Sketches of proofs of the Irrationality of , 35, Constants given as certain double integrals By Shalosh B. Ekhad ------------------------------------------------------------ Theorem number, 1, : The following constant c. 1 1 / / 1 | | 1 c = ---------------- | | ------------------------------ dx dy Beta(-2/3, -2/3) | | (5/3) (5/3) / / x (1 - x) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.105813989, yielding an irrationality measure that is approximately , 16.02729794 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (1 + n) (-2 + 3 n) (3 n + 2) (3 n + 1) X(n) - -------------------------------------------- 2 (3 n - 1) (-4 + 3 n) (3 n + 11) (2 + n) 4 3 2 3 (3 n + 1) (99 n + 132 n - 25 n + 14 n + 16) X(1 + n) + --------------------------------------------------------- + X(2 + n) = 0 2 (3 n - 1) (-4 + 3 n) (3 n + 11) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = 5/28 B(0) = 1, B(1) = 3/14 A(n) Then, ----, approximates B(n) 1 1 / / 1 | | 1 c = ---------------- | | ------------------------------ dx dy Beta(-2/3, -2/3) | | (5/3) (5/3) / / x (1 - x) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| 1 | | \ -x y + 1 / F(n) = ---------------- | | -------------------------- dx dy Beta(-2/3, -2/3) | | 5/3 5/3 | | x (1 - x) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.103750044, 2.105813989, 2.103205437, 2.101442576, 2.101832569, 2.100408640, 2.100432734, 2.097725196, 2.098353685, 2.096463108, 2.099617785, 2.096684452, 2.092654976, 2.092046948, 2.082155838, 2.088378762, 2.079630809, 2.082092497, 2.081958589, 2.088402232, 2.083343954] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.06703367475, 0.06654556288, 0.06716254587, 0.06757990772, 0.06748754776, 0.06782484663, 0.06781913749, 0.06846107785, 0.06831199825, 0.06876057485, 0.06801227565, 0.06870803716, 0.06966527367, 0.06980986484, 0.07216750750, 0.07068300219, 0.07277103875, 0.07218263893, 0.07221462941, 0.07067741110, 0.07188375945] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 2, : The following constant c. 1 1 / / 1 | | 1 c = --------------- | | ------------------------------ dx dy Beta(-2/3, 1/3) | | (2/3) (5/3) / / x (1 - x) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.107839045, yielding an irrationality measure that is approximately , 16.13613091 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 2 (1 + n) (-2 + 3 n) (1 + 3 n) (45 n + 72 n + 34) X(n) - ------------------------------------------------------ + 2 2 (45 n - 18 n + 7) (-1 + 3 n) (3 n + 8) (2 + n) 5 4 3 2 3 (1 + 3 n) (1485 n + 3366 n + 2094 n + 883 n + 566 n + 176) X(1 + n) ------------------------------------------------------------------------- 2 2 (45 n - 18 n + 7) (-1 + 3 n) (3 n + 8) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 7/20 B(0) = 1, B(1) = 3/10 A(n) Then, ----, approximates B(n) 1 1 / / 1 | | 1 c = --------------- | | ------------------------------ dx dy Beta(-2/3, 1/3) | | (2/3) (5/3) / / x (1 - x) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| 1 | | \ -x y + 1 / F(n) = --------------- | | -------------------------- dx dy Beta(-2/3, 1/3) | | 2/3 5/3 | | x (1 - x) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.098316243, 2.105724730, 2.100397536, 2.107839045, 2.093253358, 2.092890006, 2.097973109, 2.098212054, 2.101236116, 2.093988183, 2.099527409, 2.095988795, 2.093876629, 2.088748819, 2.084111732, 2.087266365, 2.085386753, 2.073243882, 2.086726325, 2.083202482, 2.077131980] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.06832087845, 0.06656666291, 0.06782747777, 0.06606708188, 0.06952301450, 0.06960939319, 0.06840226712, 0.06834558994, 0.06762880901, 0.06934836917, 0.06803369854, 0.06887317406, 0.06937487846, 0.07059485298, 0.07170047717, 0.07094806746, 0.07139624538, 0.07430067635, 0.07107679713, 0.07191753818, 0.07336897698] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 3, : The following constant c. 1 1 / / 1 | | 1 c = ---------------- | | ------------------------------ dx dy Beta(-1/3, -1/3) | | (4/3) (4/3) / / x (1 - x) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.105946388, yielding an irrationality measure that is approximately , 16.03436861 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (1 + n) (-1 + 3 n) (15 n + 14) (3 n + 2) X(n) - ---------------------------------------------- 2 (15 n - 1) (-2 + 3 n) (3 n + 10) (2 + n) 4 3 2 3 (3 n + 2) (495 n + 1122 n + 622 n + 125 n + 50) X(1 + n) + ------------------------------------------------------------- + X(2 + n) 2 (15 n - 1) (-2 + 3 n) (3 n + 10) (2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1/35 B(0) = 1, B(1) = 3/35 A(n) Then, ----, approximates B(n) 1 1 / / 1 | | 1 c = ---------------- | | ------------------------------ dx dy Beta(-1/3, -1/3) | | (4/3) (4/3) / / x (1 - x) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| 1 | | \ -x y + 1 / F(n) = ---------------- | | -------------------------- dx dy Beta(-1/3, -1/3) | | 4/3 4/3 | | x (1 - x) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.099566896, 2.105946388, 2.098000694, 2.102598870, 2.094147064, 2.092742019, 2.100221072, 2.101050044, 2.097781675, 2.095138844, 2.092610641, 2.099845030, 2.092052151, 2.086615826, 2.090176659, 2.087290401, 2.090098628, 2.088796881, 2.082586274, 2.085000898, 2.077136111] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.06802433837, 0.06651426647, 0.06839572372, 0.06730611533, 0.06931061554, 0.06964457774, 0.06786929344, 0.06767288512, 0.06844767915, 0.06907500695, 0.06967581536, 0.06795841289, 0.06980862738, 0.07110314066, 0.07025487122, 0.07094233871, 0.07027344554, 0.07058340547, 0.07206469263, 0.07148829572, 0.07336798794] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 4, : The following constant c. 1 1 / / 1 | | 1 c = -------------- | | ------------------------------ dx dy Beta(1/3, 1/3) | | (2/3) (2/3) / / x (1 - x) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.103204249, yielding an irrationality measure that is approximately , 15.88918862 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (1 + n) (1 + 3 n) (15 n + 22) (3 n + 4) X(n) - --------------------------------------------- 2 (15 n + 7) (2 + 3 n) (3 n + 8) (2 + n) 4 3 2 3 (3 n + 4) (495 n + 2046 n + 3004 n + 1843 n + 398) X(1 + n) + ---------------------------------------------------------------- 2 (15 n + 7) (2 + 3 n) (3 n + 8) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 7/5 B(0) = 1, B(1) = 3/5 A(n) Then, ----, approximates B(n) 1 1 / / 1 | | 1 c = -------------- | | ------------------------------ dx dy Beta(1/3, 1/3) | | (2/3) (2/3) / / x (1 - x) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| 1 | | \ -x y + 1 / F(n) = -------------- | | -------------------------- dx dy Beta(1/3, 1/3) | | 2/3 2/3 | | x (1 - x) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.100370707, 2.096105706, 2.103204249, 2.096980586, 2.096369228, 2.098057099, 2.092095464, 2.095924090, 2.094538706, 2.095587692, 2.093342932, 2.092912373, 2.086669530, 2.077242188, 2.080251542, 2.085567747, 2.076645787, 2.078609105, 2.080090003, 2.076113926, 2.071251211] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.06783383507, 0.06884541785, 0.06716282703, 0.06863775557, 0.06878285954, 0.06838234421, 0.06979832613, 0.06888853649, 0.06921756413, 0.06896841214, 0.06950172246, 0.06960407554, 0.07109033719, 0.07334259155, 0.07262260846, 0.07135307254, 0.07348539430, 0.07301543909, 0.07266123187, 0.07361277560, 0.07477880443] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 5, : The following constant c. 1 1 / / 1 | | 1 c = --------------- | | ------------------------------ dx dy Beta(-1/3, 2/3) | | (4/3) (1/3) / / x (1 - x) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.105796682, yielding an irrationality measure that is approximately , 