Sketches of proofs of the Irrationality of , 88, Constants given as certain double integrals By Shalosh B. Ekhad ------------------------------------------------------------ Theorem number, 1, : The following constant c. 1 1 / / | | 1 c = | | -------- dx dy | | -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 , 1.986221570, yielding an irrationality measure that is approximately , 11.46185755 We hope that the reader can find nu exactly. 2 Pi Comment: Note that this constant appears to be , --- 6 Prove it! 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) X(n) (11 n + 33 n + 25) X(1 + n) - ------------- + ---------------------------- + X(2 + n) = 0 2 2 (2 + n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -5 B(0) = 1, B(1) = -3 A(n) Then, ----, approximates B(n) 1 1 / / | | 1 c = | | -------- dx dy | | -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) = | | ---------------------- dx dy | | -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 [1.968633183, 1.967987604, 1.965104878, 1.964000580, 1.962338779, 1.959058643, 1.963586130, 1.966261733, 1.971208570, 1.965890535, 1.962000646, 1.961637726, 1.964419821, 1.967221113, 1.972897407, 1.966640898, 1.968688097, 1.972286583, 1.972300169, 1.977334404, 1.986221570] 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 [0.09999010472, 0.1001524556, 0.1008779919, 0.1011561797, 0.1015750753, 0.1024028459, 0.1012606217, 0.1005867150, 0.09934292013, 0.1006801597, 0.1016603486, 0.1017518876, 0.1010505506, 0.1003452759, 0.09891893533, 0.1004912811, 0.09997629710, 0.09907224576, 0.09906883535, 0.09780657798, 0.09558531983] 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 c = 1/2 | | --------------- dx dy | | 1/2 / / 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 , 2.000213330, yielding an irrationality measure that is approximately , 11.85701148 We hope that the reader can find nu exactly. Comment: Note that this constant appears to be , arcsinh(15/8) Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 4 (1 + n) (1 + 2 n) (10 n + 17) X(n) - ------------------------------------- 2 (10 n + 7) (3 + 2 n) (2 n + 5) 3 2 2 (220 n + 814 n + 950 n + 341) X(1 + n) + ------------------------------------------ + X(2 + n) = 0 (10 n + 7) (3 + 2 n) (2 n + 5) Subject to the initial conditions A(0) = 0, A(1) = -14/3 B(0) = 1, B(1) = -10/3 A(n) Then, ----, approximates B(n) 1 1 / / | | 1 c = 1/2 | | --------------- dx dy | | 1/2 / / 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 / / |-------------------| | | \ -x y + 1 / F(n) = 1/2 | | ---------------------- dx dy | | 1/2 | | 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 [2.000213330, 1.996296691, 1.993743910, 1.991342314, 1.997055920, 1.993141015, 1.991927118, 1.998027082, 1.997022165, 1.988825768, 1.992378963, 1.991357704, 1.989994901, 1.990912504, 1.987590676, 1.988390837, 1.987404247, 1.987169300, 1.991691714, 1.984595870, 1.980254144] 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 [0.09210637770, 0.09307799077, 0.09371219841, 0.09430951819, 0.09288951136, 0.09386208785, 0.09416400692, 0.09264851409, 0.09289788970, 0.09493612851, 0.09405160508, 0.09430568835, 0.09464492964, 0.09441648844, 0.09524392429, 0.09504449740, 0.09529039907, 0.09534897448, 0.09422257562, 0.09599097526, 0.09707582539] 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 c = -1/2 | | ----------------- dx dy | | (3/2) / / 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 , 2.002481790, yielding an irrationality measure that is approximately , 11.92365832 We hope that the reader can find nu exactly. Comment: Note that this constant appears to be , 2/3 - 2/3 ln(2) Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 4 (1 + n) (-1 + 2 n) (10 n + 11) X(n) - -------------------------------------- 2 (10 n + 1) (1 + 2 n) (2 n + 7) 4 3 2 2 (440 n + 1584 n + 1862 n + 720 n + 29) X(1 + n) + ---------------------------------------------------- + X(2 + n) = 0 2 (10 n + 1) (1 + 2 n) (2 n + 7) Subject to the initial conditions A(0) = 0, A(1) = 2/5 B(0) = 1, B(1) = 6/5 A(n) Then, ----, approximates B(n) 1 1 / / | | 1 c = -1/2 | | ----------------- dx dy | | (3/2) / / 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 / / |-------------------| | | \ -x y + 1 / F(n) = -1/2 | | ---------------------- dx dy | | 3/2 | | 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 [2.000074094, 1.991087986, 2.001563686, 1.994414234, 1.994360859, 1.995035228, 2.000469086, 1.991778579, 2.002481790, 2.000873958, 1.990135670, 1.999670114, 1.993817269, 1.993156617, 1.994653020, 1.994686865, 1.996741042, 1.994291109, 1.993402726, 1.992437424, 1.989683533] 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 [0.09214088881, 0.09437281231, 0.09177179091, 0.09354559326, 0.09355885745, 0.09339129408, 0.09204299157, 0.09420096281, 0.09154442315, 0.09194266384, 0.09460987840, 0.09224103184, 0.09369396299, 0.09385820842, 0.09348625664, 0.09347784692, 0.09296767227, 0.09357619146, 0.09379701722, 0.09403706389, 0.09472246771] 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 c = -3/2 | | ----------------- dx dy | | (5/2) / / 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 , 2.003678214, yielding an irrationality measure that is approximately , 11.95911167 We hope that the reader can find nu exactly. Comment: Note that this constant appears to be , 8/5 - 6/5 ln(2) Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 4 (1 + n) (-3 + 2 n) (2 n + 1) X(n) - ------------------------------------ 3 (-1 + 2 n) (2 n + 9) 4 3 2 2 (88 n + 176 n + 14 n - 92 n - 9) X(1 + n) + ---------------------------------------------- + X(2 + n) = 0 3 (-1 + 2 n) (2 n + 9) Subject to the initial conditions -10 A(0) = 0, A(1) = --- 21 B(0) = 1, B(1) = 2/21 A(n) Then, ----, approximates B(n) 1 1 / / | | 1 c = -3/2 | | ----------------- dx dy | | (5/2) / / 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 / / |-------------------| | | \ -x y + 1 / F(n) = -3/2 | | ---------------------- dx dy | | 5/2 | | 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 [2.000594592, 2.001334526, 2.002382625, 1.999651972, 1.993099490, 2.001937519, 1.995529156, 2.002895737, 2.003678214, 2.001864945, 2.000556870, 1.994367837, 1.998575934, 1.999386196, 1.994327508, 1.993789557, 1.987660937, 1.999894677, 1.992070806, 1.990817420, 1.992989712] 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 [0.09201188907, 0.09182855698, 0.09156897673, 0.09224552951, 0.09387241315, 0.09167920002, 0.09326859824, 0.09144194046, 0.09124827174, 0.09169717390, 0.09202123704, 0.09355712333, 0.09251236153, 0.09231142356, 0.09356714565, 0.09370085151, 0.09522640999, 0.09218536241, 0.09412826030, 0.09444015559, 0.09389971067] 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(2/3, 2/3) | | (1/3) (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 , 2.101160490, yielding an irrationality measure that is approximately , 15.78268217 We hope that the reader can find nu exactly. 1/2 Pi 3 Comment: Note that this constant appears to be , ------- 3 Prove it! 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) (15 n + 26) (3 n + 5) X(n) - --------------------------------------------- 2 (15 n + 11) (4 + 3 n) (3 n + 7) (2 + n) 3 2 3 (3 n + 5) (165 n + 616 n + 725 n + 266) X(1 + n) + ---------------------------------------------------- + X(2 + n) = 0 (15 n + 11) (4 + 3 n) (3 n + 7) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -11/2 B(0) = 1, B(1) = -3 A(n) Then, ----, approximates B(n) 1 1 / / 1 | | 1 c = -------------- | | ------------------------------ dx dy Beta(2/3, 2/3) | | (1/3) (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(2/3, 2/3) | | 1/3 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 [2.101160490, 2.097767542, 2.093462586, 2.093483246, 2.094682519, 2.092932543, 2.094264714, 2.092147113, 2.089031033, 2.096983951, 2.092274059, 2.085968414, 2.081868786, 2.083904547, 2.080946082, 2.074282070, 2.077517462, 2.081031397, 2.080021976, 2.076981428, 2.071081366] 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 [0.06764672260, 0.06845103193, 0.06947328162, 0.06946837105, 0.06918339923, 0.06959928026, 0.06928266102, 0.06978604257, 0.07052763812, 0.06863695701, 0.06975585248, 0.07125751305, 0.07223608432, 0.07174992974, 0.07245657805, 0.07405173860, 0.07327669253, 0.07243618697, 0.07267749772, 0.07340502347, 0.07481957729] 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 c = 2/9 | | ------------------------ dx dy | | (1/3) (2/3) / / 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 , 2.101001862, yielding an irrationality measure that is approximately , 15.77447527 We hope that the reader can find nu exactly. 1/2 2 Pi 3 Comment: Note that this constant appears to be , --------- 9 Prove it! 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 + 26) (1 + 3 n) (1 + n) X(n) - ----------------------------------------- (15 n + 11) (3 n + 7) (3 n + 5) (4 + 3 n) 3 2 9 (165 n + 616 n + 725 n + 266) X(1 + n) + ------------------------------------------ + X(2 + n) = 0 (3 n + 5) (3 n + 7) (15 n + 11) Subject to the initial conditions A(0) = 0, A(1) = -11/4 B(0) = 1, B(1) = -9/4 A(n) Then, ----, approximates B(n) 1 1 / / | | 1 c = 2/9 | | ------------------------ dx dy | | (1/3) (2/3) / / 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) = 2/9 | | ---------------------- dx dy | | 1/3 2/3 | | 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 [2.101001862, 2.098163569, 2.093858413, 2.093878874, 2.095077947, 2.093327772, 2.094659743, 2.092541944, 2.089425666, 2.097378385, 2.092668294, 2.086362452, 2.082262626, 2.084298189, 2.081339527, 2.074675318, 2.077910512, 2.081424251, 2.079864778, 2.077373889, 2.070924325] 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 [0.06768429890, 0.06835708998, 0.06937920737, 0.06937434495, 0.06908947073, 0.06950532598, 0.06918880986, 0.06969215011, 0.07043366260, 0.06854336034, 0.06966210704, 0.07116355122, 0.07214199805, 0.07165597602, 0.07236254748, 0.07395747532, 0.07318261263, 0.07234230120, 0.07271508703, 0.07331106208, 0.07485727918] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 7, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Pi Beta(1/6, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (2/3) (1/3) (5/6) 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 , 2.185882783, yielding an irrationality measure that is approximately , 21.85574620 We hope that the reader can find nu exactly. (2/3) Comment: Note that this constant appears to be , 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (1 + 2 n) (1 + 6 n) (5 n + 8) (6 n + 7) X(n) -1/12 -------------------------------------------- (5 n + 3) (6 n + 11) (1 + n) (2 + n) 2 (6 n + 7) (330 n + 748 n + 351) X(1 + n) + 1/6 ----------------------------------------- + X(2 + n) = 0 (5 n + 3) (6 n + 11) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -3/5 -11 B(0) = 1, B(1) = --- 30 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Pi Beta(1/6, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (2/3) (1/3) (5/6) 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 / | 1/2 | 3 F(n) = 1/2 |----------------- |Pi Beta(1/6, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 2/3 1/3 5/6 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 [2.177170220, 2.171522196, 2.177985967, 2.180808478, 2.181267519, 2.184037771, 2.184978185, 2.184958150, 2.183659155, 2.182527103, 2.182032195, 2.185882783, 2.183023279, 2.175763741, 2.179363261, 2.177846431, 2.179095616, 2.178554083, 2.181011535, 2.180604605, 2.182393273] 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 [0.04994053052, 0.05123599398, 0.04975368990, 0.04910772808, 0.04900274679, 0.04836964426, 0.04815489944, 0.04815947354, 0.04845612659, 0.04871479164, 0.04882791435, 0.04794841625, 0.04860140358, 0.05026283004, 0.04943838210, 0.04978564484, 0.04949963999, 0.04962360655, 0.04906128697, 0.04915435996, 0.04874537918] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 8, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (5/6) (1/6) (2/3) 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 , 2.198419016, yielding an irrationality measure that is approximately , 23.17528282 We hope that the reader can find nu exactly. (1/3) Comment: Note that this constant appears to be , 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (1 + 2 n) (1 + 3 n) (1 + 6 n) (3 n + 4) (10 n + 13) X(n) -1/12 -------------------------------------------------------- (3 n + 2) (10 n + 3) (1 + n) (3 n + 5) (2 + n) 2 (3 n + 4) (330 n + 539 n + 135) X(1 + n) + 1/3 ----------------------------------------- + X(2 + n) = 0 (10 n + 3) (3 n + 5) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -3/4 -7 B(0) = 1, B(1) = -- 12 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (5/6) (1/6) (2/3) 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) = 1/2 |----------------- |Pi Beta(1/3, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 2/3 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 [2.182960929, 2.186595620, 2.182052302, 2.186052749, 2.181919193, 2.171955320, 2.183915794, 2.182816988, 2.182397698, 2.186951901, 2.188556691, 2.188561255, 2.189005483, 2.194810684, 2.198419016, 2.193665461, 2.196767100, 2.192047684, 2.189717780, 2.193794569, 2.187416722] 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 [0.04861565070, 0.04778576165, 0.04882331795, 0.04790962895, 0.04885374700, 0.05113653699, 0.04839750433, 0.04864854302, 0.04874436780, 0.04770448465, 0.04733854640, 0.04733750604, 0.04723625466, 0.04591489213, 0.04509525350, 0.04617530029, 0.04547032948, 0.04654338185, 0.04707394407, 0.04614593640, 0.04759846580] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 9, : The following constant c. 1 c = ----------------- Beta(1/3, 5/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (1/6) (2/3) 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 , 2.303957650, yielding an irrationality measure that is approximately , 47.13074174 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , -3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (1 + 2 n) (1 + 3 n) (-1 + 6 n) (5 + 6 n) (15 n + 23) (2 n + 3) X(n) -1/4 ------------------------------------------------------------------- (15 n + 8) (6 n + 1) (3 n + 5) (6 n + 7) (1 + n) (2 + n) 2 (5 + 6 n) (2 n + 3) (330 n + 671 n + 256) X(1 + n) + 3/4 --------------------------------------------------- + X(2 + n) = 0 (15 n + 8) (3 n + 5) (6 n + 7) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2 B(0) = 1, B(1) = 9/8 A(n) Then, ----, approximates B(n) 1 c = ----------------- Beta(1/3, 5/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (1/6) (2/3) 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(1/3, 5/6) Pi 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/6 2/3 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 [2.296609103, 2.296149792, 2.296390472, 2.299964848, 2.295007538, 2.293217132, 2.296646290, 2.296818911, 2.293460922, 2.291257227, 2.289321219, 2.292820318, 2.295608643, 2.295531788, 2.298042883, 2.303073864, 2.301725245, 2.303957650, 2.303882663, 2.302955561, 2.300410218] 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 [0.02327402587, 0.02337397911, 0.02332160095, 0.02254435541, 0.02362263609, 0.02401263234, 0.02326593425, 0.02322837487, 0.02395951116, 0.02443989065, 0.02486228962, 0.02409910881, 0.02349176663, 0.02350849724, 0.02296213854, 0.02186926155, 0.02216199210, 0.02167751834, 0.02169378447, 0.02189493363, 0.02244759273] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 10, : The following constant c. 1 c = ----------------- Beta(1/6, 2/3) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (5/6) (1/3) 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 , 2.304316806, yielding an irrationality measure that is approximately , 47.29711586 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (15 n + 23) (2 n + 3) (6 n + 7) (1 + 6 n) (2 + 3 n) (1 + 2 n) X(n) -1/4 ------------------------------------------------------------------ (15 n + 8) (3 n + 4) (5 + 6 n) (6 n + 11) (1 + n) (2 + n) 2 (2 n + 3) (6 n + 7) (330 n + 671 n + 256) X(1 + n) + 3/4 --------------------------------------------------- + X(2 + n) = 0 (15 n + 8) (3 n + 4) (6 n + 11) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -4/5 -9 B(0) = 1, B(1) = -- 20 A(n) Then, ----, approximates B(n) 1 c = ----------------- Beta(1/6, 2/3) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (5/6) (1/3) 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(1/6, 2/3) Pi 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 5/6 1/3 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 [2.296971342, 2.296511849, 2.296752346, 2.300326539, 2.295369047, 2.293578459, 2.297007435, 2.297179875, 2.293821703, 2.291617827, 2.289681638, 2.293180556, 2.295968700, 2.295891664, 2.298402579, 2.303433379, 2.302084580, 2.304316806, 2.304241639, 2.303314357, 2.300768834] 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 [0.02319521078, 0.02329518823, 0.02324285795, 0.02246577173, 0.02354392622, 0.02393390213, 0.02318735842, 0.02314984418, 0.02388090807, 0.02436125327, 0.02478362688, 0.02402060262, 0.02341339292, 0.02343016036, 0.02288392440, 0.02179125371, 0.02208397864, 0.02159961763, 0.02161592032, 0.02181707788, 0.02236969184] 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 c = ----------------- Beta(5/6, 5/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (1/6) (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 , 2.303028120, yielding an irrationality measure that is approximately , 46.70553539 We hope that the reader can find nu exactly. 1/2 4 3 Comment: Note that this constant appears to be , ------ 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (6 n + 11) (2 n + 3) (15 n + 22) (1 + 3 n) (5 + 6 n) (1 + 2 n) X(n) -1/4 ------------------------------------------------------------------- (15 n + 7) (6 n + 13) (5 + 3 n) (6 n + 7) (1 + n) (2 + n) 2 (2 n + 3) (6 n + 11) (330 n + 649 n + 234) X(1 + n) + 3/4 ---------------------------------------------------- + X(2 + n) = 0 (15 n + 7) (6 n + 13) (5 + 3 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -5 B(0) = 1, B(1) = -15/7 A(n) Then, ----, approximates B(n) 1 c = ----------------- Beta(5/6, 5/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (1/6) (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(5/6, 5/6) Pi 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/6 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 [2.293876033, 2.295100184, 2.291601047, 2.297935003, 2.294169108, 2.295807107, 2.296230122, 2.293627188, 2.295320883, 2.285668253, 2.291973782, 2.295886176, 2.296712622, 2.293582425, 2.298129151, 2.302031621, 2.303028120, 2.302916451, 2.302495065, 2.300871130, 2.301565872] 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 [0.02386907228, 0.02360246350, 0.02436491228, 0.02298559885, 0.02380523061, 0.02344856543, 0.02335649664, 0.02392328541, 0.02355441207, 0.02566024458, 0.02428364068, 0.02343135489, 0.02325150122, 0.02393303804, 0.02294337889, 0.02209547556, 0.02187918797, 0.02190342089, 0.02199487482, 0.02234747262, 0.02219659660] 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 c = ----------------- Beta(1/6, 1/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (5/6) (5/6) 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 , 2.303622617, yielding an irrationality measure that is approximately , 46.97659403 We hope that the reader can find nu exactly. 1/2 3 Comment: Note that this constant appears to be , ---- 6 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (15 n + 14) (2 n + 3) (6 n + 7) (-1 + 3 n) (1 + 6 n) (1 + 2 n) X(n) -1/4 ------------------------------------------------------------------- (15 n - 1) (6 n + 5) (1 + 3 n) (6 n + 11) (1 + n) (2 + n) 2 (2 n + 3) (6 n + 7) (330 n + 143 n - 10) X(1 + n) + 3/4 -------------------------------------------------- + X(2 + n) = 0 (15 n - 1) (1 + 3 n) (6 n + 11) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1/5 B(0) = 1, B(1) = -3/5 A(n) Then, ----, approximates B(n) 1 c = ----------------- Beta(1/6, 1/6) Pi 1 1 / / | | 1 | | ---------------------------------------------- dx dy | | (5/6) (5/6) 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(1/6, 1/6) Pi 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 5/6 5/6 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 [2.295220970, 2.295462241, 2.292296386, 2.293423940, 2.296721817, 2.291664537, 2.295361485, 2.296482192, 2.294978650, 2.294497207, 2.290747768, 2.289138281, 2.294745924, 2.297659273, 2.294992439, 2.298379258, 2.301237843, 2.303622617, 2.302469945, 2.302742487, 2.302897360] 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 [0.02357616495, 0.02352363743, 0.02421331011, 0.02396756908, 0.02324950053, 0.02435106793, 0.02354557254, 0.02330164173, 0.02362892627, 0.02373376897, 0.02455101085, 0.02490222112, 0.02367960378, 0.02304556583, 0.02362592379, 0.02288899508, 0.02226782859, 0.02175019749, 0.02200032716, 0.02194117453, 0.02190756388] 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 c = 1/2 |----------------- |Pi Beta(1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) (1/3) 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 , 2.185290410, yielding an irrationality measure that is approximately , 21.79710223 We hope that the reader can find nu exactly. 