0.02087871728, 0.009542384467, 0.01068562967, -0.01609533271, 0.006655688351, 0.001956088044, -0.005340943898, 0.007043566605, -0.005122942330, -0.001597071019, 0.006516944121, -0.004712097589, -0.00002605120092, 0.002594072236, -0.002854255950, 0.002067165922, -0.00009958318629, -0.002108305100, 0.002397122718, -0.0006640146917, -0.0009984147843, 0.001365360000, -0.0009031007743, 0.0001662411411, 0.0006305652383, -0.0009857655592, 0.0005357667677, 0.0002447696851, -0.0006178789394] The largest is 0.3316766937 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 99, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 22, 81, 298, 1096, 4031, 14826, 54530, 200561, 737662, 2713116, 9978823, 36702046, 134989886, 496491921, 1826094050, 6716362016, 24702735727, 90856501026, 334169618698] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 2 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 6, a(2) = 22, a(3) = 81, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 2 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 6, b(2) = 22, b(3) = 81, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 2 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (548 + 12 1641 ) 20 (548 + 12 1641 ) [--------------------- + ----------------------- + 4/3, - --------------------- 6 1/2 1/3 12 3 (548 + 12 1641 ) 10 - ----------------------- + 4/3 1/2 1/3 3 (548 + 12 1641 ) / 1/2 1/3 \ 1/2 |(548 + 12 1641 ) 20 | + 1/2 I 3 |--------------------- - -----------------------|, | 6 1/2 1/3| \ 3 (548 + 12 1641 ) / 1/2 1/3 (548 + 12 1641 ) 10 - --------------------- - ----------------------- + 4/3 12 1/2 1/3 3 (548 + 12 1641 ) / 1/2 1/3 \ 1/2 |(548 + 12 1641 ) 20 | - 1/2 I 3 |--------------------- - -----------------------|] | 6 1/2 1/3| \ 3 (548 + 12 1641 ) / In floating-point [3.677993483, 0.161003259 + 0.8886732135 I, 0.161003259 - 0.8886732135 I] The largest root is, 3.677993483 and the remaining roots are [0.161003259 + 0.8886732135 I, 0.161003259 - 0.8886732135 I] whose absolute values are [0.9031401496, 0.9031401496] so the largest absolute value is, 0.9031401496 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.9031401496 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, 0.2272727273, 0.3456790123, -0.07382550336, -0.3056569343, -0.03820391962, 0.2370160529, 0.1074821199, -0.1587148050, -0.1387762959, 0.08477079491, 0.1404913185, -0.02390520681, -0.1222910804, -0.01987995289, 0.09334672877, 0.04627357969, -0.06123899744, -0.05746296284, 0.03144688262, 0.05699646381, -0.007296798496, -0.04883947376, -0.009774906612, 0.03668892558, 0.01978709428, -0.02355419390, -0.02372418740, 0.01157292103, 0.02307747722, -0.002008495369, -0.01947017283, -0.004631268919, 0.01438978388, 0.008411154854, -0.009028755097, -0.009767978464, 0.004219060897, 0.009325935226, -0.0004383162831, -0.007747952893, -0.002137373329, 0.005631463622, 0.003556742467, -0.003448077363, -0.004011403518, 0.001520768054, 0.003761647163, -0.00002915801083, -0.003077622208, -0.0009672313214, 0.002198845098, 0.001496976411, -0.001311478517, -0.001643331597, 0.0005405598782, 0.001514467155, 0.00004675407120] The largest is 0.3456790123 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 100, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 23, 88, 337, 1291, 4946, 18949, 72597, 278132, 1065573, 4082399, 15640394, 59921121, 229568433, 879517348, 3369586817, 12909484211, 49458521666, 189484360909, 725948164637] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 5 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 6, a(2) = 23, a(3) = 88, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 5 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 6, b(2) = 23, b(3) = 88, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 5 t + 5 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (316 + 12 249 ) 20 (316 + 12 249 ) [-------------------- + ---------------------- + 5/3, - -------------------- 6 1/2 1/3 12 3 (316 + 12 249 ) 10 - ---------------------- + 5/3 1/2 1/3 3 (316 + 12 249 ) / 1/2 1/3 \ 1/2 |(316 + 12 249 ) 20 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (316 + 12 249 ) / 1/2 1/3 (316 + 12 249 ) 10 - -------------------- - ---------------------- + 5/3 12 1/2 1/3 3 (316 + 12 249 ) / 1/2 1/3 \ 1/2 |(316 + 12 249 ) 20 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (316 + 12 249 ) / In floating-point [3.831177208, 0.584411397 + 0.4248482988 I, 0.584411397 - 0.4248482988 I] The largest root is, 3.831177208 and the remaining roots are [0.584411397 + 0.4248482988 I, 0.584411397 - 0.4248482988 I] whose absolute values are [0.7225183444, 0.7225183444] so the largest absolute value is, 0.7225183444 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7225183444 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, -0.3043478261, -0.4431818182, -0.3590504451, -0.1882261813, -0.03255155681, 0.06021425933, 0.08737275645, 0.07068945680, 0.03701201138, 0.006358271203, -0.01188978999, -0.01721628339, -0.01391592458, -0.007277785952, -0.001241873626, 0.002347712465, 0.003392358553, 0.002739483187, 0.001431048102, 0.0002425416779, -0.0004635657447, -0.0006684409091, -0.0005392924665, -0.0002813892761, -0.00004736586648, 0.00009153211526, 0.0001317113565, 0.0001061644730, 0.00005532981330, -5 0.9249414369 10 , -0.00001807304865, -0.00002595268848, -0.00002089937041, -5 -5 -5 -0.00001087950698, -0.1806059773 10 , 0.3568495194 10 , 0.5113760882 10 , -5 -5 -6 -6 0.4114208891 10 , 0.2139230434 10 , 0.3526294805 10 , -0.7045869872 10 , -5 -6 -6 -0.1007621470 10 , -0.8099134533 10 , -0.4206338903 10 , -7 -6 -6 -6 -0.6884512551 10 , 0.1391169176 10 , 0.1985424348 10 , 0.1594373351 10 , -7 -7 -7 0.8270833668 10 , 0.1343987745 10 , -0.2746762593 10 , -7 -7 -7 -0.3912084353 10 , -0.3138633310 10 , -0.1626269970 10 , -8 -8 -8 -0.2623520086 10 , 0.5423231888 10 , 0.7708360464 10 ] The largest is 0.1666666667 The smallest is -0.4431818182 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 101, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 25, 104, 433, 1803, 7508, 31265, 130194, 542155, 2257647, 9401315, 39149045, 163024824, 678869516, 2826954868, 11772032235, 49021208124, 204134579142, 850059147789, 3539824353997] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 3 a(n - 3) - a(n - 4) + 2 a(n - 5) - 2 a(n - 6) with initial conditions, a(1) = 6, a(2) = 25, a(3) = 104, a(4) = 433, a(5) = 1803, a(6) = 7508, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 3 b(n - 3) - b(n - 4) + 2 b(n - 5) - 2 b(n - 6) with initial conditions, b(1) = 6, b(2) = 25, b(3) = 104, b(4) = 433, b(5) = 1803, b(6) = 7508, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 3 2 t - 4 t - 3 t + t - 2 t + 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 3 2 %1 := _Z - 4 _Z - 3 _Z + _Z - 2 _Z + 2 In floating-point [0.637514941830891, 4.16420947083987, 0.219349631191768 + 0.876985437485620 I, -0.620211837527150 + 0.732943678324603 I, -0.620211837527150 - 0.732943678324603 I, 0.219349631191768 - 0.876985437485620 I] The largest root is, 4.16420947083987 and the remaining roots are [0.637514941830891, 0.219349631191768 + 0.876985437485620 I, -0.620211837527150 + 0.732943678324603 I, -0.620211837527150 - 0.732943678324603 I, 0.219349631191768 - 0.876985437485620 I] whose absolute values are [0.637514941830891, 0.904000950367758, 0.960140176747543, 0.960140176747543, 0.904000950367758] so the largest absolute value is, 0.960140176747543 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.960140176747543 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, -0.3600000000, -0.2211538462, -0.3579676674, -0.4054353855, 0.4892115077, 0.04992803454, -0.3801480867, 0.07891470152, -0.1186589400, 0.1242874002, 0.2354784389, -0.3531301221, -0.002873653853, 0.1755060184, -0.1069520512, 0.1390830334, -0.09149327924, -0.1618223523, 0.2336710865, -0.04379466418, -0.07708226533, 0.09335392522, -0.1322975244, 0.07434464844, 0.09959113337, -0.1584571671, 0.06237518223, 0.02262658142, -0.07117196348, 0.1113878298, -0.06504071986, -0.05464065255, 0.1072756410, -0.06500451507, 0.006220288470, 0.04849163022, -0.08752253073, 0.05740784469, 0.02432566877, -0.07130694026, 0.05906098728, -0.02021521117, -0.02924658339, 0.06533921668, -0.04961497211, -0.005248572378, 0.04571754703, -0.04937668935, 0.02553409774, 0.01462922683, -0.04659790827, 0.04151958829, -0.005756536847, -0.02762752486, 0.03883683186, -0.02589614160, -0.004475611013] The largest is 0.4892115077 The smallest is -0.4054353855 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 10, :The Pisot Sequence a(n), defined by, a(1) = 6, a(2) = 26, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [6, 26, 113, 491, 2133, 9266, 40253, 174866, 759648, 3300042, 14335952, 62277850, 270545730, 1175297349, 5105694548, 22180018393, 96353828316, 418577661508, 1818373610843, 7899328828724] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 11, :The Pisot Sequence a(n), defined by, a(1) = 6, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [6, 27, 122, 551, 2489, 11243, 50785, 229398, 1036201, 4680566, 21142325, 95500823, 431381468, 1948569291, 8801774215, 39758005881, 179588682125, 811210070342, 3664271993300, 16551679684181] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 102, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 28, 131, 613, 2868, 13418, 62776, 293697, 1374059, 6428524, 30075798, 140709380, 658307707, 3079887333, 14409228212, 67413458746, 315393326640, 1475565151817, 6903419740843, 32297593948716] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - a(n - 2) - 3 a(n - 3) + 3 a(n - 4) with initial conditions, a(1) = 6, a(2) = 28, a(3) = 131, a(4) = 613, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - b(n - 2) - 3 b(n - 3) + 3 b(n - 4) with initial conditions, b(1) = 6, b(2) = 28, b(3) = 131, b(4) = 613, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t + t + 3 t - 3 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z + _Z + 3 _Z - 3 In floating-point [4.67849198820004, 0.630842332243811 + 0.532983826308360 I, -0.940176652687666, 0.630842332243811 - 0.532983826308360 I] The largest root is, 4.67849198820004 and the remaining roots are [0.630842332243811 + 0.532983826308360 I, -0.940176652687666, 0.630842332243811 - 0.532983826308360 I] whose absolute values are [0.825853381210679, 0.940176652687666, 0.825853381210679] so the largest absolute value is, 0.940176652687666 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.940176652687666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, -0.1071428571, 0.4656488550, 0.3099510604, 0.4030683403, -0.01266954837, 0.0003982413661, -0.2647184003, -0.07678418467, -0.1584058798, 0.08010474070, -0.004873100855, 0.1403948397, -0.008684562488, 0.07111587248, -0.07153989681, 0.01842285007, -0.07574715775, 0.03080866903, -0.04009773770, 0.05121266591, -0.02350641311, 0.04397448876, -0.03055235391, 0.02742097874, -0.03478545800, 0.02223225928, -0.02797324354, 0.02452083324, -0.02047574209, 0.02371696475, -0.01842166448, 0.01916443884, -0.01833426186, 0.01558013956, -0.01652335029, 0.01429921107, -0.01372379864, 0.01339226531, -0.01178255889, 0.01176396935, -0.01074578621, 0.01003157222, -0.009735937424, 0.008818007335, -0.008506101181, 0.007954015688, -0.007385654655, 0.007090036582, -0.006544513040, 0.006206409249, -0.005850514427, 0.005444667482, -0.005178915030, 0.004831528398, -0.004548988708, 0.004294275595, -0.004010963600] The largest is 0.4656488550 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 103, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 29, 140, 676, 3264, 15760, 76096, 367424, 1774080, 8566016, 41360384, 199705600, 964263936, 4655878144, 22480568320, 108545785856, 524105416704, 2530604810240, 12218840907776, 58997782872064] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 4 a(n - 2) with initial conditions, a(1) = 6, a(2) = 29, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 4 b(n - 2) with initial conditions, b(1) = 6, b(2) = 29, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t - 4 = 0 whose roots are 1/2 1/2 [2 + 2 2 , 2 - 2 2 ] In floating-point [4.828427124, -0.828427124] The largest root is, 4.828427124 and the remaining roots are [-0.828427124] whose absolute values are [0.828427124] so the largest absolute value is, 0.828427124 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.828427124 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, -0.1379310345, 0.1142857143, -0.09467455621, 0.07843137255, -0.06497461929, 0.05382674516, -0.04459153458, 0.03694083694, -0.03060279131, 0.02535218242, -0.02100243559, 0.01739898732, -0.01441379304, 0.01194077713, -0.009892063663, 0.008194853858, -0.006788839219, 0.005624058555, -0.004659122658, 0.003859743587, -0.003197516282, 0.002648909220, -0.002194428249, 0.001817923885, -0.001506017457, 0.001247625712, -0.001033566981, 0.0008562349223, -0.0007093282348, 0.0005876267500, -0.0004868059390, 0.0004032832443, -0.0003340907786, 0.0002767698631, -0.0002292836619, 0.0001899448048, -0.0001573554285, 0.0001303575052, -0.0001079916932, 0.00008946324790, -0.00007411378122, 0.00006139786668, -0.00005086365816, 0.00004213683409, -0.00003490729631, 0.00002891815111, -0.00002395658078, 0.00001984628133, -5 -0.00001644119778, 0.00001362033421, -0.00001128345430, 0.9347519607 10 , -5 -5 -5 -0.7743738791 10 , 0.6415123262 10 , -0.5314462118 10 , -5 -5 0.4402644572 10 , -0.3647270184 10 ] The largest is 0.1666666667 The smallest is -0.1379310345 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 12, :The Pisot Sequence a(n), defined by, a(1) = 6, a(2) = 31, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [6, 31, 160, 826, 4264, 22012, 113632, 586600, 3028192, 15632368, 80698625, 416588714, 2150546637, 11101718992, 57310156616, 295850944680, 1527264740428, 7884164743424, 40700248002864, 210105983500192] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 104, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 32, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 32, 171, 914, 4885, 26109, 139546, 745838, 3986315, 21305843, 113874329, 608629417, 3252969923, 17386299486, 92925362660, 496662503280, 2654535157069, 14187817388225, 75830286785080, 405293656984887] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - 4 a(n - 2) + 3 a(n - 3) - 2 a(n - 4) + a(n - 5) with initial conditions, a(1) = 6, a(2) = 32, a(3) = 171, a(4) = 914, a(5) = 4885, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - 4 b(n - 2) + 3 b(n - 3) - 2 b(n - 4) + b(n - 5) with initial conditions, b(1) = 6, b(2) = 32, b(3) = 171, b(4) = 914, b(5) = 4885, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 6 t + 4 t - 3 t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 6 _Z + 4 _Z - 3 _Z + 2 _Z - 1 In floating-point [5.34474646170837, 0.509572462607406 + 0.367715945551002 I, -0.181945693461589 + 0.663860439683354 I, -0.181945693461589 - 0.663860439683354 I, 0.509572462607406 - 0.367715945551002 I] The largest root is, 5.34474646170837 and the remaining roots are [0.509572462607406 + 0.367715945551002 I, -0.181945693461589 + 0.663860439683354 I, -0.181945693461589 - 0.663860439683354 I, 0.509572462607406 - 0.367715945551002 I] whose absolute values are [0.628394073221767, 0.688342152381935, 0.688342152381935, 0.628394073221767] so the largest absolute value is, 0.688342152381935 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.688342152381935 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, -0.2187500000, 0.3567251462, -0.4387308534, -0.4767656090, 0.06794591903, 0.06631505023, -0.07000045586, 0.03336765910, 0.06649396600, -0.009192396646, -0.01471130502, 0.01124788020, -0.0008649615182, -0.009436445712, 0.0008150258140, 0.002833987745, -0.001587710679, -5 -0.0004092077067, 0.001331062371, 0.7123342054 10 , -0.0004997234504, 0.0001970577780, 0.0001312780464, -0.0001829174977, -0.00002487359493, 0.00008242355402, -0.00002021510408, -0.00002849258355, 0.00002400526923, -5 -5 -5 0.7635934380 10 , -0.00001282945912, 0.1265378493 10 , 0.5314788551 10 , -5 -5 -5 -0.2927759546 10 , -0.1734723385 10 , 0.1886847422 10 , -6 -6 -6 -6 0.1125008281 10 , -0.9062472336 10 , 0.3147427775 10 , 0.3425298536 10 , -6 -7 -6 -0.2606879227 10 , -0.6502332283 10 , 0.1444685260 10 , -7 -7 -7 -0.2547625035 10 , -0.6189587572 10 , 0.3329404816 10 , -7 -7 -9 -7 0.1695866588 10 , -0.2169079777 10 , 0.2181954238 10 , 0.1046438922 10 , -8 -8 -8 -0.3782123259 10 , -0.3555448762 10 , 0.3061779520 10 , -9 -8 -9 -9 0.5355193689 10 , -0.1671712412 10 , 0.3417608798 10 , 0.6649652300 10 ] The largest is 0.3567251462 The smallest is -0.4767656090 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 105, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 33, 182, 1004, 5539, 30558, 168585, 930064, 5131056, 28307445, 156168914, 861565913, 4753159918, 26222635860, 144667262055, 798112623850, 4403095429473, 24291370392720, 134012692890224, 739332592823657] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) + 3 a(n - 2) - a(n - 3) + a(n - 4) with initial conditions, a(1) = 6, a(2) = 33, a(3) = 182, a(4) = 1004, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) + 3 b(n - 2) - b(n - 3) + b(n - 4) with initial conditions, b(1) = 6, b(2) = 33, b(3) = 182, b(4) = 1004, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t - 3 t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z - 3 _Z + _Z - 1 In floating-point [5.51688483291115, 0.193215260440345 + 0.404141783125213 I, -0.903315353791842, 0.193215260440345 - 0.404141783125213 I] The largest root is, 5.51688483291115 and the remaining roots are [0.193215260440345 + 0.404141783125213 I, -0.903315353791842, 0.193215260440345 - 0.404141783125213 I] whose absolute values are [0.447953923673693, 0.903315353791842, 0.447953923673693] so the largest absolute value is, 0.903315353791842 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.903315353791842 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2424242424, -0.4505494505, 0.2878486056, -0.1716916411, 0.2131356764, -0.1878221669, 0.1598341620, -0.1491231045, 0.1348447732, -0.1208017813, 0.1094826787, -0.09895982843, 0.08929544831, -0.08068670374, 0.07289533330, -0.06583872146, 0.05947454465, -0.05372547816, 0.04853029791, -0.04383821103, 0.03959986141, -0.03577110210, 0.03231258267, -0.02918846542, 0.02636638442, -0.02381715894, 0.02151440663, -0.01943429353, 0.01755529560, -0.01585796817, 0.01432474611, -0.01293976308, 0.01168868668, -0.01055857014, 0.009537718520, -0.008615567580, 0.007782574477, -0.007030119017, 0.006350414447, -0.005736426873, 0.005181802470, -0.004680801732, 0.004228240072, -0.003819434177, 0.003450153535, -0.003116576661, 0.002815251549, -0.002543059949, 0.002297185098, -0.002075082569, 0.001874453945, -0.001693223029, 0.001529514359, -0.001381633804, 0.001248051029, -0.001127383657, 0.001018382967] The largest is 0.2878486056 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 106, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 34, 193, 1096, 6224, 35345, 200718, 1139842, 6472961, 36758800, 208746720, 1185435681, 6731879446, 38229152034, 217096588993, 1232853109336, 7001154630064, 39758318150705, 225780452753438, 1282167234849282] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 6, a(2) = 34, a(3) = 193, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 6, b(2) = 34, b(3) = 193, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 6 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (1404 + 12 1689 ) 20 (1404 + 12 1689 ) [---------------------- + ---------------------- + 2, - ---------------------- 6 1/2 1/3 12 (1404 + 12 1689 ) 10 - ---------------------- + 2 1/2 1/3 (1404 + 12 1689 ) / 1/2 1/3 \ 1/2 |(1404 + 12 1689 ) 20 | + 1/2 I 3 |---------------------- - ----------------------|, | 6 1/2 1/3| \ (1404 + 12 1689 ) / 1/2 1/3 (1404 + 12 1689 ) 10 - ---------------------- - ---------------------- + 2 12 1/2 1/3 (1404 + 12 1689 ) / 1/2 1/3 \ 1/2 |(1404 + 12 1689 ) 20 | - 1/2 I 3 |---------------------- - ----------------------|] | 6 1/2 1/3| \ (1404 + 12 1689 ) / In floating-point [5.678823031, 0.160588485 + 0.3876908118 I, 0.160588485 - 0.3876908118 I] The largest root is, 5.678823031 and the remaining roots are [0.160588485 + 0.3876908118 I, 0.160588485 - 0.3876908118 I] whose absolute values are [0.4196341586, 0.4196341586] so the largest absolute value is, 0.4196341586 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.4196341586 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, -0.4411764706, -0.08290155440, 0.05109489051, 0.03100899743, 0.0009619465271, -0.005151506093, -0.001823937002, 0.0003213367113, 0.0004243881737, 0.00007971861785, -0.00004912792903, -0.00002981663614, -6 -5 -5 -0.9253409536 10 , 0.4953297539 10 , 0.1753830999 10 , -6 -6 -7 -0.3089500396 10 , -0.4080646958 10 , -0.7665709670 10 , -7 -7 -9 -8 0.4723677174 10 , 0.2867012806 10 , 0.8901281949 10 , -0.4762715217 10 , -8 -9 -9 -0.1686419632 10 , 0.2970408383 10 , 0.3923690764 10 , -10 -10 -10 0.7371315016 10 , -0.4541841359 10 , -0.2756770543 10 , -12 -11 -11 -0.8562552616 10 , 0.4579465707 10 , 0.1621599329 10 , -12 -12 -13 -0.2855907026 10 , -0.3772771661 10 , -0.7088226291 10 , -13 -13 -15 0.4367005223 10 , 0.2650767309 10 , 0.8236711802 10 , -14 -14 -15 -0.4403266872 10 , -0.1559270498 10 , 0.2745819367 10 , -15 -16 -16 0.3627657440 10 , 0.6816009304 10 , -0.4198899317 10 , -16 -18 -17 -0.2548840105 10 , -0.7923269160 10 , 0.4233847432 10 , -17 -18 -18 0.1499337375 10 , -0.2639975273 10 , -0.3488124826 10 , -19 -19 -19 -0.6554246537 10 , 0.4037264564 10 , 0.2450832199 10 , -21 -20 -20 0.7621752999 10 , -0.4070946545 10 , -0.1441707879 10 , -21 -21 0.2538211167 10 , 0.3353959129 10 ] The largest is 0.05109489051 The smallest is -0.4411764706 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 107, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 35, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - a(n - 2) with initial conditions, a(1) = 6, a(2) = 35, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - b(n - 2) with initial conditions, b(1) = 6, b(2) = 35, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 6 t + 1 = 0 whose roots are 1/2 1/2 [3 + 2 2 , 3 - 2 2 ] In floating-point [5.828427124, 0.171572876] The largest root is, 5.828427124 and the remaining roots are [0.171572876] whose absolute values are [0.171572876] so the largest absolute value is, 0.171572876 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.171572876 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, 0.02857142857, 0.004901960784, 0.0008410428932, 0.0001443001443, -5 -6 -6 0.00002475799064, 0.4247799640 10 , 0.7288071977 10 , 0.1250435464 10 , -7 -8 -9 -9 0.2145408079 10 , 0.3680938327 10 , 0.6315491724 10 , 0.1083567074 10 , -10 -11 -12 0.1859107184 10 , 0.3189723649 10 , 0.5472700578 10 , -13 -13 -14 0.9389669735 10 , 0.1611012634 10 , 0.2764060697 10 , -15 -16 -16 0.4742378412 10 , 0.8136634996 10 , 0.1396025861 10 , -17 -18 -19 0.2395201709 10 , 0.4109516441 10 , 0.7050815517 10 , -19 -20 -21 0.1209728691 10 , 0.2075566298 10 , 0.3561108775 10 , -22 -22 -23 0.6109896717 10 , 0.1048292547 10 , 0.1798585664 10 , -24 -25 -26 0.3085885138 10 , 0.5294541859 10 , 0.9083997698 10 , -26 -27 -28 0.1558567604 10 , 0.2674079251 10 , 0.4587994657 10 , -29 -29 -30 0.7871754350 10 , 0.1350579527 10 , 0.2317228127 10 , -31 -32 -32 0.3975734924 10 , 0.6821282722 10 , 0.1170347089 10 , -33 -34 -35 0.2007998152 10 , 0.3445180164 10 , 0.5910994665 10 , -35 -36 -37 0.1014166350 10 , 0.1740034367 10 , 0.2985426994 10 , -38 -39 -39 0.5122182932 10 , 0.8788276533 10 , 0.1507829873 10 , -40 -41 -42 0.2587027068 10 , 0.4438636723 10 , 0.7615496648 10 , -42 -43 -44 0.1306612656 10 , 0.2241792903 10 , 0.3846308541 10 ] The largest is 0.1666666667 The smallest is -44 0.3846308541 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 108, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 37, 228, 1405, 8658, 53353, 328776, 2026009, 12484830, 76934989, 474094764, 2921503573, 18003116202, 110940200785, 683644320912, 4212806126257, 25960481078454, 159975692596981, 985814636660340, 6074863512559021] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + a(n - 2) with initial conditions, a(1) = 6, a(2) = 37, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + b(n - 2) with initial conditions, b(1) = 6, b(2) = 37, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 6 t - 1 = 0 whose roots are 1/2 1/2 [3 + 10 , 3 - 10 ] In floating-point [6.162277660, -0.162277660] The largest root is, 6.162277660 and the remaining roots are [-0.162277660] whose absolute values are [0.162277660] so the largest absolute value is, 0.162277660 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.162277660 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1666666667, -0.02702702703, 0.004385964912, -0.0007117437722, -5 -6 0.0001155001155, -0.00001874308849, 0.3041584544 10 , -0.4935812230 10 , -7 -7 -8 0.8009720597 10 , -0.1299798717 10 , 0.2109282945 10 , -9 -10 -11 -0.3422895009 10 , 0.5554593931 10 , -0.9013865064 10 , -11 -12 -13 0.1462748932 10 , -0.2373714740 10 , 0.3852008740 10 , -14 -14 -15 -0.6250949652 10 , 0.1014389483 10 , -0.1646127519 10 , -16 -17 -18 0.2671297221 10 , -0.4334918626 10 , 0.7034604516 10 , -18 -19 -20 -0.1141559161 10 , 0.1852495496 10 , -0.3006186346 10 , -21 -22 -22 0.4878368862 10 , -0.7916502844 10 , 0.1284671558 10 , -23 -24 -25 -0.2084734946 10 , 0.3383059090 10 , -0.5489949134 10 , -26 -26 -27 0.8908960999 10 , -0.1445725345 10 , 0.2346089263 10 , -28 -29 -29 -0.3807178761 10 , 0.6178200613 10 , -0.1002583939 10 , -30 -31 -32 0.1626969758 10 , -0.2640208455 10 , 0.4284468505 10 , -33 -33 -34 -0.6952735240 10 , 0.1128273607 10 , -0.1830936009 10 , -35 -36 -37 0.2971200115 10 , -0.4821594025 10 , 0.7824369966 10 , -37 -38 -39 -0.1269720450 10 , 0.2060472638 10 , -0.3343686785 10 , -40 -41 -41 0.5426056678 10 , -0.8805277816 10 , 0.1428899881 10 , -42 -43 -44 -0.2318785293 10 , 0.3762870518 10 , -0.6106298232 10 , -45 -45 0.9909157894 10 , -0.1608034957 10 ] The largest is 0.1666666667 The smallest is -0.02702702703 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 109, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 38, 241, 1528, 9688, 61425, 389454, 2469262, 15655905, 99263408, 629361520, 3990351841, 25300097494, 160410650166, 1017054447825, 6448448084776, 40885208054472, 259225198944209, 1643570057858974, 10420755953096734] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 6, a(2) = 38, a(3) = 241, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 6, b(2) = 38, b(3) = 241, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 6 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (2268 + 84 57 ) 28 (2268 + 84 57 ) [-------------------- + -------------------- + 2, - -------------------- 6 1/2 1/3 12 (2268 + 84 57 ) 14 - -------------------- + 2 1/2 1/3 (2268 + 84 57 ) / 1/2 1/3 \ 1/2 |(2268 + 84 57 ) 28 | + 1/2 I 3 |-------------------- - --------------------|, | 6 1/2 1/3| \ (2268 + 84 57 ) / 1/2 1/3 (2268 + 84 57 ) 14 - -------------------- - -------------------- + 2 12 1/2 1/3 (2268 + 84 57 ) / 1/2 1/3 \ 1/2 |(2268 + 84 57 ) 28 | - 1/2 I 3 |-------------------- - --------------------|] | 6 1/2 1/3| \ (2268 + 84 57 ) / In floating-point [6.340317472, -0.170158736 + 0.3588409309 I, -0.170158736 - 0.3588409309 I] The largest root is, 6.340317472 and the remaining roots are [-0.170158736 + 0.3588409309 I, -0.170158736 - 0.3588409309 I] whose absolute values are [0.3971407926, 0.3971407926] so the largest absolute value is, 0.3971407926 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.3971407926 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, 0.4473684211, -0.09958506224, -0.03664921466, 0.02817919075, -0.003809523810, -0.003147996939, 0.001672159536, -0.00007256048117, -5 -0.0002390407551, 0.00009279404308, 0.6122267152 10 , -0.00001671906601, -5 -5 -5 -6 0.4724181338 10 , 0.1029223167 10 , -0.1095364328 10 , 0.2104417076 10 , -6 -7 -8 -8 0.1011447579 10 , -0.6761236460 10 , 0.7057035913 10 , 0.8262244217 10 , -8 -10 -9 -0.3924827471 10 , 0.3255951926 10 , 0.6079463902 10 , -9 -10 -10 -0.2120300919 10 , -0.2372825170 10 , 0.4151669621 10 , -10 -11 -11 -0.1038641802 10 , -0.3013367392 10 , 0.2663655822 10 , -12 -12 -12 -0.4312178712 10 , -0.2733629753 10 , 0.1610422279 10 , -13 -13 -14 -0.1169045420 10 , -0.2142124464 10 , 0.9133851699 10 , -15 -14 -15 0.2701667177 10 , -0.1532540932 10 , 0.4789395414 10 , -16 -15 -16 0.7872210181 10 , -0.1023292385 10 , 0.2240831410 10 , -17 -17 -18 0.8513509472 10 , -0.6431553441 10 , 0.8460124032 10 , -18 -18 -19 0.7264770092 10 , -0.3806665793 10 , 0.1496694613 10 , -19 -19 -20 0.5494552752 10 , -0.2105952184 10 , -0.1499129827 10 , -20 -20 -21 0.3831704890 10 , -0.1067552150 10 , -0.2410329465 10 , -21 -22 -22 0.2504029114 10 , -0.4720057433 10 , -0.2343056963 10 , -22 0.1541834497 10 ] The largest is 0.4473684211 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 110, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 39, 254, 1654, 10771, 70142, 456773, 2974560, 19370688, 126144221, 821466150, 5349485139, 34836482614, 226859312510, 1477334788439, 9620579613790, 62650357136089, 407986567010304, 2656856983261632, 17301768245050633] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) - 3 a(n - 2) - a(n - 3) - a(n - 4) with initial conditions, a(1) = 6, a(2) = 39, a(3) = 254, a(4) = 1654, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) - 3 b(n - 2) - b(n - 3) - b(n - 4) with initial conditions, b(1) = 6, b(2) = 39, b(3) = 254, b(4) = 1654, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 7 t + 3 t + t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 7 _Z + 3 _Z + _Z + 1 In floating-point [0.884461565336866, 6.51211877570109, -0.198290170518979 + 0.366470457776413 I, -0.198290170518979 - 0.366470457776413 I] The largest root is, 6.51211877570109 and the remaining roots are [0.884461565336866, -0.198290170518979 + 0.366470457776413 I, -0.198290170518979 - 0.366470457776413 I] whose absolute values are [0.884461565336866, 0.416676838985921, 0.416676838985921] so the largest absolute value is, 0.884461565336866 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.884461565336866 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2564102564, -0.4645669291, -0.2581620314, -0.1688793984, -0.1994667959, -0.1668750123, -0.1426819429, -0.1298022559, -0.1142281342, -0.1006332142, -0.08926389766, -0.07891725110, -0.06976771640, -0.06172514970, -0.05459174991, -0.04828183275, -0.04270471341, -0.03777059602, -0.03340644923, -0.02954681040, -0.02613301568, -0.02311363332, -0.02044312654, -0.01808115977, -0.01599208976, -0.01414438917, -0.01251016861, -0.01106476323, -0.009786357835, -0.008655657373, -0.007655596263, -0.006771080656, -0.005988760596, -0.005296828571, -0.004684841289, -0.004143562060, -0.003664821386, -0.003241393659, -0.002866888110, -0.002535652345, -0.002242687043, -0.001983570492, -0.001754391862, -0.001551692173, -0.001372412088, -0.001213845744, -0.001073599907, -0.0009495578540, -0.0008398474259, -0.0007428127690, -0.0006569893444, -0.0005810818239, -0.0005139445396, -0.0004545641920, -0.0004020445568, -0.0003555929580, -0.0003145083043] The largest is 0.2564102564 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 111, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 6, a(2) = 40, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [6, 40, 267, 1782, 11893, 79373, 529730, 3535382, 23594899, 157470751, 1050949081, 7013962681, 46810709843, 312411493426, 2085013056884, 13915235318980, 92869334004061, 619803618167633, 4136527188586176, 27606901089894615] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) + 4 a(n - 2) + 3 a(n - 3) + 2 a(n - 4) + a(n - 5) with initial conditions, a(1) = 6, a(2) = 40, a(3) = 267, a(4) = 1782, a(5) = 11893, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) + 4 b(n - 2) + 3 b(n - 3) + 2 b(n - 4) + b(n - 5) with initial conditions, b(1) = 6, b(2) = 40, b(3) = 267, b(4) = 1782, b(5) = 11893, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 6 t - 4 t - 3 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 6 _Z - 4 _Z - 3 _Z - 2 _Z - 1 In floating-point [6.67393197996370, 0.146027932134791 + 0.644811456974029 I, -0.482993922116641 + 0.330922207340797 I, -0.482993922116641 - 0.330922207340797 I, 0.146027932134791 - 0.644811456974029 I] The largest root is, 6.67393197996370 and the remaining roots are [0.146027932134791 + 0.644811456974029 I, -0.482993922116641 + 0.330922207340797 I, -0.482993922116641 - 0.330922207340797 I, 0.146027932134791 - 0.644811456974029 I] whose absolute values are [0.661139903506461, 0.585484958058635, 0.585484958058635, 0.661139903506461] so the largest absolute value is, 0.661139903506461 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.661139903506461 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, 0.2250000000, 0.3483146067, 0.4281705948, -0.4844025898, -0.03258034848, 0.07296924848, 0.05892404272, 0.007043132501, -0.05270102509, 0.002096655337, 0.01372256275, 0.005629230253, -0.003403319168, -0.005243020176, 0.001358073863, 0.001947428222, 0.0002103961688, -0.0007630480327, -0.0004212913064, 0.0003041788433, 0.0002189842957, -0.00004895266826, -0.0001108729423, -0.00001702905939, 0.00004962330445, 0.00001808372144, -0.00001479018457, -0.00001246736939, -5 -5 -6 -5 0.2503759185 10 , 0.6573271155 10 , 0.5559077948 10 , -0.2585114408 10 , -5 -6 -6 -0.1027092824 10 , 0.8150103010 10 , 0.7114340288 10 , -6 -6 -7 -0.1669541178 10 , -0.3502577454 10 , -0.3213307928 10 , -6 -7 -7 0.1431865482 10 , 0.5733952912 10 , -0.4708547894 10 , -7 -8 -7 -8 -0.3811901659 10 , 0.9202589107 10 , 0.1934863786 10 , 0.1713705062 10 , -8 -8 -8 -0.8038962984 10 , -0.3046882448 10 , 0.2603833394 10 , -8 -9 -8 0.2094629604 10 , -0.5217570481 10 , -0.1073251569 10 , -10 -9 -9 -0.8186445449 10 , 0.4436284546 10 , 0.1656737103 10 , -9 -9 -10 -0.1452974697 10 , -0.1151850916 10 , 0.3011315705 10 ] The largest is 0.4281705948 The smallest is -0.4844025898 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 112, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 9, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 2) + a(n - 3) with initial conditions, a(1) = 7, a(2) = 9, a(3) = 12, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 2) + b(n - 3) with initial conditions, b(1) = 7, b(2) = 9, b(3) = 12, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [1.324717958, -0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, 0., 0.3333333333, -0.4375000000, 0.3333333333, -0.1071428571, -0.1081081081, 0.2244897959, -0.2153846154, 0.1162790698, 0.008771929825, -0.09933774834, 0.1250000000, -0.09056603774, 0.02564102564, 0.03440860215, -0.06493506494, 0.06004901961, -0.03052728955, -0.004888268156, 0.02952029520, -0.03541583764, 0.02463202163, -0.005895691610, -0.01078397809, 0.01873627084, -0.01667967226, 0.007952286282, 0.002056583847, -0.008727394789, 0.01000886862, -0.006670811018, 0.001281472791, 0.003338056570, -0.005389338572, 0.004619529351, -0.002051282051, -0.0007698093173, 0.002568247246, -0.002821091377, 0.001798437928, -0.0002528441374, -0.001022653455, 0.001545593789, -0.001275497592, 0.0005229403335, 0.0002700961958, -0.0007525572592, 0.0007930365292, -0.0004824610634, 0.00004047926998, 0.0003105754658, -0.0004419817935, 0.0003510547358, -0.0001314063277, -0.00009092705769, 0.0002196484081, -0.0002223333854] The largest is 0.3333333333 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 113, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 10, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 10, 14, 20, 29, 42, 61, 89, 130, 190, 278, 407, 596, 873, 1279, 1874, 2746, 4024, 5897, 8642] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 7, a(2) = 10, a(3) = 14, a(4) = 20, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 7, b(2) = 10, b(3) = 14, b(4) = 20, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t + t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [1, ------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.465571232, -0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, -0.4000000000, -0.4285714286, 0.05000000000, -0.1724137931, -0.4047619048, -0.1475409836, -0.1123595506, -0.3076923077, -0.2421052632, -0.1402877698, -0.2334152334, -0.2600671141, -0.1844215349, -0.2017200938, -0.2454642476, -0.2134013110, -0.1985586481, -0.2274037646, -0.2241379310, -0.2060007896, -0.2166909110, -0.2241011690, -0.2133647720, -0.2133125663, -0.2206663786, -0.2172807828, -0.2138410261, -0.2177537536, -0.2182799455, -0.2153657483, -0.2163638549, -0.2178878591, -0.2164974644, -0.2161050407, -0.2172365282, -0.2169775571, -0.2163261190, -0.2168061388, -0.2170271673, -0.2165967439, -0.2166463309, -0.2169169401, -0.2167571215, -0.2166468869, -0.2168072594, -0.2168078119, -0.2166981289, -0.2167488177, -0.2168000586, -0.2167416162, -0.2167338624, -0.2167773493, -0.2167623938, -0.2167396844, -0.2167604618, -0.2167662837, -0.2167493962] The largest is 0.2857142857 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 13, :The Pisot Sequence a(n), defined by, a(1) = 7, a(2) = 11, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [7, 11, 17, 26, 40, 62, 96, 149, 231, 358, 555, 860, 1333, 2066, 3202, 4963, 7692, 11922, 18478, 28639] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 114, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 12, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625, 299426] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 7, a(2) = 12, a(3) = 21, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 7, b(2) = 12, b(3) = 21, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (100 + 12 69 ) 2 (100 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (100 + 12 69 ) 1 - ----------------------- + 2/3 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (100 + 12 69 ) / 1/2 1/3 (100 + 12 69 ) 1 - ------------------- - ----------------------- + 2/3 12 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (100 + 12 69 ) / In floating-point [1.754877667, 0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] The largest root is, 1.754877667 and the remaining roots are [0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] whose absolute values are [0.7548776666, 0.7548776666] so the largest absolute value is, 0.7548776666 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7548776666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, -0.2500000000, 0.1904761905, 0.1891891892, -0.06153846154, -0.1228070175, 0.005000000000, 0.07122507122, 0.01461038961, -0.03700277521, -0.01739588824, 0.01682186843, 0.01403628894, -0.006145142411, -0.009504752376, 0.001171924490, 0.005703456367, 0.0007302348065, -0.003071062271, -0.001168903168, 0.001463490759, 0.001024822391, -0.0005827491423, -0.0007268299193, 0.0001539116948, 0.0004519041666, 0.00002306671899, -0.0002518590338, -0.00007488062000, 0.0001251645128, 0.00007335061177, -0.00005334390925, -0.00005487391748, 0.00001694668605, -6 -5 0.00003542338034, -0.9738428538 10 , -0.00002042437999, -0.4451536794 10 , -5 -5 -5 0.00001054746355, 0.5122083903 10 , -0.4754832539 10 , -0.4084285431 10 , -5 -5 -6 0.1708345581 10 , 0.2746144054 10 , -0.3003429039 10 , -5 -6 -6 -0.1638484280 10 , -0.2304816029 10 , 0.8771781708 10 , -6 -6 -6 0.3463536640 10 , -0.4149524457 10 , -0.2990803846 10 , -6 -6 -7 0.1631453405 10 , 0.2104186200 10 , -0.4138848514 10 , -6 -8 -7 -0.1300502497 10 , -0.8293394300 10 , 0.7207497597 10 , -7 0.2239309654 10 ] The largest is 0.1904761905 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 115, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 13, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 3) with initial conditions, a(1) = 7, a(2) = 13, a(3) = 24, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 3) with initial conditions, b(1) = 7, b(2) = 13, b(3) = 24, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (19 + 3 33 ) 4 (19 + 3 33 ) [----------------- + ------------------- + 1/3, - ----------------- 3 1/2 1/3 6 3 (19 + 3 33 ) 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(19 + 3 33 ) 4 | (19 + 3 33 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (19 + 3 33 ) / 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 |(19 + 3 33 ) 4 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (19 + 3 33 ) / In floating-point [1.839286755, -0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] The largest root is, 1.839286755 and the remaining roots are [-0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] whose absolute values are [0.7373527065, 0.7373527065] so the largest absolute value is, 0.7373527065 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7373527065 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, 0.3076923077, -0.3333333333, 0.1136363636, 0.08641975309, -0.1342281879, 0.06569343066, 0.01785714286, -0.05070118662, 0.03284457478, 0., -0.01785714286, 0.01498727496, -0.002869881617, -0.005739760379, 0.006377628310, -0.002232014166, -0.001594146438, 0.002551467581, -0.001274693041, -0.0003173719016, 0.0009594026349, -0.0006326623085, -5 0.9368424754 10 , 0.0003361087511, -0.0002871851327, 0.00005829204314, 0.0001072156615, -0.0001216774280, 0.00004383027661, 0.00002936851008, -5 -0.00004847864135, 0.00002472014534, 0.5610014078 10 , -0.00001814848193, -6 -5 -5 0.00001218167749, -0.3567903558 10 , -0.6323594790 10 , 0.5501292349 10 , -5 -5 -5 -0.1179092797 10 , -0.2001395238 10 , 0.2320804314 10 , -6 -6 -6 -0.8596837211 10 , -0.5402746452 10 , 0.9208459475 10 , -6 -7 -6 -0.4791124188 10 , -0.9854111662 10 , 0.3431924120 10 , -6 -7 -6 -0.2344611235 10 , 0.1019017190 10 , 0.1189214604 10 , -6 -7 -7 -0.1053494911 10 , 0.2376214118 10 , 0.3733411046 10 , -7 -7 -8 -0.4425323951 10 , 0.1684301214 10 , 0.9923883091 10 , -7 -0.1748634428 10 ] The largest is 0.3076923077 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 14, :The Pisot Sequence a(n), defined by, a(1) = 7, a(2) = 15, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [7, 15, 32, 68, 145, 309, 658, 1401, 2983, 6351, 13522, 28790, 61297, 130508, 277866, 591608, 1259600, 2681830, 5709918, 12157058] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 116, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 16, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 16, 37, 86, 200, 465, 1081, 2513, 5842, 13581, 31572, 73396, 170625, 396655, 922111, 2143648, 4983377, 11584946, 26931732, 62608681] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 7, a(2) = 16, a(3) = 37, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 7, b(2) = 16, b(3) = 37, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + ------------------- + 1, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- + 1 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- + 1 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [2.324717958, 0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] The largest root is, 2.324717958 and the remaining roots are [0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] whose absolute values are [0.6558656185, 0.6558656185] so the largest absolute value is, 0.6558656185 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6558656185 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, -0.4375000000, -0.1081081081, 0.1162790698, 0.1250000000, 0.03440860215, -0.03052728955, -0.03541583764, -0.01078397809, 0.007952286282, 0.01000886862, 0.003338056570, -0.002051282051, -0.002821091377, -0.001022653455, 0.0005229403335, 0.0007930365292, 0.0003105754658, -0.0001314063277, -0.0002223333854, -0.00009361203497, -5 0.00003242433815, 0.00006216369902, 0.00002803038580, -0.7811902501 10 , -5 -5 -5 -0.00001733278008, -0.8344149430 10 , 0.1821209362 10 , 0.4819146869 10 , -5 -6 -5 0.2470872453 10 , -0.4044670174 10 , -0.1335999089 10 , -6 -7 -6 -6 -0.7281907795 10 , 0.8295882229 10 , 0.3692589368 10 , 0.2136683863 10 , -7 -6 -7 -0.1455389243 10 , -0.1017395131 10 , -0.6244236808 10 , -8 -7 -7 -9 0.1598029464 10 , 0.2793931149 10 , 0.1817950744 10 , 0.2579288246 10 , -8 -8 -9 -0.7645916928 10 , -0.5274100990 10 , -0.2725402894 10 , -8 -8 -9 -9 0.2084664184 10 , 0.1524972140 10 , 0.1330477632 10 , -0.5661368067 10 , -9 -10 -9 -0.4395338064 10 , -0.5328004264 10 , 0.1530906782 10 , -9 -10 -10 0.1262983136 10 , 0.1943354158 10 , -0.4120532415 10 , -10 -11 -0.3618474206 10 , -0.6710036285 10 ] The largest is 0.1250000000 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 117, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) with initial conditions, a(1) = 7, a(2) = 17, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) with initial conditions, b(1) = 7, b(2) = 17, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 2 t - 1 = 0 whose roots are 1/2 1/2 [1 + 2 , 1 - 2 ] In floating-point [2.414213562, -0.414213562] The largest root is, 2.414213562 and the remaining roots are [-0.414213562] whose absolute values are [0.414213562] so the largest absolute value is, 0.414213562 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.414213562 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, -0.1176470588, 0.04878048780, -0.02020202020, 0.008368200837, -0.003466204506, 0.001435750179, -0.0005947071067, 0.0002463357556, -5 -0.0001020356104, 0.00004226453372, -0.00001750654307, 0.7251447570 10 , -5 -5 -6 -0.3003647930 10 , 0.1244151709 10 , -0.5153445117 10 , -6 -7 -7 0.2134626860 10 , -0.8841913961 10 , 0.3662440680 10 , -7 -8 -8 -0.1517032601 10 , 0.6283754779 10 , -0.2602816452 10 , -8 -9 -9 0.1078121875 10 , -0.4465727025 10 , 0.1849764699 10 , -10 -10 -10 -0.7661976257 10 , 0.3173694480 10 , -0.1314587297 10 , -11 -11 -12 0.5445198872 10 , -0.2255475222 10 , 0.9342484267 10 , -12 -12 -13 -0.3869783690 10 , 0.1602916888 10 , -0.6639499143 10 , -13 -13 -14 0.2750170592 10 , -0.1139157958 10 , 0.4718546760 10 , -14 -15 -15 -0.1954486063 10 , 0.8095746346 10 , -0.3353367934 10 , -15 -16 -16 0.1389010478 10 , -0.5753469782 10 , 0.2383165214 10 , -17 -17 -17 -0.9871393532 10 , 0.4088865080 10 , -0.1693663371 10 , -18 -18 -18 0.7015383384 10 , -0.2905866943 10 , 0.1203649498 10 , -19 -19 -20 -0.4985679465 10 , 0.2065136052 10 , -0.8554073609 10 , -20 -20 -21 0.3543213302 10 , -0.1467647004 10 , 0.6079192939 10 , -21 -21 -22 -0.2518084164 10 , 0.1043024612 10 , -0.4320349401 10 ] The largest is 0.2857142857 The smallest is -0.1176470588 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 118, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 18, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 18, 46, 118, 303, 778, 1998, 5131, 13177, 33840, 86905, 223182, 573157, 1471933, 3780093, 9707713, 24930522, 64024444, 164422126, 422254905] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) - a(n - 3) - a(n - 4) with initial conditions, a(1) = 7, a(2) = 18, a(3) = 46, a(4) = 118, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) - b(n - 3) - b(n - 4) with initial conditions, b(1) = 7, b(2) = 18, b(3) = 46, b(4) = 118, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t - 2 t + t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 2 _Z - 2 _Z + _Z + 1 In floating-point [0.775918595324392, 2.56811485944016, -0.672016727382274 + 0.224138976542088 I, -0.672016727382274 - 0.224138976542088 I] The largest root is, 2.56811485944016 and the remaining roots are [0.775918595324392, -0.672016727382274 + 0.224138976542088 I, -0.672016727382274 - 0.224138976542088 I] whose absolute values are [0.775918595324392, 0.708410024411651, 0.708410024411651] so the largest absolute value is, 0.775918595324392 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.775918595324392 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, -0.4444444444, -0.3043478261, 0.04237288136, -0.3630363036, 0.1105398458, -0.2427427427, 0.05632430325, -0.1202853457, 0.004284869976, -0.04557850526, -0.01862605407, -0.01240846749, -0.02077540214, -0.002163174292, -0.01484263080, -0.0008277403899, -0.008402165898, -0.001454007473, -0.004041975546, -0.001762059750, -0.001751897220, -0.001531930920, -0.0007636209847, -0.001077146841, -0.0003977075105, -0.0006541567970, -0.0002629607896, -0.0003593808220, -0.0001928189157, -0.0001872818888, -0.0001378599974, -0.00009808403485, -0.00009178726009, -0.00005460070361, -0.00005683189510, -0.00003299390250, -0.00003326363150, -0.00002108246928, -0.00001886640396, -0.00001364021249, -0.00001066713212, -5 -5 -5 -0.8665815969 10 , -0.6159279725 10 , -0.5342846781 10 , -5 -5 -5 -0.3671304925 10 , -0.3203207719 10 , -0.2246898782 10 , -5 -5 -5 -0.1886061295 10 , -0.1391407509 10 , -0.1104831107 10 , -6 -6 -6 -0.8595171566 10 , -0.6512277239 10 , -0.5252511444 10 , -6 -6 -6 -0.3886094728 10 , -0.3169763540 10 , -0.2346927852 10 , -6 -0.1894776613 10 ] The largest is 0.2857142857 The smallest is -0.4444444444 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 119, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 19, 52, 142, 388, 1060, 2896, 7912, 21616, 59056, 161344, 440800, 1204288, 3290176, 8988928, 24558208, 67094272, 183304960, 500798464, 1368206848] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 7, a(2) = 19, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 7, b(2) = 19, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 2 t - 2 = 0 whose roots are 1/2 1/2 [1 + 3 , 1 - 3 ] In floating-point [2.732050808, -0.732050808] The largest root is, 2.732050808 and the remaining roots are [-0.732050808] whose absolute values are [0.732050808] so the largest absolute value is, 0.732050808 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.732050808 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, 0.3157894737, -0.2307692308, 0.1690140845, -0.1237113402, 0.09056603774, -0.06629834254, 0.04853387260, -0.03552923760, 0.02600921160, -0.01904006347, 0.01393829401, -0.01020353935, 0.007469509230, -0.005468060263, 0.004002897931, -0.002930324663, 0.002145146536, -0.001570356254, 0.001149580564, -0.0008415513802, 0.0006160583675, -0.0004509860255, 0.0003301446841, -0.0002416826826, 0.0001769240030, -0.0001295173593, 0.00009481328745, -0.00006940814365, 0.00005081028761, -0.00003719571208, 0.00002722915106, -0.00001993312203, 0.00001459205808, -5 -5 -5 -0.00001068212790, 0.7819860355 10 , -0.5724535088 10 , 0.4190650534 10 , -5 -5 -5 -0.3067769108 10 , 0.2245762853 10 , -0.1644012510 10 , -5 -6 -6 0.1203500686 10 , -0.8810236488 10 , 0.6449540736 10 , -6 -6 -6 -0.4721391504 10 , 0.3456298463 10 , -0.2530186081 10 , -6 -6 -7 0.1852224764 10 , -0.1355922634 10 , 0.9926042595 10 , -7 -7 -7 -0.7266367497 10 , 0.5319350195 10 , -0.3894034606 10 , -7 -7 -7 0.2850631178 10 , -0.2086806856 10 , 0.1527648644 10 , -7 -8 -0.1118316424 10 , 0.8186644410 10 ] The largest is 0.3157894737 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 120, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 20, 57, 162, 460, 1306, 3708, 10528, 29892, 84872, 240976, 684200, 1942640, 5515712, 15660688, 44465184, 126249408, 358458272, 1017765824, 2889729024] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 4 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 7, a(2) = 20, a(3) = 57, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 4 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 7, b(2) = 20, b(3) = 57, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 4 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (19 + 3 33 ) 4 (19 + 3 33 ) [----------------- + ------------------- + 4/3, - ----------------- 3 1/2 1/3 6 3 (19 + 3 33 ) 2 - ------------------- + 4/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(19 + 3 33 ) 4 | (19 + 3 33 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (19 + 3 33 ) / 2 - ------------------- + 4/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 |(19 + 3 33 ) 4 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (19 + 3 33 ) / In floating-point [2.839286755, 0.5803566220 + 0.6062907300 I, 0.5803566220 - 0.6062907300 I] The largest root is, 2.839286755 and the remaining roots are [0.5803566220 + 0.6062907300 I, 0.5803566220 - 0.6062907300 I] whose absolute values are [0.8392867555, 0.8392867555] so the largest absolute value is, 0.8392867555 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8392867555 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, 0.4500000000, 0.4210526316, 0.1728395062, -0.09565217391, -0.2327718224, -0.2028047465, -0.07142857143, 0.05994915027, 0.1198981996, 0.09693911427, 0.02806197018, -0.03571222666, -0.06121856979, -0.04590143166, -0.01015590085, 0.02054498347, 0.03100067391, 0.02151096007, 0.003131111577, -0.01151804616, -0.01557471079, -0.009964435384, -0.0005949906867, 0.006328357208, 0.007764520812, 0.004554673043, -0.0001826766622, -0.003420357195, -0.003841376044, -0.002049428724, 0.0003270748936, 0.001823262381, 0.001885892500, 0.0009046702666, -0.0002783641740, -0.0009603527617, -0.0009186138176, -0.0003897725716, 0.0001946594606, 0.0005005004934, 0.0004438189883, 0.0001625929006, -0.0001239033641, -0.0002583470820, -0.0002125890704, -0.00006477468199, 0.00007456338981, 0.0001321741464, 0.0001008936622, 0.00002400484294, -5 -0.00004320698427, -0.00006705998446, -0.00004740231490, -0.7783290300 10 , 0.00002435612949, 0.00003375304935, 0.00002202109883] The largest is 0.4500000000 The smallest is -0.2327718224 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 121, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 22, 69, 216, 676, 2116, 6623, 20730, 64885, 203090, 635672, 1989654, 6227619, 19492454, 61011401, 190965748, 597722988, 1870873568, 5855836195, 18328773322] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 3 a(n - 2) + 2 a(n - 3) - a(n - 4) with initial conditions, a(1) = 7, a(2) = 22, a(3) = 69, a(4) = 216, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 3 b(n - 2) + 2 b(n - 3) - b(n - 4) with initial conditions, b(1) = 7, b(2) = 22, b(3) = 69, b(4) = 216, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t - 3 t - 2 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 2 _Z - 3 _Z - 2 _Z + 1 In floating-point [0.319488712753034, 3.13000103003015, -0.724744871391589 + 0.689017322998187 I, -0.724744871391589 - 0.689017322998187 I] The largest root is, 3.13000103003015 and the remaining roots are [0.319488712753034, -0.724744871391589 + 0.689017322998187 I, -0.724744871391589 - 0.689017322998187 I] whose absolute values are [0.319488712753034, 1.00000000000000, 1.00000000000000] so the largest absolute value is, 1.00000000000000 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 1.00000000000000 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, 0.4090909091, 0.1739130435, -0.3703703704, 0.4556213018, -0.2603969754, -0.06869998490, 0.3630004824, -0.4564999615, 0.2990004432, 0.02320064436, -0.3325980296, 0.4589068471, -0.3325795203, 0.02316478522, 0.2990027301, -0.4565660707, 0.3627851398, -0.06928725762, -0.2623539672, 0.4495666431, -0.3892882704, 0.1147027119, 0.2230278661, -0.4379793161, 0.4118186602, -0.1589476077, -0.1814257332, 0.4219223470, -0.4301463811, 0.2015704202, 0.1379721245, -0.4015595995, 0.4440843961, -0.2421361776, -0.09311049034, 0.3770988783, -0.4534904657, 0.2802309002, 0.04729865021, -0.3487898086, 0.4582685997, -0.3154658261, -0.001004120518, 0.3169212885, -0.4583700364, 0.3474813777, -0.04530065636, -0.2818185410, 0.4537937408, -0.3759508317, 0.09114313351, 0.2438397946, -0.4445864146, 0.4005836532, -0.1360554818, -0.2033726278, 0.4308420198] The largest is 0.4589068471 The smallest is -0.4583700364 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 122, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 23, 76, 251, 829, 2738, 9043, 29867, 98644, 325799, 1076041, 3553922, 11737807, 38767343, 128039836, 422886851, 1396700389, 4612988018, 15235664443, 50319981347] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 2) with initial conditions, a(1) = 7, a(2) = 23, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 2) with initial conditions, b(1) = 7, b(2) = 23, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 1 = 0 whose roots are 1/2 1/2 13 13 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.302775638, -0.302775638] The largest root is, 3.302775638 and the remaining roots are [-0.302775638] whose absolute values are [0.302775638] so the largest absolute value is, 0.302775638 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.302775638 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, 0.1304347826, -0.03947368421, 0.01195219124, -0.003618817853, 0.001095690285, -0.0003317483136, 0.0001004453075, -0.00003041239204, -5 -5 -6 0.9208131394 10 , -0.2787997855 10 , 0.8441378286 10 , -6 -7 -7 -0.2555843694 10 , 0.7738472043 10 , -0.2343020808 10 , -8 -8 -9 0.7094096194 10 , -0.2147919499 10 , 0.6503376962 10 , -9 -10 -10 -0.1969064107 10 , 0.5961846407 10 , -0.1805101848 10 , -11 -11 -12 0.5465408632 10 , -0.1654792584 10 , 0.5010308799 10 , -12 -13 -13 -0.1516999442 10 , 0.4593104735 10 , -0.1390680215 10 , -14 -14 -15 0.4210640890 10 , -0.1274879481 10 , 0.3860024479 10 , -15 -16 -16 -0.1168721373 10 , 0.3538603591 10 , -0.1071402959 