16.02637413 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 2 (1 + n) (2 + 3 n) (-1 + 3 n) (45 n + 126 n + 82) X(n) - ------------------------------------------------------- + 3 (2 + 3 n) 2 2 (45 n + 36 n + 1) (1 + 3 n) (3 n + 10) (2 + n) 5 4 3 2 / (1485 n + 7623 n + 14262 n + 11429 n + 3203 n - 62) X(1 + n) / ( / 2 2 (45 n + 36 n + 1) (1 + 3 n) (3 n + 10) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1/14 B(0) = 1, B(1) = 3/7 A(n) Then, ----, approximates B(n) 1 1 / / 1 | | 1 c = --------------- | | ------------------------------ dx dy Beta(-1/3, 2/3) | | (4/3) (1/3) / / x (1 - x) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| 1 | | \ -x y + 1 / F(n) = --------------- | | -------------------------- dx dy Beta(-1/3, 2/3) | | 4/3 1/3 | | x (1 - x) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.101132991, 2.105796682, 2.100064448, 2.097238785, 2.099427709, 2.101609312, 2.098772905, 2.096392251, 2.098185510, 2.095468158, 2.100456369, 2.095286414, 2.091439274, 2.091016259, 2.082289681, 2.086265874, 2.077656606, 2.087106928, 2.081698924, 2.081494559, 2.076074940] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.06765323646, 0.06654965403, 0.06790641048, 0.06857648465, 0.06805733260, 0.06754041857, 0.06821258106, 0.06877739439, 0.06835188582, 0.06899679764, 0.06781353718, 0.06903995890, 0.06995441089, 0.07005505559, 0.07213553536, 0.07118657957, 0.07324338533, 0.07098606934, 0.07227666854, 0.07232550045, 0.07362211398] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 6, : The following constant c. / 1 1 \ | 1/2 / / | |3 | | 1 | c = 1/4 |---- | | ----------------------------------- dx dy| | Pi | | (1/3) (2/3) 1/2 | | / / x (1 - x) y (-x y + 1) | \ 0 0 / 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.375661395, yielding an irrationality measure that is approximately , 158.3051843 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 2 (1 + 3 n) (2 + 3 n) (15 n + 48 n + 38) X(n) -4/9 ----------------------------------------------- 2 (15 n + 18 n + 5) (6 n + 11) (3 + 2 n) (2 + n) 5 4 3 2 (2970 n + 16434 n + 34737 n + 34809 n + 16382 n + 2860) X(1 + n) + 2/3 -------------------------------------------------------------------- 2 (15 n + 18 n + 5) (6 n + 11) (3 + 2 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2 -14 B(0) = 1, B(1) = --- 15 A(n) Then, ----, approximates B(n) / 1 1 \ | 1/2 / / | |3 | | 1 | c = 1/4 |---- | | ----------------------------------- dx dy| | Pi | | (1/3) (2/3) 1/2 | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | 1/2 / / |-------------------| | |3 | | \ -x y + 1 / | F(n) = 1/4 |---- | | ------------------------------- dx dy| | Pi | | 1/3 2/3 1/2 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.353292911, 2.361183280, 2.355743622, 2.358921974, 2.358311636, 2.369006193, 2.362507297, 2.366065851, 2.366671780, 2.363623676, 2.375552756, 2.370719926, 2.362171911, 2.371623674, 2.375661395, 2.371403112, 2.370863732, 2.370422350, 2.368654356, 2.371771564, 2.366758213] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01108684829, 0.009413375949, 0.01056648200, 0.009892410908, 0.01002178281, 0.007759670189, 0.009133107132, 0.008380600718, 0.008252580291, 0.008896912179, 0.006379934164, 0.007398123008, 0.009204087149, 0.007207563258, 0.006357069568, 0.007254063199, 0.007367795975, 0.007460884165, 0.007833929543, 0.007176386781, 0.008234321412] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem Number, 7, : , let A(n), B(n), be two sequences of rational numbers t\ hat satisfy the second-order recurrence 2 2 (2 + 3 n) (1 + 3 n) (15 n + 51 n + 43) X(n) -4/9 ----------------------------------------------- 2 (15 n + 21 n + 7) (6 n + 13) (3 + 2 n) (2 + n) 5 4 3 2 (1485 n + 9009 n + 21063 n + 23607 n + 12608 n + 2548) X(1 + n) + 4/3 ------------------------------------------------------------------- 2 (15 n + 21 n + 7) (6 n + 13) (3 + 2 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2 -32 B(0) = 1, B(1) = --- 21 A(n) Then, ----, approximates B(n) / 1 1 \ | 1/2 / / | |3 | | 1 | c = 1/4 |---- | | ----------------------------------- dx dy| | Pi | | (2/3) (1/3) 1/2 | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | 1/2 / / |-------------------| | |3 | | \ -x y + 1 / | F(n) = 1/4 |---- | | ------------------------------- dx dy| | Pi | | 2/3 1/3 1/2 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. ------------------------------------------------------------ Theorem number, 8, : The following constant c. / 1 1 \ | 1/2 / / | |3 | | 1 | c = -1/4 |---- | | ------------------------------------- dx dy| | Pi | | (1/3) (2/3) (3/2) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.381137568, yielding an irrationality measure that is approximately , 193.0905942 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 3 2 (1 + 3 n) (-1 + 2 n) (2 + 3 n) (30 n + 123 n + 164 n + 70) X(n) -4/9 ------------------------------------------------------------------ + 2/3 ( 3 2 (2 + n) (1 + 2 n) (6 n + 5) (2 n + 5) (30 n + 33 n + 8 n - 1) 7 6 5 4 3 2 11880 n + 70488 n + 166146 n + 196590 n + 120385 n + 33757 n + 1814 n / - 580) X(n + 1) / ((2 + n) (1 + 2 n) (6 n + 5) (2 n + 5) / 3 2 (30 n + 33 n + 8 n - 1)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2/3 B(0) = 1, B(1) = 10/9 A(n) Then, ----, approximates B(n) / 1 1 \ | 1/2 / / | |3 | | 1 | c = -1/4 |---- | | ------------------------------------- dx dy| | Pi | | (1/3) (2/3) (3/2) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | 1/2 / / |-------------------| | |3 | | \ -x y + 1 / | F(n) = -1/4 |---- | | ------------------------------- dx dy| | Pi | | 1/3 2/3 3/2 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.353547716, 2.362294479, 2.365700359, 2.364857380, 2.363252386, 2.359152375, 2.372551203, 2.368512586, 2.370360381, 2.375454288, 2.375214790, 2.377136461, 2.372863875, 2.374263366, 2.371682035, 2.373642887, 2.381137568, 2.372871563, 2.376671329, 2.380668015, 2.366921975] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01103271988, 0.009178146093, 0.008457837436, 0.008636023070, 0.008975454613, 0.009843581969, 0.007012064115, 0.007863854870, 0.007473954906, 0.006400659029, 0.006451070478, 0.006046724095, 0.006946178041, 0.006651383680, 0.007195260030, 0.006782062547, 0.005205876967, 0.006944558138, 0.006144564550, 0.005304482428, 0.008199728677] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem Number, 9, : , let A(n), B(n), be two sequences of rational numbers t\ hat satisfy the second-order recurrence 2 3 2 (2 + 3 n) (-1 + 2 n) (1 + 3 n) (30 n + 129 n + 181 n + 83) X(n) -4/9 ------------------------------------------------------------------ + 4/3 ( 3 2 (2 + n) (1 + 2 n) (6 n + 7) (2 n + 5) (30 n + 39 n + 13 n + 1) 7 6 5 4 3 2 5940 n + 38412 n + 99975 n + 134301 n + 99067 n + 39253 n + 7448 n / + 476) X(1 + n) / ((2 + n) (1 + 2 n) (6 n + 7) (2 n + 5) / 3 2 (30 n + 39 n + 13 n + 1)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2/3 B(0) = 1, B(1) = -16/9 A(n) Then, ----, approximates B(n) / 1 1 \ | 1/2 / / | |3 | | 1 | c = -1/4 |---- | | ------------------------------------- dx dy| | Pi | | (2/3) (1/3) (3/2) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | 1/2 / / |-------------------| | |3 | | \ -x y + 1 / | F(n) = -1/4 |---- | | ------------------------------- dx dy| | Pi | | 2/3 1/3 3/2 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. ------------------------------------------------------------ Theorem number, 10, : The following constant c. / 1 1 \ | 1/2 / / | |3 | | 1 | c = -3/4 |---- | | ------------------------------------- dx dy| | Pi | | (1/3) (2/3) (5/2) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.382325593, yielding an irrationality measure that is approximately , 202.7560942 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 3 2 (1 + 