The Apery Limits for various summands that yield potential irrationality pro\ ofs By Shalosh B. Ekhad ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(n + k, n) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(n + k, n) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = -------------------- - ---------------- n n and in Maple notation b(n) = 3*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 3 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 (2 n - 1) a(n - 1) (n - 1) a(n - 2) a(n) = -------------------- - ---------------- n n but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.34657359027997265470861606072908828403775006718012762706034000474669681098\ 48473578029316634982093438 This constant is identified as, 1/2 ln(2) The implied delta is, 0.28208944895293503567247526227370938010128280589396180\ 3590112900926868791792305346843655678313101413 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) ----------------------- This took, 0.517, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(n + k, n) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(n + k, n) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 5 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = -------------------- - ---------------- n n and in Maple notation b(n) = 5*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 5 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 5 (2 n - 1) a(n - 1) (n - 1) a(n - 2) a(n) = -------------------- - ---------------- n n but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.20273255405408219098900655773217456828599521173124709880700716207205033562\ 44571256338762139086567006 This constant is identified as, 1/2 ln(3) - 1/2 ln(2) The implied delta is, 0.40330606914792842287305513778568069708698690184753427\ 2747814797033162530601321284962333451961341788 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(3) - 1/2 ln(2) ----------------------- This took, 0.623, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(n + k, n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(n + k, n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = -------------------- - ---------------- n n and in Maple notation b(n) = 7*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 7 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 (2 n - 1) a(n - 1) (n - 1) a(n - 2) a(n) = -------------------- - ---------------- n n but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.14384103622589046371960950299691371575175485544888052825333284267464647536\ 03902321690554495895526431 This constant is identified as, ln(2) - 1/2 ln(3) The implied delta is, 0.45955185308894937690096235029189032763778189029918608\ 8580594674934576733340494643077882965851758694 Since this is positive, this suggests an Apery-style irrationality proof of, ln(2) - 1/2 ln(3) ----------------------- This took, 0.626, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(n + 2 k, n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(n + 2 k, n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 2 (3 n - 2) (93 n - 155 n + 42) b(n - 1) b(n) = 1/2 --------------------------------------- n (2 n - 1) (3 n - 4) 2 (n - 1) (18 n - 33 n + 10) b(n - 2) - ------------------------------------ n (2 n - 1) (3 n - 4) 2 (3 n - 1) (n - 1) (n - 2) b(n - 3) + ------------------------------------ n (2 n - 1) (3 n - 4) and in Maple notation b(n) = 1/2*(3*n-2)*(93*n^2-155*n+42)/n/(2*n-1)/(3*n-4)*b(n-1)-(n-1)*(18*n^2-33* n+10)/n/(2*n-1)/(3*n-4)*b(n-2)+2*(3*n-1)*(n-1)*(n-2)/n/(2*n-1)/(3*n-4)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 10, b(2) = 172 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 2 (3 n - 2) (93 n - 155 n + 42) a(n - 1) a(n) = 1/2 --------------------------------------- n (2 n - 1) (3 n - 4) 2 (n - 1) (18 n - 33 n + 10) a(n - 2) - ------------------------------------ n (2 n - 1) (3 n - 4) 2 (3 n - 1) (n - 1) (n - 2) a(n - 3) + ------------------------------------ n (2 n - 1) (3 n - 4) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.10076663134634543614411545875789754068244812489404114308721582974378462117\ 36863268713402771228922316 1/2 3 Pi This constant is identified as, ------- 54 The implied delta is, 0.06358044089477022109069338782503122229426112713571172\ 9230995627881338427074074180475573475924428287 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 54 ----------------------- This took, 4.361, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(2 n + k, n) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(2 n + k, n) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (220 n - 506 n + 334 n - 63) b(n - 1) b(n) = --------------------------------------- n (2 n - 1) (10 n - 13) (2 n - 3) (10 n - 3) (n - 1) b(n - 2) + ------------------------------------- n (2 n - 1) (10 n - 13) and in Maple notation b(n) = (220*n^3-506*n^2+334*n-63)/n/(2*n-1)/(10*n-13)*b(n-1)+(2*n-3)*(10*n-3)*( n-1)/n/(2*n-1)/(10*n-13)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 5 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (220 n - 506 n + 334 n - 63) a(n - 1) a(n) = --------------------------------------- n (2 n - 1) (10 n - 13) (2 n - 3) (10 n - 3) (n - 1) a(n - 2) + ------------------------------------- n (2 n - 1) (10 n - 13) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.19804205158855580269063774898805044802157146696007292974876571699811246341\ 99127758873895219989767679 This constant is identified as, 2/7 ln(2) The implied delta is, 0.09450081864947373486118808924334890240127476350947215\ 5465207315783771390873766377966411706938259117 Since this is positive, this suggests an Apery-style irrationality proof of, 2/7 ln(2) ----------------------- This took, 0.940, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + k, n) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + k, n) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (2717 n - 6292 n + 4175 n - 792) b(n - 1) b(n) = 1/4 ------------------------------------------- n (2 n - 1) (19 n - 25) (2 n - 3) (19 n - 6) (n - 1) b(n - 2) + 1/2 ------------------------------------- n (2 n - 1) (19 n - 25) and in Maple notation b(n) = 1/4*(2717*n^3-6292*n^2+4175*n-792)/n/(2*n-1)/(19*n-25)*b(n-1)+1/2*(2*n-3 )*(19*n-6)*(n-1)/n/(2*n-1)/(19*n-25)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 8 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (2717 n - 6292 n + 4175 n - 792) a(n - 1) a(n) = 1/4 ------------------------------------------- n (2 n - 1) (19 n - 25) (2 n - 3) (19 n - 6) (n - 1) a(n - 2) + 1/2 ------------------------------------- n (2 n - 1) (19 n - 25) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.12475849480251211753169634321979973432984320721922898388123517665972328346\ 12043850054622854822502773 This constant is identified as, 4/13 ln(3) - 4/13 ln(2) The implied delta is, 0.16077786103954935164309882348392598081857123485719659\ 5546217461153715685953105023490876165199490287 Since this is positive, this suggests an Apery-style irrationality proof of, 4/13 ln(3) - 4/13 ln(2) ----------------------- This took, 1.096, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + k, n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + k, n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (4144 n - 9620 n + 6394 n - 1215) b(n - 1) b(n) = 1/3 -------------------------------------------- n (2 n - 1) (28 n - 37) (28 n - 9) (2 n - 3) (n - 1) b(n - 2) + 1/3 ------------------------------------- n (2 n - 1) (28 n - 37) and in Maple notation b(n) = 1/3*(4144*n^3-9620*n^2+6394*n-1215)/n/(2*n-1)/(28*n-37)*b(n-1)+1/3*(28*n -9)*(2*n-3)*(n-1)/n/(2*n-1)/(28*n-37)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 11 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (4144 n - 9620 n + 6394 n - 1215) a(n - 1) a(n) = 1/3 -------------------------------------------- n (2 n - 1) (28 n - 37) (28 n - 9) (2 n - 3) (n - 1) a(n - 2) + 1/3 ------------------------------------- n (2 n - 1) (28 n - 37) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.09084697024793081919133231768226129415900306659929296521263126905767145812\ 235172558045607342498061672 12 This constant is identified as, -- ln(2) - 6/19 ln(3) 19 The implied delta is, 0.19469433665088390406866976250823815531399448640449284\ 8165186845140709040438054502212716954833707070 Since this is positive, this suggests an Apery-style irrationality proof of, 12 -- ln(2) - 6/19 ln(3) 19 ----------------------- This took, 1.255, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(2 n + k, n + k) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(2 n + k, n + k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (220 n - 506 n + 334 n - 63) b(n - 1) b(n) = --------------------------------------- n (2 n - 1) (10 n - 13) (2 n - 3) (10 n - 3) (n - 1) b(n - 2) + ------------------------------------- n (2 n - 1) (10 n - 13) and in Maple notation b(n) = (220*n^3-506*n^2+334*n-63)/n/(2*n-1)/(10*n-13)*b(n-1)+(2*n-3)*(10*n-3)*( n-1)/n/(2*n-1)/(10*n-13)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 5 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (220 n - 506 n + 334 n - 63) a(n - 1) a(n) = --------------------------------------- n (2 n - 1) (10 n - 13) (2 n - 3) (10 n - 3) (n - 1) a(n - 2) + ------------------------------------- n (2 n - 1) (10 n - 13) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.19804205158855580269063774898805044802157146696007292974876571699811246341\ 99127758873895219989767679 This constant is identified as, 2/7 ln(2) The implied delta is, 0.09450081864947373486118808924334890240127476350947215\ 5465207315783771390873766377966411706938259117 Since this is positive, this suggests an Apery-style irrationality proof of, 2/7 ln(2) ----------------------- This took, 0.282, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + k, n + k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + k, n + k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (2717 n - 6292 n + 4175 n - 792) b(n - 1) b(n) = 1/4 ------------------------------------------- n (2 n - 1) (19 n - 25) (2 n - 3) (19 n - 6) (n - 1) b(n - 2) + 1/2 ------------------------------------- n (2 n - 1) (19 n - 25) and in Maple notation b(n) = 1/4*(2717*n^3-6292*n^2+4175*n-792)/n/(2*n-1)/(19*n-25)*b(n-1)+1/2*(2*n-3 )*(19*n-6)*(n-1)/n/(2*n-1)/(19*n-25)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 8 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (2717 n - 6292 n + 4175 n - 792) a(n - 1) a(n) = 1/4 ------------------------------------------- n (2 n - 1) (19 n - 25) (2 n - 3) (19 n - 6) (n - 1) a(n - 2) + 1/2 ------------------------------------- n (2 n - 1) (19 n - 25) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.12475849480251211753169634321979973432984320721922898388123517665972328346\ 12043850054622854822502773 This constant is identified as, 4/13 ln(3) - 4/13 ln(2) The implied delta is, 0.16077786103954935164309882348392598081857123485719659\ 5546217461153715685953105023490876165199490287 Since this is positive, this suggests an Apery-style irrationality proof of, 4/13 ln(3) - 4/13 ln(2) ----------------------- This took, 0.404, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + k, n + k) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + k, n + k) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (4144 n - 9620 n + 6394 n - 1215) b(n - 1) b(n) = 1/3 -------------------------------------------- n (2 n - 1) (28 n - 37) (28 n - 9) (2 n - 3) (n - 1) b(n - 2) + 1/3 ------------------------------------- n (2 n - 1) (28 n - 37) and in Maple notation b(n) = 1/3*(4144*n^3-9620*n^2+6394*n-1215)/n/(2*n-1)/(28*n-37)*b(n-1)+1/3*(28*n -9)*(2*n-3)*(n-1)/n/(2*n-1)/(28*n-37)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 11 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (4144 n - 9620 n + 6394 n - 1215) a(n - 1) a(n) = 1/3 -------------------------------------------- n (2 n - 1) (28 n - 37) (28 n - 9) (2 n - 3) (n - 1) a(n - 2) + 1/3 ------------------------------------- n (2 n - 1) (28 n - 37) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.09084697024793081919133231768226129415900306659929296521263126905767145812\ 235172558045607342498061672 12 This constant is identified as, -- ln(2) - 6/19 ln(3) 19 The implied delta is, 0.19469433665088390406866976250823815531399448640449284\ 8165186845140709040438054502212716954833707070 Since this is positive, this suggests an Apery-style irrationality proof of, 12 -- ln(2) - 6/19 ln(3) 19 ----------------------- This took, 0.377, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(2 n + k, 2 n) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(2 n + k, 2 n) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (1207 n - 2840 n + 1897 n - 360) b(n - 1) b(n) = 1/4 ------------------------------------------- n (2 n - 1) (17 n - 23) (17 n - 6) (2 n - 3) (n - 1) b(n - 2) - 1/2 ------------------------------------- n (2 n - 1) (17 n - 23) and in Maple notation b(n) = 1/4*(1207*n^3-2840*n^2+1897*n-360)/n/(2*n-1)/(17*n-23)*b(n-1)-1/2*(17*n-\ 6)*(2*n-3)*(n-1)/n/(2*n-1)/(17*n-23)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 4 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (1207 n - 2840 n + 1897 n - 360) a(n - 1) a(n) = 1/4 ------------------------------------------- n (2 n - 1) (17 n - 23) (17 n - 6) (2 n - 3) (n - 1) a(n - 2) - 1/2 ------------------------------------- n (2 n - 1) (17 n - 23) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.25205352020361647615172077143933693384563641249463827422570182163396131707\ 98889874930412098168795227 This constant is identified as, 4/11 ln(2) The implied delta is, 0.03971455809607690407724930452436971953812899578434684\ 5186697939176116293063562374974861726912730253 Since this is positive, this suggests an Apery-style irrationality proof of, 4/11 ln(2) ----------------------- This took, 0.882, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + k, 2 n) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + k, 2 n) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (2444 n - 5734 n + 3830 n - 729) b(n - 1) b(n) = 1/3 ------------------------------------------- n (2 n - 1) (26 n - 35) (26 n - 9) (2 n - 3) (n - 1) b(n - 2) - 1/3 ------------------------------------- n (2 n - 1) (26 n - 35) and in Maple notation b(n) = 1/3*(2444*n^3-5734*n^2+3830*n-729)/n/(2*n-1)/(26*n-35)*b(n-1)-1/3*(26*n-\ 9)*(2*n-3)*(n-1)/n/(2*n-1)/(26*n-35)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 7 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (2444 n - 5734 n + 3830 n - 729) a(n - 1) a(n) = 1/3 ------------------------------------------- n (2 n - 1) (26 n - 35) (26 n - 9) (2 n - 3) (n - 1) a(n - 2) - 1/3 ------------------------------------- n (2 n - 1) (26 n - 35) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.14310533227346978187459286428153498937834956122205677562847564381556494279\ 37344416239126215825812004 This constant is identified as, 6/17 ln(3) - 6/17 ln(2) The implied delta is, 0.13149585668492110648740137365006082564166423387376784\ 4495224958840499306882347203954146688655201411 Since this is positive, this suggests an Apery-style irrationality proof of, 6/17 ln(3) - 6/17 ln(2) ----------------------- This took, 1.191, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + k, 2 n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + k, 2 n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (12565 n - 29438 n + 19657 n - 3744) b(n - 1) b(n) = 1/8 ----------------------------------------------- n (2 n - 1) (35 n - 47) (35 n - 12) (2 n - 3) (n - 1) b(n - 2) - 1/4 -------------------------------------- n (2 n - 1) (35 n - 47) and in Maple notation b(n) = 1/8*(12565*n^3-29438*n^2+19657*n-3744)/n/(2*n-1)/(35*n-47)*b(n-1)-1/4*( 35*n-12)*(2*n-3)*(n-1)/n/(2*n-1)/(35*n-47)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 10 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (12565 n - 29438 n + 19657 n - 3744) a(n - 1) a(n) = 1/8 ----------------------------------------------- n (2 n - 1) (35 n - 47) (35 n - 12) (2 n - 3) (n - 1) a(n - 2) - 1/4 -------------------------------------- n (2 n - 1) (35 n - 47) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.10006332954844553997885878469350519356643816031226471530666632533888450459\ 85323354219516171057757518 16 This constant is identified as, -- ln(2) - 8/23 ln(3) 23 The implied delta is, 0.17340163870401647072048442938485016399227287481101632\ 3874093268900625975183354722280784146320809256 Since this is positive, this suggests an Apery-style irrationality proof of, 16 -- ln(2) - 8/23 ln(3) 23 ----------------------- This took, 1.239, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(2 n + 2 k, n) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(2 n + 2 k, n) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients b(n) = 2/3 ( 6 5 4 3 2 26200 n - 171500 n + 433204 n - 534701 n + 335932 n - 100607 n + 11040 / 3 2 ) b(n - 1) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) - / 5 4 3 2 4/3 (n - 1) (4323 n - 26136 n + 57416 n - 55380 n + 22231 n - 2880) / 3 2 b(n - 2) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) / 3 2 (2 n - 5) (n - 1) (n - 2) (131 n - 268 n + 155 n - 24) b(n - 3) + 8/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578) and in Maple notation b(n) = 2/3*(26200*n^6-171500*n^5+433204*n^4-534701*n^3+335932*n^2-100607*n+ 11040)/n/(3*n-1)/(3*n-2)/(131*n^3-661*n^2+1084*n-578)*b(n-1)-4/3*(n-1)*(4323*n^ 5-26136*n^4+57416*n^3-55380*n^2+22231*n-2880)/n/(3*n-1)/(3*n-2)/(131*n^3-661*n^ 2+1084*n-578)*b(n-2)+8/3*(2*n-5)*(n-1)*(n-2)*(131*n^3-268*n^2+155*n-24)/n/(3*n-\ 1)/(3*n-2)/(131*n^3-661*n^2+1084*n-578)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 6, b(2) = 64 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. a(n) = 2/3 ( 6 5 4 3 2 26200 n - 171500 n + 433204 n - 534701 n + 335932 n - 100607 n + 11040 / 3 2 ) a(n - 1) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) - / 5 4 3 2 4/3 (n - 1) (4323 n - 26136 n + 57416 n - 55380 n + 22231 n - 2880) / 3 2 a(n - 2) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) / 3 2 (2 n - 5) (n - 1) (n - 2) (131 n - 268 n + 155 n - 24) a(n - 3) + 8/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.17328679513998632735430803036454414201887503359006381353017000237334840549\ 24236789014658317491046719 This constant is identified as, 1/4 ln(2) The implied delta is, 0.02362677832577121704651587686528684448340284092551631\ 7179168960359509353689600282504383247155132055 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(2) ----------------------- This took, 1.040, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + 2 