The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (2 x + 1)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (2 x + 1)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) 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) 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, 0.792, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |3 x + 3 x + 1| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |3 x + 3 x + 1| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 3 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 3*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) 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) 3 (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 0.30229989403903630843234637627369262204734437468212342926164748923135386352\ 10589806140208313686766948 1/2 3 Pi This constant is identified as, ------- 18 The implied delta is, 0.14497379959755737479671870841107182956397560571728623\ 7953661294351601350972231778006905872499225947 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 18 ----------------------- This took, 5.810, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |5 x + 5 x + 1| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |5 x + 5 x + 1| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 5 (2 n - 1) b(n - 1) 5 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 5*(2*n-1)/n*b(n-1)-5*(n-1)/n*b(n-2) 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) 5 (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 0.21520447048200201944471661647530271271228534127014482737850305199628060773\ 05659806807451323486099777 #Note added later by a human: this is log(phi)/sqrt(5), where phi is the Golden ratio (1+sqrt(5))/2, currently Maple, and our program was unable to detect it The implied delta is, 0.19252839668463989809019054952672840099285666374391460\ 9105451304890875469070833047189652560238065544 Since this is positive, this suggests an Apery-style irrationality proof of, 0.2152044704820020194447166164753027127122853412701448273785030519962806\ 077305659806807451323486099777 ----------------------- This took, 5.871, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (3 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (3 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) 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) 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, 0.894, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |7 x + 5 x + 1| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |7 x + 5 x + 1| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 5 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 5*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) 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) 3 (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 0.19253082576711371276367574726314191299857385545021425289061213085025553034\ 87174962331829263926134182 The implied delta is, 0.07918439240457512945210944289295953512343460149005304\ 0725966287356117338551816477755118308004127005 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1925308257671137127636757472631419129985738554502142528906121308502555\ 303487174962331829263926134182 ----------------------- This took, 4.268, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (4 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (4 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 6*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 (2 n - 1) a(n - 1) 4 (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 0.17328679513998632735430803036454414201887503359006381353017000237334840549\ 24236789014658317491046719 This constant is identified as, 1/4 ln(2) The implied delta is, 0.28201844359566526000468075844868556336700580720996580\ 6860077634584536252229215865253360926714653745 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(2) ----------------------- This took, 0.558, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |10 x + 6 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |10 x + 6 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 6*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 (2 n - 1) a(n - 1) 4 (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 0.16087527719832109670070230717933065951037764777882809571640152967837811870\ 29052721782042111753206872 1/2 1/2 (50 - 15 10 ) This constant is identified as, arcsin(------------------) 10 The implied delta is, 0.16067340533262483191783177523485056026246586371084074\ 8340060802000413725691265859410474876548527093 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (50 - 15 10 ) arcsin(------------------) 10 ----------------------- This took, 2.999, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |12 x + 6 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |12 x + 6 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = -------------------- + ------------------- n n and in Maple notation b(n) = 6*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 (2 n - 1) a(n - 1) 12 (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 0.15114994701951815421617318813684631102367218734106171463082374461567693176\ 05294903070104156843383474 1/2 3 Pi This constant is identified as, ------- 36 The implied delta is, 0.14493027260456053563520735898834553641898336141547323\ 0165005604153985803503684704639842784380864596 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 36 ----------------------- This took, 3.811, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 7 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 7 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 (2 n - 1) b(n - 1) 5 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 7*(2*n-1)/n*b(n-1)-5*(n-1)/n*b(n-2) 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) 5 (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 0.14803728593913396657830557413156260733691580883685263249088915620319565241\ 26983486401685289705691292 The implied delta is, 0.00956485545750940671873946026554331080416949980817551\ 9305723159316293262324974624891760838685330692 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1480372859391339665783055741315626073369158088368526324908891562031956\ 524126983486401685289705691292 ----------------------- This took, 5.063, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 1) (4 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 1) (4 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) 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) 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, 0.760, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 7 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 7 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 7*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) 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) 3 (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 0.14004431800470691548061360400212346398708395070826392056961046507953474826\ 80550627105973941674569784 The implied delta is, 0.15868594456653576363494758485012344144253070114462712\ 9736233792824810707447103927069625288731704176 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1400443180047069154806136040021234639870839507082639205696104650795347\ 482680550627105973941674569784 ----------------------- This took, 3.534, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 7 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 7 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 7*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) 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) 7 (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 0.13658393455025893346464769601732639443227044282595070753810721596569048088\ 04054835183669967750559749 The implied delta is, 0.26788864690158903575310546037966631193359997521914487\ 3766972625332255904390493791012759629142648643 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1365839345502589334646476960173263944322704428259507075381072159656904\ 808804054835183669967750559749 ----------------------- This took, 3.080, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 8 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 8 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)-8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 8 (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 0.13063761434511996994652465900951978293361564705086556431482098017089212419\ 28349413329767703089758148 The implied delta is, 0.00951434378572660087622240791810913692821999269665859\ 4664124310427536132518016110623162677905711527 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1306376143451199699465246590095197829336156470508655643148209801708921\ 241928349413329767703089758148 ----------------------- This took, 3.217, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 1) (5 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 1) (5 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 4 (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 0.12770640594149767080137852407591548371952769911144206754448838945917117362\ 22621994391295308198618805 This constant is identified as, 1/4 ln(5) - 1/4 ln(3) The implied delta is, 0.10596649575065411124555937123919187192353661313417166\ 0007799359035713014130826697010193290456439653 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(5) - 1/4 ln(3) ----------------------- This took, 0.795, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 8 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 8 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 4 (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 0.12248933156343207708604124060563790545707204919059203356368795733367755979\ 38210482872670788343509957 1/2 1/2 (578 - 136 17 ) This constant is identified as, arcsin(--------------------) 34 The implied delta is, 0.11469371798309472207858858381352891988603224328816120\ 0044948387359570910467091050024157806409462694 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (578 - 136 17 ) arcsin(--------------------) 34 ----------------------- This took, 2.832, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 8 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 8 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 8 (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 0.12015049158624418214020747250309315160801624089460805716794839720641948933\ 82479303980890853684306077 The implied delta is, 0.02670016187612724207713835131029575601841276139214336\ 9393765659776983891559404098395622671775299027 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1201504915862441821402074725030931516080162408946080571679483972064194\ 893382479303980890853684306077 ----------------------- This took, 4.126, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 1) (6 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 1) (6 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (2 n - 1) a(n - 1) 9 (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 0.11552453009332421823620535357636276134591668906004254235344666824889893699\ 49491192676438878327364479 This constant is identified as, 1/6 ln(2) The implied delta is, 0.28197692469263188999180798130074546532217055266304040\ 7728165468891969758894540883119075334406309515 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(2) ----------------------- This took, 0.856, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 9 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 9 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) 5 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)-5*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (2 n - 1) a(n - 1) 5 (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 0.11348594569005491581276826726732974223595549230735296517944187702881632074\ 05405428935012016224164943 The implied delta is, 0.07402060650746152252437030229936813081093256664373236\ 7243350220304580366383919633603807349822502928 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1134859456900549158127682672673297422359554923073529651794418770288163\ 207405405428935012016224164943 ----------------------- This took, 4.192, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(4 x + 1) (5 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(4 x + 1) (5 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = -------------------- - ---------------- n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (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 0.11157177565710487788314754515491725168730054277400360683564393624369587188\ 41341667092036120501711179 This constant is identified as, 1/2 ln(5) - ln(2) The implied delta is, 0.49208524974733895207987264851883876017360748086750998\ 0752669244389825092453152133202625701195068714 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(5) - ln(2) ----------------------- This took, 0.888, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 9 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 9 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (2 n - 1) a(n - 1) 3 (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 0.10976906827192259566867062901055070904877051923190917637103535838109833317\ 23414843808379049760632766 The implied delta is, 0.40903511743182374497883468784098098812098759625609746\ 5078016616638783867745510746813699798663377275 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1097690682719225956686706290105507090487705192319091763710353583810983\ 331723414843808379049760632766 ----------------------- This took, 3.815, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |27 x + 9 x + 1| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |27 x + 9 x + 1| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) 27 (n - 1) b(n - 2) b(n) = -------------------- + ------------------- n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)+27*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (2 n - 1) a(n - 1) 27 (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 0.10076663134634543614411545875789754068244812489404114308721582974378462117\ 36863268713402771228922316 1/2 3 Pi This constant is identified as, ------- 54 The implied delta is, 0.14490482306019478704113623172161445584388633721529141\ 3056950429551534328795697066058224893358150364 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 54 ----------------------- This took, 3.443, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |20 x + 10 x + 1| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |20 x + 10 x + 1| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)-20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 20 (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 0.10760223524100100972235830823765135635614267063507241368925152599814030386\ 52829903403725661743049889 The implied delta is, 0.19247336948998677574975760444645387535294026463422781\ 2186683150902973405143658074831466502437650534 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1076022352410010097223583082376513563561426706350724136892515259981403\ 038652829903403725661743049889 ----------------------- This took, 4.613, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(4 x + 1) (6 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(4 x + 1) (6 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 4 (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 0.10136627702704109549450327886608728414299760586562354940350358103602516781\ 22285628169381069543283503 This constant is identified as, 1/4 ln(3) - 1/4 ln(2) The implied delta is, 0.40322063105998097410617228978635520167305437842456454\ 3485213485255081852518226083013184063088440816 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(3) - 1/4 ln(2) ----------------------- This took, 0.891, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |26 x + 10 x + 1| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |26 x + 10 x + 1| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 4 (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 0.09869777992494037918502488259739514672379255189392605075884447012051698498\ 912189286634891401864402206 1/2 1/2 (338 - 65 26 ) This constant is identified as, arcsin(-------------------) 26 The implied delta is, 0.27279015851679530122551511258324116663957251223242409\ 3777342346874863076771218871853895144090435632 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (338 - 65 26 ) arcsin(-------------------) 26 ----------------------- This took, 4.404, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |28 x + 10 x + 1| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |28 x + 10 x + 1| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 12 (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 0.09626541288355685638183787363157095649928692772510712644530606542512776517\ 435874811659146319630670909 The implied delta is, 0.07916786443955308683937807391588653096766141858144688\ 8102825610972009669602484145480313628946192650 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0962654128835568563818378736315709564992869277251071264453060654251277\ 6517435874811659146319630670909 ----------------------- This took, 5.390, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |30 x + 10 x + 1| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |30 x + 10 x + 1| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 20 (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 0.09403433605676343518168600277404782507929297311294733417545385860544677198\ 381079303252695013504968627 The implied delta is, 0.07720156228804965939235880149890205848873639970474698\ 2173838234295769481177143391467973181913521255 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0940343360567634351816860027740478250792929731129473341754538586054467\ 7198381079303252695013504968627 ----------------------- This took, 4.583, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 11 x + 1| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 11 x + 1| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) 5 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)-5*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (2 n - 1) a(n - 1) 5 (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 0.09219327514691665863321916315870214081535549697550627500377733115791949917\ 861799563754029656875635181 The implied delta is, 0.12108264916403604724478249509327745816144064820005511\ 1805817243469931937090572814731868231417025298 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0921932751469166586332191631587021408153554969755062750037773311579194\ 9917861799563754029656875635181 ----------------------- This took, 3.285, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 1) (6 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 1) (6 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (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 0.09116077839697731310585901257725731659869466895724349197136322582835446374\ 032295892467260185848558276 This constant is identified as, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) The implied delta is, 0.51738394387588245866899222655825392097534060270982986\ 7955158319124421684607024589198054063909583979 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) ----------------------- This took, 0.867, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (3 x + 2)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (3 x + 2)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) 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) 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, 0.672, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |5 x + 6 x + 2| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |5 x + 6 x + 2| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 6*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 (2 n - 1) a(n - 1) 4 (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 0.16087527719832109670070230717933065951037764777882809571640152967837811870\ 29052721782042111753206872 1/2 1/2 (50 - 15 10 ) This constant is identified as, arcsin(------------------) 10 The implied delta is, 0.16067340533262483191783177523485056026246586371084074\ 8340060802000413725691265859410474876548527093 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (50 - 15 10 ) arcsin(------------------) 10 ----------------------- This took, 2.559, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 2) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 2) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) 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) 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, 0.786, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |7 x + 7 x + 2| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |7 x + 7 x + 2| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 7*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) 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) 7 (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 0.13658393455025893346464769601732639443227044282595070753810721596569048088\ 04054835183669967750559749 The implied delta is, 0.26788864690158903575310546037966631193359997521914487\ 3766972625332255904390493791012759629142648643 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1365839345502589334646476960173263944322704428259507075381072159656904\ 808804054835183669967750559749 ----------------------- This took, 2.608, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |7 x + 8 x + 2| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |7 x + 8 x + 2| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)-8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 8 (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 0.13063761434511996994652465900951978293361564705086556431482098017089212419\ 28349413329767703089758148 The implied delta is, 0.00951434378572660087622240791810913692821999269665859\ 4664124310427536132518016110623162677905711527 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1306376143451199699465246590095197829336156470508655643148209801708921\ 241928349413329767703089758148 ----------------------- This took, 2.549, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |9 x + 8 x + 2| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |9 x + 8 x + 2| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 8 (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 0.12015049158624418214020747250309315160801624089460805716794839720641948933\ 82479303980890853684306077 The implied delta is, 0.02670016187612724207713835131029575601841276139214336\ 9393765659776983891559404098395622671775299027 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1201504915862441821402074725030931516080162408946080571679483972064194\ 893382479303980890853684306077 ----------------------- This took, 2.329, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 2) (3 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 2) (3 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (2 n - 1) a(n - 1) 9 (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 0.11552453009332421823620535357636276134591668906004254235344666824889893699\ 49491192676438878327364479 This constant is identified as, 1/6 ln(2) The implied delta is, 0.28197692469263188999180798130074546532217055266304040\ 7728165468891969758894540883119075334406309515 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(2) ----------------------- This took, 0.412, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (5 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (5 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = -------------------- - ---------------- n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (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 0.11157177565710487788314754515491725168730054277400360683564393624369587188\ 41341667092036120501711179 This constant is identified as, 1/2 ln(5) - ln(2) The implied delta is, 0.49208524974733895207987264851883876017360748086750998\ 0752669244389825092453152133202625701195068714 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(5) - ln(2) ----------------------- This took, 0.708, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 10 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 10 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 4 (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 0.09869777992494037918502488259739514672379255189392605075884447012051698498\ 912189286634891401864402206 1/2 1/2 (338 - 65 26 ) This constant is identified as, arcsin(-------------------) 26 The implied delta is, 0.27279015851679530122551511258324116663957251223242409\ 3777342346874863076771218871853895144090435632 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (338 - 65 26 ) arcsin(-------------------) 26 ----------------------- This took, 2.783, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 10 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 10 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 20 (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 0.09403433605676343518168600277404782507929297311294733417545385860544677198\ 381079303252695013504968627 The implied delta is, 0.07720156228804965939235880149890205848873639970474698\ 2173838234295769481177143391467973181913521255 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0940343360567634351816860027740478250792929731129473341754538586054467\ 7198381079303252695013504968627 ----------------------- This took, 2.426, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 2) (3 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 2) (3 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (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 0.09116077839697731310585901257725731659869466895724349197136322582835446374\ 032295892467260185848558276 This constant is identified as, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) The implied delta is, 0.51738394387588245866899222655825392097534060270982986\ 7955158319124421684607024589198054063909583979 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) ----------------------- This took, 0.742, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 11 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 11 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (2 n - 1) a(n - 1) 7 (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 0.08921446688943169293092583111195045716518487104723565893293488689922295893\ 499162985078998393501845002 The implied delta is, 0.04540904658398761381316120489309151071581081582901689\ 2989551633036984124706114487083387730713454991 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0892144668894316929309258311119504571651848710472356589329348868992229\ 5893499162985078998393501845002 ----------------------- This took, 3.692, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 12 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 12 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 12 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 12*(2*n-1)/n*b(n-1)-8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 12 (2 n - 1) a(n - 1) 8 (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 0.08493011723284795205233176292102063059070900912711015983375237981685523117\ 059978030387807506822794549 The implied delta is, 0.12110913332415970117985294538299293645462578951696875\ 9447732321737166810967307673220076151976383945 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0849301172328479520523317629210206305907090091271101598337523798168552\ 3117059978030387807506822794549 ----------------------- This took, 4.133, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 12 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 12 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 12 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 12*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 12 (2 n - 1) a(n - 1) 8 (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 0.08183960687724986564113675264593130448776441989967400849580847995465457184\ 899226703439723968535278178 The implied delta is, 0.12722424960718222869287001285932564166835338978695181\ 2803952994841755363916385018216661705767802666 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0818396068772498656411367526459313044877644198996740084958084799546545\ 7184899226703439723968535278178 ----------------------- This took, 5.636, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 2) (4 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 2) (4 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (2 n - 1) a(n - 1) 9 (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 0.07833393820762259227515617185805701078348317480204134007489868950100031303\ 357106369790935048267940864 This constant is identified as, 1/2 ln(2) - 1/6 ln(5) The implied delta is, 0.01610305134163057059476030447407709848452163925938429\ 5868197117252961128356062199075422485134368797 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) - 1/6 ln(5) ----------------------- This took, 1.353, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 1) (7 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 1) (7 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (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 0.07707533991362915214643769253123872845704701869942824130200790340334597378\ 188279265408970838886037533 This constant is identified as, 1/2 ln(7) - 1/2 ln(2) - 1/2 ln(3) The implied delta is, 0.53598696183853215629483188639824344780581812162220287\ 2066477785480308463725655439137125446634136583 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(7) - 1/2 ln(2) - 1/2 ln(3) ----------------------- This took, 1.140, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |22 x + 13 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |22 x + 13 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (2 n - 1) a(n - 1) 7 (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 0.07588666029425126194667792957990351954635552062910919636294014340832577279\ 573527290819229322993544913 The implied delta is, 0.08341738038702336058364740517185903189403827696615325\ 6864804634845084861556551581449375207636797003 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0758866602942512619466779295799035195463555206291091963629401434083257\ 7279573527290819229322993544913 ----------------------- This took, 5.058, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 14 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 14 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)-28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 28 (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 0.07515497185030861111837468180255027686449784352277401390010827505367720667\ 693974200463726324464985163 The implied delta is, 0.10686249638390301523636562155291556457477022400357666\ 7164412126293469051547912365992267528292025175 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0751549718503086111183746818025502768644978435227740139001082750536772\ 0667693974200463726324464985163 ----------------------- This took, 5.207, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 14 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 14 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)-12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 12 (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 0.07294230569649749652305334425532049557607901455248479694567492853814165173\ 455084661999481437967006056 The implied delta is, 0.05367194919665695911732870269101876993348991642421949\ 7345170801770025825435068253223787848741071185 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0729423056964974965230533442553204955760790145524847969456749285381416\ 5173455084661999481437967006056 ----------------------- This took, 5.355, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 14 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 14 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 4 (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 0.07094852730208196140642580855127654150389087936423203618906501468172081337\ 996558047209593081712325906 1/2 1/2 (50 - 35 2 ) This constant is identified as, arcsin(-----------------) 10 The implied delta is, 0.33551462144924097388011542229878210355571808363541944\ 8570488510188258420343720921802437054447955885 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (50 - 35 2 ) arcsin(-----------------) 10 ----------------------- This took, 4.618, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 2) (5 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 2) (5 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 25 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)-25*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 25 (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 0.06931471805599453094172321214581765680755001343602552541206800094933936219\ 696947156058633269964186875 This constant is identified as, 1/10 ln(2) The implied delta is, 0.28192463446027303694078778562679520138377881802097597\ 5787561534138507219474094032037479747071724205 Since this is positive, this suggests an Apery-style irrationality proof of, 1/10 ln(2) ----------------------- This took, 0.692, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 1) (9 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 1) (9 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 9 (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 0.06757751801802739699633551924405818942866507057708236626900238735735011187\ 481904187795873796955223354 This constant is identified as, 1/6 ln(3) - 1/6 ln(2) The implied delta is, 0.40317066976122116343384728408550145692326574382398771\ 8793808639008031557783689239572916016095385712 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(3) - 1/6 ln(2) ----------------------- This took, 1.001, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(7 x + 2) (4 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(7 x + 2) (4 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (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 0.06676569631226131157317181046567498729470783674945228695132493927130050157\ 850743951496574120069226782 This constant is identified as, 3/2 ln(2) - 1/2 ln(7) The implied delta is, 0.55621365185476501964096807703410704299320585275394282\ 5728523991339290921553255210996924685007797235 Since this is positive, this suggests an Apery-style irrationality proof of, 3/2 ln(2) - 1/2 ln(7) ----------------------- This took, 1.213, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 15 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 15 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 7 (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 0.06598793397360693966380231920818052157628951826084796837646525287682068337\ 583920453938307232029484360 The implied delta is, 0.11269371495250835459980142386814902875539939787059502\ 8273710585508268614022824884706199010771120711 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0659879339736069396638023192081805215762895182608479683764652528768206\ 8337583920453938307232029484360 ----------------------- This took, 4.117, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |30 x + 15 x + 2| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |30 x + 15 x + 2| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 15 (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 0.06524176133871410934932854113492265498822323720049651191532932060199214477\ 817605188744976432300036974 The implied delta is, 0.35563101509770246590172023698722401567618661496957575\ 9764109128677597308464170617516368549119007347 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0652417613387141093493285411349226549882232372004965119153293206019921\ 4477817605188744976432300036974 ----------------------- This took, 3.465, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |4 x + 6 x + 3| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |4 x + 6 x + 3| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 6 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = -------------------- + ------------------- n n and in Maple notation b(n) = 6*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 6 (2 n - 1) a(n - 1) 12 (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 0.15114994701951815421617318813684631102367218734106171463082374461567693176\ 05294903070104156843383474 1/2 3 Pi This constant is identified as, ------- 36 The implied delta is, 0.14493027260456053563520735898834553641898336141547323\ 0165005604153985803503684704639842784380864596 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 36 ----------------------- This took, 2.574, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (4 x + 3)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (4 x + 3)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 (-1 + 2 n) b(n - 1) (n - 1) b(-2 + n) b(n) = --------------------- - ----------------- n n and in Maple notation b(n) = 7*(-1+2*n)/n*b(n-1)-(n-1)/n*b(-2+n) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 (-1 + 2 n) a(n - 1) (n - 1) a(-2 + n) 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, 0.895, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (5 x + 3)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (5 x + 3)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = -------------------- - ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 4 (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 0.12770640594149767080137852407591548371952769911144206754448838945917117362\ 22621994391295308198618805 This constant is identified as, 1/4 ln(5) - 1/4 ln(3) The implied delta is, 0.10596649575065411124555937123919187192353661313417166\ 0007799359035713014130826697010193290456439653 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(5) - 1/4 ln(3) ----------------------- This took, 0.735, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |6 x + 8 x + 3| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |6 x + 8 x + 3| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 8 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 8*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 8 (2 n - 1) a(n - 1) 8 (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 0.12015049158624418214020747250309315160801624089460805716794839720641948933\ 82479303980890853684306077 The implied delta is, 0.02670016187612724207713835131029575601841276139214336\ 9393765659776983891559404098395622671775299027 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1201504915862441821402074725030931516080162408946080571679483972064194\ 893382479303980890853684306077 ----------------------- This took, 2.535, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |7 x + 9 x + 3| |--------------| \ x / By Shalosh B. Ekhad / 2 \n |7 x + 9 x + 3| Let , b(n), be the coefficient of x^0 (i.e. constant term of, |--------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = -------------------- + ------------------ n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (2 n - 1) a(n - 1) 3 (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 0.10976906827192259566867062901055070904877051923190917637103535838109833317\ 23414843808379049760632766 The implied delta is, 0.40903511743182374497883468784098098812098759625609746\ 5078016616638783867745510746813699798663377275 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1097690682719225956686706290105507090487705192319091763710353583810983\ 331723414843808379049760632766 ----------------------- This took, 2.375, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(4 x + 3) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(4 x + 3) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 4 (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 0.10136627702704109549450327886608728414299760586562354940350358103602516781\ 22285628169381069543283503 This constant is identified as, 1/4 ln(3) - 1/4 ln(2) The implied delta is, 0.40322063105998097410617228978635520167305437842456454\ 3485213485255081852518226083013184063088440816 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(3) - 1/4 ln(2) ----------------------- This took, 0.538, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |10 x + 10 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |10 x + 10 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 20 (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 0.09403433605676343518168600277404782507929297311294733417545385860544677198\ 381079303252695013504968627 The implied delta is, 0.07720156228804965939235880149890205848873639970474698\ 2173838234295769481177143391467973181913521255 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0940343360567634351816860027740478250792929731129473341754538586054467\ 7198381079303252695013504968627 ----------------------- This took, 2.678, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 3) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 3) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (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 0.09116077839697731310585901257725731659869466895724349197136322582835446374\ 032295892467260185848558276 This constant is identified as, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) The implied delta is, 0.51738394387588245866899222655825392097534060270982986\ 7955158319124421684607024589198054063909583979 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) ----------------------- This took, 0.998, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 11 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 11 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) 11 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)+11*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (2 n - 1) a(n - 1) 11 (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 0.08829541785376189678245510944001178006628494558224065441903839600220059171\ 091478726508029336640113096 The implied delta is, 0.31912231784569922673210692882303980008803271791939056\ 6870831432237522248495256022917415812912444388 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0882954178537618967824551094400117800662849455822406544190383960022005\ 9171091478726508029336640113096 ----------------------- This took, 3.077, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 12 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 12 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 12 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 12*(2*n-1)/n*b(n-1)-12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 12 (2 n - 1) a(n - 1) 12 (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 0.08577125750874812237933268457539347784274556970909649446645428089781586169\ 125884465150975517795928491 The implied delta is, 0.07050303292430218794666305137572702833652730986717109\ 7753612487518346853564995146588401281924878919 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0857712575087481223793326845753934778427455697090964944664542808978158\ 6169125884465150975517795928491 ----------------------- This took, 3.304, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 12 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 12 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 12 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 12*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 12 (2 n - 1) a(n - 1) 12 (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 0.08112778801716469647586638613578457903013021198692975434601851207590955762\ 650195895171171860060985819 The implied delta is, 0.08063545228673139764982135747018502293708024693378942\ 7226381796204861036907559649707183538954024079 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0811277880171646964758663861357845790301302119869297543460185120759095\ 5762650195895171171860060985819 ----------------------- This took, 3.893, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 12 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 12 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 12 (2 n - 1) b(n - 1) 48 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 12*(2*n-1)/n*b(n-1)+48*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 12 (2 n - 1) a(n - 1) 48 (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 0.07557497350975907710808659406842315551183609367053085731541187230783846588\ 026474515350520784216917370 1/2 3 Pi This constant is identified as, ------- 72 The implied delta is, 0.14488677174084305763785639764352656144751798287769925\ 6234506712864263733553178233256147247065672827 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 72 ----------------------- This took, 4.298, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 13 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 13 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) 13 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)-13*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (2 n - 1) a(n - 1) 13 (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 0.07899181796594889652723542155561019426013185926486066666293376718165684237\ 124773386243062264950432773 The implied delta is, 0.33019398190422710222745578044004292872312182212316832\ 0983204045493457096152533228189535245964623369 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0789918179659488965272354215556101942601318592648606666629337671816568\ 4237124773386243062264950432773 ----------------------- This took, 4.767, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (7 x + 3)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (7 x + 3)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (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 0.07707533991362915214643769253123872845704701869942824130200790340334597378\ 188279265408970838886037533 This constant is identified as, 1/2 ln(7) - 1/2 ln(2) - 1/2 ln(3) The implied delta is, 0.53598696183853215629483188639824344780581812162220287\ 2066477785480308463725655439137125446634136583 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(7) - 1/2 ln(2) - 1/2 ln(3) ----------------------- This took, 1.139, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 14 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 14 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)-28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 28 (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 0.07515497185030861111837468180255027686449784352277401390010827505367720667\ 693974200463726324464985163 The implied delta is, 0.10686249638390301523636562155291556457477022400357666\ 7164412126293469051547912365992267528292025175 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0751549718503086111183746818025502768644978435227740139001082750536772\ 0667693974200463726324464985163 ----------------------- This took, 3.442, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (8 x + 3)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (8 x + 3)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 4 (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 0.07192051811294523185980475149845685787587742772444026412666642133732323768\ 019511608452772479477632157 This constant is identified as, 1/2 ln(2) - 1/4 ln(3) The implied delta is, 0.45946390201188529890288496005943615735436983493987986\ 5602656739857769744937606764374052531299868561 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) - 1/4 ln(3) ----------------------- This took, 1.197, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 14 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 14 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 8 (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 0.07047988118784926283706913599267386030688253675923505905708218358864090066\ 808096237060356016387612910 The implied delta is, 0.06968726033156156150684441540455075204738468655637340\ 7186077402759751252725889321369790371749605763 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0704798811878492628370691359926738603068825367592350590570821835886409\ 0066808096237060356016387612910 ----------------------- This took, 4.374, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 15 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 15 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 3 (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 0.06637271839737003644198751480960832860464328677720489495171413554157748475\ 108558336488452799144086278 The implied delta is, 0.48811403120256723261133197004712123132816429731695092\ 4318921655022823705428878616664461466161992557 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0663727183973700364419875148096083286046432867772048949517141355415774\ 8475108558336488452799144086278 ----------------------- This took, 5.986, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |20 x + 15 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |20 x + 15 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 15 (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 0.06524176133871410934932854113492265498822323720049651191532932060199214477\ 817605188744976432300036974 The implied delta is, 0.35563101509770246590172023698722401567618661496957575\ 9764109128677597308464170617516368549119007347 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0652417613387141093493285411349226549882232372004965119153293206019921\ 4477817605188744976432300036974 ----------------------- This took, 3.556, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 15 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 15 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 75 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+75*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 75 (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 0.06045997880780726168646927525473852440946887493642468585232949784627077270\ 421179612280416627373533896 1/2 3 Pi This constant is identified as, ------- 90 The implied delta is, 0.14487277314664983625359310728141584796158669850985255\ 9086113490950430595667900491616655656453361689 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 90 ----------------------- This took, 4.874, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (10 x + 3)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (10 x + 3)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 16 (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 0.06385320297074883540068926203795774185976384955572103377224419472958558681\ 113109971956476540993094025 This constant is identified as, 1/8 ln(5) - 1/8 ln(3) The implied delta is, 0.10615034289762958358759021710934598219278290513911819\ 4663861486759455333095825793683868620538599174 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(5) - 1/8 ln(3) ----------------------- This took, 0.878, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(7 x + 3) (3 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(7 x + 3) (3 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 4 (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 0.06282860707022651942128443260046791991447409651590942875249962933435218092\ 128716648989325275989816265 This constant is identified as, 1/2 ln(3) - 1/4 ln(7) The implied delta is, 0.24726794658096722576995160167459976418634699867116936\ 8019402426017252451484100726909737452714571334 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(3) - 1/4 ln(7) ----------------------- This took, 1.413, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |22 x + 16 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |22 x + 16 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 8 (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 0.06186089934397364766167805961521493911390506219717794272116055504762000878\ 355963260558777854258289550 The implied delta is, 0.18614007624007003473212386151298091116665090859195718\ 8735980949280727987067183567085883629690535105 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0618608993439736476616780596152149391139050621971779427211605550476200\ 0878355963260558777854258289550 ----------------------- This took, 3.262, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |24 x + 16 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |24 x + 16 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 32 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)+32*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 32 (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 0.06007524579312209107010373625154657580400812044730402858397419860320974466\ 912396519904454268421530384 The implied delta is, 0.02669478119360248299763676646264780151576010829662806\ 9703765788628838578923186785107340098266245593 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0600752457931220910701037362515465758040081204473040285839741986032097\ 4466912396519904454268421530384 ----------------------- This took, 3.248, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(8 x + 3) (3 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(8 x + 3) (3 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (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 0.05889151782819172726939705473526085253424035628236657055367431939740386026\ 406689346482076431910405747 This constant is identified as, ln(3) - 3/2 ln(2) The implied delta is, 0.56699175013732249851317171476635314501988343365560925\ 0888942795636346050682355683916571549465815244 Since this is positive, this suggests an Apery-style irrationality proof of, ln(3) - 3/2 ln(2) ----------------------- This took, 1.134, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 17 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 17 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) 11 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)+11*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (2 n - 1) a(n - 1) 11 (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 0.05809380401645286932845839538933593430322189548043015928749092971870836258\ 588018482326301833941239337 The implied delta is, 0.03510789040775259729732552971699613456032344320168967\ 0950605148289146553202186823376185318634015031 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0580938040164528693284583953893359343032218954804301592874909297187083\ 6258588018482326301833941239337 ----------------------- This took, 3.289, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |26 x + 18 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |26 x + 18 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)-12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 12 (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 0.05625708338586260381578117769348148903118350424572836423420400122983851411\ 098925064855857658563755154 The implied delta is, 0.27780175210518099077819661085816579282335749466746754\ 6371715412236858321612176011509442533123151172 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0562570833858626038157811776934814890311835042457283642342040012298385\ 1411098925064855857658563755154 ----------------------- This took, 3.508, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |28 x + 18 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |28 x + 18 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 12 (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 0.05488453413596129783433531450527535452438525961595458818551767919054916658\ 617074219041895248803163831 The implied delta is, 0.40895042964810924600624793751049929879378971965734642\ 6535549487656215862929214834673121682709336075 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0548845341359612978343353145052753545243852596159545881855176791905491\ 6658617074219041895248803163831 ----------------------- This took, 3.492, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 18 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 18 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 24 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)+24*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 24 (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 0.05424172927286820168242345115236634533960267583759247780555379289951543625\ 562525999492088353881231129 The implied delta is, -0.0001687129495169170274710563792711968426440872060282\ 274861652238296203059613364981549992428328208876 Since this is negative, there is no Apery-style irrationality proof of, 0.054\ 241729272868201682423451152366345339602675837592477805553792899515436255\ 62525999492088353881231129, but still a very fast way to compute it to many digits ----------------------- This took, 3.220, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 19 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 19 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) 13 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-13*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (2 n - 1) a(n - 1) 13 (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 0.05327736404510432403985627372236438761340387555738829700056631360524062650\ 906679652228965952352612634 The implied delta is, 0.02016882726220699591877030186419214602476380477659517\ 8414861259413234755737069616348711592175891304 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0532773640451043240398562737223643876134038755573882970005663136052406\ 2650906679652228965952352612634 ----------------------- This took, 3.597, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 1) (10 x + 3)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 1) (10 x + 3)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (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 0.05268025782891315061375049041965639915306018649163703628196961684629201162\ 006727324438284773106706038 This constant is identified as, 1/2 ln(2) + 1/2 ln(5) - ln(3) The implied delta is, 0.57337177129358798630972583682740462220014717381052020\ 3127971283730002664556591027268962044943194084 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(5) - ln(3) ----------------------- This took, 1.225, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |30 x + 20 x + 3| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |30 x + 20 x + 3| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 40 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-40*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 40 (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 0.05177441474570309562625581681897158600064073773170552450255681654745943689\ 051284618097664366479131960 The implied delta is, 0.04566308295447717905112811140638400015597404182898616\ 4853836664038983172727996437356568042526582583 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0517744147457030956262558168189715860006407377317055245025568165474594\ 3689051284618097664366479131960 ----------------------- This took, 5.170, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (5 x + 4)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (5 x + 4)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 9 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = -------------------- - ---------------- n n and in Maple notation b(n) = 9*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 9 (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 0.11157177565710487788314754515491725168730054277400360683564393624369587188\ 41341667092036120501711179 This constant is identified as, 1/2 ln(5) - ln(2) The implied delta is, 0.49208524974733895207987264851883876017360748086750998\ 0752669244389825092453152133202625701195068714 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(5) - ln(2) ----------------------- This took, 1.326, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |5 x + 10 x + 4| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |5 x + 10 x + 4| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)-20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 20 (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 0.10760223524100100972235830823765135635614267063507241368925152599814030386\ 52829903403725661743049889 The implied delta is, 0.19247336948998677574975760444645387535294026463422781\ 2186683150902973405143658074831466502437650534 Since this is positive, this suggests an Apery-style irrationality proof of, 0.1076022352410010097223583082376513563561426706350724136892515259981403\ 038652829903403725661743049889 ----------------------- This took, 3.447, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |7 x + 10 x + 4| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |7 x + 10 x + 4| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 12 (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 0.09626541288355685638183787363157095649928692772510712644530606542512776517\ 435874811659146319630670909 The implied delta is, 0.07916786443955308683937807391588653096766141858144688\ 8102825610972009669602484145480313628946192650 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0962654128835568563818378736315709564992869277251071264453060654251277\ 6517435874811659146319630670909 ----------------------- This took, 3.614, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |8 x + 11 x + 4| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |8 x + 11 x + 4| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (2 n - 1) a(n - 1) 7 (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 0.08921446688943169293092583111195045716518487104723565893293488689922295893\ 