3 n) (-3 + 2 n) (2 + 3 n) (30 n + 105 n + 108 n + 26) X(n) -4/9 ------------------------------------------------------------------ + 2/3 ( 3 2 (2 + n) (-1 + 2 n) (6 n - 1) (2 n + 7) (30 n + 15 n - 12 n - 7) 7 6 5 4 3 2 11880 n + 51480 n + 71538 n + 18474 n - 34233 n - 24761 n - 3550 n / + 532) X(1 + n) / ((2 + n) (-1 + 2 n) (6 n - 1) (2 n + 7) / 3 2 (30 n + 15 n - 12 n - 7)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2/5 B(0) = 1, B(1) = 2/35 A(n) Then, ----, approximates B(n) / 1 1 \ | 1/2 / / | |3 | | 1 | c = -3/4 |---- | | ------------------------------------- dx dy| | Pi | | (1/3) (2/3) (5/2) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | 1/2 / / |-------------------| | |3 | | \ -x y + 1 / | F(n) = -3/4 |---- | | ------------------------------- dx dy| | Pi | | 1/3 2/3 5/2 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.359559856, 2.358529510, 2.369326572, 2.365838591, 2.364956686, 2.371080593, 2.374892770, 2.376708014, 2.365453151, 2.374054755, 2.377493295, 2.377286214, 2.377225886, 2.374888230, 2.377133246, 2.373506863, 2.374235954, 2.379718494, 2.373510335, 2.382325593, 2.377990212] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.009757235982, 0.009975596770, 0.007692059936, 0.008428624500, 0.008615028880, 0.007322065936, 0.006518859776, 0.006136847174, 0.008510084780, 0.006695315384, 0.005971677007, 0.006015227620, 0.006027915739, 0.006519815569, 0.006047400304, 0.006810715049, 0.006657156195, 0.005503939610, 0.006809983676, 0.004956479773, 0.005867187193] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 11, : The following constant c. 1 1 / / | | 1 c = -4/9 | | ------------------------ dx dy | | (1/3) (5/3) / / x y (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.105842726, yielding an irrationality measure that is approximately , 16.02883209 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 2 9 (45 n + 126 n + 82) (-2 + 3 n) (1 + n) X(n) - ------------------------------------------------- 2 (45 n + 36 n + 1) (3 n + 10) (5 + 3 n) (1 + 3 n) 5 4 3 2 9 (1485 n + 7623 n + 14262 n + 11429 n + 3203 n - 62) X(1 + n) + ------------------------------------------------------------------ 2 (45 n + 36 n + 1) (3 n + 10) (5 + 3 n) (1 + 3 n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1/7 B(0) = 1, B(1) = 9/7 A(n) Then, ----, approximates B(n) 1 1 / / | | 1 c = -4/9 | | ------------------------ dx dy | | (1/3) (5/3) / / x y (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / F(n) = -4/9 | | ---------------------- dx dy | | 1/3 5/3 | | x y (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.101179059, 2.105842726, 2.100110469, 2.097284783, 2.099473684, 2.101655264, 2.098818833, 2.096438157, 2.098231392, 2.095514018, 2.100502205, 2.095332227, 2.091485065, 2.091062027, 2.082335426, 2.086311595, 2.077702305, 2.087152604, 2.081194722, 2.081540190, 2.075571242] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.06764232409, 0.06653876988, 0.06789550407, 0.06856557001, 0.06804643402, 0.06752953598, 0.06820169045, 0.06876649748, 0.06834100328, 0.06898590718, 0.06780267650, 0.06902907872, 0.06994351733, 0.07004416545, 0.07212460833, 0.07117567759, 0.07323244671, 0.07097518217, 0.07239715293, 0.07231459679, 0.07374278018] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 12, : The following constant c. / 1 1 \ | / / | | 1 | | 1 | c = -2/3 |---- | | --------------------------------- dx dy| | Pi | | 1/2 1/2 (5/3) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.374738617, yielding an irrationality measure that is approximately , 153.6411303 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 2 (2 n + 3) (15 n + 39 n + 23) (-2 + 3 n) (1 + 2 n) X(n) -9/4 -------------------------------------------------------- 2 (15 n + 9 n - 1) (3 n + 8) (6 n + 5) (1 + 3 n) (2 + n) 4 3 2 3 (3 n + 5) (2 n + 3) (495 n + 1122 n + 741 n + 104 n - 20) X(1 + n) + ----------------------------------------------------------------------- 2 (15 n + 9 n - 1) (3 n + 8) (6 n + 5) (1 + 3 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3/5 B(0) = 1, B(1) = 6/5 A(n) Then, ----, approximates B(n) / 1 1 \ | / / | | 1 | | 1 | c = -2/3 |---- | | --------------------------------- dx dy| | Pi | | 1/2 1/2 (5/3) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | / / |-------------------| | | 1 | | \ -x y + 1 / | F(n) = -2/3 |---- | | ------------------------------- dx dy| | Pi | | 1/2 1/2 5/3 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.355678555, 2.358662242, 2.352003883, 2.356404457, 2.354631497, 2.354534906, 2.354875616, 2.369746794, 2.364241928, 2.362596573, 2.371488166, 2.374738617, 2.370514517, 2.368234531, 2.371711960, 2.371336340, 2.371616787, 2.367245566, 2.367309199, 2.373537121, 2.372484461] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01058029093, 0.009947461635, 0.01136076633, 0.01042625674, 0.01080255620, 0.01082306508, 0.01075072686, 0.007603393357, 0.008766154701, 0.009114214729, 0.007236131198, 0.006551314172, 0.007441444572, 0.007922552890, 0.007188951582, 0.007268141240, 0.007209015143, 0.008131380985, 0.008117941749, 0.006804341272, 0.007026129070] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 13, : The following constant c. 1 1 / / | | 1 c = -1/9 | | ------------------------ dx dy | | (2/3) (4/3) / / x y (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.103471307, yielding an irrationality measure that is approximately , 15.90321210 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 9 (15 n + 22) (-1 + 3 n) (1 + n) X(n) - ---------------------------------------- (15 n + 7) (3 n + 8) (3 n + 4) (2 + 3 n) 4 3 2 9 (495 n + 2046 n + 3004 n + 1843 n + 398) X(1 + n) + ------------------------------------------------------ + X(2 + n) = 0 (15 n + 7) (3 n + 8) (3 n + 4) (2 + 3 n) Subject to the initial conditions A(0) = 0, A(1) = -7/5 B(0) = 1, B(1) = -9/5 A(n) Then, ----, approximates B(n) 1 1 / / | | 1 c = -1/9 | | ------------------------ dx dy | | (2/3) (4/3) / / x y (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / F(n) = -1/9 | | ---------------------- dx dy | | 2/3 4/3 | | x y (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.100638035, 2.096372899, 2.103471307, 2.097247510, 2.096636017, 2.098323754, 2.092361984, 2.096190477, 2.094804959, 2.095853811, 2.093608917, 2.093178224, 2.086935248, 2.077507773, 2.080516994, 2.085833065, 2.076910973, 2.078874158, 2.079805067, 2.076378714, 2.070966560] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.06777049337, 0.06878198812, 0.06709962879, 0.06857441432, 0.06871953313, 0.06831909704, 0.06973494331, 0.06882529300, 0.06915431353, 0.06890522283, 0.06943850193, 0.06954087476, 0.07102699236, 0.07327901188, 0.07255914585, 0.07128979202, 0.07342189325, 0.07295202545, 0.07272936586, 0.07354935480, 0.07484713928] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem Number, 14, : , let A(n), B(n), be two sequences of rational numbers \ that satisfy the second-order recurrence 2 2 (2 n + 3) (15 n + 42 n + 28) (-1 + 3 n) (1 + 2 n) X(n) -9/4 -------------------------------------------------------- + 3/2 (2 n + 3) 2 (15 n + 12 n + 1) (3 n + 7) (6 n + 7) (2 + 3 n) (2 + n) 5 4 3 2 / (2970 n + 13266 n + 21507 n + 15291 n + 4408 n + 308) X(1 + n) / ( / 2 (15 n + 12 n + 1) (3 n + 7) (6 n + 7) (2 + 3 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3/4 B(0) = 1, B(1) = -15/8 A(n) Then, ----, approximates B(n) / 1 1 \ | / / | | 1 | | 1 | c = -1/3 |---- | | --------------------------------- dx dy| | Pi | | 1/2 1/2 (4/3) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | / / |-------------------| | | 1 | | \ -x y + 1 / | F(n) = -1/3 |---- | | ------------------------------- dx dy| | Pi | | 1/2 1/2 4/3 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. ------------------------------------------------------------ Theorem number, 15, : The following constant c. / 1 1 \ | / / | | 1 | | 1 | c = 1/3 |---- | | --------------------------------- dx dy| | Pi | | 1/2 1/2 (2/3) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.375792608, yielding an irrationality measure that is approximately , 158.9914773 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 2 (2 n + 3) (15 n + 48 n + 38) (1 + 3 n) (1 + 2 n) X(n) -9/4 --------------------------------------------------------- + 3/2 (2 n + 3) 2 (15 n + 18 n + 5) (3 n + 5) (6 n + 11) (4 + 3 n) (2 + n) 5 4 3 2 / (2970 n + 16434 n + 34737 n + 34809 n + 16382 n + 2860) X(1 + n) / ( / 2 (15 n + 18 n + 5) (3 n + 5) (6 n + 11) (4 + 3 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3/2 -21 B(0) = 1, B(1) = --- 20 A(n) Then, ----, approximates B(n) / 1 1 \ | / / | | 1 | | 1 | c = 1/3 |---- | | --------------------------------- dx dy| | Pi | | 1/2 1/2 (2/3) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | / / |-------------------| | | 1 | | \ -x y + 1 / | F(n) = 1/3 |---- | | ------------------------------- dx dy| | Pi | | 1/2 1/2 2/3 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.352870197, 2.361315353, 2.355875629, 2.359053915, 2.358792878, 2.369138001, 2.362639038, 2.366197526, 2.366803388, 2.363755218, 2.375684232, 2.370851336, 2.362303255, 2.371754952, 2.375792608, 2.371534258, 2.370994813, 2.370553366, 2.368235450, 2.371902449, 2.366889032] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01117665853, 0.009385411657, 0.01053846779, 0.009864448043, 0.009919772318, 0.007731853372, 0.009105228621, 0.008352777714, 0.008224778504, 0.008869088807, 0.006352263334, 0.007370410077, 0.009176288737, 0.007179888637, 0.006329455343, 0.007226413852, 0.007340154090, 0.007433250880, 0.007922358875, 0.007148796714, 0.008206687295] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem Number, 16, : , let A(n), B(n), be two sequences of rational numbers \ that satisfy the second-order recurrence 2 2 (2 n + 3) (15 n + 51 n + 43) (2 + 3 n) (1 + 2 n) X(n) -9/4 --------------------------------------------------------- + 3 (2 n + 3) 2 (15 n + 21 n + 7) (3 n + 4) (6 n + 13) (5 + 3 n) (2 + n) 5 4 3 2 / (1485 n + 9009 n + 21063 n + 23607 n + 12608 n + 2548) X(1 + n) / ( / 2 (15 n + 21 n + 7) (3 n + 4) (6 n + 13) (5 + 3 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3 B(0) = 1, B(1) = -12/7 A(n) Then, ----, approximates B(n) / 1 1 \ | / / | | 1 | | 1 | c = 2/3 |---- | | --------------------------------- dx dy| | Pi | | 1/2 1/2 (1/3) | | / / x (1 - x) y (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | / / |-------------------| | | 1 | | \ -x y + 1 / | F(n) = 2/3 |---- | | ------------------------------- dx dy| | Pi | | 1/2 1/2 1/3 | | | | x (1 - x) y (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. ------------------------------------------------------------ Theorem number, 17, : The following constant c. / 1 1 \ | 1/2 / / | |3 | | 1 | c = 1/2 |---- | | ---------------------------------------------- dx dy| | 2 | | (1/3) (2/3) 1/2 1/2 | |Pi / / x (1 - x) y (1 - y) (-x y + 1) | \ 0 0 / 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.377298715, yielding an irrationality measure that is approximately , 167.3174426 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (1 + 2 n) (1 + 3 n) (-1 + 6 n) (2 + 3 n) (15 n + 39 n + 25) X(n) -1/36 ----------------------------------------------------------------- 2 2 (15 n + 9 n + 1) (3 n + 4) (1 + n) (2 + n) 5 4 3 2 (1485 n + 5841 n + 8391 n + 5355 n + 1448 n + 120) X(1 + n) + 1/3 --------------------------------------------------------------- 2 (15 n + 9 n + 1) (3 n + 4) (1 + n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1 B(0) = 1, B(1) = 2/3 A(n) Then, ----, approximates B(n) / 1 1 \ | 1/2 / / | |3 | | 1 | c = 1/2 |---- | | ---------------------------------------------- dx dy| | 2 | | (1/3) (2/3) 1/2 1/2 | |Pi / / x (1 - x) y (1 - y) (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | 1/2 / / |-------------------| | |3 | | \ -x y + 1 / | F(n) = 1/2 |---- | | ------------------------------------------ dx dy| | 2 | | 1/3 2/3 1/2 1/2 | |Pi | | x (1 - x) y (1 - y) (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.358848255, 2.362716178, 2.356921249, 2.358128182, 2.358534009, 2.364109970, 2.368948818, 2.363487934, 2.369475306, 2.364746978, 2.377298715, 2.373995306, 2.371166092, 2.368731566, 2.368900513, 2.373485284, 2.375172520, 2.370884841, 2.375231905, 2.368911689, 2.372602471] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.009908035190, 0.009088905274, 0.01031662366, 0.01006067560, 0.009974643094, 0.008794060371, 0.007771779114, 0.008925625531, 0.007660675371, 0.008659364080, 0.006012598464, 0.006707835541, 0.007304037682, 0.007817632524, 0.007781974052, 0.006815260663, 0.006459968329, 0.007363344483, 0.006447467809, 0.007779615300, 0.007001260359] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem Number, 18, : , let A(n), B(n), be two sequences of rational numbers \ that satisfy the second-order recurrence 2 (1 + 2 n) (2 + 3 n) (1 + 6 n) (1 + 3 n) (15 n + 42 n + 29) X(n) -1/36 ---------------------------------------------------------------- 2 2 (15 n + 12 n + 2) (3 n + 5) (1 + n) (2 + n) 5 4 3 2 (2970 n + 13266 n + 22065 n + 16797 n + 5702 n + 660) X(1 + n) + 1/6 ------------------------------------------------------------------ 2 (15 n + 12 n + 2) (3 n + 5) (1 + n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1 -5 B(0) = 1, B(1) = -- 12 A(n) Then, ----, approximates B(n) / 1 1 \ | 1/2 / / | |3 | | 1 | c = 1/2 |---- | | ---------------------------------------------- dx dy| | 2 | | (2/3) (1/3) 1/2 1/2 | |Pi / / x (1 - x) y (1 - y) (-x y + 1) | \ 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / 1 1 /x (1 - x) y (1 - y)\n \ | 1/2 / / |-------------------| | |3 | | \ -x y + 1 / | F(n) = 1/2 |---- | | ------------------------------------------ dx dy| | 2 | | 2/3 1/3 1/2 1/2 | |Pi | | x (1 - x) y (1 - y) (-x y + 1) | | / / | \ 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. ------------------------------------------------------------ Theorem number, 19, : The following constant c. 1 c = ----------------- Beta(2/3, 5/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (1/3) (1/6) 1/2 1/2 / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.375515021, yielding an irrationality measure that is approximately , 157.5465514 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (1 + 2 n) (5 + 6 n) (1 + 3 n) (2 + 3 n) (15 n + 48 n + 38) X(n) -1/36 ---------------------------------------------------------------- 2 2 (15 n + 18 n + 5) (3 n + 4) (1 + n) (2 + n) 5 4 3 2 (2970 n + 16434 n + 34737 n + 34809 n + 16382 n + 2860) X(1 + n) + 1/6 -------------------------------------------------------------------- 2 2 (15 n + 18 n + 5) (3 n + 4) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -5/2 B(0) = 1, B(1) = -7/6 A(n) Then, ----, approximates B(n) 1 c = ----------------- Beta(2/3, 5/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (1/3) (1/6) 1/2 1/2 / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------- Beta(2/3, 5/6) Pi 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/3 1/6 1/2 1/2 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.353145502, 2.361035945, 2.355596362, 2.358774788, 2.358164524, 2.368859155, 2.362360333, 2.365918962, 2.366524964, 2.363476934, 2.375406088, 2.370573331, 2.362025390, 2.371477226, 2.375515021, 2.371256811, 2.370717505, 2.370276196, 2.368508274, 2.371625556, 2.366612278] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01111816514, 0.009444573548, 0.01059773500, 0.009923606544, 0.01005297075, 0.007790702965, 0.009164208892, 0.008411640261, 0.008283596530, 0.008927952420, 0.006410804144, 0.007429040088, 