10 , -17 -18 -18 0.3243947142 10 , -0.9821881645 10 , 0.2973826479 10 , -19 -19 -20 -0.9004022087 10 , 0.2726198529 10 , -0.8254264983 10 , -20 -21 -21 0.2499190344 10 , -0.7566939503 10 , 0.2291084934 10 , -22 -22 -23 -0.6936847019 10 , 0.2100308280 10 , -0.6359221789 10 , -23 -24 -24 0.1925417433 10 , -0.5829694911 10 , 0.1765089594 10 , -25 -25 -26 -0.5344261276 10 , 0.1618112116 10 , -0.4899249279 10 , -26 -27 -27 0.1483373325 10 , -0.4491293044 10 , 0.1359854116 10 , -28 -28 -29 -0.4117306971 10 , 0.1246620244 10 , -0.3774462393 10 , -29 0.1142815258 10 ] The largest is 0.1304347826 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 123, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 24, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 24, 82, 280, 956, 3264, 11144, 38048, 129904, 443520, 1514272, 5170048, 17651648, 60266496, 205762688, 702517760, 2398545664, 8189147136, 27959497216, 95459694592] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 2 a(n - 2) with initial conditions, a(1) = 7, a(2) = 24, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 2 b(n - 2) with initial conditions, b(1) = 7, b(2) = 24, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t + 2 = 0 whose roots are 1/2 1/2 [2 + 2 , 2 - 2 ] In floating-point [3.414213562, 0.585786438] The largest root is, 3.414213562 and the remaining roots are [0.585786438] whose absolute values are [0.585786438] so the largest absolute value is, 0.585786438 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.585786438 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, 0.1666666667, 0.09756097561, 0.05714285714, 0.03347280335, 0.01960784314, 0.01148600144, 0.006728343146, 0.003941372090, 0.002308802309, 0.001352465079, 0.0007922557005, 0.0004640926445, 0.0002718591769, 0.0001592514188, 0.00009328732131, 0.00005464644762, -5 0.00003201114788, 0.00001875169628, 0.00001098448936, 0.6434564894 10 , -5 -5 -5 -6 0.3769280847 10 , 0.2207993600 10 , 0.1293412705 10 , 0.7576636209 10 , -6 -6 -6 -7 0.4438290734 10 , 0.2599890518 10 , 0.1522980605 10 , 0.8921413831 10 , -7 -7 -7 -7 0.5226043227 10 , 0.3061345245 10 , 0.1793294525 10 , 0.1050487612 10 , -8 -8 -8 -8 0.6153613958 10 , 0.3604703599 10 , 0.2111586480 10 , 0.1236938722 10 , -9 -9 -9 -9 0.7245819274 10 , 0.4244502660 10 , 0.2486372093 10 , 0.1456483051 10 , -10 -10 -10 0.8531880178 10 , 0.4997859696 10 , 0.2927678427 10 , -10 -10 -11 0.1714994316 10 , 0.1004620411 10 , 0.5884930118 10 , -11 -11 -11 0.3447312249 10 , 0.2019388762 10 , 0.1182930549 10 , -12 -12 -12 0.6929446723 10 , 0.4059175910 10 , 0.2377810196 10 , -12 -13 -13 0.1392888964 10 , 0.8159354643 10 , 0.4779639290 10 , -13 -13 0.2799847873 10 , 0.1640112911 10 ] The largest is 0.2857142857 The smallest is -13 0.1640112911 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 124, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 25, 89, 317, 1129, 4021, 14321, 51005, 181657, 646981, 2304257, 8206733, 29228713, 104099605, 370756241, 1320467933, 4702916281, 16749684709, 59654886689, 212464029485] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 7, a(2) = 25, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 7, b(2) = 25, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 2 = 0 whose roots are 1/2 1/2 17 17 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.561552813, -0.561552813] The largest root is, 3.561552813 and the remaining roots are [-0.561552813] whose absolute values are [0.561552813] so the largest absolute value is, 0.561552813 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.561552813 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, -0.1600000000, 0.08988764045, -0.05047318612, 0.02834366696, -0.01591643870, 0.008937923329, -0.005019115773, 0.002818498599, -0.001582735814, 0.0008887897487, -0.0004991023834, 0.0002802723473, -0.0001573877250, 0.00008838151965, -0.00004963089096, 0.00002787036642, -5 -5 -5 -0.00001565068266, 0.8788684869 10 , -0.4935310709 10 , 0.2771437611 10 , -5 -6 -6 -0.1556308586 10 , 0.8739494640 10 , -0.4907687797 10 , -6 -6 -7 0.2755925887 10 , -0.1547597934 10 , 0.8690579728 10 , -7 -7 -7 -0.4880219491 10 , 0.2740500982 10 , -0.1538936035 10 , -8 -8 -8 0.8641938593 10 , -0.4852904925 10 , 0.2725162411 10 , -8 -9 -9 -0.1530322617 10 , 0.8593569702 10 , -0.4825743238 10 , -9 -9 -10 0.2709909689 10 , -0.1521757408 10 , 0.8545471532 10 , -10 -10 -10 -0.4798733575 10 , 0.2694742337 10 , -0.1513240139 10 , -11 -11 -11 0.8497642566 10 , -0.4771875085 10 , 0.2679659877 10 , -11 -12 -12 -0.1504770541 10 , 0.8450081300 10 , -0.4745166922 10 , -12 -12 -13 0.2664661832 10 , -0.1496348347 10 , 0.8402786233 10 , -13 -13 -13 -0.4718608245 10 , 0.2649747732 10 , -0.1487973292 10 , -14 -14 -14 0.8355755877 10 , -0.4692198216 10 , 0.2634917106 10 , -14 -0.1479645113 10 ] The largest is 0.2857142857 The smallest is -0.1600000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 125, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - a(n - 2) with initial conditions, a(1) = 7, a(2) = 26, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - b(n - 2) with initial conditions, b(1) = 7, b(2) = 26, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t + 1 = 0 whose roots are 1/2 1/2 [2 + 3 , 2 - 3 ] In floating-point [3.732050808, 0.267949192] The largest root is, 3.732050808 and the remaining roots are [0.267949192] whose absolute values are [0.267949192] so the largest absolute value is, 0.267949192 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.267949192 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, -0.1153846154, -0.03092783505, -0.008287292818, -0.002220577350, -0.0005950019833, -0.0001594303024, -0.00004271922080, -0.00001144658072, -5 -6 -6 -0.3067102059 10 , -0.8218275198 10 , -0.2202080202 10 , -7 -7 -8 -0.5900456119 10 , -0.1581022452 10 , -0.4236336892 10 , -8 -9 -10 -0.1135123049 10 , -0.3041553043 10 , -0.8149816817 10 , -10 -11 -11 -0.2183736835 10 , -0.5851305213 10 , -0.1567852507 10 , -12 -12 -13 -0.4201048130 10 , -0.1125667454 10 , -0.3016216852 10 , -14 -14 -15 -0.8081928696 10 , -0.2165546267 10 , -0.5802563735 10 , -15 -16 -16 -0.1554792267 10 , -0.4166053323 10 , -0.1116290624 10 , -17 -18 -18 -0.2991091711 10 , -0.8014606085 10 , -0.2147507228 10 , -19 -19 -20 -0.5754228275 10 , -0.1541840819 10 , -0.4131350024 10 , -20 -21 -22 -0.1106991903 10 , -0.2966175863 10 , -0.7947844272 10 , -22 -23 -23 -0.2129618454 10 , -0.5706295450 10 , -0.1528997258 10 , -24 -24 -25 -0.4096935804 10 , -0.1097770640 10 , -0.2941467565 10 , -26 -26 -27 -0.7881638586 10 , -0.2111878694 10 , -0.5658761906 10 , -27 -28 -28 -0.1516260683 10 , -0.4062808255 10 , -0.1088626191 10 , -29 -30 -30 -0.2916965087 10 , -0.7815984394 10 , -0.2094286707 10 , -31 -31 -32 -0.5611624317 10 , -0.1503630204 10 , -0.4028964989 10 , -32 -0.1079557915 10 ] The largest is -32 -0.1079557915 10 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 126, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 27, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 27, 104, 401, 1546, 5960, 22976, 88573, 341451, 1316302, 5074377, 19561850, 75411420, 290712906, 1120705507, 4320347695, 16655048172, 64205626305, 247514291562, 954173770324] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 5 a(n - 2) + 2 a(n - 3) + a(n - 4) with initial conditions, a(1) = 7, a(2) = 27, a(3) = 104, a(4) = 401, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 5 b(n - 2) + 2 b(n - 3) + b(n - 4) with initial conditions, b(1) = 7, b(2) = 27, b(3) = 104, b(4) = 401, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t + 5 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z + 5 _Z - 2 _Z - 1 In floating-point [3.85502495351856, 0.706785225482515 + 0.682807291071660 I, -0.268595404483592, 0.706785225482515 - 0.682807291071660 I] The largest root is, 3.85502495351856 and the remaining roots are [0.706785225482515 + 0.682807291071660 I, -0.268595404483592, 0.706785225482515 - 0.682807291071660 I] whose absolute values are [0.982736562717083, 0.268595404483592, 0.982736562717083] so the largest absolute value is, 0.982736562717083 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.982736562717083 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, -0.4074074074, 0.1634615385, 0.3890274314, 0.4553686934, 0.2510067114, -0.08038823120, -0.3572759193, -0.4270685984, -0.2587331783, 0.04673677182, 0.3159363762, 0.4014630145, 0.2623735563, -0.01683776771, -0.2771942157, -0.3755721133, -0.2631914670, -0.009322967629, 0.2410040545, 0.3496800632, 0.2615426413, 0.03199803214, -0.2073588652, -0.3240191410, -0.2577626732, -0.05143735917, 0.1762294227, 0.2987894222, 0.2521626057, 0.06788740387, -0.1475677420, -0.2741610958, -0.2450293557, -0.08158937960, 0.1213099470, 0.2502768255, 0.2366262779, 0.09277777639, -0.09737890977, -0.2272540494, -0.2271938672, -0.1016791323, 0.07568666584, 0.2051872071, 0.2169505744, 0.1085110360, -0.05613661200, -0.1841498842, -0.2060937144, -0.1134813391, 0.03862549626, 0.1641968637, 0.1948004447, 0.1167875582, -0.02304520842, -0.1453660803, -0.1832287982] The largest is 0.4553686934 The smallest is -0.4270685984 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 127, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 29, 120, 497, 2058, 8522, 35289, 146129, 605109, 2505710, 10375953, 42966026, 177919020, 736749023, 3050821227, 12633216833, 52313182476, 216624878441, 897027015726, 3714520108370] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 4 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 7, a(2) = 29, a(3) = 120, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 4 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 7, b(2) = 29, b(3) = 120, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t - 4 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (972 + 12 2445 ) 14 (972 + 12 2445 ) [--------------------- + --------------------- + 1, - --------------------- 6 1/2 1/3 12 (972 + 12 2445 ) 7 - --------------------- + 1 1/2 1/3 (972 + 12 2445 ) / 1/2 1/3 \ 1/2 |(972 + 12 2445 ) 14 | + 1/2 I 3 |--------------------- - ---------------------|, | 6 1/2 1/3| \ (972 + 12 2445 ) / 1/2 1/3 (972 + 12 2445 ) 7 - --------------------- - --------------------- + 1 12 1/2 1/3 (972 + 12 2445 ) / 1/2 1/3 \ 1/2 |(972 + 12 2445 ) 14 | - 1/2 I 3 |--------------------- - ---------------------|] | 6 1/2 1/3| \ (972 + 12 2445 ) / In floating-point [4.140923344, -0.570461672 + 0.6317037155 I, -0.570461672 - 0.6317037155 I] The largest root is, 4.140923344 and the remaining roots are [-0.570461672 + 0.6317037155 I, -0.570461672 - 0.6317037155 I] whose absolute values are [0.8511616200, 0.8511616200] so the largest absolute value is, 0.8511616200 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8511616200 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, -0.4482758621, 0.4083333333, -0.1408450704, -0.1350826045, 0.2561605257, -0.1943948539, 0.03620773426, 0.09952421795, -0.1397811399, 0.08737664868, 0.001578037494, -0.06510271358, 0.07313395514, -0.03627487642, -0.01159694945, 0.03951151139, -0.03667789291, 0.01322151845, 0.01148751785, -0.02268505140, 0.01755947253, -0.003599234487, -0.008614967560, 0.01243657695, -0.007947842837, 0.00005787662772, 0.005691989399, -0.006536053804, 0.003333426065, 0.0009320311718, -0.003478363640, 0.003293311961, -0.001237425161, -0.0009741185594, 0.002007879560, -0.001585111042, 0.0003538294332, 0.0007446828089, -0.001105966968, 0.0007223186315, -0.00002286355020, -0.0004972170281, 0.0005838506094, -0.0003059069348, -0.00007396945087, 0.0003060157367, -0.0002955513978, 0.0001155004007, 0.00008234282117, -0.0001776241268, 0.0001430001064, -0.00003446772449, -0.00006427512819, -5 0.00009830403677, -0.00006559157592, 0.3616034727 10 , 0.00004339391079] The largest is 0.4083333333 The smallest is -0.4482758621 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 128, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 30, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 30, 129, 555, 2388, 10275, 44211, 190230, 818517, 3521895, 15153924, 65203935, 280557903, 1207177710, 5194214841, 22349541075, 96165060852, 413776681035, 1780388222619, 7660611069990] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 3 a(n - 2) with initial conditions, a(1) = 7, a(2) = 30, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 3 b(n - 2) with initial conditions, b(1) = 7, b(2) = 30, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 5 t + 3 = 0 whose roots are 1/2 1/2 13 13 [5/2 + -----, 5/2 - -----] 2 2 In floating-point [4.302775638, 0.697224362] The largest root is, 4.302775638 and the remaining roots are [0.697224362] whose absolute values are [0.697224362] so the largest absolute value is, 0.697224362 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.697224362 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, -0.3000000000, -0.2093023256, -0.1459459459, -0.1017587940, -0.07094890511, -0.04946732714, -0.03448982810, -0.02404714868, -0.01676625794, -0.01168984350, -0.008150443681, -0.005682687898, -0.003962108445, -0.002762478534, -0.001926067334, -0.001342901069, -0.0009363033413, -0.0006528135000, -0.0004551574762, -0.0003173468811, -0.0002212619768, -0.0001542692407, -0.0001075602729, -0.00007499364270, -0.00005228739471, -0.00003645604543, -0.00002541804303, -0.00001772207884, -5 -5 -0.00001235626512, -0.8615089066 10 , -0.6006649980 10 , -5 -5 -5 -0.4187982702 10 , -0.2919963568 10 , -0.2035869737 10 , -5 -6 -6 -0.1419457979 10 , -0.9896806841 10 , -0.6900294838 10 , -6 -6 -6 -0.4811053668 10 , -0.3354383825 10 , -0.2338758124 10 , -6 -6 -7 -0.1630639141 10 , -0.1136921335 10 , -0.7926892529 10 , -7 -7 -7 -0.5526822589 10 , -0.3853435355 10 , -0.2686709008 10 , -7 -7 -8 -0.1873238975 10 , -0.1306067849 10 , -0.9106223234 10 , -8 -8 -8 -0.6349080687 10 , -0.4426733733 10 , -0.3086426604 10 , -8 -8 -8 -0.2151931821 10 , -0.1500379291 10 , -0.1046100994 10 , -9 -9 -0.7293670987 10 , -0.5085325103 10 ] The largest is -9 -0.5085325103 10 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 129, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 31, 137, 605, 2672, 11801, 52120, 230192, 1016661, 4490163, 19831157, 87585860, 386829819, 1708464230, 7545566246, 33325585033, 147185059621, 650054357761, 2871016046956, 12680067510461] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 2 a(n - 2) - 3 a(n - 3) + 2 a(n - 4) with initial conditions, a(1) = 7, a(2) = 31, a(3) = 137, a(4) = 605, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 2 b(n - 2) - 3 b(n - 3) + 2 b(n - 4) with initial conditions, b(1) = 7, b(2) = 31, b(3) = 137, b(4) = 605, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t + 2 t + 3 t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z + 2 _Z + 3 _Z - 2 In floating-point [4.41657841791064, 0.697940317124258 + 0.265043397884637 I, -0.812459052159161, 0.697940317124258 - 0.265043397884637 I] The largest root is, 4.41657841791064 and the remaining roots are [0.697940317124258 + 0.265043397884637 I, -0.812459052159161, 0.697940317124258 - 0.265043397884637 I] whose absolute values are [0.746571288645461, 0.812459052159161, 0.746571288645461] so the largest absolute value is, 0.812459052159161 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.812459052159161 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, 0.4516129032, -0.2846715328, -0.03471074380, -0.3888473054, -0.1179561054, -0.2773599386, -0.05375947036, -0.1379102769, 0.01413512160, -0.04694547071, 0.04321429281, -0.008263514970, 0.04136049486, -0.0002043156139, 0.02747656274, -0.002817069586, 0.01429546315, -0.005726864539, 0.006179085248, -0.006171373297, 0.002556482930, -0.004865823576, 0.001430206645, -0.003129515007, 0.001202448266, -0.001750995743, 0.001089083064, -0.0009189680067, 0.0008848775986, -0.0005069166710, 0.0006307315961, -0.0003250774869, 0.0004036545839, -0.0002376002370, 0.0002413852999, -0.0001789917519, 0.0001423805193, -0.0001294702733, 0.00008763345030, -0.00008801726356, 0.00005781864023, -0.00005671316934, 0.00004011556411, -0.00003548628858, 0.00002811421738, -0.00002222936697, 0.00001931472436, -0.00001428287355, 0.00001287271917, -5 -5 -5 -0.9473564071 10 , 0.8364810691 10 , -0.6412723014 10 , -5 -5 -5 0.5372894098 10 , -0.4351643697 10 , 0.3463783742 10 , -5 -5 -0.2921922217 10 , 0.2263540716 10 ] The largest is 0.4516129032 The smallest is -0.3888473054 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 130, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 32, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 32, 146, 666, 3038, 13858, 63214, 288354, 1315342, 6000002, 27369326, 124846626, 569494478, 2597779138, 11849906734, 54053975394, 246570063502, 1124742366722, 5130571706606, 23403373799586] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 2 a(n - 2) with initial conditions, a(1) = 7, a(2) = 32, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 2 b(n - 2) with initial conditions, b(1) = 7, b(2) = 32, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 5 t + 2 = 0 whose roots are 1/2 1/2 17 17 [5/2 + -----, 5/2 - -----] 2 2 In floating-point [4.561552813, 0.438447187] The largest root is, 4.561552813 and the remaining roots are [0.438447187] whose absolute values are [0.438447187] so the largest absolute value is, 0.438447187 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.438447187 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, 0.1250000000, 0.05479452055, 0.02402402402, 0.01053324556, 0.004618271035, 0.002024867909, 0.0008877976376, 0.0003892523769, 0.0001706666098, 0.00007482829500, 0.00003280825547, 0.00001438468733, -5 -5 -5 -6 0.6306925697 10 , 0.2765253832 10 , 0.1212417764 10 , 0.5315811585 10 , -6 -6 -7 -7 0.2330702637 10 , 0.1021890015 10 , 0.4480448028 10 , 0.1964439835 10 , -8 -8 -8 -9 0.8613031203 10 , 0.3776359304 10 , 0.1655734115 10 , 0.7259519653 10 , -9 -9 -10 0.3182915972 10 , 0.1395540555 10 , 0.6118708310 10 , -10 -10 -11 0.2682730448 10 , 0.1176235619 10 , 0.5157171985 10 , -11 -12 -12 0.2261147551 10 , 0.9913937835 10 , 0.4346738158 10 , -12 -13 -13 0.1905815119 10 , 0.8355992781 10 , 0.3663661531 10 , -13 -14 -14 0.1606322093 10 , 0.7042874034 10 , 0.3087928310 10 , -14 -15 -15 0.1353893482 10 , 0.5936107889 10 , 0.2602669807 10 , -15 -16 -16 0.1141133256 10 , 0.5003266663 10 , 0.2193668195 10 , -17 -17 -17 0.9618076497 10 , 0.4217018586 10 , 0.1848939938 10 , -18 -18 -18 0.8106625149 10 , 0.3554326994 10 , 0.1558384673 10 , -19 -19 -19 0.6832693764 10 , 0.2995775362 10 , 0.1313489281 10 , -20 -20 -20 0.5758956806 10 , 0.2524998413 10 , 0.1107078452 10 ] The largest is 0.2857142857 The smallest is -20 0.1107078452 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 131, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 33, 156, 737, 3482, 16451, 77724, 367213, 1734926, 8196791, 38726368, 182965697, 864435474, 4084091723, 19295604708, 91163564949, 430709257366, 2034918933727, 9614130636904, 45422697863529] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 3 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 7, a(2) = 33, a(3) = 156, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 3 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 7, b(2) = 33, b(3) = 156, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t - 3 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (145 + 30 6 ) 25 (145 + 30 6 ) [------------------ + -------------------- + 4/3, - ------------------ 3 1/2 1/3 6 3 (145 + 30 6 ) 25 - -------------------- + 4/3 1/2 1/3 6 (145 + 30 6 ) / 1/2 1/3 \ 1/2 |(145 + 30 6 ) 25 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (145 + 30 6 ) / 1/2 1/3 (145 + 30 6 ) 25 - ------------------ - -------------------- + 4/3 6 1/2 1/3 6 (145 + 30 6 ) / 1/2 1/3 \ 1/2 |(145 + 30 6 ) 25 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (145 + 30 6 ) / In floating-point [4.724576727, -0.362288365 + 0.5404308335 I, -0.362288365 - 0.5404308335 I] The largest root is, 4.724576727 and the remaining roots are [-0.362288365 + 0.5404308335 I, -0.362288365 - 0.5404308335 I] whose absolute values are [0.6506291918, 0.6506291918] so the largest absolute value is, 0.6506291918 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6506291918 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, 0.4545454545, -0.1474358974, -0.08548168250, 0.1243538196, -0.05391769497, -0.01357367094, 0.03265951913, -0.01791834349, -0.0008421588400, 0.008195372207, -0.005581674690, 0.0005751001838, 0.001946121081, -0.001653564503, 0.0003743056014, 0.0004287710604, -0.0004691279594, 0.0001584125461, 0.00008380842698, -0.0001277845726, 0.00005711208286, 0.00001271146764, -0.00003338702601, 0.00001881046461, -6 -5 -5 0.5037156950 10 , -0.8327795412 10 , 0.5820894654 10 , -6 -5 -5 -0.6923762309 10 , -0.1962411786 10 , 0.1715013470 10 , -6 -6 -6 -0.4119339391 10 , -0.4275189179 10 , 0.4841494518 10 , -6 -7 -6 -0.1698268248 10 , -0.8189677958 10 , 0.1312313109 10 , -7 -7 -7 -0.6041874465 10 , -0.1177460503 10 , 0.3410796776 10 , -7 -9 -8 -0.1972943335 10 , -0.1430401964 10 , 0.8455474668 10 , -8 -9 -8 -0.6066088626 10 , 0.8159891058 10 , 0.1976639880 10 , -8 -9 -9 -0.1777650414 10 , 0.4512961974 10 , 0.4255133088 10 , -9 -9 -10 -0.4993590001 10 , 0.1816963207 10 , 0.7973489992 10 , -9 -10 -10 -0.1346894386 10 , 0.6383958680 10 , 0.1075983130 10 , -10 -10 -12 -0.3482079157 10 , 0.2067550123 10 , -0.2407071756 10 ] The largest is 0.4545454545 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 132, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 34, 165, 801, 3888, 18872, 91603, 444633, 2158210, 10475764, 50848449, 246813957, 1198013520, 5815053620, 28225765435, 137005414989, 665012390170, 3227912408536, 15668006598369, 76051143803457] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - a(n - 2) + a(n - 3) + 2 a(n - 4) with initial conditions, a(1) = 7, a(2) = 34, a(3) = 165, a(4) = 801, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - b(n - 2) + b(n - 3) + 2 b(n - 4) with initial conditions, b(1) = 7, b(2) = 34, b(3) = 165, b(4) = 801, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t + t - t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z + _Z - _Z - 2 In floating-point [4.85391318455111, 0.362722251644733 + 0.761335470504124 I, -0.579357687840571, 0.362722251644733 - 0.761335470504124 I] The largest root is, 4.85391318455111 and the remaining roots are [0.362722251644733 + 0.761335470504124 I, -0.579357687840571, 0.362722251644733 - 0.761335470504124 I] whose absolute values are [0.843326230165978, 0.579357687840571, 0.843326230165978] so the largest absolute value is, 0.843326230165978 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.843326230165978 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, -0.2647058824, 0.4909090909, 0.08988764045, -0.02057613169, -0.2314010174, -0.06485595450, 0.06631986380, 0.1239017519, 0.02553064387, -0.05964058019, -0.06719206726, -0.002985609044, 0.04368472960, 0.03493602940, -0.006374326183, -0.02909414880, -0.01679092922, 0.008637235295, 0.01813430453, 0.007055060547, -0.007803624949, -0.01066441017, -0.002194756271, 0.005997124957, 0.005908720994, 0.00002290340698, -0.004186591543, -0.003052890215, 0.0007624858609, 0.002724534790, 0.001434114789, -0.0008972554143, -0.001670885348, -0.0005739869573, 0.0007719247258, 0.0009682144095, 0.0001533896675, -0.0005773152607, -0.0005279021099, 0.00002762319760, 0.0003954821723, 0.0002672550324, -0.00008738803257, -0.0002534666278, -0.0001217257294, 0.00009196001312, 0.0001532831021, 0.00004579651224, -0.00007579198649, -5 -0.00008755331639, -0.9611879135 10 , 0.00005529495871, 0.00004694938330, -5 -5 -0.5266554136 10 , -0.00003721095354, -0.00002324891286, 0.9598601688 10 ] The largest is 0.4909090909 The smallest is -0.2647058824 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 133, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 36, 185, 951, 4889, 25134, 129212, 664269, 3414956, 17556027, 90254189, 463989867, 2385336338, 12262831260, 63042275471, 324095505540, 1666150149665, 8565550196703, 44034836948513, 226379721156558] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 6 a(n - 1) - 5 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 7, a(2) = 36, a(3) = 185, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 6 b(n - 1) - 5 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 7, b(2) = 36, b(3) = 185, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 6 t + 5 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (972 + 12 2445 ) 14 (972 + 12 2445 ) [--------------------- + --------------------- + 2, - --------------------- 6 1/2 1/3 12 (972 + 12 2445 ) 7 - --------------------- + 2 1/2 1/3 (972 + 12 2445 ) / 1/2 1/3 \ 1/2 |(972 + 12 2445 ) 14 | + 1/2 I 3 |--------------------- - ---------------------|, | 6 1/2 1/3| \ (972 + 12 2445 ) / 1/2 1/3 (972 + 12 2445 ) 7 - --------------------- - --------------------- + 2 12 1/2 1/3 (972 + 12 2445 ) / 1/2 1/3 \ 1/2 |(972 + 12 2445 ) 14 | - 1/2 I 3 |--------------------- - ---------------------|] | 6 1/2 1/3| \ (972 + 12 2445 ) / In floating-point [5.140923344, 0.429538328 + 0.6317037155 I, 0.429538328 - 0.6317037155 I] The largest root is, 5.140923344 and the remaining roots are [0.429538328 + 0.6317037155 I, 0.429538328 - 0.6317037155 I] whose absolute values are [0.7639062504, 0.7639062504] so the largest absolute value is, 0.7639062504 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7639062504 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, -0.3055555556, -0.3459459459, -0.1188222923, 0.09981591327, 0.1550887244, 0.07498529548, -0.02608431223, -0.06616629907, -0.04162035066, 0.002856454674, 0.02674158399, 0.02130617858, 0.002698515563, -0.01011504757, -0.01026432745, -0.002915180206, 0.003485413337, 0.004695398691, 0.001999784844, -0.001022044381, -0.002044994434, -0.001160390164, 0.0001964980404, 0.0008459557613, 0.0006120738742, 0.00003215855943, -0.0003295507302, -0.0003018755558, -0.00006702400563, 0.0001185815547, 0.0001409826891, 0.00005191634385, -0.00003767071802, -5 -0.00006265796021, -0.00003184513958, 0.9206809482 10 , 0.00002649267418, -6 -5 0.00001738657890, -0.5234690199 10 , -0.00001059568610, -0.8797034809 10 , -5 -5 -5 -5 -0.1374185401 10 , 0.3953003332 10 , 0.4197842571 10 , 0.1299482564 10 , -5 -5 -6 -0.1333307477 10 , -0.1903729966 10 , -0.8573947190 10 , -6 -6 -6 -7 0.3743590845 10 , 0.8219382047 10 , 0.4876496489 10 , -0.6071587668 10 , -6 -6 -7 -0.3367288904 10 , -0.2538450123 10 , -0.2157325163 10 , -6 -6 0.1295988803 10 , 0.1239245034 10 ] The largest is 0.1550887244 The smallest is -0.3459459459 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 134, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 37, 196, 1038, 5497, 29111, 154166, 816432, 4323659, 22897225, 121259080, 642163602, 3400768765, 18009784667, 95376182906, 505093005492, 2674870564367, 14165574376045, 75018021461068, 397280293374486] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) + a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 7, a(2) = 37, a(3) = 196, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) + b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 7, b(2) = 37, b(3) = 196, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 5 t - t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (188 + 12 93 ) 28 (188 + 12 93 ) [------------------- + --------------------- + 5/3, - ------------------- 3 1/2 1/3 6 3 (188 + 12 93 ) 14 - --------------------- + 5/3 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 28 | + 1/2 I 3 |------------------- - ---------------------|, | 3 1/2 1/3| \ 3 (188 + 12 93 ) / 1/2 1/3 (188 + 12 93 ) 14 - ------------------- - --------------------- + 5/3 6 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 28 | - 1/2 I 3 |------------------- - ---------------------|] | 3 1/2 1/3| \ 3 (188 + 12 93 ) / In floating-point [5.295798070, -0.147899035 + 0.7379788150 I, -0.147899035 - 0.7379788150 I] The largest root is, 5.295798070 and the remaining roots are [-0.147899035 + 0.7379788150 I, -0.147899035 - 0.7379788150 I] whose absolute values are [0.7526532109, 0.7526532109] so the largest absolute value is, 0.7526532109 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7526532109 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, 0.2702702703, 0.1632653061, -0.2013487476, -0.03292705112, 0.1238019992, -0.01796764527, -0.06481740059, 0.02935129713, 0.02803614849, -0.02492016268, -0.008510773552, 0.01663441501, -0.00009918652738, -0.009393838280, 0.002834867110, 0.004482937690, -0.002931959280, -0.001672257378, 0.002155566902, 0.0003096992921, -0.001312708771, 0.0002128561444, 0.0006806698275, -0.0003219210303, -0.0002903668907, -5 0.0002682539986, 0.00008514001160, -0.0001771466155, 0.4168929737 10 , 0.00009911806793, -0.00003168057724, -0.00004677802904, 0.00003178348137, -5 0.00001709764611, -0.00002306237518, -0.2863785691 10 , 0.00001391163470, -5 -5 -5 -0.2492737761 10 , -0.7143411180 10 , 0.3525110427 10 , -5 -5 -6 0.3003927675 10 , -0.2885484736 10 , -0.8481647246 10 , -5 -7 -5 0.1885474666 10 , -0.7724560173 10 , -0.1045247516 10 , -6 -6 -6 0.3529408174 10 , 0.4877197661 10 , -0.3442029003 10 , -6 -6 -7 -0.1744722829 10 , 0.2465949836 10 , 0.2589393442 10 , -6 -7 -7 -0.1473521929 10 , 0.2891792096 10 , 0.7491921522 10 , -7 -7 -0.3854258150 10 , -0.3103992942 10 ] The largest is 0.2702702703 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 15, :The Pisot Sequence a(n), defined by, a(1) = 7, a(2) = 38, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [7, 38, 206, 1117, 6057, 32844, 178096, 965722, 5236608, 28395401, 153973488, 834918127, 4527326671, 24549337382, 133118285843, 721831214832, 3914115175133, 21224210437854, 115087852184882, 624061552692958] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 16, :The Pisot Sequence a(n), defined by, a(1) = 7, a(2) = 39, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [7, 39, 217, 1207, 6714, 37347, 207745, 1155594, 6428061, 35756475, 198897538, 1106379491, 6154302313, 34233675938, 190426876781, 1059260929683, 5892202488010, 32775729932852, 182316964635420, 1014149056694422] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 135, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 7, a(2) = 40, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [7, 40, 229, 1311, 7505, 42963, 245945, 1407931, 8059809, 46138995, 264126217, 1512010795, 8655621809, 49549771171, 283651466905, 1623784586203, 9295479452865, 53212685348051, 304620100158249, 1743821136134795] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 7 a(n - 1) - 8 a(n - 2) + 4 a(n - 3) with initial conditions, a(1) = 7, a(2) = 40, a(3) = 229, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 7 b(n - 1) - 8 b(n - 2) + 4 b(n - 3) with initial conditions, b(1) = 7, b(2) = 40, b(3) = 229, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 7 t + 8 t - 4 = 0 whose roots are 1/2 1/3 1/2 1/3 (145 + 30 6 ) 25 (145 + 30 6 ) [------------------ + -------------------- + 7/3, - ------------------ 3 1/2 1/3 6 3 (145 + 30 6 ) 25 - -------------------- + 7/3 1/2 1/3 6 (145 + 30 6 ) / 1/2 1/3 \ 1/2 |(145 + 30 6 ) 25 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (145 + 30 6 ) / 1/2 1/3 (145 + 30 6 ) 25 - ------------------ - -------------------- + 7/3 6 1/2 1/3 6 (145 + 30 6 ) / 1/2 1/3 \ 1/2 |(145 + 30 6 ) 25 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (145 + 30 6 ) / In floating-point [5.724576727, 0.637711635 + 0.5404308335 I, 0.637711635 - 0.5404308335 I] The largest root is, 5.724576727 and the remaining roots are [0.637711635 + 0.5404308335 I, 0.637711635 - 0.5404308335 I] whose absolute values are [0.8359076595, 0.8359076595] so the largest absolute value is, 0.8359076595 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8359076595 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, 0.02500000000, 0.3318777293, 0.4057971014, 0.2856762159, 0.08081372344, -0.09654190978, -0.1795997105, -0.1616077999, -0.08062455630, 0.01009166008, 0.06920687097, 0.08121659096, 0.05522780928, 0.01368942113, -0.02113016247, -0.03651526922, -0.03180790027, -0.01505379798, 0.003025539389, 0.01437755849, 0.01622340242, 0.01064550653, 0.002241560331, -0.004579520233, -0.007407098172, -0.006247284016, -0.002792283668, 0.0008038937623, 0.002976389619, 0.003234442560, 0.002045556019, 0.0003489101264, -0.0009843070254, -0.001499206113, -0.001224346084, -0.0005140017853, 0.0001999317238, 0.0006141520121, 0.0006436031530, 0.0003917328694, 0.00004991291030, -0.0002100599713, -0.0003027916036, -0.0002394098139, -0.00009377575352, 0.00004768182209, 0.0001263395272, -5 0.0001278190998, 0.00007474476894, 0.6018693322 10 , -0.00004455089919, -0.00006102676513, -0.00004670538915, -0.00001692719970, 0.00001104565470, 0.00002591562396, 0.00002533533129] The largest is 0.4057971014 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 17, : Consider the Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 10, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 8, 10, 13, 17, 22, 28, 36, 46, 59, 76, 98, 126, 162, 208, 267, 343, 441, 567, 729, 937, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = a(n - 1) + a(n - 6), . Alas, it breaks down at the, 39, -th term. a(39), equals , 110155, while the corresponding term for the solution of the recurrence is , 110156 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03282504251949 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 136, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 11, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 11, 15, 20, 27, 36, 48, 64, 85, 113, 150, 199, 264, 350, 464, 615, 815, 1080, 1431, 1896] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) - a(n - 4) with initial conditions, a(1) = 8, a(2) = 11, a(3) = 15, a(4) = 20, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) - b(n - 4) with initial conditions, b(1) = 8, b(2) = 11, b(3) = 15, b(4) = 20, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [1, ------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.324717958, -0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.4545454545, -0.3333333333, 0.4500000000, 0., 0., 0.3333333333, -0.1093750000, 0.2235294118, 0.1150442478, 0.006666666667, 0.2311557789, 0.01515151515, 0.1314285714, 0.1400862069, 0.04065040650, 0.1656441718, 0.07500000000, 0.1006289308, 0.1350210970, 0.07006369427, 0.1301081731, 0.09956906328, 0.09467556925, 0.1241922978, 0.08877182714, 0.1134020619, 0.1075041690, 0.09671869755, 0.1154540559, 0.09877340219, 0.1067251462, 0.1087812521, 0.1000534825, 0.1100621118, 0.1033910821, 0.1046724093, 0.1080103532, 0.1026209240, 0.1072403917, 0.1051890587, 0.1044192125, 0.1069874322, 0.1041663191, 0.1059647414, 0.1057118851, 0.1046892225, 0.1062348095, 0.1049593066, 0.1054822431, 0.1057523363, 0.1049997768, 0.1057928117, 0.1053103493, 0.1053508277, 0.1056614024, 0.1052194201, 0.1055704744] The largest is 0.4545454545 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 18, :The Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 12, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [8, 12, 18, 27, 41, 62, 94, 143, 218, 332, 506, 771, 1175, 1791, 2730, 4161, 6342, 9666, 14732, 22453] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 137, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 13, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 8, a(2) = 13, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 8, b(2) = 13, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, -0.07692307692, 0.04761904762, -0.02941176471, 0.01818181818, -0.01123595506, 0.006944444444, -0.004291845494, 0.002652519894, -0.001639344262, 0.001013171226, -0.0006261740764, 0.0003869969040, -0.0002391772303, 0.0001478196600, -0.00009135757354, 0.00005646208571, -5 -0.00003489548801, 0.00002156659765, -0.00001332889037, 0.8237707281 10 , -5 -5 -5 -0.5091183089 10 , 0.3146524192 10 , -0.1944658897 10 , -5 -6 -6 0.1201865295 10 , -0.7427936022 10 , 0.4590716928 10 , -6 -6 -6 -0.2837219094 10 , 0.1753497834 10 , -0.1083721260 10 , -7 -7 -7 0.6697765733 10 , -0.4139446871 10 , 0.2558318861 10 , -7 -8 -8 -0.1581128010 10 , 0.9771908509 10 , -0.6039371593 10 , -8 -8 -8 0.3732536915 10 , -0.2306834678 10 , 0.1425702237 10 , -9 -9 -9 -0.8811324406 10 , 0.5445697969 10 , -0.3365626437 10 , -9 -9 -10 0.2080071532 10 , -0.1285554906 10 , 0.7945166260 10 , -10 -10 -10 -0.4910382795 10 , 0.3034783465 10 , -0.1875599330 10 , -10 -11 -11 0.1159184135 10 , -0.7164151948 10 , 0.4427689404 10 , -11 -11 -11 -0.2736462543 10 , 0.1691226861 10 , -0.1045235683 10 , -12 -12 -12 0.6459911781 10 , -0.3992445045 10 , 0.2467466736 10 , -12 -0.1524978309 10 ] The largest is 0.1250000000 The smallest is -0.07692307692 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 19, :The Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 14, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [8, 14, 25, 45, 81, 146, 263, 474, 854, 1539, 2773, 4996, 9001, 16217, 29218, 52642, 94845, 170882, 307878, 554704] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 138, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 15, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 15, 28, 52, 97, 181, 338, 631, 1178, 2199, 4105, 7663, 14305, 26704, 49850, 93058, 173717, 324288, 605368, 1130077] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) + a(n - 4) with initial conditions, a(1) = 8, a(2) = 15, a(3) = 28, a(4) = 52, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) + b(n - 4) with initial conditions, b(1) = 8, b(2) = 15, b(3) = 28, b(4) = 52, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 2) b(n) - b(n + 1) b(n + 2) b(n) - b(n + 1) -1/2 <= - -------------------------, - ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - 2 t + t - 1 = 0 whose roots are 1/2 1/2 1/2 1/2 [1/2 - 1/2 I (-3 + 2 5 ) , 1/2 + 1/2 I (-3 + 2 5 ) , 1/2 1/2 1/2 1/2 (3 + 2 5 ) (3 + 2 5 ) 1/2 - ---------------, 1/2 + ---------------] 2 2 In floating-point [0.5000000000 - 0.6066580490 I, 0.5000000000 + 0.6066580490 I, -0.8667603990, 1.866760399] The largest root is, 1.866760399 and the remaining roots are [0.5000000000 - 0.6066580490 I, 0.5000000000 + 0.6066580490 I, -0.8667603990] whose absolute values are [0.7861513775, 0.7861513775, 0.8667603990] so the largest absolute value is, 0.8667603990 that is less than 1, It follows that the sequence 2 b(n + 2) b(n) - b(n + 1) c(n) = - ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8667603990 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.2666666667, -0.4285714286, -0.05769230769, -0.2577319588, 0.1823204420, -0.008875739645, 0.1822503962, -0.07555178268, 0.04001819009, -0.1110840438, 0.03562573405, -0.04432016777, 0.06246255243, -0.02178535607, 0.03637516388, -0.03403236298, 0.01618314585, -0.02579422764, 0.01881907162, -0.01257736719, 0.01682263922, -0.01096802092, 0.009460396898, -0.01047921260, 0.006832234925, -0.006763947973, 0.006411713556, -0.004488020418, 0.004620142063, -0.003935377404, 0.003028979166, -0.003050204149, 0.002455111170, -0.002054134231, 0.001970914853, -0.001563485612, 0.001382274176, -0.001260500732, 0.001013399002, -0.0009189617851, 0.0008048513377, -0.0006641970581, 0.0006039666705, -0.0005158797818, 0.0004372888321, -0.0003935860643, 0.0003326743236, -0.0002878199666, 0.0002552349632, -0.0002157904616, 0.0001889133670, -0.0001652281957, 0.0001405690333, -0.0001235657621, 0.0001070100387, -0.00009177715172, 0.00008058049191] The largest is 0.2666666667 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 139, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 17, 36, 76, 160, 337, 710, 1496, 3152, 6641, 13992, 29480, 62112, 130865, 275722, 580924, 1223960, 2578785, 5433292, 11447508] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 4) with initial conditions, a(1) = 8, a(2) = 17, a(3) = 36, a(4) = 76, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 4) with initial conditions, b(1) = 8, b(2) = 17, b(3) = 36, b(4) = 76, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - 2 _Z - 1 In floating-point [2.10691934037622, 0.304876704453035 + 0.754529173144244 I, -0.716672749282287, 0.304876704453035 - 0.754529173144244 I] The largest root is, 2.10691934037622 and the remaining roots are [0.304876704453035 + 0.754529173144244 I, -0.716672749282287, 0.304876704453035 - 0.754529173144244 I] whose absolute values are [0.813796091194765, 0.716672749282287, 0.813796091194765] so the largest absolute value is, 0.813796091194765 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.813796091194765 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.2352941176, 0.4444444444, -0.1578947368, -0.1937500000, -0.1543026706, 0.1352112676, 0.1122994652, 0.03077411168, -0.09275711489, -0.05031446541, 0.01166892809, 0.05411192684, 0.01546632025, -0.01938184113, -0.02709476627, -0.00007761691559, 0.01531108642, 0.01124033091, -0.004614104659, -0.009305826251, -0.003300566111, 0.004639198685, 0.004664292708, 0.00002275916505, -0.003255047781, -0.001870896877, 0.0009224989550, 0.001867757075, 0.0004804663693, -0.0009099641381, -0.0008974293212, 0.00007289843248, 0.0006262632342, 0.0003425623304, -0.0002123046605, -0.0003517108884, -0.00007715854265, 0.0001882452451, 0.0001641858297, -0.00002333922906, -0.0001238370008, -0.00005942875648, 0.00004532831672, 0.00006731740438, 0.00001079780798, -0.00003783314052, -5 -0.00003033796432, 0.6641475740 10 , 0.00002408075946, 0.00001032837840, -5 -5 -0.9681207517 10 , -0.00001272093929, -0.1361119125 10 , -5 -5 -5 0.7606140152 10 , 0.5531072787 10 , -0.1658793720 10 , -5 -0.4678706566 10 ] The largest is 0.4444444444 The smallest is -0.1937500000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 140, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 18, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 18, 41, 93, 211, 479, 1087, 2467, 5599, 12707, 28839, 65451, 148543, 337123, 765111, 1736443, 3940911, 8944019, 20298727, 46068587] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + 2 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 8, a(2) = 18, a(3) = 41, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + 2 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 8, b(2) = 18, b(3) = 41, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 2 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (37 + 3 114 ) 7 (37 + 3 114 ) [------------------ + -------------------- + 1/3, - ------------------ 3 1/2 1/3 6 3 (37 + 3 114 ) 7 - -------------------- + 1/3 1/2 1/3 6 (37 + 3 114 ) / 1/2 1/3 \ 1/2 |(37 + 3 114 ) 7 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (37 + 3 114 ) / 1/2 1/3 (37 + 3 114 ) 7 - ------------------ - -------------------- + 1/3 6 1/2 1/3 6 (37 + 3 114 ) / 1/2 1/3 \ 1/2 |(37 + 3 114 ) 7 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (37 + 3 114 ) / In floating-point [2.269530842, -0.6347654213 + 0.6916012305 I, -0.6347654213 - 0.6916012305 I] The largest root is, 2.269530842 and the remaining roots are [-0.6347654213 + 0.6916012305 I, -0.6347654213 - 0.6916012305 I] whose absolute values are [0.9387435231, 0.9387435231] so the largest absolute value is, 0.9387435231 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.9387435231 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.3888888889, -0.04878048780, -0.2795698925, 0.3981042654, -0.2588726514, -0.02207911684, 0.2561815971, -0.3057688873, 0.1624301566, 0.06324768543, -0.2234343249, 0.2279205348, -0.09245290295, -0.08348069757, 0.1874544687, -0.1644127462, 0.04353479124, 0.08961823074, -0.1521376811, 0.1141683626, -0.01087053839, -0.08680917568, 0.1197864726, -0.07557295550, -0.009618361572, 0.07880867271, -0.09157396144, 0.04680666084, 0.02127608338, -0.06825851782, 0.06790697063, -0.02605789824, -0.02676099262, 0.05693715214, -0.04870062959, 0.01165168944, 0.02812473454, -0.04597314575, 0.03357970222, -0.002117120204, -0.02690400728, 0.03602115675, -0.02202109821, -0.003786799253, 0.02421331784, -0.02740247708, 0.01345056009, 0.007072241610, -0.02083159237, 0.02021401104, -0.007304690475, -0.008539853131, 0.01727878800, -0.01441029922, 0.003067570513, 0.008804548073, -0.01388090933] The largest is 0.3981042654 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 20, :The Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 19, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [8, 19, 45, 107, 254, 603, 1432, 3401, 8077, 19182, 45555, 108188, 256934, 610189, 1449129, 3441515, 8173203, 19410419, 46097517, 109476311] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 141, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 20, 50, 125, 313, 784, 1964, 4920, 12325, 30875, 77344, 193752, 485362, 1215865, 3045825, 7630000, 19113672, 47881056, 119945321, 300471235] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) - a(n - 3) + a(n - 4) with initial conditions, a(1) = 8, a(2) = 20, a(3) = 50, a(4) = 125, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) - b(n - 3) + b(n - 4) with initial conditions, b(1) = 8, b(2) = 20, b(3) = 50, b(4) = 125, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 3 t + t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 3 _Z + _Z + _Z - 1 In floating-point [2.50506841362147, 0.592608040497570 + 0.476565325929643 I, -0.690284494616613, 0.592608040497570 - 0.476565325929643 I] The largest root is, 2.50506841362147 and the remaining roots are [0.592608040497570 + 0.476565325929643 I, -0.690284494616613, 0.592608040497570 - 0.476565325929643 I] whose absolute values are [0.760459597572939, 0.690284494616613, 0.760459597572939] so the largest absolute value is, 0.760459597572939 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.760459597572939 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., 0., -0.5000000000, -0.2480000000, -0.2428115016, 0.02040816327, 0.05091649694, 0.1270325203, 0.06693711968, 0.04327125506, -0.01323955317, -0.02289524753, -0.03178040308, -0.01593515727, -0.006369374472, 0.005712188729, 0.007660694397, 0.007704111622, 0.003370077270, 0.0004576145201, -0.002040650936, -0.002245532976, -0.001783485242, -0.0006066572949, 0.0001683953979, 0.0006497957549, 0.0006041639192, 0.0003876433100, 0.00007736565395, -0.0001099145126, -0.0001905885825, -5 -0.0001515735788, -0.00007685198725, 0.1691686888 10 , 0.00004291204420, -5 0.00005232285421, 0.00003551284428, 0.00001299532134, -0.5937690289 10 , -5 -0.00001399838228, -0.00001353993360, -0.7688406889 10 , -5 -5 -5 -5 -0.1464595080 10 , 0.2836172967 10 , 0.4121587273 10 , 0.3304777043 10 , -5 -6 -5 0.1491975809 10 , -0.1142639211 10 , -0.1017957343 10 , -5 -6 -6 -0.1126806874 10 , -0.7562235473 10 , -0.2381703466 10 , -6 -6 -6 -6 0.1505620381 10 , 0.3192731348 10 , 0.2892041656 10 , 0.1596069772 10 , -7 -7 0.2090566941 10 , -0.6682099977 10 ] The largest is 0.1270325203 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 142, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368, 121393, 317811, 832040, 2178309, 5702887, 14930352, 39088169, 102334155, 267914296, 701408733] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) with initial conditions, a(1) = 8, a(2) = 21, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) with initial conditions, b(1) = 8, b(2) = 21, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t + 1 = 0 whose roots are 1/2 1/2 5 5 [3/2 + ----, 3/2 - ----] 2 2 In floating-point [2.618033988, 0.381966012] The largest root is, 2.618033988 and the remaining roots are [0.381966012] whose absolute values are [0.381966012] so the largest absolute value is, 0.381966012 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.381966012 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.04761904762, 0.01818181818, 0.006944444444, 0.002652519894, 0.001013171226, 0.0003869969040, 0.0001478196600, 0.00005646208571, -5 -5 -5 0.00002156659765, 0.8237707281 10 , 0.3146524192 10 , 0.1201865295 10 , -6 -6 -7 -7 0.4590716928 10 , 0.1753497834 10 , 0.6697765733 10 , 0.2558318861 10 , -8 -8 -8 -9 0.9771908509 10 , 0.3732536915 10 , 0.1425702237 10 , 0.5445697969 10 , -9 -10 -10 0.2080071532 10 , 0.7945166260 10 , 0.3034783465 10 , -10 -11 -11 0.1159184135 10 , 0.4427689404 10 , 0.1691226861 10 , -12 -12 -13 0.6459911781 10 , 0.2467466736 10 , 0.9424884271 10 , -13 -13 -14 0.3599985451 10 , 0.1375072083 10 , 0.5252307989 10 , -14 -15 -15 0.2006203132 10 , 0.7663014082 10 , 0.2927010923 10 , -15 -16 -16 0.1118018687 10 , 0.4270451385 10 , 0.1631167282 10 , -17 -17 -18 0.6230504602 10 , 0.2379840991 10 , 0.9090183708 10 , -18 -18 -19 0.3472141212 10 , 0.1326239929 10 , 0.5065785758 10 , -19 -20 -20 0.1934957980 10 , 0.7390881815 10 , 0.2823065646 10 , -20 -21 -21 0.1078315124 10 , 0.4118797270 10 , 0.1573240564 10 , -22 -22 -23 0.6009244230 10 , 0.2295327049 10 , 0.8767369176 10 , -23 -23 -24 0.3348837033 10 , 0.1279141924 10 , 0.4885887385 10 , -24 0.1866242916 10 ] The largest is 0.1250000000 The smallest is -24 0.1866242916 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 143, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 22, 61, 169, 468, 1296, 3589, 9939, 27524, 76222, 211081, 584545, 1618776, 4482864, 12414361, 34378995, 95205488, 263651830, 730128997, 2021940649] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 8, a(2) = 22, a(3) = 61, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 8, b(2) = 22, b(3) = 61, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (27 + 3 57 ) 2 (27 + 3 57 ) [----------------- + ----------------- + 1, - ----------------- 3 1/2 1/3 6 (27 + 3 57 ) 1 - ------------------- + 1 1/2 (1/3) (27 + 3 57 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(27 + 3 57 ) 2 | (27 + 3 57 ) + 1/2 I 3 |----------------- - -----------------|, - ----------------- | 3 1/2 1/3| 6 \ (27 + 3 57 ) / 1 - ------------------- + 1 1/2 (1/3) (27 + 3 57 ) / 1/2 1/3 \ 1/2 |(27 + 3 57 ) 2 | - 1/2 I 3 |----------------- - -----------------|] | 3 1/2 1/3| \ (27 + 3 57 ) / In floating-point [2.769292354, 0.1153538227 + 0.5897428050 I, 0.1153538227 - 0.5897428050 I] The largest root is, 2.769292354 and the remaining roots are [0.1153538227 + 0.5897428050 I, 0.1153538227 - 0.5897428050 I] whose absolute values are [0.6009185306, 0.6009185306] so the largest absolute value is, 0.6009185306 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6009185306 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.1363636364, 0.2131147541, 0., -0.07692307692, -0.01774691358, 0.02368347729, 0.01187242177, -0.005813108560, -0.005628296292, 0.0008006405124, 0.002217109034, 0.0002223902504, -0.0007492977704, -0.0002531745291, 0.0002121644335, 0.0001403700593, -0.00004422878461, -5 -5 -0.00006089197961, 0.1922905107 10 , 0.00002243191033, 0.4480846264 10 , -5 -5 -5 -0.7066466429 10 , -0.3248335223 10 , 0.1802307024 10 , -5 -6 -6 0.1588789866 10 , -0.2842726488 10 , -0.6393007885 10 , -7 -6 -7 -0.4483985065 10 , 0.2205085877 10 , 0.6706482535 10 , -7 -7 -7 -0.6415396235 10 , -0.3901812465 10 , 0.1416441375 10 , -7 -8 -8 0.1735740356 10 , -0.1110327729 10 , -0.6523972991 10 , -8 -8 -9 -0.1104187686 10 , 0.2101082203 10 , 0.8834613036 10 , -9 -9 -10 -0.5548859783 10 , -0.4470370360 10 , 0.9723617397 10 , -9 -11 -10 0.1838595796 10 , 0.7305528795 10 , -0.6470681922 10 , -10 -10 -10 -0.1756640688 10 , 0.1931312738 10 , 0.1079896979 10 , -11 -11 -12 -0.4482624894 10 , -0.4933717093 10 , 0.4804434027 10 , -11 -12 -12 0.1892422407 10 , 0.2631067256 10 , -0.6226588276 10 , -12 -12 -12 -0.2386608013 10 , 0.1697831492 10 , 0.1253514212 10 ] The largest is 0.2131147541 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 144, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 23, 66, 189, 541, 1549, 4435, 12698, 36356, 104092, 298029, 853296, 2443098, 6994909, 20027339, 57340890, 164174465, 470052958, 1345823075, 3853267421] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) + a(n - 3) + a(n - 4) with initial conditions, a(1) = 8, a(2) = 23, a(3) = 66, a(4) = 189, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) + b(n - 3) + b(n - 4) with initial conditions, b(1) = 8, b(2) = 23, b(3) = 66, b(4) = 189, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t - 2 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 2 _Z - 2 _Z - _Z - 1 In floating-point [2.86313074324295, -0.0254123440528567 + 0.655229102900684 I, -0.812306055137237, -0.0254123440528567 - 0.655229102900684 I] The largest root is, 2.86313074324295 and the remaining roots are [-0.0254123440528567 + 0.655229102900684 I, -0.812306055137237, -0.0254123440528567 - 0.655229102900684 I] whose absolute values are [0.655721712709207, 0.812306055137237, 0.655721712709207] so the largest absolute value is, 0.812306055137237 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.812306055137237 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.3913043478, 0.2272727273, -0.4232804233, 0.1219963031, 0.01484828922, 0.07756482525, -0.1165537880, 0.05886236110, -0.02297006494, 0.03279546621, -0.03804072678, 0.02540176448, -0.01545252411, 0.01465321978, -0.01423757113, 0.01078053764, -0.007713371309, 0.006549981319, -0.005783813467, 0.004599502039, -0.003532012845, 0.002901146240, -0.002446044638, 0.001977692397, -0.001567571087, 0.001275344223, -0.001052805970, 0.0008551978164, -0.0006874431706, 0.0005580475446, -0.0004563994051, 0.0003710509249, -0.0003000925864, 0.0002435648164, -0.0001984040203, 0.0001612799307, -0.0001307759492, 0.0001061687593, -0.00008633846937, 0.00007016456132, -0.00005695500599, 0.00004624940054, -0.00003758511895, 0.00003053811851, -0.00002479960632, 0.00002014130598, -5 -0.00001636360112, 0.00001329392190, -0.00001079765878, 0.8770231106 10 , -5 -5 -5 -0.7124534565 10 , 0.5787656206 10 , -0.4701184392 10 , -5 -5 -5 0.3818640169 10 , -0.3101966805 10 , 0.2519818542 10 , -5 -0.2046840749 10 ] The largest is 0.3913043478 The smallest is -0.4232804233 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 145, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 25, 78, 243, 757, 2358, 7345, 22879, 71266, 221987, 691469, 2153862, 6709081, 20898167, 65095858, 202767579, 631602261, 1967382646, 6128215041, 19088823247] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 2) - 2 a(n - 3) with initial conditions, a(1) = 8, a(2) = 25, a(3) = 78, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 2) - 2 b(n - 3) with initial conditions, b(1) = 8, b(2) = 25, b(3) = 78, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 2) b(n) - b(n + 1) b(n + 2) b(n) - b(n + 1) -1/2 <= - -------------------------, - ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t - t + 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 I 687 ) 8 (108 + 12 I 687 ) [---------------------- + ---------------------- + 1, - ---------------------- 6 1/2 1/3 12 (108 + 12 I 687 ) 4 - ---------------------- + 1 1/2 1/3 (108 + 12 I 687 ) / 1/2 1/3 \ 1/2 |(108 + 12 I 687 ) 8 | + 1/2 I 3 |---------------------- - ----------------------|, | 6 1/2 1/3| \ (108 + 12 I 687 ) / 1/2 1/3 (108 + 12 I 687 ) 4 - ---------------------- - ---------------------- + 1 12 1/2 1/3 (108 + 12 I 687 ) / 1/2 1/3 \ 1/2 |(108 + 12 I 687 ) 8 | - 1/2 I 3 |---------------------- - ----------------------|] | 6 1/2 1/3| \ (108 + 12 I 687 ) / In floating-point -9 -9 [3.114907542 - 0.1 10 I, -0.8608058533 + 0.1 10 I, -9 0.7458983117 + 0.1 10 I] -9 The largest root is, 3.114907542 - 0.1 10 I and the remaining roots are -9 -9 [-0.8608058533 + 0.1 10 I, 0.7458983117 + 0.1 10 I] whose absolute values are [0.8608058533, 0.7458983117] so the largest absolute value is, 0.8608058533 that is less than 1, It follows that the sequence 2 b(n + 2) b(n) - b(n + 1) c(n) = - ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8608058533 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.3600000000, 0.03846153846, 0.2263374486, -0.001321003963, 0.1454622561, -0.01756296801, 0.09541500940, -0.02224061965, 0.06381905247, -0.02161340566, 0.04346007312, -0.01887128803, 0.03007302028, -0.01557237328, 0.02109847650, -0.01242298434, 0.01497427003, -0.009697127239, 0.01072885700, -0.007459096292, 0.007745822606, -0.005679342484, 0.005625987739, -0.004293024478, 0.004105599272, -0.003228202140, 0.003007041809, -0.002418275256, 0.002208620322, -0.001806497909, 0.001625677106, -0.001346707234, 0.001198551222, -0.001002407780, 0.0008847423505, -0.0007452831728, 0.0006537083919, -0.0005536426981, 0.0004833466431, -0.0004110195527, 0.0003575733812, -0.0003049926954, 0.0002646344005, -0.0002262362561, 0.0001959110228, -0.0001677719888, 0.0001450675688, -0.0001243913279, 0.0001074375625, -0.00009221377795, 0.00007957888456, -0.00006835224935, 0.00005894969240, -0.00005066094125, 0.00004367136734, -0.00003754622403, 0.00003235457776] The largest is 0.3600000000 The smallest is -0.02224061965 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 146, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 26, 85, 278, 909, 2972, 9717, 31770, 103873, 339616, 1110385, 3630438, 11869829, 38808772, 126886477, 414859250, 1356395113, 4434775656, 14499635785, 47407006398] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 3 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 