2/3 2 Comment: Note that this constant appears to be , - ---- 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 n + 8) (5 + 6 n) (2 + 3 n) (-1 + 3 n) (1 + 2 n) X(n) -1/12 ------------------------------------------------------- (5 n + 3) (3 n + 1) (1 + n) (3 n + 4) (2 + n) 2 (2 + 3 n) (330 n + 748 n + 351) X(1 + n) + 1/6 ----------------------------------------- + X(2 + n) = 0 (5 n + 3) (3 n + 4) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -3/2 B(0) = 1, B(1) = 11/6 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) (1/3) 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) = 1/2 |----------------- |Pi Beta(1/2, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 1/3 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 [2.176574555, 2.170926832, 2.177390904, 2.180213714, 2.180673055, 2.183443607, 2.184384320, 2.184364584, 2.183065888, 2.181934134, 2.181439524, 2.185290410, 2.182431203, 2.175171963, 2.178771779, 2.177255246, 2.178504727, 2.177963490, 2.180421238, 2.180014602, 2.181803566] 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 [0.05007700506, 0.05137273638, 0.04988997798, 0.04924378006, 0.04913870293, 0.04850536779, 0.04829049909, 0.04829500609, 0.04859166755, 0.04885033136, 0.04896341518, 0.04808362189, 0.04873670990, 0.05039849716, 0.04957376853, 0.04992105286, 0.04963490647, 0.04975883720, 0.04919630504, 0.04928933472, 0.04888018106] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 14, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Pi Beta(1/2, 5/6) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/3) (2/3) (1/6) 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 , 2.198175192, yielding an irrationality measure that is approximately , 23.14810087 We hope that the reader can find nu exactly. (1/3) Comment: Note that this constant appears to be , -2 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (10 n + 13) (5 + 6 n) (-1 + 6 n) (1 + 2 n) X(n) -1/12 ----------------------------------------------- (10 n + 3) (6 n + 7) (1 + n) (2 + n) 2 (5 + 6 n) (330 n + 539 n + 135) X(1 + n) + 1/3 ----------------------------------------- + X(2 + n) = 0 (10 n + 3) (6 n + 7) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -3 B(0) = 1, B(1) = 7/6 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Pi Beta(1/2, 5/6) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/3) (2/3) (1/6) 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 / | 1/2 | 3 F(n) = 1/2 |----------------- |Pi Beta(1/2, 5/6) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/3 2/3 1/6 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 [2.182715381, 2.186350197, 2.181807002, 2.185807573, 2.181674140, 2.171710391, 2.183670988, 2.182572305, 2.182153138, 2.186707464, 2.188312377, 2.188317064, 2.188761414, 2.194566738, 2.198175192, 2.193421760, 2.196523521, 2.191804226, 2.189474445, 2.193551356, 2.187173630] 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 [0.04867176273, 0.04784175638, 0.04887939552, 0.04796558055, 0.04890977135, 0.05119277700, 0.04845342352, 0.04870446089, 0.04880026777, 0.04776024576, 0.04739424052, 0.04739317201, 0.04729188206, 0.04597035127, 0.04515059806, 0.04623073132, 0.04552565812, 0.04659879658, 0.04712938698, 0.04620125332, 0.04765390884] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 15, : The following constant c. / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/4, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/4) (3/4) (5/4) 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 , 2.287775258, yielding an irrationality measure that is approximately , 40.68279447 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 2 2 Prove it! 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 + 4 n) (3 + 4 n) (20 n + 57 n + 41) X(n) -1/16 ---------------------------------------------- 2 (20 n + 17 n + 4) (4 n + 9) (1 + n) (2 + n) 4 3 2 (3 + 4 n) (3520 n + 13552 n + 18284 n + 9947 n + 1824) X(1 + n) + 1/8 ------------------------------------------------------------------ 2 (20 n + 17 n + 4) (4 n + 9) (1 + 2 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2/5 -7 B(0) = 1, B(1) = -- 40 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/4, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/4) (3/4) (5/4) 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 / | 1/2 | 2 F(n) = 1/2 |------------------ |Pi Beta(-1/4, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/4 3/4 5/4 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 [2.269393305, 2.278924981, 2.279201018, 2.286381242, 2.287712517, 2.286035419, 2.272542320, 2.282496681, 2.287775258, 2.285540015, 2.281660600, 2.287275495, 2.283447694, 2.284313067, 2.271688665, 2.272230280, 2.277910735, 2.277391744, 2.278665074, 2.272863265, 2.266401840] 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 [0.02923050166, 0.02713651554, 0.02707600072, 0.02550440147, 0.02521354191, 0.02557998456, 0.02853775996, 0.02635405210, 0.02519983820, 0.02568827950, 0.02653710808, 0.02530900514, 0.02614591166, 0.02595658788, 0.02872546063, 0.02860636301, 0.02735892711, 0.02747277266, 0.02719350075, 0.02846720866, 0.02988944927] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 16, : The following constant c. / | 1/2 | 2 c = 1/2 |----------------- |Pi Beta(1/4, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (3/4) (1/4) (3/4) 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 , 2.288623710, yielding an irrationality measure that is approximately , 40.97672112 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 2 Prove it! 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 (4 n + 5) (20 n + 59 n + 43) (1 + 4 n) (1 + 2 n) X(n) -1/16 ------------------------------------------------------- 2 (20 n + 19 n + 4) (4 n + 7) (1 + n) (2 n + 3) (2 + n) 4 3 2 (4 n + 5) (3520 n + 13904 n + 18684 n + 9765 n + 1632) X(1 + n) + 1/8 ------------------------------------------------------------------ 2 (20 n + 19 n + 4) (4 n + 7) (2 n + 3) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2/3 -11 B(0) = 1, B(1) = --- 24 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |----------------- |Pi Beta(1/4, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (3/4) (1/4) (3/4) 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 / | 1/2 | 2 F(n) = 1/2 |----------------- |Pi Beta(1/4, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 3/4 1/4 3/4 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 [2.273767881, 2.275197647, 2.280703380, 2.283664815, 2.285256340, 2.288623710, 2.284340401, 2.282075569, 2.282471986, 2.280826111, 2.286964576, 2.284200875, 2.284362232, 2.287444223, 2.280187893, 2.275148905, 2.270662772, 2.273989039, 2.273580009, 2.279389981, 2.276125161] 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 [0.02826840476, 0.02795434738, 0.02674676706, 0.02609840399, 0.02575030093, 0.02501455777, 0.02595060897, 0.02644624442, 0.02635945802, 0.02671988062, 0.02537693321, 0.02598112898, 0.02594583380, 0.02527214603, 0.02685970917, 0.02796505072, 0.02895112345, 0.02821981342, 0.02830968633, 0.02703457900, 0.02775071549] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 17, : The following constant c. / | 1/2 | 2 c = 1/2 |----------------- |Pi Beta(1/2, 3/4) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/4) (3/4) (1/4) 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 , 2.288229093, yielding an irrationality measure that is approximately , 40.83948861 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , -2 Prove it! 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 + 2 n) (-1 + 4 n) (3 + 4 n) (20 n + 59 n + 43) X(n) -1/16 -------------------------------------------------------- 2 (20 n + 19 n + 4) (1 + n) (2 n + 3) (5 + 4 n) (2 + n) 4 3 2 (3 + 4 n) (3520 n + 13904 n + 18684 n + 9765 n + 1632) X(1 + n) + 1/8 ------------------------------------------------------------------ 2 (20 n + 19 n + 4) (2 n + 3) (5 + 4 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2 B(0) = 1, B(1) = 11/8 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |----------------- |Pi Beta(1/2, 3/4) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/4) (3/4) (1/4) 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 / | 1/2 | 2 F(n) = 1/2 |----------------- |Pi Beta(1/2, 3/4) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/4 3/4 1/4 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 [2.273372267, 2.274802233, 2.280308165, 2.283269800, 2.284861524, 2.288229093, 2.283945982, 2.281681349, 2.282077965, 2.280432287, 2.286570951, 2.283807447, 2.283969002, 2.287051190, 2.279795058, 2.274756266, 2.270270330, 2.273596794, 2.273187960, 2.278998128, 2.275733505] 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 [0.02835533784, 0.02804118344, 0.02683335561, 0.02618483944, 0.02583663423, 0.02510072380, 0.02603688913, 0.02653256440, 0.02644571983, 0.02680615987, 0.02546294340, 0.02606719746, 0.02603185305, 0.02535800930, 0.02694579517, 0.02805127911, 0.02903747401, 0.02830599798, 0.02839584288, 0.02712047903, 0.02783669212] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 18, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) (4/3) 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 , 2.191852083, yielding an irrationality measure that is approximately , 22.46479950 We hope that the reader can find nu exactly. (2/3) Comment: Note that this constant appears to be , 2 2 Prove it! 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 n + 2) (30 n + 85 n + 61) (5 + 6 n) (-1 + 3 n) (1 + 2 n) X(n) -1/12 ----------------------------------------------------------------- 2 (30 n + 25 n + 6) (3 n + 7) (1 + n) (3 n + 1) (2 + n) 4 3 2 (3 n + 2) (2970 n + 11385 n + 15384 n + 8428 n + 1566) X(1 + n) + 1/3 ------------------------------------------------------------------ 2 (30 n + 25 n + 6) (3 n + 7) (3 n + 1) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3/8 B(0) = 1, B(1) = -1/6 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) (4/3) 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) = 1/2 |------------------ |Pi Beta(-1/3, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 4/3 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 [2.183821583, 2.179350250, 2.180444175, 2.184596541, 2.188226014, 2.191852083, 2.191210271, 2.188739602, 2.188263779, 2.191306061, 2.190409201, 2.189657648, 2.186796501, 2.180661159, 2.182785586, 2.181614734, 2.185395471, 2.182245478, 2.181507123, 2.183297740, 2.188598581] 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 [0.04841902353, 0.04944135986, 0.04919105803, 0.04824203776, 0.04741392935, 0.04658790314, 0.04673401437, 0.04729685366, 0.04740531968, 0.04671220478, 0.04691643860, 0.04708764436, 0.04773993390, 0.04914142395, 0.04865571904, 0.04892335373, 0.04805963979, 0.04877916058, 0.04894795843, 0.04853869323, 0.04732899770] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 19, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (5/6) (1/6) (5/3) 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 , 2.205160127, yielding an irrationality measure that is approximately , 23.95292310 We hope that the reader can find nu exactly. (1/3) Comment: Note that this constant appears to be , 2 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/12 (1 + 2 n) (5 + 6 n) (1 + 3 n) (-2 + 3 n) (1 + 6 n) 3 2 / (90 n + 333 n + 392 n + 146) X(n) / ((2 + n) (-1 + 6 n) (3 n + 2) / 3 2 (1 + n) (3 n + 8) (90 n + 63 n - 4 n - 3)) + 1/6 (1 + 3 n) ( 6 5 4 3 2 106920 n + 484704 n + 787698 n + 532944 n + 103641 n - 28678 n - 8532) / 3 2 X(1 + n) / ((90 n + 63 n - 4 n - 3) (3 n + 8) (3 n + 2) (-1 + 6 n) / (2 + n)) + X(2 + n) = 0 Subject to the initial conditions -3 A(0) = 0, A(1) = -- 10 -1 B(0) = 1, B(1) = -- 30 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (5/6) (1/6) (5/3) 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) = 1/2 |------------------ |Pi Beta(-2/3, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 5/3 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 [2.187421036, 2.189146211, 2.190598577, 2.190086117, 2.192586893, 2.187284138, 2.188292846, 2.191218014, 2.190861825, 2.195345492, 2.193622586, 2.186909020, 2.196906807, 2.198826336, 2.205160127, 2.201329474, 2.201162808, 2.199372318, 2.201381951, 2.199804341, 2.192833459] 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 [0.04759748194, 0.04720418309, 0.04687330708, 0.04699003113, 0.04642067060, 0.04762870408, 0.04739869309, 0.04673225140, 0.04681335667, 0.04579332846, 0.04618505200, 0.04771426626, 0.04543859787, 0.04500281077, 0.04356743564, 0.04443507349, 0.04447285587, 0.04487892384, 0.04442317777, 0.04478091579, 0.04636456758] 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 |------------------ |Pi Beta(-1/2, 1/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (2/3) (1/3) (5/6) (3/2) | / / 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 , 2.192384032, yielding an irrationality measure that is approximately , 22.52072613 We hope that the reader can find nu exactly. 2/3 6 2 Comment: Note that this constant appears to be , ------ 5 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 3 2 -1/12 (90 n + 261 n + 236 n + 71) (1 + 6 n) (-5 + 6 n) (2 + 3 n) (-1 + 2 n) / 3 2 X(n) / ((90 n - 9 n - 16 n + 6) (6 n + 11) (1 + n) (-1 + 3 n) (2 + n)) / 5 4 3 2 + 1/6 (1 + 6 n) (17820 n + 39798 n + 17010 n - 4431 n - 370 n + 1332) / 3 2 X(1 + n) / ((90 n - 9 n - 16 n + 6) (6 n + 11) (-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) = 7/12 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-1/2, 1/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (2/3) (1/3) (5/6) (3/2) | / / 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 / | 1/2 | 3 F(n) = 1/2 |------------------ |Pi Beta(-1/2, 1/6) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 2/3 1/3 5/6 3/2 | | | 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 [2.187127039, 2.180790291, 2.173766211, 2.186701715, 2.185997463, 2.180981260, 2.185327329, 2.192384032, 2.189762144, 2.188326331, 2.184295402, 2.186064348, 2.190112977, 2.179120985, 2.179248480, 2.185770859, 2.184226776, 2.185880650, 2.189168567, 2.185992120, 2.188561616] 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 [0.04766453572, 0.04911188783, 0.05072091116, 0.04776155729, 0.04792224525, 0.04906821090, 0.04807519433, 0.04646683360, 0.04706383655, 0.04739105939, 0.04831080512, 0.04790698210, 0.04698391253, 0.04949383330, 0.04946465200, 0.04797395957, 0.04832647765, 0.04794890303, 0.04719908839, 0.04792346454, 0.04733742375] 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 = 1/2 |------------------ |Pi Beta(-1/2, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (2/3) (3/2) | / / 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 , 2.201499947, yielding an irrationality measure that is approximately , 23.52433314 We hope that the reader can find nu exactly. 1/3 3 2 Comment: Note that this constant appears to be , ------ 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/12 (-1 + 2 n) (5 + 6 n) (-2 + 3 n) (1 + 3 n) (1 + 6 n) 3 2 / (90 n + 279 n + 275 n + 89) X(n) / ((2 + n) (-1 + 6 n) (3 n + 2) / 3 2 (1 + n) (3 n + 5) (90 n + 9 n - 13 n + 3)) + 1/3 (1 + 3 n) 5 4 3 2 / (17820 n + 40392 n + 18981 n - 2274 n - 259 n + 414) X(1 + n) / ( / 3 2 (90 n + 9 n - 13 n + 3) (3 n + 5) (-1 + 6 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 3/4 B(0) = 1, B(1) = 2/3 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/2, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (2/3) (3/2) | / / 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) = 1/2 |------------------ |Pi Beta(-1/2, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 2/3 3/2 | | | 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 [2.184784625, 2.190982108, 2.189944550, 2.194613467, 2.192284368, 2.189593607, 2.190908687, 2.182816988, 2.187390700, 2.187436463, 2.192577304, 2.192361781, 2.194502514, 2.197021027, 2.200052966, 2.199383493, 2.201499947, 2.195167345, 2.196010270, 2.198408600, 2.190772890] 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 [0.04819909194, 0.04678596647, 0.04702228072, 0.04595972736, 0.04648951461, 0.04710223566, 0.04680268533, 0.04864854302, 0.04760440047, 0.04759396364, 0.04642285258, 0.04647189728, 0.04598495306, 0.04541265655, 0.04472452145, 0.04487638847, 0.04439643091, 0.04583381874, 0.04564226155, 0.04509761766, 0.04683360939] 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 = 1/2 |------------------ |Pi Beta(-1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (1/3) (3/2) | / / 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 , 2.190743816, yielding an irrationality measure that is approximately , 22.34916910 We hope that the reader can find nu exactly. 2/3 3 2 Comment: Note that this constant appears to be , ------ 4 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 3 2 -1/12 (-1 + 2 n) (-4 + 3 n) (2 + 3 n) (5 + 6 n) (90 n + 261 n + 236 n + 71) / 3 2 X(n) / ((2 + n) (3 n + 1) (1 + n) (3 n + 4) (90 n - 9 n - 16 n + 6)) / 5 4 3 2 (17820 n + 39798 n + 17010 n - 4431 n - 370 n + 1332) X(1 + n) + 1/6 ------------------------------------------------------------------ 3 2 (90 n - 9 n - 16 n + 6) (3 n + 4) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 3/2 B(0) = 1, B(1) = 7/3 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (1/3) (3/2) | / / 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) = 1/2 |------------------ |Pi Beta(-1/2, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 1/3 3/2 | | | 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 [2.185481024, 2.179495004, 2.173171020, 2.185058190, 2.184354767, 2.179688584, 2.184384320, 2.190743816, 2.188122753, 2.187036255, 2.183702604, 2.184427428, 2.189284829, 2.178641044, 2.178309256, 2.184137222, 2.182593957, 2.184595743, 2.188578144, 2.184361752, 2.186932062] 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 [0.04804011156, 0.04940823173, 0.05085747989, 0.04813663425, 0.04829724796, 0.04936393264, 0.04829049909, 0.04684023043, 0.04743747152, 0.04768524315, 0.04844620138, 0.04828065488, 0.04717259430, 0.04960369765, 0.04967966142, 0.04834693033, 0.04869951247, 0.04824221998, 0.04733365623, 0.04829565283, 0.04770901012] 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. / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-1/2, 5/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (1/6) (3/2) | / / 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 , 2.201950778, yielding an irrationality measure that is approximately , 23.57629328 We hope that the reader can find nu exactly. 1/3 6 2 Comment: Note that this constant appears to be , ------ 7 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 3 2 (90 n + 279 n + 275 n + 89) (5 + 6 n) (-7 + 6 n) (-1 + 2 n) X(n) -1/12 ------------------------------------------------------------------ 3 2 (90 n + 9 n - 13 n + 3) (6 n + 7) (1 + n) (2 + n) 5 4 3 2 (17820 n + 40392 n + 18981 n - 2274 n - 259 n + 414) X(1 + n) + 1/3 ----------------------------------------------------------------- 3 2 (90 n + 9 n - 13 n + 3) (6 n + 7) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 3 B(0) = 1, B(1) = 14/3 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-1/2, 5/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (1/6) (3/2) | / / 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 / | 1/2 | 3 F(n) = 1/2 |------------------ |Pi Beta(-1/2, 5/6) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/3 2/3 1/6 3/2 | | | 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 [2.185239098, 2.190736557, 2.190398565, 2.194368163, 2.192737925, 2.189348551, 2.191361788, 2.181592857, 2.186864517, 2.187191900, 2.193029494, 2.192117464, 2.194954251, 2.196776955, 2.200504249, 2.199139666, 2.201950778, 2.193949345, 2.195486721, 2.197191819, 2.191222820] 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 [0.04809533519, 0.04684188354, 0.04691886112, 0.04601550003, 0.04638630444, 0.04715807373, 0.04669951720, 0.04892835570, 0.04772441806, 0.04764974172, 0.04631996655, 0.04652749995, 0.04588225609, 0.04546809106, 0.04462217478, 0.04493171053, 0.04429425094, 0.04611073684, 0.04576123134, 0.04537386915, 0.04673115714] 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. / | 1/2 | 2 c = 1/2 |------------------- |Pi Beta(-1/2, -1/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (3/4) (5/4) (3/2) | / / 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 , 2.293102466, yielding an irrationality measure that is approximately , 42.60145699 We hope that the reader can find nu exactly. 