499162985078998393501845002 The implied delta is, 0.04540904658398761381316120489309151071581081582901689\ 2989551633036984124706114487083387730713454991 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0892144668894316929309258311119504571651848710472356589329348868992229\ 5893499162985078998393501845002 ----------------------- This took, 4.318, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 4) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 4) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (2 n - 1) a(n - 1) 9 (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 0.07833393820762259227515617185805701078348317480204134007489868950100031303\ 357106369790935048267940864 This constant is identified as, 1/2 ln(2) - 1/6 ln(5) The implied delta is, 0.01610305134163057059476030447407709848452163925938429\ 5868197117252961128356062199075422485134368797 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) - 1/6 ln(5) ----------------------- This took, 1.050, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 13 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 13 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (2 n - 1) a(n - 1) 7 (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 0.07588666029425126194667792957990351954635552062910919636294014340832577279\ 573527290819229322993544913 The implied delta is, 0.08341738038702336058364740517185903189403827696615325\ 6864804634845084861556551581449375207636797003 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0758866602942512619466779295799035195463555206291091963629401434083257\ 7279573527290819229322993544913 ----------------------- This took, 3.960, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 14 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 14 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)-20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 20 (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 0.07401864296956698328915278706578130366845790441842631624544457810159782620\ 634917432008426448528456462 The implied delta is, 0.00956302183336710747165309413534796446032722860459440\ 8537264956010296660146491527942647381783135010 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0740186429695669832891527870657813036684579044184263162454445781015978\ 2620634917432008426448528456462 ----------------------- This took, 4.447, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 14 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 14 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 12 (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 0.07002215900235345774030680200106173199354197535413196028480523253976737413\ 402753135529869708372848921 The implied delta is, 0.15865568467687828602963877791376917583094136061892326\ 2393033422056391962688103245299338902973559973 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0700221590023534577403068020010617319935419753541319602848052325397673\ 7413402753135529869708372848921 ----------------------- This took, 3.329, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(7 x + 4) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(7 x + 4) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (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 0.06676569631226131157317181046567498729470783674945228695132493927130050157\ 850743951496574120069226782 This constant is identified as, 3/2 ln(2) - 1/2 ln(7) The implied delta is, 0.55621365185476501964096807703410704299320585275394282\ 5728523991339290921553255210996924685007797235 Since this is positive, this suggests an Apery-style irrationality proof of, 3/2 ln(2) - 1/2 ln(7) ----------------------- This took, 0.969, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 15 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 15 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 15 (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 0.06524176133871410934932854113492265498822323720049651191532932060199214477\ 817605188744976432300036974 The implied delta is, 0.35563101509770246590172023698722401567618661496957575\ 9764109128677597308464170617516368549119007347 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0652417613387141093493285411349226549882232372004965119153293206019921\ 4477817605188744976432300036974 ----------------------- This took, 2.767, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 2) (5 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 2) (5 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 16 (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 0.06385320297074883540068926203795774185976384955572103377224419472958558681\ 113109971956476540993094025 This constant is identified as, 1/8 ln(5) - 1/8 ln(3) The implied delta is, 0.10615034289762958358759021710934598219278290513911819\ 4663861486759455333095825793683868620538599174 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(5) - 1/8 ln(3) ----------------------- This took, 1.045, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 16 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 16 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)+16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 16 (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 0.06124466578171603854302062030281895272853602459529601678184397866683877989\ 691052414363353941717549784 This constant is identified as, 4 3 2 arctan(RootOf(_Z + 16 _Z - 6 _Z - 16 _Z + 1, index = 1)) The implied delta is, 0.11487770571995119154588826269746259719379655721411845\ 1668796217661994791759545077606287375009554965 Since this is positive, this suggests an Apery-style irrationality proof of, 4 3 2 arctan(RootOf(_Z + 16 _Z - 6 _Z - 16 _Z + 1, index = 1)) ----------------------- This took, 2.807, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 17 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 17 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) 17 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)-17*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (2 n - 1) a(n - 1) 17 (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 0.06001943292689520099383489098000372511696421714199915636554428250224777596\ 076376507533428748823867607 The implied delta is, 0.35797743963699034751018837031627690157571159682028593\ 9152877196834485208493230993737724456797825488 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0600194329268952009938348909800037251169642171419991563655442825022477\ 7596076376507533428748823867607 ----------------------- This took, 3.119, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (9 x + 4)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (9 x + 4)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (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 0.05889151782819172726939705473526085253424035628236657055367431939740386026\ 406689346482076431910405747 This constant is identified as, ln(3) - 3/2 ln(2) The implied delta is, 0.56699175013732249851317171476635314501988343365560925\ 0888942795636346050682355683916571549465815244 Since this is positive, this suggests an Apery-style irrationality proof of, ln(3) - 3/2 ln(2) ----------------------- This took, 0.870, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 18 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 18 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)-20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 20 (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 0.05674297284502745790638413363366487111797774615367648258972093851440816037\ 027027144675060081120824713 The implied delta is, 0.07400731640432512616918974634463036891175733582857203\ 5147835789635756333392594209264497730706444462 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0567429728450274579063841336336648711179777461536764825897209385144081\ 6037027027144675060081120824713 ----------------------- This took, 3.510, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 18 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 18 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 12 (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 0.05488453413596129783433531450527535452438525961595458818551767919054916658\ 617074219041895248803163831 The implied delta is, 0.40895042964810924600624793751049929879378971965734642\ 6535549487656215862929214834673121682709336075 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0548845341359612978343353145052753545243852596159545881855176791905491\ 6658617074219041895248803163831 ----------------------- This took, 2.374, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |27 x + 18 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |27 x + 18 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 108 (n - 1) b(n - 2) b(n) = --------------------- + -------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)+108*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 108 (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 0.05038331567317271807205772937894877034122406244702057154360791487189231058\ 684316343567013856144611580 1/2 3 Pi This constant is identified as, ------- 108 The implied delta is, 0.14486133747022038890565267938018354696695665307788026\ 3392575271308456781478196981104133746412120780 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 108 ----------------------- This took, 2.885, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 19 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 19 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) 57 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-57*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (2 n - 1) a(n - 1) 57 (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 0.05569786524605326515251509718625707544688681893731607996267835076945774344\ 382920322109965172002084211 The implied delta is, 0.01513427527809917715748317869432640754106945569914584\ 0963485631195528300133495165760709887555974836 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0556978652460532651525150971862570754468868189373160799626783507694577\ 4344382920322109965172002084211 ----------------------- This took, 3.067, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (11 x + 4)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (11 x + 4)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (2 n - 1) a(n - 1) 9 (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 0.05307562185308910263504120226509993259920107514277523547608794677506545966\ 735876104291078783556312146 This constant is identified as, 1/6 ln(11) - 1/2 ln(2) The implied delta is, 0.09962659107436244254080249667527706313010630082718949\ 6730614138309097556796771747256705966726982679 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(11) - 1/2 ln(2) ----------------------- This took, 1.196, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 19 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 19 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (2 n - 1) a(n - 1) 7 (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 0.05229529713784615953269393800299452421459673663565102285382817486274567995\ 031716724406350130700857979 The implied delta is, 0.15664310493321996321377297688838661173108301235304997\ 2360238858926180656754847134529058761678817342 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0522952971378461595326939380029945242145967366356510228538281748627456\ 7995031716724406350130700857979 ----------------------- This took, 3.273, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(9 x + 4) (3 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(9 x + 4) (3 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 9 (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 0.04794701207529682123986983433230457191725161848296017608444428089154882512\ 013007738968514986318421438 This constant is identified as, 1/3 ln(2) - 1/6 ln(3) The implied delta is, 0.45941246953427346656204961556678864299476593113738310\ 9354684929340294305799729810672161877193950129 Since this is positive, this suggests an Apery-style irrationality proof of, 1/3 ln(2) - 1/6 ln(3) ----------------------- This took, 0.630, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |28 x + 21 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |28 x + 21 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 7 (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 0.04736946766082724053372186490537593726708557177871504860517232906646752194\ 541385366757701284003752490 The implied delta is, 0.47645322687075672502926149857061858978879352350445771\ 9083837341078271695508371937340849362063010437 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0473694676608272405337218649053759372670855717787150486051723290664675\ 2194541385366757701284003752490 ----------------------- This took, 3.419, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 22 x + 4| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 22 x + 4| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)-20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 20 (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 0.04609663757345832931660958157935107040767774848775313750188866557895974958\ 930899781877014828437817591 The implied delta is, 0.12106201044416215074050257583112536382114803351413277\ 5003781756586614535600757046880868705935449754 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0460966375734583293166095815793510704076777484877531375018886655789597\ 4958930899781877014828437817591 ----------------------- This took, 3.193, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |6 x + 10 x + 5| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |6 x + 10 x + 5| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 10 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 10*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 10 (2 n - 1) a(n - 1) 20 (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 0.09403433605676343518168600277404782507929297311294733417545385860544677198\ 381079303252695013504968627 The implied delta is, 0.07720156228804965939235880149890205848873639970474698\ 2173838234295769481177143391467973181913521255 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0940343360567634351816860027740478250792929731129473341754538586054467\ 7198381079303252695013504968627 ----------------------- This took, 2.322, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (6 x + 5)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (6 x + 5)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 11 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 11*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 11 (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 0.09116077839697731310585901257725731659869466895724349197136322582835446374\ 032295892467260185848558276 This constant is identified as, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) The implied delta is, 0.51738394387588245866899222655825392097534060270982986\ 7955158319124421684607024589198054063909583979 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(3) - 1/2 ln(5) ----------------------- This took, 0.701, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (7 x + 5)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (7 x + 5)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 12 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 12*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 12 (2 n - 1) a(n - 1) 4 (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 0.08411805915530323262614835255424802252787084382833586663668556461585021876\ 110287578938115512367297905 This constant is identified as, 1/4 ln(7) - 1/4 ln(5) The implied delta is, 0.19465129863063449750724901593324670032608191704430191\ 5206647384755147331521682139677859985676867757 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(7) - 1/4 ln(5) ----------------------- This took, 1.010, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |8 x + 12 x + 5| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |8 x + 12 x + 5| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 12 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 12*(2*n-1)/n*b(n-1)+16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 12 (2 n - 1) a(n - 1) 16 (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 0.08043763859916054835035115358966532975518882388941404785820076483918905935\ 145263608910210558766034361 This constant is identified as, 4 3 2 arctan(RootOf(_Z + 12 _Z - 6 _Z - 12 _Z + 1, index = 1)) The implied delta is, 0.16063795481906656264925625934045099490103846024953814\ 5272997489739037579524518608740239665466944445 Since this is positive, this suggests an Apery-style irrationality proof of, 4 3 2 arctan(RootOf(_Z + 12 _Z - 6 _Z - 12 _Z + 1, index = 1)) ----------------------- This took, 2.248, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (8 x + 5)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (8 x + 5)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (2 n - 1) a(n - 1) 9 (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 0.07833393820762259227515617185805701078348317480204134007489868950100031303\ 357106369790935048267940864 This constant is identified as, 1/2 ln(2) - 1/6 ln(5) The implied delta is, 0.01610305134163057059476030447407709848452163925938429\ 5868197117252961128356062199075422485134368797 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) - 1/6 ln(5) ----------------------- This took, 1.228, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |10 x + 14 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |10 x + 14 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 4 (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 0.07094852730208196140642580855127654150389087936423203618906501468172081337\ 996558047209593081712325906 1/2 1/2 (50 - 35 2 ) This constant is identified as, arcsin(-----------------) 10 The implied delta is, 0.33551462144924097388011542229878210355571808363541944\ 8570488510188258420343720921802437054447955885 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (50 - 35 2 ) arcsin(-----------------) 10 ----------------------- This took, 3.716, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |9 x + 15 x + 5| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |9 x + 15 x + 5| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 45 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)-45*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 45 (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 0.07173482349400067314823887215843423757076178042338160912616768399876020257\ 685532689358171078286999257 The implied delta is, 0.19244119522381735461498668356879235390638367946531955\ 9604835350201284520804025111883105010562696584 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0717348234940006731482388721584342375707617804233816091261676839987602\ 0257685532689358171078286999257 ----------------------- This took, 3.949, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 15 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 15 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 5 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)-5*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 5 (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 0.06716718454286805286641104234374010537536953243329219488761389579308495531\ 786763204057660337804084846 The implied delta is, 0.44990483778400802664462417131648416532387503986548788\ 6689015237674958402560913850014235090224811674 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0671671845428680528664110423437401053753695324332921948876138957930849\ 5531786763204057660337804084846 ----------------------- This took, 3.767, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |12 x + 15 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |12 x + 15 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 15 (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 0.06524176133871410934932854113492265498822323720049651191532932060199214477\ 817605188744976432300036974 The implied delta is, 0.35563101509770246590172023698722401567618661496957575\ 9764109128677597308464170617516368549119007347 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0652417613387141093493285411349226549882232372004965119153293206019921\ 4477817605188744976432300036974 ----------------------- This took, 3.449, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(6 x + 5) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(6 