8, a(2) = 26, a(3) = 85, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 3 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 8, b(2) = 26, b(3) = 85, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 3 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (37 + 3 114 ) 7 (37 + 3 114 ) [------------------ + -------------------- + 4/3, - ------------------ 3 1/2 1/3 6 3 (37 + 3 114 ) 7 - -------------------- + 4/3 1/2 1/3 6 (37 + 3 114 ) / 1/2 1/3 \ 1/2 |(37 + 3 114 ) 7 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (37 + 3 114 ) / 1/2 1/3 (37 + 3 114 ) 7 - ------------------ - -------------------- + 4/3 6 1/2 1/3 6 (37 + 3 114 ) / 1/2 1/3 \ 1/2 |(37 + 3 114 ) 7 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (37 + 3 114 ) / In floating-point [3.269530842, 0.3652345784 + 0.6916012305 I, 0.3652345784 - 0.6916012305 I] The largest root is, 3.269530842 and the remaining roots are [0.3652345784 + 0.6916012305 I, 0.3652345784 - 0.6916012305 I] whose absolute values are [0.7821179958, 0.7821179958] so the largest absolute value is, 0.7821179958 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7821179958 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.1153846154, 0.2235294118, 0.2338129496, 0.03410341034, -0.1181022880, -0.1071318308, -0.006011960969, 0.06114197145, 0.04833988976, -0.002090265989, -0.03109679879, -0.02143661884, 0.003363389081, 0.01556981521, 0.009315855920, -0.002719243799, -0.007684912529, -0.003950206878, 0.001815422477, 0.003742485485, 0.001623260753, -0.001103568488, -0.001799085240, -0.0006391139910, 0.0006336627809, 0.0008538226165, 0.0002360741411, -0.0003498457233, -0.0003999600836, -0.00007815488216, 0.0001875692754, 0.0001848215810, 0.00002026873347, -5 -0.00009825125832, -0.00008416807167, -0.1381044791 10 , 0.00005047751921, -5 0.00003771706787, -0.3326375731 10 , -0.00002550166812, -0.00001659340953, -5 -5 -5 0.3478614770 10 , 0.00001269135143, 0.7142742366 10 , -0.2545855295 10 , -5 -5 -5 -0.6228945414 10 , -0.2992731037 10 , 0.1624201503 10 , -5 -5 -6 0.3017108294 10 , 0.1210366594 10 , -0.9614555001 10 , -5 -6 -6 -0.1442705195 10 , -0.4657210902 10 , 0.5423202231 10 , -6 -6 -6 0.6810337737 10 , 0.1657322450 10 , -0.2955318949 10 ] The largest is 0.2338129496 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 147, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 27, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 27, 91, 307, 1036, 3496, 11797, 39808, 134329, 453283, 1529569, 5161414, 17416798, 58771657, 198320476, 669217327, 2258222851, 7620200851, 25713786832, 86769213328] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 3 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 8, a(2) = 27, a(3) = 91, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 3 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 8, b(2) = 27, b(3) = 91, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 3 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (404 + 36 109 ) 14 (404 + 36 109 ) [-------------------- + ---------------------- + 4/3, - -------------------- 6 1/2 1/3 12 3 (404 + 36 109 ) 7 - ---------------------- + 4/3 1/2 1/3 3 (404 + 36 109 ) / 1/2 1/3 \ 1/2 |(404 + 36 109 ) 14 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (404 + 36 109 ) / 1/2 1/3 (404 + 36 109 ) 7 - -------------------- - ---------------------- + 4/3 12 1/2 1/3 3 (404 + 36 109 ) / 1/2 1/3 \ 1/2 |(404 + 36 109 ) 14 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (404 + 36 109 ) / In floating-point [3.374423763, 0.312788118 + 0.8894966410 I, 0.312788118 - 0.8894966410 I] The largest root is, 3.374423763 and the remaining roots are [0.312788118 + 0.8894966410 I, 0.312788118 - 0.8894966410 I] whose absolute values are [0.9428895381, 0.9428895381] so the largest absolute value is, 0.9428895381 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.9428895381 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, -0.2962962963, -0.2967032967, 0.07817589577, 0.3127413127, 0.1261441648, -0.1991184199, -0.2367112138, 0.02894386171, 0.2285525819, 0.1172447925, -0.1298469760, -0.1854645728, -0.0005829850943, 0.1645208486, 0.1034386308, -0.08155697783, -0.1429812580, -0.01693820622, 0.1165200157, 0.08795090730, -0.04857103646, -0.1085768207, -0.02474145159, 0.08105154644, 0.07270007837, -0.02657868059, -0.08126031816, -0.02720499574, 0.05522492972, 0.05873375165, -0.01235476979, -0.05994554495, -0.02651661548, 0.03670586358, 0.04653666588, -0.003520773632, -0.04357550144, -0.02412968722, 0.02364543454, 0.03624429552, 0.001651816767, -0.03118931585, -0.02097982714, 0.01460408929, 0.02778789102, 0.004399814811, -0.02195214596, -0.01764435522, 0.008478461456, 0.02099047359, 0.005593444329, -0.01516225907, -0.01445794853, 0.004435316100, 0.01562833276, 0.005833537149, -0.01024490138] The largest is 0.3127413127 The smallest is -0.2967032967 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 148, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 28, 98, 343, 1201, 4205, 14723, 51550, 180493, 631964, 2212709, 7747405, 27126154, 94977380, 332546321, 1164351508, 4076768704, 14274076988, 49978129409, 174989487679] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 2) - 3 a(n - 4) with initial conditions, a(1) = 8, a(2) = 28, a(3) = 98, a(4) = 343, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 2) - 3 b(n - 4) with initial conditions, b(1) = 8, b(2) = 28, b(3) = 98, b(4) = 343, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 3 t - 2 t + 3 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 3 _Z - 2 _Z + 3 In floating-point [0.880834097318656, 3.50132127289189, -0.691077685105273 + 0.703667569490520 I, -0.691077685105273 - 0.703667569490520 I] The largest root is, 3.50132127289189 and the remaining roots are [0.880834097318656, -0.691077685105273 + 0.703667569490520 I, -0.691077685105273 - 0.703667569490520 I] whose absolute values are [0.880834097318656, 0.986274006147966, 0.986274006147966] so the largest absolute value is, 0.986274006147966 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.986274006147966 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., 0., -0.5000000000, 0.2507288630, -0.2481265612, -0.2428061831, 0.2758269374, -0.4103006790, 0.06514934097, 0.1032669582, -0.3873808079, 0.2752942437, -0.1443269105, -0.1921931201, 0.2969092537, -0.3195412085, 0.06817561362, 0.1420237841, -0.3283051815, 0.2577556494, -0.08787025570, -0.1741708208, 0.2866625708, -0.2616208772, 0.05207327699, 0.1554905388, -0.2893695422, 0.2277350828, -0.05175366695, -0.1662624516, 0.2658139378, -0.2182883380, 0.03202386250, 0.1582822662, -0.2585472898, 0.1957876771, -0.02580313575, -0.1606808517, 0.2419930426, -0.1827456069, 0.01315867175, 0.1560273566, -0.2315797145, 0.1655523904, -0.005978273022, -0.1549121081, 0.2180462733, -0.1523425675, -0.003000336832, 0.1510501787, -0.2069889574, 0.1381611877, 0.009506658930, -0.1483081839, 0.1950556384, -0.1259330158, -0.01620774750, 0.1444352775] The largest is 0.2969092537 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 21, :The Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 29, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [8, 29, 105, 380, 1375, 4975, 18000, 65126, 235633, 852546, 3084605, 11160439, 40379692, 146098153, 528599136, 1912529631, 6919741900, 25036384894, 90584385606, 327744239049] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 22, :The Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [8, 30, 113, 426, 1606, 6055, 22829, 86072, 324517, 1223525, 4613051, 17392566, 65575115, 247237567, 932158709, 3514513871, 13250756154, 49959267511, 188361206050, 710177424775] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 149, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 31, 120, 465, 1802, 6983, 27060, 104861, 406350, 1574659, 6102008, 23646073, 91631602, 355084351, 1375997948, 5332170645, 20662853334, 80071238587, 310286442304, 1202400237297] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 8, a(2) = 31, a(3) = 120, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 8, b(2) = 31, b(3) = 120, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (73 + 6 87 ) 13 (73 + 6 87 ) [----------------- + ------------------- + 4/3, - ----------------- 3 1/2 1/3 6 3 (73 + 6 87 ) 13 - ------------------- + 4/3 1/2 1/3 6 (73 + 6 87 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(73 + 6 87 ) 13 | (73 + 6 87 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (73 + 6 87 ) / 13 - ------------------- + 4/3 1/2 1/3 6 (73 + 6 87 ) / 1/2 1/3 \ 1/2 |(73 + 6 87 ) 13 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (73 + 6 87 ) / In floating-point [3.875129794, 0.062435102 + 0.7156909970 I, 0.062435102 - 0.7156909970 I] The largest root is, 3.875129794 and the remaining roots are [0.062435102 + 0.7156909970 I, 0.062435102 - 0.7156909970 I] whose absolute values are [0.7184091767, 0.7184091767] so the largest absolute value is, 0.7184091767 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7184091767 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, -0.4838709677, -0.1250000000, 0.2344086022, 0.09378468368, -0.1092653587, -0.06204730229, 0.04864534956, 0.03809769903, -0.02034916766, -0.02220367459, 0.007729867027, 0.01242480733, -0.002437986911, -0.006717020918, 0.0004195178941, 0.003519118673, 0.0002229149607, -0.001788423042, -0.0003383697827, 0.0008807738325, 0.0002846190287, -0.0004190372829, -0.0001992204953, 0.0001913933590, 0.0001267193656, -0.00008295688722, -0.00007576019649, 0.00003335483249, 0.00004326575201, -5 -0.00001181221744, -0.00002380495679, 0.3123894301 10 , 0.00001267609911, -7 -5 -6 -0.2941143348 10 , -0.6545956243 10 , -0.8022153172 10 , -5 -6 -5 0.3278272107 10 , 0.8233912600 10 , -0.1589137702 10 , -6 -6 -6 -0.6233978519 10 , 0.7423288141 10 , 0.4144377050 10 , -6 -6 -6 -0.3313736980 10 , -0.2552748690 10 , 0.1391496321 10 , -6 -7 -7 0.1491260013 10 , -0.5319536496 10 , -0.8360819691 10 , -7 -7 -8 0.1701457987 10 , 0.4527578649 10 , -0.3127827730 10 , -7 -8 -7 -0.2375793766 10 , -0.1352349938 10 , 0.1209288245 10 , -8 -8 -8 0.2208004407 10 , -0.5965564697 10 , -0.1884498299 10 ] The largest is 0.2344086022 The smallest is -0.4838709677 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 150, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 33, 136, 560, 2306, 9496, 39104, 161028, 663104, 2730624, 11244552, 46304416, 190678912, 785204752, 3233427840, 13315069184, 54830686240, 225789600640, 929788540928, 3828815536192] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 2 a(n - 3) with initial conditions, a(1) = 8, a(2) = 33, a(3) = 136, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 2 b(n - 3) with initial conditions, b(1) = 8, b(2) = 33, b(3) = 136, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (91 + 3 465 ) 16 (91 + 3 465 ) [------------------ + -------------------- + 4/3, - ------------------ 3 1/2 1/3 6 3 (91 + 3 465 ) 8 - -------------------- + 4/3 1/2 1/3 3 (91 + 3 465 ) / 1/2 1/3 \ 1/2 |(91 + 3 465 ) 16 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (91 + 3 465 ) / 1/2 1/3 (91 + 3 465 ) 8 - ------------------ - -------------------- + 4/3 6 1/2 1/3 3 (91 + 3 465 ) / 1/2 1/3 \ 1/2 |(91 + 3 465 ) 16 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (91 + 3 465 ) / In floating-point [4.117942271, -0.058971136 + 0.6944075680 I, -0.058971136 - 0.6944075680 I] The largest root is, 4.117942271 and the remaining roots are [-0.058971136 + 0.6944075680 I, -0.058971136 - 0.6944075680 I] whose absolute values are [0.6969070708, 0.6969070708] so the largest absolute value is, 0.6969070708 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6969070708 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.4848484848, -0.1176470588, -0.2214285714, 0.08326105811, 0.09772535805, -0.05196399345, -0.04133442631, 0.03011292346, 0.01652369568, -0.01657407071, -0.006070436133, 0.008765646827, 0.001914445877, -0.004483088758, -0.0004010613784, 0.002224646240, -0.00006759255500, -0.001072492977, 0.0001593205737, 0.0005020971850, -0.0001365972135, -0.0002277477066, 0.00009320354345, 0.00009961974673, -0.00005701642634, -0.00004165861848, 0.00003260501954, 0.00001638722548, -0.00001776833505, -5 -5 -5 -0.5863301110 10 , 0.9321246514 10 , 0.1748315961 10 , -5 -6 -5 -0.4733338377 10 , -0.2908604810 10 , 0.2333189997 10 , -6 -5 -6 -0.1339167651 10 , -0.1117388023 10 , 0.1968279043 10 , -6 -6 -6 0.5194780871 10 , -0.1568636969 10 , -0.2337989788 10 , -6 -6 -7 0.1037602590 10 , 0.1013136423 10 , -0.6234338821 10 , -7 -7 -7 -0.4185303481 10 , 0.3521514544 10 , 0.1617380532 10 , -7 -8 -8 -0.1901084834 10 , -0.5613102483 10 , 0.9895200711 10 , -8 -8 -9 0.1559106164 10 , -0.4989780311 10 , -0.1687198238 10 , -8 -9 -8 0.2443333032 10 , -0.2062284939 10 , -0.1162353623 10 , -9 0.2372515712 10 ] The largest is 0.4848484848 The smallest is -0.2214285714 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 23, :The Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [8, 34, 145, 618, 2634, 11226, 47845, 203914, 869076, 3703979, 15786261, 67280629, 286748270, 1222113580, 5208615914, 22198983944, 94611485332, 403231660481, 1718562724639, 7324469102936] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 24, :The Pisot Sequence a(n), defined by, a(1) = 8, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [8, 35, 153, 669, 2925, 12789, 55917, 244484, 1068949, 4673729, 20434785, 89346310, 390645809, 1708007282, 7467861700, 32651475762, 142760928398, 624188714336, 2729118922639, 11932433129345] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 151, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 36, 162, 729, 3281, 14767, 66463, 299135, 1346339, 6059567, 27272739, 122748423, 552462859, 2486510239, 11191219587, 50369145431, 226700118939, 1020325905615, 4592255877683, 20668703921031] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 2 a(n - 2) - 2 a(n - 3) + 4 a(n - 4) with initial conditions, a(1) = 8, a(2) = 36, a(3) = 162, a(4) = 729, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 2 b(n - 2) - 2 b(n - 3) + 4 b(n - 4) with initial conditions, b(1) = 8, b(2) = 36, b(3) = 162, b(4) = 729, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t + 2 t + 2 t - 4 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z + 2 _Z + 2 _Z - 4 In floating-point [4.50077357872737, 0.695947332251749 + 0.715019152883293 I, -0.892668243230864, 0.695947332251749 - 0.715019152883293 I] The largest root is, 4.50077357872737 and the remaining roots are [0.695947332251749 + 0.715019152883293 I, -0.892668243230864, 0.695947332251749 - 0.715019152883293 I] whose absolute values are [0.997795108355552, 0.892668243230864, 0.997795108355552] so the largest absolute value is, 0.997795108355552 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.997795108355552 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., 0., -0.5000000000, -0.2496570645, -0.2480950930, 0.2589557798, 0.2899056618, 0.4291574038, 0.05568285551, -0.1238900403, -0.4295084553, -0.2944984149, -0.1429636594, 0.2376352813, 0.3450667318, 0.3579967558, 0.05272511520, -0.1919602742, -0.4009781855, -0.2944335864, -0.07539055207, 0.2460296869, 0.3658839694, 0.3104072318, 0.02664663790, -0.2352304654, -0.3867241885, -0.2748243604, -0.02362594255, 0.2640455234, 0.3702314692, 0.2710207425, -0.007954042884, -0.2660925439, -0.3756702424, -0.2461750682, 0.02083406023, 0.2834907464, 0.3654547786, 0.2339240069, -0.04493477445, -0.2894684575, -0.3635016383, -0.2130057000, 0.06117259393, 0.3010038162, 0.3546787397, 0.1970180781, -0.08158434552, -0.3073000984, -0.3486529985, -0.1774237923, 0.09844985020, 0.3152024390, 0.3393480853, 0.1597406789, -0.1165982535, -0.3203590398] The largest is 0.4291574038 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 152, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 37, 171, 790, 3650, 16864, 77916, 359992, 1663256, 7684672, 35505168, 164043040, 757921184, 3501791488, 16179180480, 74751989632, 345373485440, 1595714642944, 7372613501184, 34063377232384] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 2 a(n - 2) + 4 a(n - 3) with initial conditions, a(1) = 8, a(2) = 37, a(3) = 171, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 2 b(n - 2) + 4 b(n - 3) with initial conditions, b(1) = 8, b(2) = 37, b(3) = 171, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t - 2 t - 4 = 0 whose roots are 1/2 1/3 1/2 1/3 (154 + 66 3 ) 22 (154 + 66 3 ) [------------------ + -------------------- + 4/3, - ------------------ 3 1/2 1/3 6 3 (154 + 66 3 ) 11 - -------------------- + 4/3 1/2 1/3 3 (154 + 66 3 ) / 1/2 1/3 \ 1/2 |(154 + 66 3 ) 22 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (154 + 66 3 ) / 1/2 1/3 (154 + 66 3 ) 11 - ------------------ - -------------------- + 4/3 6 1/2 1/3 3 (154 + 66 3 ) / 1/2 1/3 \ 1/2 |(154 + 66 3 ) 22 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (154 + 66 3 ) / In floating-point [4.620258098, -0.310129049 + 0.8772527990 I, -0.310129049 - 0.8772527990 I] The largest root is, 4.620258098 and the remaining roots are [-0.310129049 + 0.8772527990 I, -0.310129049 - 0.8772527990 I] whose absolute values are [0.9304582207, 0.9304582207] so the largest absolute value is, 0.9304582207 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.9304582207 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, 0.2972972973, -0.2923976608, -0.07594936709, 0.3002739726, -0.1204933586, -0.1852251142, 0.2192048712, 0.02439552300, -0.2049086806, 0.1059757836, 0.1116678647, -0.1610116970, 0.003192075838, 0.1374163683, -0.08799716289, -0.06438761156, 0.1161207013, -0.01628106962, -0.09043332217, 0.07018737714, 0.03475858573, -0.08232419151, 0.02096991395, 0.05826561569, -0.05429447540, -0.01676701442, 0.05740545429, -0.02109011324, -0.03661760205, 0.04097118247, 0.006289072814, -0.03937175201, 0.01897586750, 0.02231625725, -0.03027024401, -0.0005449915345, 0.02654457485, -0.01599265972, -0.01306145534, 0.02194715859, -0.002304915226, -0.01757116509, 0.01289414356, 0.007214583147, -0.01563804064, 0.003453577948, 0.01139656310, -0.01005875429, -0.003627579188, 0.01095842705, -0.003656467365, -0.007219332120, 0.007643444972, 0.001509246192, -0.007553453767, 0.003378457203, 0.004443906045] The largest is 0.3002739726 The smallest is -0.2923976608 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 153, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 38, 181, 862, 4105, 19549, 93097, 443350, 2111338, 10054693, 47882836, 228029437, 1085930335, 5171458159, 24627711952, 117283013290, 558529563493, 2659850429698, 12666839449141, 60322497776137] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 3 a(n - 2) + 3 a(n - 3) with initial conditions, a(1) = 8, a(2) = 38, a(3) = 181, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 3 b(n - 2) + 3 b(n - 3) with initial conditions, b(1) = 8, b(2) = 38, b(3) = 181, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t - 3 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (1268 + 36 469 ) 50 (1268 + 36 469 ) [--------------------- + ----------------------- + 4/3, - --------------------- 6 1/2 1/3 12 3 (1268 + 36 469 ) 25 - ----------------------- + 4/3 1/2 1/3 3 (1268 + 36 469 ) / 1/2 1/3 \ 1/2 |(1268 + 36 469 ) 50 | + 1/2 I 3 |--------------------- - -----------------------|, | 6 1/2 1/3| \ 3 (1268 + 36 469 ) / 1/2 1/3 (1268 + 36 469 ) 25 - --------------------- - ----------------------- + 4/3 12 1/2 1/3 3 (1268 + 36 469 ) / 1/2 1/3 \ 1/2 |(1268 + 36 469 ) 50 | - 1/2 I 3 |--------------------- - -----------------------|] | 6 1/2 1/3| \ 3 (1268 + 36 469 ) / In floating-point [4.762237497, -0.381118749 + 0.6962072135 I, -0.381118749 - 0.6962072135 I] The largest root is, 4.762237497 and the remaining roots are [-0.381118749 + 0.6962072135 I, -0.381118749 - 0.6962072135 I] whose absolute values are [0.7936976660, 0.7936976660] so the largest absolute value is, 0.7936976660 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7936976660 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.1315789474, 0.2154696133, -0.2470997680, 0.05261875761, 0.1155557829, -0.1212283962, 0.01960978911, 0.06142124094, -0.05917087672, 0.006409582757, 0.03238942348, -0.02872618804, 0.001492266545, 0.01695877249, -0.01386667451, -0.0001135809331, 0.008821970210, -0.006652885492, -0.0004863741346, 0.004561757618, -0.003170748407, -0.0004568431768, 0.002345654927, -0.001500155045, -0.0003341849293, 0.001199759928, -0.0007039802078, -0.0002191958338, 0.0006105558268, -0.0003273048176, -0.0001351392912, 0.0003091958628, -0.0001505488752, -0.00008002578593, 0.0001558378192, -0.00006837270645, -0.00004605472596, 0.00007817643444, -0.00003057655947, -0.00002594111244, 0.00003903517514, -5 -0.00001341231517, -0.00001436707258, 0.00001940028960, -0.5737004840 10 , -5 -5 -5 -0.7848368293 10 , 0.9596381113 10 , -0.2370594949 10 , -5 -5 -6 -0.4238341334 10 , 0.4723993154 10 , -0.9308362322 10 , -5 -5 -6 -0.2266389470 10 , 0.2313912887 10 , -0.3360255590 10 , -5 -5 -6 -0.1201531985 10 , 0.1127534042 10 , -0.1025364627 10 ] The largest is 0.2154696133 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 154, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 8, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [8, 39, 190, 926, 4513, 21995, 107197, 522446, 2546245, 12409634, 60480832, 294765425, 1436598223, 7001548619, 34123446820, 166307439431, 810532551298, 3950292416030, 19252539761857, 93831101155391] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 3 a(n - 3) with initial conditions, a(1) = 8, a(2) = 39, a(3) = 190, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 3 b(n - 3) with initial conditions, b(1) = 8, b(2) = 39, b(3) = 190, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 5 t + 3 = 0 whose roots are 1/2 1/3 (676 + 36 I 419 ) 50 [---------------------- + ------------------------ + 5/3, 6 1/2 1/3 3 (676 + 36 I 419 ) 1/2 1/3 (676 + 36 I 419 ) 25 - ---------------------- - ------------------------ + 5/3 12 1/2 1/3 3 (676 + 36 I 419 ) / 1/2 1/3 \ 1/2 |(676 + 36 I 419 ) 50 | + 1/2 I 3 |---------------------- - ------------------------|, | 6 1/2 1/3| \ 3 (676 + 36 I 419 ) / 1/2 1/3 (676 + 36 I 419 ) 25 - ---------------------- - ------------------------ + 5/3 12 1/2 1/3 3 (676 + 36 I 419 ) / 1/2 1/3 \ 1/2 |(676 + 36 I 419 ) 50 | - 1/2 I 3 |---------------------- - ------------------------|] | 6 1/2 1/3| \ 3 (676 + 36 I 419 ) / In floating-point -9 -9 [4.873699903 - 0.1 10 I, -0.7239564893 + 0.1 10 I, -9 0.8502565877 + 0.1 10 I] -9 The largest root is, 4.873699903 - 0.1 10 I and the remaining roots are -9 -9 [-0.7239564893 + 0.1 10 I, 0.8502565877 + 0.1 10 I] whose absolute values are [0.7239564893, 0.8502565877] so the largest absolute value is, 0.8502565877 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8502565877 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1250000000, -0.3589743590, 0.03157894737, -0.2170626350, -0.007976955462, -0.1346215049, -0.02191292667, -0.08563373057, -0.02430402416, -0.05578133892, -0.02200550085, -0.03711543170, -0.01823314172, -0.02514920607, -0.01439973522, -0.01729925093, -0.01104863644, -0.01204397656, -0.008322130036, -0.008464740846, -0.006191774538, -0.005992482580, -0.004568190362, -0.004265628195, -0.003350693235, -0.003048895089, -0.002447590861, -0.002185874599, -0.001782687729, -0.001570666064, -0.001295706524, -0.001130469431, -0.0009403489619, -0.0008146252382, -0.0006817178979, -0.0005875426038, -0.0004938373046, -0.0004240328293, -0.0003575363352, -0.0003061697624, -0.0002587503238, -0.0002211426131, -0.0001872037782, -0.0001597679199, -0.0001354117601, -0.0001154474657, -0.00009793356912, -0.00008343256536, -0.00007082042957, -0.00006030144050, -0.00005120950643, -0.00004358624344, -0.00003702689571, -0.00003150595927, -0.00002677106600, -0.00002277464285, -0.00001935533647, -0.00001646348438] The largest is 0.1250000000 The smallest is -0.3589743590 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 25, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 11, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47] is a trivial linear sequence ------------------------------------------------------------------------- Theorem , 155, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 12, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 2) + a(n - 3) with initial conditions, a(1) = 9, a(2) = 12, a(3) = 16, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 2) + b(n - 3) with initial conditions, b(1) = 9, b(2) = 12, b(3) = 16, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [1.324717958, -0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., 0.3333333333, -0.4375000000, 0.3333333333, -0.1071428571, -0.1081081081, 0.2244897959, -0.2153846154, 0.1162790698, 0.008771929825, -0.09933774834, 0.1250000000, -0.09056603774, 0.02564102564, 0.03440860215, -0.06493506494, 0.06004901961, -0.03052728955, -0.004888268156, 0.02952029520, -0.03541583764, 0.02463202163, -0.005895691610, -0.01078397809, 0.01873627084, -0.01667967226, 0.007952286282, 0.002056583847, -0.008727394789, 0.01000886862, -0.006670811018, 0.001281472791, 0.003338056570, -0.005389338572, 0.004619529351, -0.002051282051, -0.0007698093173, 0.002568247246, -0.002821091377, 0.001798437928, -0.0002528441374, -0.001022653455, 0.001545593789, -0.001275497592, 0.0005229403335, 0.0002700961958, -0.0007525572592, 0.0007930365292, -0.0004824610634, 0.00004047926998, 0.0003105754658, -0.0004419817935, 0.0003510547358, -0.0001314063277, -0.00009092705769, 0.0002196484081, -0.0002223333854, 0.0001287213504] The largest is 0.3333333333 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 156, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 13, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 3) with initial conditions, a(1) = 9, a(2) = 13, a(3) = 19, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 3) with initial conditions, b(1) = 9, b(2) = 13, b(3) = 19, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / In floating-point [1.465571232, -0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2222222222, -0.2307692308, 0.2631578947, 0.03571428571, -0.1951219512, 0.06666666667, 0.1022727273, -0.09302325581, -0.02645502646, 0.07581227437, -0.01724137931, -0.04369747899, 0.03211009174, 0.01486697966, -0.02883075280, 0.003278688525, 0.01814566244, -0.01068521031, -0.007406550168, 0.01073910297, 0.00005387931034, -0.007352670858, 0.003386429199, 0.003440308087, -0.003912363067, -0.0005259341307, 0.002914373953, -0.0009979891816, -0.001523923318, 0.001390450627, 0.0003924614404, -0.001131461878, 0.0002589887477, 0.0006514501880, -0.0004800116899, -0.0002210229423, 0.0004304272457, -0.00004958444422, -0.0002706073865, 0.0001598198592, 0.0001102354150, -0.0001603719715, -6 -0.5521122886 10 , 0.0001096833027, -0.00005068866879, -0.00005124078108, -5 0.00005844252163, 0.7753852834 10 , -0.00004348692825, 0.00001495559338, -5 0.00002270944621, -0.00002077748203, -0.5821888655 10 , 0.00001688755756, -5 -5 -5 -0.3889924476 10 , -0.9711813131 10 , 0.7175744428 10 , -5 0.3285819952 10 ] The largest is 0.2631578947 The smallest is -0.2307692308 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 157, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 14, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 14, 22, 35, 56, 90, 145, 234, 378, 611, 988, 1598, 2585, 4182, 6766, 10947, 17712, 28658, 46369, 75026] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) with initial conditions, a(1) = 9, a(2) = 14, a(3) = 22, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) with initial conditions, b(1) = 9, b(2) = 14, b(3) = 22, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + 1 = 0 whose roots are 1/2 1/2 5 5 [1, 1/2 - ----, ---- + 1/2] 2 2 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [-0.6180339880, 1.618033988] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2222222222, -0.4285714286, -0.3181818182, -0.4000000000, -0.3571428571, -0.3888888889, -0.3724137931, -0.3846153846, -0.3783068783, -0.3829787234, -0.3805668016, -0.3823529412, -0.3814313346, -0.3821138211, -0.3817617499, -0.3820224719, -0.3818879855, -0.3819875776, -0.3819362074, -0.3819742489, -0.3819546271, -0.3819691578, -0.3819616629, -0.3819672131, -0.3819643503, -0.3819664703, -0.3819653768, -0.3819661866, -0.3819657689, -0.3819660782, -0.3819659187, -0.3819660368, -0.3819659759, -0.3819660210, -0.3819659977, -0.3819660150, -0.3819660061, -0.3819660127, -0.3819660093, -0.3819660118, -0.3819660105, -0.3819660115, -0.3819660110, -0.3819660113, -0.3819660111, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660112] The largest is -0.2222222222 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 26, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 15, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 15, 25, 42, 71, 120, 203, 343, 580, 981, 1659, 2806, 4746, 8027, 13576, 22961, 38834, 65680, 111085, 187879] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 158, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 16, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 16, 28, 49, 86, 151, 265, 465, 816, 1432, 2513, 4410, 7739, 13581, 23833, 41824, 73396, 128801, 226030, 396655] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 9, a(2) = 16, a(3) = 28, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 9, b(2) = 16, b(3) = 28, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (100 + 12 69 ) 2 (100 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (100 + 12 69 ) 1 - ----------------------- + 2/3 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (100 + 12 69 ) / 1/2 1/3 (100 + 12 69 ) 1 - ------------------- - ----------------------- + 2/3 12 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (100 + 12 69 ) / In floating-point [1.754877667, 0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] The largest root is, 1.754877667 and the remaining roots are [0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] whose absolute values are [0.7548776666, 0.7548776666] so the largest absolute value is, 0.7548776666 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7548776666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4444444444, 0., -0.2500000000, -0.06122448980, 0.1279069767, 0.06622516556, -0.05660377358, -0.05161290323, 0.01960784314, 0.03421787709, -0.002785515320, -0.02018140590, -0.003359607184, 0.01067668066, 0.004531531910, -0.004973221117, -0.003801297073, 0.001902159145, 0.002632393930, -0.0004386683642, -0.001607571533, -0.0001440807769, 0.0008807416143, 0.0002979924718, -0.0004288374476, -0.0002749257528, 0.0001769784138, 0.0002000451328, -0.00005181390101, -0.0001266945210, -5 -0.1530008233 10 , 0.00007182060353, 0.00001847669429, -0.00003639722319, -5 -0.00001945053714, 0.00001597284320, 0.00001499900035, -0.5425379648 10 , -5 -6 -5 -5 -0.9876916443 10 , 0.6705471088 10 , 0.5792631012 10 , 0.1037798473 10 , -5 -5 -5 -0.3046486958 10 , -0.1338141377 10 , 0.1408002678 10 , -5 -6 -6 0.1107659774 10 , -0.5308245068 10 , -0.7613061097 10 , -6 -6 -7 -6 0.1158720611 10 , 0.4622257251 10 , 0.4727327944 10 , -0.2518071051 10 , -7 -6 -7 -0.8866176457 10 , 0.1217568554 10 , 0.8036837027 10 , -7 -7 -7 -0.4968187944 10 , -0.5797527374 10 , 0.1409970224 10 ] The largest is 0.4444444444 The smallest is -0.2500000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 27, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 17, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 17, 32, 60, 113, 213, 401, 755, 1422, 2678, 5043, 9497, 17885, 33682, 63432, 119459, 224972, 423680, 797898, 1502646] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 28, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 19, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 19, 40, 84, 176, 369, 774, 1624, 3407, 7148, 14997, 31465, 66016, 138507, 290599, 609700, 1279199, 2683861, 5630953, 11814185] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 159, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 20, 44, 97, 214, 472, 1041, 2296, 5064, 11169, 24634, 54332, 119833, 264300, 582932, 1285697, 2835694, 6254320, 13794337, 30424368] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 3) with initial conditions, a(1) = 9, a(2) = 20, a(3) = 44, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 3) with initial conditions, b(1) = 9, b(2) = 20, b(3) = 44, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (172 + 12 177 ) 8 (172 + 12 177 ) [-------------------- + ---------------------- + 2/3, - -------------------- 6 1/2 1/3 12 3 (172 + 12 177 ) 4 - ---------------------- + 2/3 1/2 1/3 3 (172 + 12 177 ) / 1/2 1/3 \ 1/2 |(172 + 12 177 ) 8 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (172 + 12 177 ) / 1/2 1/3 (172 + 12 177 ) 4 - -------------------- - ---------------------- + 2/3 12 1/2 1/3 3 (172 + 12 177 ) / 1/2 1/3 \ 1/2 |(172 + 12 177 ) 8 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (172 + 12 177 ) / In floating-point [2.205569431, -0.1027847152 + 0.6654569515 I, -0.1027847152 - 0.6654569515 I] The largest root is, 2.205569431 and the remaining roots are [-0.1027847152 + 0.6654569515 I, -0.1027847152 - 0.6654569515 I] whose absolute values are [0.6733480912, 0.6733480912] so the largest absolute value is, 0.6733480912 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6733480912 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4444444444, -0.2000000000, -0.1590909091, 0.1237113402, 0.04672897196, -0.06567796610, -0.007684918348, 0.03135888502, -0.002962085308, -0.01360909661, 0.004140618657, 0.005319148936, -0.002970801031, -0.001800983731, 0.001717181421, 0.0004635617879, -0.0008738601556, -0.00003053889152, 0.0004024840048, -0.00006889214593, -0.0001683231834, 0.00006583763810, 0.00006278313027, -0.00004275692283, -0.00001967620755, -5 -6 0.00002343071517, 0.4104507522 10 , -0.00001146719251, 0.4963301608 10 , -5 -5 -5 0.5097167843 10 , -0.1272856820 10 , -0.2049383478 10 , -6 -6 -6 0.9984008870 10 , 0.7239449544 10 , -0.6014935694 10 , -6 -6 -7 -0.2045862517 10 , 0.3147724510 10 , 0.2805133261 10 , -6 -7 -7 -0.1484835865 10 , 0.1780527797 10 , 0.6366188856 10 , -7 -7 -7 -0.2115980939 10 , -0.2451434080 10 , 0.1463320696 10 , -8 -8 -8 0.8106604528 10 , -0.8301131744 10 , -0.1969056531 10 , -8 -10 -8 0.4168491467 10 , 0.3585119014 10 , -0.1897354150 10 , -9 -9 -9 0.3737831665 10 , 0.7834175232 10 , -0.3305191039 10 , -9 -9 -10 -0.2872550413 10 , 0.2089074407 10 , 0.8729577750 10 , -9 -10 -0.1126634863 10 , -0.1641953181 10 ] The largest is 0.4444444444 The smallest is -0.2000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 160, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 21, 49, 114, 265, 616, 1432, 3329, 7739, 17991, 41824, 97229, 226030, 525456, 1221537, 2839729, 6601569, 15346786, 35676949, 82938844] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 9, a(2) = 21, a(3) = 49, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 9, b(2) = 21, b(3) = 49, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + ------------------- + 1, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- + 1 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- + 1 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [2.324717958, 0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] The largest root is, 2.324717958 and the remaining roots are [0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] whose absolute values are [0.6558656185, 0.6558656185] so the largest absolute value is, 0.6558656185 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6558656185 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., 0.3333333333, 0.2244897959, 0.008771929825, -0.09056603774, -0.06493506494, -0.004888268156, 0.02463202163, 0.01873627084, 0.002056583847, -0.006670811018, -0.005389338572, -0.0007698093173, 0.001798437928, 0.001545593789, 0.0002700961958, -0.0004824610634, -0.0004419817935, -0.00009092705769, 0.0001287213504, 0.0001260363731, 0.00002973936087, -5 -5 -0.00003413331322, -0.00003584228830, -0.9520877576 10 , 0.8988630647 10 , -5 -5 -5 0.00001016535879, 0.2997937507 10 , -0.2348274416 10 , -0.2875339470 10 , -6 -6 -6 -6 -0.9315320716 10 , 0.6078083096 10 , 0.8111496018 10 , 0.2863001145 10 , -6 -6 -7 -0.1555905505 10 , -0.2282222787 10 , -0.8718562064 10 , -7 -7 -7 -8 0.3929714499 10 , 0.6404039755 10 , 0.2634128202 10 , -0.9759804042 10 , -7 -8 -8 -0.1792157862 10 , -0.7903845753 10 , 0.2371815938 10 , -8 -8 -9 0.5001560701 10 , 0.2357204473 10 , -0.5596920437 10 , -8 -9 -9 -0.1391924377 10 , -0.6991845699 10 , 0.1266030003 10 , -9 -9 -10 0.3862537638 10 , 0.2063707209 10 , -0.2679236467 10 , -9 -10 -11 -0.1068647720 10 , -0.6063886573 10 , 0.5020582097 10 , -10 -10 0.2947470577 10 , 0.1774408739 10 ] The largest is 0.3333333333 The smallest is -0.09056603774 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 29, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 22, 54, 133, 328, 809, 1995, 4920, 12134, 29926, 73806, 182027, 448931, 1107193, 2730656, 6734582, 16609414, 40963587, 101027975, 249164013] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 30, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 23, 59, 151, 386, 987, 2524, 6454, 16503, 42198, 107900, 275900, 705476, 1803901, 4612572, 11794339, 30158105, 77114224, 197180942, 504191339] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 31, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 24, 64, 171, 457, 1221, 3262, 8715, 23284, 62208, 166201, 444039, 1186338, 3169537, 8468046, 22624062, 60444662, 161489885, 431452209, 1152710021] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 161, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 25, 69, 190, 523, 1440, 3965, 10918, 30064, 82785, 227959, 627714, 1728490, 4759616, 13106205, 36089594, 99377264, 273647872, 753524044, 2074923809] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 3) + a(n - 5) with initial conditions, a(1) = 9, a(2) = 25, a(3) = 69, a(4) = 190, a(5) = 523, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 3) + b(n - 5) with initial conditions, b(1) = 9, b(2) = 25, b(3) = 69, b(4) = 190, b(5) = 523, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 2 t - 3 t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 2 %1 := _Z - 3 _Z + 2 _Z - 1 In floating-point [2.75362654389286, 0.753691611774574 + 0.381030390567161 I, -0.630504883721005 + 0.334115050746452 I, -0.630504883721005 - 0.334115050746452 I, 0.753691611774574 - 0.381030390567161 I] The largest root is, 2.75362654389286 and the remaining roots are [0.753691611774574 + 0.381030390567161 I, -0.630504883721005 + 0.334115050746452 I, -0.630504883721005 - 0.334115050746452 I, 0.753691611774574 - 0.381030390567161 I] whose absolute values are [0.844532535900849, 0.713560982349331, 0.713560982349331, 0.844532535900849] so the largest absolute value is, 0.844532535900849 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.844532535900849 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4444444444, 0.4400000000, 0.1884057971, -0.3736842105, -0.1816443595, -0.4826388889, -0.2612862547, -0.2320937901, -0.1048097392, 0.02646614725, 0.06091446269, 0.1310724311, 0.1081903858, 0.09793248027, 0.05811873078, 0.01888987723, -0.008122904249, -0.03241579017, -0.03709464512, -0.03691939611, -0.02703673075, -0.01504380625, -0.003708416719, 0.005853566218, 0.01072891505, 0.01256684785, 0.01094960486, 0.007682567756, 0.003767573786, 0.0001324266901, -0.002401007592, -0.003788565488, -0.003947982087, -0.003274357291, -0.002113514206, -0.0008455860375, 0.0002233909810, 0.0009492192690, 0.001264472591, 0.001233121606, 0.0009553402426, 0.0005604665258, 0.0001643756343, -0.0001530809907, -0.0003470544177, -0.0004145742793, -0.0003770943306, -0.0002727985219, -0.0001423279979, -0.00001984975024, 0.00007147351380, 0.0001219822066, 0.0001328475983, 0.0001132677694, 0.00007598914496, 0.00003374575207, -5 -0.3316076117 10 , -0.00002907891997] The largest is 0.4444444444 The smallest is -0.4826388889 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 162, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 26, 75, 216, 622, 1791, 5157, 14849, 42756, 123111, 354484, 1020696, 2938977, 8462447, 24366645, 70160958, 202020427, 581694636, 1674922950, 4822748423] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 3) with initial conditions, a(1) = 9, a(2) = 26, a(3) = 75, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 3) with initial conditions, b(1) = 9, b(2) = 26, b(3) = 75, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (4 + 4 I 3 ) 2 (4 + 4 I 3 ) [----------------- + ----------------- + 1, - ----------------- 2 1/2 1/3 4 (4 + 4 I 3 ) 1 - ------------------- + 1 1/2 (1/3) (4 + 4 I 3 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(4 + 4 I 3 ) 2 | (4 + 4 I 3 ) + 1/2 I 3 |----------------- - -----------------|, - ----------------- | 2 1/2 1/3| 4 \ (4 + 4 I 3 ) / 1 - ------------------- + 1 1/2 (1/3) (4 + 4 I 3 ) / 1/2 1/3 \ 1/2 |(4 + 4 I 3 ) 2 | - 1/2 I 3 |----------------- - -----------------|] | 2 1/2 1/3| \ (4 + 4 I 3 ) / In floating-point -9 -9 [2.879385242 - 0.1 10 I, -0.5320888864 + 0.2732050808 10 I, -10 0.6527036446 - 0.732050808 10 I] -9 The largest root is, 2.879385242 - 0.1 10 I and the remaining roots are -9 -10 [-0.5320888864 + 0.2732050808 10 I, 0.6527036446 - 0.732050808 10 I] whose absolute values are [0.5320888864, 0.6527036446] so the largest absolute value is, 0.6527036446 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6527036446 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1111111111, 0.3461538462, 0.08000000000, 0.1296296296, 0.04340836013, 0.05025125628, 0.02113631957, 0.02000134689, 0.009753017120, 0.008122751013, 0.004366910777, 0.003347715676, 0.001920396111, 0.001394277565, 0.0008351170216, 0.0005849549546, 0.0003605872984, 0.0002466448737, 0.0001549796664, 0.0001043517007, 0.00006641022846, 0.00004425101899, -5 0.00002840135627, 0.00001879384034, 0.00001213050203, 0.7990149825 10 , -5 -5 -5 -5 0.5176609134 10 , 0.3399325373 10 , 0.2207826294 10 , 0.1446869748 10 , -6 -6 -6 -6 0.9412838702 10 , 0.6160253166 10 , 0.4012062022 10 , 0.2623347364 10 , -6 -6 -7 -7 0.1709788926 10 , 0.1117304757 10 , 0.7285669058 10 , 0.4759117911 10 , -7 -7 -7 -8 0.3104306167 10 , 0.2027249444 10 , 0.1322630420 10 , 0.8635850917 10 , -8 -8 -8 -8 0.5635058314 10 , 0.3678870746 10 , 0.2400761321 10 , 0.1567225649 10 , -8 -9 -9 -9 0.1022806200 10 , 0.6676572784 10 , 0.4357461867 10 , 0.2844323605 10 , -9 -9 -10 0.1856398030 10 , 0.1211732221 10 , 0.7908730588 10 , -10 -10 -10 0.5162211470 10 , 0.3369312198 10 , 0.2199206006 10 , -10 -11 0.1435406547 10 , 0.9369074444 10 ] The largest is 0.3461538462 The smallest is -11 0.9369074444 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 163, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 28, 87, 270, 838, 2601, 8073, 25057, 77772, 241389, 749224, 2325444, 7217721, 22402387, 69532605, 215815536, 669848995, 2079079590, 6453054306, 20029011913] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 3) with initial conditions, a(1) = 9, a(2) = 28, a(3) = 87, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 3) with initial conditions, b(1) = 9, b(2) = 28, b(3) = 87, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (12 + 4 5 ) 2 (12 + 4 5 ) [---------------- + ---------------- + 1, - ---------------- 2 1/2 1/3 4 (12 + 4 5 ) 1 - ------------------ + 1 1/2 (1/3) (12 + 4 5 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(12 + 4 5 ) 2 | (12 + 4 5 ) + 1/2 I 3 |---------------- - ----------------|, - ---------------- | 2 1/2 1/3| 4 \ (12 + 4 5 ) / 1 - ------------------ + 1 1/2 (1/3) (12 + 4 5 ) / 1/2 1/3 \ 1/2 |(12 + 4 5 ) 2 | - 1/2 I 3 |---------------- - ----------------|] | 2 1/2 1/3| \ (12 + 4 5 ) / In floating-point [3.103803402, -0.051901701 + 0.5652358515 I, -0.051901701 - 0.5652358515 I] The largest root is, 3.103803402 and the remaining roots are [-0.051901701 + 0.5652358515 I, -0.051901701 - 0.5652358515 I] whose absolute values are [0.5676137370, 0.5676137370] so the largest absolute value is, 0.5676137370 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5676137370 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1111111111, 0.3214285714, -0.06896551724, -0.09629629630, 0.03221957041, 0.02768166090, -0.01325405673, -0.007542802410, 0.005053232526, 0.001905637788, -0.001825889187, -0.0004244350756, 0.0006323325604, -6 0.00007110849393, -0.0002111095938, -0.9962211432 10 , 0.00006811983050, -5 -5 -5 -0.6750102337 10 , -0.00002124652816, 0.4380246034 10 , 0.6390635765 10 , -5 -5 -6 -0.2074620860 10 , -0.1843616546 10 , 0.8597861282 10 , -6 -6 -6 0.5047375248 10 , -0.3294039712 10 , -0.1284257853 10 , -6 -7 -7 0.1194601688 10 , 0.2897653518 10 , -0.4149617980 10 , -8 -7 -9 -0.5028370600 10 , 0.1389142338 10 , 0.1780903489 10 , -8 -9 -8 -0.4494099553 10 , 0.4091247236 10 , 0.1405464520 10 , -9 -9 -9 -0.2777059937 10 , -0.4239932576 10 , 0.1334847469 10 , -9 -10 -10 0.1227482471 10 , -0.5574851626 10 , -0.3376080184 10 , -10 -11 -11 0.2146584159 10 , 0.8649008513 10 , -0.7813776299 10 , -11 -11 -12 -0.1975487307 10 , 0.2722546593 10 , 0.3538634781 10 , -12 -13 -12 -0.9138968725 10 , -0.1914402507 10 , 0.2964314029 10 , -13 -13 -13 -0.2460266379 10 , -0.9295201645 10 , 0.1757535356 10 , -13 -14 -14 0.2812339688 10 , -0.8581825811 10 , -0.8170123876 10 , -14 0.3613025251 10 ] The largest is 0.3214285714 The smallest is -0.09629629630 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 164, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 29, 93, 298, 955, 3060, 9805, 31418, 100672, 322581, 1033639, 3312066, 10612778, 34006284, 108965565, 349155890, 1118792304, 3584920820, 11487080524, 36807791745] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 3) + a(n - 5) with initial conditions, a(1) = 9, a(2) = 29, a(3) = 93, a(4) = 298, a(5) = 955, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 3) + b(n - 5) with initial conditions, b(1) = 9, b(2) = 29, b(3) = 93, b(4) = 298, b(5) = 955, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 2 t - 3 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 2 %1 := _Z - 3 _Z - 2 _Z - 1 In floating-point [3.20427733296502, 0.292959648460615 + 0.678807270244302 I, -0.395098314943126 + 0.644084883565252 I, -0.395098314943126 - 0.644084883565252 I, 0.292959648460615 - 0.678807270244302 I] The largest root is, 3.20427733296502 and the remaining roots are [0.292959648460615 + 0.678807270244302 I, -0.395098314943126 + 0.644084883565252 I, -0.395098314943126 - 0.644084883565252 I, 0.292959648460615 - 0.678807270244302 I] whose absolute values are [0.739327170989061, 0.755611021431108, 0.755611021431108, 0.739327170989061] so the largest absolute value is, 0.755611021431108 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.755611021431108 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4444444444, 0.2413793103, -0.1182795699, 0.4865771812, -0.1832460733, -0.3447712418, 0.1799082101, 0.05493666051, -0.03821320725, 0.06192243189, -0.04913320802, -0.04391820694, 0.04702689531, 0.004601061380, -0.01211079849, 0.008588186784, -0.008951523857, -0.004049273256, 0.009629615181, -0.001125000660, -0.002885361707, 0.001651621385, -0.001344410422, -0.0001743394973, 0.001655223617, -0.0006085116977, -0.0005225927032, 0.0003982587037, -0.0001965867815, 0.00002027786649, 0.0002488393092, -0.0001692483385, -0.00006893057874, 0.00009430010074, -5 -0.00003531850826, 0.5022626976 10 , 0.00003441974392, -0.00003630836349, -5 -5 -6 -0.4579735783 10 , 0.00001978177224, -0.8248783299 10 , 0.5139224586 10 , -5 -5 -6 -5 0.4796948357 10 , -0.6686457312 10 , 0.7502452167 10 , 0.3595849064 10 , -5 -7 -6 -0.2071444975 10 , 0.8310386527 10 , 0.7545524109 10 , -5 -6 -6 -0.1128987500 10 , 0.3750942935 10 , 0.5629427274 10 , -6 -7 -6 -0.4860429532 10 , 0.4661213828 10 , 0.1367343693 10 , -6 -7 -7 -0.1867885051 10 , 0.9580148868 10 , 0.7483025140 10 ] The largest is 0.4865771812 The smallest is -0.3447712418 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 32, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 30, 100, 333, 1109, 3693, 12298, 40953, 136376, 454140, 1512313, 5036092, 16770485, 55846710, 185972858, 619300652, 2062307918, 6867607736, 22869541257, 76156929372] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 165, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 31, 107, 369, 1273, 4392, 15153, 52280, 180373, 622311, 2147056, 7407630, 25557313, 88176144, 304219476, 1049597832, 3621252733, 12493805681, 43105298609, 148719038507] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 2) - 2 a(n - 3) + a(n - 4) + 2 a(n - 5) with initial conditions, a(1) = 9, a(2) = 31, a(3) = 107, a(4) = 369, a(5) = 1273, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 2) - 2 b(n - 3) + b(n - 4) + 2 b(n - 5) with initial conditions, b(1) = 9, b(2) = 31, b(3) = 107, b(4) = 369, b(5) = 1273, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 3 t - 2 t + 2 t - t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 3 _Z - 2 _Z + 2 _Z - _Z - 2 In floating-point [3.45013358696405, 0.598541772328301 + 0.666708275435757 I, -0.823608565810327 + 0.209270834834283 I, -0.823608565810327 - 0.209270834834283 I, 0.598541772328301 - 0.666708275435757 I] The largest root is, 3.45013358696405 and the remaining roots are [0.598541772328301 + 0.666708275435757 I, -0.823608565810327 + 0.209270834834283 I, -0.823608565810327 - 0.209270834834283 I, 0.598541772328301 - 0.666708275435757 I] whose absolute values are [0.895964384200860, 0.849779590239952, 0.849779590239952, 0.895964384200860] so the largest absolute value is, 0.895964384200860 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.895964384200860 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2222222222, 0.3225806452, -0.4672897196, -0.3224932249, -0.08248232522, -0.07991803279, 0.4178050551, 0.0009372609028, 0.2707334246, -0.2664166309, -0.001689755647, -0.2428240881, 0.07358942624, 0.01354978734, 0.1389533522, 0.05057717859, -0.009520085601, -0.04458396426, -0.08789349429, -0.005324357170, -0.01095785997, 0.06864055886, 0.01759324816, 0.03086523638, -0.03110548656, -0.02004764432, -0.02921001298, 0.0005373782341, 0.01391238358, 0.01897331573, 0.01436465626, -0.004666814638, -0.008230622796, -0.01595672730, -0.002686510483, 0.0005507574392, 0.01062845388, 0.005941924582, 0.003381201542, -0.004051717489, -0.005546629794, -0.004306895109, -0.0006454592302, 0.003253777274, 0.003434138807, 0.002700734732, -0.0007963221855, -0.001892915893, -0.002731168159, -0.0008176795453, 0.0004756041106, 0.001768229295, 0.001374255253, 0.0004280002650, -0.0006637022697, -0.001164179269, -0.0007652290316, -0.0001201303216] The largest is 0.4178050551 The smallest is -0.4672897196 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 166, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 32, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 32, 114, 406, 1446, 5150, 18342, 65326, 232662, 828638, 2951238, 10510990, 37435446, 133328318, 474855846, 1691224174, 6023384214, 21452600990, 76404571398, 272118916174] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 9, a(2) = 32, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 9, b(2) = 32, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 2 = 0 whose roots are 1/2 1/2 17 17 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.561552813, -0.561552813] The largest root is, 3.561552813 and the remaining roots are [-0.561552813] whose absolute values are [0.561552813] so the largest absolute value is, 0.561552813 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.561552813 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2222222222, 0.1250000000, -0.07017543860, 0.03940886700, -0.02213001383, 0.01242718447, -0.006978519245, 0.003918807213, -0.002200617204, 0.001235762782, -0.0006939460660, 0.0003896873653, -0.0002188300361, 0.0001228846223, -0.00006900620531, 0.00003875062869, -0.00002176052454, -5 -5 -5 0.00001221968376, -0.6861997789 10 , 0.3853374160 10 , -0.2163873098 10 , -5 -6 -6 0.1215129025 10 , -0.6823591219 10 , 0.3831806842 10 , -6 -6 -7 -0.2151761910 10 , 0.1208327953 10 , -0.6785399610 10 , -7 -7 -7 0.3810360237 10 , -0.2139718509 10 , 0.1201564947 10 , -8 -8 -8 -0.6747421759 10 , 0.3789033668 10 , -0.2127742514 10 , -8 -9 -9 0.1194839794 10 , -0.6709656470 10 , 0.3767826464 10 , -9 -9 -10 -0.2115833549 10 , 0.1188152281 10 , -0.6672102553 10 , -10 -10 -10 0.3746737956 10 , -0.2103991238 10 , 0.1181502198 10 , -11 -11 -11 -0.6634758826 10 , 0.3725767481 10 , -0.2092215209 10 , -11 -12 -12 0.1174889336 10 , -0.6597624111 10 , 0.3704914377 10 , -12 -12 -13 -0.2080505090 10 , 0.1168313485 10 , -0.6560697239 10 , -13 -13 -13 0.3684177988 10 , -0.2068860512 10 , 0.1161774440 10 , -14 -14 -14 -0.6523977046 10 , 0.3663557661 10 , -0.2057281109 10 , -14 0.1155271994 10 ] The largest is 0.1250000000 The smallest is -0.2222222222 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 33, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 33, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 33, 121, 444, 1629, 5977, 21930, 80463, 295225, 1083203, 3974354, 14582206, 53503219, 196307366, 720266606, 2642712774, 9696313487, 35576509170, 130532908865, 478935137097] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 34, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 34, 128, 482, 1815, 6834, 25732, 96888, 364810, 1373610, 5172019, 19474072, 73325230, 276089631, 1039553294, 3914203685, 14738051985, 55492813812, 208945686167, 786737899356] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 35, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 35, 136, 528, 2050, 7959, 30900, 119966, 465755, 1808243, 7020306, 27255571, 105816776, 410822069, 1594971788, 6192303667, 24040942287, 93336331215, 362368106070, 1406854571927] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 36, :The Pisot Sequence a(n), defined by, a(1) = 9, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [9, 37, 152, 624, 2562, 10519, 43189, 177326, 728068, 2989314, 12273576, 50393056, 206904662, 849512662, 3487943461, 14320857277, 58798818112, 241417182260, 991214751619, 4069746298211] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 167, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 38, 160, 674, 2839, 11958, 50368, 212154, 893609, 3763950, 15854048, 66778474, 281276087, 1184756590, 4990286208, 21019470706, 88535633097, 372918920630, 1570763279136, 6616176178226] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 4 a(n - 3) - a(n - 4) with initial conditions, a(1) = 9, a(2) = 38, a(3) = 160, a(4) = 674, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 4 b(n - 3) - b(n - 4) with initial conditions, b(1) = 9, b(2) = 38, b(3) = 160, b(4) = 674, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - 4 t - 4 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - 4 _Z - 4 _Z + 1 In floating-point [0.237412553236626, 4.21207718954655, -0.224744871391589 + 0.974417642894041 I, -0.224744871391589 - 0.974417642894041 I] The largest root is, 4.21207718954655 and the remaining roots are [0.237412553236626, -0.224744871391589 + 0.974417642894041 I, -0.224744871391589 - 0.974417642894041 I] whose absolute values are [0.237412553236626, 1., 1.] so the largest absolute value is, 1. that equals 1. It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 1. for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) is bounded. we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4444444444, -0.3157894737, 0.2250000000, 0.3397626113, -0.3480098626, -0.1762836595, 0.4289231258, -0.01611565184, -0.4215848318, 0.2056361004, 0.3291588369, -0.3535883584, -0.1702241968, 0.4301024601, -0.02310243044, -0.4197181506, 0.2117614349, 0.3245335579, -0.3576359403, -0.1637798710, 0.4312533125, -0.03006406945, -0.4177398216, 0.2178338344, 0.3198257474, -0.3615922274, -0.1572937501, 0.4322941546, -0.03701803824, -0.4156549262, 0.2238506641, 0.3150363487, -0.3654562714, -0.1507675033, 0.4332247177, -0.04396256365, -0.4134639963, 0.2298103890, 0.3101665837, -0.3692270869, -0.1442027953, 0.4340447643, -0.05089587411, -0.4111675909, 0.2357114888, 0.3052176945, -0.3729037118, -0.1376013010, 0.4347540851, -0.05781620094, -0.4087662959, 0.2415524581, 0.3001909436, -0.3764852081, -0.1309647042, 0.4353524994, -0.06472177874, -0.4062607237] The largest is 0.4444444444 The smallest is -0.4215848318 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 168, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 39, 169, 732, 3171, 13737, 59510, 257803, 1116827, 4838200, 20959539, 90798701, 393348542, 1704023007, 7381988487, 31979470816, 138538085703, 600160062081, 2599950030270, 11263228906739] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 3 a(n - 2) + 2 a(n - 4) with initial conditions, a(1) = 9, a(2) = 39, a(3) = 169, a(4) = 732, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 3 b(n - 2) + 2 b(n - 4) with initial conditions, b(1) = 9, b(2) = 39, b(3) = 169, b(4) = 732, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 5 t + 3 