k, n) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + 2 k, n) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 5 4 3 2 b(n) = 2/3 (84952 n - 556780 n + 1407916 n - 1739017 n + 1092652 n / - 326923 n + 35760) b(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (259 n - 1309 n + 2150 n - 1148)) - 2/3 (n - 1) 5 4 3 2 (16835 n - 101920 n + 224499 n - 217604 n + 88102 n - 11520) b(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148)) / 3 2 (2 n - 5) (n - 1) (n - 2) (259 n - 532 n + 309 n - 48) b(n - 3) + 4/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148) and in Maple notation b(n) = 2/3*(84952*n^6-556780*n^5+1407916*n^4-1739017*n^3+1092652*n^2-326923*n+ 35760)/n/(3*n-1)/(3*n-2)/(259*n^3-1309*n^2+2150*n-1148)*b(n-1)-2/3*(n-1)*(16835 *n^5-101920*n^4+224499*n^3-217604*n^2+88102*n-11520)/n/(3*n-1)/(3*n-2)/(259*n^3 -1309*n^2+2150*n-1148)*b(n-2)+4/3*(2*n-5)*(n-1)*(n-2)*(259*n^3-532*n^2+309*n-48 )/n/(3*n-1)/(3*n-2)/(259*n^3-1309*n^2+2150*n-1148)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 10, b(2) = 178 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 5 4 3 2 a(n) = 2/3 (84952 n - 556780 n + 1407916 n - 1739017 n + 1092652 n / - 326923 n + 35760) a(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (259 n - 1309 n + 2150 n - 1148)) - 2/3 (n - 1) 5 4 3 2 (16835 n - 101920 n + 224499 n - 217604 n + 88102 n - 11520) a(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148)) / 3 2 (2 n - 5) (n - 1) (n - 2) (259 n - 532 n + 309 n - 48) a(n - 3) + 4/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.10136627702704109549450327886608728414299760586562354940350358103602516781\ 22285628169381069543283503 This constant is identified as, 1/4 ln(3) - 1/4 ln(2) The implied delta is, 0.07160227753194817374680632792013916548785650294048345\ 0510932389745246108697191885894659462301350069 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(3) - 1/4 ln(2) ----------------------- This took, 1.197, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + 2 k, n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + 2 k, n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 5 4 3 2 b(n) = 2 (58824 n - 385700 n + 975628 n - 1205239 n + 757140 n - 226381 n / + 24720) b(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (387 n - 1957 n + 3216 n - 1718)) - 4/9 (n - 1) 5 4 3 2 (37539 n - 227368 n + 501278 n - 486692 n + 197617 n - 25920) b(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718)) / 3 2 (2 n - 5) (n - 1) (n - 2) (387 n - 796 n + 463 n - 72) b(n - 3) + 8/9 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718) and in Maple notation b(n) = 2*(58824*n^6-385700*n^5+975628*n^4-1205239*n^3+757140*n^2-226381*n+24720 )/n/(3*n-1)/(3*n-2)/(387*n^3-1957*n^2+3216*n-1718)*b(n-1)-4/9*(n-1)*(37539*n^5-\ 227368*n^4+501278*n^3-486692*n^2+197617*n-25920)/n/(3*n-1)/(3*n-2)/(387*n^3-\ 1957*n^2+3216*n-1718)*b(n-2)+8/9*(2*n-5)*(n-1)*(n-2)*(387*n^3-796*n^2+463*n-72) /n/(3*n-1)/(3*n-2)/(387*n^3-1957*n^2+3216*n-1718)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 14, b(2) = 348 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 5 4 3 2 a(n) = 2 (58824 n - 385700 n + 975628 n - 1205239 n + 757140 n - 226381 n / + 24720) a(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (387 n - 1957 n + 3216 n - 1718)) - 4/9 (n - 1) 5 4 3 2 (37539 n - 227368 n + 501278 n - 486692 n + 197617 n - 25920) a(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718)) / 3 2 (2 n - 5) (n - 1) (n - 2) (387 n - 796 n + 463 n - 72) a(n - 3) + 8/9 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.07192051811294523185980475149845685787587742772444026412666642133732323768\ 019511608452772479477632157 This constant is identified as, 1/2 ln(2) - 1/4 ln(3) The implied delta is, 0.09518263135642492669137578921452533782491720108863642\ 2067157992831004535255119175129378724936380376 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) - 1/4 ln(3) ----------------------- This took, 1.388, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(2 n + 2 k, n + 2 k) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(2 n + 2 k, n + 2 k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients b(n) = 2/3 ( 6 5 4 3 2 26200 n - 171500 n + 433204 n - 534701 n + 335932 n - 100607 n + 11040 / 3 2 ) b(n - 1) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) - / 5 4 3 2 4/3 (n - 1) (4323 n - 26136 n + 57416 n - 55380 n + 22231 n - 2880) / 3 2 b(n - 2) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) / 3 2 (2 n - 5) (n - 1) (n - 2) (131 n - 268 n + 155 n - 24) b(n - 3) + 8/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578) and in Maple notation b(n) = 2/3*(26200*n^6-171500*n^5+433204*n^4-534701*n^3+335932*n^2-100607*n+ 11040)/n/(3*n-1)/(3*n-2)/(131*n^3-661*n^2+1084*n-578)*b(n-1)-4/3*(n-1)*(4323*n^ 5-26136*n^4+57416*n^3-55380*n^2+22231*n-2880)/n/(3*n-1)/(3*n-2)/(131*n^3-661*n^ 2+1084*n-578)*b(n-2)+8/3*(2*n-5)*(n-1)*(n-2)*(131*n^3-268*n^2+155*n-24)/n/(3*n-\ 1)/(3*n-2)/(131*n^3-661*n^2+1084*n-578)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 6, b(2) = 64 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. a(n) = 2/3 ( 6 5 4 3 2 26200 n - 171500 n + 433204 n - 534701 n + 335932 n - 100607 n + 11040 / 3 2 ) a(n - 1) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) - / 5 4 3 2 4/3 (n - 1) (4323 n - 26136 n + 57416 n - 55380 n + 22231 n - 2880) / 3 2 a(n - 2) / (n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578)) / 3 2 (2 n - 5) (n - 1) (n - 2) (131 n - 268 n + 155 n - 24) a(n - 3) + 8/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (131 n - 661 n + 1084 n - 578) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.17328679513998632735430803036454414201887503359006381353017000237334840549\ 24236789014658317491046719 This constant is identified as, 1/4 ln(2) The implied delta is, 0.02362677832577121704651587686528684448340284092551631\ 7179168960359509353689600282504383247155132055 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(2) ----------------------- This took, 0.648, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + 2 k, n + 2 k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + 2 k, n + 2 k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 5 4 3 2 b(n) = 2/3 (84952 n - 