1/2 8 2 Comment: Note that this constant appears to be , ------ 5 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 3 2 -1/16 (80 n + 192 n + 127 n + 27) (3 + 4 n) (-1 + 4 n) (-5 + 4 n) (1 + 4 n) / 3 2 X(n) / ((80 n - 48 n - 17 n + 12) (4 n + 9) (1 + n) (-3 + 4 n) (2 + n)) / + 1/8 (-1 + 4 n) 6 5 4 3 2 (56320 n + 92928 n - 7424 n - 33216 n + 5428 n + 7617 n + 864) / 3 2 X(1 + n) / ((80 n - 48 n - 17 n + 12) (4 n + 9) (-1 + 2 n) (-3 + 4 n) / (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 2/5 B(0) = 1, B(1) = 3/8 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |------------------- |Pi Beta(-1/2, -1/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (3/4) (5/4) (3/2) | / / 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 / | 1/2 | 2 F(n) = 1/2 |------------------- |Pi Beta(-1/2, -1/4) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/4 3/4 5/4 3/2 | | | 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 [2.270729331, 2.281113949, 2.287184648, 2.291463010, 2.289882986, 2.286184702, 2.287535681, 2.290594121, 2.293102466, 2.287698214, 2.290677203, 2.291235289, 2.288368018, 2.285478275, 2.284144941, 2.281606899, 2.278022157, 2.282122938, 2.285186157, 2.276195299, 2.275417678] 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 [0.02893647965, 0.02665683004, 0.02532885202, 0.02439501331, 0.02473968721, 0.02554735590, 0.02525216788, 0.02458452816, 0.02403761965, 0.02521666598, 0.02456640398, 0.02444467514, 0.02507038738, 0.02570177742, 0.02599336453, 0.02654886789, 0.02733448894, 0.02643587329, 0.02576564659, 0.02773532026, 0.02790603318] 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. / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/2, 1/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/4) (1/4) (3/4) (3/2) | / / 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 , 2.295235921, yielding an irrationality measure that is approximately , 43.42157667 We hope that the reader can find nu exactly. 1/2 4 2 Comment: Note that this constant appears to be , ------ 3 Prove it! 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 -1/16 (-1 + 2 n) (3 + 4 n) (-3 + 4 n) (1 + 4 n) (80 n + 240 n + 227 n + 71) / 3 X(n) / ((2 + n) (-1 + 4 n) (2 n + 1) (1 + n) (4 n + 7) (80 n - 13 n + 4) / ) + 1/8 (1 + 4 n) 6 5 4 3 2 (56320 n + 154880 n + 119808 n + 17280 n - 6108 n + 2395 n + 1184) / 3 X(1 + n) / ((80 n - 13 n + 4) (4 n + 7) (2 n + 1) (-1 + 4 n) (2 + n)) / + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 2/3 B(0) = 1, B(1) = 5/8 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/2, 1/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/4) (1/4) (3/4) (3/2) | / / 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 / | 1/2 | 2 F(n) = 1/2 |------------------ |Pi Beta(-1/2, 1/4) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 3/4 1/4 3/4 3/2 | | | 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 [2.278301602, 2.276769046, 2.286675528, 2.290954147, 2.292283372, 2.293147282, 2.284625561, 2.290639183, 2.294353691, 2.290681322, 2.295235921, 2.290074519, 2.292202315, 2.286687023, 2.289268289, 2.285054084, 2.277145681, 2.287536039, 2.284681116, 2.283885389, 2.277723592] 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 [0.02727320342, 0.02760940062, 0.02544009108, 0.02450599377, 0.02421614709, 0.02402785348, 0.02588823853, 0.02457469789, 0.02376502626, 0.02456550544, 0.02357290979, 0.02469789295, 0.02423381744, 0.02543757924, 0.02487384280, 0.02579452586, 0.02752675771, 0.02525208968, 0.02587608837, 0.02605014529, 0.02739997578] 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. / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/2, 3/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (3/4) (1/4) (3/2) | / / 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 , 2.294842170, yielding an irrationality measure that is approximately , 43.26784751 We hope that the reader can find nu exactly. 1/2 4 2 Comment: Note that this constant appears to be , ------ 5 Prove it! 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 + 2 n) (-5 + 4 n) (3 + 4 n) (80 n + 240 n + 227 n + 71) X(n) -1/16 ------------------------------------------------------------------- + 1/8 3 (2 + n) (2 n + 1) (1 + n) (4 n + 5) (80 n - 13 n + 4) 6 5 4 3 2 (56320 n + 154880 n + 119808 n + 17280 n - 6108 n + 2395 n + 1184) / 3 X(1 + n) / ((80 n - 13 n + 4) (4 n + 5) (2 n + 1) (2 + n)) + X(2 + n) = / 0 Subject to the initial conditions A(0) = 0, A(1) = 2 B(0) = 1, B(1) = 25/8 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/2, 3/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (3/4) (1/4) (3/2) | / / 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 / | 1/2 | 2 F(n) = 1/2 |------------------ |Pi Beta(-1/2, 3/4) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/4 3/4 1/4 3/2 | | | 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 [2.277905861, 2.276373504, 2.286280187, 2.290559005, 2.291888429, 2.292752538, 2.284231016, 2.290244836, 2.293959543, 2.290287372, 2.294842170, 2.289680966, 2.291808959, 2.286293865, 2.288875327, 2.284661320, 2.275946782, 2.287143668, 2.284288942, 2.283493411, 2.277331810] 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 [0.02735999615, 0.02769620650, 0.02552648690, 0.02459218885, 0.02430225001, 0.02411388137, 0.02597453576, 0.02466073107, 0.02385088016, 0.02465145046, 0.02365864502, 0.02478377354, 0.02431957730, 0.02552349754, 0.02495962387, 0.02588041782, 0.02778987160, 0.02533780496, 0.02596186493, 0.02613590808, 0.02748592137] 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 = 1/2 |------------------- |Pi Beta(-2/3, -2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (5/3) (5/3) | / / 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 , 2.191740202, yielding an irrationality measure that is approximately , 22.45307219 We hope that the reader can find nu exactly. 2/3 27 2 Comment: Note that this constant appears to be , ------- 20 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/108 (3 n + 1) (30 n + 59 n + 20) (1 + 6 n) (-5 + 6 n) (5 + 6 n) (-2 + 3 n) / 2 X(n) / ((30 n - n - 9) (3 n + 8) (1 + n) (2 n - 1) (-4 + 3 n) (2 + n)) / + 1/9 4 3 2 (3 n + 1) (1 + 6 n) (2970 n + 3861 n - 1836 n - 665 n + 432) X(1 + n) ------------------------------------------------------------------------ 2 (30 n - n - 9) (3 n + 8) (2 n - 1) (-4 + 3 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 2/5 11 B(0) = 1, B(1) = -- 27 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------- |Pi Beta(-2/3, -2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (5/3) (5/3) | / / 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 F(n) = 1/2 |------------------- |Pi Beta(-2/3, -2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 5/3 5/3 | | | 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 [2.182845844, 2.183180478, 2.182264558, 2.178715139, 2.184450585, 2.184156488, 2.181149460, 2.187564562, 2.189085446, 2.188727372, 2.189541458, 2.182205065, 2.180409161, 2.185272693, 2.185726233, 2.188081521, 2.186280317, 2.186658013, 2.191029668, 2.191740202, 2.185908855] 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 [0.04864194890, 0.04856548498, 0.04877479935, 0.04958673490, 0.04827536679, 0.04834253022, 0.04902974452, 0.04756475016, 0.04721803104, 0.04729964127, 0.04711411775, 0.04878839812, 0.04919906776, 0.04808766623, 0.04798414447, 0.04744687218, 0.04785770102, 0.04777152727, 0.04677513676, 0.04661337039, 0.04794246627] 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 = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (2/3) (5/3) | / / 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 , 2.202210289, yielding an irrationality measure that is approximately , 23.60630722 We hope that the reader can find nu exactly. 2/3 7 2 Comment: Note that this constant appears to be , ------ 27 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/12 (-2 + 3 n) (1 + 6 n) (-3 + 2 n) (1 + 3 n) (5 + 6 n) 3 2 / (90 n + 225 n + 164 n + 43) X(n) / ((2 + n) (-1 + 3 n) (6 n - 1) / 3 2 (1 + n) (3 n + 5) (90 n - 45 n - 16 n + 14)) + 1/6 (1 + 3 n) 6 5 4 3 2 (106920 n + 142560 n - 49518 n - 20448 n + 28689 n + 5678 n - 476) / 3 2 X(1 + n) / ((90 n - 45 n - 16 n + 14) (3 n + 5) (6 n - 1) (-1 + 3 n) / (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1 33 B(0) = 1, B(1) = -- 28 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (2/3) (5/3) | / / 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 F(n) = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 2/3 5/3 | | | 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 [2.181824632, 2.188221166, 2.190654395, 2.196817139, 2.196887472, 2.194194647, 2.187042608, 2.185227791, 2.188706073, 2.186032902, 2.192913193, 2.192459904, 2.197986130, 2.198311949, 2.199961531, 2.199599822, 2.202210289, 2.200243771, 2.191372547, 2.200545681, 2.196295307] 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 [0.04887536496, 0.04741503461, 0.04686059487, 0.04545896392, 0.04544298931, 0.04605495447, 0.04768379403, 0.04809791634, 0.04730449602, 0.04791415801, 0.04634642639, 0.04644956771, 0.04519351646, 0.04511955545, 0.04474526047, 0.04482731035, 0.04423544235, 0.04468124625, 0.04669706769, 0.04461277940, 0.04557750193] 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 = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (2/3) (5/3) | / / 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 , 2.192355474, yielding an irrationality measure that is approximately , 22.51771660 We hope that the reader can find nu exactly. 2/3 6 2 Comment: Note that this constant appears to be , ------ 5 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/108 (-2 + 3 n) (5 + 6 n) (-5 + 6 n) (1 + 3 n) (1 + 6 n) 3 2 / (90 n + 261 n + 236 n + 71) X(n) / ((2 + n) (-1 + 3 n) (2 n + 1) / 3 2 (1 + n) (3 n + 5) (90 n - 9 n - 16 n + 6)) + 1/18 (1 + 3 n) (1 + 6 n) 5 4 3 2 / (17820 n + 39798 n + 17010 n - 4431 n - 370 n + 1332) X(1 + n) / ( / 3 2 (90 n - 9 n - 16 n + 6) (3 n + 5) (2 n + 1) (-1 + 3 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1 35 B(0) = 1, B(1) = -- 36 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (2/3) (5/3) | / / 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 F(n) = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 2/3 5/3 | | | 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 [2.187098380, 2.180761646, 2.173737581, 2.186673099, 2.185968862, 2.180952673, 2.185298756, 2.192355474, 2.189733600, 2.188297802, 2.184266887, 2.186035847, 2.190084491, 2.179092513, 2.179220023, 2.185742416, 2.184198347, 2.185852235, 2.189140166, 2.185963734, 2.188533243] 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 [0.04767107263, 0.04911843960, 0.05072747961, 0.04776808560, 0.04792877213, 0.04907474886, 0.04808171672, 0.04647333258, 0.04707033975, 0.04739756324, 0.04831731721, 0.04791348597, 0.04699040153, 0.04950035024, 0.04947116515, 0.04798045103, 0.04833297029, 0.04795538780, 0.04720556069, 0.04792994237, 0.04734389138] 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. / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-2/3, 5/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (1/6) (5/3) | / / 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 , 2.190438829, yielding an irrationality measure that is approximately , 22.31755702 We hope that the reader can find nu exactly. 2/3 3 2 Comment: Note that this constant appears to be , ------ 4 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -4/27 (-2 + 3 n) (1 + 3 n) (-4 + 3 n) (5 + 6 n) (2 + 3 n) 3 2 / (90 n + 261 n + 236 n + 71) X(n) / ((2 + n) (1 + 6 n) (2 n + 1) (1 + n) / 3 2 (6 n + 7) (90 n - 9 n - 16 n + 6)) + 2/9 (1 + 3 n) 5 4 3 2 / (17820 n + 39798 n + 17010 n - 4431 n - 370 n + 1332) X(1 + n) / ( / 3 2 (90 n - 9 n - 16 n + 6) (6 n + 7) (2 n + 1) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 4 B(0) = 1, B(1) = 56/9 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-2/3, 5/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (1/6) (5/3) | / / 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 |------------------ |Pi Beta(-2/3, 5/6) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/3 2/3 1/6 5/3 | | | 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 [2.185174958, 2.179189093, 2.172865264, 2.184752588, 2.184049319, 2.179383290, 2.184079180, 2.190438829, 2.187817920, 2.186731575, 2.183398077, 2.184123054, 2.188980608, 2.178336976, 2.178005340, 2.183833458, 2.182290346, 2.184292283, 2.188274836, 2.184058596, 2.186629058] 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 [0.04810997719, 0.04947824441, 0.05092765051, 0.04820640680, 0.04836700673, 0.04943379820, 0.04836018661, 0.04690969041, 0.04750697567, 0.04775474528, 0.04851576956, 0.04835016615, 0.04724192382, 0.04967331443, 0.04974925347, 0.04841631105, 0.04876890490, 0.04831151742, 0.04740279894, 0.04836488789, 0.04777813304] 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. / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-2/3, 1/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (5/6) (5/3) | / / 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 , 2.201328023, yielding an irrationality measure that is approximately , 23.50457846 We hope that the reader can find nu exactly. 2/3 7 2 Comment: Note that this constant appears to be , ------ 24 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -4/27 (-2 + 3 n) (1 + 3 n) (-4 + 3 n) (1 + 6 n) (2 + 3 n) 3 2 / (90 n + 225 n + 164 n + 43) X(n) / ((2 + n) (-1 + 2 n) (6 n - 1) / 3 2 (1 + n) (6 n + 11) (90 n - 45 n - 16 n + 14)) + 2/9 (1 + 3 n) 6 5 4 3 2 (106920 n + 142560 n - 49518 n - 20448 n + 28689 n + 5678 n - 476) / 3 2 X(1 + n) / ((90 n - 45 n - 16 n + 14) (6 n + 11) (6 n - 1) (-1 + 2 n) / (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 4/5 88 B(0) = 1, B(1) = --- 105 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-2/3, 1/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (5/6) (5/3) | / / 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 |------------------ |Pi Beta(-2/3, 1/6) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/3 2/3 5/6 5/3 | | | 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 [2.180935237, 2.187682117, 2.190815061, 2.195929089, 2.195999870, 2.193656684, 2.186853933, 2.184341529, 2.187820256, 2.185496022, 2.193073212, 2.191575423, 2.197102092, 2.197776146, 2.199773612, 2.198717113, 2.201328023, 2.199709041, 2.191531925, 2.199664739, 2.195414806] 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 [0.04907873658, 0.04753794271, 0.04682400590, 0.04566070712, 0.04564462455, 0.04617729656, 0.04772683246, 0.04830027109, 0.04750644301, 0.04803668821, 0.04631002054, 0.04665088096, 0.04539424607, 0.04524118843, 0.04478788631, 0.04502759771, 0.04443540241, 0.04480253381, 0.04666078337, 0.04481258367, 0.04577757522] 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. / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-1/3, 1/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (2/3) (1/3) (5/6) (4/3) | / / 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 , 2.201659443, yielding an irrationality measure that is approximately , 23.54268951 We hope that the reader can find nu exactly. 1/3 3 2 Comment: Note that this constant appears to be , ------ 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -4/27 (-1 + 3 n) (2 + 3 n) (-2 + 3 n) (1 + 6 n) (1 + 3 n) 3 2 / (90 n + 279 n + 275 n + 89) X(n) / ((2 + n) (-1 + 6 n) (2 n + 1) / 3 2 (1 + n) (6 n + 11) (90 n + 9 n - 13 n + 3)) + 4/9 (2 + 3 n) (1 + 3 n) 5 4 3 2 / (17820 n + 40392 n + 18981 n - 2274 n - 259 n + 414) X(1 + n) / ( / 3 2 (90 n + 9 n - 13 n + 3) (6 n + 11) (2 n + 1) (-1 + 6 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 2/5 16 B(0) = 1, B(1) = -- 45 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-1/3, 1/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (2/3) (1/3) (5/6) (4/3) | / / 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 |------------------ |Pi Beta(-1/3, 1/6) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 2/3 1/3 5/6 4/3 | | | 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 [2.184945410, 2.191142813, 2.190105173, 2.194774009, 2.192444829, 2.189753988, 2.191068986, 2.182977206, 2.187550839, 2.187596520, 2.192737281, 2.192521678, 2.194662331, 2.197180764, 2.200212622, 2.199543069, 2.201659443, 2.195326761, 2.196169607, 2.198567857, 2.190932067] 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 [0.04816238218, 0.04674937384, 0.04698569024, 0.04592322952, 0.04645299822, 0.04706569473, 0.04676618398, 0.04861193133, 0.04756787967, 0.04755746227, 0.04638645097, 0.04643551046, 0.04594861828, 0.04537637969, 0.04468831070, 0.04484018534, 0.04436027917, 0.04579758560, 0.04560605963, 0.04506147160, 0.04679736136] 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 = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (1/3) (4/3) | / / 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 , 2.202385457, yielding an irrationality measure that is approximately , 23.62660965 We hope that the reader can find nu exactly. 1/3 6 2 Comment: Note that this constant appears to be , ------ 7 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/108 (-1 + 3 n) (1 + 6 n) (-7 + 6 n) (2 + 3 n) (5 + 6 n) 3 2 / (90 n + 279 n + 275 n + 89) X(n) / ((2 + n) (1 + 3 n) (2 n + 1) (1 + n) / 3 2 (3 n + 4) (90 n + 9 n - 13 n + 3)) + 1/9 (2 + 3 n) 5 4 3 2 / (17820 n + 40392 n + 18981 n - 2274 n - 259 n + 414) X(1 + n) / ( / 3 2 (90 n + 9 n - 13 n + 3) (3 n + 4) (2 n + 1) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1 B(0) = 1, B(1) = 14/9 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (1/3) (4/3) | / / 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 F(n) = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 1/3 4/3 | | | 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 [2.185677290, 2.191174527, 2.190836314, 2.194805692, 2.193175233, 2.189785639, 2.191798655, 2.182029505, 2.187300946, 2.187628108, 2.193465484, 2.192553234, 2.195389802, 2.197212288, 2.200939364, 2.199574563, 2.202385457, 2.194383806, 2.195920965, 2.197625846, 2.191656630] 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 [0.04799531486, 0.04674215284, 0.04681916608, 0.04591602696, 0.04628681111, 0.04705848372, 0.04660006469, 0.04882852928, 0.04762487060, 0.04755025885, 0.04622078564, 0.04642832975, 0.04578325791, 0.04536922077, 0.04452351384, 0.04483304057, 0.04419575073, 0.04601194324, 0.04566255305, 0.04527531321, 0.04663239452] 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 = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (1/3) (4/3) | / / 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 , 2.187633591, yielding an irrationality measure that is approximately , 22.03093275 We hope that the reader can find nu exactly. 1/3 2 Comment: Note that this constant appears to be , ---- 3 Prove it! 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 + 3 n) (5 + 6 n) (-1 + 2 n) (2 + 3 n) (15 n + 35 n + 19) X(n) -1/12 ------------------------------------------------------------------ 2 (15 n + 5 n - 1) (3 n + 4) (1 + n) (1 + 3 n) (2 + n) 2 2 (165 n + 220 n - 31) (2 + 3 n) X(1 + n) + 1/3 ----------------------------------------- + X(2 + n) = 0 2 (15 n + 5 n - 1) (3 n + 4) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = 1 B(0) = 1, B(1) = 2 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (1/3) (4/3) | / / 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 F(n) = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 1/3 4/3 | | | 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 [2.180128330, 2.174204213, 2.175014753, 2.175991907, 2.180119319, 2.181560650, 2.183277964, 2.187633591, 2.185319971, 2.182774003, 2.185752691, 2.184390480, 2.184512858, 2.177625096, 2.181276773, 2.180937869, 2.171264675, 2.182181233, 2.182372363, 2.182312469, 2.182726480] 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 [0.04926331451, 0.05062043269, 0.05043454399, 0.05021053157, 0.04926537612, 0.04893571961, 0.04854321151, 0.04754900850, 0.04807687394, 0.04865836603, 0.04797810599, 0.04828909237, 0.04826114650, 0.04983633645, 0.04900063065, 0.04907813463, 0.05129513669, 0.04879384569, 0.04875015843, 0.04876384822, 0.04866922629] 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. / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-1/3, 5/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (2/3) (1/3) (1/6) (4/3) | / / 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 , 2.187473386, yielding an irrationality measure that is approximately , 22.01478592 We hope that the reader can find nu exactly. 