t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 5 _Z + 3 _Z - 2 In floating-point [4.33209437704822, 0.618607934207216 + 0.654410956421808 I, -0.569310245462654, 0.618607934207216 - 0.654410956421808 I] The largest root is, 4.33209437704822 and the remaining roots are [0.618607934207216 + 0.654410956421808 I, -0.569310245462654, 0.618607934207216 - 0.654410956421808 I] whose absolute values are [0.900516227587834, 0.569310245462654, 0.900516227587834] so the largest absolute value is, 0.900516227587834 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.900516227587834 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., 0.3333333333, -0.4437869822, -0.3319672131, -0.3282876064, 0.02103807236, 0.2023021341, 0.2844381175, 0.1587058694, -0.01770885867, -0.1600577665, -0.1782860638, -0.09384528239, 0.03021406213, 0.1124906245, 0.1152388085, 0.05103160416, -0.03013028041, -0.07876496550, -0.07295636933, -0.02642374179, 0.02648983822, 0.05419048547, 0.04557017403, 0.01243193015, -0.02157119488, -0.03677079391, -0.02800003685, -0.004823942223, 0.01673800968, 0.02462028724, 0.01688733345, 0.0009279211140, -0.01254637544, -0.01627506607, -0.009961537105, 0.0008733549041, 0.009158634955, 0.01062297793, 0.005715910579, -0.001542671090, -0.006543817278, -0.006845117257, -0.003162313295, 0.001638443117, 0.004591520913, 0.004352040701, 0.001661014174, -0.001474164996, -0.003170825677, -0.002727551998, -0.0008032546069, 0.001218052966, 0.002158377296, 0.001682623587, 0.0003314768323, -0.0009543806670, -0.001449579240] The largest is 0.3333333333 The smallest is -0.4437869822 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 169, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 9, a(2) = 40, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [9, 40, 178, 792, 3524, 15680, 69768, 310432, 1381264, 6145920, 27346208, 121676672, 541399104, 2408949760, 10718597248, 47692288512, 212206348544, 944209971200, 4201252581888, 18693430269952] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 9, a(2) = 40, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 9, b(2) = 40, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t - 2 = 0 whose roots are 1/2 1/2 [2 + 6 , 2 - 6 ] In floating-point [4.449489743, -0.449489743] The largest root is, 4.449489743 and the remaining roots are [-0.449489743] whose absolute values are [0.449489743] so the largest absolute value is, 0.449489743 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.449489743 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2222222222, 0.1000000000, -0.04494382022, 0.02020202020, -0.009080590238, 0.004081632653, -0.001834651989, 0.0008246572518, -0.0003706749760, 0.0001666145996, -0.00007489155352, 0.00003366298513, -0.00001513116653, -5 -5 -5 0.6801304150 10 , -0.3057116453 10 , 0.1374142488 10 , -6 -6 -6 -0.6176629535 10 , 0.2776331621 10 , -0.1247932586 10 , -7 -7 -7 0.5609328972 10 , -0.2521335837 10 , 0.1133314597 10 , -8 -8 -8 -0.5094132866 10 , 0.2289760472 10 , -0.1029223845 10 , -9 -9 -10 0.4626255615 10 , -0.2079454447 10 , 0.9346934443 10 , -10 -10 -11 -0.4201351159 10 , 0.1888464252 10 , -0.8488453108 10 , -11 -11 -12 0.3815472604 10 , -0.1715015799 10 , 0.7708820105 10 , -12 -12 -13 -0.3465035566 10 , 0.1557497945 10 , -0.7000793509 10 , -13 -13 -14 0.3146784873 10 , -0.1414447523 10 , 0.6357796535 10 , -14 -14 -15 -0.2857764329 10 , 0.1284535753 10 , -0.5773856453 10 , -15 -15 -16 0.2595289252 10 , -0.1166555898 10 , 0.5243549107 10 , -16 -16 -17 -0.2356921539 10 , 0.1059412056 10 , -0.4761948527 10 , -17 -18 -18 0.2140447019 10 , -0.9621089799 10 , 0.4324581179 10 , -18 -19 -19 -0.1943854882 10 , 0.8737428308 10 , -0.3927384403 10 , -19 -20 -20 0.1765319005 10 , -0.7934927855 10 , 0.3566668681 10 ] The largest is 0.1000000000 The smallest is -0.2222222222 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 37, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 12, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48 ] is a trivial linear sequence ------------------------------------------------------------------------- Fact, 38, : Consider the Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 13, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 10, 13, 17, 22, 28, 36, 46, 59, 76, 98, 126, 162, 208, 267, 343, 441, 567, 729, 937, 1204, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = a(n - 1) + a(n - 6), . Alas, it breaks down at the, 38, -th term. a(38), equals , 110155, while the corresponding term for the solution of the recurrence is , 110156 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03282504251949 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 170, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 14, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 14, 20, 29, 42, 61, 89, 130, 190, 278, 407, 596, 873, 1279, 1874, 2746, 4024, 5897, 8642, 12665] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 10, a(2) = 14, a(3) = 20, a(4) = 29, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 10, b(2) = 14, b(3) = 20, b(4) = 29, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t + t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [1, ------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.465571232, -0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4000000000, -0.4285714286, 0.05000000000, -0.1724137931, -0.4047619048, -0.1475409836, -0.1123595506, -0.3076923077, -0.2421052632, -0.1402877698, -0.2334152334, -0.2600671141, -0.1844215349, -0.2017200938, -0.2454642476, -0.2134013110, -0.1985586481, -0.2274037646, -0.2241379310, -0.2060007896, -0.2166909110, -0.2241011690, -0.2133647720, -0.2133125663, -0.2206663786, -0.2172807828, -0.2138410261, -0.2177537536, -0.2182799455, -0.2153657483, -0.2163638549, -0.2178878591, -0.2164974644, -0.2161050407, -0.2172365282, -0.2169775571, -0.2163261190, -0.2168061388, -0.2170271673, -0.2165967439, -0.2166463309, -0.2169169401, -0.2167571215, -0.2166468869, -0.2168072594, -0.2168078119, -0.2166981289, -0.2167488177, -0.2168000586, -0.2167416162, -0.2167338624, -0.2167773493, -0.2167623938, -0.2167396844, -0.2167604618, -0.2167662837, -0.2167493962, -0.2167532861] The largest is 0.05000000000 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 39, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 15, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 15, 23, 35, 53, 80, 121, 183, 277, 419, 634, 959, 1451, 2195, 3320, 5022, 7597, 11492, 17384, 26297] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 171, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 16, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 10, a(2) = 16, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 10, b(2) = 16, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4000000000, 0.2500000000, -0.1538461538, 0.09523809524, -0.05882352941, 0.03636363636, -0.02247191011, 0.01388888889, -0.008583690987, 0.005305039788, -0.003278688525, 0.002026342452, -0.001252348153, 0.0007739938080, -0.0004783544607, 0.0002956393200, -0.0001827151471, 0.0001129241714, -0.00006979097603, 0.00004313319531, -0.00002665778074, -5 -5 0.00001647541456, -0.00001018236618, 0.6293048384 10 , -0.3889317794 10 , -5 -5 -6 0.2403730590 10 , -0.1485587204 10 , 0.9181433855 10 , -6 -6 -6 -0.5674438188 10 , 0.3506995667 10 , -0.2167442521 10 , -6 -7 -7 0.1339553146 10 , -0.8278893743 10 , 0.5116637722 10 , -7 -7 -7 -0.3162256020 10 , 0.1954381702 10 , -0.1207874319 10 , -8 -8 -8 0.7465073831 10 , -0.4613669356 10 , 0.2851404475 10 , -8 -8 -9 -0.1762264881 10 , 0.1089139594 10 , -0.6731252874 10 , -9 -9 -9 0.4160143063 10 , -0.2571109811 10 , 0.1589033252 10 , -10 -10 -10 -0.9820765590 10 , 0.6069566930 10 , -0.3751198660 10 , -10 -10 -11 0.2318368270 10 , -0.1432830390 10 , 0.8855378808 10 , -11 -11 -11 -0.5472925087 10 , 0.3382453722 10 , -0.2090471365 10 , -11 -12 -12 0.1291982356 10 , -0.7984890090 10 , 0.4934933472 10 ] The largest is 0.2500000000 The smallest is -0.4000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 172, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 17, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 17, 29, 49, 83, 141, 240, 409, 697, 1188, 2025, 3452, 5885, 10033, 17105, 29162, 49718, 84764, 144514, 246382] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) + a(n - 5) - a(n - 6) with initial conditions, a(1) = 10, a(2) = 17, a(3) = 29, a(4) = 49, a(5) = 83, a(6) = 141, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) + b(n - 5) - b(n - 6) with initial conditions, b(1) = 10, b(2) = 17, b(3) = 29, b(4) = 49, b(5) = 83, b(6) = 141, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 3 t - 2 t + t - t + 1 = 0 whose roots are [1, RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 %1 := _Z - _Z - _Z - 1 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.70490277604165, 0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] The largest root is, 1.70490277604165 and the remaining roots are [0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] whose absolute values are [0.829475823582982, 0.923307702773949, 0.923307702773949, 0.829475823582982] so the largest absolute value is, 0.923307702773949 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.923307702773949 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.1000000000, 0.4705882353, -0.2068965517, -0.4081632653, -0.4698795181, -0.4893617021, 0.004166666667, -0.2029339853, -0.1162123386, -0.2954545455, -0.4054320988, -0.2001738123, -0.3119796092, -0.1316655038, -0.2423852675, -0.2826966600, -0.2284082224, -0.3262233967, -0.1894210942, -0.2611513828, -0.2363869665, -0.2290618245, -0.2947863021, -0.2163822362, -0.2754326428, -0.2313142787, -0.2389210400, -0.2681338343, -0.2265492575, -0.2732278565, -0.2342034935, -0.2494644815, -0.2549138947, -0.2340397150, -0.2652935457, -0.2366488323, -0.2545189369, -0.2491937408, -0.2408644694, -0.2584638325, -0.2390892107, -0.2551840565, -0.2465790844, -0.2457396866, -0.2538946798, -0.2418356534, -0.2540264661, -0.2455532802, -0.2484315092, -0.2509915455, -0.2443707845, -0.2525008724, -0.2455370134, -0.2495814714, -0.2492221067, -0.2462864389, -0.2511214943, -0.2460570229] The largest is 0.4705882353 The smallest is -0.4893617021 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 40, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 18, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 18, 32, 57, 102, 183, 328, 588, 1054, 1889, 3386, 6069, 10878, 19498, 34949, 62644, 112286, 201267, 360761, 646646] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 173, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 19, 36, 68, 128, 241, 454, 855, 1610, 3032, 5710, 10753, 20250, 38135, 71816, 135244, 254692, 479637, 903254, 1701011] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + 2 a(n - 3) - a(n - 4) with initial conditions, a(1) = 10, a(2) = 19, a(3) = 36, a(4) = 68, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + 2 b(n - 3) - b(n - 4) with initial conditions, b(1) = 10, b(2) = 19, b(3) = 36, b(4) = 68, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t + t - 2 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 2 _Z + _Z - 2 _Z + 1 In floating-point [0.531010056459569, 1.88320350591353, -0.207106781186548 + 0.978318343478516 I, -0.207106781186548 - 0.978318343478516 I] The largest root is, 1.88320350591353 and the remaining roots are [0.531010056459569, -0.207106781186548 + 0.978318343478516 I, -0.207106781186548 - 0.978318343478516 I] whose absolute values are [0.531010056459569, 1., 1.] so the largest absolute value is, 1. that equals 1. It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 1. for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) is bounded. we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1000000000, 0.2105263158, 0.4444444444, -0.05882352941, -0.2421875000, 0.2531120332, 0.1872246696, -0.3040935673, -0.04720496894, 0.3311345646, -0.08598949212, -0.2934064912, 0.2086419753, 0.2075783401, -0.2943076752, -0.08550471740, 0.3298140499, -0.05106153195, -0.3086385446, 0.1789171263, 0.2345357204, -0.2760612410, -0.1201854173, 0.3258447323, -0.01478332818, -0.3197209780, 0.1472162520, 0.2587420941, -0.2543906918, -0.1533699957, 0.3179186367, 0.02168379134, -0.3269003536, 0.1137227706, 0.2797948407, -0.2296175877, -0.1846841214, 0.3061162557, 0.05788661671, -0.3300936774, 0.07884266134, 0.2974359778, -0.2020446773, -0.2137463322, 0.2905813070, 0.09338361392, -0.3292620664, 0.04300119955, 0.3114503863, -0.1720081736, -0.2402022680, 0.2715032107, 0.1277419559, -0.3244156613, 0.006635410860, 0.3216671841, -0.1398743211, -0.2637293433] The largest is 0.4444444444 The smallest is -0.3300936774 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 41, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 21, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 21, 44, 92, 192, 401, 838, 1751, 3659, 7646, 15977, 33385, 69760, 145768, 304592, 636465, 1329935, 2778986, 5806873, 12133841] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 42, : Consider the Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 10, 22, 48, 105, 230, 504, 1104, 2418, 5296, 11600, 25408, 55652, 121896, 266992, 584800, 1280904, 2805600, 6145184, 13459968, 29481744, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) + 2 a(n - 4), . Alas, it breaks down at the, 23, -th term. a(23), equals , 309799425, while the corresponding term for the solution of the recurrence is , 309799424 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.01573535617675 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 43, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 23, 53, 122, 281, 647, 1490, 3431, 7901, 18195, 41901, 96493, 222212, 511728, 1178449, 2713829, 6249628, 14392156, 33143437, 76325425] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 174, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 24, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 24, 58, 140, 338, 816, 1970, 4756, 11482, 27720, 66922, 161564, 390050, 941664, 2273378, 5488420, 13250218, 31988856, 77227930, 186444716] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) with initial conditions, a(1) = 10, a(2) = 24, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) with initial conditions, b(1) = 10, b(2) = 24, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 2 t - 1 = 0 whose roots are 1/2 1/2 [1 + 2 , 1 - 2 ] In floating-point [2.414213562, -0.414213562] The largest root is, 2.414213562 and the remaining roots are [-0.414213562] whose absolute values are [0.414213562] so the largest absolute value is, 0.414213562 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.414213562 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4000000000, 0.1666666667, -0.06896551724, 0.02857142857, -0.01183431953, 0.004901960784, -0.002030456853, 0.0008410428932, -0.0003483713639, 0.0001443001443, -0.00005977107678, 0.00002475799064, -0.00001025509550, -5 -5 -6 0.4247799640 10 , -0.1759496221 10 , 0.7288071977 10 , -6 -6 -7 -0.3018818256 10 , 0.1250435464 10 , -0.5179473281 10 , -7 -8 -8 0.2145408079 10 , -0.8886571232 10 , 0.3680938327 10 , -8 -9 -9 -0.1524694577 10 , 0.6315491724 10 , -0.2615962325 10 , -9 -10 -10 0.1083567074 10 , -0.4488281777 10 , 0.1859107184 10 , -11 -11 -11 -0.7700674094 10 , 0.3189723649 10 , -0.1321226796 10 , -12 -12 -13 0.5472700578 10 , -0.2266866802 10 , 0.9389669735 10 , -13 -13 -14 -0.3889328550 10 , 0.1611012634 10 , -0.6673032822 10 , -14 -14 -15 0.2764060697 10 , -0.1144911428 10 , 0.4742378412 10 , -15 -16 -16 -0.1964357456 10 , 0.8136634996 10 , -0.3370304568 10 , -16 -17 -17 0.1396025861 10 , -0.5782528451 10 , 0.2395201709 10 , -18 -18 -18 -0.9921250326 10 , 0.4109516441 10 , -0.1702217445 10 , -19 -19 -19 0.7050815517 10 , -0.2920543413 10 , 0.1209728691 10 , -20 -20 -21 -0.5010860306 10 , 0.2075566298 10 , -0.8597277103 10 , -21 -21 -22 0.3561108775 10 , -0.1475059552 10 , 0.6109896717 10 ] The largest is 0.1666666667 The smallest is -0.4000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 175, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 25, 63, 159, 401, 1011, 2549, 6427, 16205, 40859, 103021, 259755, 654941, 1651355, 4163693, 10498251, 26470077, 66741115, 168279693, 424297003] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 10, a(2) = 25, a(3) = 63, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 10, b(2) = 25, b(3) = 63, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 2 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (27 + 3 78 ) 1 (27 + 3 78 ) [----------------- + ------------------- + 1, - ----------------- 3 1/2 (1/3) 6 (27 + 3 78 ) 1 - --------------------- + 1 1/2 (1/3) 2 (27 + 3 78 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(27 + 3 78 ) 1 | (27 + 3 78 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 (1/3)| 6 \ (27 + 3 78 ) / 1 - --------------------- + 1 1/2 (1/3) 2 (27 + 3 78 ) / 1/2 1/3 \ 1/2 |(27 + 3 78 ) 1 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 (1/3)| \ (27 + 3 78 ) / In floating-point [2.521379707, 0.2393101464 + 0.8578736270 I, 0.2393101464 - 0.8578736270 I] The largest root is, 2.521379707 and the remaining roots are [0.2393101464 + 0.8578736270 I, 0.2393101464 - 0.8578736270 I] whose absolute values are [0.8906270297, 0.8906270297] so the largest absolute value is, 0.8906270297 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8906270297 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, -0.2400000000, 0.2857142857, 0.3270440252, -0.06982543641, -0.2927794263, -0.08473911338, 0.1916913023, 0.1589632829, -0.07596857485, -0.1624523155, -0.01749340725, 0.1204871889, 0.07154367171, -0.06133017012, -0.08610348524, 0.007437228082, 0.07185831402, 0.02849351526, -0.04336162610, -0.04335528080, 0.01364444032, 0.04092063036, 0.008762448829, -0.02826503359, -0.02047873770, 0.01261875173, 0.02228366341, 0.0006560113662, -0.01736178926, -0.008830063711, 0.009545410126, 0.01157277927, -0.002032609855, -0.01015256786, -0.003246925319, 0.006499140050, 0.005686135071, -0.002433725524, -0.005675166615, -0.0007857786537, 0.004125546221, 0.002597862739, -0.002029061530, -0.003031817628, 0.0001583956542, 0.002480699159, 0.001061670911, -0.001459594275, -0.001540726330, 0.0004203513823, 0.001423318257, 0.0003477993471, -0.0009625357088, -0.0007365693059, 0.0004109621942, 0.0007809537767, 0.00004779833003] The largest is 0.3270440252 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 176, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 26, 68, 178, 466, 1220, 3194, 8362, 21892, 57314, 150050, 392836, 1028458, 2692538, 7049156, 18454930, 48315634, 126491972, 331160282, 866988874] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) with initial conditions, a(1) = 10, a(2) = 26, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) with initial conditions, b(1) = 10, b(2) = 26, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t + 1 = 0 whose roots are 1/2 1/2 5 5 [3/2 + ----, 3/2 - ----] 2 2 In floating-point [2.618033988, 0.381966012] The largest root is, 2.618033988 and the remaining roots are [0.381966012] whose absolute values are [0.381966012] so the largest absolute value is, 0.381966012 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.381966012 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4000000000, -0.1538461538, -0.05882352941, -0.02247191011, -0.008583690987, -0.003278688525, -0.001252348153, -0.0004783544607, -0.0001827151471, -5 -0.00006979097603, -0.00002665778074, -0.00001018236618, -0.3889317794 10 , -5 -6 -6 -0.1485587204 10 , -0.5674438188 10 , -0.2167442521 10 , -7 -7 -7 -0.8278893743 10 , -0.3162256020 10 , -0.1207874319 10 , -8 -8 -9 -0.4613669356 10 , -0.1762264881 10 , -0.6731252874 10 , -9 -10 -10 -0.2571109811 10 , -0.9820765590 10 , -0.3751198660 10 , -10 -11 -11 -0.1432830390 10 , -0.5472925087 10 , -0.2090471365 10 , -12 -12 -12 -0.7984890090 10 , -0.3049956618 10 , -0.1164979764 10 , -13 -13 -14 -0.4449826736 10 , -0.1699682569 10 , -0.6492209713 10 , -14 -15 -15 -0.2479803448 10 , -0.9472006318 10 , -0.3617984472 10 , -15 -16 -16 -0.1381947098 10 , -0.5278568206 10 , -0.2016233643 10 , -17 -17 -17 -0.7701327223 10 , -0.2941645241 10 , -0.1123608499 10 , -18 -18 -19 -0.4291802566 10 , -0.1639322707 10 , -0.6261655556 10 , -19 -20 -20 -0.2391739597 10 , -0.9135632337 10 , -0.3489501044 10 , -20 -21 -21 -0.1332870795 10 , -0.5091113411 10 , -0.1944632282 10 , -22 -22 -22 -0.7427834362 10 , -0.2837180264 10 , -0.1083706428 10 , -23 -23 -24 -0.4139390219 10 , -0.1581106371 10 , -0.6039288938 10 ] The largest is -24 -0.6039288938 10 The smallest is -0.4000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 44, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 27, 73, 197, 532, 1437, 3882, 10487, 28330, 76532, 206747, 558516, 1508801, 4075945, 11010947, 29745483, 80355828, 217076962, 586421777, 1584186997] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 45, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 28, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 28, 78, 217, 604, 1681, 4678, 13018, 36227, 100814, 280549, 780722, 2172622, 6046053, 16825180, 46821733, 130297250, 362596006, 1009045575, 2808009343] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 177, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 29, 84, 243, 703, 2034, 5885, 17027, 49264, 142535, 412395, 1193178, 3452209, 9988239, 28898864, 83612771, 241915927, 699932738, 2025107829, 5859222603] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 10, a(2) = 29, a(3) = 84, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 10, b(2) = 29, b(3) = 84, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (324 + 12 633 ) 4 (324 + 12 633 ) [-------------------- + -------------------- + 1, - -------------------- 6 1/2 1/3 12 (324 + 12 633 ) 2 - -------------------- + 1 1/2 1/3 (324 + 12 633 ) / 1/2 1/3 \ 1/2 |(324 + 12 633 ) 4 | + 1/2 I 3 |-------------------- - --------------------|, | 6 1/2 1/3| \ (324 + 12 633 ) / 1/2 1/3 (324 + 12 633 ) 2 - -------------------- - -------------------- + 1 12 1/2 1/3 (324 + 12 633 ) / 1/2 1/3 \ 1/2 |(324 + 12 633 ) 4 | - 1/2 I 3 |-------------------- - --------------------|] | 6 1/2 1/3| \ (324 + 12 633 ) / In floating-point [2.893289196, 0.0533554020 + 0.8297035535 I, 0.0533554020 - 0.8297035535 I] The largest root is, 2.893289196 and the remaining roots are [0.0533554020 + 0.8297035535 I, 0.0533554020 - 0.8297035535 I] whose absolute values are [0.8314173354, 0.8314173354] so the largest absolute value is, 0.8314173354 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8314173354 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1000000000, 0.3103448276, -0.03571428571, -0.2181069959, 0.001422475107, 0.1509341200, 0.01512319456, -0.1027192107, -0.02141523222, 0.06871996352, 0.02213654385, -0.04514079207, -0.02011900207, 0.02905687379, 0.01700803879, -0.01827076153, -0.01370657584, 0.01116711160, 0.01066638758, -0.006581100534, -0.008075465977, 0.003687477767, 0.005975698210, -0.001911315091, -0.004334687948, 0.0008586476664, 0.003088000766, -0.0002640212641, -0.002162769226, -0.00004828488028, 0.001489872057, 0.0001923625991, -0.001009354020, -0.0002406805457, 0.0006720375811, 0.0002380852491, -0.0004391429251, -0.0002114388621, 0.0002809968371, 0.0001761435234, -0.0001754439912, -0.0001404818226, 0.0001062855701, 0.0001084505506, -0.00006189756346, -0.00008157210074, 0.00003408236246, -5 0.00006002406121, -0.00001715438031, -0.00004332247721, 0.7235071102 10 , -5 -5 0.00003071892990, -0.1723235817 10 , -0.00002141849515, -0.1094389828 10 , -5 -5 0.00001468885403, 0.2323961624 10 , -0.9905748816 10 ] The largest is 0.3103448276 The smallest is -0.2181069959 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 46, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 31, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 31, 96, 297, 919, 2844, 8801, 27235, 84280, 260808, 807081, 2497545, 7728755, 23916948, 74011972, 229033069, 708752183, 2193262567, 6787140559, 21003083562] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 178, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 32, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 32, 102, 325, 1036, 3302, 10524, 33542, 106905, 340727, 1085963, 3461175, 11031437, 35159332, 112059619, 357155768, 1138324793, 3628062180, 11563338744, 36854606199] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 3) - a(n - 5) with initial conditions, a(1) = 10, a(2) = 32, a(3) = 102, a(4) = 325, a(5) = 1036, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 3) - b(n - 5) with initial conditions, b(1) = 10, b(2) = 32, b(3) = 102, b(4) = 325, b(5) = 1036, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 2 t - 3 t - 2 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 2 %1 := _Z - 3 _Z - 2 _Z + 1 In floating-point [0.592778961608464, 3.18719420185137, -0.107663024412886 + 0.962185739698060 I, -0.564647114634058, -0.107663024412886 - 0.962185739698060 I] The largest root is, 3.18719420185137 and the remaining roots are [0.592778961608464, -0.107663024412886 + 0.962185739698060 I, -0.564647114634058, -0.107663024412886 - 0.962185739698060 I] whose absolute values are [0.592778961608464, 0.968190438139126, 0.564647114634058, 0.968190438139126] so the largest absolute value is, 0.968190438139126 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.968190438139126 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4000000000, 0.1250000000, -0.4607843137, 0.4492307692, 0.3281853282, -0.3355542096, -0.2333713417, 0.4171188361, 0.1310883495, -0.4016588060, -0.03518167746, 0.3900031059, -0.05042670325, -0.3527318153, 0.1234696059, 0.3047370888, -0.1812554670, -0.2464004859, 0.2230045355, 0.1830330664, -0.2484388615, -0.1180520464, 0.2583104795, 0.05504918020, -0.2539896186, 0.003090964817, 0.2374233012, -0.05401981296, -0.2109266895, 0.09605615265, 0.1770378672, -0.1281630785, -0.1383571174, 0.1499310717, 0.09741090522, -0.1615193863, -0.05653293706, 0.1635801167, 0.01777050570, -0.1571652622, 0.01718383289, 0.1436254471, -0.04703429972, -0.1245057391, 0.07089893930, 0.1014443856, -0.08830376857, -0.07607912737, 0.09915712811, 0.04996490789, -0.1037079167, -0.02450572520, 0.1024917675, 0.0009023411755, -0.09626933477, 0.01988344743, 0.08596074983, -0.03714818757] The largest is 0.4492307692 The smallest is -0.4607843137 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 179, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930, 72171863277] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 2) with initial conditions, a(1) = 10, a(2) = 33, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 2) with initial conditions, b(1) = 10, b(2) = 33, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 1 = 0 whose roots are 1/2 1/2 13 13 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.302775638, -0.302775638] The largest root is, 3.302775638 and the remaining roots are [-0.302775638] whose absolute values are [0.302775638] so the largest absolute value is, 0.302775638 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.302775638 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.1000000000, 0.03030303030, -0.009174311927, 0.002777777778, -0.0008410428932, 0.0002546473135, -0.00007710100231, 0.00002334430516, -5 -5 -6 -0.7068086881 10 , 0.2140044513 10 , -0.6479533422 10 , -6 -7 -7 0.1961844864 10 , -0.5939988298 10 , 0.1798483745 10 , -8 -8 -9 -0.5445370629 10 , 0.1648725565 10 , -0.4991939343 10 , -9 -10 -10 0.1511437618 10 , -0.4576264887 10 , 0.1385581520 10 , -11 -11 -12 -0.4195203283 10 , 0.1270205349 10 , -0.3845872347 10 , -12 -13 -13 0.1164436452 10 , -0.3525629895 10 , 0.1067474840 10 , -14 -15 -15 -0.3232053754 10 , 0.9785871365 10 , -0.2962923443 10 , -16 -16 -17 0.8971010351 10 , -0.2716203380 10 , 0.8224002107 10 , -17 -18 -18 -0.2490027483 10 , 0.7539196590 10 , -0.2282685055 10 , -19 -19 -20 0.6911414234 10 , -0.2092607852 10 , 0.6335906770 10 , -20 -21 -21 -0.1918358213 10 , 0.5808321313 10 , -0.1758618190 10 , -22 -22 -23 0.5324667439 10 , -0.1612179580 10 , 0.4881287004 10 , -23 -24 -24 -0.1477934786 10 , 0.4474826472 10 , -0.1354868439 10 , -25 -25 -26 0.4102211556 10 , -0.1242049720 10 , 0.3760623961 10 , -26 -27 -27 -0.1138625318 10 , 0.3447480068 10 , -0.1043812976 10 , -28 -29 -29 0.3160411395 10 , -0.9568955757 10 , 0.2897246682 10 , -30 -30 -0.8772157117 10 , 0.2655995465 10 ] The largest is 0.03030303030 The smallest is -0.1000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 180, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 34, 116, 396, 1352, 4616, 15760, 53808, 183712, 627232, 2141504, 7311552, 24963200, 85229696, 290992384, 993510144, 3392055808, 11581202944, 39540700160, 135000394752] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 2 a(n - 2) with initial conditions, a(1) = 10, a(2) = 34, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 2 b(n - 2) with initial conditions, b(1) = 10, b(2) = 34, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 4 t + 2 = 0 whose roots are 1/2 1/2 [2 + 2 , 2 - 2 ] In floating-point [3.414213562, 0.585786438] The largest root is, 3.414213562 and the remaining roots are [0.585786438] whose absolute values are [0.585786438] so the largest absolute value is, 0.585786438 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.585786438 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4000000000, -0.2352941176, -0.1379310345, -0.08080808081, -0.04733727811, -0.02772963605, -0.01624365482, -0.009515313708, -0.005573941822, -0.003265139534, -0.001912674457, -0.001120418757, -0.0006563261120, -0.0003844669351, -0.0002252155163, -0.0001319281950, -0.00007728174736, -5 -0.00004527059948, -0.00002651890320, -0.00001553441384, -0.9099848941 10 , -5 -5 -5 -0.5330568094 10 , -0.3122574494 10 , -0.1829161789 10 , -5 -6 -6 -0.1071498168 10 , -0.6276690950 10 , -0.3676800432 10 , -6 -6 -7 -0.2153819827 10 , -0.1261678444 10 , -0.7390741209 10 , -7 -7 -7 -0.4329395964 10 , -0.2536101439 10 , -0.1485613827 10 , -8 -8 -8 -0.8702524316 10 , -0.5097820718 10 , -0.2986234238 10 , -8 -8 -9 -0.1749295516 10 , -0.1024713589 10 , -0.6002633227 10 , -9 -9 -9 -0.3516261135 10 , -0.2059778084 10 , -0.1206590066 10 , -10 -10 -10 -0.7068040965 10 , -0.4140362538 10 , -0.2425368221 10 , -10 -11 -11 -0.1420747810 10 , -0.8322547986 10 , -0.4875235737 10 , -11 -11 -12 -0.2855846975 10 , -0.1672916426 10 , -0.9799717535 10 , -12 -12 -12 -0.5740541624 10 , -0.3362731428 10 , -0.1969842464 10 , -12 -13 -13 -0.1153907000 10 , -0.6759430707 10 , -0.3959582834 10 , -13 -0.2319469923 10 ] The largest is -13 -0.2319469923 10 The smallest is -0.4000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 181, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 35, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 35, 123, 432, 1517, 5327, 18706, 65687, 230663, 809984, 2844297, 9987883, 35072922, 123160219, 432482915, 1518684144, 5332930965, 18726838487, 65760176162, 230919958639] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 10, a(2) = 35, a(3) = 123, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 10, b(2) = 35, b(3) = 123, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (332 + 12 321 ) 20 (332 + 12 321 ) [-------------------- + ---------------------- + 4/3, - -------------------- 6 1/2 1/3 12 3 (332 + 12 321 ) 10 - ---------------------- + 4/3 1/2 1/3 3 (332 + 12 321 ) / 1/2 1/3 \ 1/2 |(332 + 12 321 ) 20 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (332 + 12 321 ) / 1/2 1/3 (332 + 12 321 ) 10 - -------------------- - ---------------------- + 4/3 12 1/2 1/3 3 (332 + 12 321 ) / 1/2 1/3 \ 1/2 |(332 + 12 321 ) 20 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (332 + 12 321 ) / In floating-point [3.511547142, 0.244226429 + 0.4744767784 I, 0.244226429 - 0.4744767784 I] The largest root is, 3.511547142 and the remaining roots are [0.244226429 + 0.4744767784 I, 0.244226429 - 0.4744767784 I] whose absolute values are [0.5336429161, 0.5336429161] so the largest absolute value is, 0.5336429161 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5336429161 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2571428571, 0.2682926829, 0.05787037037, -0.04812129202, -0.03998498217, -0.005827007377, 0.008540502687, 0.005831017545, -5 0.0004160576011, -0.001457302103, -0.0008303060819, 0.9437480002 10 , 0.0002410599806, 0.0001150588804, -0.00001244695948, -0.00003884561817, -5 -5 -5 -0.00001542967331, 0.3525583621 10 , 0.6116062935 10 , 0.1983411187 10 , -6 -6 -6 -0.7728975005 10 , -0.9423494412 10 , -0.2401915767 10 , -6 -6 -7 -7 0.1510350753 10 , 0.1421740132 10 , 0.2643432557 10 , -0.2757564882 10 , -7 -8 -8 -0.2099723325 10 , -0.2403309778 10 , 0.4805578562 10 , -8 -9 -9 0.3031700553 10 , 0.1123353118 10 , -0.8084812981 10 , -9 -10 -9 -0.4268952626 10 , 0.2171685770 10 , 0.1321766578 10 , -10 -11 -10 0.5837765334 10 , -0.9125844580 10 , -0.2108202718 10 , -11 -11 -11 -0.7698766221 10 , 0.2243144898 10 , 0.3288084852 10 , -12 -12 -12 0.9672833908 10 , -0.4638912426 10 , -0.5020469004 10 , -12 -13 -13 -0.1131217255 10 , 0.8771565620 10 , 0.7505917540 10 , -13 -13 -13 0.1168366368 10 , -0.1566803986 10 , -0.1098031139 10 , -15 -14 -14 -0.9015021699 10 , 0.2686574247 10 , 0.1568989935 10 , -17 -15 -15 0.1309078047 10 , -0.4461693118 10 , -0.2183054680 10 ] The largest is 0.2682926829 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 47, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 36, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 36, 130, 469, 1692, 6104, 22021, 79444, 286606, 1033974, 3730216, 13457313, 48549273, 175148777, 631875457, 2279585390, 8223945863, 29669116961, 107035785000, 386147632423] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 48, :The Pisot Sequence a(n), defined by, a(1) = 10, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [10, 37, 137, 507, 1876, 6942, 25688, 95055, 351738, 1301558, 4816236, 17821818, 65947183, 244028468, 902993737, 3341403959, 12364405156, 45752778394, 169301855153, 626478197923] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 182, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 38, 144, 546, 2070, 7848, 29754, 112806, 427680, 1621458, 6147414, 23306616, 88362090, 335006118, 1270104624, 4815332226, 18256310550, 69214928328, 262413716634, 994885934886] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 3 a(n - 2) with initial conditions, a(1) = 10, a(2) = 38, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 3 b(n - 2) with initial conditions, b(1) = 10, b(2) = 38, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t - 3 = 0 whose roots are 1/2 1/2 21 21 [3/2 + -----, 3/2 - -----] 2 2 In floating-point [3.791287848, -0.791287848] The largest root is, 3.791287848 and the remaining roots are [-0.791287848] whose absolute values are [0.791287848] so the largest absolute value is, 0.791287848 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.791287848 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4000000000, -0.3157894737, 0.2500000000, -0.1978021978, 0.1565217391, -0.1238532110, 0.09800362976, -0.07754906654, 0.06136363636, -0.04855629933, 0.03842200965, -0.03040286930, 0.02405742100, -0.01903634488, 0.01506322837, -0.01191934955, 0.009431636449, -0.007463139304, 0.005905491435, -0.004672943606, 0.003697643487, -0.002925900356, 0.002315229394, -0.001832012884, 0.001449649532, -0.001147090057, 0.0009076784224, -0.0007182349050, 0.0005683305520, -0.0004497130591, 0.0003558524786, -0.0002815817418, 0.0002228122103, -0.0001763085943, 0.0001395108481, -0.0001103932387, 0.00008735282821, -0.00006912123141, 0.00005469479041, -0.00004327932297, 0.00003424640232, -0.00002709876197, 0.00002144292103, -5 -0.00001696752283, 0.00001342619461, -0.00001062398464, 0.8406629934 10 , -5 -5 -5 -0.6652064105 10 , 0.5263697487 10 , -0.4165099854 10 , -5 -5 -5 0.3295792898 10 , -0.2607920868 10 , 0.2063616090 10 , -5 -5 -5 -0.1632914334 10 , 0.1292105269 10 , -0.1022427197 10 , -6 -6 0.8090342156 10 , -0.6401789430 10 ] The largest is 0.4000000000 The smallest is -0.3157894737 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 183, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 10, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [10, 39, 152, 592, 2306, 8982, 34985, 136267, 530762, 2067326, 8052266, 31363698, 122162079, 475823149, 1853338376, 7218781060, 28117261622, 109517160106, 426570998233, 1661500502363] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 5 a(n - 1) - 5 a(n - 2) + 3 a(n - 3) - a(n - 4) - a(n - 5) + 2 a(n - 6) with initial conditions, a(1) = 10, a(2) = 39, a(3) = 152, a(4) = 592, a(5) = 2306, a(6) = 8982, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 5 b(n - 1) - 5 b(n - 2) + 3 b(n - 3) - b(n - 4) - b(n - 5) + 2 b(n - 6) with initial conditions, b(1) = 10, b(2) = 39, b(3) = 152, b(4) = 592, b(5) = 2306, b(6) = 8982, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 2 t - 5 t + 5 t - 3 t + t + t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 3 2 %1 := _Z - 5 _Z + 5 _Z - 3 _Z + _Z + _Z - 2 In floating-point [3.89501515397323, 0.867858748597485 + 0.462544571094097 I, 0.00461647496971950 + 0.910824982777052 I, -0.639965601107643, 0.00461647496971950 - 0.910824982777052 I, 0.867858748597485 - 0.462544571094097 I] The largest root is, 3.89501515397323 and the remaining roots are [0.867858748597485 + 0.462544571094097 I, 0.00461647496971950 + 0.910824982777052 I, -0.639965601107643, 0.00461647496971950 - 0.910824982777052 I, 0.867858748597485 - 0.462544571094097 I] whose absolute values are [0.983425791692396, 0.910836681898551, 0.639965601107643, 0.910836681898551, 0.983425791692396] so the largest absolute value is, 0.983425791692396 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.983425791692396 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1000000000, 0.4102564103, -0.3157894737, 0.4932432432, 0.3963573287, 0.003451347139, -0.3796198371, -0.08364460948, -0.03092911701, -0.2886782249, -0.3708036967, -0.03324738684, 0.1770802951, 0.09154543257, 0.07020721617, 0.2512444288, 0.2943820569, 0.09118928761, -0.06982261853, -0.06027413989, -0.08390179733, -0.2106886297, -0.2473591367, -0.1225825936, -0.003652473146, 0.02661533922, 0.07383545281, 0.1737076195, 0.2107236442, 0.1384584290, 0.05204104412, 0.005771614249, -0.05273222031, -0.1381628748, -0.1788906144, -0.1427311591, -0.08264865368, -0.03382099247, 0.02753387732, 0.1041244118, 0.1490882789, 0.1384282956, 0.1000631264, 0.05613871679, -0.002482097329, -0.07218244236, -0.1204004389, -0.1278815268, -0.1074831330, -0.07226737489, -0.01994710289, 0.04306904179, 0.09284238076, 0.1130128406, 0.1073076362, 0.08234443165, 0.03841687069, -0.01743203396] The largest is 0.4932432432 The smallest is -0.3796198371 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 49, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 13, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49 ] is a trivial linear sequence ------------------------------------------------------------------------- Fact, 50, : Consider the Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 14, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 11, 14, 18, 23, 29, 37, 47, 60, 77, 99, 127, 163, 209, 268, 344, 442, 568, 730, 938, 1205, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 2) + a(n - 6) - a(n - 7), . Alas, it breaks down at the, 30, -th term. a(30), equals , 14799, while the corresponding term for the solution of the recurrence is , 14800 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03282504251949 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 184, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 15, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 15, 20, 27, 36, 48, 64, 85, 113, 150, 199, 264, 350, 464, 615, 815, 1080, 1431, 1896, 2512] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) - a(n - 4) with initial conditions, a(1) = 11, a(2) = 15, a(3) = 20, a(4) = 27, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) - b(n - 4) with initial conditions, b(1) = 11, b(2) = 15, b(3) = 20, b(4) = 27, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [1, ------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.324717958, -0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4545454545, -0.3333333333, 0.4500000000, 0., 0., 0.3333333333, -0.1093750000, 0.2235294118, 0.1150442478, 0.006666666667, 0.2311557789, 0.01515151515, 0.1314285714, 0.1400862069, 0.04065040650, 0.1656441718, 0.07500000000, 0.1006289308, 0.1350210970, 0.07006369427, 0.1301081731, 0.09956906328, 0.09467556925, 0.1241922978, 0.08877182714, 0.1134020619, 0.1075041690, 0.09671869755, 0.1154540559, 0.09877340219, 0.1067251462, 0.1087812521, 0.1000534825, 0.1100621118, 0.1033910821, 0.1046724093, 0.1080103532, 0.1026209240, 0.1072403917, 0.1051890587, 0.1044192125, 0.1069874322, 0.1041663191, 0.1059647414, 0.1057118851, 0.1046892225, 0.1062348095, 0.1049593066, 0.1054822431, 0.1057523363, 0.1049997768, 0.1057928117, 0.1053103493, 0.1053508277, 0.1056614024, 0.1052194201, 0.1055704744, 0.1054390677] The largest is 0.4545454545 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 51, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 16, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 16, 23, 33, 47, 67, 96, 138, 198, 284, 407, 583, 835, 1196, 1713, 2453, 3513, 5031, 7205, 10318] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 52, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 17, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 17, 26, 40, 62, 96, 149, 231, 358, 555, 860, 1333, 2066, 3202, 4963, 7692, 11922, 18478, 28639, 44388] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 185, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 18, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 11, a(2) = 18, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 11, b(2) = 18, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4545454545, -0.2777777778, 0.1724137931, -0.1063829787, 0.06578947368, -0.04065040650, 0.02512562814, -0.01552795031, 0.009596928983, -0.005931198102, 0.003665689150, -0.002265518804, 0.001400168020, -0.0008653513326, 0.0005348165579, -0.0003305348053, 0.0002042817454, -0.0001262530616, 0.00007802868334, -0.00004822437839, 0.00002980430493, -5 -5 -0.00001842007346, 0.00001138423147, -0.7035841986 10 , 0.4348389487 10 , -5 -5 -5 -0.2687452499 10 , 0.1660936988 10 , -0.1026515512 10 , -6 -6 -6 0.6344214761 10 , -0.3920940354 10 , 0.2423274407 10 , -6 -7 -7 -0.1497665948 10 , 0.9256084594 10 , -0.5720574882 10 , -7 -7 -7 0.3535509712 10 , -0.2185065170 10 , 0.1350444542 10 , -8 -8 -8 -0.8346206272 10 , 0.5158239153 10 , -0.3187967119 10 , -8 -8 -9 0.1970272034 10 , -0.1217695084 10 , 0.7525769500 10 , -9 -9 -9 -0.4651181343 10 , 0.2874588158 10 , -0.1776593185 10 , -9 -10 -10 0.1097994973 10 , -0.6785982125 10 , 0.4193967600 10 , -10 -10 -11 -0.2592014525 10 , 0.1601953076 10 , -0.9900614491 10 , -11 -11 -11 0.6118916265 10 , -0.3781698226 10 , 0.2337218039 10 , -11 -12 -12 -0.1444480187 10 , 0.8927378517 10 , -0.5517423354 10 ] The largest is 0.4545454545 The smallest is -0.2777777778 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 186, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 19, 33, 57, 98, 168, 288, 494, 847, 1452, 2489, 4267, 7315, 12540, 21497, 36852, 63175, 108300, 185657, 318269] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 4) - a(n - 6) with initial conditions, a(1) = 11, a(2) = 19, a(3) = 33, a(4) = 57, a(5) = 98, a(6) = 168, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 4) - b(n - 6) with initial conditions, b(1) = 11, b(2) = 19, b(3) = 33, b(4) = 57, b(5) = 98, b(6) = 168, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 2 t - t - t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 2 %1 := _Z - _Z - _Z - _Z + 1 In floating-point [0.768085398012510, 1.71428532914257, 0.112851094827483 + 0.965281519117545 I, -0.854036458405025 + 0.273331145162114 I, -0.854036458405025 - 0.273331145162114 I, 0.112851094827483 - 0.965281519117545 I] The largest root is, 1.71428532914257 and the remaining roots are [0.768085398012510, 0.112851094827483 + 0.965281519117545 I, -0.854036458405025 + 0.273331145162114 I, -0.854036458405025 - 0.273331145162114 I, 0.112851094827483 - 0.965281519117545 I] whose absolute values are [0.768085398012510, 0.971855843607290, 0.896709644868745, 0.896709644868745, 0.971855843607290] so the largest absolute value is, 0.971855843607290 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.971855843607290 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.1818181818, 0.3157894737, 0.4545454545, 0.4912280702, 0., -0.2857142857, 0.3472222222, 0.2449392713, 0.1428571429, -0.3877410468, 0.1020490157, 0.2449027420, 0.1428571429, -0.2448963317, -0.1428571429, 0.2448985130, 0.1428571429, -0.1020406279, -0.2448978493, 0.1428571429, 0.1836735068, -6 -0.02040796689, -0.2244894777, 0.3638087428 10 , 0.2040822563, 0.04081754933, -0.1632631780, -0.1020372950, 0.1632712643, 0.1020511557, -0.1020230130, -0.1428267016, 0.08168472803, 0.1429464771, -0.04066307203, -0.1425944522, 0.0004502168399, 0.1436289433, 0.02173136012, -0.1201806259, -0.05733597690, 0.1087067928, 0.07265195915, -0.08245081730, -0.08886619517, 0.05757040619, 0.09869214706, -0.03489505682, -0.09772106408, 0.007405102586, 0.09724238074, 0.01218202032, -0.08698881008, -0.03250663035, 0.07546800440, 0.04773829179, -0.06102489463, -0.05797525351] The largest is 0.4912280702 The smallest is -0.3877410468 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 187, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 20, 36, 65, 117, 211, 381, 688, 1242, 2242, 4047, 7305, 13186, 23802, 42965, 77556, 139996, 252706, 456158, 823408] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) + a(n - 6) with initial conditions, a(1) = 11, a(2) = 20, a(3) = 36, a(4) = 65, a(5) = 117, a(6) = 211, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) + b(n - 6) with initial conditions, b(1) = 11, b(2) = 20, b(3) = 36, b(4) = 65, b(5) = 117, b(6) = 211, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 t - 2 t + t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 3 %1 := _Z - 2 _Z + _Z - _Z - 1 In floating-point [1.80509425159131, 0.647170796637930 + 0.726361507447275 I, -0.202712494937727 + 0.895728484592812 I, -0.694010854991715, -0.202712494937727 - 0.895728484592812 I, 0.647170796637930 - 0.726361507447275 I] The largest root is, 1.80509425159131 and the remaining roots are [0.647170796637930 + 0.726361507447275 I, -0.202712494937727 + 0.895728484592812 I, -0.694010854991715, -0.202712494937727 - 0.895728484592812 I, 0.647170796637930 - 0.726361507447275 I] whose absolute values are [0.972846894183278, 0.918380026848806, 0.694010854991715, 0.918380026848806, 0.972846894183278] so the largest absolute value is, 0.972846894183278 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.972846894183278 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3636363636, -0.2000000000, 0.3611111111, -0.4000000000, -0.4786324786, -0.03317535545, 0.3727034121, 0.09883720930, 0.1529790660, 0.1779661017, -0.1771682728, -0.4125941136, -0.09752768087, 0.1391899840, 0.1162806936, 0.1738098922, 0.1933483814, -0.08343292205, -0.2839323217, -0.1518943221, 0.01299106725, 0.06775379767, 0.1639705778, 0.1897454118, -0.0006583536292, -0.1789858690, -0.1545769096, -0.06307251040, 0.01341659383, 0.1250741996, 0.1730009409, 0.05535840629, -0.09178683851, -0.1290036528, -0.09744546703, -0.03259992018, 0.07624291475, 0.1429986889, 0.08536770444, -0.02502401813, -0.08986251879, -0.1019332352, -0.06278505496, 0.02949929541, 0.1052181150, 0.09312786159, 0.02067438476, -0.04849421221, -0.08732000255, -0.07597211272, -0.007900320041, 0.06597933168, 0.08456125544, 0.04674864695, -0.01240463241, -0.06296876904, -0.07468457877, -0.03282568923] The largest is 0.3727034121 The smallest is -0.4786324786 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 53, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 21, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 21, 40, 76, 144, 273, 518, 983, 1865, 3538, 6712, 12733, 24155, 45823, 86928, 164906, 312833, 593456, 1125808, 2135699] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 188, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 23, 48, 100, 208, 433, 901, 1875, 3902, 8120, 16898, 35165, 73179, 152287, 316912, 659500, 1372432, 2856057, 5943509, 12368555] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + 2 a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 11, a(2) = 23, a(3) = 48, a(4) = 100, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + 2 b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 11, b(2) = 23, b(3) = 48, b(4) = 100, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - 2 t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - _Z - 2 _Z - _Z + 1 In floating-point [0.480533816184295, 2.08101899662454, -0.780776406404415 + 0.624810533843826 I, -0.780776406404415 - 0.624810533843826 I] The largest root is, 2.08101899662454 and the remaining roots are [0.480533816184295, -0.780776406404415 + 0.624810533843826 I, -0.780776406404415 - 0.624810533843826 I] whose absolute values are [0.480533816184295, 1.00000000000000, 1.00000000000000] so the largest absolute value is, 1.00000000000000 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 1.00000000000000 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.09090909091, 0.1739130435, 0.3333333333, -0.3600000000, 0.3894230769, -0.1709006928, -0.08546059933, 0.3221333333, -0.4090210149, 0.3206896552, -0.08977393774, -0.1795535333, 0.3706117875, -0.3989572321, 0.2524864947, 0.004736921911, -0.2598591406, 0.4010585223, -0.3664093047, 0.1711116618, 0.09921070579, -0.3260337960, 0.4099085846, -0.3140599632, 0.08051270364, 0.1883351576, -0.3746079828, 0.3966349993, -0.2447585123, -0.01443165414, 0.2672943033, -0.4029625166, 0.3619529482, -0.1622461276, -0.1085970512, 0.3318261584, -0.4095670198, 0.3077343734, -0.07097645660, -0.1969008880, 0.3784475921, -0.3940650139, 0.2369057388, 0.02412419105, -0.2745769372, 0.4046421976, -0.3572932246, 0.1532900423, 0.1179227279, -0.3374326097, 0.4089961130, -0.3012364209, 0.06140046742, 0.2053563483, -0.3820752507, 0.3912743342, -0.2289202864, -0.03380321705] The largest is 0.4099085846 The smallest is -0.4095670198 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 189, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 24, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 24, 52, 113, 246, 536, 1168, 2545, 5545, 12081, 26321, 57346, 124941, 272212, 593075, 1292147, 2815232, 6133614, 13363453, 29115278] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 4) with initial conditions, a(1) = 11, a(2) = 24, a(3) = 52, a(4) = 113, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 4) with initial conditions, b(1) = 11, b(2) = 24, b(3) = 52, b(4) = 113, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 3 t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 3 _Z + 2 _Z - 1 In floating-point [2.17872417610522, 0.667076110379437 + 0.670769076539608 I, -0.512876396864097, 0.667076110379437 - 0.670769076539608 I] The largest root is, 2.17872417610522 and the remaining roots are [0.667076110379437 + 0.670769076539608 I, -0.512876396864097, 0.667076110379437 - 0.670769076539608 I] whose absolute values are [0.946003007966020, 0.512876396864097, 0.946003007966020] so the largest absolute value is, 0.946003007966020 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.946003007966020 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3636363636, -0.3333333333, -0.4423076923, -0.4601769912, -0.1300813008, 0.1940298507, 0.3981164384, 0.3457760314, 0.1110910730, -0.1643075904, -0.3170472246, -0.2767586231, -0.08508816161, 0.1339433971, 0.2549576360, 0.2202272652, 0.06567842366, -0.1094759142, -0.2048270009, -0.1753019154, -0.05057331886, 0.08940795846, 0.1645435108, 0.1395127000, 0.03887775956, -0.07298416291, -0.1321644971, -0.1110124654, -0.02983064247, 0.05954884047, 0.1061433093, 0.08831978147, 0.02284208343, -0.04856447220, -0.08523427419, -0.07025409670, -0.01745165828, 0.03958874634, 0.06843528141, 0.05587425484, 0.01330054341, -0.03225813310, -0.05494020471, -0.04443009310, -0.01010932647, 0.02627407369, 0.04410066931, 0.03532376743, 0.007660637210, -0.02139154954, -0.03539525373, -0.02807889468, -0.005785539369, 0.01740962171, 0.02840469015, 0.02231593233, 0.004352877342, -0.01416361093 ] The largest is 0.3981164384 The smallest is -0.4601769912 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 190, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 25, 57, 130, 296, 674, 1535, 3496, 7962, 18133, 41297, 94052, 214199, 487828, 1111005, 2530261, 5762549, 13123931, 29889128, 68071066] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 3) + a(n - 4) with initial conditions, a(1) = 11, a(2) = 25, a(3) = 57, a(4) = 130, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 3) + b(n - 4) with initial conditions, b(1) = 11, b(2) = 25, b(3) = 57, b(4) = 130, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - 2 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - 2 _Z - _Z - 1 In floating-point [2.27745239043720, 0.139932506403605 + 0.876514201643070 I, -0.557317403244406, 0.139932506403605 - 0.876514201643070 I] The largest root is, 2.27745239043720 and the remaining roots are [0.139932506403605 + 0.876514201643070 I, -0.557317403244406, 0.139932506403605 - 0.876514201643070 I] whose absolute values are [0.887613796665185, 0.557317403244406, 0.887613796665185] so the largest absolute value is, 0.887613796665185 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.887613796665185 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.1818181818, -0.04000000000, 0.4912280702, -0.03076923077, -0.2837837838, -0.1172106825, 0.2254071661, 0.1361556064, -0.1287364984, -0.1492858325, 0.06298278325, 0.1333836601, -0.01125588822, -0.1088149922, -0.02126363068, 0.07960048390, 0.03913007941, -0.05181846811, -0.04530008370, 0.02813039537, 0.04357240198, -0.009973747918, -0.03711718417, -0.002531570997, 0.02853551206, 0.009980092037, -0.01968857109, -0.01337320112, 0.01176920186, 0.01382992466, -0.005401922883, -0.01240784503, 0.0007834364724, 0.009994874726, 0.002179981543, -0.007264445468, -0.003750579736, 0.004673696796, 0.004262929669, -0.002489165867, -0.004055214674, 0.0008261971174, 0.003426158036, 0.0003079355320, -0.002613146492, -0.0009739378309, 0.001786217907, 0.001267224853, -0.001052634617, -0.001292989159, 0.0004674644423, 0.001149519120, -0.00004658553498, -0.0009186957863, -0.0002204080098, 0.0006621175658, 0.0003589538104, -0.0004211961753] The largest is 0.4912280702 The smallest is -0.2837837838 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 54, : Consider the Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 26, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 11, 26, 61, 143, 335, 785, 1839, 4308, 10092, 23642, 55385, 129748, 303955, 712062, 1668116, 3907821, 9154678, 21446256, 50241188, 117697792, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 3) + a(n - 6), . Alas, it breaks down at the, 78, -th term. a(78), equals , 326479880122546620131473983538, while the corresponding term for the solution of the recurrence is , 326479880122546620131473983537 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.01165724459672 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 55, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 27, 66, 161, 393, 959, 2340, 5710, 13933, 33998, 82959, 202429, 493949, 1205290, 2941040, 7176461, 17511354, 42729629, 104264993, 254418047] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 56, : Consider the Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 28, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 11, 28, 71, 180, 456, 1155, 2925, 7407, 18757, 47499, 120283, 304596, 771337, 1953278, 4946340, 12525754, 31719314, 80323698, 203405927, 515090467, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 3 a(n - 1) - 2 a(n - 2) + 3 a(n - 3) - 3 a(n - 4) + 2 a(n - 5) - a(n - 6), . Alas, it breaks down at the, 56, -th term. a(56), equals , 173220874359291434997149, while the corresponding term for the solution of the recurrence is , 173220874359291434997150 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.01224792866759 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 191, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) with initial conditions, a(1) = 11, a(2) = 29, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) with initial conditions, b(1) = 11, b(2) = 29, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t + 1 = 0 whose roots are 1/2 1/2 5 5 [3/2 + ----, 3/2 - ----] 2 2 In floating-point [2.618033988, 0.381966012] The largest root is, 2.618033988 and the remaining roots are [0.381966012] whose absolute values are [0.381966012] so the largest absolute value is, 0.381966012 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.381966012 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4545454545, 0.1724137931, 