x + 5) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 16 (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 0.06385320297074883540068926203795774185976384955572103377224419472958558681\ 113109971956476540993094025 This constant is identified as, 1/8 ln(5) - 1/8 ln(3) The implied delta is, 0.10615034289762958358759021710934598219278290513911819\ 4663861486759455333095825793683868620538599174 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(5) - 1/8 ln(3) ----------------------- This took, 0.934, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 16 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 16 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 4 (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 0.06217749727338071751567742458193551278658509588490204495755705955786113371\ 378337931185529715667666516 1/2 1/2 (8450 - 1040 65 ) This constant is identified as, arcsin(----------------------) 130 The implied delta is, 0.24784947239424543281039077715088539761606818623457026\ 8317431564895143195941304894277394009079862295 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (8450 - 1040 65 ) arcsin(----------------------) 130 ----------------------- This took, 3.901, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(7 x + 5) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(7 x + 5) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (2 n - 1) a(n - 1) 9 (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 0.05944582398978872981877311854019741299400279317448529792898962517166545782\ 088053540805645108362112856 This constant is identified as, 1/6 ln(2) + 1/6 ln(5) - 1/6 ln(7) The implied delta is, 0.07611198593963543775096609279350010649069975182356418\ 1844947486669166758373378389861083147387485543 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(2) + 1/6 ln(5) - 1/6 ln(7) ----------------------- This took, 1.282, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 17 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 17 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) 11 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)+11*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (2 n - 1) a(n - 1) 11 (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 0.05809380401645286932845839538933593430322189548043015928749092971870836258\ 588018482326301833941239337 The implied delta is, 0.03510789040775259729732552971699613456032344320168967\ 0950605148289146553202186823376185318634015031 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0580938040164528693284583953893359343032218954804301592874909297187083\ 6258588018482326301833941239337 ----------------------- This took, 4.069, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 17 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 17 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) 51 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)+51*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (2 n - 1) a(n - 1) 51 (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 0.05568905703830166191559200070822420551981044978755486015085540951056441662\ 158426225699299791687749345 The implied delta is, 0.02248546426488827660316269814530081012484033151887880\ 7435156488296057788081333567215200787225129255 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0556890570383016619155920007082242055198104497875548601508554095105644\ 1662158426225699299791687749345 ----------------------- This took, 4.063, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(8 x + 5) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(8 x + 5) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 4 (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 0.05578588782855243894157377257745862584365027138700180341782196812184793594\ 206708335460180602508555893 This constant is identified as, 1/4 ln(5) - 1/2 ln(2) The implied delta is, 0.49199742178760878483436315027683384791232232715975328\ 9110863942641162777431703575817615425245818238 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(5) - 1/2 ln(2) ----------------------- This took, 0.690, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 18 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 18 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)+36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 36 (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 0.05362509239944036556690076905977688650345921592627603190546717655945937290\ 096842405940140372510689574 This constant is identified as, arctan( 6 5 4 3 2 RootOf(_Z + 18 _Z - 15 _Z - 60 _Z + 15 _Z + 18 _Z - 1, index = 1)) The implied delta is, 0.16061722484890563230786124818828382971672445401129761\ 7274250617734415105448505431680291314028864036 Since this is positive, this suggests an Apery-style irrationality proof of, arctan( 6 5 4 3 2 RootOf(_Z + 18 _Z - 15 _Z - 60 _Z + 15 _Z + 18 _Z - 1, index = 1)) ----------------------- This took, 3.686, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(9 x + 5) (2 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(9 x + 5) (2 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (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 0.05268025782891315061375049041965639915306018649163703628196961684629201162\ 006727324438284773106706038 This constant is identified as, 1/2 ln(2) + 1/2 ln(5) - ln(3) The implied delta is, 0.57337177129358798630972583682740462220014717381052020\ 3127971283730002664556591027268962044943194084 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(5) - ln(3) ----------------------- This took, 0.929, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 19 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 19 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) 19 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)+19*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (2 n - 1) a(n - 1) 19 (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 0.05173632351176964518572279583252761388906410025418529486272107390704629984\ 161983760655745255281660317 The implied delta is, 0.37985557990797613666219089783982035976906391067131271\ 4693255810111080445187484333796874552972560161 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0517363235117696451857227958325276138890641002541852948627210739070462\ 9984161983760655745255281660317 ----------------------- This took, 3.444, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 20 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 20 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 80 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-80*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 80 (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 0.05380111762050050486117915411882567817807133531753620684462576299907015193\ 264149517018628308715249443 The implied delta is, 0.19241837374136457755047606181877356209171493884639992\ 6784811711772314974523730911590576170017932817 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0538011176205005048611791541188256781780713353175362068446257629990701\ 5193264149517018628308715249443 ----------------------- This took, 3.525, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 20 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 20 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 40 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-40*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 40 (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 0.05177441474570309562625581681897158600064073773170552450255681654745943689\ 051284618097664366479131960 The implied delta is, 0.04566308295447717905112811140638400015597404182898616\ 4853836664038983172727996437356568042526582583 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0517744147457030956262558168189715860006407377317055245025568165474594\ 3689051284618097664366479131960 ----------------------- This took, 2.774, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 20 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 20 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 20 (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 0.05085926239597355181597417460398648523816492448139593109953058748373214349\ 501271889362196536309674172 The implied delta is, 0.13355733679220129512447085613880841717184023506996683\ 5345945528848101960806098428008666415740788844 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0508592623959735518159741746039864852381649244813959310995305874837321\ 4349501271889362196536309674172 ----------------------- This took, 2.907, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 20 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 20 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 20 (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 0.04919080716889344574097636756153234522285727490773021655302300944044905350\ 211374891483921052889318277 The implied delta is, 0.13863272796629144134638102593430955758525488588670898\ 2509250910100860016474186419275088494114249089 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0491908071688934457409763675615323452228572749077302165530230094404490\ 5350211374891483921052889318277 ----------------------- This took, 3.346, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |22 x + 20 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |22 x + 20 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 40 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+40*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 40 (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 0.04842670411701946246901902500592326026755587116454064533465773417374428298\ 822154940811526578349125121 The implied delta is, 0.05773499370554175330731823631097035020848250640477256\ 1852740643560102740298910950674333307825105535 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0484267041170194624690190250059232602675558711645406453346577341737442\ 8298822154940811526578349125121 ----------------------- This took, 3.635, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |24 x + 20 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |24 x + 20 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 80 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+80*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 80 (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 0.04701716802838171759084300138702391253964648655647366708772692930272338599\ 190539651626347506752484313 The implied delta is, 0.07718308342065558358695422994012552592438240455617915\ 8195030857272987552219894370995019698213439525 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0470171680283817175908430013870239125396464865564736670877269293027233\ 8599190539651626347506752484313 ----------------------- This took, 3.333, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 21 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 21 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 21 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)-21*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 21 (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 0.04839726432379090629296210200095405235968357244556184368796730619374097974\ 318039341226944590436921713 The implied delta is, 0.38449708080617744030381255205049812806089007313320668\ 9805507300773481011458692277230135664412103898 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0483972643237909062929621020009540523596835724455618436879673061937409\ 7974318039341226944590436921713 ----------------------- This took, 2.825, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (11 x + 5)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (11 x + 5)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (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 0.04765508990216243002197606164038254611030268265432209959261990408150050711\ 794211641952875145651824654 This constant is identified as, 1/2 ln(11) - 1/2 ln(2) - 1/2 ln(5) The implied delta is, 0.58448163333107330960348282540393061902711881756838552\ 1535141988810736997507168010457014961147704158 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(11) - 1/2 ln(2) - 1/2 ln(5) ----------------------- This took, 1.476, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |22 x + 22 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |22 x + 22 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 44 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)-44*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 44 (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 0.04691233882745658088469094517599776535573223426848000037555554869189588873\ 318733018462329129940292162 The implied delta is, 0.17365920458969177782584436581682405206744689026287059\ 9984227227928722096106320149129986935460878405 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0469123388274565808846909451759977653557322342684800003755555486918958\ 8873318733018462329129940292162 ----------------------- This took, 2.794, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (12 x + 5)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (12 x + 5)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 4 (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 0.04558038919848865655292950628862865829934733447862174598568161291417723187\ 016147946233630092924279138 This constant is identified as, 1/4 ln(2) + 1/4 ln(3) - 1/4 ln(5) The implied delta is, 0.51729668886995206530084709053887995949953745192577735\ 9815781195706547758164390443811084687132348320 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(2) + 1/4 ln(3) - 1/4 ln(5) ----------------------- This took, 1.187, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 22 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 22 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)+16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 16 (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 0.04496337494811956764713824931402705900324338420729802976366825749832865266\ 146984585305414017909871408 This constant is identified as, 4 3 2 arctan(RootOf(_Z + 22 _Z - 6 _Z - 22 _Z + 1, index = 1)) The implied delta is, 0.00931239097512787661404518740398702849387333760540556\ 8470894784748821445574618664273885486759785495 Since this is positive, this suggests an Apery-style irrationality proof of, 4 3 2 arctan(RootOf(_Z + 22 _Z - 6 _Z - 22 _Z + 1, index = 1)) ----------------------- This took, 3.226, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 23 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 23 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 69 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)-69*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 69 (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 0.04553191063783334348531137715469670756730225101952805845053581354545885630\ 928038531951499397864176865 The implied delta is, 0.04562433460760107869697011320796169233352579285082243\ 2697574933564964326084472901735013283272446499 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0455319106378333434853113771546967075673022510195280584505358135454588\ 5630928038531951499397864176865 ----------------------- This took, 5.333, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (13 x + 5)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (13 x + 5)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 9 (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 0.04372737741124850867258266448015906620069440935523902339761960161598200960\ 784294986059624305848186507 This constant is identified as, 1/6 ln(13) - 1/6 ln(2) - 1/6 ln(5) The implied delta is, 0.13483860817151004634809823925040754975159289838216717\ 2217996162065164183767152605569956017751349947 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(13) - 1/6 ln(2) - 1/6 ln(5) ----------------------- This took, 1.192, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |27 x + 23 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |27 x + 23 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 11 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)+11*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 11 (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 0.04318060420506946996695508989112974642538934363218783432261890072916620644\ 069688938742683754227846868 The implied delta is, 0.09424908812473310387972776499766723099521381573971991\ 5157830531015081719707215612869907653723893316 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0431806042050694699669550898911297464253893436321878343226189007291662\ 0644069688938742683754227846868 ----------------------- This took, 4.164, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(9 x + 5) (3 x + 1)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(9 x + 5) (3 x + 1)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 36 (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 0.04256880198049922360045950802530516123984256637048068918149612981972372454\ 075406647970984360662062683 This constant is identified as, 1/12 ln(5) - 1/12 ln(3) The implied delta is, 0.10593562707994796705398063442339156177884302503791325\ 5167835526414470865707643963934009810273033480 Since this is positive, this suggests an Apery-style irrationality proof of, 1/12 ln(5) - 1/12 ln(3) ----------------------- This took, 1.013, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (14 x + 5)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (14 x + 5)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 16 (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 0.04205902957765161631307417627712401126393542191416793331834278230792510938\ 055143789469057756183648952 This constant is identified as, 1/8 ln(7) - 1/8 ln(5) The implied delta is, 0.19461879274227127765289610441958127788741542827729278\ 5340052753021540455403016448549792561298027677 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(7) - 1/8 ln(5) ----------------------- This took, 1.685, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 24 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 24 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 4 (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 0.04157061594422061495533415732539064876556567031400971311731373319799870316\ 836930147086841894450747869 This constant is identified as, 1/2 arctan(1/12) The implied delta is, 0.31130241218905106709386590504843215305733025052140308\ 8860957485128865708755294993420496372911533017 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 arctan(1/12) ----------------------- This took, 3.887, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |30 x + 24 x + 5| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |30 x + 24 x + 5| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 24 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+24*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 24 (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 0.04110201346700524332646682755080302988958301481227112460848308056441555358\ 699858757241274207689709603 The implied delta is, 0.15872249030202671343471562268121857169174306272504912\ 2716760813338163056076735287433594279115223953 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0411020134670052433264668275508030298895830148122711246084830805644155\ 5358699858757241274207689709603 ----------------------- This took, 3.662, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (7 x + 6)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (7 x + 6)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 13 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 13*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 13 (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 0.07707533991362915214643769253123872845704701869942824130200790340334597378\ 188279265408970838886037533 This constant is identified as, 1/2 ln(7) - 1/2 ln(2) - 1/2 ln(3) The implied delta is, 0.53598696183853215629483188639824344780581812162220287\ 2066477785480308463725655439137125446634136583 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(7) - 1/2 ln(2) - 1/2 ln(3) ----------------------- This took, 1.325, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |7 x + 14 x + 6| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |7 x + 14 x + 6| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)-28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 28 (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 