0.009235099517, 0.007238437651, 0.006387876266, 0.007284909448, 0.007398633584, 0.007491712075, 0.007864765090, 0.007207166504, 0.008265150409] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 20, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | 1/2 (1/6) (2/3) (1/3) | / / x (1 - x) y (1 - y) (-x y + 1) | 0 0 / 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.375718363, yielding an irrationality measure that is approximately , 158.6024191 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (2 + 3 n) (5 + 6 n) (1 + 2 n) (1 + 3 n) (15 n + 48 n + 38) X(n) -1/36 ---------------------------------------------------------------- 2 2 (15 n + 18 n + 5) (4 + 3 n) (1 + n) (2 + n) 5 4 3 2 (2970 n + 16434 n + 34737 n + 34809 n + 16382 n + 2860) X(1 + n) + 1/6 -------------------------------------------------------------------- 2 2 (15 n + 18 n + 5) (4 + 3 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -5/3 B(0) = 1, B(1) = -7/6 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | 1/2 (1/6) (2/3) (1/3) | / / x (1 - x) y (1 - y) (-x y + 1) | 0 0 / / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral / | 1/2 | 3 F(n) = 1/2 |----------------- |Beta(1/2, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/2 1/6 2/3 1/3 | | | x (1 - x) y (1 - y) (-x y + 1) | / / | 0 0 / Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.352795428, 2.361240622, 2.355800935, 2.358979259, 2.358718260, 2.369063420, 2.362564495, 2.366123021, 2.366728920, 2.363680787, 2.375609839, 2.370776980, 2.362228936, 2.371680671, 2.375718363, 2.371460051, 2.370920643, 2.370479233, 2.368161354, 2.371828390, 2.366815011] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01119254569, 0.009401234531, 0.01055431898, 0.009880270043, 0.009935587998, 0.007747592790, 0.009121002928, 0.008368520475, 0.008240509453, 0.008884832042, 0.006367920119, 0.007386090756, 0.009192017856, 0.007195547574, 0.006345080271, 0.007242058634, 0.007355794604, 0.007448886481, 0.007938001851, 0.007164407888, 0.008222323255] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 21, : The following constant c. 1 c = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 2/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (3/7) (5/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.326673290, yielding an irrationality measure that is approximately , 60.61683738 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (7 n + 9) (245 n + 693 n + 470) (6 + 7 n) (-1 + 7 n) (2 + 7 n) X(n) -1/7 -------------------------------------------------------------------- + 2 (245 n + 203 n + 22) (7 n + 15) (10 + 7 n) (7 n + 5) (2 + n) 4 3 2 (6 + 7 n) (7 n + 9) (18865 n + 74921 n + 99477 n + 48367 n + 5330) / 2 X(n + 1) / ((245 n + 203 n + 22) (7 n + 15) (10 + 7 n) (7 n + 5) (2 + n) / ) + X(2 + n) = 0 Subject to the initial conditions 11 A(0) = 0, A(1) = -- 28 B(0) = 1, B(1) = -1/4 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 2/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (3/7) (5/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 2/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/7 3/7 5/7 5/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.313984926, 2.311147138, 2.306617247, 2.315256922, 2.315866195, 2.316380982, 2.316132035, 2.312661246, 2.316310740, 2.313038199, 2.318152234, 2.317381106, 2.321168021, 2.326673290, 2.322727479, 2.317750428, 2.325416596, 2.322930024, 2.321018956, 2.315448738, 2.318460956] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.01950706349, 0.02012038094, 0.02110093504, 0.01923239243, 0.01910088023, 0.01898978938, 0.01904350903, 0.01979305228, 0.01900494615, 0.01971159305, 0.01860773880, 0.01877403220, 0.01795790669, 0.01677378479, 0.01762220480, 0.01869438131, 0.01704384293, 0.01757861953, 0.01799000726, 0.01919098509, 0.01854117826] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 22, : The following constant c. 1 c = ----------------------------- Beta(4/7, 4/7) Beta(2/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (3/7) (4/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.292473517, yielding an irrationality measure that is approximately , 42.36556316 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (2 + 7 n) (4 + 7 n) (-1 + 7 n) (3 + 7 n) (7 n + 11) (35 n + 46) X(n) -1/7 -------------------------------------------------------------------- (35 n + 11) (7 n + 1) (7 n + 12) (7 n + 8) (5 + 7 n) (2 + n) 3 2 (7 n + 11) (2695 n + 7007 n + 5380 n + 1092) X(n + 1) + ------------------------------------------------------- + X(2 + n) = 0 (35 n + 11) (7 n + 12) (7 n + 8) (2 + n) Subject to the initial conditions -44 A(0) = 0, A(1) = --- 35 B(0) = 1, B(1) = 4/5 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(4/7, 4/7) Beta(2/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (3/7) (4/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(4/7, 4/7) Beta(2/7, 3/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 3/7 3/7 4/7 5/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.275709036, 2.276538058, 2.275631927, 2.288305120, 2.288951699, 2.289698644, 2.278838411, 2.288623884, 2.283490974, 2.292473517, 2.287543215, 2.283278352, 2.290884045, 2.271691288, 2.289908744, 2.285409374, 2.287562110, 2.284871710, 2.278243159, 2.280825445, 2.280069910] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02784206405, 0.02766009160, 0.02785899295, 0.02508412191, 0.02494295123, 0.02477991556, 0.02715549551, 0.02501451978, 0.02613644132, 0.02417469807, 0.02525052218, 0.02618296798, 0.02452128455, 0.02872488379, 0.02473406638, 0.02571684133, 0.02524639486, 0.02583440670, 0.02728602004, 0.02672002652, 0.02688556249] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 23, : The following constant c. 1 c = ----------------------------- Beta(3/7, 5/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (4/7) (3/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.292444896, yielding an irrationality measure that is approximately , 42.35489068 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (35 n + 46) (5 + 7 n) (-2 + 7 n) (3 + 7 n) (2 + 7 n) X(n) -1/7 --------------------------------------------------------- (35 n + 11) (7 n + 1) (7 n + 8) (6 + 7 n) (2 + n) 3 2 (5 + 7 n) (2695 n + 7007 n + 5380 n + 1092) X(n + 1) + ------------------------------------------------------ + X(2 + n) = 0 (35 n + 11) (7 n + 8) (6 + 7 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -11/7 B(0) = 1, B(1) = 2 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 5/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (4/7) (3/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(3/7, 5/7) Beta(2/7, 4/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/7 4/7 3/7 5/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.275680284, 2.276509321, 2.275603205, 2.288276412, 2.288923005, 2.289669965, 2.278809746, 2.288595234, 2.283462338, 2.292444896, 2.287514608, 2.283249760, 2.290855467, 2.271662724, 2.289880195, 2.285380839, 2.287533589, 2.284843203, 2.278214667, 2.280796967, 2.280041446] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02784837634, 0.02766639837, 0.02786529886, 0.02509039076, 0.02494921530, 0.02478617437, 0.02716178030, 0.02502077512, 0.02614270730, 0.02418093685, 0.02525677101, 0.02618922490, 0.02452751817, 0.02873116558, 0.02474029626, 0.02572308010, 0.02525262485, 0.02584064078, 0.02729226849, 0.02672626501, 0.02689179993] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 24, : The following constant c. 1 c = ----------------------------- Beta(3/7, 6/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (1/7) (3/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.303741241, yielding an irrationality measure that is approximately , 47.03105716 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (35 n + 58) (4 + 7 n) (1 + 7 n) X(n) -1/7 ------------------------------------ (35 n + 23) (7 n + 12) (2 + n) 3 2 (2695 n + 9086 n + 9539 n + 3124) X(n + 1) + -------------------------------------------- + X(2 + n) = 0 (35 n + 23) (7 n + 12) (2 + n) Subject to the initial conditions -46 A(0) = 0, A(1) = --- 35 B(0) = 1, B(1) = -2/5 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 6/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (1/7) (3/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(3/7, 6/7) Beta(2/7, 4/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 4/7 1/7 3/7 5/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.285806218, 2.295690820, 2.296744471, 2.292521250, 2.294575713, 2.288366479, 2.289629580, 2.289616901, 2.293968768, 2.295673091, 2.296345403, 2.303741241, 2.300618409, 2.303233041, 2.295814164, 2.294754333, 2.298179592, 2.296887503, 2.299988424, 2.292955041, 2.288502699] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02563008488, 0.02347387809, 0.02324457141, 0.02416429345, 0.02371667144, 0.02507072343, 0.02479498798, 0.02479775507, 0.02384887059, 0.02347773734, 0.02333140872, 0.02172446304, 0.02240236669, 0.02183472173, 0.02344702935, 0.02367777258, 0.02293241043, 0.02321345119, 0.02253923274, 0.02406974740, 0.02504097942] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 25, : The following constant c. 1 c = ----------------------------- Beta(4/7, 5/7) Beta(2/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (3/7) (1/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.303582816, yielding an irrationality measure that is approximately , 46.95834869 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (4 + 7 n) (-1 + 7 n) (6 + 7 n) (5 + 7 n) (35 n + 58) X(n) -1/7 --------------------------------------------------------- (35 n + 23) (7 n + 3) (7 n + 10) (8 + 7 n) (2 + n) 3 2 (6 + 7 n) (2695 n + 9086 n + 9539 n + 3124) X(n + 1) + ------------------------------------------------------ + X(2 + n) = 0 (35 n + 23) (7 n + 10) (8 + 7 n) (2 + n) Subject to the initial conditions -46 A(0) = 0, A(1) = --- 21 B(0) = 1, B(1) = 2/3 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(4/7, 5/7) Beta(2/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (3/7) (1/7) (5/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(4/7, 5/7) Beta(2/7, 6/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/7 3/7 1/7 5/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.285646914, 2.295531595, 2.296585327, 2.292362186, 2.294416730, 2.288207576, 2.289470756, 2.289458157, 2.293810105, 2.295514507, 2.296186899, 2.303582816, 2.300460065, 2.303074776, 2.295655978, 2.294596226, 2.298021565, 2.296729554, 2.299830555, 2.292797251, 2.288344987] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02566490952, 0.02350853925, 0.02327919942, 0.02419896631, 0.02375129635, 0.02510542253, 0.02482965117, 0.02483240099, 0.02388343473, 0.02351225922, 0.02336590332, 0.02175883226, 0.02243676393, 0.02186906365, 0.02348146253, 0.02371220408, 0.02296677441, 0.02324781709, 0.02257353598, 0.02410413618, 0.02507541643] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 26, : The following constant c. 1 c = ----------------------------- Beta(3/7, 5/7) Beta(3/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (4/7) (1/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.303513406, yielding an irrationality measure that is approximately , 46.92656405 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (3 + 7 n) (-1 + 7 n) (6 + 7 n) (5 + 7 n) (35 n + 58) (7 n + 10) X(n) -1/7 -------------------------------------------------------------------- (35 n + 23) (7 n + 2) (8 + 7 n) (7 n + 11) (9 + 7 n) (2 + n) 3 2 (7 n + 10) (6 + 7 n) (2695 n + 9086 n + 9539 n + 3124) X(n + 1) + ----------------------------------------------------------------- (35 n + 23) (8 + 7 n) (7 n + 11) (9 + 7 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions -69 A(0) = 0, A(1) = --- 28 B(0) = 1, B(1) = 3/4 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 5/7) Beta(3/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (4/7) (1/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(3/7, 5/7) Beta(3/7, 6/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/7 4/7 1/7 4/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.285577118, 2.295461835, 2.296515601, 2.292292496, 2.294347075, 2.288137956, 2.289401172, 2.289388607, 2.293740590, 2.295445027, 2.296117454, 2.303513406, 2.300390689, 2.303005435, 2.295586672, 2.294526955, 2.297952328, 2.296660353, 2.299761388, 2.292728118, 2.288275889] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02568016802, 0.02352372582, 0.02329437178, 0.02421415811, 0.02376646725, 0.02512062595, 0.02484483855, 0.02484758103, 0.02389857905, 0.02352738494, 0.02338101710, 0.02177389101, 0.02245183528, 0.02188411067, 0.02349654946, 0.02372729019, 0.02298183116, 0.02326287428, 0.02258856595, 0.02411920377, 0.02509050497] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 27, : The following constant c. 1 c = ----------------------------- Beta(4/7, 5/7) Beta(3/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (2/7) (4/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.303601834, yielding an irrationality measure that is approximately , 46.96706505 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (3 + 7 n) (5 + 7 n) (1 + 7 n) (4 + 7 n) (35 n + 58) (7 n + 10) X(n) -1/7 ------------------------------------------------------------------- (35 n + 23) (7 n + 2) (7 n + 13) (7 n + 9) (6 + 7 n) (2 + n) 3 2 (7 n + 10) (2695 n + 9086 n + 9539 n + 3124) X(n + 1) + ------------------------------------------------------- + X(2 + n) = 0 (35 n + 23) (7 n + 13) (7 n + 9) (2 + n) Subject to the initial conditions -23 A(0) = 0, A(1) = --- 14 B(0) = 1, B(1) = -1/2 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(4/7, 5/7) Beta(3/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (2/7) (4/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(4/7, 5/7) Beta(3/7, 3/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 3/7 2/7 4/7 4/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.285666037, 2.295550709, 2.296604431, 2.292381280, 2.294435814, 2.288226651, 2.289489822, 2.289477213, 2.293829151, 2.295533544, 2.296205926, 2.303601834, 2.300479072, 2.303093774, 2.295674967, 2.294615205, 2.298040534, 2.296748514, 2.299849505, 2.292816192, 2.288363919] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02566072902, 0.02350437827, 0.02327504247, 0.02419480406, 0.02374713992, 0.02510125706, 0.02482548991, 0.02482824189, 0.02387928551, 0.02350811497, 0.02336176243, 0.02175470631, 0.02243263490, 0.02186494115, 0.02347732897, 0.02370807084, 0.02296264935, 0.02324369172, 0.02256941823, 0.02410000806, 0.02507128243] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 28, : The following constant c. 1 c = ----------------------------- Beta(2/7, 6/7) Beta(3/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (1/7) (4/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.303863637, yielding an irrationality measure that is approximately , 47.08738461 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (35 n + 58) (7 n + 9) (7 n + 10) (2 + 7 n) (3 + 7 n) X(n) -1/7 --------------------------------------------------------- (35 n + 23) (7 n + 12) (8 + 7 n) (7 n + 11) (2 + n) 3 2 (7 n + 10) (7 n + 9) (2695 n + 9086 n + 9539 n + 3124) X(n + 1) + ----------------------------------------------------------------- (35 n + 23) (7 n + 12) (8 + 7 n) (7 n + 11) (2 + n) + X(2 + n) = 0 Subject to the initial conditions -69 A(0) = 0, A(1) = --- 70 B(0) = 1, B(1) = -3/5 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(2/7, 6/7) Beta(3/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (1/7) (4/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(2/7, 