0.06578947368, 0.02512562814, 0.009596928983, 0.003665689150, 0.001400168020, 0.0005348165579, 0.0002042817454, -5 0.00007802868334, 0.00002980430493, 0.00001138423147, 0.4348389487 10 , -5 -6 -6 -7 0.1660936988 10 , 0.6344214761 10 , 0.2423274407 10 , 0.9256084594 10 , -7 -7 -8 -8 0.3535509712 10 , 0.1350444542 10 , 0.5158239153 10 , 0.1970272034 10 , -9 -9 -9 -10 0.7525769500 10 , 0.2874588158 10 , 0.1097994973 10 , 0.4193967600 10 , -10 -11 -11 0.1601953076 10 , 0.6118916265 10 , 0.2337218039 10 , -12 -12 -12 0.8927378517 10 , 0.3409955163 10 , 0.1302486972 10 , -13 -13 -14 0.4975057535 10 , 0.1900302882 10 , 0.7258511121 10 , -14 -14 -15 0.2772504541 10 , 0.1059002501 10 , 0.4045029610 10 , -15 -16 -16 0.1545063826 10 , 0.5901618666 10 , 0.2254217742 10 , -17 -17 -17 0.8610345594 10 , 0.3288859362 10 , 0.1256232492 10 , -18 -18 -19 0.4798381142 10 , 0.1832818505 10 , 0.7000743738 10 , -19 -19 -20 0.2674046161 10 , 0.1021394746 10 , 0.3901380771 10 , -20 -21 -21 0.1490194851 10 , 0.5692037834 10 , 0.2174164987 10 , -22 -22 -22 0.8304571280 10 , 0.3172063967 10 , 0.1211620621 10 , -23 -23 -24 0.4627978957 10 , 0.1767730662 10 , 0.6752130301 10 ] The largest is 0.4545454545 The smallest is -24 0.6752130301 10 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 192, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 30, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 30, 82, 224, 612, 1672, 4568, 12480, 34096, 93152, 254496, 695296, 1899584, 5189760, 14178688, 38736896, 105831168, 289136128, 789934592, 2158141440] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) with initial conditions, a(1) = 11, a(2) = 30, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) with initial conditions, b(1) = 11, b(2) = 30, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 2 t - 2 = 0 whose roots are 1/2 1/2 [1 + 3 , 1 - 3 ] In floating-point [2.732050808, -0.732050808] The largest root is, 2.732050808 and the remaining roots are [-0.732050808] whose absolute values are [0.732050808] so the largest absolute value is, 0.732050808 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.732050808 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.1818181818, 0.1333333333, -0.09756097561, 0.07142857143, -0.05228758170, 0.03827751196, -0.02802101576, 0.02051282051, -0.01501642421, 0.01099278598, -0.008047277757, 0.005891016200, -0.004312523163, 0.003156986065, -0.002311074198, 0.001691823733, -0.001238500930, 0.0009066456060, -0.0006637106481, 0.0004858699159, -0.0003556814643, 0.0002603769032, -0.0001906091223, 0.0001395355619, -0.0001021471208, 0.00007477688224, -0.00005474047703, 0.00004007281042, -0.00002933533323, 0.00002147495438, -5 -5 -0.00001572075770, 0.00001150839337, -0.8424728658 10 , 0.6167329418 10 , -5 -5 -5 -0.4514798481 10 , 0.3305061874 10 , -0.2419473214 10 , -5 -5 -6 0.1771177320 10 , -0.1296591788 10 , 0.9491710652 10 , -6 -6 -6 -0.6948414448 10 , 0.5086592408 10 , -0.3723644080 10 , -6 -6 -6 0.2725896656 10 , -0.1995494848 10 , 0.1460803615 10 , -6 -7 -7 -0.1069382466 10 , 0.7828422979 10 , -0.5730803364 10 , -7 -7 -7 0.4195239231 10 , -0.3071128267 10 , 0.2248221928 10 , -7 -7 -8 -0.1645812678 10 , 0.1204818500 10 , -0.8819883559 10 , -8 -8 -8 0.6456602882 10 , -0.4726561354 10 , 0.3460083056 10 ] The largest is 0.1333333333 The smallest is -0.1818181818 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 57, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 31, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 31, 87, 244, 684, 1917, 5373, 15060, 42212, 118317, 331633, 929541, 2605430, 7302814, 20469209, 57373571, 160813574, 450747707, 1263410111, 3541238444] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 58, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 32, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 32, 93, 270, 784, 2277, 6613, 19206, 55780, 162002, 470503, 1366484, 3968686, 11526274, 33475813, 97223965, 282368030, 820082830, 2381770514, 6917387578] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 59, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 34, 105, 324, 1000, 3086, 9523, 29387, 90685, 279844, 863568, 2664876, 8223515, 25376865, 78310221, 241656750, 745726217, 2301229288, 7101341102, 21913959513] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 60, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 35, 111, 352, 1116, 3538, 11216, 35556, 112717, 357327, 1132771, 3591025, 11383996, 36088684, 114405619, 362680048, 1149740881, 3644821657, 11554538184, 36629323794] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 193, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 36, 118, 387, 1269, 4161, 13644, 44739, 146700, 481032, 1577313, 5172039, 16959213, 55609578, 182344851, 597912192, 1960565310, 6428730483, 21079928025, 69121480005] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 3 a(n - 3) with initial conditions, a(1) = 11, a(2) = 36, a(3) = 118, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 3 b(n - 3) with initial conditions, b(1) = 11, b(2) = 36, b(3) = 118, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t - 3 = 0 whose roots are 1/2 1/3 1/2 1/3 (20 + 4 21 ) 2 (20 + 4 21 ) [----------------- + ----------------- + 1, - ----------------- 2 1/2 1/3 4 (20 + 4 21 ) 1 - ------------------- + 1 1/2 (1/3) (20 + 4 21 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(20 + 4 21 ) 2 | (20 + 4 21 ) + 1/2 I 3 |----------------- - -----------------|, - ----------------- | 2 1/2 1/3| 4 \ (20 + 4 21 ) / 1 - ------------------- + 1 1/2 (1/3) (20 + 4 21 ) / 1/2 1/3 \ 1/2 |(20 + 4 21 ) 2 | - 1/2 I 3 |----------------- - -----------------|] | 2 1/2 1/3| \ (20 + 4 21 ) / In floating-point [3.279018786, -0.139509393 + 0.9462795415 I, -0.139509393 - 0.9462795415 I] The largest root is, 3.279018786 and the remaining roots are [-0.139509393 + 0.9462795415 I, -0.139509393 - 0.9462795415 I] whose absolute values are [0.9565081502, 0.9565081502] so the largest absolute value is, 0.9565081502 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.9565081502 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.1818181818, -0.2222222222, 0.2288135593, 0.1395348837, -0.2482269504, -0.05839942322, 0.2434036939, -0.01448400724, -0.2186503067, 0.07425909295, 0.1793252195, -0.1179753285, -0.1311487155, 0.1445295089, 0.07966254007, -0.1544585262, -0.02978705208, 0.1496264640, -0.01449618683, -0.1328497167, 0.05033024171, 0.1075021646, -0.07604265629, -0.07713724375, 0.09109476265, 0.04515631908, -0.09594277401, -0.01454403408, 0.09183685502, -0.01231775699, -0.08058537319, 0.03375444547, 0.06431006544, -0.04882592325, -0.04521443333, 0.05728689635, 0.02538291932, -0.05949454203, -0.006622937026, 0.05627994688, -0.009643785445, -0.04880016741, 0.02243933840, 0.03838665887, -0.03124052563, -0.02640356167, 0.03594929160, 0.01412629791, -0.03683179128, -0.002647499038, 0.03443639662, -0.007186183955, -0.02950104898, 0.01480604293, 0.02285957692, -0.01992441618, -0.01535511976, 0.02251337148 ] The largest is 0.2434036939 The smallest is -0.2482269504 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 61, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 37, 124, 416, 1396, 4685, 15723, 52767, 177088, 594314, 1994540, 6693751, 22464479, 75391633, 253017144, 849135012, 2849728905, 9563796943, 32096460757, 107716924488] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 62, :The Pisot Sequence a(n), defined by, a(1) = 11, a(2) = 38, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [11, 38, 131, 452, 1560, 5384, 18582, 64133, 221345, 763938, 2636614, 9099866, 31406782, 108395657, 374110867, 1291185871, 4456328593, 15380329800, 53082832610, 183207197410] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 194, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 39, 138, 488, 1726, 6105, 21594, 76380, 270163, 955591, 3380012, 11955409, 42287366, 149574249, 529057685, 1871325017, 6619046313, 23412167152, 82810958684, 292909871762] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 2 a(n - 2) + a(n - 3) + a(n - 4) with initial conditions, a(1) = 11, a(2) = 39, a(3) = 138, a(4) = 488, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 2 b(n - 2) + b(n - 3) + b(n - 4) with initial conditions, b(1) = 11, b(2) = 39, b(3) = 138, b(4) = 488, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 4 t + 2 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 4 _Z + 2 _Z - _Z - 1 In floating-point [3.53709069930844, 0.431287054905263 + 0.722066348142627 I, -0.399664809118964, 0.431287054905263 - 0.722066348142627 I] The largest root is, 3.53709069930844 and the remaining roots are [0.431287054905263 + 0.722066348142627 I, -0.399664809118964, 0.431287054905263 - 0.722066348142627 I] whose absolute values are [0.841063811401302, 0.399664809118964, 0.841063811401302] so the largest absolute value is, 0.841063811401302 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.841063811401302 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2727272727, 0.3076923077, -0.3188405797, -0.3360655738, -0.1268829664, 0.1533169533, 0.2120033343, 0.07841057869, -0.08393081214, -0.1272249320, -0.05062437648, 0.04643212123, 0.07582148768, 0.03257239821, -0.02554563781, -0.04507373879, -0.02080979366, 0.01393506333, 0.02674046405, 0.01320819711, -0.007522870007, -0.01583234685, -0.008334986245, 0.004010075832, 0.009355058955, 0.005232751055, -0.002104024101, -0.005516463728, -0.003269996702, 0.001081667604, 0.003246175990, 0.002034908321, -0.0005410477953, -0.001906164229, -0.001261477014, 0.0002602809275, 0.001116865714, 0.0007792597567, -0.0001178884864, -0.0006529268179, -0.0004798048287, 0.00004800559152, 0.0003808167192, 0.0002945240471, -5 -0.00001533648712, -0.0002215717320, -0.0001802731874, 0.1238274405 10 , -5 0.0001285912533, 0.0001100435450, 0.3956760299 10 , -0.00007443052104, -5 -0.00006700080650, -0.5141878690 10 , 0.00004296033751, 0.00004069377989, -5 0.4711759347 10 , -0.00002472206357] The largest is 0.3076923077 The smallest is -0.3360655738 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 195, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 11, a(2) = 40, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [11, 40, 145, 526, 1908, 6921, 25105, 91065, 330326, 1198213, 4346356, 15765820, 57188385, 207443151, 752472043, 2729490816, 9900859685, 35914032730, 130273308376, 472548850273] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 11, a(2) = 40, a(3) = 145, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 11, b(2) = 40, b(3) = 145, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (540 + 60 21 ) 10 (540 + 60 21 ) [------------------- + ------------------- + 1, - ------------------- 6 1/2 1/3 12 (540 + 60 21 ) 5 - ------------------- + 1 1/2 1/3 (540 + 60 21 ) / 1/2 1/3 \ 1/2 |(540 + 60 21 ) 10 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (540 + 60 21 ) / 1/2 1/3 (540 + 60 21 ) 5 - ------------------- - ------------------- + 1 12 1/2 1/3 (540 + 60 21 ) / 1/2 1/3 \ 1/2 |(540 + 60 21 ) 10 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (540 + 60 21 ) / In floating-point [3.627365085, -0.313682542 + 0.4210528072 I, -0.313682542 - 0.4210528072 I] The largest root is, 3.627365085 and the remaining roots are [-0.313682542 + 0.4210528072 I, -0.313682542 - 0.4210528072 I] whose absolute values are [0.5250544768, 0.5250544768] so the largest absolute value is, 0.5250544768 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5250544768 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4545454545, -0.3750000000, 0.1103448276, 0.03422053232, -0.05188679245, 0.02311804652, -0.0001991635132, -0.006248284193, 0.003974861198, -0.0007711483684, -0.0006120069318, 0.0005965436622, -0.0002055312455, -5 -0.00003551334409, 0.00007894113881, -0.00003973451728, 0.3165381694 10 , -5 -5 -5 -6 0.8968249331 10 , -0.6499005902 10 , 0.1604862650 10 , 0.7848254764 10 , -6 -6 -7 -0.9348041733 10 , 0.3701010826 10 , 0.2552037746 10 , -6 -7 -8 -0.1180408758 10 , 0.6701921014 10 , -0.9503743690 10 , -7 -7 -8 -0.1251368657 10 , 0.1047066304 10 , -0.3119127715 10 , -9 -8 -9 -0.9297436375 10 , 0.1443176698 10 , -0.6490848965 10 , -11 -9 -9 0.9355068744 10 , 0.1730721111 10 , -0.1111584256 10 , -10 -10 -10 0.2202401412 10 , 0.1682730224 10 , -0.1662849064 10 , -11 -12 -11 0.5793146678 10 , 0.9497609882 10 , -0.2192914325 10 , -11 -13 -12 0.1113925679 10 , -0.9429062450 10 , -0.2479348403 10 , -12 -13 -13 0.1815399092 10 , -0.4554057740 10 , -0.2147675403 10 , -13 -13 -15 0.2602849234 10 , -0.1040860843 10 , -0.6455946201 10 , -14 -14 -15 0.3274491631 10 , -0.1876322772 10 , 0.2744203261 10 , -15 -15 -16 0.3451070654 10 , -0.2921609236 10 , 0.8815168627 10 , -16 0.2524027710 10 ] The largest is 0.4545454545 The smallest is -0.3750000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 63, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 14, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50 ] is a trivial linear sequence Fact, 64, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 15, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 15, 19, 24, 30, 38, 48, 61, 78, 100, 128, 164, 210, 269, 345, 442, 566, 725, 929, 1190] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 196, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 16, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 2) + a(n - 3) with initial conditions, a(1) = 12, a(2) = 16, a(3) = 21, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 2) + b(n - 3) with initial conditions, b(1) = 12, b(2) = 16, b(3) = 21, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [1.324717958, -0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.4375000000, 0.3333333333, -0.1071428571, -0.1081081081, 0.2244897959, -0.2153846154, 0.1162790698, 0.008771929825, -0.09933774834, 0.1250000000, -0.09056603774, 0.02564102564, 0.03440860215, -0.06493506494, 0.06004901961, -0.03052728955, -0.004888268156, 0.02952029520, -0.03541583764, 0.02463202163, -0.005895691610, -0.01078397809, 0.01873627084, -0.01667967226, 0.007952286282, 0.002056583847, -0.008727394789, 0.01000886862, -0.006670811018, 0.001281472791, 0.003338056570, -0.005389338572, 0.004619529351, -0.002051282051, -0.0007698093173, 0.002568247246, -0.002821091377, 0.001798437928, -0.0002528441374, -0.001022653455, 0.001545593789, -0.001275497592, 0.0005229403335, 0.0002700961958, -0.0007525572592, 0.0007930365292, -0.0004824610634, 0.00004047926998, 0.0003105754658, -0.0004419817935, 0.0003510547358, -0.0001314063277, -0.00009092705769, 0.0002196484081, -5 -0.0002223333854, 0.0001287213504, -0.2684977277 10 ] The largest is 0.3333333333 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 65, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 17, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 17, 24, 34, 48, 68, 96, 136, 193, 274, 389, 552, 783, 1111, 1576, 2236, 3172, 4500, 6384, 9057] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 66, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 18, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 18, 27, 41, 62, 94, 143, 218, 332, 506, 771, 1175, 1791, 2730, 4161, 6342, 9666, 14732, 22453, 34221] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 67, : Consider the Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 19, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 12, 19, 30, 47, 74, 117, 185, 293, 464, 735, 1164, 1843, 2918, 4620, 7315, 11582, 18338, 29035, 45972, 72789, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 2) + a(n - 4) - a(n - 6) + a(n - 7), . Alas, it breaks down at the, 55, -th term. a(55), equals , 703162155202, while the corresponding term for the solution of the recurrence is , 703162155201 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.02968394081287 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 197, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, 4180, 6764, 10945, 17710, 28656, 46367, 75024, 121392] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) with initial conditions, a(1) = 12, a(2) = 20, a(3) = 33, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) with initial conditions, b(1) = 12, b(2) = 20, b(3) = 33, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + 1 = 0 whose roots are 1/2 1/2 5 5 [1, 1/2 - ----, ---- + 1/2] 2 2 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [-0.6180339880, 1.618033988] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.4500000000, 0.3636363636, 0.4074074074, 0.3750000000, 0.3916083916, 0.3793103448, 0.3856382979, 0.3809523810, 0.3833671400, 0.3815789474, 0.3825009679, 0.3818181818, 0.3821703134, 0.3819095477, 0.3820440429, 0.3819444444, 0.3819958160, 0.3819577735, 0.3819773955, 0.3819628647, 0.3819703596, 0.3819648094, 0.3819676722, 0.3819655522, 0.3819666457, 0.3819658359, 0.3819662536, 0.3819659443, 0.3819661038, 0.3819659857, 0.3819660466, 0.3819660015, 0.3819660248, 0.3819660075, 0.3819660164, 0.3819660098, 0.3819660132, 0.3819660107, 0.3819660120, 0.3819660110, 0.3819660115, 0.3819660112, 0.3819660114, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660112, 0.3819660112, 0.3819660112] The largest is 0.4500000000 The smallest is 0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 198, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625, 299426, 525456] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 12, a(2) = 21, a(3) = 37, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 12, b(2) = 21, b(3) = 37, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (100 + 12 69 ) 2 (100 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (100 + 12 69 ) 1 - ----------------------- + 2/3 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (100 + 12 69 ) / 1/2 1/3 (100 + 12 69 ) 1 - ------------------- - ----------------------- + 2/3 12 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (100 + 12 69 ) / In floating-point [1.754877667, 0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] The largest root is, 1.754877667 and the remaining roots are [0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] whose absolute values are [0.7548776666, 0.7548776666] so the largest absolute value is, 0.7548776666 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7548776666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2500000000, 0.1904761905, 0.1891891892, -0.06153846154, -0.1228070175, 0.005000000000, 0.07122507122, 0.01461038961, -0.03700277521, -0.01739588824, 0.01682186843, 0.01403628894, -0.006145142411, -0.009504752376, 0.001171924490, 0.005703456367, 0.0007302348065, -0.003071062271, -0.001168903168, 0.001463490759, 0.001024822391, -0.0005827491423, -0.0007268299193, 0.0001539116948, 0.0004519041666, 0.00002306671899, -0.0002518590338, -0.00007488062000, 0.0001251645128, 0.00007335061177, -0.00005334390925, -0.00005487391748, 0.00001694668605, -6 -5 0.00003542338034, -0.9738428538 10 , -0.00002042437999, -0.4451536794 10 , -5 -5 -5 0.00001054746355, 0.5122083903 10 , -0.4754832539 10 , -0.4084285431 10 , -5 -5 -6 0.1708345581 10 , 0.2746144054 10 , -0.3003429039 10 , -5 -6 -6 -0.1638484280 10 , -0.2304816029 10 , 0.8771781708 10 , -6 -6 -6 0.3463536640 10 , -0.4149524457 10 , -0.2990803846 10 , -6 -6 -7 0.1631453405 10 , 0.2104186200 10 , -0.4138848514 10 , -6 -8 -7 -0.1300502497 10 , -0.8293394300 10 , 0.7207497597 10 , -7 -7 0.2239309654 10 , -0.3558217720 10 ] The largest is 0.1904761905 The smallest is -0.2500000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 68, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 22, 40, 73, 133, 242, 440, 800, 1455, 2646, 4812, 8751, 15914, 28940, 52628, 95705, 174041, 316496, 575552, 1046649] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 199, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 23, 44, 84, 160, 305, 581, 1107, 2109, 4018, 7655, 14584, 27785, 52935, 100850, 192136, 366051, 697388, 1328640, 2531280] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + 2 a(n - 2) - a(n - 4) with initial conditions, a(1) = 12, a(2) = 23, a(3) = 44, a(4) = 84, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + 2 b(n - 2) - b(n - 4) with initial conditions, b(1) = 12, b(2) = 23, b(3) = 44, b(4) = 84, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - 2 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - _Z - 2 _Z + 1 In floating-point [0.671043606703789, 1.90516616775402, -0.788104887228904 + 0.401357867370711 I, -0.788104887228904 - 0.401357867370711 I] The largest root is, 1.90516616775402 and the remaining roots are [0.671043606703789, -0.788104887228904 + 0.401357867370711 I, -0.788104887228904 - 0.401357867370711 I] whose absolute values are [0.671043606703789, 0.884419273294318, 0.884419273294318] so the largest absolute value is, 0.884419273294318 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.884419273294318 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.08333333333, 0.1739130435, 0.3636363636, -0.2380952381, 0.4062500000, -0.2426229508, 0.2065404475, -0.04065040650, -0.03366524419, 0.1276754604, -0.1461789680, 0.1498217224, -0.1088716934, 0.06309625012, -0.008468021815, -0.03209705625, 0.05983865636, -0.06745169117, 0.06069364162, -0.04211268607, 0.01943594092, 0.002662260477, -0.01915949895, 0.02827770819, -0.02947723061, 0.02441592530, -0.01537903697, 0.005175105438, 0.003894262111, -0.01017145231, 0.01299610888, -0.01252190119, 0.009576054456, -0.005296295603, 0.0008597044324, 0.002789014411, -0.005067631180, 0.005806693246, -0.005188273546, 0.003636098535, -0.001672817377, -0.0002073135524, 0.001635325240, -0.002415400400, 0.002528067457, -0.002095419791, 0.001325389883, -0.0004500492981, -0.0003273369888, 0.0008679842056, -0.001112079655, 0.001073938054, -0.0008228842669, 0.0004570076362, -0.00007668124262, -0.0002366040246, 0.0004329177570, -0.0004972979284] The largest is 0.4062500000 The smallest is -0.2426229508 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 69, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 25, 52, 108, 224, 465, 965, 2003, 4158, 8632, 17920, 37202, 77232, 160335, 332858, 691018, 1434563, 2978173, 6182729, 12835432] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 70, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 26, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 26, 56, 121, 261, 563, 1214, 2618, 5646, 12176, 26258, 56626, 122115, 263343, 567903, 1224691, 2641064, 5695493, 12282414, 26487206] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 71, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 27, 61, 138, 312, 705, 1593, 3600, 8136, 18387, 41554, 93911, 212237, 479651, 1084001, 2449819, 5536538, 12512456, 28277880, 63907397] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 200, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 28, 65, 151, 351, 816, 1897, 4410, 10252, 23833, 55405, 128801, 299426, 696081, 1618192, 3761840, 8745217, 20330163, 47261895, 109870576] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 12, a(2) = 28, a(3) = 65, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 12, b(2) = 28, b(3) = 65, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + ------------------- + 1, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- + 1 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- + 1 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [2.324717958, 0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] The largest root is, 2.324717958 and the remaining roots are [0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] whose absolute values are [0.6558656185, 0.6558656185] so the largest absolute value is, 0.6558656185 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6558656185 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.1071428571, -0.2153846154, -0.09933774834, 0.02564102564, 0.06004901961, 0.02952029520, -0.005895691610, -0.01667967226, -0.008727394789, 0.001281472791, 0.004619529351, 0.002568247246, -0.0002528441374, -0.001275497592, -0.0007525572592, 0.00004047926998, -5 0.0003510547358, 0.0002196484081, -0.2684977277 10 , -0.00009629701225, -5 -0.00006387267410, -0.1708975075 10 , 0.00002632141072, 0.00001850950822, -5 -5 -5 0.1176728145 10 , -0.7167421285 10 , -0.5346211923 10 , -6 -5 -5 -6 -0.5270650543 10 , 0.1943807399 10 , 0.1539340381 10 , 0.2033412922 10 , -6 -6 -7 -0.5248494873 10 , -0.4418906650 10 , -0.7263172821 10 , -6 -6 -7 -7 0.1410366581 10 , 0.1264827656 10 , 0.2474325256 10 , -0.3769911553 10 , -7 -8 -7 -0.3610108606 10 , -0.8161774577 10 , 0.1001773287 10 , -7 -8 -8 0.1027566169 10 , 0.2629744763 10 , -0.2644356227 10 , -8 -9 -9 -0.2916896517 10 , -0.8322323331 10 , 0.6927398069 10 , -9 -9 -9 0.8257875702 10 , 0.2596507635 10 , -0.1798830429 10 , -9 -10 -10 -0.2331630855 10 , -0.8007240731 10 , 0.4622590625 10 , -10 -10 -10 0.6565944783 10 , 0.2445412367 10 , -0.1173061838 10 , -10 -0.1844065467 10 ] The largest is 0.3333333333 The smallest is -0.2153846154 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 201, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) with initial conditions, a(1) = 12, a(2) = 29, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) with initial conditions, b(1) = 12, b(2) = 29, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 2 t - 1 = 0 whose roots are 1/2 1/2 [1 + 2 , 1 - 2 ] In floating-point [2.414213562, -0.414213562] The largest root is, 2.414213562 and the remaining roots are [-0.414213562] whose absolute values are [0.414213562] so the largest absolute value is, 0.414213562 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.414213562 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.08333333333, -0.03448275862, 0.01428571429, -0.005917159763, 0.002450980392, -0.001015228426, 0.0004205214466, -0.0001741856819, 0.00007215007215, -5 -5 -0.00002988553839, 0.00001237899532, -0.5127547750 10 , 0.2123899820 10 , -6 -6 -6 -0.8797481105 10 , 0.3644035988 10 , -0.1509409128 10 , -7 -7 -7 0.6252177321 10 , -0.2589736641 10 , 0.1072704040 10 , -8 -8 -9 -0.4443285616 10 , 0.1840469164 10 , -0.7623472887 10 , -9 -9 -10 0.3157745862 10 , -0.1307981163 10 , 0.5417835369 10 , -10 -11 -11 -0.2244140888 10 , 0.9295535919 10 , -0.3850337047 10 , -11 -12 -12 0.1594861825 10 , -0.6606133979 10 , 0.2736350289 10 , -12 -13 -13 -0.1133433401 10 , 0.4694834868 10 , -0.1944664275 10 , -14 -14 -14 0.8055063171 10 , -0.3336516411 10 , 0.1382030349 10 , -15 -15 -16 -0.5724557140 10 , 0.2371189206 10 , -0.9821787280 10 , -16 -16 -17 0.4068317498 10 , -0.1685152284 10 , 0.6980129306 10 , -17 -17 -18 -0.2891264226 10 , 0.1197600855 10 , -0.4960625163 10 , -18 -19 -19 0.2054758220 10 , -0.8511087223 10 , 0.3525407758 10 , -19 -20 -20 -0.1460271706 10 , 0.6048643455 10 , -0.2505430153 10 , -20 -21 -21 0.1037783149 10 , -0.4298638551 10 , 0.1780554388 10 , -22 -22 -22 -0.7375297759 10 , 0.3054948358 10 , -0.1265401042 10 ] The largest is 0.08333333333 The smallest is -0.03448275862 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 202, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 30, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 30, 75, 188, 471, 1180, 2956, 7405, 18550, 46469, 116408, 291610, 730503, 1829960, 4584175, 11483672, 28767384, 72064265, 180525914, 452229765] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) - a(n - 3) + a(n - 4) with initial conditions, a(1) = 12, a(2) = 30, a(3) = 75, a(4) = 188, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) - b(n - 3) + b(n - 4) with initial conditions, b(1) = 12, b(2) = 30, b(3) = 75, b(4) = 188, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 3 t + t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 3 _Z + _Z + _Z - 1 In floating-point [2.50506841362147, 0.592608040497570 + 0.476565325929643 I, -0.690284494616613, 0.592608040497570 - 0.476565325929643 I] The largest root is, 2.50506841362147 and the remaining roots are [0.592608040497570 + 0.476565325929643 I, -0.690284494616613, 0.592608040497570 - 0.476565325929643 I] whose absolute values are [0.760459597572939, 0.690284494616613, 0.760459597572939] so the largest absolute value is, 0.760459597572939 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.760459597572939 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.5000000000, 0.2533333333, 0.005319148936, 0.2632696391, 0.03050847458, 0.07611637348, -0.06009453072, -0.02366576819, -0.05651079214, -0.009655693767, -0.008885154830, 0.01584524636, 0.009565782859, 0.01208156320, 0.001948505670, 0.00004341722556, -0.004334034351, -0.002912462750, -0.002498265456, -0.0002048820404, 0.0004620477338, 0.001176827947, 0.0007750526928, 0.0004814003569, -0.00004563183561, -0.0002165206092, -0.0003102776561, -0.0001872801665, -0.00008067406990, 0.00003901500372, 0.00007472159150, 0.00007854367414, 0.00004122035731, -5 0.9410810010 10 , -0.00001681000992, -0.00002251752295, -0.00001893301162, -5 -6 -5 -5 -0.8060691989 10 , 0.4584486813 10 , 0.5851526708 10 , 0.6223811809 10 , -5 -5 -6 0.4300768047 10 , 0.1285414306 10 , -0.8168102298 10 , -5 -5 -6 -0.1812801234 10 , -0.1606239730 10 , -0.9036934217 10 , -6 -6 -6 -6 -0.1088495307 10 , 0.3705833262 10 , 0.5180532007 10 , 0.3887323848 10 , -6 -7 -6 0.1687110967 10 , -0.3006896921 10 , -0.1295971884 10 , -6 -7 -7 -0.1387013078 10 , -0.8772666919 10 , -0.2495048059 10 ] The largest is 0.2632696391 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 72, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 31, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 31, 80, 206, 530, 1364, 3510, 9032, 23241, 59803, 153883, 395966, 1018885, 2621757, 6746208, 17359093, 44667776, 114937469, 295752844, 761020279] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 203, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 32, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 32, 85, 226, 601, 1598, 4249, 11298, 30041, 79878, 212393, 564746, 1501641, 3992814, 10616761, 28229618, 75061625, 199586390, 530693641, 1411096922] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 12, a(2) = 32, a(3) = 85, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 12, b(2) = 32, b(3) = 85, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (44 + 3 177 ) 7 (44 + 3 177 ) [------------------ + -------------------- + 2/3, - ------------------ 3 1/2 1/3 6 3 (44 + 3 177 ) 7 - -------------------- + 2/3 1/2 1/3 6 (44 + 3 177 ) / 1/2 1/3 \ 1/2 |(44 + 3 177 ) 7 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (44 + 3 177 ) / 1/2 1/3 (44 + 3 177 ) 7 - ------------------ - -------------------- + 2/3 6 1/2 1/3 6 (44 + 3 177 ) / 1/2 1/3 \ 1/2 |(44 + 3 177 ) 7 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (44 + 3 177 ) / In floating-point [2.658967082, -0.3294835411 + 0.8022545575 I, -0.3294835411 - 0.8022545575 I] The largest root is, 2.658967082 and the remaining roots are [-0.3294835411 + 0.8022545575 I, -0.3294835411 - 0.8022545575 I] whose absolute values are [0.8672783745, 0.8672783745] so the largest absolute value is, 0.8672783745 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8672783745 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.2187500000, -0.1058823529, 0.2345132743, -0.07487520799, -0.1270337922, 0.1400329489, 0.003274915914, -0.1074864352, 0.06836675931, 0.03579684830, -0.07501248349, 0.02250537912, 0.04159196997, -0.04433565002, -0.002068572093, 0.03471114568, -0.02131758082, -0.01206116016, 0.02398239020, -0.006731541398, -0.01360301291, 0.01402721319, 0.0009883306755, -0.01120215128, 0.006638454498, 0.004051419067, -0.007663009928, 0.002002308205, 0.004444444615, -0.004434822423, -0.0004205838211, 0.003612899164, -0.002064430338, -0.001357129155, 0.002447109679, -0.0005917704736, -0.001450689578, 0.001401069728, 0.0001679089305, -0.001164491568, 0.0006410652513, 0.0004534567960, -0.0007810042921, 0.0001735787144, 0.0004730667288, -0.0004422964122, -0.00006436866680, 0.0003750997117, -0.0001987620678, -0.0001511617576, 0.0002491138404, -0.00005045821240, -0.0001541260995, 0.0001395172694, 0.00002399201453, -0.0001207509005, 0.00006152475231] The largest is 0.3333333333 The smallest is -0.2187500000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 73, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 33, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 33, 91, 251, 692, 1908, 5261, 14506, 39997, 110283, 304081, 838436, 2311802, 6374283, 17575676, 48461041, 133620607, 368429284, 1015862301, 2801015716] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 204, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 34, 96, 271, 765, 2160, 6099, 17221, 48625, 137297, 387670, 1094620, 3090755, 8727016, 24641490, 69577394, 196457834, 554715811, 1566288423, 4422551828] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) + 2 a(n - 3) - 2 a(n - 4) + a(n - 5) with initial conditions, a(1) = 12, a(2) = 34, a(3) = 96, a(4) = 271, a(5) = 765, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) + 2 b(n - 3) - 2 b(n - 4) + b(n - 5) with initial conditions, b(1) = 12, b(2) = 34, b(3) = 96, b(4) = 271, b(5) = 765, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 3 t + t - 2 t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 3 _Z + _Z - 2 _Z + 2 _Z - 1 In floating-point [2.82358712677536, 0.492840476802846 + 0.382116749982025 I, -0.404634040190526 + 0.864250752203957 I, -0.404634040190526 - 0.864250752203957 I, 0.492840476802846 - 0.382116749982025 I] The largest root is, 2.82358712677536 and the remaining roots are [0.492840476802846 + 0.382116749982025 I, -0.404634040190526 + 0.864250752203957 I, -0.404634040190526 - 0.864250752203957 I, 0.492840476802846 - 0.382116749982025 I] whose absolute values are [0.623622438813808, 0.954284061045774, 0.954284061045774, 0.623622438813808] so the largest absolute value is, 0.954284061045774 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.954284061045774 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, 0.05882352941, 0.01041666667, -0.4981549815, -0.1764705882, 0.2041666667, -0.1695359895, -0.05876546077, 0.2562262211, -0.09643327968, -0.1198184022, 0.1974246771, -0.05199182724, -0.1439442760, 0.1582118614, -0.00007154622664, -0.1449067233, 0.1176718235, 0.03741110267, -0.1368970085, 0.07698341941, 0.06241910169, -0.1206705131, 0.04074131740, 0.07686882139, -0.09933066339, 0.01038195112, 0.08206101164, -0.07585656837, -0.01333666633, 0.07987402704, -0.05249446146, -0.03025659569, 0.07228949275, -0.03094856937, -0.04078544230, 0.06118995790, -0.01237740394, -0.04570642283, 0.04826036647, 0.002567356260, -0.04602637757, 0.03490968570, 0.01366299141, -0.04284781266, 0.02223305339, 0.02102720669, -0.03726335577, 0.01100744953, 0.02502619829, -0.03027692619, 0.001711840432, 0.02618658923, -0.02275087215, -0.005435474170, 0.02511702106, -0.01537654500, -0.01042927084] The largest is 0.3333333333 The smallest is -0.4981549815 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 74, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 35, 102, 297, 865, 2519, 7336, 21364, 62217, 181191, 527672, 1536708, 4475264, 13033047, 37955373, 110535191, 321905108, 937465233, 2730124628, 7950780703] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 205, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 37, 114, 351, 1081, 3329, 10252, 31572, 97229, 299426, 922111, 2839729, 8745217, 26931732, 82938844, 255418101, 786584466, 2422362079, 7459895657, 22973462017] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 3 a(n - 2) + a(n - 3) with initial conditions, a(1) = 12, a(2) = 37, a(3) = 114, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 3 b(n - 2) + b(n - 3) with initial conditions, b(1) = 12, b(2) = 37, b(3) = 114, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 3 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (388 + 12 69 ) 26 (388 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (388 + 12 69 ) 13 - --------------------- + 2/3 1/2 1/3 3 (388 + 12 69 ) / 1/2 1/3 \ 1/2 |(388 + 12 69 ) 26 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (388 + 12 69 ) / 1/2 1/3 (388 + 12 69 ) 13 - ------------------- - --------------------- + 2/3 12 1/2 1/3 3 (388 + 12 69 ) / 1/2 1/3 \ 1/2 |(388 + 12 69 ) 26 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (388 + 12 69 ) / In floating-point [3.079595625, -0.5397978113 + 0.1825822549 I, -0.5397978113 - 0.1825822549 I] The largest root is, 3.079595625 and the remaining roots are [-0.5397978113 + 0.1825822549 I, -0.5397978113 - 0.1825822549 I] whose absolute values are [0.5698402907, 0.5698402907] so the largest absolute value is, 0.5698402907 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.5698402907 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.08333333333, 0.2432432432, -0.2894736842, 0.2336182336, -0.1581868640, 0.09492340042, -0.05111197815, 0.02435702521, -0.009698752430, 0.002561567800, 0.0003839017212, -0.001246245680, 0.001220781600, -0.0009132721208, 0.0005895548773, -0.0003399250079, 0.0001755424954, -5 -5 -0.00007913515558, 0.00002843216712, -0.4998637119 10 , -0.3835928469 10 , -5 -5 -5 0.5764398822 10 , -0.4977624883 10 , 0.3502018229 10 , -5 -5 -6 -0.2164439370 10 , 0.1199551065 10 , -0.5921977494 10 , -6 -7 -8 -7 0.2498183270 10 , -0.7740552904 10 , 0.2446173554 10 , 0.2249408701 10 , -7 -7 -7 -0.2507883437 10 , 0.1977076584 10 , -0.1320088441 10 , -8 -8 -8 0.7831694338 10 , -0.4168498712 10 , 0.1957201179 10 , -9 -9 -10 -0.7593994407 10 , 0.1843059438 10 , 0.4761474476 10 , -9 -9 -10 -0.1112521196 10 , 0.1046459388 10 , -0.7684973654 10 , -10 -10 -10 0.4898622374 10 , -0.2793082330 10 , 0.1424728809 10 , -11 -11 -12 -0.6311669982 10 , 0.2187701005 10 , -0.3123198467 10 , -12 -12 -12 -0.3732066617 10 , 0.5043281412 10 , -0.4232835492 10 , -12 -12 -13 0.2932106635 10 , -0.1791011796 10 , 0.9814608197 10 , -13 -13 -14 -0.4780071138 10 , 0.1973564356 10 , -0.5784765056 10 ] The largest is 0.2432432432 The smallest is -0.2894736842 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 206, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 12, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [12, 38, 120, 379, 1197, 3780, 11937, 37696, 119041, 375922, 1187132, 3748869, 11838632, 37385464, 118060340, 372825221, 1177352576, 3717986365, 11741107033, 37077487873] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) + a(n - 2) - 2 a(n - 3) + a(n - 4) + a(n - 5) with initial conditions, a(1) = 12, a(2) = 38, a(3) = 120, a(4) = 379, a(5) = 1197, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) + b(n - 2) - 2 b(n - 3) + b(n - 4) + b(n - 5) with initial conditions, b(1) = 12, b(2) = 38, b(3) = 120, b(4) = 379, b(5) = 1197, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 3 t - t + 2 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 3 _Z - _Z + 2 _Z - _Z - 1 In floating-point [3.15792094977000, 0.604926762024975 + 0.580720180564985 I, -0.552109297197246, -0.815665176622700, 0.604926762024975 - 0.580720180564985 I] The largest root is, 3.15792094977000 and the remaining roots are [0.604926762024975 + 0.580720180564985 I, -0.552109297197246, -0.815665176622700, 0.604926762024975 - 0.580720180564985 I] whose absolute values are [0.838553704618523, 0.552109297197246, 0.815665176622700, 0.838553704618523] so the largest absolute value is, 0.838553704618523 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.838553704618523 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3333333333, -0.05263157895, 0.008333333333, 0.4986807388, -0.1578947368, 0.2880952381, -0.3351763425, 0.1052896859, -0.2547693652, 0.1415346801, -0.08782763838, 0.1577030299, -0.04726762349, 0.07832070775, -0.07400452176, 0.02071778025, -0.05805719068, 0.02560833598, -0.01835155726, 0.03338130405, -0.006763727504, 0.01734438135, -0.01423641282, 0.003192344704, -0.01273056487, 0.004054129580, -0.003844896997, 0.006936500215, -0.0006818756760, 0.003904231893, -0.002632947845, 0.0004607429280, -0.002804558307, 0.0005353199128, -0.0008488003773, 0.001425830478, 0.00001423585295, 0.0008969003973, -0.0004602043764, 0.00006484566345, -0.0006194018493, 0.00003818511842, -0.0001978418000, 0.0002881047042, 0.00003554589001, 0.0002092092433, -0.00007269247006, 0.00001030295738, -5 -0.0001365514903, -0.9211440095 10 , -0.00004827495210, 0.00005667717156, -5 -5 0.00001393090986, 0.00004925687492, -0.9139200697 10 , 0.2379672584 10 , -5 -0.00002990585137, -0.5871695343 10 ] The largest is 0.4986807388 The smallest is -0.3351763425 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 75, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 39, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 39, 127, 414, 1350, 4402, 14354, 46805, 152620, 497658, 1622746, 5291394, 17253994, 56261225, 183454650, 598202556, 1950598134, 6360442700, 20739910818, 67627981420] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 76, :The Pisot Sequence a(n), defined by, a(1) = 12, a(2) = 40, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [12, 40, 133, 442, 1469, 4882, 16225, 53923, 179210, 595594, 1979422, 6578494, 21863243, 72661219, 241485344, 802562525, 2667269971, 8864516940, 29460707553, 97910951651] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 77, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 15, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51 ] is a trivial linear sequence Fact, 78, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 16, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 16, 20, 25, 31, 38, 47, 58, 72, 89, 110, 136, 168, 208, 258, 320, 397, 493, 612, 760] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 79, : Consider the Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 17, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 13, 17, 22, 28, 36, 46, 59, 76, 98, 126, 162, 208, 267, 343, 441, 567, 729, 937, 1204, 1547, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = a(n - 1) + a(n - 6), . Alas, it breaks down at the, 37, -th term. a(37), equals , 110155, while the corresponding term for the solution of the recurrence is , 110156 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03282504251949 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 80, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 18, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 18, 25, 35, 49, 69, 97, 136, 191, 268, 376, 528, 741, 1040, 1460, 2050, 2878, 4040, 5671, 7960] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 207, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 3) with initial conditions, a(1) = 13, a(2) = 19, a(3) = 28, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 3) with initial conditions, b(1) = 13, b(2) = 19, b(3) = 28, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / In floating-point [1.465571232, -0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2307692308, 0.2631578947, 0.03571428571, -0.1951219512, 0.06666666667, 0.1022727273, -0.09302325581, -0.02645502646, 0.07581227437, -0.01724137931, -0.04369747899, 0.03211009174, 0.01486697966, -0.02883075280, 0.003278688525, 0.01814566244, -0.01068521031, -0.007406550168, 0.01073910297, 0.00005387931034, -0.007352670858, 0.003386429199, 0.003440308087, -0.003912363067, -0.0005259341307, 0.002914373953, -0.0009979891816, -0.001523923318, 0.001390450627, 0.0003924614404, -0.001131461878, 0.0002589887477, 0.0006514501880, -0.0004800116899, -0.0002210229423, 0.0004304272457, -0.00004958444422, -0.0002706073865, -6 0.0001598198592, 0.0001102354150, -0.0001603719715, -0.5521122886 10 , 0.0001096833027, -0.00005068866879, -0.00005124078108, 0.00005844252163, -5 0.7753852834 10 , -0.00004348692825, 0.00001495559338, 0.00002270944621, -5 -5 -0.00002077748203, -0.5821888655 10 , 0.00001688755756, -0.3889924476 10 , -5 -5 -5 -0.9711813131 10 , 0.7175744428 10 , 0.3285819952 10 , -5 -0.6425993179 10 ] The largest is 0.2631578947 The smallest is -0.2307692308 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 208, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 20, 31, 48, 74, 114, 176, 272, 420, 649, 1003, 1550, 2395, 3701, 5719, 8837, 13655, 21100, 32604, 50380] ] The sequence a(n) satisfies, for n>=, 8, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) - a(n - 5) + a(n - 7) with initial conditions, a(1) = 13, a(2) = 20, a(3) = 31, a(4) = 48, a(5) = 74, a(6) = 114, a(7) = 176, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) - b(n - 5) + b(n - 7) with initial conditions, b(1) = 13, b(2) = 20, b(3) = 31, b(4) = 48, b(5) = 74, b(6) = 114, b(7) = 176, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 7, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 7 6 5 2 t - t - t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6), RootOf(%1, index = 7)] 7 6 5 2 %1 := _Z - _Z - _Z + _Z - 1 In floating-point [1.54521564973276, 0.808712727401677 + 0.424846366714998 I, -0.212614366617129 + 0.953977000479370 I, -0.868706185650926 + 0.239062189977012 I, -0.868706185650926 - 0.239062189977012 I, -0.212614366617129 - 0.953977000479370 I, 0.808712727401677 - 0.424846366714998 I] The largest root is, 1.54521564973276 and the remaining roots are [0.808712727401677 + 0.424846366714998 I, -0.212614366617129 + 0.953977000479370 I, -0.868706185650926 + 0.239062189977012 I, -0.868706185650926 - 0.239062189977012 I, -0.212614366617129 - 0.953977000479370 I, 0.808712727401677 - 0.424846366714998 I] whose absolute values are [0.913515577739315, 0.977382722548142, 0.901000093043717, 0.901000093043717, 0.977382722548142, 0.913515577739315] so the largest absolute value is, 0.977382722548142 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.977382722548142 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2307692308, 0.05000000000, 0.3225806452, 0.08333333333, -0.3783783784, -0.2807017544, 0.3636363636, -0.4705882353, -0.1404761905, 0.09090909091, 0.3140578265, -0.3387096774, 0.1653444676, 0.3307214266, -0.06574575975, -0.1895439629, 0.1742951300, 0.1334597156, -0.3616734143, 0.002858277094, 0.1614428116, -0.07574069764, -0.2373020869, 0.2229266764, 0.1162258318, -0.1839651323, 0.01085969776, 0.2256394347, -0.06216821224, -0.1900567146, 0.1546667973, 0.06997621712, -0.1849615564, -0.04195743144, 0.1887771568, -0.07001528585, -0.1412710621, 0.1283420053, 0.09900459172, -0.1463921168, -0.01932967065, 0.1643264312, -0.05336053061, -0.1293097532, 0.09206383831, 0.08108834752, -0.1375663622, -0.02244715476, 0.1336226674, -0.03424885629, -0.1110242896, 0.08435705468, 0.07686826738, -0.1099637076, -0.02129373867, 0.1133895107, -0.02651013890, -0.1010131851] The largest is 0.3636363636 The smallest is -0.4705882353 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 209, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 13, a(2) = 21, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 13, b(2) = 21, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.07692307692, 0.04761904762, -0.02941176471, 0.01818181818, -0.01123595506, 0.006944444444, -0.004291845494, 0.002652519894, -0.001639344262, 0.001013171226, -0.0006261740764, 0.0003869969040, -0.0002391772303, 0.0001478196600, -0.00009135757354, 0.00005646208571, -0.00003489548801, -5 -5 0.00002156659765, -0.00001332889037, 0.8237707281 10 , -0.5091183089 10 , -5 -5 -5 0.3146524192 10 , -0.1944658897 10 , 0.1201865295 10 , -6 -6 -6 -0.7427936022 10 , 0.4590716928 10 , -0.2837219094 10 , -6 -6 -7 0.1753497834 10 , -0.1083721260 10 , 0.6697765733 10 , -7 -7 -7 -0.4139446871 10 , 0.2558318861 10 , -0.1581128010 10 , -8 -8 -8 0.9771908509 10 , -0.6039371593 10 , 0.3732536915 10 , -8 -8 -9 -0.2306834678 10 , 0.1425702237 10 , -0.8811324406 10 , -9 -9 -9 0.5445697969 10 , -0.3365626437 10 , 0.2080071532 10 , -9 -10 -10 -0.1285554906 10 , 0.7945166260 10 , -0.4910382795 10 , -10 -10 -10 0.3034783465 10 , -0.1875599330 10 , 0.1159184135 10 , -11 -11 -11 -0.7164151948 10 , 0.4427689404 10 , -0.2736462543 10 , -11 -11 -12 0.1691226861 10 , -0.1045235683 10 , 0.6459911781 10 , -12 -12 -12 -0.3992445045 10 , 0.2467466736 10 , -0.1524978309 10 , -13 0.9424884271 10 ] The largest is 0.04761904762 The smallest is -0.07692307692 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 210, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 22, 37, 62, 104, 174, 291, 487, 815, 1364, 2283, 3821, 6395, 10703, 17913, 29980, 50176, 83977, 140548, 235228] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) - a(n - 4) + a(n - 5) with initial conditions, a(1) = 13, a(2) = 22, a(3) = 37, a(4) = 62, a(5) = 104, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) - b(n - 4) + b(n - 5) with initial conditions, b(1) = 13, b(2) = 22, b(3) = 37, b(4) = 62, b(5) = 104, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 2 t + t - t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 2 _Z + _Z - _Z + _Z - 1 In floating-point [1.67364854629984, 0.610459929296212 + 0.637106092426574 I, -0.447284202446133 + 0.753239695112345 I, -0.447284202446133 - 0.753239695112345 I, 0.610459929296212 - 0.637106092426574 I] The largest root is, 1.67364854629984 and the remaining roots are [0.610459929296212 + 0.637106092426574 I, -0.447284202446133 + 0.753239695112345 I, -0.447284202446133 - 0.753239695112345 I, 0.610459929296212 - 0.637106092426574 I] whose absolute values are [0.882363586217946, 0.876032645539429, 0.876032645539429, 0.882363586217946] so the largest absolute value is, 0.882363586217946 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.882363586217946 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2307692308, 0.2272727273, -0.1081081081, 0.4516129032, 0.1153846154, -0.3275862069, 0.01374570447, -0.08829568788, -0.1828220859, 0.1796187683, 0.1121331581, -0.03611619995, 0.08975762314, -0.03466317855, -0.1277284654, 0.01721147432, 0.001614317602, -0.01729044858, 0.07408145260, 0.02212746782, -0.03151980370, 0.007819128028, -0.02208636134, -0.03155768526, 0.02043739191, 0.01100717434, -0.00007523884867, 0.01875106345, -0.003410537037, -0.01621715889, 0.0008096958817, -0.004400288682, -0.003665831650, 0.01068494312, 0.003608574427, -0.001923641349, 0.002494628960, -0.003829301070, -0.005000503760, 0.001855138287, 0.0004632089544, 0.0003947058924, 0.003352543807, -0.00008205137040, -0.001730011323, 0.00004307559336, -0.001223726775, -0.0007859452895, 0.001342871742, 0.0004748750824, 0.00008773550163, 0.0006056861774, -0.0005303050963, -0.0007105642084, 0.0001020024376, -0.0001336866885, 0.00005605125074, 0.0005280507397] The largest is 0.4516129032 The smallest is -0.3275862069 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 211, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 23, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 23, 41, 73, 130, 232, 414, 739, 1319, 2354, 4201, 7497, 13379, 23876, 42609, 76040, 135701, 242172, 432180, 771268] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) + a(n - 5) with initial conditions, a(1) = 13, a(2) = 23, a(3) = 41, a(4) = 73, a(5) = 130, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) + b(n - 5) with initial conditions, b(1) = 13, b(2) = 23, b(3) = 41, b(4) = 73, b(5) = 130, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 2 t - 2 t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 2 %1 := _Z - 2 _Z + _Z - 1 In floating-point [1.78459893336865, 0.748895482504927 + 0.616266848955766 I, -0.641194949189250 + 0.429636187896782 I, -0.641194949189250 - 0.429636187896782 I, 0.748895482504927 - 0.616266848955766 I] The largest root is, 1.78459893336865 and the remaining roots are [0.748895482504927 + 0.616266848955766 I, -0.641194949189250 + 0.429636187896782 I, -0.641194949189250 - 0.429636187896782 I, 0.748895482504927 - 0.616266848955766 I] whose absolute values are [0.969860439876870, 0.771827841436342, 0.771827841436342, 0.969860439876870] so the largest absolute value is, 0.969860439876870 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.969860439876870 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3076923077, 0.08695652174, -0.02439024390, -0.4931506849, 0.03076923077, -0.2241379310, 0.1328502415, 0.2097428958, 0.1493555724, 0.1966864911, -0.04046655558, -0.09750566893, -0.1820016444, -0.1741916569, -0.05419042925, 0.03315360337, 0.1429908402, 0.1581685744, 0.1089916239, 0.02080210770, -0.08341082996, -0.1328225983, -0.1282787833, -0.06415510872, 0.02531449087, 0.09549692707, 0.1223263587, 0.09105944209, 0.02246684878, -0.05207817042, -0.09971885629, -0.09957820289, -0.05601879328, 0.01014811854, 0.06779626953, 0.09189247602, 0.07405863060, 0.02430219839, -0.03313996070, -0.07254228246, -0.07749428731, -0.04778998332, 0.001264514216, 0.04688335504, 0.06901441094, 0.05927002036, 0.02386670235, -0.02001649202, -0.05241964935, -0.05969159011, -0.04009666784, -0.003906983974, 0.03186113014, 0.05139927877, 0.04701395142, 0.02207010486, -0.01116605303, -0.03748492734] The largest is 0.2097428958 The smallest is -0.4931506849 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 212, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 24, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 3) with initial conditions, a(1) = 13, a(2) = 24, a(3) = 44, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 3) with initial conditions, b(1) = 13, b(2) = 24, b(3) = 44, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (19 + 3 33 ) 4 (19 + 3 33 ) [----------------- + ------------------- + 1/3, - ----------------- 3 1/2 1/3 6 3 (19 + 3 33 ) 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(19 + 3 33 ) 4 | (19 + 3 33 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (19 + 3 33 ) / 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 |(19 + 3 33 ) 4 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (19 + 3 33 ) / In floating-point [1.839286755, -0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] The largest root is, 1.839286755 and the remaining roots are [-0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] whose absolute values are [0.7373527065, 0.7373527065] so the largest absolute value is, 0.7373527065 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7373527065 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3076923077, -0.3333333333, 0.1136363636, 0.08641975309, -0.1342281879, 0.06569343066, 0.01785714286, -0.05070118662, 0.03284457478, 0., -0.01785714286, 0.01498727496, -0.002869881617, -0.005739760379, 0.006377628310, -0.002232014166, -0.001594146438, 0.002551467581, -0.001274693041, -0.0003173719016, 0.0009594026349, -0.0006326623085, -5 0.9368424754 10 , 0.0003361087511, -0.0002871851327, 0.00005829204314, 0.0001072156615, -0.0001216774280, 0.00004383027661, 0.00002936851008, -5 -0.00004847864135, 0.00002472014534, 0.5610014078 10 , -0.00001814848193, -6 -5 -5 0.00001218167749, -0.3567903558 10 , -0.6323594790 10 , 0.5501292349 10 , -5 -5 -5 -0.1179092797 10 , -0.2001395238 10 , 0.2320804314 10 , -6 -6 -6 -0.8596837211 10 , -0.5402746452 10 , 0.9208459475 10 , -6 -7 -6 -0.4791124188 10 , -0.9854111662 10 , 0.3431924120 10 , -6 -7 -6 -0.2344611235 10 , 0.1019017190 10 , 0.1189214604 10 , -6 -7 -7 -0.1053494911 10 , 0.2376214118 10 , 0.3733411046 10 , -7 -7 -8 -0.4425323951 10 , 0.1684301214 10 , 0.9923883091 10 , -7 -8 -0.1748634428 10 , 0.9280550949 10 ] The largest is 0.3076923077 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 81, : Consider the Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 13, 25, 48, 92, 176, 337, 645, 1234, 2361, 4517, 8642, 16534, 31633, 60521, 115790, 221532, 423840, 810900, 1551432, 2968234, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 4) + a(n - 5) - a(n - 6) + a(n - 7), . Alas, it breaks down at the, 58, -th term. a(58), equals , 151210910848492306, while the corresponding term for the solution of the recurrence is , 151210910848492305 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.01075655064021 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 82, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 27, 56, 116, 240, 497, 1029, 2130, 4409, 9126, 18890, 39101, 80936, 167531, 346776, 717799, 1485787, 3075461, 6365960, 13177032] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 213, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 28, 60, 129, 277, 595, 1278, 2745, 5896, 12664, 27201, 58425, 125491, 269542, 578949, 1243524, 2670964, 5736961, 12322413, 26467299] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 13, a(2) = 28, a(3) = 60, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 13, b(2) = 28, b(3) = 60, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (188 + 12 93 ) 14 (188 + 12 93 ) [------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (188 + 12 93 ) 7 - --------------------- + 1/3 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 14 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (188 + 12 93 ) / 1/2 1/3 (188 + 12 93 ) 7 - ------------------- - --------------------- + 1/3 12 1/2 1/3 3 (188 + 12 93 ) / 1/2 1/3 \ 1/2 |(188 + 12 93 ) 14 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (188 + 12 93 ) / In floating-point [2.147899035, -0.5739495177 + 0.3689894078 I, -0.5739495177 - 0.3689894078 I] The largest root is, 2.147899035 and the remaining roots are [-0.5739495177 + 0.3689894078 I, -0.5739495177 - 0.3689894078 I] whose absolute values are [0.6823278039, 0.6823278039] so the