0.07515497185030861111837468180255027686449784352277401390010827505367720667\ 693974200463726324464985163 The implied delta is, 0.10686249638390301523636562155291556457477022400357666\ 7164412126293469051547912365992267528292025175 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0751549718503086111183746818025502768644978435227740139001082750536772\ 0667693974200463726324464985163 ----------------------- This took, 2.519, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |10 x + 15 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |10 x + 15 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 15 (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 0.06524176133871410934932854113492265498822323720049651191532932060199214477\ 817605188744976432300036974 The implied delta is, 0.35563101509770246590172023698722401567618661496957575\ 9764109128677597308464170617516368549119007347 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0652417613387141093493285411349226549882232372004965119153293206019921\ 4477817605188744976432300036974 ----------------------- This took, 2.859, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 16 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 16 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 8 (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 0.06186089934397364766167805961521493911390506219717794272116055504762000878\ 355963260558777854258289550 The implied delta is, 0.18614007624007003473212386151298091116665090859195718\ 8735980949280727987067183567085883629690535105 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0618608993439736476616780596152149391139050621971779427211605550476200\ 0878355963260558777854258289550 ----------------------- This took, 2.682, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(4 x + 3) (3 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(4 x + 3) (3 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (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 0.05889151782819172726939705473526085253424035628236657055367431939740386026\ 406689346482076431910405747 This constant is identified as, ln(3) - 3/2 ln(2) The implied delta is, 0.56699175013732249851317171476635314501988343365560925\ 0888942795636346050682355683916571549465815244 Since this is positive, this suggests an Apery-style irrationality proof of, ln(3) - 3/2 ln(2) ----------------------- This took, 1.416, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 18 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 18 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)-12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 12 (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 0.05625708338586260381578117769348148903118350424572836423420400122983851411\ 098925064855857658563755154 The implied delta is, 0.27780175210518099077819661085816579282335749466746754\ 6371715412236858321612176011509442533123151172 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0562570833858626038157811776934814890311835042457283642342040012298385\ 1411098925064855857658563755154 ----------------------- This took, 2.294, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 2) (5 x + 3)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 2) (5 x + 3)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (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 0.05268025782891315061375049041965639915306018649163703628196961684629201162\ 006727324438284773106706038 This constant is identified as, 1/2 ln(2) + 1/2 ln(5) - ln(3) The implied delta is, 0.57337177129358798630972583682740462220014717381052020\ 3127971283730002664556591027268962044943194084 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(5) - ln(3) ----------------------- This took, 1.373, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 20 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 20 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 40 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-40*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 40 (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 0.05177441474570309562625581681897158600064073773170552450255681654745943689\ 051284618097664366479131960 The implied delta is, 0.04566308295447717905112811140638400015597404182898616\ 4853836664038983172727996437356568042526582583 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0517744147457030956262558168189715860006407377317055245025568165474594\ 3689051284618097664366479131960 ----------------------- This took, 2.137, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 20 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 20 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 8 (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 0.04967061039839491930313833651041929130113370413537299811086621361777858867\ 016696802748552520455447858 The implied delta is, 0.23040620853875572752475260410122484253806905942047892\ 8867151614179424617761342654894984426168613891 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0496706103983949193031383365104192913011337041353729981108662136177785\ 8867016696802748552520455447858 ----------------------- This took, 3.736, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 21 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 21 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 15 (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 0.04708990665173687087705945368512562031565589942280528803395248311257030595\ 223218813682855133287388630 The implied delta is, 0.14796367788986981861737279434105200048337713023360029\ 2537281745665031843769974796842741861385712397 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0470899066517368708770594536851256203156558994228052880339524831125703\ 0595223218813682855133287388630 ----------------------- This took, 3.491, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 22 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 22 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 20 (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 0.04484352888786998944070963521251547985643569820521711762243084916499771848\ 169704411768773960615650350 The implied delta is, 0.04345228710234823610804653089872902925199841868863775\ 6911169567427881177512877642957319168077290959 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0448435288878699894407096352125154798564356982052171176224308491649977\ 1848169704411768773960615650350 ----------------------- This took, 3.499, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(11 x + 6) (2 x + 1)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(11 x + 6) (2 x + 1)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (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 0.04350568849481488308388295093687477048839198630292139237874332174685395662\ 238084250514385040196733623 This constant is identified as, ln(2) + 1/2 ln(3) - 1/2 ln(11) The implied delta is, 0.59061022211924122523032694961230388993126871412053070\ 1772420840244819687524482681229208270558273191 Since this is positive, this suggests an Apery-style irrationality proof of, ln(2) + 1/2 ln(3) - 1/2 ln(11) ----------------------- This took, 1.018, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 23 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 23 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 23 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)+23*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 23 (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 0.04286408522941130351756596778358269951577516600040348070000305839552437419\ 944580503842240868061966763 The implied delta is, 0.39238365592280191149827103141409923270590587393104123\ 4282443420931110369066409731132018459518468236 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0428640852294113035175659677835826995157751660004034807000030583955243\ 7419944580503842240868061966763 ----------------------- This took, 3.297, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 24 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 24 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 24 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-24*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 24 (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 0.04226028299326606124120455269664863144684882830331269412586315622938280884\ 682510078775479255433948579 The implied delta is, 0.15544797538878460596532542601012849379755283393335943\ 0877827755493118711916148593604169767544585855 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0422602829932660612412045526966486314468488283033126941258631562293828\ 0884682510078775479255433948579 ----------------------- This took, 3.442, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 24 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 24 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 24 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+24*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 24 (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 0.04110201346700524332646682755080302988958301481227112460848308056441555358\ 699858757241274207689709603 The implied delta is, 0.15872249030202671343471562268121857169174306272504912\ 2716760813338163056076735287433594279115223953 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0411020134670052433264668275508030298895830148122711246084830805644155\ 5358699858757241274207689709603 ----------------------- This took, 1.998, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 3) (5 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 3) (5 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 25 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)-25*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 25 (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 0.04054651081081643819780131154643491365719904234624941976140143241441006712\ 489142512677524278173134012 This constant is identified as, 1/10 ln(3) - 1/10 ln(2) The implied delta is, 0.40310774359081769831902773956023441949947542760807547\ 5452008370382755255741916214559013866579750046 Since this is positive, this suggests an Apery-style irrationality proof of, 1/10 ln(3) - 1/10 ln(2) ----------------------- This took, 0.579, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (13 x + 6)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (13 x + 6)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (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 0.04002135383676821291188898086321988200338855910847357822149557901959156508\ 320589065711612731696001245 This constant is identified as, 1/2 ln(13) - ln(2) - 1/2 ln(3) The implied delta is, 0.60013843912498329579659016153611624482487887296542625\ 4327249019701506377029369398348550371353894390 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(13) - ln(2) - 1/2 ln(3) ----------------------- This took, 1.531, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |27 x + 26 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |27 x + 26 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)-28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 28 (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 0.03900616621219270031235406676281317733458068887816479548306424170916916862\ 094441445724560992903497071 The implied delta is, -0.0021907515132339270980877411521375595648536769275683\ 707725341174918791698726231998390580736950490735 Since this is negative, there is no Apery-style irrationality proof of, 0.039\ 006166212192700312354066762813177334580688878164795483064241709169168620\ 94441445724560992903497071, but still a very fast way to compute it to many digits ----------------------- This took, 3.021, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 26 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 26 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 20 (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 0.03808882748097714898361105284226208870283548820614541897504716670357796138\ 563352270709944941834037385 The implied delta is, 0.07441987974677790809813059955740279997471963228806469\ 4169707171818900634894966422366856958894409602 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0380888274809771489836110528422620887028354882061454189750471667035779\ 6138563352270709944941834037385 ----------------------- This took, 3.689, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 27 x + 6| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 27 x + 6| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) 33 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)-33*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (2 n - 1) a(n - 1) 33 (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 0.03761158252848053508404908973300588578509691865752796012018828400019744557\ 831826254183051661597772393 The implied delta is, 0.01354937434329701380729492484408917691211671616471934\ 7404593157654871473576490476273032748723062422 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0376115825284805350840490897330058857850969186575279601201882840001974\ 4557831826254183051661597772393 ----------------------- This took, 4.917, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |8 x + 14 x + 7| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |8 x + 14 x + 7| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 14 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 14*(2*n-1)/n*b(n-1)+28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 14 (2 n - 1) a(n - 1) 28 (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 0.06829196727512946673232384800866319721613522141297535376905360798284524044\ 020274175918349838752798746 The implied delta is, 0.26781986603328533130929186802726998229938812736201754\ 1678137799909458304201247253716008369923799243 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0682919672751294667323238480086631972161352214129753537690536079828452\ 4044020274175918349838752798746 ----------------------- This took, 4.388, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (8 x + 7)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (8 x + 7)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (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 0.06676569631226131157317181046567498729470783674945228695132493927130050157\ 850743951496574120069226782 This constant is identified as, 3/2 ln(2) - 1/2 ln(7) The implied delta is, 0.55621365185476501964096807703410704299320585275394282\ 5728523991339290921553255210996924685007797235 Since this is positive, this suggests an Apery-style irrationality proof of, 3/2 ln(2) - 1/2 ln(7) ----------------------- This took, 1.235, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |9 x + 15 x + 7| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |9 x + 15 x + 7| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 15 (2 n - 1) b(n - 1) 27 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 15*(2*n-1)/n*b(n-1)+27*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 15 (2 n - 1) a(n - 1) 27 (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 0.06417694192237123758789191575438063766619128515007141763020404361675184344\ 957249874439430879753780606 The implied delta is, 0.07951528682693472215196613877780250919344358808659570\ 6742859327337976383839172758121834451624476093 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0641769419223712375878919157543806376661912851500714176302040436167518\ 4344957249874439430879753780606 ----------------------- This took, 5.282, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |8 x + 16 x + 7| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |8 x + 16 x + 7| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 32 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)-32*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 32 (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 0.06531880717255998497326232950475989146680782352543278215741049008544606209\ 641747066648838515448790741 The implied delta is, 0.00951238899503296468987235071777560668702736455958250\ 4540453547961093502177815919224934468006407511 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0653188071725599849732623295047598914668078235254327821574104900854460\ 6209641747066648838515448790741 ----------------------- This took, 4.689, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (9 x + 7)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (9 x + 7)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 4 (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 0.06282860707022651942128443260046791991447409651590942875249962933435218092\ 128716648989325275989816265 This constant is identified as, 1/2 ln(3) - 1/4 ln(7) The implied delta is, 0.24726794658096722576995160167459976418634699867116936\ 8019402426017252451484100726909737452714571334 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(3) - 1/4 ln(7) ----------------------- This took, 0.933, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (10 x + 7)\n |------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (10 x + 7)\n |------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (2 n - 1) a(n - 1) 9 (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 0.05944582398978872981877311854019741299400279317448529792898962517166545782\ 088053540805645108362112856 This constant is identified as, 1/6 ln(2) + 1/6 ln(5) - 1/6 ln(7) The implied delta is, 0.07611198593963543775096609279350010649069975182356418\ 1844947486669166758373378389861083147387485543 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(2) + 1/6 ln(5) - 1/6 ln(7) ----------------------- This took, 0.917, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |12 x + 18 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |12 x + 18 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 12 (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 0.05488453413596129783433531450527535452438525961595458818551767919054916658\ 617074219041895248803163831 The implied delta is, 0.40895042964810924600624793751049929879378971965734642\ 6535549487656215862929214834673121682709336075 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0548845341359612978343353145052753545243852596159545881855176791905491\ 6658617074219041895248803163831 ----------------------- This took, 2.721, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 19 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 19 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (2 n - 1) a(n - 1) 3 (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 0.05248650776240679728306214326101844901148990474195033232100166577072078208\ 066243352258553222515643976 The implied delta is, 0.33810164621336086873031754496543126090298209622663550\ 8681673884252151133271768712316879291330002369 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0524865077624067972830621432610184490114899047419503323210016657707207\ 8208066243352258553222515643976 ----------------------- This took, 4.297, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 20 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 20 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 8 (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 0.05033739137987531680396072210624443545811636207155249781971542770630075388\ 783720756788099599986168438 The implied delta is, 0.22659790410276788169731799456865555953489559468620292\ 7274515510405818366802895587855798182465179745 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0503373913798753168039607221062444354581163620715524978197154277063007\ 5388783720756788099599986168438 ----------------------- This took, 3.927, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 20 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 20 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 20 (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 0.04919080716889344574097636756153234522285727490773021655302300944044905350\ 211374891483921052889318277 The implied delta is, 0.13863272796629144134638102593430955758525488588670898\ 2509250910100860016474186419275088494114249089 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0491908071688934457409763675615323452228572749077302165530230094404490\ 5350211374891483921052889318277 ----------------------- This took, 2.290, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 20 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 20 