1/3 2 Comment: Note that this constant appears to be , ---- 4 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -4/27 (15 n + 35 n + 19) (1 + 3 n) (5 + 6 n) (-2 + 3 n) (2 + 3 n) (-1 + 3 n) / 2 X(n) / ((6 n + 1) (15 n + 5 n - 1) (1 + n) (6 n + 7) (1 + 2 n) (2 + n)) / 2 2 (1 + 3 n) (165 n + 220 n - 31) (2 + 3 n) X(1 + n) + 4/9 --------------------------------------------------- + X(2 + n) = 0 2 (15 n + 5 n - 1) (6 n + 7) (1 + 2 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = 2 B(0) = 1, B(1) = 16/3 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |------------------ |Pi Beta(-1/3, 5/6) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (2/3) (1/3) (1/6) (4/3) | / / 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 |------------------ |Pi Beta(-1/3, 5/6) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 2/3 1/3 1/6 4/3 | | | 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 [2.179267410, 2.174043523, 2.174154702, 2.175831379, 2.179260135, 2.181400283, 2.182419645, 2.187473386, 2.184462516, 2.182613959, 2.184896097, 2.184230597, 2.183657124, 2.177465374, 2.180421898, 2.180778307, 2.170410657, 2.182021831, 2.181519200, 2.182153226, 2.181874169] 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 [0.04946031941, 0.05065729304, 0.05063178964, 0.05024732603, 0.04946198446, 0.04897238799, 0.04873935158, 0.04758554304, 0.04827264227, 0.04869494120, 0.04817364084, 0.04832560501, 0.04845659055, 0.04987292007, 0.04919615406, 0.04911462884, 0.05149131956, 0.04883028354, 0.04894519703, 0.04880024765, 0.04886404000] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 36, : The following constant c. / | | 1 c = 1/2 |------------------- |Pi Beta(-1/3, -1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (4/3) (4/3) | / / 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 , 2.192272986, yielding an irrationality measure that is approximately , 22.50902829 We hope that the reader can find nu exactly. 1/3 5 2 Comment: Note that this constant appears to be , ------ 12 Prove it! 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 + 3 n) (5 + 6 n) (-1 + 2 n) (3 n + 2) (30 n + 67 n + 32) X(n) -1/12 ------------------------------------------------------------------ 2 (30 n + 7 n - 5) (3 n + 7) (1 + n) (-2 + 3 n) (2 + n) 4 3 2 (3 n + 2) (5940 n + 13266 n + 5874 n - 751 n - 410) X(1 + n) + 1/6 --------------------------------------------------------------- 2 (30 n + 7 n - 5) (3 n + 7) (-2 + 3 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1/4 11 B(0) = 1, B(1) = -- 40 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------- |Pi Beta(-1/3, -1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (4/3) (4/3) | / / 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 F(n) = 1/2 |------------------- |Pi Beta(-1/3, -1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 4/3 4/3 | | | 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 [2.183534251, 2.183789966, 2.180603948, 2.182718941, 2.187502685, 2.191298626, 2.191263323, 2.188976086, 2.189421292, 2.189874173, 2.192272986, 2.188128506, 2.187639150, 2.187159193, 2.186445336, 2.185644800, 2.181322945, 2.187785550, 2.187035887, 2.187816104, 2.182132884] 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 [0.04848465992, 0.04842624552, 0.04915451024, 0.04867094917, 0.04757886125, 0.04671389755, 0.04672193531, 0.04724295441, 0.04714149846, 0.04703831365, 0.04649210493, 0.04743615988, 0.04754774082, 0.04765720174, 0.04782004914, 0.04800273027, 0.04899007246, 0.04751435681, 0.04768532709, 0.04750738976, 0.04880489756] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 37, : The following constant c. / | | 1 c = 1/2 |------------------- |Pi Beta(-1/3, -1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (4/3) (4/3) | / / 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 , 2.202501267, yielding an irrationality measure that is approximately , 23.64005152 We hope that the reader can find nu exactly. 1/3 27 2 Comment: Note that this constant appears to be , ------- 14 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/108 (3 n + 2) (30 n + 55 n + 16) (5 + 6 n) (-7 + 6 n) (1 + 6 n) (-1 + 3 n) / 2 X(n) / ((30 n - 5 n - 9) (3 n + 7) (1 + n) (2 n - 1) (-2 + 3 n) (2 + n)) / + 1/18 (3 n + 2) 5 4 3 2 / (35640 n + 47520 n - 15282 n - 13296 n + 2393 n + 702) X(1 + n) / ( / 2 (30 n - 5 n - 9) (3 n + 7) (2 n - 1) (-2 + 3 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1/4 49 B(0) = 1, B(1) = --- 216 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------- |Pi Beta(-1/3, -1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (4/3) (4/3) | / / 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 F(n) = 1/2 |------------------- |Pi Beta(-1/3, -1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 4/3 4/3 | | | 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 [2.182862047, 2.186733534, 2.192577440, 2.191781410, 2.198109841, 2.193639896, 2.186535416, 2.182772499, 2.181989923, 2.187976278, 2.190145566, 2.196144067, 2.199755101, 2.196491982, 2.202501267, 2.199536361, 2.196769496, 2.196911587, 2.189498053, 2.195069526, 2.191959107] 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 [0.04863824625, 0.04775429837, 0.04642282163, 0.04660399015, 0.04516543279, 0.04618111490, 0.04779949700, 0.04865870973, 0.04883757773, 0.04747086774, 0.04697648898, 0.04561186224, 0.04479208537, 0.04553282259, 0.04416951080, 0.04484170714, 0.04546978526, 0.04543751223, 0.04712400773, 0.04585605294, 0.04656354264] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 38, : The following constant c. / | | 1 c = 1/2 |------------------- |Pi Beta(-1/2, -1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (3/2) (4/3) | / / 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 , 2.315047928, yielding an irrationality measure that is approximately , 52.87391450 We hope that the reader can find nu exactly. 1/2 16 3 Comment: Note that this constant appears to be , ------- 7 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/36 (-1 + 3 n) (1 + 6 n) (-7 + 6 n) (-1 + 2 n) (5 + 6 n) 3 2 / (180 n + 432 n + 277 n + 45) X(n) / ((2 + n) (-5 + 6 n) (3 n - 2) / 3 2 7 (1 + n) (2 n + 5) (180 n - 108 n - 47 n + 20)) + 1/12 (855360 n 6 5 4 3 2 + 1197504 n - 582768 n - 743904 n + 155388 n + 111900 n - 25005 n / 3 2 - 3200) X(1 + n) / ((180 n - 108 n - 47 n + 20) (2 n + 5) (3 n - 2) / (-5 + 6 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 2/9 B(0) = 1, B(1) = 7/36 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------- |Pi Beta(-1/2, -1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (3/2) (4/3) | / / 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 F(n) = 1/2 |------------------- |Pi Beta(-1/2, -1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 3/2 4/3 | | | 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 [2.305654973, 2.299688157, 2.295787952, 2.300490131, 2.304093973, 2.304077040, 2.301378648, 2.305111696, 2.307182628, 2.302101134, 2.300469365, 2.291907198, 2.304400187, 2.308762929, 2.307588189, 2.307461390, 2.298683852, 2.315047928, 2.311303593, 2.308667682, 2.312242615] 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 [0.02130947462, 0.02260447951, 0.02345273489, 0.02243023248, 0.02164794857, 0.02165162140, 0.02223725136, 0.02142724873, 0.02097844799, 0.02208038497, 0.02243474362, 0.02429815779, 0.02158153404, 0.02063623927, 0.02089060327, 0.02091806638, 0.02282277130, 0.01927751182, 0.02008654786, 0.02065685818, 0.01988353335] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 39, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Pi Beta(1/6, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (5/6) (2/3) | / / 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 , 2.197476754, yielding an irrationality measure that is approximately , 23.07058946 We hope that the reader can find nu exactly. 2/3 2 Comment: Note that this constant appears to be , ---- 4 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -4/27 (3 n + 4) (30 n + 71 n + 40) (2 + 3 n) (1 + 6 n) (-1 + 3 n) (1 + 3 n) / 2 X(n) / ((30 n + 11 n - 1) (6 n + 11) (1 + n) (6 n + 5) (1 + 2 n) (2 + n) / 3 2 (2 + 3 n) (3 n + 4) (1980 n + 3366 n + 734 n - 71) X(1 + n) ) + 2/9 ------------------------------------------------------------- 2 (30 n + 11 n - 1) (6 n + 11) (1 + 2 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2/5 -14 B(0) = 1, B(1) = --- 15 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Pi Beta(1/6, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (2/3) (5/6) (2/3) | / / 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 |----------------- |Pi Beta(1/6, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/3 2/3 5/6 2/3 | | | 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 [2.182725992, 2.184436729, 2.185617834, 2.184699168, 2.188734138, 2.182841097, 2.177674271, 2.177848277, 2.185495622, 2.187202588, 2.179270125, 2.187108146, 2.190681123, 2.193867949, 2.197476754, 2.191856547, 2.196812372, 2.195700930, 2.188396381, 2.197003388, 2.191758529] 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 [0.04866933781, 0.04827853091, 0.04800888498, 0.04821860418, 0.04729809908, 0.04864303367, 0.04982507364, 0.04978522208, 0.04803677951, 0.04764730396, 0.04945969801, 0.04766884507, 0.04685450784, 0.04612924783, 0.04530916593, 0.04658688703, 0.04546004665, 0.04571255182, 0.04737509020, 0.04541666258, 0.04660919856] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 40, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (2/3) (2/3) | / / 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 , 2.197868057, yielding an irrationality measure that is approximately , 23.11395151 We hope that the reader can find nu exactly. 2/3 2 Comment: Note that this constant appears to be , ---- 6 Prove it! 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 + 3 n) (1 + 6 n) (-1 + 2 n) (3 n + 4) (30 n + 71 n + 40) X(n) -1/12 ----------------------------------------------------------------- 2 (30 n + 11 n - 1) (3 n + 5) (1 + n) (2 + 3 n) (2 + n) 3 2 (3 n + 4) (1980 n + 3366 n + 734 n - 71) X(1 + n) + 1/6 --------------------------------------------------- + X(2 + n) = 0 2 (30 n + 11 n - 1) (3 n + 5) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1/2 B(0) = 1, B(1) = -7/4 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (2/3) (2/3) | / / 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 F(n) = 1/2 |----------------- |Pi Beta(1/3, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 2/3 2/3 | | | 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 [2.183120061, 2.184830599, 2.186011506, 2.185092641, 2.189127413, 2.183234174, 2.178067150, 2.178240958, 2.185888105, 2.187594875, 2.179662214, 2.187500039, 2.191072819, 2.194259448, 2.197868057, 2.192247654, 2.197203282, 2.196091645, 2.188786900, 2.197393713, 2.192148658] 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 [0.04857928945, 0.04818859509, 0.04791904060, 0.04812876926, 0.04720846699, 0.04855321647, 0.04973509919, 0.04969529980, 0.04794720169, 0.04755783740, 0.04936996696, 0.04757946468, 0.04676531120, 0.04604021953, 0.04522032164, 0.04649787004, 0.04537126597, 0.04562377256, 0.04728607310, 0.04532802210, 0.04652040036] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 41, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (2/3) (2/3) | / / 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 , 2.185998816, yielding an irrationality measure that is approximately , 21.86727026 We hope that the reader can find nu exactly. (2/3) Comment: Note that this constant appears to be , 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 n + 8) (3 n + 4) (6 n + 7) (1 + 6 n) (5 + 6 n) (1 + 3 n) X(n) -1/108 ---------------------------------------------------------------- (5 n + 3) (3 n + 5) (1 + n) (2 n + 3) (2 + 3 n) (2 + n) 2 (3 n + 4) (6 n + 7) (330 n + 748 n + 351) X(1 + n) + 1/18 --------------------------------------------------- + X(2 + n) = 0 (5 n + 3) (3 n + 5) (2 n + 3) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1/2 -11 B(0) = 1, B(1) = --- 36 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (2/3) (2/3) | / / 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 F(n) = 1/2 |----------------- |Pi Beta(1/3, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 2/3 2/3 | | | 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 [2.177286897, 2.171638815, 2.178102527, 2.180924978, 2.181383961, 2.184154154, 2.185094510, 2.185074416, 2.183775363, 2.182643253, 2.182148286, 2.185998816, 2.183139254, 2.175879658, 2.179479119, 2.177962231, 2.179211358, 2.178669767, 2.181127161, 2.180720173, 2.182508784] 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 [0.04991380247, 0.05120921325, 0.04972699815, 0.04908108290, 0.04897612020, 0.04834306328, 0.04812834258, 0.04813292992, 0.04842958119, 0.04868824639, 0.04880137687, 0.04792193649, 0.04857490407, 0.05023625978, 0.04941186704, 0.04975912551, 0.04947314839, 0.04959712196, 0.04903484401, 0.04912792558, 0.04871897843] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 42, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (5/6) (1/6) 1/2 (2/3) | / / 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 , 2.304432490, yielding an irrationality measure that is approximately , 47.35095530 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (1 + 3 n) (1 + 6 n) (15 n + 23) (6 n + 7) X(n) -1/36 ---------------------------------------------- (15 n + 8) (1 + n) (3 n + 5) (2 + n) 2 (6 n + 7) (330 n + 671 n + 256) X(1 + n) + 1/4 ----------------------------------------- + X(2 + n) = 0 (15 n + 8) (3 n + 5) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2/3 B(0) = 1, B(1) = -3/8 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/3, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (5/6) (1/6) 1/2 (2/3) | / / 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 F(n) = 1/2 |----------------- |Pi Beta(1/3, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 1/2 2/3 | | | 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 [2.297088019, 2.296628467, 2.296868906, 2.300443040, 2.295485489, 2.293694843, 2.297123760, 2.297296141, 2.293937911, 2.291733976, 2.289797729, 2.293296588, 2.296084675, 2.296007580, 2.298518437, 2.303549179, 2.302200322, 2.304432490, 2.304357265, 2.303429925, 2.300884345] 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 [0.02316982707, 0.02326981239, 0.02321749733, 0.02244046243, 0.02351857636, 0.02390854559, 0.02316205167, 0.02312455212, 0.02385559249, 0.02433592679, 0.02475829217, 0.02399531848, 0.02338815117, 0.02340493062, 0.02285873420, 0.02176612987, 0.02205885299, 0.02157452837, 0.02159084284, 0.02179200310, 0.02234460232] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 43, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/6, 1/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (5/6) (1/3) (2/3) | / / 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 , 2.188227157, yielding an irrationality measure that is approximately , 22.09096440 We hope that the reader can find nu exactly. 1/3 2 Comment: Note that this constant appears to be , ---- 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (6 n + 7) (15 n + 35 n + 19) (-1 + 2 n) (1 + 6 n) X(n) -1/12 ------------------------------------------------------- 2 (15 n + 5 n - 1) (6 n + 11) (1 + n) (2 + n) 2 (2 + 3 n) (6 n + 7) (165 n + 220 n - 31) X(1 + n) + 1/3 -------------------------------------------------- + X(2 + n) = 0 2 (15 n + 5 n - 1) (6 n + 11) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1/5 B(0) = 1, B(1) = -2/5 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/6, 1/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (5/6) (1/3) (2/3) | / / 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(1/6, 1/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 5/6 1/3 2/3 | | | 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 [2.180723995, 2.174799578, 2.175609817, 2.176586670, 2.180713783, 2.182154814, 2.183871828, 2.188227157, 2.185913239, 2.183366972, 2.186345362, 2.184982853, 2.185104934, 2.178216875, 2.181868255, 2.181529055, 2.171855565, 2.182771826, 2.182962661, 2.182902471, 2.183316187] 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 [0.04912705138, 0.05048388566, 0.05029811427, 0.05007422901, 0.04912938716, 0.04879988466, 0.04840754671, 0.04741366877, 0.04794146579, 0.04852287591, 0.04784285958, 0.04815383372, 0.04812596286, 0.04970081426, 0.04886539202, 0.04894294370, 0.05115944177, 0.04865886353, 0.04861525493, 0.04862900884, 0.04853447863] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 44, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (2/3) (1/3) (2/3) | / / 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 , 2.185130569, yielding an irrationality measure that is approximately , 21.78133211 We hope that the reader can find nu exactly. 2/3 2 Comment: Note that this constant appears to be , - ---- 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (1 + 3 n) (-1 + 3 n) (2 + 3 n) (5 + 6 n) (5 n + 8) (3 n + 4) X(n) -4/27 ----------------------------------------------------------------- (5 n + 3) (6 n + 1) (2 n + 3) (6 n + 7) (1 + n) (2 + n) 2 (2 + 3 n) (3 n + 4) (330 n + 748 n + 351) X(1 + n) + 2/9 --------------------------------------------------- + X(2 + n) = 0 (5 n + 3) (2 n + 3) (6 n + 7) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2 B(0) = 1, B(1) = 22/9 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (2/3) (1/3) (2/3) | / / 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(1/3, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 2/3 1/3 2/3 | | | 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 [2.176413826, 2.170766185, 2.177230337, 2.180053228, 2.180512650, 2.183283283, 2.184224077, 2.184204421, 2.182905806, 2.181774133, 2.181279603, 2.185130569, 2.182271443, 2.175012282, 2.178612179, 2.177095725, 2.178345287, 2.177804129, 2.180261957, 2.179855401, 2.181644445] 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 [0.05011383623, 0.05140963967, 0.04992675891, 0.04928049721, 0.04917539419, 0.04854199626, 0.04832709405, 0.04833158309, 0.04862824675, 0.04888691009, 0.04899998350, 0.04812011063, 0.04877322562, 0.05043511058, 0.04961030593, 0.04995759635, 0.04967141149, 0.04979533275, 0.04923274319, 0.04932576103, 0.04891656067] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 45, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (5/3) (2/3) | / / 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 , 2.204609168, yielding an irrationality measure that is approximately , 23.88741265 We hope that the reader can find nu exactly. 2/3 7 2 Comment: Note that this constant appears to be , ------ 12 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/12 (1 + 3 n) (1 + 6 n) (-1 + 2 n) (-2 + 3 n) (5 + 6 n) 3 2 / (90 n + 279 n + 263 n + 67) X(n) / ((2 + n) (-1 + 3 n) (6 n - 1) / 3 2 (1 + n) (3 n + 8) (90 n + 9 n - 25 n - 7)) + 1/3 (1 + 3 n) 6 5 4 3 2 (53460 n + 156816 n + 117639 n - 30204 n - 52575 n - 9236 n + 1064) / 3 2 X(1 + n) / ((90 n + 9 n - 25 n - 7) (3 n + 8) (6 n - 1) (-1 + 3 n) / (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/5 -2 B(0) = 1, B(1) = -- 35 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (5/3) (2/3) | / / 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 F(n) = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 5/3 2/3 | | | 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 [2.191197149, 2.186531067, 2.195337391, 2.193438438, 2.193874797, 2.186858759, 2.191131499, 2.187725687, 2.191118779, 2.196618311, 2.191690903, 2.195921311, 2.196780645, 2.198903682, 2.204609168, 2.196816164, 2.203680231, 2.204108536, 2.198437744, 2.202709991, 2.201127980] 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 [0.04673700207, 0.04780048923, 0.04579516963, 0.04622693764, 0.04612769044, 0.04772573156, 0.04675194996, 0.04752800728, 0.04675484623, 0.04550412600, 0.04662459261, 0.04566247443, 0.04546725293, 0.04498525866, 0.04369213836, 0.04545918537, 0.04390245920, 0.04380547604, 0.04509100275, 0.04412222198, 0.04448075156] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 46, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (5/3) (2/3) | / / 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 , 2.186749203, yielding an irrationality measure that is approximately , 21.94209093 We hope that the reader can find nu exactly. 