556780 n + 1407916 n - 1739017 n + 1092652 n / - 326923 n + 35760) b(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (259 n - 1309 n + 2150 n - 1148)) - 2/3 (n - 1) 5 4 3 2 (16835 n - 101920 n + 224499 n - 217604 n + 88102 n - 11520) b(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148)) / 3 2 (2 n - 5) (n - 1) (n - 2) (259 n - 532 n + 309 n - 48) b(n - 3) + 4/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148) and in Maple notation b(n) = 2/3*(84952*n^6-556780*n^5+1407916*n^4-1739017*n^3+1092652*n^2-326923*n+ 35760)/n/(3*n-1)/(3*n-2)/(259*n^3-1309*n^2+2150*n-1148)*b(n-1)-2/3*(n-1)*(16835 *n^5-101920*n^4+224499*n^3-217604*n^2+88102*n-11520)/n/(3*n-1)/(3*n-2)/(259*n^3 -1309*n^2+2150*n-1148)*b(n-2)+4/3*(2*n-5)*(n-1)*(n-2)*(259*n^3-532*n^2+309*n-48 )/n/(3*n-1)/(3*n-2)/(259*n^3-1309*n^2+2150*n-1148)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 10, b(2) = 178 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 5 4 3 2 a(n) = 2/3 (84952 n - 556780 n + 1407916 n - 1739017 n + 1092652 n / - 326923 n + 35760) a(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (259 n - 1309 n + 2150 n - 1148)) - 2/3 (n - 1) 5 4 3 2 (16835 n - 101920 n + 224499 n - 217604 n + 88102 n - 11520) a(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148)) / 3 2 (2 n - 5) (n - 1) (n - 2) (259 n - 532 n + 309 n - 48) a(n - 3) + 4/3 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (259 n - 1309 n + 2150 n - 1148) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.10136627702704109549450327886608728414299760586562354940350358103602516781\ 22285628169381069543283503 This constant is identified as, 1/4 ln(3) - 1/4 ln(2) The implied delta is, 0.07160227753194817374680632792013916548785650294048345\ 0510932389745246108697191885894659462301350069 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(3) - 1/4 ln(2) ----------------------- This took, 0.740, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + 2 k, n + 2 k) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + 2 k, n + 2 k) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 5 4 3 2 b(n) = 2 (58824 n - 385700 n + 975628 n - 1205239 n + 757140 n - 226381 n / + 24720) b(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (387 n - 1957 n + 3216 n - 1718)) - 4/9 (n - 1) 5 4 3 2 (37539 n - 227368 n + 501278 n - 486692 n + 197617 n - 25920) b(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718)) / 3 2 (2 n - 5) (n - 1) (n - 2) (387 n - 796 n + 463 n - 72) b(n - 3) + 8/9 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718) and in Maple notation b(n) = 2*(58824*n^6-385700*n^5+975628*n^4-1205239*n^3+757140*n^2-226381*n+24720 )/n/(3*n-1)/(3*n-2)/(387*n^3-1957*n^2+3216*n-1718)*b(n-1)-4/9*(n-1)*(37539*n^5-\ 227368*n^4+501278*n^3-486692*n^2+197617*n-25920)/n/(3*n-1)/(3*n-2)/(387*n^3-\ 1957*n^2+3216*n-1718)*b(n-2)+8/9*(2*n-5)*(n-1)*(n-2)*(387*n^3-796*n^2+463*n-72) /n/(3*n-1)/(3*n-2)/(387*n^3-1957*n^2+3216*n-1718)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 14, b(2) = 348 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 5 4 3 2 a(n) = 2 (58824 n - 385700 n + 975628 n - 1205239 n + 757140 n - 226381 n / + 24720) a(n - 1) / (n (3 n - 1) (3 n - 2) / 3 2 (387 n - 1957 n + 3216 n - 1718)) - 4/9 (n - 1) 5 4 3 2 (37539 n - 227368 n + 501278 n - 486692 n + 197617 n - 25920) a(n - 2) / 3 2 / (n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718)) / 3 2 (2 n - 5) (n - 1) (n - 2) (387 n - 796 n + 463 n - 72) a(n - 3) + 8/9 ----------------------------------------------------------------- 3 2 n (3 n - 1) (3 n - 2) (387 n - 1957 n + 3216 n - 1718) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.07192051811294523185980475149845685787587742772444026412666642133732323768\ 019511608452772479477632157 This constant is identified as, 1/2 ln(2) - 1/4 ln(3) The implied delta is, 0.09518263135642492669137578921452533782491720108863642\ 2067157992831004535255119175129378724936380376 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) - 1/4 ln(3) ----------------------- This took, 0.755, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(2 n + 2 k, 2 n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(2 n + 2 k, 2 n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 4 3 2 (16974 n - 77625 n + 113376 n - 61061 n + 9400) b(n - 1) b(n) = 1/2 ----------------------------------------------------------- 2 (2 n - 1) (82 n - 293 n + 239) n 4 3 2 (n - 1) (3444 n - 19194 n + 36438 n - 26143 n + 4640) b(n - 2) + 1/2 ----------------------------------------------------------------- 2 n (2 n - 1) (2 n - 3) (82 n - 293 n + 239) 2 (2 n - 5) (n - 1) (n - 2) (82 n - 129 n + 28) b(n - 3) + 1/2 ------------------------------------------------------- 2 n (2 n - 1) (2 n - 3) (82 n - 293 n + 239) and in Maple notation b(n) = 1/2*(16974*n^4-77625*n^3+113376*n^2-61061*n+9400)/(2*n-1)/(82*n^2-293*n+ 239)/n*b(n-1)+1/2*(n-1)*(3444*n^4-19194*n^3+36438*n^2-26143*n+4640)/n/(2*n-1)/( 2*n-3)/(82*n^2-293*n+239)*b(n-2)+1/2*(2*n-5)*(n-1)*(n-2)*(82*n^2-129*n+28)/n/(2 *n-1)/(2*n-3)/(82*n^2-293*n+239)*b(n-3) Of course, the initial conditions are b(0) = 1, b(1) = 19, b(2) = 721 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 4 3 2 (16974 n - 77625 n + 113376 n - 61061 n + 9400) a(n - 1) a(n) = 1/2 ----------------------------------------------------------- 2 (2 n - 1) (82 n - 293 n + 239) n 4 3 2 (n - 1) (3444 n - 19194 n + 36438 n - 26143 n + 4640) a(n - 2) + 1/2 ----------------------------------------------------------------- 2 n (2 n - 1) (2 n - 3) (82 n - 293 n + 239) 2 (2 n - 5) (n - 1) (n - 2) (82 n - 129 n + 28) a(n - 3) + 1/2 ------------------------------------------------------- 2 n (2 n - 1) (2 n - 3) (82 n - 293 n + 239) but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.05252216310905004074275805302065255858571534068864051837028967165876169841\ 421110129102435875978287949 The implied delta is, 0.11321836162580310591806431133341644615010850653410484\ 7732350089962552802605049292633329731068528935 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0525221631090500407427580530206525585857153406886405183702896716587616\ 9841421110129102435875978287949 ----------------------- This took, 