largest absolute value is, 0.6823278039 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6823278039 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3076923077, -0.4285714286, 0.3500000000, -0.2015503876, 0.06859205776, 0.01512605042, -0.04929577465, 0.04954462659, -0.03392130258, 0.01587176248, -0.002426381383, -0.004604193410, 0.006414802655, -0.005219965720, 0.003005446076, -0.001019682773, -0.0002287563591, 0.0007373241687, -0.0007398713223, 0.0005060206559, -0.0002363978200, 0.00003577216951, 0.00006899718539, -0.00009585629560, 0.00007791024469, -0.00004480516112, -5 0.00001515903265, 0.3458955094 10 , -0.00001102814072, 0.00001104880212, -5 -5 -6 -0.7548524235 10 , 0.3520939277 10 , -0.5273070744 10 , -5 -5 -5 -0.1033952755 10 , 0.1432372374 10 , -0.1162840209 10 , -6 -6 -7 0.6679517843 10 , -0.2253562603 10 , -0.5229290111 10 , -6 -6 -6 0.1649463625 10 , -0.1649957001 10 , 0.1126041238 10 , -7 -8 -7 -0.5244091381 10 , 0.7771633758 10 , 0.1549392995 10 , -7 -7 -8 -0.2140371635 10 , 0.1735577731 10 , -0.9957725445 10 , -8 -9 -8 -8 0.3350112816 10 , 0.7904392311 10 , -0.2467060582 10 , 0.2463930696 10 , -8 -9 -9 -0.1679751236 10 , 0.7810495747 10 , -0.1145222013 10 , -9 -9 -9 -0.2321742882 10 , 0.3198308839 10 , -0.2590398938 10 ] The largest is 0.3500000000 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 214, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 29, 65, 146, 328, 737, 1656, 3721, 8361, 18787, 42214, 94854, 213135, 478910, 1076101, 2417977, 5433145, 12208166, 27431500, 61638021] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) - a(n - 3) with initial conditions, a(1) = 13, a(2) = 29, a(3) = 65, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) - b(n - 3) with initial conditions, b(1) = 13, b(2) = 29, b(3) = 65, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (28 + 84 I 3 ) 14 (28 + 84 I 3 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (28 + 84 I 3 ) 7 - --------------------- + 2/3 1/2 1/3 3 (28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(28 + 84 I 3 ) 14 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (28 + 84 I 3 ) / 1/2 1/3 (28 + 84 I 3 ) 7 - ------------------- - --------------------- + 2/3 12 1/2 1/3 3 (28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(28 + 84 I 3 ) 14 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (28 + 84 I 3 ) / In floating-point -9 -9 [2.246979604 + 0.1 10 I, -0.8019377358 - 0.1866025404 10 I, -10 0.5549581322 - 0.1339745960 10 I] -9 The largest root is, 2.246979604 + 0.1 10 I and the remaining roots are -9 -10 [-0.8019377358 - 0.1866025404 10 I, 0.5549581322 - 0.1339745960 10 I] whose absolute values are [0.8019377358, 0.5549581322] so the largest absolute value is, 0.8019377358 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8019377358 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3076923077, -0.3103448276, -0.06153846154, -0.1232876712, 0.003048780488, -0.05563093623, 0.01509661836, -0.02848696587, 0.01375433561, -0.01607494544, 0.01009143886, -0.009646403947, 0.006873577779, -0.005990687185, 0.004538607436, -0.003787050084, 0.002955194459, -0.002415268600, 0.001911707344, -0.001547048371, 0.001232879201, -0.0009929973134, 0.0007939329454, -0.0006380106234, 0.0005109090120, -0.0004101255449, 0.0003286685456, -0.0002636974657, 0.0002113991590, -0.0001695676932, 0.0001359612383, -0.0001090443757, 0.00008744018015, -0.00007012525368, 0.00005623404847, -0.00004509733689, 0.00003616462836, -0.00002900212863, 0.00002325770800, -0.00001865134100, 0.00001495715462, -5 -5 -5 -0.00001199473976, 0.9619016103 10 , -0.7713862173 10 , 0.6186031517 10 , -5 -5 -5 -0.4960815243 10 , 0.3978263204 10 , -0.3190320352 10 , -5 -5 -5 0.2558437744 10 , -0.2051708069 10 , 0.1645341958 10 , -5 -5 -6 -0.1319461896 10 , 0.1058126235 10 , -0.8485513850 10 , -6 -6 -6 0.6804853607 10 , -0.5457068981 10 , 0.4376229495 10 , -6 -0.3509463599 10 ] The largest is 0.01509661836 The smallest is -0.3103448276 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 83, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 30, 69, 159, 366, 842, 1937, 4456, 10251, 23582, 54249, 124797, 287089, 660433, 1519291, 3495048, 8040172, 18495988, 42549037, 97881797] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 84, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 31, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 31, 74, 177, 423, 1011, 2416, 5774, 13799, 32978, 78814, 188357, 450153, 1075817, 2571086, 6144617, 14684969, 35095485, 83874407, 200450746] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 215, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 32, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 32, 79, 195, 481, 1186, 2924, 7209, 17773, 43817, 108025, 266321, 656578, 1618703, 3990690, 9838498, 24255465, 59798516, 147425024, 363456138] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) + a(n - 3) - a(n - 4) - a(n - 5) - a(n - 6) with initial conditions, a(1) = 13, a(2) = 32, a(3) = 79, a(4) = 195, a(5) = 481, a(6) = 1186, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) + b(n - 3) - b(n - 4) - b(n - 5) - b(n - 6) with initial conditions, b(1) = 13, b(2) = 32, b(3) = 79, b(4) = 195, b(5) = 481, b(6) = 1186, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 2 t - 2 t - t - t + t + t + 1 = 0 whose roots are [1, RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - _Z - 2 _Z - 3 _Z - 2 _Z - 1 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [2.46536258199649, -0.142126711788500 + 0.814294434570713 I, -0.590554579209747 + 0.494859189477967 I, -0.590554579209747 - 0.494859189477967 I, -0.142126711788500 - 0.814294434570713 I] The largest root is, 2.46536258199649 and the remaining roots are [-0.142126711788500 + 0.814294434570713 I, -0.590554579209747 + 0.494859189477967 I, -0.590554579209747 - 0.494859189477967 I, -0.142126711788500 - 0.814294434570713 I] whose absolute values are [0.826604759468907, 0.770480582777004, 0.770480582777004, 0.826604759468907] so the largest absolute value is, 0.826604759468907 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.826604759468907 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2307692308, 0.03125000000, 0.3291139240, 0.4666666667, 0.3160083160, -0.08263069140, 0.4887140903, 0.3850742128, 0.06549260114, 0.3050870667, 0.3387271465, 0.2569117719, 0.2183883712, 0.2767684992, 0.3195322112, 0.2334967187, 0.2492673713, 0.2994951915, 0.2670654610, 0.2530960889, 0.2704565420, 0.2788153569, 0.2653553228, 0.2603258040, 0.2742041961, 0.2717215315, 0.2633458415, 0.2681209269, 0.2714239038, 0.2680630445, 0.2663993506, 0.2690973496, 0.2697664221, 0.2674216691, 0.2678208110, 0.2692699686, 0.2685192950, 0.2678439287, 0.2684682188, 0.2687872125, 0.2682764979, 0.2681752349, 0.2685827378, 0.2685178485, 0.2682617404, 0.2683851217, 0.2685153617, 0.2684017643, 0.2683416853, 0.2684357859, 0.2684527975, 0.2683808184, 0.2683914090, 0.2684371985, 0.2684163559, 0.2683919176, 0.2684123648, 0.2684235771] The largest is 0.4887140903 The smallest is -0.2307692308 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 216, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 33, 84, 214, 545, 1388, 3535, 9003, 22929, 58396, 148724, 378773, 964666, 2456829, 6257097, 15935689, 40585304, 103363394, 263247781, 670444260] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) + a(n - 3) with initial conditions, a(1) = 13, a(2) = 33, a(3) = 84, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) + b(n - 3) with initial conditions, b(1) = 13, b(2) = 33, b(3) = 84, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (244 + 36 29 ) 14 (244 + 36 29 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (244 + 36 29 ) 7 - --------------------- + 2/3 1/2 1/3 3 (244 + 36 29 ) / 1/2 1/3 \ 1/2 |(244 + 36 29 ) 14 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (244 + 36 29 ) / 1/2 1/3 (244 + 36 29 ) 7 - ------------------- - --------------------- + 2/3 12 1/2 1/3 3 (244 + 36 29 ) / 1/2 1/3 \ 1/2 |(244 + 36 29 ) 14 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (244 + 36 29 ) / In floating-point [2.546818277, -0.2734091383 + 0.5638210935 I, -0.2734091383 - 0.5638210935 I] The largest root is, 2.546818277 and the remaining roots are [-0.2734091383 + 0.5638210935 I, -0.2734091383 - 0.5638210935 I] whose absolute values are [0.6266153385, 0.6266153385] so the largest absolute value is, 0.6266153385 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6266153385 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2307692308, -0.1818181818, 0.1904761905, -0.03271028037, -0.05688073394, 0.04394812680, -0.001697312588, -0.01632789070, 0.009594836234, 0.001164463319, -0.004404131142, 0.001951036637, 0.0006624054336, -0.001128283653, 0.0003568747616, 0.0002478713032, -0.0002756662855, -5 0.00005341349376, 0.00007903200521, -0.00006418878133, 0.4067936305 10 , -5 -5 0.00002297909649, -0.00001416265205, -0.1278271296 10 , 0.6259901853 10 , -5 -6 -5 -0.2921119635 10 , -0.8606087114 10 , 0.1617564796 10 , -6 -6 -6 -0.5465987537 10 , -0.3362414227 10 , 0.3984831969 10 , -7 -6 -7 -0.8587378260 10 , -0.1095057910 10 , 0.9359783232 10 , -8 -7 -7 -0.8183908972 10 , -0.3227577663 10 , 0.2086237008 10 , -8 -8 -8 -8 0.1265054561 10 , -0.8883297427 10 , 0.4360829790 10 , 0.1103416714 10 , -8 -9 -9 -0.2315634209 10 , 0.8329780863 10 , 0.4537386779 10 , -9 -9 -9 -0.5751787668 10 , 0.1363592306 10 , 0.1512783722 10 , -9 -10 -10 -0.1362627918 10 , 0.1511201906 10 , 0.4523961846 10 , -10 -12 -10 -0.3067153586 10 , -0.9914342046 10 , 0.1258521419 10 , -11 -11 -11 -0.6492541688 10 , -0.1391303390 10 , 0.3310065721 10 , -11 -12 -0.1263713636 10 , -0.6086649401 10 ] The largest is 0.1904761905 The smallest is -0.2307692308 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 217, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437, 1134903170] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) with initial conditions, a(1) = 13, a(2) = 34, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) with initial conditions, b(1) = 13, b(2) = 34, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - 3 t + 1 = 0 whose roots are 1/2 1/2 5 5 [3/2 + ----, 3/2 - ----] 2 2 In floating-point [2.618033988, 0.381966012] The largest root is, 2.618033988 and the remaining roots are [0.381966012] whose absolute values are [0.381966012] so the largest absolute value is, 0.381966012 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.381966012 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.07692307692, -0.02941176471, -0.01123595506, -0.004291845494, -0.001639344262, -0.0006261740764, -0.0002391772303, -0.00009135757354, -5 -0.00003489548801, -0.00001332889037, -0.5091183089 10 , -5 -6 -6 -0.1944658897 10 , -0.7427936022 10 , -0.2837219094 10 , -6 -7 -7 -0.1083721260 10 , -0.4139446871 10 , -0.1581128010 10 , -8 -8 -9 -0.6039371593 10 , -0.2306834678 10 , -0.8811324406 10 , -9 -9 -10 -0.3365626437 10 , -0.1285554906 10 , -0.4910382795 10 , -10 -11 -11 -0.1875599330 10 , -0.7164151948 10 , -0.2736462543 10 , -11 -12 -12 -0.1045235683 10 , -0.3992445045 10 , -0.1524978309 10 , -13 -13 -14 -0.5824898820 10 , -0.2224913368 10 , -0.8498412846 10 , -14 -14 -15 -0.3246104857 10 , -0.1239901724 10 , -0.4736003159 10 , -15 -16 -16 -0.1808992236 10 , -0.6909735488 10 , -0.2639284103 10 , -16 -17 -17 -0.1008116821 10 , -0.3850663611 10 , -0.1470822620 10 , -18 -18 -19 -0.5618042495 10 , -0.2145901283 10 , -0.8196613536 10 , -19 -19 -20 -0.3130827778 10 , -0.1195869798 10 , -0.4567816168 10 , -20 -21 -21 -0.1744750522 10 , -0.6664353975 10 , -0.2545556705 10 , -22 -22 -22 -0.9723161412 10 , -0.3713917181 10 , -0.1418590132 10 , -23 -23 -24 -0.5418532142 10 , -0.2069695109 10 , -0.7905531854 10 , -24 -24 -0.3019644469 10 , -0.1153401553 10 ] The largest is -24 -0.1153401553 10 The smallest is -0.07692307692 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 218, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 35, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 35, 94, 252, 676, 1813, 4862, 13039, 34968, 93777, 251491, 674448, 1808733, 4850656, 13008478, 34886106, 93557478, 250902227, 672869009, 1804498544] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) + a(n - 3) - 2 a(n - 4) + a(n - 5) with initial conditions, a(1) = 13, a(2) = 35, a(3) = 94, a(4) = 252, a(5) = 676, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) + b(n - 3) - 2 b(n - 4) + b(n - 5) with initial conditions, b(1) = 13, b(2) = 35, b(3) = 94, b(4) = 252, b(5) = 676, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 3 t + t - t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 3 _Z + _Z - _Z + 2 _Z - 1 In floating-point [2.68179767508346, 0.617204494100616 + 0.265082191819291 I, -0.458103331642344 + 0.785207530925176 I, -0.458103331642344 - 0.785207530925176 I, 0.617204494100616 - 0.265082191819291 I] The largest root is, 2.68179767508346 and the remaining roots are [0.617204494100616 + 0.265082191819291 I, -0.458103331642344 + 0.785207530925176 I, -0.458103331642344 - 0.785207530925176 I, 0.617204494100616 - 0.265082191819291 I] whose absolute values are [0.671721635767166, 0.909070695316611, 0.909070695316611, 0.671721635767166] so the largest absolute value is, 0.909070695316611 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.909070695316611 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2307692308, 0.4571428571, -0.4255319149, 0.3968253968, 0.3801775148, -0.3656922228, 0.2272727273, 0.2086816474, -0.3305593686, 0.1384667882, 0.1344024239, -0.2559115010, 0.1061306450, 0.1012127020, -0.1887421419, 0.08491695806, 0.07653292022, -0.1403550994, 0.06601572580, 0.05635913885, -0.1054422910, 0.05057283280, 0.04113337732, -0.07931754378, 0.03873054506, 0.03005459964, -0.05957821175, 0.02970977504, 0.02198350262, -0.04471613315, 0.02278889612, 0.01606856230, -0.03355657258, 0.01746638499, 0.01173036446, -0.02518009266, 0.01337745000, 0.008553464572, -0.01889149288, 0.01024005657, 0.006230134499, -0.01417062510, 0.007834497117, 0.004532644921, -0.01062739988, 0.005991037260, 0.003293537245, -0.007968618126, 0.004579090310, 0.002389951906, -0.005973889949, 0.003498242056, 0.001731769277, -0.004477637677, 0.002671291552, 0.001252907548, -0.003355503082, 0.002038919390] The largest is 0.4571428571 The smallest is -0.4255319149 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 85, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 36, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 36, 100, 278, 773, 2149, 5974, 16607, 46165, 128332, 356744, 991696, 2756769, 7663412, 21303157, 59219640, 164621880, 457624588, 1272128975, 3536331245] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 219, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 37, 105, 298, 846, 2402, 6820, 19364, 54980, 156104, 443224, 1258440, 3573072, 10144976, 28804496, 81784224, 232208864, 659307552, 1871963200, 5315040320] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 4 a(n - 1) - 4 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 13, a(2) = 37, a(3) = 105, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 4 b(n - 1) - 4 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 13, b(2) = 37, b(3) = 105, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 4 t + 4 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (19 + 3 33 ) 4 (19 + 3 33 ) [----------------- + ------------------- + 4/3, - ----------------- 3 1/2 1/3 6 3 (19 + 3 33 ) 2 - ------------------- + 4/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(19 + 3 33 ) 4 | (19 + 3 33 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (19 + 3 33 ) / 2 - ------------------- + 4/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 |(19 + 3 33 ) 4 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (19 + 3 33 ) / In floating-point [2.839286755, 0.5803566220 + 0.6062907300 I, 0.5803566220 - 0.6062907300 I] The largest root is, 2.839286755 and the remaining roots are [0.5803566220 + 0.6062907300 I, 0.5803566220 - 0.6062907300 I] whose absolute values are [0.8392867555, 0.8392867555] so the largest absolute value is, 0.8392867555 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8392867555 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3076923077, -0.02702702703, -0.2476190476, -0.2684563758, -0.1371158392, 0.02997502082, 0.1313782991, 0.1313778145, 0.05994907239, -0.02295905294, -0.06887713662, -0.06377419662, -0.02550634300, 0.01531713826, 0.03574553084, 0.03070088432, 0.01045569044, -0.009489713838, -0.01837984849, -0.01464915773, -0.004056664634, 0.005610275406, 0.009369444697, 0.006923347895, 0.001436163604, -0.003209847770, -0.004737349705, -0.003237680532, -0.0004210188498, 0.001791947321, 0.002376503617, 0.001496187487, 0.00006263011984, -0.0009812222338, -0.001183034441, -0.0006819885877, 0.00004173894410, 0.0005288412460, 0.0005844320322, 0.0003058410329, -0.00005668150513, -0.0002812260877, -0.0002864962646, -0.0001344437179, 0.00004575801154, 0.0001478143884, 0.0001393380718, 0.00005761075654, -0.00003128048417, -0.00007688881924, -0.00006721182721, -0.00002385300019, 0.00001965766956, 0.00003961902460, -5 0.00003213941979, 0.9396919859 10 , -0.00001173195051, -0.00002023664191] The largest is 0.3076923077 The smallest is -0.2684563758 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 220, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 13, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [13, 38, 111, 324, 946, 2762, 8064, 23544, 68740, 200696, 585960, 1710792, 4994896, 14583296, 42577968, 124312320, 362947168, 1059674912, 3093868800, 9032981760] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 13, a(2) = 38, a(3) = 111, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 13, b(2) = 38, b(3) = 111, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 2 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (53 + 3 201 ) 10 (53 + 3 201 ) [------------------ + -------------------- + 2/3, - ------------------ 3 1/2 1/3 6 3 (53 + 3 201 ) 5 - -------------------- + 2/3 1/2 1/3 3 (53 + 3 201 ) / 1/2 1/3 \ 1/2 |(53 + 3 201 ) 10 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (53 + 3 201 ) / 1/2 1/3 (53 + 3 201 ) 5 - ------------------ - -------------------- + 2/3 6 1/2 1/3 3 (53 + 3 201 ) / 1/2 1/3 \ 1/2 |(53 + 3 201 ) 10 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (53 + 3 201 ) / In floating-point [2.919639566, -0.4598197833 + 0.6881728200 I, -0.4598197833 - 0.6881728200 I] The largest root is, 2.919639566 and the remaining roots are [-0.4598197833 + 0.6881728200 I, -0.4598197833 - 0.6881728200 I] whose absolute values are [0.8276569720, 0.8276569720] so the largest absolute value is, 0.8276569720 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8276569720 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.07692307692, 0.2368421053, -0.2702702703, 0.08641975309, 0.1057082452, -0.1564083997, 0.07142857143, 0.04145429834, -0.08705266221, 0.05166022243, 0.01212369445, -0.04653751011, 0.03449281026, 0.0001579889759, -0.02377342197, 0.02175475448, -0.003721357043, -0.01148004908, 0.01310669670, -0.004189418844, -0.005125542444, 0.007583470830, -0.003462980917, -0.002010105063, 0.004220769700, -0.002504632559, -5 -0.0005879358428, 0.002256402597, -0.001672331609, -0.7729708546 10 , 0.001152682560, -0.001054757514, 0.0001803906739, 0.0005566314387, -0.0006354708038, 0.0002031026176, 0.0002485265050, -0.0003676833625, 0.0001678915200, 0.00009746932480, -0.0002046450355, 0.0001214316185, -6 0.00002851181557, -0.0001094032029, 0.00008108046237, 0.3781500864 10 , -5 -0.00005588918088, 0.00005113886315, -0.8744335274 10 , -0.00002698930600, -5 0.00003081044377, -0.9846395002 10 , -0.00001205051446, 0.00001782706862, -5 -5 -5 -0.8139681678 10 , -0.4726255031 10 , 0.9922263820 10 , -5 -0.5887345779 10 ] The largest is 0.2368421053 The smallest is -0.2702702703 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 86, :The Pisot Sequence a(n), defined by, a(1) = 13, a(2) = 40, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [13, 40, 123, 378, 1162, 3572, 10980, 33752, 103752, 318929, 980373, 3013621, 9263731, 28476279, 87534760, 269077790, 827132639, 2542567346, 7815734024, 24025203671] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 87, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 16, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52 ] is a trivial linear sequence Fact, 88, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 17, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 17, 21, 26, 32, 39, 48, 59, 73, 90, 111, 137, 169, 208, 256, 315, 388, 478, 589, 726] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 89, : Consider the Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 18, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 14, 18, 23, 29, 37, 47, 60, 77, 99, 127, 163, 209, 268, 344, 442, 568, 730, 938, 1205, 1548, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 2) + a(n - 6) - a(n - 7), . Alas, it breaks down at the, 29, -th term. a(29), equals , 14799, while the corresponding term for the solution of the recurrence is , 14800 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03282504251949 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Theorem , 221, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 19, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657, 907, 1252, 1728, 2385, 3292, 4544, 6272] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 4) with initial conditions, a(1) = 14, a(2) = 19, a(3) = 26, a(4) = 36, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 4) with initial conditions, b(1) = 14, b(2) = 19, b(3) = 26, b(4) = 36, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - _Z - 1 In floating-point [1.38027756909761, 0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] The largest root is, 1.38027756909761 and the remaining roots are [0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] whose absolute values are [0.940435682699417, 0.819172513396165, 0.940435682699417] so the largest absolute value is, 0.940435682699417 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.940435682699417 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2142857143, -0.4210526316, -0.1538461538, 0.4444444444, 0.2200000000, -0.2028985507, -0.3578947368, 0.08396946565, 0.3038674033, 0.1000000000, -0.2579710145, -0.1743697479, 0.1293759513, 0.2293274531, -0.02875399361, -0.2031250000, -0.07379454927, 0.1555285541, 0.1267605634, -0.07637117347, -0.1501674945, 0.005356096744, 0.1321166556, 0.05574346585, -0.09442428871, -0.08906873256, 0.04304757450, 0.09879098137, 0.004366467633, -0.08470226525, -0.04165477757, 0.05713618859, 0.06150263548, -0.02319964717, -0.06485442654, -0.007718248118, 0.05378438726, 0.03058473642, -0.03426969099, -0.04198793988, 0.01179644653, 0.04238118289, 0.008111491454, -0.03387644844, -0.02208000207, 0.02030118078, 0.02841267220, -0.005463776279, -0.02754377835, -0.007242597591, 0.02117007461, 0.01570629833, -0.01183748002, -0.01908007762, 0.002089996993, 0.01779629532, 0.005958815294, -0.01312126232] The largest is 0.4444444444 The smallest is -0.4210526316 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 222, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 20, 29, 42, 61, 89, 130, 190, 278, 407, 596, 873, 1279, 1874, 2746, 4024, 5897, 8642, 12665, 18561] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 14, a(2) = 20, a(3) = 29, a(4) = 42, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 14, b(2) = 20, b(3) = 29, b(4) = 42, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t + t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [1, ------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.465571232, -0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, 0.05000000000, -0.1724137931, -0.4047619048, -0.1475409836, -0.1123595506, -0.3076923077, -0.2421052632, -0.1402877698, -0.2334152334, -0.2600671141, -0.1844215349, -0.2017200938, -0.2454642476, -0.2134013110, -0.1985586481, -0.2274037646, -0.2241379310, -0.2060007896, -0.2166909110, -0.2241011690, -0.2133647720, -0.2133125663, -0.2206663786, -0.2172807828, -0.2138410261, -0.2177537536, -0.2182799455, -0.2153657483, -0.2163638549, -0.2178878591, -0.2164974644, -0.2161050407, -0.2172365282, -0.2169775571, -0.2163261190, -0.2168061388, -0.2170271673, -0.2165967439, -0.2166463309, -0.2169169401, -0.2167571215, -0.2166468869, -0.2168072594, -0.2168078119, -0.2166981289, -0.2167488177, -0.2168000586, -0.2167416162, -0.2167338624, -0.2167773493, -0.2167623938, -0.2167396844, -0.2167604618, -0.2167662837, -0.2167493962, -0.2167532861, -0.2167629979] The largest is 0.05000000000 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 90, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 21, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 21, 32, 49, 75, 115, 176, 269, 411, 628, 960, 1468, 2245, 3433, 5250, 8029, 12279, 18779, 28720, 43923] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 223, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 22, 35, 56, 90, 145, 234, 378, 611, 988, 1598, 2585, 4182, 6766, 10947, 17712, 28658, 46369, 75026, 121394] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) with initial conditions, a(1) = 14, a(2) = 22, a(3) = 35, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) with initial conditions, b(1) = 14, b(2) = 22, b(3) = 35, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + 1 = 0 whose roots are 1/2 1/2 5 5 [1, 1/2 - ----, ---- + 1/2] 2 2 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [-0.6180339880, 1.618033988] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4285714286, -0.3181818182, -0.4000000000, -0.3571428571, -0.3888888889, -0.3724137931, -0.3846153846, -0.3783068783, -0.3829787234, -0.3805668016, -0.3823529412, -0.3814313346, -0.3821138211, -0.3817617499, -0.3820224719, -0.3818879855, -0.3819875776, -0.3819362074, -0.3819742489, -0.3819546271, -0.3819691578, -0.3819616629, -0.3819672131, -0.3819643503, -0.3819664703, -0.3819653768, -0.3819661866, -0.3819657689, -0.3819660782, -0.3819659187, -0.3819660368, -0.3819659759, -0.3819660210, -0.3819659977, -0.3819660150, -0.3819660061, -0.3819660127, -0.3819660093, -0.3819660118, -0.3819660105, -0.3819660115, -0.3819660110, -0.3819660113, -0.3819660111, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660113, -0.3819660112, -0.3819660112, -0.3819660112] The largest is -0.3181818182 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 91, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 23, 38, 63, 104, 172, 284, 469, 775, 1281, 2117, 3499, 5783, 9558, 15797, 26109, 43152, 71320, 117875, 194819] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 92, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 24, 41, 70, 120, 206, 354, 608, 1044, 1793, 3079, 5287, 9078, 15587, 26763, 45952, 78899, 135469, 232599, 399370] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 93, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 25, 45, 81, 146, 263, 474, 854, 1539, 2773, 4996, 9001, 16217, 29218, 52642, 94845, 170882, 307878, 554704, 999411] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 94, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 26, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 26, 48, 89, 165, 306, 567, 1051, 1948, 3611, 6694, 12409, 23003, 42641, 79044, 146525, 271616, 503499, 933344, 1730154] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 95, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 27, 52, 100, 192, 369, 709, 1362, 2616, 5025, 9652, 18540, 35612, 68404, 131391, 252377, 484768, 931147, 1788556, 3435475] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 224, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 29, 60, 124, 256, 529, 1093, 2258, 4665, 9638, 19912, 41138, 84991, 175591, 362770, 749481, 1548424, 3199036, 6609192, 13654557] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + 2 a(n - 3) + a(n - 4) with initial conditions, a(1) = 14, a(2) = 29, a(3) = 60, a(4) = 124, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + 2 b(n - 3) + b(n - 4) with initial conditions, b(1) = 14, b(2) = 29, b(3) = 60, b(4) = 124, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - _Z - _Z - 2 _Z - 1 In floating-point [2.06599489201647, -0.269448305785011 + 0.919611948070651 I, -0.527098280446453, -0.269448305785011 - 0.919611948070651 I] The largest root is, 2.06599489201647 and the remaining roots are [-0.269448305785011 + 0.919611948070651 I, -0.527098280446453, -0.269448305785011 - 0.919611948070651 I] whose absolute values are [0.958273721086366, 0.527098280446453, 0.958273721086366] so the largest absolute value is, 0.958273721086366 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.958273721086366 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.07142857143, 0.1379310345, 0.2666666667, -0.4838709677, 0.1289062500, 0.3156899811, -0.2570905764, -0.1678476528, 0.3352625938, -0.03112678979, -0.2886701487, 0.1828722835, 0.1672059395, -0.2583902364, -0.01411086915, 0.2447827230, -0.1189028328, -0.1607321706, 0.1958195495, 0.04206441849, -0.2024832113, 0.07048813273, 0.1479533065, -0.1444605654, -0.05801420511, 0.1639199751, -0.03506205442, -0.1316310550, 0.1031326356, 0.06529744683, -0.1298940820, 0.01003758105, 0.1138710283, -0.07058210778, -0.06652999935, 0.1006675306, 0.006844343970, -0.09613023195, 0.04551917378, 0.06374516033, -0.07615178582, -0.01749850988, 0.07935919873, -0.02669772247, -0.05848732933, 0.05603483577, 0.02351126023, -0.06412628514, 0.01296731731, 0.05189838840, -0.03987560433, -0.02616886645, 0.05071962333, -0.003302063389, -0.04479577730, 0.02717253952, 0.02649225877, -0.03922881970] The largest is 0.3352625938 The smallest is -0.4838709677 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 96, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 30, 64, 137, 293, 627, 1342, 2872, 6146, 13152, 28144, 60225, 128875, 275779, 590138, 1262833, 2702329, 5782698, 12374362, 26479826] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 225, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 31, 69, 154, 344, 768, 1715, 3830, 8553, 19100, 42653, 95250, 212706, 475001, 1060741, 2368777, 5289797, 11812827, 26379629, 58909254] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 3) - a(n - 4) + a(n - 6) with initial conditions, a(1) = 14, a(2) = 31, a(3) = 69, a(4) = 154, a(5) = 344, a(6) = 768, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 3) - b(n - 4) + b(n - 6) with initial conditions, b(1) = 14, b(2) = 31, b(3) = 69, b(4) = 154, b(5) = 344, b(6) = 768, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 2 t - 3 t + 2 t - t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 3 2 %1 := _Z - 3 _Z + 2 _Z - _Z + _Z - 1 In floating-point [2.23313428742442, 0.873142899848906 + 0.467754628298683 I, -0.199095014422082 + 0.863470624391878 I, -0.581230058278069, -0.199095014422082 - 0.863470624391878 I, 0.873142899848906 - 0.467754628298683 I] The largest root is, 2.23313428742442 and the remaining roots are [0.873142899848906 + 0.467754628298683 I, -0.199095014422082 + 0.863470624391878 I, -0.581230058278069, -0.199095014422082 - 0.863470624391878 I, 0.873142899848906 - 0.467754628298683 I] whose absolute values are [0.990541728475584, 0.886126595896674, 0.581230058278069, 0.886126595896674, 0.990541728475584] so the largest absolute value is, 0.990541728475584 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.990541728475584 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3571428571, -0.4193548387, -0.2898550725, 0.4155844156, -0.3953488372, -0.2799479167, 0.2944606414, 0.2112271540, -0.1296621069, 0.1784816754, 0.3160856212, -0.02954330709, -0.1182148129, 0.05327146680, -0.07904662872, -0.2538736234, -0.1159560565, -0.001982336658, -0.06707672045, -0.006076447683, 0.1508511806, 0.1457384219, 0.08055712017, 0.1051398084, 0.08211570560, -0.03519024934, -0.09436829047, -0.09001005360, -0.1180424147, -0.1281453695, -0.06187733681, 0.007436118243, 0.04159178301, 0.08616109164, 0.1265707493, 0.1134003599, 0.06975155291, 0.03029971483, -0.02018256773, -0.07863484821, -0.1084204979, -0.1050737200, -0.08708089178, -0.05058117019, 0.005582483163, 0.05726776990, 0.08871756704, 0.1027070946, 0.09529054482, 0.06132607211, 0.01296913746, -0.03389351177, -0.07286571590, -0.09645996417, -0.09522056511, -0.07038789901, -0.03134767762, 0.01407865244] The largest is 0.4155844156 The smallest is -0.4193548387 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 97, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 32, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 32, 73, 167, 382, 874, 2000, 4577, 10474, 23969, 54851, 125522, 287247, 657342, 1504275, 3442414, 7877691, 18027470, 41254433, 94407493] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 226, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 33, 78, 184, 434, 1024, 2416, 5700, 13448, 31728, 74856, 176608, 416672, 983056, 2319328, 5472000, 12910112, 30458880, 71861760, 169543744] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 3) with initial conditions, a(1) = 14, a(2) = 33, a(3) = 78, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 3) with initial conditions, b(1) = 14, b(2) = 33, b(3) = 78, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (35 + 3 129 ) 4 (35 + 3 129 ) [------------------ + -------------------- + 2/3, - ------------------ 3 1/2 1/3 6 3 (35 + 3 129 ) 2 - -------------------- + 2/3 1/2 1/3 3 (35 + 3 129 ) / 1/2 1/3 \ 1/2 |(35 + 3 129 ) 4 | + 1/2 I 3 |------------------ - --------------------|, | 3 1/2 1/3| \ 3 (35 + 3 129 ) / 1/2 1/3 (35 + 3 129 ) 2 - ------------------ - -------------------- + 2/3 6 1/2 1/3 3 (35 + 3 129 ) / 1/2 1/3 \ 1/2 |(35 + 3 129 ) 4 | - 1/2 I 3 |------------------ - --------------------|] | 3 1/2 1/3| \ 3 (35 + 3 129 ) / In floating-point [2.359304086, -0.1796520430 + 0.9030131455 I, -0.1796520430 - 0.9030131455 I] The largest root is, 2.359304086 and the remaining roots are [-0.1796520430 + 0.9030131455 I, -0.1796520430 - 0.9030131455 I] whose absolute values are [0.9207103766, 0.9207103766] so the largest absolute value is, 0.9207103766 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.9207103766 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2142857143, 0.3636363636, 0.05128205128, -0.3260869565, 0.07373271889, 0.2500000000, -0.1523178808, -0.1571929825, 0.1856038073, 0.06656580938, -0.1812546756, 0.008697227759, 0.1505260733, -0.06145733305, -0.1055202196, 0.09001169591, 0.05710872222, -0.09682299546, -0.01362259984, 0.08697224476, -0.01970150152, -0.06664820271, 0.04064808409, 0.04189316514, -0.04951007514, -0.01772398210, 0.04833836607, -0.002343418139, -0.04013480049, 0.01640713117, 0.02812742605, -0.02401474886, -0.01521523539, 0.02582438132, 0.003619264923, -0.02319194094, 0.005264880767, 0.01776829138, -0.01084729912, -0.01116483671, 0.01320690934, 0.004719220425, -0.01289123257, 0.0006313535250, 0.01070114790, -0.004380169345, -0.007497631641, 0.006407032519, 0.004053726347, -0.006887810588, -0.0009615561376, 0.006184340419, -0.001406940338, -0.004736992950, 0.002894694937, 0.002975509198, -0.003522967505, -0.001256545136] The largest is 0.3636363636 The smallest is -0.3260869565 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 98, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 34, 83, 203, 496, 1212, 2962, 7239, 17692, 43239, 105676, 258272, 631216, 1542690, 3770330, 9214676, 22520642, 55040385, 134518544, 328762938] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 227, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 35, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 35, 88, 221, 555, 1394, 3501, 8793, 22084, 55465, 139303, 349866, 878705, 2206909, 5542756, 13920893, 34962979, 87811170, 220541893, 553901361] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + 3 a(n - 2) + 2 a(n - 3) with initial conditions, a(1) = 14, a(2) = 35, a(3) = 88, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + 3 b(n - 2) + 2 b(n - 3) with initial conditions, b(1) = 14, b(2) = 35, b(3) = 88, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 3 t - 2 = 0 whose roots are 1/2 1/3 1/2 1/3 (332 + 12 321 ) 20 (332 + 12 321 ) [-------------------- + ---------------------- + 1/3, - -------------------- 6 1/2 1/3 12 3 (332 + 12 321 ) 10 - ---------------------- + 1/3 1/2 1/3 3 (332 + 12 321 ) / 1/2 1/3 \ 1/2 |(332 + 12 321 ) 20 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (332 + 12 321 ) / 1/2 1/3 (332 + 12 321 ) 10 - -------------------- - ---------------------- + 1/3 12 1/2 1/3 3 (332 + 12 321 ) / 1/2 1/3 \ 1/2 |(332 + 12 321 ) 20 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (332 + 12 321 ) / In floating-point [2.511547142, -0.7557735707 + 0.4744767784 I, -0.7557735707 - 0.4744767784 I] The largest root is, 2.511547142 and the remaining roots are [-0.7557735707 + 0.4744767784 I, -0.7557735707 - 0.4744767784 I] whose absolute values are [0.8923687037, 0.8923687037] so the largest absolute value is, 0.8923687037 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8923687037 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2571428571, 0.01136363636, -0.2217194570, 0.3261261261, -0.3163558106, 0.2185089974, -0.07835778460, -0.05556058685, 0.1463806004, -0.1770170061, 0.1510035271, -0.08728640442, 0.01169010594, 0.05183793044, -0.08766456290, 0.09122944015, -0.06808838784, 0.03027080664, 0.008464523343, -0.03689983242, 0.04903535089, -0.04473509970, 0.02857128813, -0.007563309175, -0.01131964418, 0.02313300456, -0.02595254632, 0.02080717900, -0.01078445086, -0.0002680065130, 0.008992998911, -0.01337992234, 0.01306306136, -0.009090707841, 0.003338631566, 0.002192630770, -0.005972890215, 0.007282265225, -0.006251143882, 0.003649871363, -0.0005390298324, -0.002091703508, 0.003590949720, -0.003762220468, 0.002827221678, -0.001277540284, -0.0003203161838, 0.001501506322, -0.002014522796, 0.001849363802, -0.001191191942, 0.0003278538725, 0.0004530056507, -0.0009458166163, 0.001068908081, -0.0008625304664, 0.0004525605438] The largest is 0.3261261261 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 99, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 36, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 36, 93, 240, 619, 1597, 4120, 10629, 27421, 70741, 182498, 470809, 1214595, 3133417, 8083602, 20854110, 53799520, 138792226, 358056763, 923716329] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 100, :The Pisot Sequence a(n), defined by, a(1) = 14, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [14, 37, 98, 260, 690, 1831, 4859, 12895, 34221, 90816, 241008, 639588, 1697341, 4504410, 11953820, 31723092, 84186860, 223415403, 592900630, 1573441904] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 228, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 38, 103, 279, 756, 2049, 5553, 15049, 40784, 110528, 299540, 811778, 2199985, 5962140, 16157889, 43789206, 118672344, 321611797, 871594379, 2362092338] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) + 2 a(n - 3) + a(n - 4) + a(n - 5) with initial conditions, a(1) = 14, a(2) = 38, a(3) = 103, a(4) = 279, a(5) = 756, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) + 2 b(n - 3) + b(n - 4) + b(n - 5) with initial conditions, b(1) = 14, b(2) = 38, b(3) = 103, b(4) = 279, b(5) = 756, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 2 t - t - 2 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 2 _Z - _Z - 2 _Z - _Z - 1 In floating-point [2.71008211491908, 0.147141025706037 + 0.802357948988671 I, -0.502182083165575 + 0.549846871951780 I, -0.502182083165575 - 0.549846871951780 I, 0.147141025706037 - 0.802357948988671 I] The largest root is, 2.71008211491908 and the remaining roots are [0.147141025706037 + 0.802357948988671 I, -0.502182083165575 + 0.549846871951780 I, -0.502182083165575 - 0.549846871951780 I, 0.147141025706037 - 0.802357948988671 I] whose absolute values are [0.815738168624671, 0.744659940676061, 0.744659940676061, 0.815738168624671] so the largest absolute value is, 0.815738168624671 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.815738168624671 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1428571429, 0.1842105263, -0.2621359223, -0.4838709677, 0.4404761905, 0.1991215227, -0.2072753467, -0.08080271114, -0.01412318556, 0.1159525188, 0.04802029779, -0.1043327117, -0.02366607045, 0.04620505389, 0.02405140919, -0.009336684479, -0.03021063610, 0.0008838450662, 0.02314014808, 0.001457593738, -0.01172429489, -0.005037490911, 0.005139903915, 0.006391468960, -0.002418841138, -0.004928191288, 0.0006101272100, 0.002985753730, 0.0006978799156, -0.001745264445, -0.001139205593, 0.0009679651392, 0.0009898294399, -0.0003781716975, -0.0007150537155, -6 0.1392970073 10 , 0.0004866760625, 0.0001550417333, -0.0002961872900, -0.0001788951403, 0.0001429212554, 0.0001562905862, -0.00004343340968, -5 -0.0001198161528, -0.6458427856 10 , 0.00007961201369, 0.00002599047044, -0.00004457346362, -0.00003020701007, 0.00002014704295, 0.00002654263272, -5 -5 -0.5764704919 10 , -0.00001947316490, -0.1685736382 10 , 0.00001231562818, -5 -5 -5 0.4777117980 10 , -0.6739478444 10 , -0.5229483836 10 ] The largest is 0.4404761905 The smallest is -0.4838709677 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 229, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 39, 109, 305, 853, 2386, 6674, 18668, 52217, 146058, 408544, 1142753, 3196435, 8940862, 25008803, 69953012, 195668057, 547310079, 1530900481, 4282136165] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + 2 a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 14, a(2) = 39, a(3) = 109, a(4) = 305, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + 2 b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 14, b(2) = 39, b(3) = 109, b(4) = 305, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t - 2 t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 2 _Z - 2 _Z - _Z + 1 In floating-point [0.451027959780797, 2.79713555395954, -0.624081756870166 + 0.634960473669366 I, -0.624081756870166 - 0.634960473669366 I] The largest root is, 2.79713555395954 and the remaining roots are [0.451027959780797, -0.624081756870166 + 0.634960473669366 I, -0.624081756870166 - 0.634960473669366 I] whose absolute values are [0.451027959780797, 0.890310531433038, 0.890310531433038] so the largest absolute value is, 0.890310531433038 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.890310531433038 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3571428571, -0.3589743590, 0.4403669725, -0.3967213115, 0.08675263775, 0.1793797150, -0.3047647588, 0.2327512321, -0.05140088477, -0.1214449054, 0.1918251155, -0.1433914415, 0.02682332036, 0.08013377234, -0.1213023670, 0.08787757416, -0.01353913378, -0.05275925861, 0.07658315642, -0.05376891232, 0.006408363364, 0.03462131712, -0.04829270777, 0.03283449438, -0.002703473022, -0.02265198217, 0.03041629177, -0.02000934821, 0.0008653779533, 0.01478033341, -0.01913421724, 0.01216695851, -0.00001956201036, -0.009619757663, 0.01202253641, -0.007380963032, -0.0003170489032, 0.006246270200, -0.007545056846, 0.004466340836, 0.0004058870832, -0.004046871207, 0.004729429435, -0.002695337297, -0.0003845740135, 0.002616478021, -0.002960958716, 0.001621801894, 0.0003227383915, -0.001688356166, 0.001851525061, -0.0009727257122, -0.0002534958596, 0.001087438083, -0.001156366326, 0.0005813733684, 0.0001909480286, -0.0006991616152] The largest is 0.4403669725 The smallest is -0.3967213115 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 230, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 14, a(2) = 40, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [14, 40, 114, 325, 927, 2644, 7541, 21508, 61344, 174962, 499017, 1423269, 4059370, 11577913, 33021890, 94183228, 268624250, 766155389, 2185186483, 6232469332] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) + 2 a(n - 3) - a(n - 4) with initial conditions, a(1) = 14, a(2) = 40, a(3) = 114, a(4) = 325, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) + 2 b(n - 3) - b(n - 4) with initial conditions, b(1) = 14, b(2) = 40, b(3) = 114, b(4) = 325, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 3 t + t - 2 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 3 _Z + _Z - 2 _Z + 1 In floating-point [0.476888865617442, 2.85214528850330, -0.164517077060369 + 0.841512767036371 I, -0.164517077060369 - 0.841512767036371 I] The largest root is, 2.85214528850330 and the remaining roots are [0.476888865617442, -0.164517077060369 + 0.841512767036371 I, -0.164517077060369 - 0.841512767036371 I] whose absolute values are [0.476888865617442, 0.857443645804024, 0.857443645804024] so the largest absolute value is, 0.857443645804024 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.857443645804024 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2857142857, -0.1000000000, -0.4649122807, 0.08923076923, 0.2470334412, -0.1781391831, -0.1379127437, 0.1692393528, 0.04231872718, -0.1399675358, 0.01416985794, 0.09787538406, -0.04279752770, -0.05796070501, 0.05049632229, 0.02597923274, -0.04568250633, -0.004073402139, 0.03492444311, -0.008497513935, -0.02288128286, 0.01377595370, 0.01228967299, -0.01417198652, -0.004372442298, 0.01185805191, -0.0006870480191, -0.008492094035, 0.003299312026, 0.005157882167, -0.004122805575, -0.002435580806, 0.003832515466, 0.0005296338882, -0.002991969838, 0.0005950683357, 0.002003927155, -0.001096860435, -0.001112401949, 0.001172440562, 0.0004320756099, -0.001004157196, 0.00001273587476, 0.0007340754783, -0.0002508994415, -0.0004571448575, 0.0003348799510, 0.0002259103490, -0.0003205391773, -0.00006062312156, 0.0002556105597, -0.00003953390300, -0.0001749193345, 0.00008662014050, 0.0001001013903, -0.00009662073563, -0.00004180398167, 0.00008479143069] The largest is 0.2857142857 The smallest is -0.4649122807 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 101, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 17, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53 ] is a trivial linear sequence Fact, 102, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 18, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 18, 22, 27, 33, 40, 48, 58, 70, 84, 101, 121, 145, 174, 209, 251, 301, 361, 433, 519] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 103, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 19, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 19, 24, 30, 38, 48, 61, 78, 100, 128, 164, 210, 269, 345, 442, 566, 725, 929, 1190, 1524] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 231, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 20, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 20, 27, 36, 48, 64, 85, 113, 150, 199, 264, 350, 464, 615, 815, 1080, 1431, 1896, 2512, 3328] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) - a(n - 4) with initial conditions, a(1) = 15, a(2) = 20, a(3) = 27, a(4) = 36, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) - b(n - 4) with initial conditions, b(1) = 15, b(2) = 20, b(3) = 27, b(4) = 36, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [1, ------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.324717958, -0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.3333333333, 0.4500000000, 0., 0., 0.3333333333, -0.1093750000, 0.2235294118, 0.1150442478, 0.006666666667, 0.2311557789, 0.01515151515, 0.1314285714, 0.1400862069, 0.04065040650, 0.1656441718, 0.07500000000, 0.1006289308, 0.1350210970, 0.07006369427, 0.1301081731, 0.09956906328, 0.09467556925, 0.1241922978, 0.08877182714, 0.1134020619, 0.1075041690, 0.09671869755, 0.1154540559, 0.09877340219, 0.1067251462, 0.1087812521, 0.1000534825, 0.1100621118, 0.1033910821, 0.1046724093, 0.1080103532, 0.1026209240, 0.1072403917, 0.1051890587, 0.1044192125, 0.1069874322, 0.1041663191, 0.1059647414, 0.1057118851, 0.1046892225, 0.1062348095, 0.1049593066, 0.1054822431, 0.1057523363, 0.1049997768, 0.1057928117, 0.1053103493, 0.1053508277, 0.1056614024, 0.1052194201, 0.1055704744, 0.1054390677, 0.1053481404] The largest is 0.4500000000 The smallest is -0.3333333333 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 232, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 21, 29, 40, 55, 76, 105, 145, 200, 276, 381, 526, 726, 1002, 1383, 1909, 2635, 3637, 5020, 6929] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 4) with initial conditions, a(1) = 15, a(2) = 21, a(3) = 29, a(4) = 40, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 4) with initial conditions, b(1) = 15, b(2) = 21, b(3) = 29, b(4) = 40, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - _Z - 1 In floating-point [1.38027756909761, 0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] The largest root is, 1.38027756909761 and the remaining roots are [0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] whose absolute values are [0.940435682699417, 0.819172513396165, 0.940435682699417] so the largest absolute value is, 0.940435682699417 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.940435682699417 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4000000000, 0.04761904762, 0.1724137931, -0.3750000000, 0.01818181818, 0.06578947368, 0.2380952381, -0.1379310345, -0.1200000000, -0.05434782609, 0.1837270341, 0.04562737643, -0.07438016529, -0.1287425150, 0.05495300072, 0.1005762179, 0.02618595825, -0.1025570525, -0.04760956175, 0.05296579593, 0.07915098285, -0.02340731763, -0.07101695845, -0.01805168986, 0.06109926831, 0.03769174580, -0.03332526914, -0.05137699100, 0.009722222222, 0.04741396659, 0.01408867282, -0.03728831975, -0.02756610552, 0.01984785791, 0.03393652954, -0.003351792723, -0.03091789826, -0.01107004145, 0.02286648799, 0.01951469495, -0.01140320347, -0.02247324496, 0.0003932429124, 0.01990793787, 0.008504734346, -0.01396851062, -0.01357526772, 0.006332670134, 0.01483740448, 0.0008688938494, -0.01270637387, -0.006373703742, 0.008463700736, 0.009332594585, -0.003373779290, -0.009747483032, -0.001283782297, 0.008048812288] The largest is 0.4000000000 The smallest is -0.3750000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 233, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 22, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 22, 32, 47, 69, 101, 148, 217, 318, 466, 683, 1001, 1467, 2150, 3151, 4618, 6768, 9919, 14537, 21305] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 3) with initial conditions, a(1) = 15, a(2) = 22, a(3) = 32, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 3) with initial conditions, b(1) = 15, b(2) = 22, b(3) = 32, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / In floating-point [1.465571232, -0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2666666667, -0.4545454545, 0.03125000000, 0.2978723404, -0.1594202899, -0.1287128713, 0.1689189189, 0.009216589862, -0.1194968553, 0.04935622318, 0.05856515373, -0.06093906094, -0.01158827539, 0.04697674419, -0.01396382101, -0.02555218709, 0.02142434988, 0.007460429479, -0.01809176584, 0.003332551044, 0.01079297976, -0.007298791547, -0.003966242209, 0.006826737206, -0.0004720550361, -0.004438297247, 0.002388439822, 0.001916384759, -0.002521912500, -0.0001334726925, 0.001782912066, -0.0007390004374, -0.0008724731303, 0.0009104389357, 0.0001714384981, -0.0007010346322, 0.0002094043034, 0.0003808428015, -0.0003201918307, -0.0001107875273, 0.0002700552742, -0.00005013655650, -0.0001609240838, 0.0001091311904, 0.00005899463392, -0.0001019294499, -5 0.7201740545 10 , 0.00006619637446, -0.00003573307541, -0.00002853133487, -5 0.00003766503959, 0.1931964179 10 , -0.00002659937069, 0.00001106566890, -5 0.00001299763308, -0.00001360173761, -0.2536068703 10 , 0.00001046156438] The largest is 0.2978723404 The smallest is -0.4545454545 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 104, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 23, 35, 53, 80, 121, 183, 277, 419, 634, 959, 1451, 2195, 3320, 5022, 7597, 11492, 17384, 26297, 39780] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 105, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 24, 38, 60, 95, 150, 237, 374, 590, 931, 1469, 2318, 3658, 5773, 9111, 14379, 22693, 35814, 56522, 89204] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 106, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 25, 42, 71, 120, 203, 343, 580, 981, 1659, 2806, 4746, 8027, 13576, 22961, 38834, 65680, 111085, 187879, 317761] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 234, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 26, 45, 78, 135, 234, 406, 704, 1221, 2118, 3674, 6373, 11055, 19177, 33266, 57706, 100102, 173646, 301222, 522527] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 3) - a(n - 5) - a(n - 6) with initial conditions, a(1) = 15, a(2) = 26, a(3) = 45, a(4) = 78, a(5) = 135, a(6) = 234, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 3) - b(n - 5) - b(n - 6) with initial conditions, b(1) = 15, b(2) = 26, b(3) = 45, b(4) = 78, b(5) = 135, b(6) = 234, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 t - t - t - t + t + 1 = 0 whose roots are [1, RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 3 2 %1 := _Z - _Z - 2 _Z - 2 _Z - 1 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.73469134569247, -0.166159654583804 + 0.938712793124580 I, -0.701186018262431 + 0.377711778281479 I, -0.701186018262431 - 0.377711778281479 I, -0.166159654583804 - 0.938712793124580 I] The largest root is, 1.73469134569247 and the remaining roots are [-0.166159654583804 + 0.938712793124580 I, -0.701186018262431 + 0.377711778281479 I, -0.701186018262431 - 0.377711778281479 I, -0.166159654583804 - 0.938712793124580 I] whose absolute values are [0.953305165614432, 0.796447122952478, 0.796447122952478, 0.953305165614432] so the largest absolute value is, 0.953305165614432 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.953305165614432 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.06666666667, -0.1153846154, 0.2000000000, -0.3461538462, -0.4000000000, 0.4273504274, -0.2709359606, -0.3281250000, -0.02457002457, 0.1237016053, -0.2561241154, -0.3131962969, 0.1536861149, -0.06288783438, -0.3215294896, -0.09829133886, 0.08663163573, -0.1736751782, -0.2761318894, 0.02124483520, -0.008739848834, -0.2519671018, -0.1524183100, 0.03668214102, -0.1128161161, -0.2410572529, -0.05648423356, -0.005972159131, -0.1877774715, -0.1740998849, -0.01397614096, -0.07831200845, -0.2039316411, -0.1024701592, -0.02283645186, -0.1411622262, -0.1741806878, -0.05593571618, -0.06487682976, -0.1696866226, -0.1265004905, -0.04572102888, -0.1117917380, -0.1632007115, -0.08615002596, -0.06495536231, -0.1420845804, -0.1356772017, -0.06772469494, -0.09613573962, -0.1484322480, -0.1052527399, -0.07205894543, -0.1223420367, -0.1357932874, -0.08562628191, -0.09007661810, -0.1341845021] The largest is 0.4273504274 The smallest is -0.4000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 235, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 27, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 27, 49, 89, 162, 295, 537, 978, 1781, 3243, 5905, 10752, 19578, 35649, 64912, 118196, 215219, 391885, 713570, 1299315] ] The sequence a(n) satisfies, for n>=, 8, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) + a(n - 5) - a(n - 6) + a(n - 7) with initial conditions, a(1) = 15, a(2) = 27, a(3) = 49, a(4) = 89, a(5) = 162, a(6) = 295, a(7) = 537, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) + b(n - 5) - b(n - 6) + b(n - 7) with initial conditions, b(1) = 15, b(2) = 27, b(3) = 49, b(4) = 89, b(5) = 162, b(6) = 295, b(7) = 537, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 7, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 7 6 5 4 2 t - 2 t + t - t - t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6), RootOf(%1, index = 7)] 7 6 5 4 2 %1 := _Z - 2 _Z + _Z - _Z - _Z + _Z - 1 In floating-point [1.82086565458103, 0.609947304872010 + 0.571389385608650 I, 0.187685145696476 + 0.951618951449115 I, -0.708065277859001 + 0.578212135481728 I, -0.708065277859001 - 0.578212135481728 I, 0.187685145696476 - 0.951618951449115 I, 0.609947304872010 - 0.571389385608650 I] The largest root is, 1.82086565458103 and the remaining roots are [0.609947304872010 + 0.571389385608650 I, 0.187685145696476 + 0.951618951449115 I, -0.708065277859001 + 0.578212135481728 I, -0.708065277859001 - 0.578212135481728 I, 0.187685145696476 - 0.951618951449115 I, 0.609947304872010 - 0.571389385608650 I] whose absolute values are [0.835776013479006, 0.969950690845788, 0.914158471671015, 0.914158471671015, 0.969950690845788, 0.835776013479006] so the largest absolute value is, 0.969950690845788 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.969950690845788 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4000000000, -0.07407407407, -0.3469387755, -0.1235955056, 0.1913580247, -0.4779661017, 0.1620111732, 0.3139059305, 0.1370016844, 0.08911501696, -0.4379339543, 0.003348214286, 0.2076310144, -0.04095486549, -0.02021197930, -0.1818927882, 0.1458746672, 0.2197711063, -0.1334627297, -0.1124484825, -0.07429884381, 0.1379434654, 0.1297409765, -0.1601195893, -0.07125133628, 