2/3 3 2 Comment: Note that this constant appears to be , ------ 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/108 (6 n + 7) (15 n + 35 n + 17) (-2 + 3 n) (1 + 6 n) (5 + 6 n) (1 + 3 n) / 2 X(n) / ((15 n + 5 n - 3) (3 n + 8) (1 + n) (2 n + 1) (-1 + 3 n) (2 + n)) / 2 (3 n + 5) (1 + 3 n) (6 n + 7) (165 n + 220 n + 72) X(1 + n) + 1/9 ------------------------------------------------------------ 2 (15 n + 5 n - 3) (3 n + 8) (2 n + 1) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/5 -2 B(0) = 1, B(1) = -- 45 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (5/3) (2/3) | / / 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 F(n) = 1/2 |------------------ |Pi Beta(-2/3, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 5/3 2/3 | | | 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 [2.177971366, 2.179712335, 2.171090900, 2.177124579, 2.173962032, 2.183049405, 2.178649026, 2.185167221, 2.182602241, 2.184817222, 2.186749203, 2.182444139, 2.182791204, 2.183103082, 2.179005887, 2.180998883, 2.179682471, 2.183926589, 2.186070773, 2.178321359, 2.171882277] 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 [0.04975703356, 0.04935849769, 0.05133504991, 0.04995098621, 0.05067598708, 0.04859543385, 0.04960187028, 0.04811174343, 0.04869761924, 0.04819164932, 0.04775072381, 0.04873375328, 0.04865443521, 0.04858316898, 0.04952017854, 0.04906418048, 0.04936533149, 0.04839503864, 0.04790551593, 0.04967689021, 0.05115330832] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 47, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) 1/2 (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 , 2.304189033, yielding an irrationality measure that is approximately , 47.23779232 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , -3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (15 n + 23) (5 + 6 n) (-1 + 6 n) (2 + 3 n) X(n) -1/36 ----------------------------------------------- (15 n + 8) (1 + n) (3 n + 4) (2 + n) 2 (5 + 6 n) (330 n + 671 n + 256) X(1 + n) + 1/4 ----------------------------------------- + X(2 + n) = 0 (15 n + 8) (3 n + 4) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -4/3 B(0) = 1, B(1) = 3/4 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) 1/2 (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 F(n) = 1/2 |----------------- |Pi Beta(1/2, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 1/2 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 [2.296842471, 2.296383043, 2.296623605, 2.300197864, 2.295240437, 2.293449913, 2.296878954, 2.297051458, 2.293693351, 2.291489539, 2.289553415, 2.293052397, 2.295840606, 2.295763635, 2.298274614, 2.303305478, 2.301956743, 2.304189033, 2.304113930, 2.303186712, 2.300641253] 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 [0.02322324883, 0.02332321761, 0.02327087033, 0.02249372723, 0.02357192656, 0.02396190987, 0.02321531119, 0.02317778098, 0.02390887064, 0.02438922813, 0.02481161063, 0.02404853076, 0.02344127381, 0.02345802801, 0.02291174840, 0.02181900440, 0.02211173134, 0.02162733015, 0.02164361984, 0.02184477442, 0.02239740445] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 48, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (3/2) (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 , 2.311032962, yielding an irrationality measure that is approximately , 50.63993009 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 4 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 3 2 -1/36 (2 + 3 n) (-1 + 6 n) (-1 + 2 n) (5 + 6 n) (180 n + 648 n + 745 n + 261) / 3 2 X(n) / ((2 + n) (3 n + 1) (1 + n) (2 n + 5) (180 n + 108 n - 11 n - 16) / ) + 1/12 (5 + 6 n) 5 4 3 2 / (23760 n + 85536 n + 98448 n + 29232 n - 11061 n - 5248) X(1 + n) / ( / 3 2 (180 n + 108 n - 11 n - 16) (2 n + 5) (3 n + 1) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -4/9 -7 B(0) = 1, B(1) = -- 72 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/2, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (3/2) (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 F(n) = 1/2 |------------------ |Pi Beta(-1/2, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 3/2 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 [2.299746492, 2.297082881, 2.297290485, 2.293255005, 2.299761183, 2.296479744, 2.299211318, 2.304535257, 2.299090329, 2.298850734, 2.294295827, 2.293922874, 2.304052095, 2.301401487, 2.306433306, 2.301564408, 2.302770786, 2.311032962, 2.306370491, 2.304204663, 2.308919295] 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 [0.02259180290, 0.02317094484, 0.02312578248, 0.02400437955, 0.02258861050, 0.02330217443, 0.02270811174, 0.02155224156, 0.02273440983, 0.02278649203, 0.02377762938, 0.02385886819, 0.02165703212, 0.02223229181, 0.02114079130, 0.02219691449, 0.02193503290, 0.02014507269, 0.02115440274, 0.02162394010, 0.02060239122] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 49, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-3/2, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (5/2) (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 , 2.311575799, yielding an irrationality measure that is approximately , 50.93087270 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 8 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/36 (2 + 3 n) (1 + 6 n) (-1 + 6 n) (-3 + 2 n) (5 + 6 n) 3 2 / (180 n + 540 n + 439 n + 39) X(n) / ((2 + n) (-5 + 6 n) (3 n - 2) / 3 (1 + n) (2 n + 7) (180 n - 101 n - 40)) + 1/12 (5 + 6 n) ( 6 5 4 3 2 142560 n + 308880 n - 16272 n - 349464 n - 110058 n + 38109 n + 12160) / 3 X(1 + n) / ((180 n - 101 n - 40) (2 n + 7) (3 n - 2) (-5 + 6 n) (2 + n)) / + X(2 + n) = 0 Subject to the initial conditions -4 A(0) = 0, A(1) = -- 15 -1 B(0) = 1, B(1) = -- 60 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-3/2, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (5/2) (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 F(n) = 1/2 |------------------ |Pi Beta(-3/2, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 5/2 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 [2.305527962, 2.299158104, 2.298637808, 2.298571161, 2.297086998, 2.297733238, 2.301094130, 2.304700347, 2.308300925, 2.300109326, 2.299775601, 2.298207097, 2.305082648, 2.303624171, 2.305620836, 2.306598477, 2.305455783, 2.311575799, 2.310365558, 2.309085000, 2.306997270] 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 [0.02133700622, 0.02271967814, 0.02283278148, 0.02284727119, 0.02317004919, 0.02302947891, 0.02229903921, 0.02151644095, 0.02073626091, 0.02251296359, 0.02258547743, 0.02292642952, 0.02143354666, 0.02174986035, 0.02131687420, 0.02110500198, 0.02135265280, 0.02002768920, 0.02028942969, 0.02056652404, 0.02101860167] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 50, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (4/3) (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 , 2.200592125, yielding an irrationality measure that is approximately , 23.42039476 We hope that the reader can find nu exactly. (1/3) Comment: Note that this constant appears to be , 3 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence -1/108 (2 + 3 n) (1 + 6 n) (-1 + 6 n) (-1 + 3 n) (5 + 6 n) 3 2 / (90 n + 333 n + 398 n + 149) X(n) / ((2 + n) (1 + 3 n) (2 n + 1) / 3 2 (1 + n) (3 n + 7) (90 n + 63 n + 2 n - 6)) + 1/18 (5 + 6 n) (2 + 3 n) 5 4 3 2 / (17820 n + 59994 n + 61668 n + 14493 n - 4087 n - 1206) X(1 + n) / ( / 3 2 (90 n + 63 n + 2 n - 6) (3 n + 7) (2 n + 1) (1 + 3 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/2 -11 B(0) = 1, B(1) = --- 72 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (4/3) (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 F(n) = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 4/3 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 [2.185327598, 2.185762709, 2.189411482, 2.192168097, 2.192826246, 2.187638574, 2.187073051, 2.179450714, 2.183050233, 2.189443931, 2.191122426, 2.192693429, 2.195042215, 2.197792177, 2.200592125, 2.200006588, 2.199251900, 2.186687732, 2.195348202, 2.194490045, 2.188885664] 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 [0.04807513292, 0.04797581961, 0.04714373380, 0.04651597620, 0.04636620872, 0.04754787217, 0.04767685005, 0.04941836763, 0.04859524465, 0.04713633989, 0.04675401583, 0.04639642892, 0.04586226088, 0.04523754881, 0.04460224767, 0.04473504067, 0.04490624496, 0.04776474729, 0.04579271254, 0.04598778802, 0.04726356267] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 51, : The following constant c. / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (4/3) (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 , 2.191298626, yielding an irrationality measure that is approximately , 22.40690571 We hope that the reader can find nu exactly. 1/3 2 Comment: Note that this constant appears to be , ---- 2 Prove it! 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 + 44 n + 31) X(n) -1/12 ----------------------------------------------------------------- 2 (15 n + 14 n + 2) (3 n + 7) (1 + n) (1 + 3 n) (2 + n) 4 3 2 (2 + 3 n) (2970 n + 11682 n + 15141 n + 7195 n + 950) X(1 + n) + 1/6 ----------------------------------------------------------------- 2 (15 n + 14 n + 2) (3 n + 7) (1 + 3 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/2 B(0) = 1, B(1) = -7/8 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (4/3) (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 F(n) = 1/2 |------------------ |Pi Beta(-1/3, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 4/3 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 [2.177768486, 2.178795676, 2.176836778, 2.177255146, 2.187672278, 2.191298626, 2.188874133, 2.184858376, 2.186743508, 2.189649029, 2.188270702, 2.182921786, 2.178090032, 2.182999488, 2.178220988, 2.173799935, 2.179512772, 2.182044193, 2.177989342, 2.183352877, 2.184672556] 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 [0.04980349576, 0.04956829799, 0.05001692202, 0.04992107577, 0.04754018636, 0.04671389755, 0.04726619078, 0.04818225310, 0.04775202301, 0.04708960812, 0.04740374139, 0.04862459516, 0.04972985939, 0.04860683984, 0.04969987247, 0.05071317413, 0.04940416552, 0.04882517164, 0.04975291702, 0.04852609613, 0.04822468059] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 52, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/6, 2/3) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/3) (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 , 2.198375287, yielding an irrationality measure that is approximately , 23.17040313 We hope that the reader can find nu exactly. (1/3) Comment: Note that this constant appears to be , 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (3 n + 5) (3 n + 4) (10 n + 13) (1 + 6 n) (1 + 3 n) (2 + 3 n) X(n) -4/27 ------------------------------------------------------------------ (10 n + 3) (2 n + 3) (5 + 6 n) (6 n + 11) (1 + n) (2 + n) 2 (3 n + 4) (3 n + 5) (330 n + 539 n + 135) X(1 + n) + 4/9 --------------------------------------------------- + X(2 + n) = 0 (10 n + 3) (2 n + 3) (6 n + 11) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -4/5 -28 B(0) = 1, B(1) = --- 45 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Beta(1/6, 2/3) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/3) (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(1/6, 2/3) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/3 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 [2.182916892, 2.186551605, 2.182008310, 2.186008779, 2.181875244, 2.171911393, 2.183871890, 2.182773105, 2.182353837, 2.186908063, 2.188512875, 2.188517461, 2.188961711, 2.194766933, 2.198375287, 2.193621755, 2.196723416, 2.192004021, 2.189674139, 2.193750951, 2.187373125] 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 [0.04862571349, 0.04779580350, 0.04883337444, 0.04791966291, 0.04886379424, 0.05114662296, 0.04840753255, 0.04865857125, 0.04875439283, 0.04771448457, 0.04734853431, 0.04734748892, 0.04724623059, 0.04592483813, 0.04510517892, 0.04618524101, 0.04548025181, 0.04655331978, 0.04708388707, 0.04615585654, 0.04760840872] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 53, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Beta(2/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (1/3) (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 , 2.186555245, yielding an irrationality measure that is approximately , 21.92270246 We hope that the reader can find nu exactly. 1/3 3 2 Comment: Note that this constant appears to be , ------ 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (10 n + 17) (3 n + 5) (3 n + 4) (5 + 6 n) (1 + 3 n) (2 + 3 n) X(n) -4/27 ------------------------------------------------------------------ (10 n + 7) (6 n + 13) (3 + 2 n) (6 n + 7) (1 + n) (2 + n) 2 (3 n + 5) (3 n + 4) (330 n + 781 n + 377) X(1 + n) + 4/9 --------------------------------------------------- + X(2 + n) = 0 (10 n + 7) (6 n + 13) (3 + 2 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -4 -44 B(0) = 1, B(1) = --- 21 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Beta(2/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (1/3) (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(2/3, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 1/3 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 [2.170874292, 2.174043736, 2.171383882, 2.178740854, 2.181945959, 2.186555245, 2.184285093, 2.183447589, 2.181115708, 2.184914175, 2.185472722, 2.184375257, 2.180964998, 2.178273129, 2.176882120, 2.176766638, 2.180125056, 2.180360243, 2.177156348, 2.182297306, 2.181050350] 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 [0.05138480541, 0.05065724418, 0.05126775858, 0.04958084803, 0.04884762809, 0.04779497304, 0.04831315942, 0.04850445808, 0.04903746319, 0.04816951342, 0.04804200654, 0.04829256875, 0.04907193007, 0.04968793346, 0.05000653352, 0.05003299249, 0.04926406356, 0.04921025835, 0.04994370837, 0.04876731404, 0.04905241007] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 54, : The following constant c. / | 1/2 | 3 c = 1/2 |----------------- |Beta(5/6, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (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 , 2.195625971, yielding an irrationality measure that is approximately , 22.86768106 We hope that the reader can find nu exactly. (2/3) Comment: Note that this constant appears to be , 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 + 6 n) (1 + 2 n) (1 + 3 n) (6 n + 11) (5 n + 7) X(n) -1/12 ------------------------------------------------------- (5 n + 2) (3 n + 7) (6 n + 7) (1 + n) (2 + n) 2 (6 n + 11) (3 n + 4) (330 n + 572 n + 175) X(1 + n) + 1/6 ---------------------------------------------------- + X(2 + n) = 0 (5 n + 2) (3 n + 7) (6 n + 7) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -5 B(0) = 1, B(1) = -25/8 A(n) Then, ----, approximates B(n) / | 1/2 | 3 c = 1/2 |----------------- |Beta(5/6, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (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(5/6, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 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 [2.169847543, 2.182456665, 2.175001111, 2.180633100, 2.179975953, 2.178426341, 2.182215741, 2.178851305, 2.179916593, 2.181724905, 2.179450413, 2.182334626, 2.180842493, 2.192380858, 2.190544513, 2.193474354, 2.195625971, 2.187384833, 2.189204923, 2.190970663, 2.187961524] 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 [0.05162071675, 0.04873089039, 0.05043767209, 0.04914784205, 0.04929817765, 0.04965285323, 0.04878595780, 0.04955556351, 0.04931175957, 0.04889816489, 0.04941843652, 0.04875878382, 0.04909994823, 0.04646755591, 0.04688562012, 0.04621876805, 0.04572958593, 0.04760573853, 0.04719080334, 0.04678857260, 0.04747423176] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 55, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(2/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (1/3) (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 , 2.187269026, yielding an irrationality measure that is approximately , 21.99422310 We hope that the reader can find nu exactly. (1/3) Comment: Note that this constant appears to be , 2 Prove it! 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 n) (5 + 6 n) (1 + 2 n) (1 + 6 n) (10 n + 17) (3 n + 5) X(n) -1/12 ------------------------------------------------------------------ (10 n + 7) (3 n + 1) (1 + n) (6 n + 13) (4 + 3 n) (2 + n) 2 (3 n + 5) (6 n + 7) (330 n + 781 n + 377) X(1 + n) + 1/3 --------------------------------------------------- + X(2 + n) = 0 (10 n + 7) (6 n + 13) (4 + 3 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2 B(0) = 1, B(1) = -11/7 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(2/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/6) (1/6) (1/3) (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 F(n) = 1/2 |----------------- |Pi Beta(2/3, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/6 1/6 1/3 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 [2.171589875, 2.174758958, 2.172098743, 2.179455355, 2.182660100, 2.187269026, 2.184998514, 2.184160652, 2.181828412, 2.185626521, 2.186184710, 2.185086887, 2.181676270, 2.178984045, 2.177592680, 2.177476841, 2.180834903, 2.181069735, 2.177865485, 2.183006088, 2.181758778] 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 [0.05122045181, 0.05049320074, 0.05110360733, 0.04941730551, 0.04868439624, 0.04763215076, 0.04815025826, 0.04834157922, 0.04887450078, 0.04800690226, 0.04787951662, 0.04813008286, 0.04890928436, 0.04952517819, 0.04984376098, 0.04987029352, 0.04910168420, 0.04904797682, 0.04978128120, 0.04860533173, 0.04889042062] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 56, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(2/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (1/3) (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 , 2.198406922, yielding an irrationality measure that is approximately , 23.17393305 We hope that the reader can find nu exactly. (1/3) Comment: Note that this constant appears to be , -2 2 Prove it! 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 n) (1 + 6 n) (-1 + 6 n) (5 + 6 n) (3 n + 5) (10 n + 13) X(n) -1/108 ------------------------------------------------------------------- (3 n + 1) (10 n + 3) (1 + n) (2 n + 3) (4 + 3 n) (2 + n) 2 (5 + 6 n) (3 n + 5) (330 n + 539 n + 135) X(1 + n) + 1/9 --------------------------------------------------- + X(2 + n) = 0 (10 n + 3) (2 n + 3) (4 + 3 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2 B(0) = 1, B(1) = 7/9 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(2/3, 2/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (1/3) (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 F(n) = 1/2 |----------------- |Pi Beta(2/3, 2/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 1/3 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 [2.182948750, 2.186583447, 2.182040136, 2.186040589, 2.181907038, 2.171943171, 2.183903652, 2.182804851, 2.182385568, 2.186939777, 2.188544573, 2.188549143, 2.188993377, 2.194798584, 2.198406922, 2.193653374, 2.196755019, 2.192035609, 2.189705711, 2.193782507, 2.187404666] 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 [0.04861843367, 0.04778853886, 0.04882609906, 0.04791240385, 0.04885652575, 0.05113932647, 0.04840027769, 0.04865131657, 0.04874714025, 0.04770725025, 0.04734130870, 0.04734026696, 0.04723901368, 0.04591764283, 0.04509799853, 0.04617804940, 0.04547307353, 0.04654613016, 0.04707669381, 0.04614867967, 0.04760121532] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 57, : The following constant c. / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/4, 3/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/4) (1/4) (5/4) (1/4) | / / 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 , 2.282859251, yielding an irrationality measure that is approximately , 39.05944133 We hope that the reader can find nu exactly. 1/2 2 Comment: Note that this constant appears to be , ---- 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (3 + 4 n) (-1 + 4 n) (4 n + 7) (5 n + 6) X(n) -1/16 --------------------------------------------- (5 n + 1) (4 n + 9) (1 + n) (2 + n) 3 2 (4 n + 7) (880 n + 1936 n + 1159 n + 174) X(1 + n) + 1/8 ---------------------------------------------------- + X(2 + n) = 0 (5 n + 1) (4 n + 9) (1 + 2 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -3/5 -9 B(0) = 1, B(1) = -- 10 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |------------------ |Pi Beta(-1/4, 3/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/4) (1/4) (5/4) (1/4) | / / 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 | 2 F(n) = 1/2 |------------------ |Pi Beta(-1/4, 3/4) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 3/4 1/4 5/4 1/4 | | | 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 [2.266258238, 2.267448495, 2.274951976, 2.277499077, 2.280614055, 2.281891454, 2.279572895, 2.268496633, 2.276311173, 2.279031293, 2.280256748, 2.281719492, 2.282859251, 2.279942047, 2.272760353, 2.264479819, 2.271502972, 2.273925606, 2.274893022, 2.273004041, 2.265426872] 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 [0.02992110256, 0.02965880047, 0.02800829696, 0.02744922609, 0.02676633620, 0.02648655717, 0.02699448643, 0.02942792837, 0.02770988703, 0.02711320821, 0.02684462175, 0.02652421182, 0.02627468941, 0.02691358226, 0.02848983010, 0.03031326956, 0.02876629967, 0.02823375002, 0.02802124416, 0.02843626582, 0.03010439357] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 58, : The following constant c. / | 1/2 | 2 c = 1/2 |----------------- |Beta(3/4, 3/4) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (1/4) (3/4) (1/4) | / / 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 , 2.288608954, yielding an irrationality measure that is approximately , 40.97157295 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 2 Prove it! 