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 48 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+48*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 48 (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 0.04813270644177842819091893681578547824964346386255356322265303271256388258\ 717937405829573159815335455 The implied delta is, 0.07915134337277481450301140113410009605059350261577846\ 1191855582302426473464757202043497910650013574 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0481327064417784281909189368157854782496434638625535632226530327125638\ 8258717937405829573159815335455 ----------------------- This took, 3.685, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 21 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 21 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 21 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)-21*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 21 (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 0.04839726432379090629296210200095405235968357244556184368796730619374097974\ 318039341226944590436921713 The implied delta is, 0.38449708080617744030381255205049812806089007313320668\ 9805507300773481011458692277230135664412103898 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0483972643237909062929621020009540523596835724455618436879673061937409\ 7974318039341226944590436921713 ----------------------- This took, 2.876, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 21 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 21 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 7 (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 0.04736946766082724053372186490537593726708557177871504860517232906646752194\ 541385366757701284003752490 The implied delta is, 0.47645322687075672502926149857061858978879352350445771\ 9083837341078271695508371937340849362063010437 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0473694676608272405337218649053759372670855717787150486051723290664675\ 2194541385366757701284003752490 ----------------------- This took, 2.290, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 21 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 21 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 35 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+35*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 35 (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 0.04641606960181795731040868117732920831619658168627882575097598230260823302\ 056115143988339498112492415 The implied delta is, 0.05347060584557212259453506654463495528385947076782319\ 7735880259453372906568550313980675333852415110 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0464160696018179573104086811773292083161965816862788257509759823026082\ 3302056115143988339498112492415 ----------------------- This took, 3.854, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 21 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 21 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 63 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+63*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 63 (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 0.04552797818341964448821589867244213147742348094198356917936907198856349362\ 680182783945566559168532497 The implied delta is, 0.26777964817509272096295917061879860566963611085557654\ 1141044853441927398674100451480252456054554957 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0455279781834196444882158986724421314774234809419835691793690719885634\ 9362680182783945566559168532497 ----------------------- This took, 3.381, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 22 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 22 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)-8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 8 (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 0.04570749711227201789419289608849915234290663792375540448106860035403689302\ 223516102909869524074786934 The implied delta is, 0.15489992372465127330943026355012734828332075253898229\ 3636603789912556910515242093394227431711411326 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0457074971122720178941928960884991523429066379237554044810686003540368\ 9302223516102909869524074786934 ----------------------- This took, 3.030, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 22 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 22 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 20 (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 0.04484352888786998944070963521251547985643569820521711762243084916499771848\ 169704411768773960615650350 The implied delta is, 0.04345228710234823610804653089872902925199841868863775\ 6911169567427881177512877642957319168077290959 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0448435288878699894407096352125154798564356982052171176224308491649977\ 1848169704411768773960615650350 ----------------------- This took, 3.241, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 23 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 23 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 3 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)+3*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 3 (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 0.04339634987455255922668311420094238044412723245470428141932122283952084842\ 125590101595337698462241385 The implied delta is, 0.36487984298091085561406785326930283758229338988423902\ 1123463286360790591656158699718324981558726355 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0433963498745525592266831142009423804441272324547042814193212228395208\ 4842125590101595337698462241385 ----------------------- This took, 3.627, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 24 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 24 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 72 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-72*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 72 (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 0.04354587144837332331550821966983992764453854901695518810494032672363070806\ 427831377765892343632527161 The implied delta is, 0.00951124588810055322066934407254334040943213456616871\ 2183173665783512395879721797945611171417473241 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0435458714483733233155082196698399276445385490169551881049403267236307\ 0806427831377765892343632527161 ----------------------- This took, 2.832, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(10 x + 7) (2 x + 1)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(10 x + 7) (2 x + 1)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 16 (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 0.04205902957765161631307417627712401126393542191416793331834278230792510938\ 055143789469057756183648952 This constant is identified as, 1/8 ln(7) - 1/8 ln(5) The implied delta is, 0.19461879274227127765289610441958127788741542827729278\ 5340052753021540455403016448549792561298027677 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(7) - 1/8 ln(5) ----------------------- This took, 1.336, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 24 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 24 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 12 (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 0.04138087874759555854750255912629560197490166810915253825978838623457859858\ 818800592617251070827507078 The implied delta is, 0.22520406875254449947497850877965205111116966994924880\ 2617516904180470851052549112772911751586691552 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0413808787475955585475025591262956019749016681091525382597883862345785\ 9858818800592617251070827507078 ----------------------- This took, 3.984, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(11 x + 7) (2 x + 1)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(11 x + 7) (2 x + 1)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 9 (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 0.04019367613614801174344021448937116631514633500078321122691707505006664346\ 808787838641118626360923719 This constant is identified as, 1/6 ln(2) + 1/6 ln(7) - 1/6 ln(11) The implied delta is, 0.15123323781287010337951270302675968596573262243711156\ 7255500385633944528650135517398966804449329108 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(2) + 1/6 ln(7) - 1/6 ln(11) ----------------------- This took, 1.826, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 25 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 25 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 19 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)+19*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 19 (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 0.03960190311069918023614469905113510477384104314966901036802489457913516504\ 052605987847964810358230289 The implied delta is, 0.00255659741973053342003488598622043579538160613834932\ 5163917723063133676315427934046151719383043753 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0396019031106991802361446990511351047738410431496690103680248945791351\ 6504052605987847964810358230289 ----------------------- This took, 3.673, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 25 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 25 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 75 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)+75*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 75 (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 0.03850616515342274255273514945262838259971477109004285057812242617005110606\ 974349924663658527852268364 The implied delta is, 0.07914602624433848185397796027500464387879427655087758\ 6557232384787269862054728054144772170511022615 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0385061651534227425527351494526283825997147710900428505781224261700511\ 0606974349924663658527852268364 ----------------------- This took, 4.814, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(12 x + 7) (2 x + 1)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(12 x + 7) (2 x + 1)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 4 (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 0.03853766995681457607321884626561936422852350934971412065100395170167298689\ 094139632704485419443018767 This constant is identified as, 1/4 ln(7) - 1/4 ln(2) - 1/4 ln(3) The implied delta is, 0.53590017179839788262054806460577699854650105562621891\ 3278321531220658297230698171910187851611065684 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(7) - 1/4 ln(2) - 1/4 ln(3) ----------------------- This took, 1.159, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |26 x + 26 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |26 x + 26 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 52 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)+52*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 52 (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 0.03751850165322590252698129214550130546753619616645005977428264638936114568\ 751259475342993192572040525 The implied delta is, 0.20194164999205287483007940581543500666596839230011704\ 1541842281546554549064206272877922319564519915 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0375185016532259025269812921455013054675361961664500597742826463893611\ 4568751259475342993192572040525 ----------------------- This took, 5.093, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(13 x + 7) (2 x + 1)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(13 x + 7) (2 x + 1)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (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 0.03705398607686093923454871166801884645365845959095466308051232438375440869\ 867690199697358107190036288 This constant is identified as, 1/2 ln(2) + 1/2 ln(7) - 1/2 ln(13) The implied delta is, 0.60494236728464687620594805518205321956339755800831773\ 2934297810296642755586658327001457976860588016 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(7) - 1/2 ln(13) ----------------------- This took, 1.600, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |27 x + 27 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |27 x + 27 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) 27 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)+27*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (2 n - 1) a(n - 1) 27 (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 0.03658968942397419855622354300351690301625683974396972545701178612703277772\ 411382812694596832535442554 The implied delta is, 0.40890090672493247147189725800542258937191393590072546\ 1267106096477368730827102622690966959294156401 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0365896894239741985562235430035169030162568397439697254570117861270327\ 7772411382812694596832535442554 ----------------------- This took, 5.066, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |24 x + 28 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |24 x + 28 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 112 (n - 1) b(n - 2) b(n) = --------------------- - -------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-112*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 112 (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 0.03757748592515430555918734090127513843224892176138700695005413752683860333\ 846987100231863162232492581 The implied delta is, 0.10683736851672785521746065039135789124345008689791451\ 3966800507361426222283124340893324262678990464 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0375774859251543055591873409012751384322489217613870069500541375268386\ 0333846987100231863162232492581 ----------------------- This took, 3.094, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |26 x + 28 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |26 x + 28 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 56 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-56*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 56 (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 0.03660303806605900133075137642080336005426892669479481601059981887942434063\ 514341334867673600291314363 The implied delta is, 0.09055938412400127852616928981575939585676901176033805\ 5393153080350367728073750651073382222076708088 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0366030380660590013307513764208033600542689266947948160105998188794243\ 4063514341334867673600291314363 ----------------------- This took, 4.044, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |27 x + 28 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |27 x + 28 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 28 (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 0.03614880563811591080602061503973709952991715464460921841704403334450803805\ 599532754739165331561488091 The implied delta is, 0.17130762618968464513307172464965950195365570520173940\ 0992292071258006948364516725923146333895250951 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0361488056381159108060206150397370995299171546446092184170440333445080\ 3805599532754739165331561488091 ----------------------- This took, 3.624, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 28 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 28 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 28 (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 0.03529800028832599690042268840457293642799046228255136958082098154443489875\ 228313948949196222738056566 The implied delta is, 0.17501347826858049690567236948552213360780143729452252\ 5312256327636097980824511585344542563911298130 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0352980002883259969004226884045729364279904622825513695808209815444348\ 9875228313948949196222738056566 ----------------------- This took, 2.903, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |30 x + 28 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |30 x + 28 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 56 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+56*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 56 (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 0.03489862698628779901503432896696407856305315525689241589229888341007549137\ 445117500163539028108535097 The implied delta is, 0.09853845222615003892283687190987722507020990657857969\ 9491746469521925964992532836471108774208542425 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0348986269862877990150343289669640785630531552568924158922988834100754\ 9137445117500163539028108535097 ----------------------- This took, 3.613, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 29 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 29 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) 29 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-29*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (2 n - 1) a(n - 1) 29 (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 0.03488752014155825041019112472437711598711295278301410924374609210353679125\ 225831910796780961580415917 The implied delta is, 0.41313226022802055804076661499817859123872832214118372\ 9627534454498192359066097352507613092811093200 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0348875201415582504101911247243771159871129527830141092437460921035367\ 9125225831910796780961580415917 ----------------------- This took, 2.850, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(2 x + 1) (15 x + 7)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(2 x + 1) (15 x + 7)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (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 0.03449643574347572573670985262367852323025352407457536553363603284034989810\ 225137405511390366131074253 This constant is identified as, 1/2 ln(3) + 1/2 ln(5) - 1/2 ln(2) - 1/2 ln(7) The implied delta is, 0.61160364913601387555756460059385440034977790821303047\ 2575466750919779724547038506469358114765993755 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(3) + 1/2 ln(5) - 1/2 ln(2) - 1/2 ln(7) ----------------------- This took, 1.804, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |30 x + 30 x + 7| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |30 x + 30 x + 7| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 60 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-60*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 60 (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 0.03410519202747599285289283866324677857230240163234248312183066455215426010\ 342716356232240718469065848 The implied delta is, 0.21449668881929892212985124471994174435050344873837277\ 3112054043963511093357556763599800523359493280 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0341051920274759928528928386632467785723024016323424831218306645521542\ 6010342716356232240718469065848 ----------------------- This took, 2.820, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |9 x + 16 x + 8| |---------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |9 x + 16 x + 8| |---------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 16 (2 n - 1) b(n - 1) 32 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 16*(2*n-1)/n*b(n-1)+32*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 16 (2 n - 1) a(n - 1) 32 (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 0.06007524579312209107010373625154657580400812044730402858397419860320974466\ 912396519904454268421530384 The implied delta is, 0.02669478119360248299763676646264780151576010829662806\ 9703765788628838578923186785107340098266245593 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0600752457931220910701037362515465758040081204473040285839741986032097\ 4466912396519904454268421530384 ----------------------- This took, 3.088, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (9 x + 8)\n |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (9 x + 8)\n |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 17 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 17*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 17 (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 0.05889151782819172726939705473526085253424035628236657055367431939740386026\ 406689346482076431910405747 This constant is identified as, ln(3) - 3/2 ln(2) The implied delta is, 0.56699175013732249851317171476635314501988343365560925\ 