0.05204478990, 0.1150151485, 0.02423283979, -0.1664217817, -0.02345202476, 0.1069271082, 0.06260348182, -0.04390968820, -0.1191352231, 0.03544532067, 0.1100735274, -0.002209140018, -0.07863254861, -0.05760448279, 0.07188529854, 0.07823551497, -0.04985609817, -0.07241229392, 0.002085951103, 0.07758533068, 0.02941814998, -0.07286939423, -0.04189228852, 0.04314511401, 0.05840020797, -0.01431821619, -0.06859373986, -0.004073799873, 0.05829593223, 0.02543472992, -0.04107358241, -0.04516127821, 0.02638747971] The largest is 0.3139059305 The smallest is -0.4779661017 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 236, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 28, 52, 97, 181, 338, 631, 1178, 2199, 4105, 7663, 14305, 26704, 49850, 93058, 173717, 324288, 605368, 1130077, 2109583] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) + a(n - 4) with initial conditions, a(1) = 15, a(2) = 28, a(3) = 52, a(4) = 97, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) + b(n - 4) with initial conditions, b(1) = 15, b(2) = 28, b(3) = 52, b(4) = 97, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - 2 t + t - 1 = 0 whose roots are 1/2 1/2 1/2 1/2 [1/2 - 1/2 I (-3 + 2 5 ) , 1/2 + 1/2 I (-3 + 2 5 ) , 1/2 1/2 1/2 1/2 (3 + 2 5 ) (3 + 2 5 ) 1/2 - ---------------, 1/2 + ---------------] 2 2 In floating-point [0.5000000000 - 0.6066580490 I, 0.5000000000 + 0.6066580490 I, -0.8667603990, 1.866760399] The largest root is, 1.866760399 and the remaining roots are [0.5000000000 - 0.6066580490 I, 0.5000000000 + 0.6066580490 I, -0.8667603990] whose absolute values are [0.7861513775, 0.7861513775, 0.8667603990] so the largest absolute value is, 0.8667603990 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8667603990 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2666666667, -0.4285714286, -0.05769230769, -0.2577319588, 0.1823204420, -0.008875739645, 0.1822503962, -0.07555178268, 0.04001819009, -0.1110840438, 0.03562573405, -0.04432016777, 0.06246255243, -0.02178535607, 0.03637516388, -0.03403236298, 0.01618314585, -0.02579422764, 0.01881907162, -0.01257736719, 0.01682263922, -0.01096802092, 0.009460396898, -0.01047921260, 0.006832234925, -0.006763947973, 0.006411713556, -0.004488020418, 0.004620142063, -0.003935377404, 0.003028979166, -0.003050204149, 0.002455111170, -0.002054134231, 0.001970914853, -0.001563485612, 0.001382274176, -0.001260500732, 0.001013399002, -0.0009189617851, 0.0008048513377, -0.0006641970581, 0.0006039666705, -0.0005158797818, 0.0004372888321, -0.0003935860643, 0.0003326743236, -0.0002878199666, 0.0002552349632, -0.0002157904616, 0.0001889133670, -0.0001652281957, 0.0001405690333, -0.0001235657621, 0.0001070100387, -0.00009177715172, 0.00008058049191, -0.00006941481690] The largest is 0.2666666667 The smallest is -0.4285714286 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 237, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 29, 56, 108, 208, 401, 773, 1490, 2872, 5536, 10671, 20569, 39648, 76424, 147312, 283953, 547337, 1055026, 2033628, 3919944] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 3) + a(n - 4) with initial conditions, a(1) = 15, a(2) = 29, a(3) = 56, a(4) = 108, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 3) + b(n - 4) with initial conditions, b(1) = 15, b(2) = 29, b(3) = 56, b(4) = 108, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - _Z - _Z - _Z - 1 In floating-point [1.92756197548293, -0.0763789311337457 + 0.814703647170387 I, -0.774804113215434, -0.0763789311337457 - 0.814703647170387 I] The largest root is, 1.92756197548293 and the remaining roots are [-0.0763789311337457 + 0.814703647170387 I, -0.774804113215434, -0.0763789311337457 - 0.814703647170387 I] whose absolute values are [0.818276098779540, 0.774804113215434, 0.818276098779540] so the largest absolute value is, 0.818276098779540 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.818276098779540 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.06666666667, 0.1379310345, 0.2857142857, -0.4074074074, 0.08173076923, 0.09725685786, 0.05692108668, -0.1718120805, 0.06406685237, 0.04642341040, -0.004404460688, -0.06572998201, 0.04035512510, 0.01664398618, -0.01313538612, -0.02186629477, 0.02199741658, 0.003639720727, -0.009364544548, -0.005593702359, 0.01067889017, -0.0006396360254, -0.004918992778, -0.0004734409950, 0.004646820369, -0.001385249429, -0.002130862833, 0.0006572671112, 0.001787975217, -0.001070869934, -0.0007564904386, 0.0006178819561, 0.0005784968009, -0.0006309816157, -0.0001910932973, 0.0003743038439, 0.0001307257318, -0.0003170453373, -5 -5 -0.3109058973 10 , 0.0001848751794, -0.4553485117 10 , -0.0001398327020, 0.00003737993330, 0.00007786892557, -0.00002913732827, -0.00005372117143, 0.00003239035917, 0.00002740078503, -0.00002306735550, -0.00001699738272, -5 -5 0.00001972640598, 0.7062452788 10 , -0.00001327587946, -0.3484403418 10 , -6 -5 -6 0.00001002857589, 0.3307458006 10 , -0.6400961187 10 , 0.4739570845 10 ] The largest is 0.2857142857 The smallest is -0.4074074074 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 238, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 31, 64, 132, 272, 560, 1153, 2374, 4888, 10064, 20721, 42663, 87840, 180856, 372369, 766680, 1578537, 3250090, 6691693, 13777697] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 4) - a(n - 5) with initial conditions, a(1) = 15, a(2) = 31, a(3) = 64, a(4) = 132, a(5) = 272, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 4) - b(n - 5) with initial conditions, b(1) = 15, b(2) = 31, b(3) = 64, b(4) = 132, b(5) = 272, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 t - 2 t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 %1 := _Z - 2 _Z - _Z + 1 In floating-point [0.695208900696950, 2.05892544882736, 0.0727669163136779 + 0.878203366434257 I, -0.899668182151670, 0.0727669163136779 - 0.878203366434257 I] The largest root is, 2.05892544882736 and the remaining roots are [0.695208900696950, 0.0727669163136779 + 0.878203366434257 I, -0.899668182151670, 0.0727669163136779 - 0.878203366434257 I] whose absolute values are [0.695208900696950, 0.881212901021236, 0.899668182151670, 0.881212901021236] so the largest absolute value is, 0.899668182151670 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.899668182151670 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.06666666667, 0.1290322581, 0.2500000000, 0.4848484848, -0.05882352941, -0.05535714286, 0.01040763226, 0.2561078349, -0.03109656301, -0.05872416534, -0.05168669466, 0.1423247310, -0.002550091075, -0.03272769496, -0.05841785970, 0.07717561434, 0.009476496275, -0.01122461224, -0.04813938715, 0.03931469824, 0.01093028013, 0.001159451704, -0.03459587148, 0.01826234242, 0.008140266764, 0.006509705100, -0.02273591298, 0.007386387933, 0.004650700209, 0.007670838755, -0.01390394058, 0.002314419766, 0.001893151809, 0.006806442163, -0.007961895003, 0.0002945703352, 0.0001678727130, 0.005249035780, -0.004270265606, -0.0002840658735, -0.0006948293692, 0.003691504329, -0.002136292728, -0.0002863857241, -0.0009835349438, 0.002419263811, -0.0009892694361, -0.0001286318680, -0.0009544129557, 0.001493972843, -0.0004205875609, 0.00001946244641, -0.0007868561949, 0.0008746734088, -0.0001652135861, 0.0001096228350, -0.0005870729713, 0.0004873836612] The largest is 0.4848484848 The smallest is -0.05882352941 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 107, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 32, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 32, 68, 145, 309, 658, 1401, 2983, 6351, 13522, 28790, 61297, 130508, 277866, 591608, 1259600, 2681830, 5709918, 12157058, 25883745] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 239, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 33, 73, 161, 355, 783, 1727, 3809, 8401, 18529, 40867, 90135, 198799, 438465, 967065, 2132929, 4704323, 10375711, 22884351, 50473025] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 3) with initial conditions, a(1) = 15, a(2) = 33, a(3) = 73, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 3) with initial conditions, b(1) = 15, b(2) = 33, b(3) = 73, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (172 + 12 177 ) 8 (172 + 12 177 ) [-------------------- + ---------------------- + 2/3, - -------------------- 6 1/2 1/3 12 3 (172 + 12 177 ) 4 - ---------------------- + 2/3 1/2 1/3 3 (172 + 12 177 ) / 1/2 1/3 \ 1/2 |(172 + 12 177 ) 8 | + 1/2 I 3 |-------------------- - ----------------------|, | 6 1/2 1/3| \ 3 (172 + 12 177 ) / 1/2 1/3 (172 + 12 177 ) 4 - -------------------- - ---------------------- + 2/3 12 1/2 1/3 3 (172 + 12 177 ) / 1/2 1/3 \ 1/2 |(172 + 12 177 ) 8 | - 1/2 I 3 |-------------------- - ----------------------|] | 6 1/2 1/3| \ 3 (172 + 12 177 ) / In floating-point [2.205569431, -0.1027847152 + 0.6654569515 I, -0.1027847152 - 0.6654569515 I] The largest root is, 2.205569431 and the remaining roots are [-0.1027847152 + 0.6654569515 I, -0.1027847152 - 0.6654569515 I] whose absolute values are [0.6733480912, 0.6733480912] so the largest absolute value is, 0.6733480912 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6733480912 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4000000000, 0.4848484848, 0.08219178082, -0.2360248447, 0.01126760563, 0.1047254151, -0.02663578460, -0.04200577579, 0.02071182002, 0.01478763020, -0.01243056745, -0.004149331558, 0.006488966242, 0.0005473640998, -0.003054603362, 0.0003797594763, 0.001306883052, -0.0004408372592, -0.0005019150423, 0.0003030529674, 0.0001652686755, -0.0001713776912, -6 -0.00003970241500, 0.00008586384555, 0.3499998984 10 , -0.00003900241520, -5 -5 -5 0.7859015145 10 , 0.00001606803019, -0.6866354824 10 , -0.5873694502 10 , -5 -5 -5 0.4320641185 10 , 0.1774927546 10 , -0.2323839411 10 , -6 -5 -7 -0.3270376368 10 , 0.1120852272 10 , -0.8213486664 10 , -6 -6 -6 -0.4913073701 10 , 0.1382375319 10 , 0.1943401971 10 , -6 -7 -7 -0.1026269759 10 , -0.6701641997 10 , 0.6030735715 10 , -7 -7 -8 0.1798773837 10 , -0.3104094323 10 , -0.1774529315 10 , -7 -8 -8 0.1443867974 10 , -0.2163583746 10 , -0.6101696808 10 , -8 -8 -8 0.2235286127 10 , 0.2306988507 10 , -0.1487719794 10 , -9 -9 -9 -0.7401534606 10 , 0.8266815858 10 , 0.1656433781 10 , -9 -11 -9 -0.4088667044 10 , 0.8948176943 10 , 0.1835397319 10 , -10 -0.4178724056 10 ] The largest is 0.4848484848 The smallest is -0.4000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 108, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 34, 77, 174, 393, 888, 2006, 4532, 10239, 23133, 52264, 118079, 266774, 602718, 1361711, 3076492, 6950669, 15703535, 35478745, 80156560] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 109, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 35, 82, 192, 450, 1055, 2473, 5797, 13589, 31855, 74674, 175050, 410350, 961937, 2254960, 5286047, 12391481, 29047945, 68093806, 159624594] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 110, : Consider the Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 36, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 15, 36, 86, 205, 489, 1166, 2780, 6628, 15802, 37674, 89820, 214143, 510546, 1217211, 2901996, 6918752, 16495243, 39326896, 93760653, 223538111, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) + a(n - 2) - a(n - 4) + a(n - 5) + a(n - 6) - a(n - 7), . Alas, it breaks down at the, 40, -th term. a(40), equals , 7869925867584729, while the corresponding term for the solution of the recurrence is , 7869925867584728 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.02168368889029 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 111, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 37, 91, 224, 551, 1355, 3332, 8194, 20151, 49556, 121870, 299707, 737050, 1812579, 4457557, 10962178, 26958566, 66297435, 163040938, 400955896] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 240, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 15, a(2) = 38, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [15, 38, 96, 243, 615, 1556, 3937, 9961, 25202, 63763, 161325, 408164, 1032685, 2612769, 6610498, 16725047, 42315601, 107061588, 270873705, 685330429] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - a(n - 2) - 2 a(n - 4) + 2 a(n - 5) with initial conditions, a(1) = 15, a(2) = 38, a(3) = 96, a(4) = 243, a(5) = 615, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - b(n - 2) - 2 b(n - 4) + 2 b(n - 5) with initial conditions, b(1) = 15, b(2) = 38, b(3) = 96, b(4) = 243, b(5) = 615, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 t - 3 t + t + 2 t - 2 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 %1 := _Z - 3 _Z + _Z + 2 _Z - 2 In floating-point [2.53007367037607, 0.828616128606492 + 0.426151215131703 I, -0.593652963794526 + 0.747034176337147 I, -0.593652963794526 - 0.747034176337147 I, 0.828616128606492 - 0.426151215131703 I] The largest root is, 2.53007367037607 and the remaining roots are [0.828616128606492 + 0.426151215131703 I, -0.593652963794526 + 0.747034176337147 I, -0.593652963794526 - 0.747034176337147 I, 0.828616128606492 - 0.426151215131703 I] whose absolute values are [0.931777627304411, 0.954192801292142, 0.954192801292142, 0.931777627304411] so the largest absolute value is, 0.954192801292142 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.954192801292142 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2666666667, -0.4736842105, 0.09375000000, 0.4814814815, -0.1934959350, 0.4196658098, 0.3167386335, -0.2448549342, 0.2983493374, -0.08636670169, -0.3516441965, 0.1546167717, -0.2709141703, -0.1979275627, 0.2076864708, -0.1915351867, 0.06876988466, 0.2518715956, -0.1243831659, 0.1734222208, 0.1240396856, -0.1675065861, 0.1259500790, -0.05025395035, -0.1779468595, 0.09950591518, -0.1104487253, -0.07844403225, 0.1305024468, -0.08495417659, 0.03454430428, 0.1245777034, -0.07870415232, 0.07022308654, 0.05037645020, -0.09916053416, 0.05870565874, -0.02257696733, -0.08674328806, 0.06142107187, -0.04472588212, -0.03303346609, 0.07395812531, -0.04142087784, 0.01407314915, 0.06025549323, -0.04728985225, 0.02863295631, 0.02220066721, -0.05439564285, 0.02970309521, -0.008340688648, -0.04186058295, 0.03595155993, -0.01848221339, -0.01531063239, 0.03959010482, -0.02154333889] The largest is 0.4814814815 The smallest is -0.4736842105 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 112, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 39, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 39, 101, 262, 680, 1765, 4581, 11890, 30861, 80101, 207905, 539625, 1400616, 3635349, 9435679, 24490644, 63566347, 164988739, 428234204, 1111497273] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 113, :The Pisot Sequence a(n), defined by, a(1) = 15, a(2) = 40, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [15, 40, 107, 286, 764, 2041, 5452, 14564, 38905, 103927, 277620, 741606, 1981051, 5291979, 14136457, 37762700, 100875454, 269468476, 719830808, 1922883151] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 114, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 18, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54 ] is a trivial linear sequence Fact, 115, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 19, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 19, 23, 28, 34, 41, 49, 59, 71, 85, 102, 122, 146, 175, 210, 252, 302, 362, 434, 520] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 116, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 20, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 20, 25, 31, 38, 47, 58, 72, 89, 110, 136, 168, 208, 258, 320, 397, 493, 612, 760, 944] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 241, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 21, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 2) + a(n - 3) with initial conditions, a(1) = 16, a(2) = 21, a(3) = 28, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 2) + b(n - 3) with initial conditions, b(1) = 16, b(2) = 21, b(3) = 28, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [1.324717958, -0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 + 0.5622795125 I, -0.6623589786 - 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4375000000, 0.3333333333, -0.1071428571, -0.1081081081, 0.2244897959, -0.2153846154, 0.1162790698, 0.008771929825, -0.09933774834, 0.1250000000, -0.09056603774, 0.02564102564, 0.03440860215, -0.06493506494, 0.06004901961, -0.03052728955, -0.004888268156, 0.02952029520, -0.03541583764, 0.02463202163, -0.005895691610, -0.01078397809, 0.01873627084, -0.01667967226, 0.007952286282, 0.002056583847, -0.008727394789, 0.01000886862, -0.006670811018, 0.001281472791, 0.003338056570, -0.005389338572, 0.004619529351, -0.002051282051, -0.0007698093173, 0.002568247246, -0.002821091377, 0.001798437928, -0.0002528441374, -0.001022653455, 0.001545593789, -0.001275497592, 0.0005229403335, 0.0002700961958, -0.0007525572592, 0.0007930365292, -0.0004824610634, 0.00004047926998, 0.0003105754658, -0.0004419817935, 0.0003510547358, -0.0001314063277, -0.00009092705769, 0.0002196484081, -0.0002223333854, -5 0.0001287213504, -0.2684977277 10 , -0.00009361203497] The largest is 0.3333333333 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 117, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 22, 30, 41, 56, 76, 103, 140, 190, 258, 350, 475, 645, 876, 1190, 1617, 2197, 2985, 4056, 5511] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 118, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 23, 33, 47, 67, 96, 138, 198, 284, 407, 583, 835, 1196, 1713, 2453, 3513, 5031, 7205, 10318, 14776] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 119, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 24, 36, 54, 81, 122, 184, 278, 420, 635, 960, 1451, 2193, 3314, 5008, 7568, 11437, 17284, 26120, 39473] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 242, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 25, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 25, 39, 61, 95, 148, 231, 361, 564, 881, 1376, 2149, 3356, 5241, 8185, 12783, 19964, 31179, 48694, 76048] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 4) - a(n - 5) + a(n - 6) with initial conditions, a(1) = 16, a(2) = 25, a(3) = 39, a(4) = 61, a(5) = 95, a(6) = 148, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 4) - b(n - 5) + b(n - 6) with initial conditions, b(1) = 16, b(2) = 25, b(3) = 39, b(4) = 61, b(5) = 95, b(6) = 148, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 2 t - 2 t + t - t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 2 %1 := _Z - 2 _Z + _Z - _Z + _Z - 1 In floating-point [1.56175206772030, 0.713504817792455 + 0.658512530611071 I, -0.0290284808449552 + 0.853775991281275 I, -0.930704741615296, -0.0290284808449552 - 0.853775991281275 I, 0.713504817792455 - 0.658512530611071 I] The largest root is, 1.56175206772030 and the remaining roots are [0.713504817792455 + 0.658512530611071 I, -0.0290284808449552 + 0.853775991281275 I, -0.930704741615296, -0.0290284808449552 - 0.853775991281275 I, 0.713504817792455 - 0.658512530611071 I] whose absolute values are [0.970941747987406, 0.854269334571064, 0.930704741615296, 0.854269334571064, 0.970941747987406] so the largest absolute value is, 0.970941747987406 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.970941747987406 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.06250000000, -0.1600000000, 0.4102564103, -0.04918032787, -0.4315789474, -0.4527027027, 0.1601731602, 0.1523545706, 0.1719858156, 0.1214528944, 0.2507267442, -0.08050255933, -0.2321215733, -0.2820072505, -0.03066585217, 0.01087381679, 0.1515227409, 0.1617755541, 0.1912350598, -0.01977698296, -0.1208069514, -0.2007105658, -0.09963304716, -0.04779261427, 0.09425283452, 0.1366175880, 0.1592527775, 0.03301772699, -0.05080494111, -0.1400554673, -0.1124179876, -0.07439798255, 0.03905212780, 0.09626943812, 0.1303192858, 0.06233366924, -0.004619825231, -0.08875399190, -0.09978618330, -0.08253455334, -0.001917132164, 0.05689979156, 0.1000646987, 0.07172724378, 0.02422102677, -0.04700281986, -0.07697889156, -0.07839262658, -0.02724787995, 0.02440026383, 0.07029336269, 0.06776990666, 0.03941230570, -0.01568977806, -0.05214664290, -0.06672669995, -0.03937099526, 0.0006525323269] The largest is 0.4102564103 The smallest is -0.4527027027 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 243, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 16, a(2) = 26, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 16, b(2) = 26, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2500000000, -0.1538461538, 0.09523809524, -0.05882352941, 0.03636363636, -0.02247191011, 0.01388888889, -0.008583690987, 0.005305039788, -0.003278688525, 0.002026342452, -0.001252348153, 0.0007739938080, -0.0004783544607, 0.0002956393200, -0.0001827151471, 0.0001129241714, -0.00006979097603, 0.00004313319531, -0.00002665778074, 0.00001647541456, -5 -5 -5 -0.00001018236618, 0.6293048384 10 , -0.3889317794 10 , 0.2403730590 10 , -5 -6 -6 -0.1485587204 10 , 0.9181433855 10 , -0.5674438188 10 , -6 -6 -6 0.3506995667 10 , -0.2167442521 10 , 0.1339553146 10 , -7 -7 -7 -0.8278893743 10 , 0.5116637722 10 , -0.3162256020 10 , -7 -7 -8 0.1954381702 10 , -0.1207874319 10 , 0.7465073831 10 , -8 -8 -8 -0.4613669356 10 , 0.2851404475 10 , -0.1762264881 10 , -8 -9 -9 0.1089139594 10 , -0.6731252874 10 , 0.4160143063 10 , -9 -9 -10 -0.2571109811 10 , 0.1589033252 10 , -0.9820765590 10 , -10 -10 -10 0.6069566930 10 , -0.3751198660 10 , 0.2318368270 10 , -10 -11 -11 -0.1432830390 10 , 0.8855378808 10 , -0.5472925087 10 , -11 -11 -11 0.3382453722 10 , -0.2090471365 10 , 0.1291982356 10 , -12 -12 -12 -0.7984890090 10 , 0.4934933472 10 , -0.3049956618 10 ] The largest is 0.2500000000 The smallest is -0.1538461538 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 120, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 27, 46, 78, 132, 223, 377, 637, 1076, 1818, 3072, 5191, 8772, 14823, 25048, 42326, 71522, 120857, 204223, 345094] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 244, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 28, 49, 86, 151, 265, 465, 816, 1432, 2513, 4410, 7739, 13581, 23833, 41824, 73396, 128801, 226030, 396655, 696081] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) with initial conditions, a(1) = 16, a(2) = 28, a(3) = 49, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) with initial conditions, b(1) = 16, b(2) = 28, b(3) = 49, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (100 + 12 69 ) 2 (100 + 12 69 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (100 + 12 69 ) 1 - ----------------------- + 2/3 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (100 + 12 69 ) / 1/2 1/3 (100 + 12 69 ) 1 - ------------------- - ----------------------- + 2/3 12 1/2 (1/3) 3 (100 + 12 69 ) / 1/2 1/3 \ 1/2 |(100 + 12 69 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (100 + 12 69 ) / In floating-point [1.754877667, 0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] The largest root is, 1.754877667 and the remaining roots are [0.1225611669 + 0.7448617670 I, 0.1225611669 - 0.7448617670 I] whose absolute values are [0.7548776666, 0.7548776666] so the largest absolute value is, 0.7548776666 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7548776666 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., -0.2500000000, -0.06122448980, 0.1279069767, 0.06622516556, -0.05660377358, -0.05161290323, 0.01960784314, 0.03421787709, -0.002785515320, -0.02018140590, -0.003359607184, 0.01067668066, 0.004531531910, -0.004973221117, -0.003801297073, 0.001902159145, 0.002632393930, -0.0004386683642, -0.001607571533, -0.0001440807769, 0.0008807416143, 0.0002979924718, -0.0004288374476, -0.0002749257528, 0.0001769784138, -5 0.0002000451328, -0.00005181390101, -0.0001266945210, -0.1530008233 10 , 0.00007182060353, 0.00001847669429, -0.00003639722319, -0.00001945053714, -5 -5 0.00001597284320, 0.00001499900035, -0.5425379648 10 , -0.9876916443 10 , -6 -5 -5 -5 0.6705471088 10 , 0.5792631012 10 , 0.1037798473 10 , -0.3046486958 10 , -5 -5 -5 -0.1338141377 10 , 0.1408002678 10 , 0.1107659774 10 , -6 -6 -6 -0.5308245068 10 , -0.7613061097 10 , 0.1158720611 10 , -6 -7 -6 0.4622257251 10 , 0.4727327944 10 , -0.2518071051 10 , -7 -6 -7 -0.8866176457 10 , 0.1217568554 10 , 0.8036837027 10 , -7 -7 -7 -0.4968187944 10 , -0.5797527374 10 , 0.1409970224 10 , -7 0.3649279877 10 ] The largest is 0.1279069767 The smallest is -0.2500000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 245, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 29, 53, 97, 178, 327, 601, 1105, 2032, 3737, 6873, 12641, 23250, 42763, 78653, 144665, 266080, 489397, 900141, 1655617] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 4) with initial conditions, a(1) = 16, a(2) = 29, a(3) = 53, a(4) = 97, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 4) with initial conditions, b(1) = 16, b(2) = 29, b(3) = 53, b(4) = 97, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - 2 t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (19 + 3 33 ) 4 (19 + 3 33 ) [1, ----------------- + ------------------- + 1/3, - ----------------- 3 1/2 1/3 6 3 (19 + 3 33 ) 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(19 + 3 33 ) 4 | (19 + 3 33 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (19 + 3 33 ) / 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 |(19 + 3 33 ) 4 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (19 + 3 33 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.839286755, -0.4196433777 - 0.6062907300 I, -0.4196433777 + 0.6062907300 I] The largest root is, 1.839286755 and the remaining roots are [-0.4196433777 - 0.6062907300 I, -0.4196433777 + 0.6062907300 I] whose absolute values are [0.7373527065, 0.7373527065] so the largest absolute value is, 0.7373527065 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7373527065 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4375000000, -0.1379310345, -0.4716981132, -0.3608247423, -0.2752808989, -0.4097859327, -0.3444259567, -0.3266968326, -0.3774606299, -0.3446614932, -0.3446820893, -0.3625504311, -0.3475698925, -0.3504431401, -0.3561847609, -0.3498081775, -0.3520407396, -0.3536351878, -0.3510838857, -0.3523586675, -0.3526760880, -0.3517167118, -0.3523493884, -0.3523400278, -0.3520039233, -0.3522911107, -0.3522328199, -0.3521256050, -0.3522472828, -0.3522034527, -0.3521740843, -0.3522225630, -0.3521978429, -0.3521922329, -0.3522103814, -0.3521981997, -0.3521985565, -0.3522048801, -0.3521993788, -0.3522005579, -0.3522025593, -0.3522002385, -0.3522010982, -0.3522016384, -0.3522007176, -0.3522011967, -0.3522012952, -0.3522009520, -0.3522011865, -0.3522011763, -0.3522010574, -0.3522011627, -0.3522011390, -0.3522011017, -0.3522011459, -0.3522011291, -0.3522011191, -0.3522011366] The largest is -0.1379310345 The smallest is -0.4716981132 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 121, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 30, 56, 105, 197, 370, 695, 1305, 2450, 4600, 8637, 16217, 30449, 57171, 107344, 201549, 378428, 710536, 1334102, 2504909] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 246, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 31, 60, 116, 224, 433, 837, 1618, 3128, 6047, 11690, 22599, 43688, 84457, 163271, 315633, 610177, 1179585, 2280356, 4408350] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 4) + a(n - 5) with initial conditions, a(1) = 16, a(2) = 31, a(3) = 60, a(4) = 116, a(5) = 224, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 4) + b(n - 5) with initial conditions, b(1) = 16, b(2) = 31, b(3) = 60, b(4) = 116, b(5) = 224, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 t - 2 t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 %1 := _Z - 2 _Z + _Z - 1 In floating-point [1.93318498189952, 0.644722826787147 + 0.477341634328316 I, -0.611315317736907 + 0.655836101734178 I, -0.611315317736907 - 0.655836101734178 I, 0.644722826787147 - 0.477341634328316 I] The largest root is, 1.93318498189952 and the remaining roots are [0.644722826787147 + 0.477341634328316 I, -0.611315317736907 + 0.655836101734178 I, -0.611315317736907 - 0.655836101734178 I, 0.644722826787147 - 0.477341634328316 I] whose absolute values are [0.802198578435313, 0.896564225272043, 0.896564225272043, 0.802198578435313] so the largest absolute value is, 0.896564225272043 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.896564225272043 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.06250000000, 0.1290322581, 0.2666666667, -0.4482758621, 0.004464285714, -0.05773672055, -0.2532855436, 0.2088998764, -0.03548593350, -0.008764676699, 0.1780153978, -0.1061551396, 0.03206830251, 0.03741548954, -0.1119488458, 0.06027253171, -0.01767847690, -0.04070414595, 0.06795605598, -0.03630927671, 0.005332453431, 0.03369057568, -0.04127905029, 0.02170723182, 0.001772733457, -0.02481265533, 0.02534431530, -0.01229765152, -0.004660804670, 0.01726377945, -0.01562941173, 0.006383143351, 0.005129439854, -0.01166570441, 0.009561782360, -0.002888990365, -0.004524277233, 0.007746589801, -0.005734307170, 0.0009821583852, 0.003599603639, -0.005071659757, 0.003337577457, -0.00004131064079, -0.002700066535, 0.003271130326, -0.001866976563, -0.0003550650271, 0.001948625840, -0.002073945181, 0.0009902165264, 0.0004685215174, -0.001366647832, 0.001289275357, -0.0004856109942, -0.0004495269793, 0.0009361153909, -0.0007836924070] The largest is 0.2666666667 The smallest is -0.4482758621 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 247, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 33, 68, 140, 288, 592, 1217, 2502, 5144, 10576, 21744, 44705, 91912, 188968, 388512, 798768, 1642241, 3376394, 6941756, 14272024] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 5) with initial conditions, a(1) = 16, a(2) = 33, a(3) = 68, a(4) = 140, a(5) = 288, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 5) with initial conditions, b(1) = 16, b(2) = 33, b(3) = 68, b(4) = 140, b(5) = 288, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 %1 := _Z - 2 _Z - 1 In floating-point [2.05596739671282, 0.554186402433763 + 0.694592654552445 I, -0.582170100790172 + 0.526390166427287 I, -0.582170100790172 - 0.526390166427287 I, 0.554186402433763 - 0.694592654552445 I] The largest root is, 2.05596739671282 and the remaining roots are [0.554186402433763 + 0.694592654552445 I, -0.582170100790172 + 0.526390166427287 I, -0.582170100790172 - 0.526390166427287 I, 0.554186402433763 - 0.694592654552445 I] whose absolute values are [0.888583999631261, 0.784862174885111, 0.784862174885111, 0.888583999631261] so the largest absolute value is, 0.888583999631261 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.888583999631261 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.06250000000, 0.1212121212, 0.2352941176, 0.4571428571, -0.1111111111, -0.1604729730, -0.2004930156, -0.1662669864, 0.1244167963, 0.1376701967, 0.1148362767, 0.02916899676, -0.1079293237, -0.09144405402, -0.04521868050, 0.02439882419, 0.07796663218, 0.04800387632, 0.004563686767, -0.03609130702, -0.04778379144, -0.01760095205, 0.01280197213, 0.03016763101, 0.02424395494, 0.0007041184238, -0.01619271520, -0.01958345827, -0.008999285538, 0.006245383866, 0.01319488616, 0.01019705711, 0.0008106559444, -0.007377973649, -0.008510563431, -0.003826240707, 0.002544575696, 0.005899807337, 0.004421641025, 0.0003327186176, -0.003160803471, -0.003777031247, -0.001654255156, 0.001113130712, 0.002558980041, 0.001957156611, 0.0001372819755, -0.001379691205, -0.001646251699, -0.0007335233566, 0.0004901098977, 0.001117501771, 0.0008553123366, 0.00006437297425, -0.0006047774081, -0.0007194449186, -0.0003213880661, 0.0002125362043] The largest is 0.4571428571 The smallest is -0.2004930156 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 122, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 34, 72, 152, 321, 678, 1432, 3025, 6390, 13498, 28513, 60230, 127228, 268753, 567707, 1199210, 2533181, 5351028, 11303377, 23876969] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 248, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 35, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 35, 77, 169, 371, 814, 1786, 3919, 8599, 18868, 41400, 90840, 199321, 437350, 959633, 2105626, 4620163, 10137558, 22243822, 48807377] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + 2 a(n - 2) + a(n - 3) + a(n - 4) - a(n - 6) with initial conditions, a(1) = 16, a(2) = 35, a(3) = 77, a(4) = 169, a(5) = 371, a(6) = 814, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + 2 b(n - 2) + b(n - 3) + b(n - 4) - b(n - 6) with initial conditions, b(1) = 16, b(2) = 35, b(3) = 77, b(4) = 169, b(5) = 371, b(6) = 814, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 2 t - t - 2 t - t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 3 2 %1 := _Z - _Z - 2 _Z - _Z - _Z + 1 In floating-point [0.630274169792734, 2.19419922780442, -0.0188246726966071 + 0.898489867130679 I, -0.893412026101969 + 0.311662101565087 I, -0.893412026101969 - 0.311662101565087 I, -0.0188246726966071 - 0.898489867130679 I] The largest root is, 2.19419922780442 and the remaining roots are [0.630274169792734, -0.0188246726966071 + 0.898489867130679 I, -0.893412026101969 + 0.311662101565087 I, -0.893412026101969 - 0.311662101565087 I, -0.0188246726966071 - 0.898489867130679 I] whose absolute values are [0.630274169792734, 0.898687047663779, 0.946212615607925, 0.946212615607925, 0.898687047663779] so the largest absolute value is, 0.946212615607925 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.946212615607925 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4375000000, 0.4000000000, -0.07792207792, 0.4437869822, -0.02695417790, -0.3316953317, 0.4182530795, -0.2273539168, 0.3284102803, -0.4833580666, 0.3913043478, -0.1426574196, 0.06676165582, -0.08325368698, -0.02949356681, 0.2114625294, -0.2553204292, 0.1975148256, -0.1979187300, 0.2365067068, -0.1876427922, 0.07350418817, -0.007872989028, -0.009515523473, 0.05851862448, -0.1313879300, 0.1559035988, -0.1313733483, 0.1154375327, -0.1132779717, 0.08360871963, -0.02749510947, -0.01402170798, 0.03269216934, -0.05467516916, 0.08247032379, -0.09181827281, 0.07863448441, -0.06318519861, 0.05204365185, -0.03283536459, 0.004230901131, 0.01923689800, -0.03172749689, 0.04132703426, -0.05070381224, 0.05229502198, -0.04374396626, 0.03223240173, -0.02193682415, 0.009752000774, 0.005070600181, -0.01742484268, 0.02427550056, -0.02798398557, 0.03014959720, -0.02871971684, 0.02280039236] The largest is 0.4437869822 The smallest is -0.4833580666 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 249, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 36, 81, 182, 409, 919, 2065, 4640, 10426, 23427, 52640, 118281, 265775, 597191, 1341876, 3015168, 6775021, 15223334, 34206521, 76861355] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) - a(n - 3) with initial conditions, a(1) = 16, a(2) = 36, a(3) = 81, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) - b(n - 3) with initial conditions, b(1) = 16, b(2) = 36, b(3) = 81, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (28 + 84 I 3 ) 14 (28 + 84 I 3 ) [------------------- + --------------------- + 2/3, - ------------------- 6 1/2 1/3 12 3 (28 + 84 I 3 ) 7 - --------------------- + 2/3 1/2 1/3 3 (28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(28 + 84 I 3 ) 14 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (28 + 84 I 3 ) / 1/2 1/3 (28 + 84 I 3 ) 7 - ------------------- - --------------------- + 2/3 12 1/2 1/3 3 (28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(28 + 84 I 3 ) 14 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (28 + 84 I 3 ) / In floating-point -9 -9 [2.246979604 + 0.1 10 I, -0.8019377358 - 0.1866025404 10 I, -10 0.5549581322 - 0.1339745960 10 I] -9 The largest root is, 2.246979604 + 0.1 10 I and the remaining roots are -9 -10 [-0.8019377358 - 0.1866025404 10 I, 0.5549581322 - 0.1339745960 10 I] whose absolute values are [0.8019377358, 0.5549581322] so the largest absolute value is, 0.8019377358 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8019377358 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0., 0.2500000000, -0.06172839506, 0.1263736264, -0.05867970660, 0.07072905332, -0.04358353511, 0.04224137931, -0.02982927297, 0.02616638921, -0.01973784195, 0.01651998208, -0.01286426489, 0.01052929465, -0.008325657512, 0.006742244545, -0.005370463058, 0.004326975944, -0.003458755715, 0.002779927572, -0.002225876514, 0.001786930259, -0.001431943569, 0.001148919635, -0.0009210345569, 0.0007387940905, -0.0005923660113, 0.0004750966248, -0.0003809668523, 0.0003055289315, -0.0002450056140, 0.0001964845558, -0.0001575654338, 0.0001263593022, -0.0001013313854, 0.00008126196526, -0.00006516675700, 0.00005225983663, -0.00004190904900, 0.00003360849562, -0.00002695189438, 0.00002161375586, -5 -0.00001733287828, 0.00001389989369, -0.00001114684676, 0.8939078447 10 , -5 -5 -5 -0.7168583556 10 , 0.5748758095 10 , -0.4610145812 10 , -5 -5 -5 0.3697050027 10 , -0.2964803854 10 , 0.2377588131 10 , -5 -5 -5 -0.1906677620 10 , 0.1529036746 10 , -0.1226192259 10 , -6 -6 -6 0.9833298476 10 , -0.7885693094 10 , 0.6323834877 10 ] The largest is 0.2500000000 The smallest is -0.06172839506 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 250, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 16, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [16, 37, 86, 200, 465, 1081, 2513, 5842, 13581, 31572, 73396, 170625, 396655, 922111, 2143648, 4983377, 11584946, 26931732, 62608681, 145547525] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 3) with initial conditions, a(1) = 16, a(2) = 37, a(3) = 86, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 3) with initial conditions, b(1) = 16, b(2) = 37, b(3) = 86, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 3 t + 2 t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [------------------- + ------------------- + 1, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- + 1 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- + 1 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / In floating-point [2.324717958, 0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] The largest root is, 2.324717958 and the remaining roots are [0.3376410214 + 0.5622795125 I, 0.3376410214 - 0.5622795125 I] whose absolute values are [0.6558656185, 0.6558656185] so the largest absolute value is, 0.6558656185 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6558656185 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4375000000, -0.1081081081, 0.1162790698, 0.1250000000, 0.03440860215, -0.03052728955, -0.03541583764, -0.01078397809, 0.007952286282, 0.01000886862, 0.003338056570, -0.002051282051, -0.002821091377, -0.001022653455, 0.0005229403335, 0.0007930365292, 0.0003105754658, -0.0001314063277, -0.0002223333854, -0.00009361203497, 0.00003242433815, -5 0.00006216369902, 0.00002803038580, -0.7811902501 10 , -0.00001733278008, -5 -5 -5 -5 -0.8344149430 10 , 0.1821209362 10 , 0.4819146869 10 , 0.2470872453 10 , -6 -5 -6 -0.4044670174 10 , -0.1335999089 10 , -0.7281907795 10 , -7 -6 -6 -7 0.8295882229 10 , 0.3692589368 10 , 0.2136683863 10 , -0.1455389243 10 , -6 -7 -8 -0.1017395131 10 , -0.6244236808 10 , 0.1598029464 10 , -7 -7 -9 -8 0.2793931149 10 , 0.1817950744 10 , 0.2579288246 10 , -0.7645916928 10 , -8 -9 -8 -0.5274100990 10 , -0.2725402894 10 , 0.2084664184 10 , -8 -9 -9 0.1524972140 10 , 0.1330477632 10 , -0.5661368067 10 , -9 -10 -9 -0.4395338064 10 , -0.5328004264 10 , 0.1530906782 10 , -9 -10 -10 0.1262983136 10 , 0.1943354158 10 , -0.4120532415 10 , -10 -11 -10 -0.3618474206 10 , -0.6710036285 10 , 0.1103405110 10 ] The largest is 0.1250000000 The smallest is -0.4375000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 123, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 38, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 38, 90, 213, 504, 1193, 2824, 6685, 15825, 37462, 88683, 209937, 496978, 1176482, 2785053, 6592978, 15607372, 36946894, 87463346, 207049526] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 124, : Consider the Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 39, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 16, 39, 95, 231, 562, 1367, 3325, 8088, 19674, 47857, 116412, 283172, 688815, 1675540, 4075745, 9914235, 24116341, 58662913, 142697326, 347110735, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) + a(n - 2) + a(n - 6) + 2 a(n - 7), . Alas, it breaks down at the, 32, -th term. a(32), equals , 14896854169592, while the corresponding term for the solution of the recurrence is , 14896854169591 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.03999622704587 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 125, :The Pisot Sequence a(n), defined by, a(1) = 16, a(2) = 40, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [16, 40, 100, 250, 625, 1563, 3909, 9776, 24449, 61145, 152919, 382439, 956451, 2392012, 5982242, 14961137, 37416678, 93576297, 234027279, 585284619] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 126, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 19, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55 ] is a trivial linear sequence Fact, 127, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 20, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 20, 24, 29, 35, 42, 50, 60, 72, 86, 103, 123, 147, 176, 211, 253, 303, 363, 435, 521] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 128, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 21, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 21, 26, 32, 39, 48, 59, 73, 90, 111, 137, 169, 208, 256, 315, 388, 478, 589, 726, 895] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 129, : Consider the Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 17, 22, 28, 36, 46, 59, 76, 98, 126, 162, 208, 267, 343, 441, 567, 729, 937, 1204, 1547, 1988, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = a(n - 1) + a(n - 6), . Alas, it breaks down at the, 36, -th term. a(36), equals , 110155, while the corresponding term for the solution of the recurrence is , 110156 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03282504251949 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 130, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 23, 31, 42, 57, 77, 104, 140, 188, 252, 338, 453, 607, 813, 1089, 1459, 1955, 2620, 3511, 4705] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 131, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 24, 34, 48, 68, 96, 136, 193, 274, 389, 552, 783, 1111, 1576, 2236, 3172, 4500, 6384, 9057, 12849] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 132, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 25, 37, 55, 82, 122, 182, 272, 407, 609, 911, 1363, 2039, 3050, 4562, 6824, 10208, 15270, 22842, 34169] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 133, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 26, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 26, 40, 62, 96, 149, 231, 358, 555, 860, 1333, 2066, 3202, 4963, 7692, 11922, 18478, 28639, 44388, 68798] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 134, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 27, 43, 68, 108, 172, 274, 436, 694, 1105, 1759, 2800, 4457, 7095, 11294, 17978, 28618, 45555, 72516, 115433] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 135, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 28, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 28, 46, 76, 126, 209, 347, 576, 956, 1587, 2634, 4372, 7257, 12046, 19995, 33189, 55089, 91440, 151778, 251931] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 251, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 17, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [17, 29, 49, 83, 141, 240, 409, 697, 1188, 2025, 3452, 5885, 10033, 17105, 29162, 49718, 84764, 144514, 246382, 420057] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) + a(n - 5) - a(n - 6) with initial conditions, a(1) = 17, a(2) = 29, a(3) = 49, a(4) = 83, a(5) = 141, a(6) = 240, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) + b(n - 5) - b(n - 6) with initial conditions, b(1) = 17, b(2) = 29, b(3) = 49, b(4) = 83, b(5) = 141, b(6) = 240, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 3 t - 2 t + t - t + 1 = 0 whose roots are [1, RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 %1 := _Z - _Z - _Z - 1 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.70490277604165, 0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] The largest root is, 1.70490277604165 and the remaining roots are [0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] whose absolute values are [0.829475823582982, 0.923307702773949, 0.923307702773949, 0.829475823582982] so the largest absolute value is, 0.923307702773949 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.923307702773949 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4705882353, -0.2068965517, -0.4081632653, -0.4698795181, -0.4893617021, 0.004166666667, -0.2029339853, -0.1162123386, -0.2954545455, -0.4054320988, -0.2001738123, -0.3119796092, -0.1316655038, -0.2423852675, -0.2826966600, -0.2284082224, -0.3262233967, -0.1894210942, -0.2611513828, -0.2363869665, -0.2290618245, -0.2947863021, -0.2163822362, -0.2754326428, -0.2313142787, -0.2389210400, -0.2681338343, -0.2265492575, -0.2732278565, -0.2342034935, -0.2494644815, -0.2549138947, -0.2340397150, -0.2652935457, -0.2366488323, -0.2545189369, -0.2491937408, -0.2408644694, -0.2584638325, -0.2390892107, -0.2551840565, -0.2465790844, -0.2457396866, -0.2538946798, -0.2418356534, -0.2540264661, -0.2455532802, -0.2484315092, -0.2509915455, -0.2443707845, -0.2525008724, -0.2455370134, -0.2495814714, -0.2492221067, -0.2462864389, -0.2511214943, -0.2460570229, -0.2498720649] The largest is 0.4705882353 The smallest is -0.4893617021 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 252, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 17, a(2) = 30, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [17, 30, 53, 94, 167, 297, 528, 939, 1670, 2970, 5282, 9394, 16707, 29713, 52844, 93982, 167145, 297264, 528678, 940243] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 3) - a(n - 5) with initial conditions, a(1) = 17, a(2) = 30, a(3) = 53, a(4) = 94, a(5) = 167, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 3) - b(n - 5) with initial conditions, b(1) = 17, b(2) = 30, b(3) = 53, b(4) = 94, b(5) = 167, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - t - t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - _Z - _Z - _Z + 1 In floating-point [0.730440478358655, 1.77847961614338, -0.337778460538625 + 0.899782138378692 I, -0.833363173424790, -0.337778460538625 - 0.899782138378692 I] The largest root is, 1.77847961614338 and the remaining roots are [0.730440478358655, -0.337778460538625 + 0.899782138378692 I, -0.833363173424790, -0.337778460538625 - 0.899782138378692 I] whose absolute values are [0.730440478358655, 0.961094264341004, 0.833363173424790, 0.961094264341004] so the largest absolute value is, 0.961094264341004 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.961094264341004 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.05882352941, -0.3666666667, -0.2830188679, -0.3085106383, 0.1976047904, -0.3333333333, -0.07386363636, 0.07454739084, -0.02395209581, -0.2208754209, 0.1631957592, -0.007770917607, -0.1400011971, 0.03937670380, 0.1124820226, -0.1513374902, 0.008292201382, 0.1094380752, -0.07298393351, -0.06773568110, 0.1200559980, -0.02895581980, -0.08607357575, 0.07801053827, 0.03071682331, -0.09740221089, 0.04028097058, 0.05966915881, -0.07546261965, -0.006229313592, 0.07537943650, -0.04659346731, -0.03711250322, 0.06713608562, -0.01034057132, -0.05569642542, 0.04769255620, 0.01876806268, -0.05637189216, 0.02042929803, 0.03852189396, -0.04511325636, -0.004930127047, 0.04485040271, -0.02562227873, -0.02422389703, 0.04011748332, -0.004798565392, -0.03375538181, 0.02718581484, 0.01285576467, -0.03383128562, 0.01100885928, 0.02378872014, -0.02621952104, -0.004277706288, 0.02712277843, -0.01438330818] The largest is 0.1976047904 The smallest is -0.3666666667 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 253, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 17, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [17, 31, 57, 105, 193, 355, 653, 1201, 2209, 4063, 7473, 13745, 25281, 46499, 85525, 157305, 289329, 532159, 978793, 1800281] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 3) with initial conditions, a(1) = 17, a(2) = 31, a(3) = 57, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 3) with initial conditions, b(1) = 17, b(2) = 31, b(3) = 57, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (19 + 3 33 ) 4 (19 + 3 33 ) [----------------- + ------------------- + 1/3, - ----------------- 3 1/2 1/3 6 3 (19 + 3 33 ) 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(19 + 3 33 ) 4 | (19 + 3 33 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (19 + 3 33 ) / 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 |(19 + 3 33 ) 4 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (19 + 3 33 ) / In floating-point [1.839286755, -0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] The largest root is, 1.839286755 and the remaining roots are [-0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] whose absolute values are [0.7373527065, 0.7373527065] so the largest absolute value is, 0.7373527065 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7373527065 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4705882353, -0.1935483871, 0.4210526316, -0.2476190476, -0.02072538860, 0.1521126761, -0.1163859112, 0.01498751041, 0.05070167497, -0.05070145213, 0.01498728757, 0.01498726810, -0.02072702820, 0.009247510699, 0.003507746273, -0.007971774578, 0.004783481780, 0.0003194533964, -0.002868839479, 0.002234095677, -0.0003152904069, -0.0009500342101, 0.0009687710596, -0.0002965535575, -0.0002778167079, 0.0003944007942, -0.0001799694712, -0.00006338538491, 0.0001510459381, -0.00009230891796, -5 -5 -0.4648364740 10 , 0.00005408865543, -0.00004286862727, 0.6571663416 10 , -5 -5 0.00001779169157, -0.00001850527228, 0.5858082704 10 , 0.5144501993 10 , -5 -5 -5 -0.7502687587 10 , 0.3499897111 10 , 0.1141711517 10 , -5 -5 -7 -0.2861078959 10 , 0.1780529669 10 , 0.6116222640 10 , -5 -6 -6 -0.1019387064 10 , 0.8223048308 10 , -0.1359200068 10 , -6 -6 -6 -0.3330022401 10 , 0.3533825839 10 , -0.1155396630 10 , -7 -6 -7 -0.9515931925 10 , 0.1426836016 10 , -0.6801538069 10 , -7 -7 -7 -0.2049109832 10 , 0.5417712260 10 , -0.3432935641 10 , -9 -7 -0.6433321419 10 , 0.1920443404 10 ] The largest is 0.4210526316 The smallest is -0.4705882353 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 136, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 32, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 32, 60, 113, 213, 401, 755, 1422, 2678, 5043, 9497, 17885, 33682, 63432, 119459, 224972, 423680, 797898, 1502646, 2829867] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 137, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 33, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 33, 64, 124, 240, 465, 901, 1746, 3383, 6555, 12701, 24610, 47685, 92396, 179029, 346891, 672145, 1302366, 2523499, 4889599] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 138, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 35, 72, 148, 304, 624, 1281, 2630, 5400, 11087, 22763, 46735, 95952, 197000, 404463, 830408, 1704921, 3500395, 7186706, 14755119] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 254, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 17, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [17, 36, 76, 160, 337, 710, 1496, 3152, 6641, 13992, 29480, 62112, 130865, 275722, 580924, 1223960, 2578785, 5433292, 11447508, 24118976] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 4) with initial conditions, a(1) = 17, a(2) = 36, a(3) = 76, a(4) = 160, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 4) with initial conditions, b(1) = 17, b(2) = 36, b(3) = 76, b(4) = 160, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - 2 _Z - 1 In floating-point [2.10691934037622, 0.304876704453035 + 0.754529173144244 I, -0.716672749282287, 0.304876704453035 - 0.754529173144244 I] The largest root is, 2.10691934037622 and the remaining roots are [0.304876704453035 + 0.754529173144244 I, -0.716672749282287, 0.304876704453035 - 0.754529173144244 I] whose absolute values are [0.813796091194765, 0.716672749282287, 0.813796091194765] so the largest absolute value is, 0.813796091194765 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.813796091194765 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2352941176, 0.4444444444, -0.1578947368, -0.1937500000, -0.1543026706, 0.1352112676, 0.1122994652, 0.03077411168, -0.09275711489, -0.05031446541, 0.01166892809, 0.05411192684, 0.01546632025, -0.01938184113, -0.02709476627, -0.00007761691559, 0.01531108642, 0.01124033091, -0.004614104659, -0.009305826251, -0.003300566111, 0.004639198685, 0.004664292708, 0.00002275916505, -0.003255047781, -0.001870896877, 0.0009224989550, 0.001867757075, 0.0004804663693, -0.0009099641381, -0.0008974293212, 0.00007289843248, 0.0006262632342, 0.0003425623304, -0.0002123046605, -0.0003517108884, -0.00007715854265, 0.0001882452451, 0.0001641858297, -0.00002333922906, -0.0001238370008, -0.00005942875648, 0.00004532831672, 0.00006731740438, 0.00001079780798, -0.00003783314052, -0.00003033796432, -5 -5 0.6641475740 10 , 0.00002408075946, 0.00001032837840, -0.9681207517 10 , -5 -5 -5 -0.00001272093929, -0.1361119125 10 , 0.7606140152 10 , 0.5531072787 10 , -5 -5 -5 -0.1658793720 10 , -0.4678706566 10 , -0.1751272980 10 ] The largest is 0.4444444444 The smallest is -0.1937500000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 139, : Consider the Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 17, 37, 81, 177, 387, 846, 1849, 4041, 8832, 19303, 42188, 92205, 201521, 440439, 962612, 2103860, 4598142, 10049580, 21964102, 48004173, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) + a(n - 3) - a(n - 5) + a(n - 6), . Alas, it breaks down at the, 45, -th term. a(45), equals , 14805308865224287, while the corresponding term for the solution of the recurrence is , 14805308865224286 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.01326626733001 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Fact, 140, : Consider the Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 38, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 17, 38, 85, 190, 425, 951, 2128, 4762, 10656, 23845, 53358, 119399, 267179, 597866, 1337844, 2993692, 6698981, 14990302, 33543781, 75060879, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) + a(n - 2) - a(n - 3) - a(n - 4) + 2 a(n - 5), . Alas, it breaks down at the, 49, -th term. a(49), equals , 1046311428155518025, while the corresponding term for the solution of the recurrence is , 1046311428155518024 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.01884404554226 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 141, :The Pisot Sequence a(n), defined by, a(1) = 17, a(2) = 39, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [17, 39, 89, 203, 463, 1056, 2409, 5496, 12539, 28607, 65265, 148898, 339701, 775006, 1768126, 4033865, 9203002, 20996054, 47901140, 109283355] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 255, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 17, a(2) = 40, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [17, 40, 94, 221, 520, 1224, 2881, 6781, 15960, 37564, 88412, 208090, 489769, 1152740, 2713135, 6385743, 15029740, 35374597, 83259066, 195961867] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) + a(n - 2) - a(n - 3) + a(n - 4) + a(n - 5) with initial conditions, a(1) = 17, a(2) = 40, a(3) = 94, a(4) = 221, a(5) = 520, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) + b(n - 2) - b(n - 3) + b(n - 4) + b(n - 5) with initial conditions, b(1) = 17, b(2) = 40, b(3) = 94, b(4) = 221, b(5) = 520, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 2 t - 2 t - t + t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 2 %1 := _Z - 2 _Z - _Z + _Z - _Z - 1 In floating-point [2.35363998628043, 0.531292779430148 + 0.731552807185123 I, -0.708112772570363 + 0.135413437149099 I, -0.708112772570363 - 0.135413437149099 I, 0.531292779430148 - 0.731552807185123 I] The largest root is, 2.35363998628043 and the remaining roots are [0.531292779430148 + 0.731552807185123 I, -0.708112772570363 + 0.135413437149099 I, -0.708112772570363 - 0.135413437149099 I, 0.531292779430148 - 0.731552807185123 I] whose absolute values are [0.904124729876938, 0.720944170957654, 0.720944170957654, 0.904124729876938] so the largest absolute value is, 0.904124729876938 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.904124729876938 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.1176470588, -0.1000000000, -0.4148936170, -0.4705882353, 0.1076923077, 0.1772875817, 