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 (4 n + 7) (20 n + 61 n + 46) (1 + 4 n) (1 + 2 n) (3 + 4 n) X(n) -1/16 ----------------------------------------------------------------- 2 (20 n + 21 n + 5) (4 n + 9) (3 + 2 n) (4 n + 5) (1 + n) (2 + n) 4 3 2 (4 n + 7) (3520 n + 14256 n + 19740 n + 10763 n + 1926) X(1 + n) + 1/8 ------------------------------------------------------------------- 2 (20 n + 21 n + 5) (4 n + 9) (3 + 2 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3 -21 B(0) = 1, B(1) = --- 10 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |----------------- |Beta(3/4, 3/4) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (1/4) (3/4) (1/4) | / / 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 | 2 F(n) = 1/2 |----------------- |Beta(3/4, 3/4) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/4 1/4 3/4 1/4 | | | 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 [2.273645929, 2.266129142, 2.279974901, 2.283284918, 2.276535098, 2.288608954, 2.279065934, 2.283472249, 2.283506331, 2.284778069, 2.286560119, 2.281114122, 2.285211975, 2.283663323, 2.277474449, 2.278859774, 2.277371107, 2.281603227, 2.272754093, 2.277958012, 2.273307614] 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 [0.02829520119, 0.02994956004, 0.02690638252, 0.02618153112, 0.02766074121, 0.02501777954, 0.02710561392, 0.02614053862, 0.02613308102, 0.02585488496, 0.02546531048, 0.02665679215, 0.02576000138, 0.02609873044, 0.02745462885, 0.02715081173, 0.02747730010, 0.02654967202, 0.02849120616, 0.02734855771, 0.02836954622] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 59, : The following constant c. / | 1/2 | 2 c = 1/2 |------------------- |Beta(-1/4, -1/4) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/4) (5/4) (3/4) (1/4) | / / 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 , 2.286111768, yielding an irrationality measure that is approximately , 40.11858511 We hope that the reader can find nu exactly. 1/2 3 2 Comment: Note that this constant appears to be , ------ 8 Prove it! 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 + 4 n) (20 n + 45 n + 22) (-1 + 4 n) X(n) -1/16 ---------------------------------------------- 2 (20 n + 5 n - 3) (4 n + 9) (1 + n) (2 + n) 4 3 2 (3 + 4 n) (3520 n + 7920 n + 3692 n - 333 n - 234) X(1 + n) + 1/8 -------------------------------------------------------------- 2 (20 n + 5 n - 3) (-1 + 2 n) (4 n + 9) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1/5 B(0) = 1, B(1) = 7/30 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |------------------- |Beta(-1/4, -1/4) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/4) (5/4) (3/4) (1/4) | / / 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 | 2 F(n) = 1/2 |------------------- |Beta(-1/4, -1/4) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/4 5/4 3/4 1/4 | | | 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 [2.274551177, 2.273202425, 2.281810530, 2.280819731, 2.280247065, 2.280120466, 2.277697066, 2.277587546, 2.285183894, 2.283026619, 2.284117692, 2.281923362, 2.286111768, 2.282452933, 2.277082818, 2.275805228, 2.274959557, 2.271097072, 2.275656855, 2.274614535, 2.271232407] 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 [0.02809632495, 0.02839266380, 0.02650427681, 0.02672127824, 0.02684674345, 0.02687448412, 0.02740579436, 0.02742981869, 0.02576614141, 0.02623805849, 0.02599932535, 0.02647957055, 0.02556329676, 0.02636362890, 0.02754055046, 0.02782094635, 0.02800663208, 0.02885557961, 0.02785352007, 0.02808240855, 0.02882581021] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 60, : The following constant c. / | 1/2 | 2 c = 1/2 |------------------ |Beta(-1/4, 3/4) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (5/4) (3/4) (1/4) | / / 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 , 2.287698836, yielding an irrationality measure that is approximately , 40.65652670 We hope that the reader can find nu exactly. 1/2 2 Comment: Note that this constant appears to be , ---- 4 Prove it! 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 + 4 n) (-1 + 4 n) (-1 + 2 n) (20 n + 47 n + 26) X(n) -1/16 --------------------------------------------------------- 2 (20 n + 7 n - 1) (1 + 2 n) (4 n + 5) (1 + n) (2 + n) 4 3 2 (3 + 4 n) (3520 n + 8272 n + 5052 n + 557 n - 118) X(1 + n) + 1/8 -------------------------------------------------------------- 2 (20 n + 7 n - 1) (1 + 2 n) (4 n + 5) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = 1 B(0) = 1, B(1) = 5/2 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |------------------ |Beta(-1/4, 3/4) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (5/4) (3/4) (1/4) | / / 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 | 2 F(n) = 1/2 |------------------ |Beta(-1/4, 3/4) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/4 5/4 3/4 1/4 | | | 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 [2.264971260, 2.272088944, 2.277684477, 2.280659621, 2.281035497, 2.287181516, 2.281919831, 2.283100910, 2.287698836, 2.282437335, 2.285278656, 2.286527813, 2.276024787, 2.282629889, 2.281407099, 2.276055282, 2.271771614, 2.275118322, 2.270959782, 2.277957887, 2.271341577] 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 [0.03020486988, 0.02863743922, 0.02740855583, 0.02675635360, 0.02667401409, 0.02532953626, 0.02648034370, 0.02622179966, 0.02521653012, 0.02636704348, 0.02574542159, 0.02547237027, 0.02777274830, 0.02632489287, 0.02659262383, 0.02776605433, 0.02870721890, 0.02797176661, 0.02888578081, 0.02734858513, 0.02880179753] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 61, : The following constant c. / | 1/2 | 2 c = 1/2 |----------------- |Pi Beta(1/4, 1/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (3/4) (3/4) (3/4) | / / 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 , 2.288092857, yielding an irrationality measure that is approximately , 40.79232421 We hope that the reader can find nu exactly. 1/2 2 Comment: Note that this constant appears to be , ---- 4 Prove it! 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 + 4 n) (-1 + 2 n) (4 n + 5) (20 n + 47 n + 26) X(n) -1/16 -------------------------------------------------------- 2 (20 n + 7 n - 1) (4 n + 7) (1 + n) (1 + 2 n) (2 + n) 4 3 2 (4 n + 5) (3520 n + 8272 n + 5052 n + 557 n - 118) X(1 + n) + 1/8 -------------------------------------------------------------- 2 (20 n + 7 n - 1) (4 n + 7) (1 + 2 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/3 B(0) = 1, B(1) = -5/6 A(n) Then, ----, approximates B(n) / | 1/2 | 2 c = 1/2 |----------------- |Pi Beta(1/4, 1/4) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/4) (3/4) (3/4) (3/4) | / / 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 | 2 F(n) = 1/2 |----------------- |Pi Beta(1/4, 1/4) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/4 3/4 3/4 3/4 | | | 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 [2.265366874, 2.272484358, 2.278079692, 2.281054637, 2.281430313, 2.287576133, 2.282314249, 2.283495130, 2.288092857, 2.282831158, 2.285672281, 2.286921240, 2.276418017, 2.283022922, 2.281799934, 2.276447921, 2.272164056, 2.275510567, 2.271351831, 2.278349740, 2.271733233] 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 [0.03011762383, 0.02855050241, 0.02732187025, 0.02666982163, 0.02658753980, 0.02524333176, 0.02639398916, 0.02613553194, 0.02513047478, 0.02628083823, 0.02565936401, 0.02538640176, 0.02768643692, 0.02623886760, 0.02650659703, 0.02767987378, 0.02862092376, 0.02788563806, 0.02879954215, 0.02726264694, 0.02871565938] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 62, : The following constant c. 1 c = --------------------------- Pi Beta(3/5, 3/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/5) (2/5) (1/5) (4/5) / / 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 , 0.004522144010, yielding an irrationality measure that is approximately , 2.003766042 We hope that the reader can find nu exactly. 1/2 5 Comment: Note that this constant appears to be , - ---- - 1/2 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 n + 8) (4 + 5 n) (-1 + 5 n) (3 + 5 n) X(n) -1/25 --------------------------------------------- (5 n + 2) (5 n + 7) (1 + n) (2 + n) (4 + 5 n) (5 n + 8) X(1 + n) + 11/5 ---------------------------- + X(2 + n) = 0 (5 n + 7) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -3/2 B(0) = 1, B(1) = 9/10 A(n) Then, ----, approximates B(n) 1 c = --------------------------- Pi Beta(3/5, 3/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/5) (2/5) (1/5) (4/5) / / 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) = --------------------------- Pi Beta(3/5, 3/5) Pi csc(----) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/5 2/5 1/5 4/5 | | 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 [0.004522144010, 0.004520116150, 0.004518090208, 0.004516066181, 0.004514044066, 0.004512023860, 0.004510005561, 0.004507989166, 0.004505974672, 0.004503962076, 0.004501951376, 0.004499942569, 0.004497935652, 0.004495930623, 0.004493927478, 0.004491926216, 0.004489926833, 0.004487929327, 0.004485933694, 0.004483939933, 0.004481948041] 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 [0.9962480883, 0.9962497678, 0.9962514455, 0.9962531216, 0.9962547961, 0.9962564689, 0.9962581401, 0.9962598104, 0.9962614783, 0.9962631453, 0.9962648107, 0.9962664736, 0.9962681356, 0.9962697960, 0.9962714556, 0.9962731127, 0.9962747682, 0.9962764228, 0.9962780750, 0.9962797263, 0.9962813760] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 63, : The following constant c. 1 c = --------------------------- Pi Beta(1/5, 2/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/5) (4/5) (1/5) (4/5) / / 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 , 0.009446944600, yielding an irrationality measure that is approximately , 2.007883583 We hope that the reader can find nu exactly. 1/2 5 Comment: Note that this constant appears to be , - 1/6 + ---- 6 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (5 n + 6) (25 n + 60 n + 34) (2 + 5 n) (-3 + 5 n) (1 + 5 n) X(n) -1/25 ----------------------------------------------------------------- 2 (25 n + 10 n - 1) (3 + 5 n) (5 n + 8) (1 + n) (2 + n) 2 (2 + 5 n) (5 n + 6) (275 n + 385 n - 31) X(1 + n) + 1/5 -------------------------------------------------- + X(2 + n) = 0 2 (25 n + 10 n - 1) (5 n + 8) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1/3 B(0) = 1, B(1) = -7/5 A(n) Then, ----, approximates B(n) 1 c = --------------------------- Pi Beta(1/5, 2/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/5) (4/5) (1/5) (4/5) / / 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) = --------------------------- Pi Beta(1/5, 2/5) Pi csc(----) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 3/5 4/5 1/5 4/5 | | 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 [0.008896351003, 0.009446944600, 0.009442687549, 0.009438434534, 0.009434185550, 0.008876483519, 0.009425699652, 0.008868562732, 0.009417229808, 0.009413000891, 0.009408775971, 0.009404555041, 0.008848825902, 0.009396125128, 0.009391916134, 0.009387711108, 0.009383510043, 0.008829181593, 0.009375119776, 0.008821349627, 0.009366745286] 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 [0.9926322857, 0.9921780822, 0.9921815931, 0.9921851008, 0.9921886052, 0.9926486785, 0.9921956040, 0.9926552144, 0.9922025898, 0.9922060777, 0.9922095624, 0.9922130437, 0.9926715004, 0.9922199966, 0.9922234681, 0.9922269363, 0.9922304013, 0.9926877098, 0.9922373213, 0.9926941726, 0.9922442290] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 64, : The following constant c. 1 c = --------------------------- 2 Pi Pi csc(----) Beta(1/5, 1/5) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/5) (3/5) (4/5) (4/5) / / 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 , 0.009370140660, yielding an irrationality measure that is approximately , 2.007819238 We hope that the reader can find nu exactly. 1/2 5 Comment: Note that this constant appears to be , ---- - 1/4 4 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (5 n + 6) (25 n + 60 n + 34) (3 + 5 n) (-2 + 5 n) (1 + 5 n) X(n) -1/25 ----------------------------------------------------------------- 2 (25 n + 10 n - 1) (5 n + 9) (1 + n) (5 n + 4) (2 + n) 2 (5 n + 6) (3 + 5 n) (275 n + 385 n - 31) X(1 + n) + 1/5 -------------------------------------------------- + X(2 + n) = 0 2 (25 n + 10 n - 1) (5 n + 9) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1/4 -7 B(0) = 1, B(1) = -- 10 A(n) Then, ----, approximates B(n) 1 c = --------------------------- 2 Pi Pi csc(----) Beta(1/5, 1/5) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/5) (3/5) (4/5) (4/5) / / 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) = --------------------------- 2 Pi Pi csc(----) Beta(1/5, 1/5) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/5 3/5 4/5 4/5 | | 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 [0.009024288662, 0.009370140660, 0.009365922329, 0.009361707995, 0.009357497653, 0.009004098763, 0.009000072215, 0.009344890522, 0.009340696090, 0.009336505621, 0.009332319110, 0.008979996330, 0.008975992485, 0.009319783264, 0.009315612526, 0.009311445717, 0.009307282831, 0.008956029631, 0.008952048296, 0.009294817657, 0.009290670408] 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 [0.9925267269, 0.9922414279, 0.9922449078, 0.9922483835, 0.9922518559, 0.9925433844, 0.9925467069, 0.9922622543, 0.9922657144, 0.9922691704, 0.9922726239, 0.9925632708, 0.9925665743, 0.9922829640, 0.9922864036, 0.9922898406, 0.9922932745, 0.9925830445, 0.9925863300, 0.9923035562, 0.9923069776] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 65, : The following constant c. 1 c = --------------------------- Pi Pi csc(----) Beta(2/5, 2/5) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/5) (1/5) (3/5) (3/5) / / 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 , 0.004804747633, yielding an irrationality measure that is approximately , 2.004001865 We hope that the reader can find nu exactly. 1/2 5 Comment: Note that this constant appears to be , ---- + 1/2 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (2 + 5 n) (1 + 5 n) (5 n + 7) (5 n + 6) X(n) -1/25 -------------------------------------------- (5 n + 3) (1 + n) (5 n + 8) (2 + n) (5 n + 7) (5 n + 6) X(1 + n) + 11/5 ---------------------------- + X(2 + n) = 0 (5 n + 8) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2/3 B(0) = 1, B(1) = -2/5 A(n) Then, ----, approximates B(n) 1 c = --------------------------- Pi Pi csc(----) Beta(2/5, 2/5) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/5) (1/5) (3/5) (3/5) / / 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) = --------------------------- Pi Pi csc(----) Beta(2/5, 2/5) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 4/5 1/5 3/5 3/5 | | 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 [0.004804747633, 0.004802577091, 0.004800408610, 0.004798242187, 0.004796077821, 0.004793915507, 0.004791755243, 0.004789597026, 0.004787440853, 0.004785286720, 0.004783134626, 0.004780984566, 0.004778836539, 0.004776690540, 0.004774546568, 0.004772404619, 0.004770264691, 0.004768126780, 0.004765990884, 0.004763856999, 0.004761725123] 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 [0.9960140860, 0.9960158835, 0.9960176784, 0.9960194725, 0.9960212642, 0.9960230541, 0.9960248433, 0.9960266300, 0.9960284150, 0.9960301984, 0.9960319801, 0.9960337602, 0.9960355386, 0.9960373154, 0.9960390905, 0.9960408640, 0.9960426358, 0.9960444059, 0.9960461744, 0.9960479413, 0.9960497065] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 66, : The following constant c. 1 c = --------------------------- 2 Pi Beta(2/5, 4/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/5) (3/5) (2/5) (3/5) / / 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 , 0.004356021867, yielding an irrationality measure that is approximately , 2.003627444 We hope that the reader can find nu exactly. 1/2 5 Comment: Note that this constant appears to be , - ---- - 1/2 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 n + 7) (4 + 5 n) (-1 + 5 n) (2 + 5 n) X(n) -1/25 --------------------------------------------- (5 n + 1) (5 n + 6) (1 + n) (2 + n) (4 + 5 n) (5 n + 7) X(1 + n) + 11/5 ---------------------------- + X(2 + n) = 0 (5 n + 6) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2 B(0) = 1, B(1) = 6/5 A(n) Then, ----, approximates B(n) 1 c = --------------------------- 2 Pi Beta(2/5, 4/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/5) (3/5) (2/5) (3/5) / / 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) = --------------------------- 2 Pi Beta(2/5, 4/5) Pi csc(----) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/5 3/5 2/5 3/5 | | 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 [0.004356021867, 0.004354077860, 0.004352135686, 0.004350195342, 0.004348256826, 0.004346320135, 0.004344385267, 0.004342452219, 0.004340520988, 0.004338591571, 0.004336663967, 0.004334738172, 0.004332814185, 0.004330892001, 0.004328971619, 0.004327053036, 0.004325136250, 0.004323221258, 0.004321308057, 0.004319396645, 0.004317487019] 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 [0.9963856666, 0.9963872767, 0.9963888851, 0.9963904927, 0.9963920978, 0.9963937021, 0.9963953048, 0.9963969058, 0.9963985051, 0.9964001028, 0.9964016997, 0.9964032949, 0.9964048884, 0.9964064803, 0.9964080705, 0.9964096600, 0.9964112477, 0.9964128338, 0.9964144183, 0.9964160011, 0.9964175831] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 67, : The following constant c. 1 c = --------------------------- 2 Pi Beta(4/5, 4/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/5) (1/5) (3/5) (2/5) / / 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 , 0.004356123898, yielding an irrationality measure that is approximately , 2.003627529 We hope that the reader can find nu exactly. 1/2 3 5 Comment: Note that this constant appears to be , - 3/2 + ------ 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (4 + 5 n) (2 + 5 n) (5 n + 9) (5 n + 7) X(n) -1/25 -------------------------------------------- (5 n + 11) (5 n + 6) (1 + n) (2 + n) (5 n + 9) (5 n + 7) X(1 + n) + 11/5 ---------------------------- + X(2 + n) = 0 (5 n + 11) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -4 -32 B(0) = 1, B(1) = --- 15 A(n) Then, ----, approximates B(n) 1 c = --------------------------- 2 Pi Beta(4/5, 4/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/5) (1/5) (3/5) (2/5) / / 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) = --------------------------- 2 Pi Beta(4/5, 4/5) Pi csc(----) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/5 1/5 3/5 2/5 | | 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 [0.004356123898, 0.004354179788, 0.004352237511, 0.003796281790, 0.004348358446, 0.004346421652, 0.004344486682, 0.003789653537, 0.004340622199, 0.004338692681, 0.004336764975, 0.003783049883, 0.004332914989, 0.004330992705, 0.004329072222, 0.003776470687, 0.004325236651, 0.004323321558, 0.004321408257, 0.003769915810, 0.004317587019] 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 [0.9963855821, 0.9963871922, 0.9963888006, 0.9968493695, 0.9963920142, 0.9963936177, 0.9963952203, 0.9968548616, 0.9963984215, 0.9964000191, 0.9964016160, 0.9968603338, 0.9964048048, 0.9964063967, 0.9964079877, 0.9968657854, 0.9964111641, 0.9964127502, 0.9964143355, 0.9968712171, 0.9964175002] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 68, : The following constant c. 1 c = --------------------------- 2 Pi Beta(1/5, 3/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/5) (2/5) (3/5) (2/5) / / 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 , 0.004727853907, yielding an irrationality measure that is approximately , 2.003937694 We hope that the reader can find nu exactly. 