6/7) Beta(3/7, 3/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 5/7 1/7 4/7 4/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.285929294, 2.295813833, 2.296867422, 2.292644139, 2.294698541, 2.288489245, 2.289752284, 2.289739543, 2.294091349, 2.295795610, 2.296467860, 2.303863637, 2.300740744, 2.303355314, 2.295936376, 2.294876483, 2.298301681, 2.297009531, 2.300110391, 2.293076947, 2.288624543] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02560318148, 0.02344710140, 0.02321782020, 0.02413750765, 0.02368992233, 0.02504391706, 0.02476820955, 0.02477099003, 0.02382216838, 0.02345106798, 0.02330476047, 0.02169791166, 0.02237579331, 0.02180819130, 0.02342042841, 0.02365117314, 0.02290586297, 0.02318690242, 0.02251273220, 0.02404318076, 0.02501437590] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 29, : The following constant c. 1 c = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (1/7) (5/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.319108032, yielding an irrationality measure that is approximately , 55.34281494 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (6 + 7 n) (2 + 7 n) (4 + 7 n) (35 n + 61) (7 n + 9) X(n) -1/7 -------------------------------------------------------- (35 n + 26) (7 n + 15) (7 n + 8) (5 + 7 n) (2 + n) 3 2 (7 n + 9) (2695 n + 8932 n + 9047 n + 2918) X(n + 1) + ------------------------------------------------------ + X(2 + n) = 0 (35 n + 26) (7 n + 15) (5 + 7 n) (2 + n) Subject to the initial conditions -39 A(0) = 0, A(1) = --- 14 B(0) = 1, B(1) = -3/2 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (1/7) (5/7) (4/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 3/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 3/7 1/7 5/7 4/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.302899768, 2.310502566, 2.305698930, 2.312791176, 2.303934038, 2.313588154, 2.307445006, 2.306328632, 2.302936232, 2.317199287, 2.313922329, 2.317068365, 2.307900718, 2.316451229, 2.308892021, 2.319108032, 2.309117793, 2.308506242, 2.309425900, 2.299673092, 2.306329447] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02190704131, 0.02025979204, 0.02129994661, 0.01976497304, 0.02168264018, 0.01959277156, 0.02092161506, 0.02116347341, 0.02189912820, 0.01881324930, 0.01952058433, 0.01884149013, 0.02082291964, 0.01897463198, 0.02060829497, 0.01840169673, 0.02055942623, 0.02069180834, 0.02049274348, 0.02260775329, 0.02116329680] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 30, : The following constant c. 1 c = ----------------------------- Beta(3/7, 5/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (2/7) (5/7) (3/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.303793639, yielding an irrationality measure that is approximately , 47.05515456 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (4 + 7 n) (5 + 7 n) (2 + 7 n) (35 n + 58) (7 n + 9) X(n) -1/7 -------------------------------------------------------- (35 n + 23) (7 n + 13) (8 + 7 n) (6 + 7 n) (2 + n) 3 2 (7 n + 9) (2695 n + 9086 n + 9539 n + 3124) X(n + 1) + ------------------------------------------------------ + X(2 + n) = 0 (35 n + 23) (7 n + 13) (8 + 7 n) (2 + n) Subject to the initial conditions -23 A(0) = 0, A(1) = --- 21 B(0) = 1, B(1) = -2/3 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 5/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (2/7) (5/7) (3/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(3/7, 5/7) Beta(2/7, 4/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 4/7 2/7 5/7 3/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.285858908, 2.295743483, 2.296797107, 2.292573860, 2.294628297, 2.288419036, 2.289682110, 2.289669405, 2.294021246, 2.295725542, 2.296397828, 2.303793639, 2.300670782, 2.303285387, 2.295866484, 2.294806626, 2.298231859, 2.296939744, 2.300040639, 2.293007230, 2.288554861] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02561856711, 0.02346241457, 0.02323311890, 0.02415282602, 0.02370521968, 0.02505924726, 0.02478352387, 0.02478629658, 0.02383743897, 0.02346631988, 0.02332000018, 0.02171309617, 0.02239099015, 0.02182336368, 0.02343564109, 0.02366638506, 0.02292104514, 0.02320208532, 0.02252788749, 0.02405837383, 0.02502959015] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 31, : The following constant c. 1 c = ----------------------------- Beta(5/7, 6/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (1/7) (5/7) (3/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.326611270, yielding an irrationality measure that is approximately , 60.56951757 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (3 + 7 n) (2 + 7 n) (5 + 7 n) (245 n + 770 n + 592) X(n) -1/7 --------------------------------------------------------- 2 (245 n + 280 n + 67) (7 n + 16) (7 n + 8) (2 + n) 4 3 2 (7 n + 9) (18865 n + 80850 n + 118979 n + 69330 n + 12720) X(n + 1) + ---------------------------------------------------------------------- 2 (245 n + 280 n + 67) (7 n + 16) (7 n + 8) (2 + n) + X(2 + n) = 0 Subject to the initial conditions -268 A(0) = 0, A(1) = ---- 63 B(0) = 1, B(1) = -8/3 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(5/7, 6/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (1/7) (5/7) (3/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(5/7, 6/7) Beta(2/7, 4/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/7 1/7 5/7 3/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.296412813, 2.309337109, 2.303162966, 2.313273376, 2.310342211, 2.317013915, 2.304429899, 2.314832458, 2.308568144, 2.315842140, 2.312472690, 2.326611270, 2.325366836, 2.318609740, 2.316790680, 2.322637984, 2.322447469, 2.313637513, 2.308504314, 2.309114440, 2.308565559] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02331673925, 0.02051195938, 0.02184992702, 0.01966077808, 0.02029448030, 0.01885323586, 0.02157509029, 0.01932403348, 0.02067840689, 0.01910607188, 0.01983380396, 0.01678710926, 0.01705453909, 0.01850910349, 0.01890139401, 0.01764146425, 0.01768246577, 0.01958210857, 0.02069222575, 0.02056015196, 0.02067896652] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 32, : The following constant c. 1 c = ----------------------------- Beta(3/7, 5/7) Beta(5/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (2/7) (1/7) (2/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.260367241, yielding an irrationality measure that is approximately , 33.02941878 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (7 n + 10) (35 n + 61) (7 n + 12) (6 + 7 n) (3 + 7 n) (5 + 7 n) X(n) -1/7 -------------------------------------------------------------------- (35 n + 26) (7 n + 9) (8 + 7 n) (7 n + 15) (11 + 7 n) (2 + n) 3 2 (7 n + 10) (7 n + 12) (2695 n + 9702 n + 10949 n + 3870) X(n + 1) + ------------------------------------------------------------------- (35 n + 26) (7 n + 9) (7 n + 15) (11 + 7 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions -195 A(0) = 0, A(1) = ---- 56 B(0) = 1, B(1) = -15/8 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 5/7) Beta(5/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (2/7) (1/7) (2/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(3/7, 5/7) Beta(5/7, 6/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 4/7 2/7 1/7 2/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.248964028, 2.251564434, 2.258036340, 2.255293554, 2.257825986, 2.257148547, 2.252717650, 2.251954514, 2.253928188, 2.256340739, 2.260084939, 2.260367241, 2.256686461, 2.254620619, 2.238976264, 2.243452244, 2.254256963, 2.255802522, 2.248022709, 2.251428493, 2.251207927] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.03374743623, 0.03317028288, 0.03173665422, 0.03234373826, 0.03178318837, 0.03193307879, 0.03291453581, 0.03308376122, 0.03264621270, 0.03211187165, 