0.4168691427, 0.01710662144, -0.08922305764, -0.2932595038, -0.09871963082, 0.03249074920, 0.1874046744, 0.1235360966, 0.01000650539, -0.1100844491, -0.1138030997, -0.03675640460, 0.05631114094, 0.08959103252, 0.04836206092, -0.02055549104, -0.06278521735, -0.04858581323, -0.001448259349, 0.03910945531, 0.04201575610, 0.01321819627, -0.02069137924, -0.03251912235, -0.01782260880, 0.007760991666, 0.02274531392, 0.01786372671, 0.0003700445211, -0.01420311530, -0.01539360720, -0.004751333601, 0.007540612123, 0.01189042707, 0.006476077364, -0.002842971129, -0.008311013445, -0.006510036189, -0.0001216102586, 0.005190862973, 0.005616167302, 0.001723758202, -0.002758825715, -0.004340807816, -0.002357169275, 0.001043604852, 0.003035780733, 0.002372702061, 0.00003960291112, -0.001897437272, -0.002048588109, -0.0006257336069] The largest is 0.4168691427 The smallest is -0.4705882353 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 142, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 20, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56 ] is a trivial linear sequence Fact, 143, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 21, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 21, 25, 30, 36, 43, 51, 60, 71, 84, 99, 117, 138, 163, 193, 229, 272, 323, 384, 457] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 144, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 22, 27, 33, 40, 48, 58, 70, 84, 101, 121, 145, 174, 209, 251, 301, 361, 433, 519, 622] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 145, : Consider the Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 18, 23, 29, 37, 47, 60, 77, 99, 127, 163, 209, 268, 344, 442, 568, 730, 938, 1205, 1548, 1989, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 2) + a(n - 6) - a(n - 7), . Alas, it breaks down at the, 28, -th term. a(28), equals , 14799, while the corresponding term for the solution of the recurrence is , 14800 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03282504251949 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 146, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 24, 32, 43, 58, 78, 105, 141, 189, 253, 339, 454, 608, 814, 1090, 1460, 1956, 2621, 3512, 4706] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 147, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 25, 35, 49, 69, 97, 136, 191, 268, 376, 528, 741, 1040, 1460, 2050, 2878, 4040, 5671, 7960, 11173] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 148, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 26, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 26, 38, 56, 83, 123, 182, 269, 398, 589, 872, 1291, 1911, 2829, 4188, 6200, 9179, 13589, 20118, 29784] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 149, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 27, 41, 62, 94, 143, 218, 332, 506, 771, 1175, 1791, 2730, 4161, 6342, 9666, 14732, 22453, 34221, 52157] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 256, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 18, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [18, 28, 44, 69, 108, 169, 264, 412, 643, 1004, 1568, 2449, 3825, 5974, 9330, 14571, 22756, 35539, 55503, 86682] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 4) - a(n - 5) + a(n - 6) with initial conditions, a(1) = 18, a(2) = 28, a(3) = 44, a(4) = 69, a(5) = 108, a(6) = 169, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 4) - b(n - 5) + b(n - 6) with initial conditions, b(1) = 18, b(2) = 28, b(3) = 44, b(4) = 69, b(5) = 108, b(6) = 169, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 2 t - 2 t + t - t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 2 %1 := _Z - 2 _Z + _Z - _Z + _Z - 1 In floating-point [1.56175206772030, 0.713504817792455 + 0.658512530611071 I, -0.0290284808449552 + 0.853775991281275 I, -0.930704741615296, -0.0290284808449552 - 0.853775991281275 I, 0.713504817792455 - 0.658512530611071 I] The largest root is, 1.56175206772030 and the remaining roots are [0.713504817792455 + 0.658512530611071 I, -0.0290284808449552 + 0.853775991281275 I, -0.930704741615296, -0.0290284808449552 - 0.853775991281275 I, 0.713504817792455 - 0.658512530611071 I] whose absolute values are [0.970941747987406, 0.854269334571064, 0.930704741615296, 0.854269334571064, 0.970941747987406] so the largest absolute value is, 0.970941747987406 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.970941747987406 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4444444444, 0.1428571429, 0.2045454545, 0.04347826087, 0.4537037037, 0.4023668639, -0.03030303030, -0.4830097087, -0.3234836703, -0.1713147410, 0.0006377551020, 0.1220906492, 0.3728104575, 0.2922664881, 0.06012861736, -0.2218790749, -0.2525487783, -0.2416781564, -0.09015728880, 0.07161809834, 0.2628530907, 0.2430755260, 0.1222682560, -0.07844236132, -0.1780754221, -0.2258680812, -0.1316152929, 0.005002285027, 0.1642550625, 0.1972727830, 0.1464678360, 0.006412366642, -0.1060056210, -0.1804036040, -0.1413514770, -0.04508203924, 0.08523724656, 0.1475709158, 0.1429510904, 0.05419709846, -0.04558908484, -0.1281236382, -0.1300407704, -0.07314097880, 0.02692371986, 0.09865096365, 0.1228719904, 0.07586917055, -0.001109721012, -0.07950234759, -0.1067502275, -0.08234996370, -0.01205660101, 0.05571330564, 0.09512561134, 0.07943583329, 0.02728919038, -0.03943750956] The largest is 0.4537037037 The smallest is -0.4830097087 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 257, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 18, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761] ] The sequence a(n) satisfies, for n>=, 3, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) with initial conditions, a(1) = 18, a(2) = 29, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) with initial conditions, b(1) = 18, b(2) = 29, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 2, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 2 t - t - 1 = 0 whose roots are 1/2 1/2 5 5 [---- + 1/2, 1/2 - ----] 2 2 In floating-point [1.618033988, -0.6180339880] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2777777778, 0.1724137931, -0.1063829787, 0.06578947368, -0.04065040650, 0.02512562814, -0.01552795031, 0.009596928983, -0.005931198102, 0.003665689150, -0.002265518804, 0.001400168020, -0.0008653513326, 0.0005348165579, -0.0003305348053, 0.0002042817454, -0.0001262530616, 0.00007802868334, -0.00004822437839, 0.00002980430493, -0.00001842007346, -5 -5 -5 0.00001138423147, -0.7035841986 10 , 0.4348389487 10 , -0.2687452499 10 , -5 -5 -6 0.1660936988 10 , -0.1026515512 10 , 0.6344214761 10 , -6 -6 -6 -0.3920940354 10 , 0.2423274407 10 , -0.1497665948 10 , -7 -7 -7 0.9256084594 10 , -0.5720574882 10 , 0.3535509712 10 , -7 -7 -8 -0.2185065170 10 , 0.1350444542 10 , -0.8346206272 10 , -8 -8 -8 0.5158239153 10 , -0.3187967119 10 , 0.1970272034 10 , -8 -9 -9 -0.1217695084 10 , 0.7525769500 10 , -0.4651181343 10 , -9 -9 -9 0.2874588158 10 , -0.1776593185 10 , 0.1097994973 10 , -10 -10 -10 -0.6785982125 10 , 0.4193967600 10 , -0.2592014525 10 , -10 -11 -11 0.1601953076 10 , -0.9900614491 10 , 0.6118916265 10 , -11 -11 -11 -0.3781698226 10 , 0.2337218039 10 , -0.1444480187 10 , -12 -12 -12 0.8927378517 10 , -0.5517423354 10 , 0.3409955163 10 ] The largest is 0.1724137931 The smallest is -0.2777777778 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 150, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 30, 50, 83, 138, 229, 380, 631, 1048, 1741, 2892, 4804, 7980, 13256, 22020, 36578, 60761, 100932, 167661, 278506] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 151, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 31, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 31, 53, 91, 156, 267, 457, 782, 1338, 2289, 3916, 6699, 11460, 19605, 33539, 57376, 98155, 167917, 287261, 491427] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 152, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 32, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 32, 57, 102, 183, 328, 588, 1054, 1889, 3386, 6069, 10878, 19498, 34949, 62644, 112286, 201267, 360761, 646646, 1159081] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 153, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 33, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 33, 61, 113, 209, 387, 717, 1328, 2460, 4557, 8442, 15639, 28972, 53672, 99430, 184199, 341238, 632161, 1171111, 2169544] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 154, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 34, 64, 120, 225, 422, 791, 1483, 2780, 5211, 9768, 18310, 34322, 64336, 120597, 226058, 423744, 794305, 1488919, 2790968] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 155, : Consider the Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 18, 35, 68, 132, 256, 496, 961, 1862, 3608, 6991, 13546, 26247, 50857, 98542, 190938, 369967, 716859, 1389007, 2691381, 5214899, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 3) + 2 a(n - 4) - a(n - 5), . Alas, it breaks down at the, 26, -th term. a(26), equals , 275975333, while the corresponding term for the solution of the recurrence is , 275975332 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.09047641696283 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 156, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 37, 76, 156, 320, 656, 1345, 2758, 5655, 11595, 23774, 48745, 99944, 204920, 420157, 861467, 1766305, 3621536, 7425401, 15224639] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 157, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 38, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 38, 80, 168, 353, 742, 1560, 3280, 6896, 14498, 30480, 64080, 134719, 283227, 595443, 1251831, 2631790, 5532950, 11632211, 24455007] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 258, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 18, a(2) = 39, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [18, 39, 85, 185, 403, 878, 1913, 4168, 9081, 19785, 43106, 93916, 204617, 445804, 971284, 2116160, 4610529, 10045071, 21885439, 47682335] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 3 a(n - 1) - 2 a(n - 2) + a(n - 4) with initial conditions, a(1) = 18, a(2) = 39, a(3) = 85, a(4) = 185, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 3 b(n - 1) - 2 b(n - 2) + b(n - 4) with initial conditions, b(1) = 18, b(2) = 39, b(3) = 85, b(4) = 185, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 3 t + 2 t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 3 _Z + 2 _Z - 1 In floating-point [2.17872417610522, 0.667076110379437 + 0.670769076539608 I, -0.512876396864097, 0.667076110379437 - 0.670769076539608 I] The largest root is, 2.17872417610522 and the remaining roots are [0.667076110379437 + 0.670769076539608 I, -0.512876396864097, 0.667076110379437 - 0.670769076539608 I] whose absolute values are [0.946003007966020, 0.512876396864097, 0.946003007966020] so the largest absolute value is, 0.946003007966020 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.946003007966020 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.5000000000, 0.2564102564, -0.3529411765, -0.1135135135, -0.1364764268, 0.07403189066, 0.1416623105, 0.1633877159, 0.07036669970, -0.04164771291, -0.1240198580, -0.1253779974, -0.05772736381, 0.03592610205, 0.09921300052, 0.1004087593, 0.04507291896, -0.02967266234, -0.07995082941, -0.08009840542, -0.03532063829, 0.02456223353, 0.06437714762, 0.06390857035, 0.02765077753, -0.02030257460, -0.05183213123, -0.05098267415, -0.02163298245, 0.01676382633, 0.04172531268, 0.04066561122, 0.01691322585, -0.01382771856, -0.03358429471, -0.03243183578, -0.01321369207, 0.01139487678, 0.02702771977, 0.02586156998, 0.01031557833, -0.009381528203, -0.02174802149, -0.02061943808, -0.008046692938, 0.007717269143, 0.01749717182, 0.01643753909, 0.006271580682, -0.006343066980, -0.01407519049, -0.01310189842, -0.004883733598, 0.005209529065, 0.01132086390, 0.01044163515, 0.003799444062, -0.004275409056] The largest is 0.2564102564 The smallest is -0.5000000000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 158, :The Pisot Sequence a(n), defined by, a(1) = 18, a(2) = 40, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [18, 40, 89, 198, 440, 978, 2174, 4833, 10744, 23884, 53094, 118028, 262376, 583261, 1296587, 2882308, 6407360, 14243538, 31663333, 70387474] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 159, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 21, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57 ] is a trivial linear sequence Fact, 160, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76 ] is a trivial linear sequence Fact, 161, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 23, 28, 34, 41, 49, 59, 71, 85, 102, 122, 146, 175, 210, 252, 302, 362, 434, 520, 623] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 162, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 24, 30, 38, 48, 61, 78, 100, 128, 164, 210, 269, 345, 442, 566, 725, 929, 1190, 1524, 1952] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 163, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 25, 33, 44, 59, 79, 106, 142, 190, 254, 340, 455, 609, 815, 1091, 1460, 1954, 2615, 3500, 4685] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 259, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 19, a(2) = 26, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657, 907, 1252, 1728, 2385, 3292, 4544, 6272, 8657] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 4) with initial conditions, a(1) = 19, a(2) = 26, a(3) = 36, a(4) = 50, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 4) with initial conditions, b(1) = 19, b(2) = 26, b(3) = 36, b(4) = 50, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - _Z - 1 In floating-point [1.38027756909761, 0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] The largest root is, 1.38027756909761 and the remaining roots are [0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] whose absolute values are [0.940435682699417, 0.819172513396165, 0.940435682699417] so the largest absolute value is, 0.940435682699417 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.940435682699417 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.4210526316, -0.1538461538, 0.4444444444, 0.2200000000, -0.2028985507, -0.3578947368, 0.08396946565, 0.3038674033, 0.1000000000, -0.2579710145, -0.1743697479, 0.1293759513, 0.2293274531, -0.02875399361, -0.2031250000, -0.07379454927, 0.1555285541, 0.1267605634, -0.07637117347, -0.1501674945, 0.005356096744, 0.1321166556, 0.05574346585, -0.09442428871, -0.08906873256, 0.04304757450, 0.09879098137, 0.004366467633, -0.08470226525, -0.04165477757, 0.05713618859, 0.06150263548, -0.02319964717, -0.06485442654, -0.007718248118, 0.05378438726, 0.03058473642, -0.03426969099, -0.04198793988, 0.01179644653, 0.04238118289, 0.008111491454, -0.03387644844, -0.02208000207, 0.02030118078, 0.02841267220, -0.005463776279, -0.02754377835, -0.007242597591, 0.02117007461, 0.01570629833, -0.01183748002, -0.01908007762, 0.002089996993, 0.01779629532, 0.005958815294, -0.01312126232, -0.01103126533] The largest is 0.4444444444 The smallest is -0.4210526316 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 164, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 27, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 27, 38, 53, 74, 103, 143, 199, 277, 386, 538, 750, 1046, 1459, 2035, 2838, 3958, 5520, 7698, 10735] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 260, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 19, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 3) with initial conditions, a(1) = 19, a(2) = 28, a(3) = 41, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 3) with initial conditions, b(1) = 19, b(2) = 28, b(3) = 41, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / In floating-point [1.465571232, -0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 + 0.7925519930 I, -0.2327856159 - 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2631578947, 0.03571428571, -0.1951219512, 0.06666666667, 0.1022727273, -0.09302325581, -0.02645502646, 0.07581227437, -0.01724137931, -0.04369747899, 0.03211009174, 0.01486697966, -0.02883075280, 0.003278688525, 0.01814566244, -0.01068521031, -0.007406550168, 0.01073910297, 0.00005387931034, -0.007352670858, 0.003386429199, 0.003440308087, -0.003912363067, -0.0005259341307, 0.002914373953, -0.0009979891816, -0.001523923318, 0.001390450627, 0.0003924614404, -0.001131461878, 0.0002589887477, 0.0006514501880, -0.0004800116899, -0.0002210229423, 0.0004304272457, -0.00004958444422, -0.0002706073865, -6 0.0001598198592, 0.0001102354150, -0.0001603719715, -0.5521122886 10 , 0.0001096833027, -0.00005068866879, -0.00005124078108, 0.00005844252163, -5 0.7753852834 10 , -0.00004348692825, 0.00001495559338, 0.00002270944621, -5 -5 -0.00002077748203, -0.5821888655 10 , 0.00001688755756, -0.3889924476 10 , -5 -5 -5 -0.9711813131 10 , 0.7175744428 10 , 0.3285819952 10 , -5 -6 -0.6425993179 10 , 0.7497512488 10 ] The largest is 0.2631578947 The smallest is -0.1951219512 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Fact, 165, : Consider the Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 29, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 19, 29, 44, 67, 102, 155, 236, 359, 546, 830, 1262, 1919, 2918, 4437, 6747, 10260, 15602, 23725, 36077, 54860, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 2) + a(n - 4) - a(n - 5) + a(n - 7), . Alas, it breaks down at the, 24, -th term. a(24), equals , 293326, while the corresponding term for the solution of the recurrence is , 293327 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03130091592356 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- ------------------------------------------------------------------------- Fact, 166, : Consider the Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 19, 30, 47, 74, 117, 185, 293, 464, 735, 1164, 1843, 2918, 4620, 7315, 11582, 18338, 29035, 45972, 72789, 115249, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) - a(n - 2) + a(n - 4) - a(n - 6) + a(n - 7), . Alas, it breaks down at the, 54, -th term. a(54), equals , 703162155202, while the corresponding term for the solution of the recurrence is , 703162155201 So the difference of the former from the latter is, 1 Note that the Pisot Indicator is, 1.02968394081287 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 167, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 31, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 31, 51, 84, 138, 227, 373, 613, 1007, 1654, 2717, 4463, 7331, 12042, 19780, 32490, 53367, 87659, 143986, 236507] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 168, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 32, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 32, 54, 91, 153, 257, 432, 726, 1220, 2050, 3445, 5789, 9728, 16347, 27470, 46161, 77570, 130350, 219042, 368081] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 261, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 19, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [19, 33, 57, 98, 168, 288, 494, 847, 1452, 2489, 4267, 7315, 12540, 21497, 36852, 63175, 108300, 185657, 318269, 545604] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 4) - a(n - 6) with initial conditions, a(1) = 19, a(2) = 33, a(3) = 57, a(4) = 98, a(5) = 168, a(6) = 288, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 4) - b(n - 6) with initial conditions, b(1) = 19, b(2) = 33, b(3) = 57, b(4) = 98, b(5) = 168, b(6) = 288, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 2 t - t - t - t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 2 %1 := _Z - _Z - _Z - _Z + 1 In floating-point [0.768085398012510, 1.71428532914257, 0.112851094827483 + 0.965281519117545 I, -0.854036458405025 + 0.273331145162114 I, -0.854036458405025 - 0.273331145162114 I, 0.112851094827483 - 0.965281519117545 I] The largest root is, 1.71428532914257 and the remaining roots are [0.768085398012510, 0.112851094827483 + 0.965281519117545 I, -0.854036458405025 + 0.273331145162114 I, -0.854036458405025 - 0.273331145162114 I, 0.112851094827483 - 0.965281519117545 I] whose absolute values are [0.768085398012510, 0.971855843607290, 0.896709644868745, 0.896709644868745, 0.971855843607290] so the largest absolute value is, 0.971855843607290 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.971855843607290 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.3157894737, 0.4545454545, 0.4912280702, 0., -0.2857142857, 0.3472222222, 0.2449392713, 0.1428571429, -0.3877410468, 0.1020490157, 0.2449027420, 0.1428571429, -0.2448963317, -0.1428571429, 0.2448985130, 0.1428571429, -0.1020406279, -0.2448978493, 0.1428571429, 0.1836735068, -0.02040796689, -6 -0.2244894777, 0.3638087428 10 , 0.2040822563, 0.04081754933, -0.1632631780, -0.1020372950, 0.1632712643, 0.1020511557, -0.1020230130, -0.1428267016, 0.08168472803, 0.1429464771, -0.04066307203, -0.1425944522, 0.0004502168399, 0.1436289433, 0.02173136012, -0.1201806259, -0.05733597690, 0.1087067928, 0.07265195915, -0.08245081730, -0.08886619517, 0.05757040619, 0.09869214706, -0.03489505682, -0.09772106408, 0.007405102586, 0.09724238074, 0.01218202032, -0.08698881008, -0.03250663035, 0.07546800440, 0.04773829179, -0.06102489463, -0.05797525351, 0.04345666633] The largest is 0.4912280702 The smallest is -0.3877410468 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 169, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 34, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 34, 61, 109, 195, 349, 625, 1119, 2003, 3585, 6416, 11483, 20552, 36783, 65832, 117822, 210870, 377401, 675447, 1208870] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 170, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 35, 64, 117, 214, 391, 714, 1304, 2382, 4351, 7948, 14519, 26523, 48452, 88512, 161694, 295383, 539606, 985753, 1800775] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 262, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 19, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [19, 36, 68, 128, 241, 454, 855, 1610, 3032, 5710, 10753, 20250, 38135, 71816, 135244, 254692, 479637, 903254, 1701011, 3203350] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + 2 a(n - 3) - a(n - 4) with initial conditions, a(1) = 19, a(2) = 36, a(3) = 68, a(4) = 128, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + 2 b(n - 3) - b(n - 4) with initial conditions, b(1) = 19, b(2) = 36, b(3) = 68, b(4) = 128, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t + t - 2 t + 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 2 %1 := _Z - 2 _Z + _Z - 2 _Z + 1 In floating-point [0.531010056459569, 1.88320350591353, -0.207106781186548 + 0.978318343478516 I, -0.207106781186548 - 0.978318343478516 I] The largest root is, 1.88320350591353 and the remaining roots are [0.531010056459569, -0.207106781186548 + 0.978318343478516 I, -0.207106781186548 - 0.978318343478516 I] whose absolute values are [0.531010056459569, 1., 1.] so the largest absolute value is, 1. that equals 1. It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 1. for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) is bounded. we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2105263158, 0.4444444444, -0.05882352941, -0.2421875000, 0.2531120332, 0.1872246696, -0.3040935673, -0.04720496894, 0.3311345646, -0.08598949212, -0.2934064912, 0.2086419753, 0.2075783401, -0.2943076752, -0.08550471740, 0.3298140499, -0.05106153195, -0.3086385446, 0.1789171263, 0.2345357204, -0.2760612410, -0.1201854173, 0.3258447323, -0.01478332818, -0.3197209780, 0.1472162520, 0.2587420941, -0.2543906918, -0.1533699957, 0.3179186367, 0.02168379134, -0.3269003536, 0.1137227706, 0.2797948407, -0.2296175877, -0.1846841214, 0.3061162557, 0.05788661671, -0.3300936774, 0.07884266134, 0.2974359778, -0.2020446773, -0.2137463322, 0.2905813070, 0.09338361392, -0.3292620664, 0.04300119955, 0.3114503863, -0.1720081736, -0.2402022680, 0.2715032107, 0.1277419559, -0.3244156613, 0.006635410860, 0.3216671841, -0.1398743211, -0.2637293433, 0.2491145919] The largest is 0.4444444444 The smallest is -0.3300936774 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 171, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 37, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 37, 72, 140, 272, 528, 1025, 1990, 3864, 7503, 14569, 28289, 54929, 106656, 207095, 402118, 780796, 1516078, 2943781, 5715964] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Fact, 172, : Consider the Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 39, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) [BTW, the first, 20, terms are:, 19, 39, 80, 164, 336, 688, 1409, 2886, 5911, 12107, 24798, 50792, 104034, 213086, 436450, 893952, 1831023, 3750364, 7681624, 15733765, ]. At first sight it seems to satisfy the following linear recurrence: a(n) = 2 a(n - 1) + a(n - 3) - 2 a(n - 4) + a(n - 5) - a(n - 6) + a(n - 7), . Alas, it breaks down at the, 29, -th term. a(29), equals , 9982727584, while the corresponding term for the solution of the recurrence is , 9982727585 So the difference of the former from the latter is, -1 Note that the Pisot Indicator is, 1.03507181408160 Since it is larger than 1, it is not at all suprising, that it does not go o\ n for ever. ------------------------------------------------------------------------- Fact, 173, :The Pisot Sequence a(n), defined by, a(1) = 19, a(2) = 40, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [19, 40, 84, 176, 369, 774, 1624, 3407, 7148, 14997, 31465, 66016, 138507, 290599, 609700, 1279199, 2683861, 5630953, 11814185, 24787095] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 174, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 22, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58 ] is a trivial linear sequence Fact, 175, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 23, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77 ] is a trivial linear sequence Fact, 176, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 24, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 24, 29, 35, 42, 50, 60, 72, 86, 103, 123, 147, 176, 211, 253, 303, 363, 435, 521, 624] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 177, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 25, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 25, 31, 38, 47, 58, 72, 89, 110, 136, 168, 208, 258, 320, 397, 493, 612, 760, 944, 1173] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 178, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 26, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 26, 34, 44, 57, 74, 96, 125, 163, 213, 278, 363, 474, 619, 808, 1055, 1378, 1800, 2351, 3071] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 263, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 27, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 27, 36, 48, 64, 85, 113, 150, 199, 264, 350, 464, 615, 815, 1080, 1431, 1896, 2512, 3328, 4409] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) - a(n - 4) with initial conditions, a(1) = 20, a(2) = 27, a(3) = 36, a(4) = 48, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) - b(n - 4) with initial conditions, b(1) = 20, b(2) = 27, b(3) = 36, b(4) = 48, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (108 + 12 69 ) 2 (108 + 12 69 ) [1, ------------------- + -------------------, - ------------------- 6 1/2 1/3 12 (108 + 12 69 ) 1 - --------------------- 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | + 1/2 I 3 |------------------- - -------------------|, | 6 1/2 1/3| \ (108 + 12 69 ) / 1/2 1/3 (108 + 12 69 ) 1 - ------------------- - --------------------- 12 1/2 (1/3) (108 + 12 69 ) / 1/2 1/3 \ 1/2 |(108 + 12 69 ) 2 | - 1/2 I 3 |------------------- - -------------------|] | 6 1/2 1/3| \ (108 + 12 69 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.324717958, -0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] The largest root is, 1.324717958 and the remaining roots are [-0.6623589786 - 0.5622795125 I, -0.6623589786 + 0.5622795125 I] whose absolute values are [0.8688369621, 0.8688369621] so the largest absolute value is, 0.8688369621 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8688369621 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4500000000, 0., 0., 0.3333333333, -0.1093750000, 0.2235294118, 0.1150442478, 0.006666666667, 0.2311557789, 0.01515151515, 0.1314285714, 0.1400862069, 0.04065040650, 0.1656441718, 0.07500000000, 0.1006289308, 0.1350210970, 0.07006369427, 0.1301081731, 0.09956906328, 0.09467556925, 0.1241922978, 0.08877182714, 0.1134020619, 0.1075041690, 0.09671869755, 0.1154540559, 0.09877340219, 0.1067251462, 0.1087812521, 0.1000534825, 0.1100621118, 0.1033910821, 0.1046724093, 0.1080103532, 0.1026209240, 0.1072403917, 0.1051890587, 0.1044192125, 0.1069874322, 0.1041663191, 0.1059647414, 0.1057118851, 0.1046892225, 0.1062348095, 0.1049593066, 0.1054822431, 0.1057523363, 0.1049997768, 0.1057928117, 0.1053103493, 0.1053508277, 0.1056614024, 0.1052194201, 0.1055704744, 0.1054390677, 0.1053481404, 0.1055677887] The largest is 0.4500000000 The smallest is -0.1093750000 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 264, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 28, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 28, 39, 54, 75, 104, 144, 199, 275, 380, 525, 725, 1001, 1382, 1908, 2634, 3636, 5019, 6928, 9563] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 4) - a(n - 5) with initial conditions, a(1) = 20, a(2) = 28, a(3) = 39, a(4) = 54, a(5) = 75, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 4) - b(n - 5) with initial conditions, b(1) = 20, b(2) = 28, b(3) = 39, b(4) = 54, b(5) = 75, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 t - 2 t + t - t + 1 = 0 whose roots are [1, RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4)] 4 3 %1 := _Z - _Z - 1 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.38027756909761, 0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] The largest root is, 1.38027756909761 and the remaining roots are [0.219447472149275 + 0.914473662967727 I, -0.819172513396165, 0.219447472149275 - 0.914473662967727 I] whose absolute values are [0.940435682699417, 0.819172513396165, 0.940435682699417] so the largest absolute value is, 0.940435682699417 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.940435682699417 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.2000000000, 0.3214285714, -0.2307692308, 0.1666666667, 0.2133333333, 0.3846153846, 0.006944444444, 0.02512562814, 0.09090909091, 0.3289473684, 0.1904761905, 0.07034482759, 0.01598401598, 0.1997105644, 0.2452830189, 0.1708428246, 0.04207920792, 0.09703128113, 0.1976039261, 0.2237791488, 0.1212121212, 0.07360043908, 0.1265656686, 0.2057154380, 0.1823057979, 0.1112875180, 0.09323547133, 0.1543345582, 0.1920259439, 0.1587002708, 0.1073230799, 0.1170452359, 0.1644591454, 0.1785477366, 0.1412593472, 0.1136932100, 0.1335410531, 0.1674775678, 0.1641257582, 0.1332078482, 0.1221378011, 0.1450042857, 0.1645189771, 0.1531157704, 0.1306425234, 0.1310357653, 0.1509437023, 0.1594484359, 0.1454799247, 0.1319046566, 0.1382373265, 0.1530747308, 0.1539436245, 0.1412372505, 0.1348635467, 0.1433272474, 0.1526598419, 0.1492860626] The largest is 0.3846153846 The smallest is -0.2307692308 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 265, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 29, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 29, 42, 61, 89, 130, 190, 278, 407, 596, 873, 1279, 1874, 2746, 4024, 5897, 8642, 12665, 18561, 27202] ] The sequence a(n) satisfies, for n>=, 5, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) - a(n - 4) with initial conditions, a(1) = 20, a(2) = 29, a(3) = 42, a(4) = 61, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) - b(n - 4) with initial conditions, b(1) = 20, b(2) = 29, b(3) = 42, b(4) = 61, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 4, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 4 3 2 t - 2 t + t - t + 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (116 + 12 93 ) 2 (116 + 12 93 ) [1, ------------------- + --------------------- + 1/3, - ------------------- 6 1/2 1/3 12 3 (116 + 12 93 ) 1 - ----------------------- + 1/3 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | + 1/2 I 3 |------------------- - ---------------------|, | 6 1/2 1/3| \ 3 (116 + 12 93 ) / 1/2 1/3 (116 + 12 93 ) 1 - ------------------- - ----------------------- + 1/3 12 1/2 (1/3) 3 (116 + 12 93 ) / 1/2 1/3 \ 1/2 |(116 + 12 93 ) 2 | - 1/2 I 3 |------------------- - ---------------------|] | 6 1/2 1/3| \ 3 (116 + 12 93 ) / Since 1 is a root, let's remove it, and the remaining roots are In floating-point [1.465571232, -0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] The largest root is, 1.465571232 and the remaining roots are [-0.2327856159 - 0.7925519930 I, -0.2327856159 + 0.7925519930 I] whose absolute values are [0.8260313581, 0.8260313581] so the largest absolute value is, 0.8260313581 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.8260313581 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.05000000000, -0.1724137931, -0.4047619048, -0.1475409836, -0.1123595506, -0.3076923077, -0.2421052632, -0.1402877698, -0.2334152334, -0.2600671141, -0.1844215349, -0.2017200938, -0.2454642476, -0.2134013110, -0.1985586481, -0.2274037646, -0.2241379310, -0.2060007896, -0.2166909110, -0.2241011690, -0.2133647720, -0.2133125663, -0.2206663786, -0.2172807828, -0.2138410261, -0.2177537536, -0.2182799455, -0.2153657483, -0.2163638549, -0.2178878591, -0.2164974644, -0.2161050407, -0.2172365282, -0.2169775571, -0.2163261190, -0.2168061388, -0.2170271673, -0.2165967439, -0.2166463309, -0.2169169401, -0.2167571215, -0.2166468869, -0.2168072594, -0.2168078119, -0.2166981289, -0.2167488177, -0.2168000586, -0.2167416162, -0.2167338624, -0.2167773493, -0.2167623938, -0.2167396844, -0.2167604618, -0.2167662837, -0.2167493962, -0.2167532861, -0.2167629979, -0.2167558222] The largest is 0.05000000000 The smallest is -0.4047619048 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 179, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 30, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 30, 45, 68, 103, 156, 236, 357, 540, 817, 1236, 1870, 2829, 4280, 6475, 9796, 14820, 22421, 33920, 51316] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 266, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 31, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 31, 48, 74, 114, 176, 272, 420, 649, 1003, 1550, 2395, 3701, 5719, 8837, 13655, 21100, 32604, 50380, 77848] ] The sequence a(n) satisfies, for n>=, 8, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) - a(n - 5) + a(n - 7) with initial conditions, a(1) = 20, a(2) = 31, a(3) = 48, a(4) = 74, a(5) = 114, a(6) = 176, a(7) = 272, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) - b(n - 5) + b(n - 7) with initial conditions, b(1) = 20, b(2) = 31, b(3) = 48, b(4) = 74, b(5) = 114, b(6) = 176, b(7) = 272, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 7, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 7 6 5 2 t - t - t + t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6), RootOf(%1, index = 7)] 7 6 5 2 %1 := _Z - _Z - _Z + _Z - 1 In floating-point [1.54521564973276, 0.808712727401677 + 0.424846366714998 I, -0.212614366617129 + 0.953977000479370 I, -0.868706185650926 + 0.239062189977012 I, -0.868706185650926 - 0.239062189977012 I, -0.212614366617129 - 0.953977000479370 I, 0.808712727401677 - 0.424846366714998 I] The largest root is, 1.54521564973276 and the remaining roots are [0.808712727401677 + 0.424846366714998 I, -0.212614366617129 + 0.953977000479370 I, -0.868706185650926 + 0.239062189977012 I, -0.868706185650926 - 0.239062189977012 I, -0.212614366617129 - 0.953977000479370 I, 0.808712727401677 - 0.424846366714998 I] whose absolute values are [0.913515577739315, 0.977382722548142, 0.901000093043717, 0.901000093043717, 0.977382722548142, 0.913515577739315] so the largest absolute value is, 0.977382722548142 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.977382722548142 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.05000000000, 0.3225806452, 0.08333333333, -0.3783783784, -0.2807017544, 0.3636363636, -0.4705882353, -0.1404761905, 0.09090909091, 0.3140578265, -0.3387096774, 0.1653444676, 0.3307214266, -0.06574575975, -0.1895439629, 0.1742951300, 0.1334597156, -0.3616734143, 0.002858277094, 0.1614428116, -0.07574069764, -0.2373020869, 0.2229266764, 0.1162258318, -0.1839651323, 0.01085969776, 0.2256394347, -0.06216821224, -0.1900567146, 0.1546667973, 0.06997621712, -0.1849615564, -0.04195743144, 0.1887771568, -0.07001528585, -0.1412710621, 0.1283420053, 0.09900459172, -0.1463921168, -0.01932967065, 0.1643264312, -0.05336053061, -0.1293097532, 0.09206383831, 0.08108834752, -0.1375663622, -0.02244715476, 0.1336226674, -0.03424885629, -0.1110242896, 0.08435705468, 0.07686826738, -0.1099637076, -0.02129373867, 0.1133895107, -0.02651013890, -0.1010131851, 0.06679743823] The largest is 0.3636363636 The smallest is -0.4705882353 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 180, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 32, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 32, 51, 81, 129, 205, 326, 518, 823, 1308, 2079, 3304, 5251, 8345, 13262, 21076, 33494, 53229, 84592, 134434] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 267, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 33, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, 4180, 6764, 10945, 17710, 28656, 46367, 75024, 121392, 196417] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 3) with initial conditions, a(1) = 20, a(2) = 33, a(3) = 54, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 3) with initial conditions, b(1) = 20, b(2) = 33, b(3) = 54, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - 2 t + 1 = 0 whose roots are 1/2 1/2 5 5 [1, 1/2 - ----, ---- + 1/2] 2 2 Since 1 is a root, let's remove it, and the remaining roots are In floating-point [-0.6180339880, 1.618033988] The largest root is, 1.618033988 and the remaining roots are [-0.6180339880] whose absolute values are [0.6180339880] so the largest absolute value is, 0.6180339880 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.6180339880 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4500000000, 0.3636363636, 0.4074074074, 0.3750000000, 0.3916083916, 0.3793103448, 0.3856382979, 0.3809523810, 0.3833671400, 0.3815789474, 0.3825009679, 0.3818181818, 0.3821703134, 0.3819095477, 0.3820440429, 0.3819444444, 0.3819958160, 0.3819577735, 0.3819773955, 0.3819628647, 0.3819703596, 0.3819648094, 0.3819676722, 0.3819655522, 0.3819666457, 0.3819658359, 0.3819662536, 0.3819659443, 0.3819661038, 0.3819659857, 0.3819660466, 0.3819660015, 0.3819660248, 0.3819660075, 0.3819660164, 0.3819660098, 0.3819660132, 0.3819660107, 0.3819660120, 0.3819660110, 0.3819660115, 0.3819660112, 0.3819660114, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660113, 0.3819660112, 0.3819660112, 0.3819660112, 0.3819660112, 0.3819660112] The largest is 0.4500000000 The smallest is 0.3636363636 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 268, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 34, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 34, 58, 99, 169, 288, 491, 837, 1427, 2433, 4148, 7072, 12057, 20556, 35046, 59750, 101868, 173675, 296099, 504820] ] The sequence a(n) satisfies, for n>=, 6, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 5) with initial conditions, a(1) = 20, a(2) = 34, a(3) = 58, a(4) = 99, a(5) = 169, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 5) with initial conditions, b(1) = 20, b(2) = 34, b(3) = 58, b(4) = 99, b(5) = 169, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 5, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 5 4 3 t - t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5)] 5 4 3 %1 := _Z - _Z - _Z - 1 In floating-point [1.70490277604165, 0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] The largest root is, 1.70490277604165 and the remaining roots are [0.428538420565535 + 0.710200650526218 I, -0.780989808586358 + 0.492495718647332 I, -0.780989808586358 - 0.492495718647332 I, 0.428538420565535 - 0.710200650526218 I] whose absolute values are [0.829475823582982, 0.923307702773949, 0.923307702773949, 0.829475823582982] so the largest absolute value is, 0.923307702773949 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.923307702773949 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, -0.05882352941, -0.01724137931, 0.4949494949, -0.2071005917, 0.08680555556, -0.1792260692, -0.1099163680, 0.2053258584, -0.1117961365, 0.1803278689, -0.1107183258, -0.04030853446, 0.05429071804, -0.09781430120, 0.1368033473, -0.07173008207, 0.02476464661, 0.007325252703, -0.06572441662, 0.07840412516, -0.05905038787, 0.04411838185, -0.007606753983, -0.02921278968, 0.04158458025, -0.04667859766, 0.03902436425, -0.01526098744, -0.005449412879, 0.02087417992, -0.03125383064, 0.02864471352, -0.01787010456, 0.005325196085, 0.008329271445, -0.01759936311, 0.01937462186, -0.01609484581, 0.008604972128, 0.0008393977603, -0.008154993222, 0.01205902639, -0.01219081264, 0.008473185880, -0.002878229001, -0.002560036343, 0.006620761049, -0.008130087936, 0.006963858993, -0.004044457944, 0.0003593647062, 0.002935667812, -0.004835055418, 0.005064471387, -0.003815041975, 0.001608794118, 0.0007294199550] The largest is 0.4949494949 The smallest is -0.2071005917 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 181, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 35, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 35, 61, 106, 184, 319, 553, 959, 1663, 2884, 5001, 8672, 15038, 26077, 45219, 78412, 135970, 235778, 408849, 708961] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 ------------------------------------------------------------------------- Theorem , 269, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 36, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 36, 65, 117, 211, 381, 688, 1242, 2242, 4047, 7305, 13186, 23802, 42965, 77556, 139996, 252706, 456158, 823408, 1486329] ] The sequence a(n) satisfies, for n>=, 7, the linear recurrence equation with constant coefficient a(n) = 2 a(n - 1) - a(n - 2) + a(n - 3) + a(n - 6) with initial conditions, a(1) = 20, a(2) = 36, a(3) = 65, a(4) = 117, a(5) = 211, a(6) = 381, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = 2 b(n - 1) - b(n - 2) + b(n - 3) + b(n - 6) with initial conditions, b(1) = 20, b(2) = 36, b(3) = 65, b(4) = 117, b(5) = 211, b(6) = 381, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 6, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 6 5 4 3 t - 2 t + t - t - 1 = 0 whose roots are [RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3), RootOf(%1, index = 4), RootOf(%1, index = 5), RootOf(%1, index = 6)] 6 5 4 3 %1 := _Z - 2 _Z + _Z - _Z - 1 In floating-point [1.80509425159131, 0.647170796637930 + 0.726361507447275 I, -0.202712494937727 + 0.895728484592812 I, -0.694010854991715, -0.202712494937727 - 0.895728484592812 I, 0.647170796637930 - 0.726361507447275 I] The largest root is, 1.80509425159131 and the remaining roots are [0.647170796637930 + 0.726361507447275 I, -0.202712494937727 + 0.895728484592812 I, -0.694010854991715, -0.202712494937727 - 0.895728484592812 I, 0.647170796637930 - 0.726361507447275 I] whose absolute values are [0.972846894183278, 0.918380026848806, 0.694010854991715, 0.918380026848806, 0.972846894183278] so the largest absolute value is, 0.972846894183278 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.972846894183278 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [-0.2000000000, 0.3611111111, -0.4000000000, -0.4786324786, -0.03317535545, 0.3727034121, 0.09883720930, 0.1529790660, 0.1779661017, -0.1771682728, -0.4125941136, -0.09752768087, 0.1391899840, 0.1162806936, 0.1738098922, 0.1933483814, -0.08343292205, -0.2839323217, -0.1518943221, 0.01299106725, 0.06775379767, 0.1639705778, 0.1897454118, -0.0006583536292, -0.1789858690, -0.1545769096, -0.06307251040, 0.01341659383, 0.1250741996, 0.1730009409, 0.05535840629, -0.09178683851, -0.1290036528, -0.09744546703, -0.03259992018, 0.07624291475, 0.1429986889, 0.08536770444, -0.02502401813, -0.08986251879, -0.1019332352, -0.06278505496, 0.02949929541, 0.1052181150, 0.09312786159, 0.02067438476, -0.04849421221, -0.08732000255, -0.07597211272, -0.007900320041, 0.06597933168, 0.08456125544, 0.04674864695, -0.01240463241, -0.06296876904, -0.07468457877, -0.03282568923, 0.03062568671] The largest is 0.3727034121 The smallest is -0.4786324786 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. ------------------------------------------------------------------------- Theorem , 270, : Let a(n) be the Pisot Sequence with parameter, 1/2, and with initial conditions, a(1) = 20, a(2) = 37, i.e. for n>1 2 a(n + 1) a(n + 2) = trunc(1/2 + ---------) a(n) [Here are the first, 20, terms [20, 37, 68, 125, 230, 423, 778, 1431, 2632, 4841, 8904, 16377, 30122, 55403, 101902, 187427, 344732, 634061, 1166220, 2145013] ] The sequence a(n) satisfies, for n>=, 4, the linear recurrence equation with constant coefficient a(n) = a(n - 1) + a(n - 2) + a(n - 3) with initial conditions, a(1) = 20, a(2) = 37, a(3) = 68, . Proof: Let b(n) be the solution of the above recurrence, i.e. the sequence d\ efined by b(n) = b(n - 1) + b(n - 2) + b(n - 3) with initial conditions, b(1) = 20, b(2) = 37, b(3) = 68, . We have to prove that a(n)=b(n), but it is easier to prove the equivalent st\ atement that b(n)=a(n). It is readily checked that b(n)=a(n) for n from 1 to, 3, . It remains to prove that b(n) satisfies the Pisot-recurrence 2 b(n + 1) b(n + 2) = trunc(1/2 + ---------), . b(n) But this is equivalent to the INEQALITIES: 2 2 b(n + 1) b(n + 1) -1/2 <= --------- - b(n + 2), --------- - b(n + 2) <= 1/2 b(n) b(n) In other words 2 2 b(n + 1) - b(n + 2) b(n) b(n + 1) - b(n + 2) b(n) -1/2 <= -------------------------, ------------------------- <= 1/2 b(n) b(n) The characteristic equation in t, of the recurrence satisfied by b(n) is 3 2 t - t - t - 1 = 0 whose roots are 1/2 1/3 1/2 1/3 (19 + 3 33 ) 4 (19 + 3 33 ) [----------------- + ------------------- + 1/3, - ----------------- 3 1/2 1/3 6 3 (19 + 3 33 ) 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 1/3 1/2 |(19 + 3 33 ) 4 | (19 + 3 33 ) + 1/2 I 3 |----------------- - -------------------|, - ----------------- | 3 1/2 1/3| 6 \ 3 (19 + 3 33 ) / 2 - ------------------- + 1/3 1/2 1/3 3 (19 + 3 33 ) / 1/2 1/3 \ 1/2 |(19 + 3 33 ) 4 | - 1/2 I 3 |----------------- - -------------------|] | 3 1/2 1/3| \ 3 (19 + 3 33 ) / In floating-point [1.839286755, -0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] The largest root is, 1.839286755 and the remaining roots are [-0.4196433777 + 0.6062907300 I, -0.4196433777 - 0.6062907300 I] whose absolute values are [0.7373527065, 0.7373527065] so the largest absolute value is, 0.7373527065 that is less than 1, It follows that the sequence 2 b(n + 1) - b(n + 2) b(n) c(n) = ------------------------- b(n) satisfies the inequality n c(n) <= A 0.7373527065 for some fixed constant, A, (independent of n), that can be easily determine\ d, if desired. In particular the sequence c(n) tends to 0 we have to show that for all n, c(n) is between , -1/2, and , 1/2 For the first, 58, terms the sequence c(n) is [0.4500000000, -0.02702702703, -0.2205882353, 0.2000000000, -0.04782608696, -0.06855791962, 0.08354755784, -0.03284416492, -0.01785714286, 0.03284445363, -0.01785714286, -0.002869878488, 0.01211738928, -0.008609642077, 0.0006378677553, 0.004145614026, -0.003826160612, 0.0009573211410, 0.001276774537, -0.001592064943, 0.0006420307333, 0.0003267403263, -0.0006232938838, 0.0003454771758, 0.00004892361838, -0.0002288930896, 0.0001655077047, -0.00001446176653, -0.00007784715143, 0.00007319878669, -0.00001911013127, -0.00002375849601, 0.00003033015942, -5 -5 -0.00001253846785, -0.5966804434 10 , 0.00001182488714, -0.6680385146 10 , -6 -5 -5 -0.8223024411 10 , 0.4322199552 10 , -0.3180488035 10 , -6 -5 -5 -6 0.3194090758 10 , 0.1461120593 10 , -0.1399958366 10 , 0.3805713022 10 , -6 -6 -6 -6 0.4417335286 10 , -0.5776535355 10 , 0.2446512954 10 , 0.1087312885 10 , -6 -6 -7 -0.2242709516 10 , 0.1291116323 10 , 0.1357196928 10 , -7 -7 -8 -0.8158734997 10 , 0.6109625164 10 , -0.6919129045 10 , -7 -7 -8 -0.2741022737 10 , 0.2676689523 10 , -0.7562461186 10 , -8 -0.8205793328 10 ] The largest is 0.4500000000 The smallest is -0.2205882353 and as n goes to infinity, c(n) tends to zero (and it is possible (but a was\ te of time) to find an N0 such that |c(n)| is guaranteed to be less than , 1/2, for n>=N0 (or any epsilon for that matter), and check that \ for for n<=N0. c(n)>= , -1/2, and c(n) < , 1/2 QED. Fact, 182, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 38, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 38, 72, 136, 257, 486, 919, 1738, 3287, 6217, 11759, 22241, 42067, 79566, 150492, 284642, 538375, 1018288, 1926000, 3642855] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 Fact, 183, :The Pisot Sequence a(n), defined by, a(1) = 20, a(2) = 39, 2 a(n + 1) and for n>1, a(n + 2) = trunc(1/2 + ---------) a(n) whose , 20, first terms [20, 39, 76, 148, 288, 560, 1089, 2118, 4119, 8010, 15577, 30293, 58912, 114569, 222808, 433306, 842672, 1638787, 3187032, 6197982] does not satisfy a linear recurrence equation with constant coefficients of order<=, 6 To sum up, out of the, 453, cases, the following, 270, had (rigorously!) provable recurrences that are listed above Here they are {[2, 5], [2, 7], [2, 9], [2, 11], [2, 13], [2, 15], [2, 17], [2, 19], [2, 21], [2, 23], [2, 25], [2, 27], [2, 29], [2, 31], [2, 33], [2, 35], [2, 37], [2, 39] , [3, 5], [3, 7], [3, 8], [3, 10], [3, 11], [3, 13], [3, 14], [3, 16], [3, 17], [3, 19], [3, 20], [3, 22], [3, 23], [3, 25], [3, 26], [3, 28], [3, 29], [3, 31] , [3, 32], [3, 34], [3, 35], [3, 37], [3, 38], [3, 40], [4, 6], [4, 7], [4, 9], [4, 10], [4, 11], [4, 14], [4, 15], [4, 17], [4, 18], [4, 19], [4, 21], [4, 23] , [4, 25], [4, 26], [4, 27], [4, 30], [4, 31], [4, 33], [4, 34], [4, 37], [4, 39], [5, 7], [5, 8], [5, 9], [5, 11], [5, 12], [5, 13], [5, 14], [5, 16], [5, 17], [5, 18], [5, 19], [5, 22], [5, 23], [5, 24], [5, 26], [5, 27], [5, 28], [5 , 31], [5, 33], [5, 34], [5, 36], [5, 37], [5, 38], [5, 39], [6, 8], [6, 9], [6 , 10], [6, 11], [6, 13], [6, 14], [6, 15], [6, 17], [6, 19], [6, 20], [6, 21], [6, 22], [6, 23], [6, 25], [6, 28], [6, 29], [6, 32], [6, 33], [6, 34], [6, 35] , [6, 37], [6, 38], [6, 39], [6, 40], [7, 9], [7, 10], [7, 12], [7, 13], [7, 16 ], [7, 17], [7, 18], [7, 19], [7, 20], [7, 22], [7, 23], [7, 24], [7, 25], [7, 26], [7, 27], [7, 29], [7, 30], [7, 31], [7, 32], [7, 33], [7, 34], [7, 36], [7 , 37], [7, 40], [8, 11], [8, 13], [8, 15], [8, 17], [8, 18], [8, 20], [8, 21], [8, 22], [8, 23], [8, 25], [8, 26], [8, 27], [8, 28], [8, 31], [8, 33], [8, 36] , [8, 37], [8, 38], [8, 39], [9, 12], [9, 13], [9, 14], [9, 16], [9, 20], [9, 21], [9, 25], [9, 26], [9, 28], [9, 29], [9, 31], [9, 32], [9, 38], [9, 39], [9 , 40], [10, 14], [10, 16], [10, 17], [10, 19], [10, 24], [10, 25], [10, 26], [ 10, 29], [10, 32], [10, 33], [10, 34], [10, 35], [10, 38], [10, 39], [11, 15], [11, 18], [11, 19], [11, 20], [11, 23], [11, 24], [11, 25], [11, 29], [11, 30], [11, 36], [11, 39], [11, 40], [12, 16], [12, 20], [12, 21], [12, 23], [12, 28], [12, 29], [12, 30], [12, 32], [12, 34], [12, 37], [12, 38], [13, 19], [13, 20], [13, 21], [13, 22], [13, 23], [13, 24], [13, 28], [13, 29], [13, 32], [13, 33], [13, 34], [13, 35], [13, 37], [13, 38], [14, 19], [14, 20], [14, 22], [14, 29], [14, 31], [14, 33], [14, 35], [14, 38], [14, 39], [14, 40], [15, 20], [15, 21], [15, 22], [15, 26], [15, 27], [15, 28], [15, 29], [15, 31], [15, 33], [15, 38], [16, 21], [16, 25], [16, 26], [16, 28], [16, 29], [16, 31], [16, 33], [16, 35], [16, 36], [16, 37], [17, 29], [17, 30], [17, 31], [17, 36], [17, 40], [18, 28], [18, 29], [18, 39], [19, 26], [19, 28], [19, 33], [19, 36], [20, 27], [20, 28], [20, 29], [20, 31], [20, 33], [20, 34], [20, 36], [20, 37]} while the following , 14, give linear trivial (linear) sequences. {[9, 11], [10, 12], [11, 13], [12, 14], [13, 15], [14, 16], [15, 17], [16, 18], [17, 19], [18, 20], [19, 21], [19, 22], [20, 22], [20, 23]} while the following , 24, satisfy recurrences of order <=, 6 that eventually break down. Here they are where each line lists the recurrences, the first time it breaks down, and the Piso\ t index that is always larger than 1, explaining the break-down. PISOT(4, 13) <> [[4, 13, 42, 136, 440, 1424], [3, 1, -1, 1, -1, 1]], 24, 1.07253346951707 PISOT(4, 29) <> [[4, 29, 210, 1521, 11016, 79785], [7, 2, -2, 2, -2, 2]], 22, 1.07574660319045 PISOT(5, 29) <> [[5, 29, 168, 973, 5635, 32634, 188993], [6, -1, -1, -1, -1, -1, -1]], 33, 1.05843714354031 PISOT(6, 16) <> [[6, 16, 43, 116, 313], [3, -1, 0, 1, 1]], 46, 1.00714432316835 PISOT(8, 10) <> [[8, 10, 13, 17, 22, 28], [1, 0, 0, 0, 0, 1]], 38, 1.03282504251949 PISOT(10, 13) <> [[10, 13, 17, 22, 28, 36], [1, 0, 0, 0, 0, 1]], 37, 1.03282504251949 PISOT(10, 22) <> [[10, 22, 48, 105], [2, 0, 0, 2]], 22, 1.01573535617675 PISOT(11, 14) <> [[11, 14, 18, 23, 29, 37, 47], [2, -1, 0, 0, 0, 1, -1]], 29, 1.03282504251949 PISOT(11, 26) <> [[11, 26, 61, 143, 335, 785], [3, -2, 1, 0, 0, 1]], 77, 1.01165724459672 PISOT(11, 28) <> [[11, 28, 71, 180, 456, 1155], [3, -2, 3, -3, 2, -1]], 55, 1.01224792866759 PISOT(12, 19) <> [[12, 19, 30, 47, 74, 117, 185], [2, -1, 0, 1, 0, -1, 1]], 54, 1.02968394081287 PISOT(13, 17) <> [[13, 17, 22, 28, 36, 46], [1, 0, 0, 0, 0, 1]], 36, 1.03282504251949 PISOT(13, 25) <> [[13, 25, 48, 92, 176, 337, 645], [2, 0, 0, -1, 1, -1, 1]], 57, 1.01075655064021 PISOT(14, 18) <> [[14, 18, 23, 29, 37, 47, 60], [2, -1, 0, 0, 0, 1, -1]], 28, 1.03282504251949 PISOT(15, 36) <> [[15, 36, 86, 205, 489, 1166, 2780], [2, 1, 0, -1, 1, 1, -1]], 39, 1.02168368889029 PISOT(16, 39) <> [[16, 39, 95, 231, 562, 1367, 3325], [2, 1, 0, 0, 0, 1, 2]], 31, 1.03999622704587 PISOT(17, 22) <> [[17, 22, 28, 36, 46, 59], [1, 0, 0, 0, 0, 1]], 35, 1.03282504251949 PISOT(17, 37) <> [[17, 37, 81, 177, 387, 846], [2, 0, 1, 0, -1, 1]], 44, 1.01326626733001 PISOT(17, 38) <> [[17, 38, 85, 190, 425], [2, 1, -1, -1, 2]], 48, 1.01884404554226 PISOT(18, 23) <> [[18, 23, 29, 37, 47, 60, 77], [2, -1, 0, 0, 0, 1, -1]], 27, 1.03282504251949 PISOT(18, 35) <> [[18, 35, 68, 132, 256], [2, 0, -1, 2, -1]], 25, 1.09047641696283 PISOT(19, 29) <> [[19, 29, 44, 67, 102, 155, 236], [2, -1, 0, 1, -1, 0, 1]], 23, 1.03130091592356 PISOT(19, 30) <> [[19, 30, 47, 74, 117, 185, 293], [2, -1, 0, 1, 0, -1, 1]], 53, 1.02968394081287 PISOT(19, 39) <> [[19, 39, 80, 164, 336, 688, 1409], [2, 0, 1, -2, 1, -1, 1]], 28, 1.03507181408160 Among the false recurrences the PISOT sequence , PISOT(11, 26, 1/2), lasted longest, up to n=, 77 Finally, the following , 145, pairs do definitely not produce (as initial v\ alues), Pisot sequences with parameter, 1/2 that have a recurrence of order <= , 6 {[4, 22], [4, 35], [4, 38], [5, 21], [5, 32], [6, 26], [6, 27], [6, 31], [7, 11 ], [7, 15], [7, 38], [7, 39], [8, 12], [8, 14], [8, 19], [8, 29], [8, 30], [8, 34], [8, 35], [9, 15], [9, 17], [9, 19], [9, 22], [9, 23], [9, 24], [9, 30], [9 , 33], [9, 34], [9, 35], [9, 37], [10, 15], [10, 18], [10, 21], [10, 23], [10, 27], [10, 28], [10, 31], [10, 36], [10, 37], [11, 16], [11, 17], [11, 21], [11, 27], [11, 31], [11, 32], [11, 34], [11, 35], [11, 37], [11, 38], [12, 15], [12, 17], [12, 18], [12, 22], [12, 25], [12, 26], [12, 27], [12, 31], [12, 33], [12, 35], [12, 39], [12, 40], [13, 16], [13, 18], [13, 27], [13, 30], [13, 31], [13, 36], [13, 40], [14, 17], [14, 21], [14, 23], [14, 24], [14, 25], [14, 26], [14, 27], [14, 30], [14, 32], [14, 34], [14, 36], [14, 37], [15, 18], [15, 19], [15, 23], [15, 24], [15, 25], [15, 32], [15, 34], [15, 35], [15, 37], [15, 39], [15, 40], [16, 19], [16, 20], [16, 22], [16, 23], [16, 24], [16, 27], [16, 30], [16, 34], [16, 38], [16, 40], [17, 20], [17, 21], [17, 23], [17, 24], [17, 25], [17, 26], [17, 27], [17, 28], [17, 32], [17, 33], [17, 35], [17, 39], [18, 21], [18, 22], [18, 24], [18, 25], [18, 26], [18, 27], [18, 30], [18, 31], [18, 32], [18, 33], [18, 34], [18, 37], [18, 38], [18, 40], [19, 23], [19, 24], [19, 25], [19, 27], [19, 31], [19, 32], [19, 34], [19, 35], [19, 37], [19, 40], [20, 24], [20, 25], [20, 26], [20, 30], [20, 32], [20, 35], [20, 38], [20, 39]} To sum-up out of the, 453, pairs of integers a<=, 20, a+2<=b<=, 40 270, satisy provable recurrences of order <=, 6 14, are trivial, linear, sequences 24, gets your hope up but then FAIL 145, do not satisy linear recurrences of order <=, 6 This concludes this article that contained, 270, theorems and, 183, facts, and took , 17.926, to generate I hope, dear readers, that you enjoyed reading it as much as I did generatin\ g it. tam v'nishlam shevach le-el bore olam .