1/2 5 Comment: Note that this constant appears to be , ---- + 1/2 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (3 + 5 n) (1 + 5 n) (5 n + 8) (5 n + 6) X(n) -1/25 -------------------------------------------- (4 + 5 n) (5 n + 9) (1 + n) (2 + n) (5 n + 8) (5 n + 6) X(1 + n) + 11/5 ---------------------------- + X(2 + n) = 0 (5 n + 9) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -3/4 -9 B(0) = 1, B(1) = -- 20 A(n) Then, ----, approximates B(n) 1 c = --------------------------- 2 Pi Beta(1/5, 3/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/5) (2/5) (3/5) (2/5) / / 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) = --------------------------- 2 Pi Beta(1/5, 3/5) Pi csc(----) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 4/5 2/5 3/5 2/5 | | 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 [0.004727853907, 0.004725722174, 0.004723592465, 0.004721464775, 0.004719339101, 0.004717215442, 0.004715093793, 0.004712974152, 0.004710856516, 0.004708740883, 0.004706627248, 0.004704515610, 0.004702405966, 0.004700298312, 0.004698192646, 0.004696088965, 0.004693987266, 0.004691887546, 0.004689789802, 0.004687694032, 0.004685600232] 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 [0.9960777505, 0.9960795158, 0.9960812794, 0.9960830405, 0.9960848008, 0.9960865594, 0.9960883156, 0.9960900709, 0.9960918237, 0.9960935758, 0.9960953261, 0.9960970740, 0.9960988211, 0.9961005665, 0.9961023095, 0.9961040516, 0.9961057920, 0.9961075300, 0.9961092671, 0.9961110026, 0.9961127365] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 69, : The following constant c. 1 c = --------------------------- Pi Beta(3/5, 4/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/5) (1/5) (4/5) (1/5) / / 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 , 0.004522246040, yielding an irrationality measure that is approximately , 2.003766127 We hope that the reader can find nu exactly. 1/2 Comment: Note that this constant appears to be , 5 - 1 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (4 + 5 n) (3 + 5 n) (1 + 5 n) (5 n + 9) (5 n + 8) X(n) -1/25 ------------------------------------------------------ (5 n + 2) (5 n + 11) (5 n + 7) (1 + n) (2 + n) (5 n + 6) (5 n + 8) (5 n + 9) X(1 + n) + 11/5 -------------------------------------- + X(2 + n) = 0 (5 n + 11) (5 n + 7) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -2 B(0) = 1, B(1) = -8/5 A(n) Then, ----, approximates B(n) 1 c = --------------------------- Pi Beta(3/5, 4/5) Pi csc(----) 5 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/5) (1/5) (4/5) (1/5) / / 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) = --------------------------- Pi Beta(3/5, 4/5) Pi csc(----) 5 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/5 1/5 4/5 1/5 | | 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 [0.004522246040, 0.004170320460, 0.004518192032, 0.004516167903, 0.004164777146, 0.004512125377, 0.004510106976, 0.004159249421, 0.004506075882, 0.004504063185, 0.004153737217, 0.004500043475, 0.004498036457, 0.004148240468, 0.004494028081, 0.004492026718, 0.004142759108, 0.004488029627, 0.004486033895, 0.004137293070, 0.004482048041] 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 [0.9962480039, 0.9965394830, 0.9962513611, 0.9962530372, 0.9965440746, 0.9962563853, 0.9962580564, 0.9965486538, 0.9962613946, 0.9962630617, 0.9965532198, 0.9962663908, 0.9962680528, 0.9965577734, 0.9962713720, 0.9962730291, 0.9965623137, 0.9962763392, 0.9962779922, 0.9965668416, 0.9962812931] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 70, : The following constant c. / | | 1 c = 1/2 |----------------- |Beta(1/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | 1/2 (1/6) (5/6) (1/6) | / / 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 , 2.303606886, yielding an irrationality measure that is approximately , 46.96938102 We hope that the reader can find nu exactly. 1/2 2 3 Comment: Note that this constant appears to be , ------ 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 + 6 n) (2 + 3 n) (1 + 6 n) (6 n + 11) (15 n + 22) X(n) -1/36 --------------------------------------------------------- (15 n + 7) (6 n + 13) (4 + 3 n) (1 + n) (2 + n) 2 (6 n + 11) (6 n + 7) (330 n + 649 n + 234) X(1 + n) + 1/4 ---------------------------------------------------- + X(2 + n) = 0 (15 n + 7) (6 n + 13) (4 + 3 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -5/3 B(0) = 1, B(1) = -10/7 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Beta(1/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | 1/2 (1/6) (5/6) (1/6) | / / 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) = 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 5/6 1/6 | | | 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 [2.294459476, 2.295683333, 2.292183901, 2.298517563, 2.294751375, 2.296389081, 2.296811802, 2.294208576, 2.295901979, 2.286249056, 2.292554293, 2.296466396, 2.297292551, 2.294162063, 2.298708498, 2.302610677, 2.303606886, 2.303494928, 2.303073252, 2.301449028, 2.302143481] 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 [0.02374198646, 0.02347550785, 0.02423783175, 0.02285892422, 0.02367841673, 0.02332190365, 0.02322992162, 0.02379663375, 0.02342791520, 0.02553329073, 0.02415709102, 0.02330507905, 0.02312533306, 0.02380676516, 0.02281741323, 0.02196978177, 0.02175361029, 0.02177789996, 0.02186939435, 0.02222196831, 0.02207119207] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 71, : The following constant c. / | | 1 c = 1/2 |----------------- |Beta(2/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (1/6) (5/6) (1/6) | / / 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 , 2.195307062, yielding an irrationality measure that is approximately , 22.83307779 We hope that the reader can find nu exactly. 2/3 3 2 Comment: Note that this constant appears to be , ------ 4 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (6 n + 11) (3 n + 5) (5 n + 7) (1 + 6 n) (2 + 3 n) (5 + 6 n) X(n) -1/108 ----------------------------------------------------------------- (5 n + 2) (3 n + 7) (3 + 2 n) (3 n + 4) (1 + n) (2 + n) 2 (3 n + 5) (6 n + 11) (330 n + 572 n + 175) X(1 + n) + 1/18 ---------------------------------------------------- + X(2 + n) = 0 (5 n + 2) (3 n + 7) (3 + 2 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -5/2 -25 B(0) = 1, B(1) = --- 12 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Beta(2/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/3) (1/6) (5/6) (1/6) | / / 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) = 1/2 |----------------- |Beta(2/3, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/3 1/6 5/6 1/6 | | | 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 [2.169526057, 2.182135342, 2.174679950, 2.180312100, 2.179655116, 2.178105665, 2.181895227, 2.178530952, 2.179596401, 2.181404874, 2.179130542, 2.182014916, 2.180522944, 2.192061469, 2.190225284, 2.193155286, 2.195307062, 2.187066083, 2.188886334, 2.190652233, 2.187643253] 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 [0.05169460486, 0.04880433569, 0.05051131951, 0.04922127188, 0.04937159123, 0.04972627958, 0.04885922588, 0.04962890230, 0.04938502744, 0.04897133821, 0.04949164582, 0.04883186431, 0.04917303945, 0.04654024435, 0.04695833021, 0.04629134891, 0.04580206279, 0.04767843942, 0.04726340997, 0.04686108725, 0.04754680518] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 72, : The following constant c. / | | 1 c = 1/2 |------------------ |Beta(-1/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (4/3) (1/6) (5/6) (1/6) | / / 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 , 2.195874564, yielding an irrationality measure that is approximately , 22.89472751 We hope that the reader can find nu exactly. 2/3 2 Comment: Note that this constant appears to be , ---- 2 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/108 (6 n + 11) (15 n + 37 n + 23) (-1 + 3 n) (1 + 6 n) (2 + 3 n) (5 + 6 n) / 2 X(n) / ((15 n + 7 n + 1) (3 n + 4) (1 + 2 n) (3 n + 7) (1 + n) (2 + n)) / 3 2 (6 n + 11) (2 + 3 n) (495 n + 891 n + 368 n + 52) X(1 + n) + 1/9 ------------------------------------------------------------ 2 (15 n + 7 n + 1) (1 + 2 n) (3 n + 7) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -5/8 B(0) = 1, B(1) = -5/6 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Beta(-1/3, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (4/3) (1/6) (5/6) (1/6) | / / 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) = 1/2 |------------------ |Beta(-1/3, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 4/3 1/6 5/6 1/6 | | | 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 [2.177961426, 2.174406908, 2.187979374, 2.180372098, 2.187435889, 2.179572455, 2.176198586, 2.177699086, 2.176599440, 2.186885939, 2.184320377, 2.184854587, 2.184471942, 2.187615574, 2.195874564, 2.189288498, 2.191682197, 2.190574680, 2.189263256, 2.190336456, 2.187093780] 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 [0.04975930986, 0.05057394058, 0.04747016183, 0.04920754635, 0.04759409455, 0.04939050729, 0.05016316268, 0.04981939022, 0.05007130288, 0.04771953136, 0.04830510155, 0.04818311819, 0.04827048979, 0.04755311713, 0.04567309640, 0.04717175822, 0.04662657444, 0.04687874957, 0.04717751031, 0.04693300766, 0.04767212186] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 73, : The following constant c. / | | 1 c = 1/2 |------------------ |Beta(-3/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/2) (1/6) (5/6) (1/6) | / / 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 , 2.306325779, yielding an irrationality measure that is approximately , 48.24984264 We hope that the reader can find nu exactly. 1/2 5 3 Comment: Note that this constant appears to be , ------ 12 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 (6 n + 11) (30 n + 53 n + 18) (-3 + 2 n) (2 + 3 n) (5 + 6 n) X(n) -1/36 ------------------------------------------------------------------ 2 (30 n - 7 n - 5) (-2 + 3 n) (2 n + 7) (1 + n) (2 + n) 4 3 2 (6 n + 11) (3960 n + 6996 n + 234 n - 2757 n - 530) X(1 + n) + 1/12 --------------------------------------------------------------- 2 (30 n - 7 n - 5) (-2 + 3 n) (2 n + 7) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/3 -2 B(0) = 1, B(1) = -- 15 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Beta(-3/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/2) (1/6) (5/6) (1/6) | / / 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) = 1/2 |------------------ |Beta(-3/2, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/2 1/6 5/6 1/6 | | | 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 [2.299389014, 2.294016387, 2.291368677, 2.293951444, 2.288939119, 2.292454858, 2.296131614, 2.299278932, 2.302994229, 2.293268415, 2.294013286, 2.295687125, 2.294310814, 2.303374948, 2.299436358, 2.301020739, 2.305114571, 2.303839094, 2.297036042, 2.306325779, 2.300221539] 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 [0.02266949031, 0.02383849743, 0.02441558505, 0.02385264445, 0.02494569751, 0.02417876533, 0.02337793533, 0.02269341580, 0.02188654242, 0.02400145745, 0.02383917293, 0.02347468241, 0.02377436509, 0.02180393130, 0.02265920080, 0.02231497851, 0.02142662540, 0.02170323566, 0.02318113479, 0.02116409165, 0.02248858357] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 74, : The following constant c. / | | 1 c = 1/2 |------------------ |Beta(-1/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/2) (1/6) (5/6) (1/6) | / / 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 , 2.307097258, yielding an irrationality measure that is approximately , 48.62598489 We hope that the reader can find nu exactly. 1/2 3 Comment: Note that this constant appears to be , ---- 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 + 6 n) (2 + 3 n) (-1 + 2 n) (6 n + 11) (5 n + 6) X(n) -1/36 -------------------------------------------------------- (5 n + 1) (1 + 3 n) (2 n + 5) (1 + n) (2 + n) 3 2 (6 n + 11) (660 n + 1452 n + 843 n + 134) X(1 + n) + 1/12 ---------------------------------------------------- + X(2 + n) = 0 (5 n + 1) (1 + 3 n) (2 n + 5) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -5/9 B(0) = 1, B(1) = -10/9 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Beta(-1/2, 5/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/2) (1/6) (5/6) (1/6) | / / 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) = 1/2 |------------------ |Beta(-1/2, 5/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 3/2 1/6 5/6 1/6 | | | 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 [2.296897430, 2.290484238, 2.291157223, 2.291195574, 2.291886752, 2.290704182, 2.293089393, 2.296898304, 2.294050549, 2.291661333, 2.283352163, 2.287464102, 2.297251165, 2.297124480, 2.300333927, 2.297700590, 2.299753434, 2.307097258, 2.293758977, 2.298466034, 2.301796991] 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 [0.02321129139, 0.02460849993, 0.02446170103, 0.02445333676, 0.02430261565, 0.02456051869, 0.02404046852, 0.02321110124, 0.02383105582, 0.02435176657, 0.02616681593, 0.02526780358, 0.02313433588, 0.02316189504, 0.02246416668, 0.02303657958, 0.02259029438, 0.02099694111, 0.02389457327, 0.02287012767, 0.02214641472] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 75, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(1/6, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (2/3) (5/6) | / / 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 , 2.188343381, yielding an irrationality measure that is approximately , 22.10275730 We hope that the reader can find nu exactly. 1/3 2 Comment: Note that this constant appears to be , ---- 4 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/108 (6 n + 7) (15 n + 35 n + 19) (5 + 6 n) (1 + 3 n) (-2 + 3 n) (1 + 6 n) / 2 X(n) / ((15 n + 5 n - 1) (3 n + 5) (1 + n) (3 n + 2) (1 + 2 n) (2 + n)) / 2 (3 n + 2) (1 + 3 n) (6 n + 7) (165 n + 220 n - 31) X(1 + n) + 1/9 ------------------------------------------------------------ 2 (15 n + 5 n - 1) (3 n + 5) (1 + 2 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/4 B(0) = 1, B(1) = -2/3 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/6, 1/3) | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (1/6) (5/6) (2/3) (5/6) | / / 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) = 1/2 |----------------- |Pi Beta(1/6, 1/3) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 2/3 5/6 | | | 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 [2.180140481, 2.174916153, 2.175026892, 2.176703129, 2.180131446, 2.182271155, 2.183290078, 2.188343381, 2.185332073, 2.183483079, 2.185764781, 2.185098844, 2.184524935, 2.178332749, 2.181288838, 2.181644813, 2.171276729, 2.182887469, 2.182384405, 2.183017998, 2.182738509] 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 [0.04926053452, 0.05045715333, 0.05043176054, 0.05004754410, 0.04926260161, 0.04877329145, 0.04854044379, 0.04738717248, 0.04807411141, 0.04849635020, 0.04797534673, 0.04812735316, 0.04825838872, 0.04967428227, 0.04899787171, 0.04891647653, 0.05129236820, 0.04863243697, 0.04874740607, 0.04860261029, 0.04866647732] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 76, : The following constant c. / | | 1 c = 1/2 |----------------- |Pi Beta(1/6, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) 1/2 (5/6) | / / 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 , 2.303738218, yielding an irrationality measure that is approximately , 47.02966766 We hope that the reader can find nu exactly. 1/2 3 Comment: Note that this constant appears to be , ---- 12 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (6 n + 7) (15 n + 14) (-2 + 3 n) (1 + 6 n) X(n) -1/36 ----------------------------------------------- (15 n - 1) (1 + n) (2 + 3 n) (2 + n) 2 (6 n + 7) (330 n + 143 n - 10) X(1 + n) + 1/4 ---------------------------------------- + X(2 + n) = 0 (15 n - 1) (2 + 3 n) (2 + n) Subject to the initial conditions A(0) = 0, A(1) = -1/3 B(0) = 1, B(1) = -2 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |----------------- |Pi Beta(1/6, 1/2) | \ 1 1 \ / / | | | 1 | | | ------------------------------------------------ dx dy| | | (1/6) (5/6) 1/2 (5/6) | / / 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) = 1/2 |----------------- |Pi Beta(1/6, 1/2) | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 1/6 5/6 1/2 5/6 | | | 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 [2.295337563, 2.295578774, 2.292412861, 2.293540356, 2.296838175, 2.291780835, 2.295477725, 2.296598374, 2.295094773, 2.294613272, 2.290863775, 2.289254229, 2.294861815, 2.297775105, 2.295108214, 2.298494974, 2.301353501, 2.303738218, 2.302585488, 2.302857972, 2.303012788] 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 [0.02355078062, 0.02349826877, 0.02418791989, 0.02394220390, 0.02322418353, 0.02432570947, 0.02352026657, 0.02327636045, 0.02360364166, 0.02370849180, 0.02452570597, 0.02487691176, 0.02365436718, 0.02302037332, 0.02360071510, 0.02286383550, 0.02224271215, 0.02172511884, 0.02197524883, 0.02191611168, 0.02188251505] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 77, : The following constant c. / | | 1 c = 1/2 |------------------ |Beta(-2/3, 1/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/3) (5/6) (1/6) (5/6) | / / 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 , 2.186684984, yielding an irrationality measure that is approximately , 21.93566766 We hope that the reader can find nu exactly. 1/3 2 2 Comment: Note that this constant appears to be , ------ 3 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence (5 n + 7) (3 n + 1) (6 n + 7) (5 + 6 n) (-2 + 3 n) (1 + 6 n) X(n) -1/108 ----------------------------------------------------------------- (5 n + 2) (3 n - 1) (-1 + 2 n) (3 n + 8) (1 + n) (2 + n) 3 2 (3 n + 1) (6 n + 7) (990 n + 1716 n + 485 n - 116) X(1 + n) + 1/18 ------------------------------------------------------------- (5 n + 2) (3 n - 1) (-1 + 2 n) (3 n + 8) (2 + n) + X(2 + n) = 0 Subject to the initial conditions -1 A(0) = 0, A(1) = -- 10 -1 B(0) = 1, B(1) = -- 60 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Beta(-2/3, 1/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (5/3) (5/6) (1/6) (5/6) | / / 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) = 1/2 |------------------ |Beta(-2/3, 1/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 5/3 5/6 1/6 5/6 | | | 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 [2.179235552, 2.175034882, 2.176849257, 2.176960769, 2.175797015, 2.178466686, 2.180098057, 2.184265776, 2.183115414, 2.180964977, 2.181409203, 2.186684984, 2.180136746, 2.178184372, 2.181458084, 2.184555577, 2.184162665, 2.183460403, 2.183357500, 2.185739985, 2.178092594] 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 [0.04946761091, 0.05042992845, 0.05001406288, 0.04998851441, 0.05025520291, 0.04964361602, 0.04927024065, 0.04831757093, 0.04858035124, 0.04907193487, 0.04897034834, 0.04776537421, 0.04926138904, 0.04970825680, 0.04895917133, 0.04825139167, 0.04834111948, 0.04850153064, 0.04852503993, 0.04798100586, 0.04972927271] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 78, : The following constant c. / | | 1 c = 1/2 |------------------ |Beta(-1/2, 1/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/2) (5/6) (1/6) (5/6) | / / 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 , 2.307254517, yielding an irrationality measure that is approximately , 48.70337879 We hope that the reader can find nu exactly. 1/2 7 3 Comment: Note that this constant appears to be , ------ 24 Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/36 (6 n + 7) (30 n + 55 n + 18) (-1 + 2 n) (5 + 6 n) (-2 + 3 n) (1 + 6 n) / 2 X(n) / ((30 n - 5 n - 7) (6 n - 1) (-1 + 3 n) (2 n + 5) (1 + n) (2 + n)) / 5 4 3 2 + 1/12 (6 n + 7) (23760 n + 39600 n + 2856 n - 13620 n - 3303 n + 322) / 2 X(1 + n) / ((30 n - 5 n - 7) (6 n - 1) (-1 + 3 n) (2 n + 5) (2 + n)) / + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -1/9 -2 B(0) = 1, B(1) = -- 63 A(n) Then, ----, approximates B(n) / | | 1 c = 1/2 |------------------ |Beta(-1/2, 1/6) Pi | \ 1 1 \ / / | | | 1 | | | -------------------------------------------------- dx dy| | | (3/2) (5/6) (1/6) (5/6) | / / 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) = 1/2 |------------------ |Beta(-1/2, 1/6) Pi | | \ 1 1 /x (1 - x) y (1 - y)\n \ / / |-------------------| | | | \ -x y + 1 / | | | ------------------------------------------ dx dy| | | 3/2 5/6 1/6 5/6 | | | 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 [2.289171920, 2.288463343, 2.302781024, 2.295291749, 2.300810566, 2.303425865, 2.298093903, 2.296213917, 2.294216234, 2.290806494, 2.294680932, 2.295550389, 2.292281750, 2.295322182, 2.296749006, 2.296482210, 2.306279934, 2.298059947, 2.304162465, 2.307254517, 2.303783219] 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 [0.02489487820, 0.02504957273, 0.02193281100, 0.02356075498, 0.02236062732, 0.02179288398, 0.02295104376, 0.02336002334, 0.02379496571, 0.02453820065, 0.02369375708, 0.02350444793, 0.02421650068, 0.02355412926, 0.02324358467, 0.02330163782, 0.02117402626, 0.02295842778, 0.02163309264, 0.02096287570, 0.02171535659] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 79, : The following constant c. 1 c = ----------------------------- Beta(4/7, 4/7) Beta(2/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (3/7) (2/7) (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 , 2.207861379, yielding an irrationality measure that is approximately , 24.27937929 We hope that the reader can find nu exactly. 