4.361, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, n) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, n) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 5 4 3 2 b(n) = 1/2 (126360 n - 533520 n + 835371 n - 597030 n + 191987 n - 21760) / 2 b(n - 1) / (n (3 n - 1) (3 n - 2) (270 n - 735 n + 497)) / 2 (n - 1) (3 n - 4) (3 n - 5) (270 n - 195 n + 32) b(n - 2) - 1/4 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (270 n - 735 n + 497) and in Maple notation b(n) = 1/2*(126360*n^5-533520*n^4+835371*n^3-597030*n^2+191987*n-21760)/n/(3*n-\ 1)/(3*n-2)/(270*n^2-735*n+497)*b(n-1)-1/4*(n-1)*(3*n-4)*(3*n-5)*(270*n^2-195*n+ 32)/n/(3*n-1)/(3*n-2)/(270*n^2-735*n+497)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 11 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 5 4 3 2 a(n) = 1/2 (126360 n - 533520 n + 835371 n - 597030 n + 191987 n - 21760) / 2 a(n - 1) / (n (3 n - 1) (3 n - 2) (270 n - 735 n + 497)) / 2 (n - 1) (3 n - 4) (3 n - 5) (270 n - 195 n + 32) a(n - 2) - 1/4 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (270 n - 735 n + 497) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.09094544480930789876142350253405961941801654358037253030594713812577958981\ 284057972360802119266842645 24 24 This constant is identified as, --- ln(3) - --- ln(2) 107 107 The implied delta is, 0.03661746617527283524134384554671528675530807862065396\ 1752763393969921416699834074598545930295214481 Since this is positive, this suggests an Apery-style irrationality proof of, 24 24 --- ln(3) - --- ln(2) 107 107 ----------------------- This took, 1.485, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients b(n) = 2/27 5 4 3 2 (2579174 n - 10911890 n + 17111599 n - 12243496 n + 3940893 n - 447120) / 2 b(n - 1) / (n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107)) / 2 (n - 1) (3 n - 4) (3 n - 5) (598 n - 437 n + 72) b(n - 2) - 1/9 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107) and in Maple notation b(n) = 2/27*(2579174*n^5-10911890*n^4+17111599*n^3-12243496*n^2+3940893*n-\ 447120)/n/(3*n-1)/(3*n-2)/(598*n^2-1633*n+1107)*b(n-1)-1/9*(n-1)*(3*n-4)*(3*n-5 )*(598*n^2-437*n+72)/n/(3*n-1)/(3*n-2)/(598*n^2-1633*n+1107)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 15 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. a(n) = 2/27 5 4 3 2 (2579174 n - 10911890 n + 17111599 n - 12243496 n + 3940893 n - 447120) / 2 a(n - 1) / (n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107)) / 2 (n - 1) (3 n - 4) (3 n - 5) (598 n - 437 n + 72) a(n - 2) - 1/9 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.06667309833646424927775891126037202275188637076600470837493539488781896711\ 983753250754501526039349983 108 54 This constant is identified as, --- ln(2) - --- ln(3) 233 233 The implied delta is, 0.06597544597881529603803186023005591329157253066585191\ 0079595969038781224778004356707050934970947861 Since this is positive, this suggests an Apery-style irrationality proof of, 108 54 --- ln(2) - --- ln(3) 233 233 ----------------------- This took, 1.823, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(3 n + k, n + k) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(3 n + k, n + k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (693 n - 1554 n + 989 n - 160) b(n - 1) b(n) = 9/4 ----------------------------------------- (2 n - 1) (33 n - 41) n (3 n - 4) (33 n - 8) (3 n - 5) (n - 1) b(n - 2) + 3/8 ----------------------------------------------- n (33 n - 41) (2 n - 1) (2 n - 3) and in Maple notation b(n) = 9/4*(693*n^3-1554*n^2+989*n-160)/(2*n-1)/(33*n-41)/n*b(n-1)+3/8*(3*n-4)* (33*n-8)*(3*n-5)*(n-1)/n/(33*n-41)/(2*n-1)/(2*n-3)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 9 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (693 n - 1554 n + 989 n - 160) a(n - 1) a(n) = 9/4 ----------------------------------------- (2 n - 1) (33 n - 41) n (3 n - 4) (33 n - 8) (3 n - 5) (n - 1) a(n - 2) + 3/8 ----------------------------------------------- n (33 n - 41) (2 n - 1) (2 n - 3) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.11090354888959124950675713943330825089208002149764084065930880151894297951\ 51511544969381323194269900 This constant is identified as, 4/25 ln(2) The implied delta is, 0.14104379627041509051379936500361885164023421843144384\ 9713953691753590882640847436845046990892375390 Since this is positive, this suggests an Apery-style irrationality proof of, 4/25 ln(2) ----------------------- This took, 1.057, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, n + k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, n + k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (18527 n - 41638 n + 26567 n - 4320) b(n - 1) b(n) = 5/12 ----------------------------------------------- (2 n - 1) (97 n - 121) n (97 n - 24) (3 n - 4) (3 n - 5) (n - 1) b(n - 2) + 1/8 ------------------------------------------------ n (97 n - 121) (2 n - 1) (2 n - 3) and in Maple notation b(n) = 5/12*(18527*n^3-41638*n^2+26567*n-4320)/(2*n-1)/(97*n-121)/n*b(n-1)+1/8* (97*n-24)*(3*n-4)*(3*n-5)*(n-1)/n/(97*n-121)/(2*n-1)/(2*n-3)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 15 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (18527 n - 41638 n + 26567 n - 4320) a(n - 1) a(n) = 5/12 ----------------------------------------------- (2 n - 1) (97 n - 121) n (97 n - 24) (3 n - 4) (3 n - 5) (n - 1) a(n - 2) + 1/8 ------------------------------------------------ n (97 n - 121) (2 n - 1) (2 n - 3) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.06665179859312291210597475870646835121731349426780726536120783410587956239\ 708179472894560457270905226 12 12 This constant is identified as, -- ln(3) - -- ln(2) 73 73 The implied delta is, 0.18295008592784481698449457030965390123418993890389950\ 8264614267749168239229918118894395728233750683 Since this is positive, this suggests an Apery-style irrationality proof of, 12 12 -- ln(3) - -- ln(2) 73 73 ----------------------- This took, 1.427, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, n + k) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, n + k) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (73919 n - 166222 n + 106127 n - 17280) b(n - 1) b(n) = 7/24 -------------------------------------------------- (2 n - 1) (193 