0.03128368606, 0.03122129711, 0.03203534511, 0.03249279382, 0.03597020195, 0.03497289674, 0.03257336179, 0.03223103009, 0.03395651852, 0.03320043867, 0.03324937056] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 33, : The following constant c. 1 c = ----------------------------- Beta(2/7, 6/7) Beta(5/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (1/7) (2/7) (2/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.260581142, yielding an irrationality measure that is approximately , 33.07798301 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 + 7 n) (6 + 7 n) (4 + 7 n) (2 + 7 n) (35 n + 61) (7 n + 12) X(n) -1/7 ------------------------------------------------------------------- (35 n + 26) (7 n + 3) (8 + 7 n) (7 n + 15) (10 + 7 n) (2 + n) 3 2 (7 n + 12) (2695 n + 9702 n + 10949 n + 3870) X(n + 1) + -------------------------------------------------------- + X(2 + n) = 0 (35 n + 26) (7 n + 15) (10 + 7 n) (2 + n) Subject to the initial conditions -65 A(0) = 0, A(1) = --- 28 B(0) = 1, B(1) = -5/3 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(2/7, 6/7) Beta(5/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (1/7) (2/7) (2/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(2/7, 6/7) Beta(5/7, 5/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 5/7 1/7 2/7 2/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.249179118, 2.251779415, 2.258251212, 2.255508319, 2.258040642, 2.257363094, 2.252932090, 2.252168846, 2.254142412, 2.256554856, 2.260298948, 2.260581142, 2.256900255, 2.254834305, 2.239189843, 2.243665717, 2.254470329, 2.256015780, 2.248235861, 2.251641539, 2.251420866] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.03369967310, 0.03312259725, 0.03168912493, 0.03229617673, 0.03173570258, 0.03188560332, 0.03286699372, 0.03303622750, 0.03259874316, 0.03206447491, 0.03123638922, 0.03117402985, 0.03198802688, 0.03244545757, 0.03592257017, 0.03492538021, 0.03252608904, 0.03218381258, 0.03390916658, 0.03315317946, 0.03320213062] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 34, : The following constant c. 1 c = ----------------------------- Beta(6/7, 6/7) Beta(4/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (1/7) (3/7) (2/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.294739865, yielding an irrationality measure that is approximately , 43.22808335 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (6 + 7 n) (4 + 7 n) (7 n + 13) (7 n + 11) X(n) -1/7 ---------------------------------------------- (7 n + 15) (7 n + 10) (9 + 7 n) (2 + n) 2 (7 n + 13) (7 n + 11) (77 n + 187 n + 114) X(n + 1) + ---------------------------------------------------- + X(2 + n) = 0 (7 n + 15) (7 n + 10) (9 + 7 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -25/7 B(0) = 1, B(1) = -2 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(6/7, 6/7) Beta(4/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (1/7) (3/7) (2/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(6/7, 6/7) Beta(4/7, 5/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/7 1/7 3/7 2/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.277435014, 2.275849385, 2.281708806, 2.283034695, 2.289386710, 2.285881175, 2.286141448, 2.286525170, 2.284231230, 2.273243393, 2.278349629, 2.283862087, 2.285102631, 2.279074821, 2.275430891, 2.276112570, 2.279774610, 2.289309775, 2.294129780, 2.292026190, 2.294739865] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02746328002, 0.02781125255, 0.02652655183, 0.02623629100, 0.02484799508, 0.02561369973, 0.02555680967, 0.02547294784, 0.02597448895, 0.02838366006, 0.02726267128, 0.02605524325, 0.02578391032, 0.02710366565, 0.02790313203, 0.02775347925, 0.02695027654, 0.02486478751, 0.02381379712, 0.02427221461, 0.02368092323] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 35, : The following constant c. 1 c = ----------------------------- Beta(5/7, 6/7) Beta(3/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (1/7) (4/7) (1/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 1/2 5 5 ln(11/2 + ------) + nu 2 is irrational, with an irrationality measure, 1 + ----------------------, 1/2 5 5 ln(11/2 + ------) - nu 2 for a certain number nu that is approximately , 2.294892691, yielding an irrationality measure that is approximately , 43.28751113 We hope that the reader can find nu exactly. We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (6 + 7 n) (5 + 7 n) (3 + 7 n) (7 n + 13) (7 n + 12) X(n) -1/7 -------------------------------------------------------- (7 n + 4) (7 n + 15) (7 n + 11) (9 + 7 n) (2 + n) 2 (7 n + 13) (7 n + 12) (77 n + 187 n + 114) X(n + 1) + ---------------------------------------------------- + X(2 + n) = 0 (7 n + 15) (7 n + 11) (9 + 7 n) (2 + n) Subject to the initial conditions -75 A(0) = 0, A(1) = --- 28 B(0) = 1, B(1) = -15/8 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(5/7, 6/7) Beta(3/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (1/7) (4/7) (1/7) / / x (1 - x) y (1 - y) (-x y + 1) 0 0 / 1 \n with an error that is OMEGA of, |----------------| , that in floating point is, |/ 1/2\2| || 5 5 | | ||11/2 + ------| | \\ 2 / / 0.0081306187557833487477241098899035253829951106830425825503257512106745\ n 44960365266103603769583487438335 Proof: Consider the Beukers type-integral 1 F(n) = ----------------------------- Beta(5/7, 6/7) Beta(3/7, 6/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/7 1/7 4/7 1/7 | | x (1 - x) y (1 - y) (-x y + 1) / / 0 0 Then , F(0) = B(0) c - A(0), F(1) = c B(1) - A(1) and F(n) also satisfies the above recurrence, thanks to the amazing multivar\ iable Almkvist-Zeilberger algorithm Hence, F(n) = B(n) c - A(n) 1 By a simple bound of the integrand, F(n) is OMEGA of, ----------------, / 1/2\n | 5 5 | |11/2 + ------| \ 2 / and by the Poincare lemma, B(n) (and for that matter, A(n)) are OMEGA of / 1/2\n | 5 5 | , |11/2 + ------| \ 2 / 1 Dividing by B(n) gives that A(n)/B(n)-c is OMEGA of , --------------------, / 1/2\(2 n) | 5 5 | |11/2 + ------| \ 2 / QED. we now claim that the sequence of RATIONAL numbers A(n),B(n), can be multipl\ ied by another sequence of rational numbers E(n) such that both A(n)E(n) and B(n)E(n) are integers Lemma: There exists a sequence of rational numbers, whose prime factorizatio\ ns consists of small primes, that hopefully can be described (and proved) explicity, that we leave to the expert reader \ such that A1(n):=E(n)A(n), B1(n):=E(n)B(n) are BOTH integers Furthermore there exists a contant, nu, that hopefully the learned reader ca\ n determine such that E(n) is OMEGA of , exp(nu n) The empircal values of nu for E(n) from, 1980, to , 2001, are [2.277589384, 2.276003677, 2.281863019, 2.283188831, 2.289540768, 2.286035155, 2.286295351, 2.286678996, 2.284384978, 2.273397064, 2.278503223, 2.284015604, 2.285256071, 2.279228184, 2.275584177, 2.276265779, 2.279927742, 2.289462830, 2.294282759, 2.292179093, 2.294892691] Multiplying F(n) by E(n) we get E(n) F(n) = B1(n) c - A1(n) and this implies that | A1(n) | CONSTANT | c - ----- | <= ---------------- | B1(n) | (delta + 1) B1(n) 1/2 5 5 ln(11/2 + ------) - nu 2 where , delta = ---------------------- 1/2 5 5 ln(11/2 + ------) + nu 2 Using the above values of nu for E(n) from, 1980, to , 2001, the estimated deltas are [0.02742941550, 0.02777738221, 0.02649278341, 0.02620255852, 0.02481437083, 0.02558004227, 0.02552317277, 0.02543933327, 0.02594085855, 0.02834988846, 0.02722899014, 0.02602165809, 0.02575035975, 0.02707004558, 0.02786947650, 0.02771985043, 0.02691671714, 0.02483138107, 0.02378047571, 0.02423887992, 0.02364764379] As you can see, they are all positive We leave it to the reader to fill-in the details. ----------------------------------------------- This ends this paper that took, 146.234, seconds to generate