1/2 1/3 (-28 + 84 I 3 ) Comment: Note that this constant appears to be , - -------------------- 12 7 - ---------------------- - 2/3 1/2 1/3 3 (-28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(-28 + 84 I 3 ) 14 | + 1/2 I 3 |-------------------- - ----------------------| | 6 1/2 1/3| \ 3 (-28 + 84 I 3 ) / Prove it! 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) (5 + 7 n) (7 n + 11) (5 n + 6) X(n) -1/49 ------------------------------------------------------------------ (7 n + 3) (5 n + 1) (8 + 7 n) (7 n + 10) (1 + n) (2 + n) 3 2 (7 n + 11) (2695 n + 5929 n + 3676 n + 516) X(1 + n) + 1/7 ------------------------------------------------------ + X(2 + n) = (5 n + 1) (8 + 7 n) (7 n + 10) (2 + n) 0 Subject to the initial conditions A(0) = 0, A(1) = -4/3 B(0) = 1, B(1) = 4/7 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(4/7, 4/7) Beta(2/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (3/7) (2/7) (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, 5/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 3/7 3/7 2/7 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 [2.197354452, 2.190180038, 2.190918288, 2.194089152, 2.195060812, 2.195770436, 2.197917607, 2.206330100, 2.207222164, 2.197007311, 2.204112528, 2.205009215, 2.193924550, 2.193657722, 2.195120163, 2.199635186, 2.207861379, 2.200062733, 2.201665778, 2.203831452, 2.203023537] 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 [0.04533693736, 0.04696863661, 0.04680049904, 0.04607894360, 0.04585803367, 0.04569675739, 0.04520907253, 0.04330272561, 0.04310098356, 0.04541577162, 0.04380457219, 0.04360158973, 0.04611637562, 0.04617706047, 0.04584454306, 0.04481928783, 0.04295647180, 0.04472230617, 0.04435884331, 0.04386821544, 0.04405119259] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 80, : The following constant c. 1 c = ----------------------------- Beta(3/7, 6/7) Beta(2/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (4/7) (2/7) (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 , 2.171496889, yielding an irrationality measure that is approximately , 20.51531539 We hope that the reader can find nu exactly. 1/2 1/3 (-28 + 84 I 3 ) Comment: Note that this constant appears to be , - -------------------- 12 7 - ---------------------- + 1/3 1/2 1/3 3 (-28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(-28 + 84 I 3 ) 14 | + 1/2 I 3 |-------------------- - ----------------------| | 6 1/2 1/3| \ 3 (-28 + 84 I 3 ) / Prove it! 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) (6 + 7 n) (35 n + 102 n + 73) X(n) -1/49 ------------------------------------------------------------------ 2 (35 n + 32 n + 6) (7 n + 11) (7 n + 8) (1 + n) (2 + n) 4 3 2 (5 + 7 n) (18865 n + 73843 n + 97769 n + 49682 n + 7656) X(1 + n) + 1/7 -------------------------------------------------------------------- 2 (35 n + 32 n + 6) (7 n + 11) (7 n + 8) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3 33 B(0) = 1, B(1) = -- 14 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 6/7) Beta(2/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (4/7) (2/7) (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, 5/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/7 4/7 2/7 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 [2.160249877, 2.160370570, 2.151912890, 2.157818471, 2.158429612, 2.151203914, 2.155764615, 2.155896759, 2.162901504, 2.162527719, 2.159478617, 2.163794145, 2.171496889, 2.170168207, 2.165919935, 2.167589471, 2.161029676, 2.166959325, 2.167918009, 2.166833758, 2.167040007] 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 [0.05383105872, 0.05380320544, 0.05575862119, 0.05439248726, 0.05425131428, 0.05592286616, 0.05486720318, 0.05483664734, 0.05321946078, 0.05330563133, 0.05400908325, 0.05301373276, 0.05124180573, 0.05154702791, 0.05252412289, 0.05213991609, 0.05365112431, 0.05228489730, 0.05206434335, 0.05231379197, 0.05226633211] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 81, : The following constant c. 1 c = ----------------------------- Beta(3/7, 4/7) Beta(2/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (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 , 2.188049332, yielding an irrationality measure that is approximately , 22.07294537 We hope that the reader can find nu exactly. 1/2 1/3 (28 + 84 I 3 ) Comment: Note that this constant appears to be , - ------------------- 12 7 - --------------------- - 1/3 1/2 1/3 3 (28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(28 + 84 I 3 ) 14 | + 1/2 I 3 |------------------- - ---------------------| | 6 1/2 1/3| \ 3 (28 + 84 I 3 ) / Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/49 (35 n + 107 n + 82) (3 + 7 n) (6 + 7 n) (-1 + 7 n) (4 + 7 n) (2 + 7 n) / 2 X(n) / ((7 n + 1) (35 n + 37 n + 10) (1 + n) (7 n + 12) (8 + 7 n) / (2 + n)) + 4 3 2 (6 + 7 n) (18865 n + 76538 n + 106995 n + 59538 n + 11520) X(1 + n) 1/7 ---------------------------------------------------------------------- 2 (35 n + 37 n + 10) (7 n + 12) (8 + 7 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2 38 B(0) = 1, B(1) = -- 35 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 4/7) Beta(2/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (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(3/7, 4/7) Beta(2/7, 6/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 4/7 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 [2.152999935, 2.165132953, 2.168844793, 2.172152193, 2.178526763, 2.175377201, 2.170711196, 2.169919548, 2.172820424, 2.178592866, 2.181973569, 2.185265049, 2.188049332, 2.183091324, 2.180365877, 2.174088831, 2.182845647, 2.184652662, 2.172338741, 2.175064287, 2.174179595] 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 [0.05550689006, 0.05270532673, 0.05185121617, 0.05109133584, 0.04962986136, 0.05035144168, 0.05142227215, 0.05160416905, 0.05093794202, 0.04961472745, 0.04884131630, 0.04808941117, 0.04745421120, 0.04858585559, 0.04920896949, 0.05064689967, 0.04864199392, 0.04822922311, 0.05104850885, 0.05042318602, 0.05062607959] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 82, : The following constant c. 1 c = ----------------------------- Beta(3/7, 4/7) Beta(3/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (4/7) (2/7) (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 , 2.207792143, yielding an irrationality measure that is approximately , 24.27090079 We hope that the reader can find nu exactly. 1/2 1/3 (-28 + 84 I 3 ) Comment: Note that this constant appears to be , - -------------------- 12 7 - ---------------------- - 2/3 1/2 1/3 3 (-28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(-28 + 84 I 3 ) 14 | + 1/2 I 3 |-------------------- - ----------------------| | 6 1/2 1/3| \ 3 (-28 + 84 I 3 ) / Prove it! 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) (5 + 7 n) (7 n + 10) (5 n + 6) X(n) -1/49 -------------------------------------------------------- (5 n + 1) (7 n + 9) (1 + n) (8 + 7 n) (2 + n) 3 2 (7 n + 10) (2695 n + 5929 n + 3676 n + 516) X(1 + n) + 1/7 ------------------------------------------------------ + X(2 + n) = (5 n + 1) (7 n + 9) (8 + 7 n) (2 + n) 0 Subject to the initial conditions A(0) = 0, A(1) = -3/2 B(0) = 1, B(1) = 9/14 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 4/7) Beta(3/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (4/7) (2/7) (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, 4/7) Beta(3/7, 5/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 3/7 4/7 2/7 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 [2.197284656, 2.190110278, 2.190848563, 2.194019462, 2.194991157, 2.195700816, 2.197848022, 2.206260550, 2.207152650, 2.196937831, 2.204043083, 2.204939804, 2.193855174, 2.193588382, 2.195050858, 2.199565916, 2.207792143, 2.199993531, 2.201596611, 2.203762320, 2.202954439] 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 [0.04535278679, 0.04698452735, 0.04681637671, 0.04609479142, 0.04587386685, 0.04571257773, 0.04522487017, 0.04331845776, 0.04311670149, 0.04543155166, 0.04382029571, 0.04361729944, 0.04613215317, 0.04619283166, 0.04586029627, 0.04483500223, 0.04297212252, 0.04473800223, 0.04437452052, 0.04388387000, 0.04406684493] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 83, : The following constant c. 1 c = ----------------------------- Beta(6/7, 6/7) Beta(3/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (1/7) (3/7) (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 , 2.187376971, yielding an irrationality measure that is approximately , 22.00507980 We hope that the reader can find nu exactly. 1/2 1/3 5 (28 + 84 I 3 ) Comment: Note that this constant appears to be , - --------------------- 12 35 - --------------------- + 10/3 1/2 1/3 3 (28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(28 + 84 I 3 ) 14 | - 5/2 I 3 |------------------- - ---------------------| | 6 1/2 1/3| \ 3 (28 + 84 I 3 ) / Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/49 (7 n + 13) (35 n + 103 n + 76) (4 + 7 n) (2 + 7 n) (6 + 7 n) (3 + 7 n) / 2 X(n) / ((35 n + 33 n + 8) (7 n + 15) (12 + 7 n) (7 n + 8) (1 + n) / (2 + n)) + 4 3 2 (7 n + 13) (18865 n + 74382 n + 100527 n + 53666 n + 9960) X(1 + n) 1/7 ---------------------------------------------------------------------- 2 (35 n + 33 n + 8) (7 n + 15) (12 + 7 n) (2 + n) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -6 B(0) = 1, B(1) = -15/7 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(6/7, 6/7) Beta(3/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (1/7) (1/7) (3/7) (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(6/7, 6/7) Beta(3/7, 4/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 1/7 1/7 3/7 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 [2.155395617, 2.165643677, 2.163968603, 2.174354848, 2.178526835, 2.174170092, 2.160927006, 2.173507269, 2.174836258, 2.178652155, 2.184324606, 2.187376971, 2.187241453, 2.175755676, 2.178815777, 2.173851236, 2.182845719, 2.180614358, 2.177228144, 2.174167959, 2.173486519] 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 [0.05495253646, 0.05258772462, 0.05297353461, 0.05058588118, 0.04962984487, 0.05062825942, 0.05367481135, 0.05078032198, 0.05047547426, 0.04960115395, 0.04830413577, 0.04760753158, 0.04763843957, 0.05026467874, 0.04956369647, 0.05070140476, 0.04864197746, 0.04915212906, 0.04992726128, 0.05062874869, 0.05078508308] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 84, : The following constant c. 1 c = ----------------------------- Beta(2/7, 5/7) Beta(3/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (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 , 2.208060744, yielding an irrationality measure that is approximately , 24.30382625 We hope that the reader can find nu exactly. Comment: Note that this constant appears to be , 1/2 1/3 (28 + 84 I 3 ) 14 ------------------- + --------------------- + 2/3 6 1/2 1/3 3 (28 + 84 I 3 ) Prove it! 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) (2 + 7 n) (7 n + 10) (5 n + 6) X(n) -1/49 ----------------------------------------------------------------- (7 n + 4) (5 n + 1) (1 + n) (7 n + 11) (6 + 7 n) (2 + n) 3 2 (7 n + 10) (2695 n + 5929 n + 3676 n + 516) X(1 + n) + 1/7 ------------------------------------------------------ + X(2 + n) = (5 n + 1) (7 n + 11) (6 + 7 n) (2 + n) 0 Subject to the initial conditions A(0) = 0, A(1) = -3/4 -9 B(0) = 1, B(1) = -- 28 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(2/7, 5/7) Beta(3/7, 3/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (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(2/7, 5/7) Beta(3/7, 3/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 5/7 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 [2.197555428, 2.190380913, 2.191119061, 2.194289823, 2.195261383, 2.195970906, 2.198117975, 2.206530368, 2.207422331, 2.197207377, 2.204312494, 2.205209080, 2.194124314, 2.193857387, 2.195319728, 2.199834651, 2.208060744, 2.200261997, 2.201864943, 2.204030517, 2.203222503] 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 [0.04529130200, 0.04692288167, 0.04675478202, 0.04603331278, 0.04581244485, 0.04565120557, 0.04516358634, 0.04325742781, 0.04305572611, 0.04537033602, 0.04375929918, 0.04355635718, 0.04607094778, 0.04613164988, 0.04579918406, 0.04477404040, 0.04291140816, 0.04467711272, 0.04431370374, 0.04382314091, 0.04400612468] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 85, : The following constant c. 1 c = ----------------------------- Beta(3/7, 4/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (3/7) (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 , 2.208019407, yielding an irrationality measure that is approximately , 24.29875330 We hope that the reader can find nu exactly. Comment: Note that this constant appears to be , 1/2 1/3 (28 + 84 I 3 ) 14 ------------------- + --------------------- + 2/3 6 1/2 1/3 3 (28 + 84 I 3 ) Prove it! 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) (2 + 7 n) (7 n + 11) (5 n + 6) X(n) -1/49 ------------------------------------------------------- (5 n + 1) (7 n + 12) (1 + n) (6 + 7 n) (2 + n) 3 2 (7 n + 11) (2695 n + 5929 n + 3676 n + 516) X(1 + n) + 1/7 ------------------------------------------------------ + X(2 + n) = (5 n + 1) (7 n + 12) (6 + 7 n) (2 + n) 0 Subject to the initial conditions A(0) = 0, A(1) = -4/5 -12 B(0) = 1, B(1) = --- 35 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(3/7, 4/7) Beta(2/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (4/7) (3/7) (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, 4/7) Beta(2/7, 4/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 4/7 3/7 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 [2.197513757, 2.190339263, 2.191077432, 2.194248215, 2.195219796, 2.195929340, 2.198076430, 2.206488844, 2.207380828, 2.197165895, 2.204271033, 2.205167639, 2.194082895, 2.193815987, 2.195278349, 2.199793293, 2.208019407, 2.200220681, 2.201823647, 2.203989243, 2.203181248] 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 [0.04530076385, 0.04693236830, 0.04676426082, 0.04604277374, 0.04582189705, 0.04566065009, 0.04517301728, 0.04326681963, 0.04306510955, 0.04537975639, 0.04376868577, 0.04356573560, 0.04608036645, 0.04614106532, 0.04580858874, 0.04478342189, 0.04292075148, 0.04468648295, 0.04432306291, 0.04383248631, 0.04401546905] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 86, : The following constant c. 1 c = ----------------------------- Beta(5/7, 6/7) Beta(3/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (1/7) (4/7) (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 , 2.208961092, yielding an irrationality measure that is approximately , 24.41484664 We hope that the reader can find nu exactly. 1/2 1/3 (-28 + 84 I 3 ) Comment: Note that this constant appears to be , - -------------------- 3 28 - ---------------------- + 4/3 1/2 1/3 3 (-28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(-28 + 84 I 3 ) 14 | - 2 I 3 |-------------------- - ----------------------| | 6 1/2 1/3| \ 3 (-28 + 84 I 3 ) / Prove it! 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) (3 + 7 n) (5 + 7 n) (5 n + 9) (7 n + 10) X(n) -1/49 ------------------------------------------------------- (5 n + 4) (7 n + 15) (7 n + 9) (1 + n) (2 + n) 3 2 (7 n + 10) (2695 n + 10241 n + 12300 n + 4680) X(1 + n) + 1/7 --------------------------------------------------------- + X(2 + n) (5 n + 4) (7 n + 15) (7 n + 9) (2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -3 -93 B(0) = 1, B(1) = --- 56 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(5/7, 6/7) Beta(3/7, 4/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (2/7) (1/7) (4/7) (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(3/7, 4/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 2/7 1/7 4/7 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 [2.196934582, 2.193461500, 2.192286924, 2.189347211, 2.196895501, 2.203018893, 2.205171187, 2.208961092, 2.201745364, 2.198982613, 2.204043156, 2.202662088, 2.194825232, 2.192419965, 2.198192043, 2.204461918, 2.199654570, 2.198954402, 2.200752639, 2.202328989, 2.196377723] 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 [0.04543228957, 0.04622169185, 0.04648893292, 0.04715837908, 0.04544116573, 0.04405224455, 0.04356493266, 0.04270794574, 0.04434080515, 0.04496734748, 0.04382027918, 0.04413307458, 0.04591158495, 0.04645865640, 0.04514677293, 0.04372547162, 0.04481489052, 0.04497374911, 0.04456585086, 0.04420854558, 0.04555877874] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 87, : The following constant c. 1 c = ----------------------------- Beta(2/7, 5/7) Beta(5/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (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 , 2.169272217, yielding an irrationality measure that is approximately , 20.32256889 We hope that the reader can find nu exactly. Comment: Note that this constant appears to be , 1/2 1/3 (-28 + 84 I 3 ) 28 -------------------- + ---------------------- - 4/3 3 1/2 1/3 3 (-28 + 84 I 3 ) Prove it! We need two lemmas Lemma: , let A(n), B(n), be two sequences of rational numbers that satisfy t\ he second-order recurrence 2 -1/49 (7 n + 12) (35 n + 108 n + 82) (2 + 7 n) (6 + 7 n) (3 + 7 n) (5 + 7 n) / 2 X(n) / ((35 n + 38 n + 9) (7 n + 8) (1 + n) (7 n + 16) (11 + 7 n) / (2 + n)) + 1/7 (7 n + 9) (7 n + 12) 4 3 2 / (18865 n + 77077 n + 107471 n + 58958 n + 10464) X(1 + n) / ( / 2 (35 n + 38 n + 9) (7 n + 8) (7 n + 16) (11 + 7 n) (2 + n)) + X(2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -5 -65 B(0) = 1, B(1) = --- 21 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(2/7, 5/7) Beta(5/7, 6/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (5/7) (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(2/7, 5/7) Beta(5/7, 6/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 5/7 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 [2.158820295, 2.159816141, 2.158787947, 2.160667909, 2.153908114, 2.157222016, 2.155907679, 2.161703303, 2.155078607, 2.162527863, 2.154705607, 2.165781810, 2.165068809, 2.169272217, 2.165070658, 2.167792856, 2.165259028, 2.163914935, 2.163056870, 2.166284320, 2.162892113] 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 [0.05416108669, 0.05393116755, 0.05416855682, 0.05373459245, 0.05529667162, 0.05453030425, 0.05483412237, 0.05349573798, 0.05502585818, 0.05330559813, 0.05511214298, 0.05255592187, 0.05272009873, 0.05175295302, 0.05271967291, 0.05209313069, 0.05267629356, 0.05298590032, 0.05318364762, 0.05244024380, 0.05322162556] As you can see, they are all positive We leave it to the reader to fill-in the details. ------------------------------------------------------------ Theorem number, 88, : The following constant c. 1 c = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (1/7) (5/7) (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 , 2.209030642, yielding an irrationality measure that is approximately , 24.42346496 We hope that the reader can find nu exactly. 1/2 1/3 (-28 + 84 I 3 ) Comment: Note that this constant appears to be , - -------------------- 4 7 - -------------------- + 1 1/2 1/3 (-28 + 84 I 3 ) / 1/2 1/3 \ 1/2 |(-28 + 84 I 3 ) 14 | - 3/2 I 3 |-------------------- - ----------------------| | 6 1/2 1/3| \ 3 (-28 + 84 I 3 ) / Prove it! 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) (5 n + 9) (7 n + 11) X(n) -1/49 ----------------------------------------------------------------- (5 n + 4) (7 n + 3) (7 n + 15) (7 n + 10) (1 + n) (2 + n) 3 2 (7 n + 11) (2695 n + 10241 n + 12300 n + 4680) X(1 + n) + 1/7 --------------------------------------------------------- + X(2 + n) (5 n + 4) (7 n + 15) (7 n + 10) (2 + n) = 0 Subject to the initial conditions A(0) = 0, A(1) = -2 -31 B(0) = 1, B(1) = --- 21 A(n) Then, ----, approximates B(n) 1 c = ----------------------------- Beta(4/7, 6/7) Beta(2/7, 5/7) 1 1 / / | | 1 | | -------------------------------------------------- dx dy | | (3/7) (1/7) (5/7) (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(4/7, 6/7) Beta(2/7, 5/7) 1 1 /x (1 - x) y (1 - y)\n / / |-------------------| | | \ -x y + 1 / | | ------------------------------------------ dx dy | | 3/7 1/7 5/7 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 [2.197004378, 2.193531261, 2.192356649, 2.189416901, 2.196965156, 2.203088513, 2.205240772, 2.209030642, 2.201814879, 2.199052093, 2.204112601, 2.203079638, 2.194894608, 2.192489305, 2.198261349, 2.204531189, 2.200418343, 2.199717792, 2.200821806, 2.202398122, 2.196446821] 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 [0.04541643774, 0.04620582403, 0.04647306518, 0.04714249901, 0.04542534565, 0.04403647441, 0.04354918515, 0.04269223199, 0.04432505008, 0.04495158145, 0.04380455567, 0.04403848472, 0.04589581405, 0.04644287719, 0.04513104098, 0.04370979035, 0.04464165589, 0.04480054868, 0.04455016791, 0.04419288106, 0.04554308163] 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, 413.737, seconds to generate