n - 241) n (3 n - 4) (193 n - 48) (3 n - 5) (n - 1) b(n - 2) + 1/16 ------------------------------------------------- n (193 n - 241) (2 n - 1) (2 n - 3) and in Maple notation b(n) = 7/24*(73919*n^3-166222*n^2+106127*n-17280)/(2*n-1)/(193*n-241)/n*b(n-1)+ 1/16*(3*n-4)*(193*n-48)*(3*n-5)*(n-1)/n/(193*n-241)/(2*n-1)/(2*n-3)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 21 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (73919 n - 166222 n + 106127 n - 17280) a(n - 1) a(n) = 7/24 -------------------------------------------------- (2 n - 1) (193 n - 241) n (3 n - 4) (193 n - 48) (3 n - 5) (n - 1) a(n - 2) + 1/16 ------------------------------------------------- n (193 n - 241) (2 n - 1) (2 n - 3) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.04761634302650167074856038719897833349023609007962941624937914791988297115\ 378435271803214882964501290 48 24 This constant is identified as, --- ln(2) - --- ln(3) 145 145 The implied delta is, 0.20124894395767552079065262049454976483484034674708539\ 5423690240436845230481520645425960184154037223 Since this is positive, this suggests an Apery-style irrationality proof of, 48 24 --- ln(2) - --- ln(3) 145 145 ----------------------- This took, 1.673, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ ) binomial(n, k) binomial(3 n + k, 2 n) / ----- k By Shalosh B. Ekhad Let ----- \ b(n) = ) binomial(n, k) binomial(3 n + k, 2 n) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (693 n - 1554 n + 989 n - 160) b(n - 1) b(n) = 9/4 ----------------------------------------- (2 n - 1) (33 n - 41) n (3 n - 4) (33 n - 8) (3 n - 5) (n - 1) b(n - 2) + 3/8 ----------------------------------------------- n (33 n - 41) (2 n - 1) (2 n - 3) and in Maple notation b(n) = 9/4*(693*n^3-1554*n^2+989*n-160)/(2*n-1)/(33*n-41)/n*b(n-1)+3/8*(3*n-4)* (33*n-8)*(3*n-5)*(n-1)/n/(33*n-41)/(2*n-1)/(2*n-3)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 9 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (693 n - 1554 n + 989 n - 160) a(n - 1) a(n) = 9/4 ----------------------------------------- (2 n - 1) (33 n - 41) n (3 n - 4) (33 n - 8) (3 n - 5) (n - 1) a(n - 2) + 3/8 ----------------------------------------------- n (33 n - 41) (2 n - 1) (2 n - 3) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.11090354888959124950675713943330825089208002149764084065930880151894297951\ 51511544969381323194269900 This constant is identified as, 4/25 ln(2) The implied delta is, 0.14104379627041509051379936500361885164023421843144384\ 9713953691753590882640847436845046990892375390 Since this is positive, this suggests an Apery-style irrationality proof of, 4/25 ln(2) ----------------------- This took, 0.382, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, 2 n) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, 2 n) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (18527 n - 41638 n + 26567 n - 4320) b(n - 1) b(n) = 5/12 ----------------------------------------------- (2 n - 1) (97 n - 121) n (97 n - 24) (3 n - 4) (3 n - 5) (n - 1) b(n - 2) + 1/8 ------------------------------------------------ n (97 n - 121) (2 n - 1) (2 n - 3) and in Maple notation b(n) = 5/12*(18527*n^3-41638*n^2+26567*n-4320)/(2*n-1)/(97*n-121)/n*b(n-1)+1/8* (97*n-24)*(3*n-4)*(3*n-5)*(n-1)/n/(97*n-121)/(2*n-1)/(2*n-3)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 15 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (18527 n - 41638 n + 26567 n - 4320) a(n - 1) a(n) = 5/12 ----------------------------------------------- (2 n - 1) (97 n - 121) n (97 n - 24) (3 n - 4) (3 n - 5) (n - 1) a(n - 2) + 1/8 ------------------------------------------------ n (97 n - 121) (2 n - 1) (2 n - 3) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.06665179859312291210597475870646835121731349426780726536120783410587956239\ 708179472894560457270905226 12 12 This constant is identified as, -- ln(3) - -- ln(2) 73 73 The implied delta is, 0.18295008592784481698449457030965390123418993890389950\ 8264614267749168239229918118894395728233750683 Since this is positive, this suggests an Apery-style irrationality proof of, 12 12 -- ln(3) - -- ln(2) 73 73 ----------------------- This took, 0.559, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, 2 n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, 2 n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 2 (73919 n - 166222 n + 106127 n - 17280) b(n - 1) b(n) = 7/24 -------------------------------------------------- (2 n - 1) (193 n - 241) n (3 n - 4) (193 n - 48) (3 n - 5) (n - 1) b(n - 2) + 1/16 ------------------------------------------------- n (193 n - 241) (2 n - 1) (2 n - 3) and in Maple notation b(n) = 7/24*(73919*n^3-166222*n^2+106127*n-17280)/(2*n-1)/(193*n-241)/n*b(n-1)+ 1/16*(3*n-4)*(193*n-48)*(3*n-5)*(n-1)/n/(193*n-241)/(2*n-1)/(2*n-3)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 21 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 3 2 (73919 n - 166222 n + 106127 n - 17280) a(n - 1) a(n) = 7/24 -------------------------------------------------- (2 n - 1) (193 n - 241) n (3 n - 4) (193 n - 48) (3 n - 5) (n - 1) a(n - 2) + 1/16 ------------------------------------------------- n (193 n - 241) (2 n - 1) (2 n - 3) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.04761634302650167074856038719897833349023609007962941624937914791988297115\ 378435271803214882964501290 48 24 This constant is identified as, --- ln(2) - --- ln(3) 145 145 The implied delta is, 0.20124894395767552079065262049454976483484034674708539\ 5423690240436845230481520645425960184154037223 Since this is positive, this suggests an Apery-style irrationality proof of, 48 24 --- ln(2) - --- ln(3) 145 145 ----------------------- This took, 0.526, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, 2 n + k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, 2 n + k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 5 4 3 2 b(n) = 1/2 (126360 n - 533520 n + 835371 n - 597030 n + 191987 n - 21760) / 2 b(n - 1) / (n (3 n - 1) (3 n - 2) (270 n - 735 n + 497)) / 2 (n - 1) (3 n - 4) (3 n - 5) (270 n - 195 n + 32) b(n - 2) - 1/4 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (270 n - 735 n + 497) and in Maple notation b(n) = 1/2*(126360*n^5-533520*n^4+835371*n^3-597030*n^2+191987*n-21760)/n/(3*n-\ 1)/(3*n-2)/(270*n^2-735*n+497)*b(n-1)-1/4*(n-1)*(3*n-4)*(3*n-5)*(270*n^2-195*n+ 32)/n/(3*n-1)/(3*n-2)/(270*n^2-735*n+497)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 11 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 5 4 3 2 a(n) = 1/2 (126360 n - 533520 n + 835371 n - 597030 n + 191987 n - 21760) / 2 a(n - 1) / (n (3 n - 1) (3 n - 2) (270 n - 735 n + 497)) / 2 (n - 1) (3 n - 4) (3 n - 5) (270 n - 195 n + 32) a(n - 2) - 1/4 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (270 n - 735 n + 497) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.09094544480930789876142350253405961941801654358037253030594713812577958981\ 284057972360802119266842645 24 24 This constant is identified as, --- ln(3) - --- ln(2) 107 107 The implied delta is, 0.03661746617527283524134384554671528675530807862065396\ 1752763393969921416699834074598545930295214481 Since this is positive, this suggests an Apery-style irrationality proof of, 24 24 --- ln(3) - --- ln(2) 107 107 ----------------------- This took, 0.474, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, 2 n + k) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, 2 n + k) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients b(n) = 2/27 5 4 3 2 (2579174 n - 10911890 n + 17111599 n - 12243496 n + 3940893 n - 447120) / 2 b(n - 1) / (n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107)) / 2 (n - 1) (3 n - 4) (3 n - 5) (598 n - 437 n + 72) b(n - 2) - 1/9 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107) and in Maple notation b(n) = 2/27*(2579174*n^5-10911890*n^4+17111599*n^3-12243496*n^2+3940893*n-\ 447120)/n/(3*n-1)/(3*n-2)/(598*n^2-1633*n+1107)*b(n-1)-1/9*(n-1)*(3*n-4)*(3*n-5 )*(598*n^2-437*n+72)/n/(3*n-1)/(3*n-2)/(598*n^2-1633*n+1107)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 15 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. a(n) = 2/27 5 4 3 2 (2579174 n - 10911890 n + 17111599 n - 12243496 n + 3940893 n - 447120) / 2 a(n - 1) / (n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107)) / 2 (n - 1) (3 n - 4) (3 n - 5) (598 n - 437 n + 72) a(n - 2) - 1/9 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (598 n - 1633 n + 1107) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.06667309833646424927775891126037202275188637076600470837493539488781896711\ 983753250754501526039349983 108 54 This constant is identified as, --- ln(2) - --- ln(3) 233 233 The implied delta is, 0.06597544597881529603803186023005591329157253066585191\ 0079595969038781224778004356707050934970947861 Since this is positive, this suggests an Apery-style irrationality proof of, 108 54 --- ln(2) - --- ln(3) 233 233 ----------------------- This took, 0.962, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, 3 n) 2 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, 3 n) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients b(n) = 2/27 5 4 3 2 (1419550 n - 6065350 n + 9581057 n - 6889280 n + 2224239 n - 252720) / 2 b(n - 1) / (n (3 n - 1) (3 n - 2) (550 n - 1525 n + 1047)) / 2 (n - 1) (3 n - 4) (3 n - 5) (550 n - 425 n + 72) b(n - 2) - 1/9 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (550 n - 1525 n + 1047) and in Maple notation b(n) = 2/27*(1419550*n^5-6065350*n^4+9581057*n^3-6889280*n^2+2224239*n-252720)/ n/(3*n-1)/(3*n-2)/(550*n^2-1525*n+1047)*b(n-1)-1/9*(n-1)*(3*n-4)*(3*n-5)*(550*n ^2-425*n+72)/n/(3*n-1)/(3*n-2)/(550*n^2-1525*n+1047)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 9 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. a(n) = 2/27 5 4 3 2 (1419550 n - 6065350 n + 9581057 n - 6889280 n + 2224239 n - 252720) / 2 a(n - 1) / (n (3 n - 1) (3 n - 2) (550 n - 1525 n + 1047)) / 2 (n - 1) (3 n - 4) (3 n - 5) (550 n - 425 n + 72) a(n - 2) - 1/9 ---------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (550 n - 1525 n + 1047) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.11114271998903998287722186921357793591313443079682582066577042387706312815\ 96008607535971121935782927 54 54 This constant is identified as, --- ln(3) - --- ln(2) 197 197 The implied delta is, 0.00754066363959366106776293275714364875608275434040344\ 4744055943321384290623104622467065850901857503 Since this is positive, this suggests an Apery-style irrationality proof of, 54 54 --- ln(3) - --- ln(2) 197 197 ----------------------- This took, 2.151, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ k ) binomial(n, k) binomial(3 n + k, 3 n) 3 / ----- k By Shalosh B. Ekhad Let ----- \ k b(n) = ) binomial(n, k) binomial(3 n + k, 3 n) 3 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients b(n) = 1/8 5 4 3 2 (2191860 n - 9351936 n + 14757885 n - 10605642 n + 3423577 n - 389120) / 2 b(n - 1) / (n (3 n - 1) (3 n - 2) (990 n - 2739 n + 1877)) / 2 (n - 1) (3 n - 4) (3 n - 5) (990 n - 759 n + 128) b(n - 2) - 1/16 ----------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (990 n - 2739 n + 1877) and in Maple notation b(n) = 1/8*(2191860*n^5-9351936*n^4+14757885*n^3-10605642*n^2+3423577*n-389120) /n/(3*n-1)/(3*n-2)/(990*n^2-2739*n+1877)*b(n-1)-1/16*(n-1)*(3*n-4)*(3*n-5)*(990 *n^2-759*n+128)/n/(3*n-1)/(3*n-2)/(990*n^2-2739*n+1877)*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 13 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. a(n) = 1/8 5 4 3 2 (2191860 n - 9351936 n + 14757885 n - 10605642 n + 3423577 n - 389120) / 2 a(n - 1) / (n (3 n - 1) (3 n - 2) (990 n - 2739 n + 1877)) / 2 (n - 1) (3 n - 4) (3 n - 5) (990 n - 759 n + 128) a(n - 2) - 1/16 ----------------------------------------------------------- 2 n (3 n - 1) (3 n - 2) (990 n - 2739 n + 1877) but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.07692891073919489981661566734096778112628671934870490647532007184827889490\ 026441386200180033758804313 192 96 This constant is identified as, --- ln(2) - --- ln(3) 359 359 The implied delta is, 0.04663380674550632575040519244580784418043645591024757\ 8786344233409452078499680160077879310851784527 Since this is positive, this suggests an Apery-style irrationality proof of, 192 96 --- ln(2) - --- ln(3) 359 359 ----------------------- This took, 1.841, seconds. ------------------------------- the whole thing took, 38.297, seconds.