0888942795636346050682355683916571549465815244 Since this is positive, this suggests an Apery-style irrationality proof of, ln(3) - 3/2 ln(2) ----------------------- This took, 0.865, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 4) (3 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 4) (3 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)-36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 36 (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 0.05776226504666210911810267678818138067295834453002127117672333412444946849\ 747455963382194391636822396 This constant is identified as, 1/12 ln(2) The implied delta is, 0.28190597596439956554136714580428907419922638411860722\ 1830270121115996219375698008870475492074548350 Since this is positive, this suggests an Apery-style irrationality proof of, 1/12 ln(2) ----------------------- This took, 0.484, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (11 x + 8)\n |------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (11 x + 8)\n |------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (2 n - 1) a(n - 1) 9 (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 0.05307562185308910263504120226509993259920107514277523547608794677506545966\ 735876104291078783556312146 This constant is identified as, 1/6 ln(11) - 1/2 ln(2) The implied delta is, 0.09962659107436244254080249667527706313010630082718949\ 6730614138309097556796771747256705966726982679 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(11) - 1/2 ln(2) ----------------------- This took, 1.507, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 20 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 20 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 16 (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 0.04934888996247018959251244129869757336189627594696302537942223506025849249\ 456094643317445700932201103 This constant is identified as, 4 3 2 arctan(RootOf(_Z + 20 _Z - 6 _Z - 20 _Z + 1, index = 1)) The implied delta is, 0.27273820446094881879969805417687087466055799722034728\ 8780038503273229697916901398126013771964744010 Since this is positive, this suggests an Apery-style irrationality proof of, 4 3 2 arctan(RootOf(_Z + 20 _Z - 6 _Z - 20 _Z + 1, index = 1)) ----------------------- This took, 2.307, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 20 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 20 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 80 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+80*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 80 (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 0.04701716802838171759084300138702391253964648655647366708772692930272338599\ 190539651626347506752484313 The implied delta is, 0.07718308342065558358695422994012552592438240455617915\ 8195030857272987552219894370995019698213439525 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0470171680283817175908430013870239125396464865564736670877269293027233\ 8599190539651626347506752484313 ----------------------- This took, 2.784, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 21 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 21 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 7 (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 0.04736946766082724053372186490537593726708557177871504860517232906646752194\ 541385366757701284003752490 The implied delta is, 0.47645322687075672502926149857061858978879352350445771\ 9083837341078271695508371937340849362063010437 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0473694676608272405337218649053759372670855717787150486051723290664675\ 2194541385366757701284003752490 ----------------------- This took, 2.207, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 4) (3 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 4) (3 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 4 (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 0.04558038919848865655292950628862865829934733447862174598568161291417723187\ 016147946233630092924279138 This constant is identified as, 1/4 ln(2) + 1/4 ln(3) - 1/4 ln(5) The implied delta is, 0.51729668886995206530084709053887995949953745192577735\ 9815781195706547758164390443811084687132348320 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(2) + 1/4 ln(3) - 1/4 ln(5) ----------------------- This took, 0.720, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 23 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 23 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 17 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)-17*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 17 (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 0.04395319321657698457024573472416317847240457702111933944446163784792889863\ 484993059030103215113819523 The implied delta is, 0.00102142337058840818989308138522412657666141446895326\ 7463999015358922437667417104003203583076544629 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0439531932165769845702457347241631784724045770211193394444616378479288\ 9863484993059030103215113819523 ----------------------- This took, 5.087, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 23 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 23 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 15 (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 0.04307416621863015356048332928876320597702267911094787079736633667472939004\ 633603888269039391634340113 The implied delta is, 0.03333019401807212862761464282519140457580705698630357\ 8955818808934579073808523535969361183640288774 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0430741662186301535604833292887632059770226791109478707973663366747293\ 9004633603888269039391634340113 ----------------------- This took, 4.932, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 24 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 24 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 32 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-32*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 32 (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 0.04246505861642397602616588146051031529535450456355507991687618990842761558\ 529989015193903753411397275 The implied delta is, 0.12108701682398821099545433787177560167510318168170254\ 5524543245613559628731023579385458069785675281 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0424650586164239760261658814605103152953545045635550799168761899084276\ 1558529989015193903753411397275 ----------------------- This took, 4.176, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 24 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 24 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 32 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+32*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 32 (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 0.04091980343862493282056837632296565224388220994983700424790423997732728592\ 449613351719861984267639089 The implied delta is, 0.12720116832348865277300314052760239681016985677412708\ 4819775337961986654292661314680237750520905937 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0409198034386249328205683763229656522438822099498370042479042399773272\ 8592449613351719861984267639089 ----------------------- This took, 4.158, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 25 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 25 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 17 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)-17*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 17 (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 0.04036870286708172875412792168523650789616092042704124023389654526825688229\ 302800447645123787536298533 The implied delta is, 0.01974142010460521459088804527841624516176739607683819\ 4782318950876063510531865017851693064887811131 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0403687028670817287541279216852365078961609204270412402338965452682568\ 8229302800447645123787536298533 ----------------------- This took, 3.180, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |20 x + 25 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |20 x + 25 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 15 (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 0.03968453045189532147432668567759031502719236015744145360172697912997143110\ 035955414660503628935818368 The implied delta is, 0.25649525891783396468389747770110005966362539540892215\ 3712282901928783672291550522568259165330188929 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0396845304518953214743266856775903150271923601574414536017269791299714\ 3110035955414660503628935818368 ----------------------- This took, 5.300, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 2) (7 x + 4)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 2) (7 x + 4)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 4 (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 0.03853766995681457607321884626561936422852350934971412065100395170167298689\ 094139632704485419443018767 This constant is identified as, 1/4 ln(7) - 1/4 ln(2) - 1/4 ln(3) The implied delta is, 0.53590017179839788262054806460577699854650105562621891\ 3278321531220658297230698171910187851611065684 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(7) - 1/4 ln(2) - 1/4 ln(3) ----------------------- This took, 1.089, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 27 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 27 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) 7 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)+7*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (2 n - 1) a(n - 1) 7 (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 0.03691916975158553339823189310895593295058813441158463607910671203647727898\ 467446260672648262800987024 The implied delta is, 0.21563368964114787979177235415928445834591493558104718\ 3263668092291311335218349849809664663153927647 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0369191697515855333982318931089559329505881344115846360791067120364772\ 7898467446260672648262800987024 ----------------------- This took, 4.912, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 28 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 28 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 112 (n - 1) b(n - 2) b(n) = --------------------- - -------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-112*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 112 (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 0.03757748592515430555918734090127513843224892176138700695005413752683860333\ 846987100231863162232492581 The implied delta is, 0.10683736851672785521746065039135789124345008689791451\ 3966800507361426222283124340893324262678990464 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0375774859251543055591873409012751384322489217613870069500541375268386\ 0333846987100231863162232492581 ----------------------- This took, 2.236, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 28 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 28 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 48 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-48*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 48 (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 0.03647115284824874826152667212766024778803950727624239847283746426907082586\ 727542330999740718983503028 The implied delta is, 0.05366251543367305591475544991216817799755023721944335\ 3102004041723588881595497264120932669176108381 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0364711528482487482615266721276602477880395072762423984728374642690708\ 2586727542330999740718983503028 ----------------------- This took, 4.171, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 28 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 28 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 16 (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 0.03547426365104098070321290427563827075194543968211601809453250734086040668\ 998279023604796540856162953 This constant is identified as, 4 3 2 arctan(RootOf(_Z + 28 _Z - 6 _Z - 28 _Z + 1, index = 1)) The implied delta is, 0.33568803906926503620920701968673519934804388993189854\ 5965357732538520999766207302618934021726951350 Since this is positive, this suggests an Apery-style irrationality proof of, 4 3 2 arctan(RootOf(_Z + 28 _Z - 6 _Z - 28 _Z + 1, index = 1)) ----------------------- This took, 2.398, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(13 x + 8) (2 x + 1)\n |--------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(13 x + 8) (2 x + 1)\n |--------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (2 n - 1) a(n - 1) 9 (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 0.03460656079637408360257350737789794458278876544680231667727908788501830342\ 572811383731310742419754357 This constant is identified as, 2/3 ln(2) - 1/6 ln(13) The implied delta is, 0.17522030265818860356785395470584241849907664784203849\ 1426654627989669007376983660394887316878375752 Since this is positive, this suggests an Apery-style irrationality proof of, 2/3 ln(2) - 1/6 ln(13) ----------------------- This took, 2.712, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 29 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 29 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) 87 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)+87*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (2 n - 1) a(n - 1) 87 (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 0.03336245289339451391744987953301863016442040495091236499040867855717205043\ 756342500175152321317282453 The implied delta is, 0.09930376776428443853895795274102942373214536302065366\ 1553302634644078499408820535395518340494391058 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0333624528933945139174498795330186301644204049509123649904086785571720\ 5043756342500175152321317282453 ----------------------- This took, 2.924, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(5 x + 4) (5 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(5 x + 4) (5 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 100 (n - 1) b(n - 2) b(n) = --------------------- - -------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-100*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 100 (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 0.03465735902799726547086160607290882840377500671801276270603400047466968109\ 848473578029316634982093438 This constant is identified as, 1/20 ln(2) The implied delta is, 0.28185371203997313819239025540618325739906485830611690\ 9480968969417082104786606660790191154315941595 Since this is positive, this suggests an Apery-style irrationality proof of, 1/20 ln(2) ----------------------- This took, 0.485, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(3 x + 2) (9 x + 4)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(3 x + 2) (9 x + 4)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 36 (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 0.03378875900901369849816775962202909471433253528854118313450119367867505593\ 740952093897936898477611677 This constant is identified as, 1/12 ln(3) - 1/12 ln(2) The implied delta is, 0.40308528902474551883329689697930003019415445884711973\ 6033812013825034198238587911490972588548740514 Since this is positive, this suggests an Apery-style irrationality proof of, 1/12 ln(3) - 1/12 ln(2) ----------------------- This took, 1.186, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |29 x + 30 x + 8| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |29 x + 30 x + 8| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 28 (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 0.03299396698680346983190115960409026078814475913042398418823262643841034168\ 791960226969153616014742180 The implied delta is, 0.11267590341120074869251145962194528229916937856697567\ 4199746484490747941597624114228887969264799195 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0329939669868034698319011596040902607881447591304239841882326264384103\ 4168791960226969153616014742180 ----------------------- This took, 2.964, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |10 x + 18 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |10 x + 18 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 18 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 18*(2*n-1)/n*b(n-1)+36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 18 (2 n - 1) a(n - 1) 36 (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 0.05362509239944036556690076905977688650345921592627603190546717655945937290\ 096842405940140372510689574 This constant is identified as, arctan( 6 5 4 3 2 RootOf(_Z + 18 _Z - 15 _Z - 60 _Z + 15 _Z + 18 _Z - 1, index = 1)) The implied delta is, 0.16061722484890563230786124818828382971672445401129761\ 7274250617734415105448505431680291314028864036 Since this is positive, this suggests an Apery-style irrationality proof of, arctan( 6 5 4 3 2 RootOf(_Z + 18 _Z - 15 _Z - 60 _Z + 15 _Z + 18 _Z - 1, index = 1)) ----------------------- This took, 2.923, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (10 x + 9)\n |------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (10 x + 9)\n |------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 19 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 19*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 19 (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 0.05268025782891315061375049041965639915306018649163703628196961684629201162\ 006727324438284773106706038 This constant is identified as, 1/2 ln(2) + 1/2 ln(5) - ln(3) The implied delta is, 0.57337177129358798630972583682740462220014717381052020\ 3127971283730002664556591027268962044943194084 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(5) - ln(3) ----------------------- This took, 0.901, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |10 x + 20 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |10 x + 20 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 40 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-40*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 40 (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 0.05177441474570309562625581681897158600064073773170552450255681654745943689\ 051284618097664366479131960 The implied delta is, 0.04566308295447717905112811140638400015597404182898616\ 4853836664038983172727996437356568042526582583 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0517744147457030956262558168189715860006407377317055245025568165474594\ 3689051284618097664366479131960 ----------------------- This took, 2.249, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (11 x + 9)\n |------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (11 x + 9)\n |------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 4 (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 0.05016767386553779031786327603001947263168143457297956793729476046389625936\ 900469483195579959379265346 This constant is identified as, 1/4 ln(11) - 1/2 ln(3) The implied delta is, 0.28622392004942569421410481226959564773405359383798251\ 4624287232397243935591407361876586379704386555 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(11) - 1/2 ln(3) ----------------------- This took, 1.601, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 21 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 21 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 45 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)-45*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 45 (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 0.04934576197971132219276852471052086911230526961228421083029638540106521747\ 089944954672284299018970975 The implied delta is, 0.00956194955787056689620559486119690023940072961524927\ 1144486116288211919964089412149735034849515003 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0493457619797113221927685247105208691123052696122842108302963854010652\ 1747089944954672284299018970975 ----------------------- This took, 2.853, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 21 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 21 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 27 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+27*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 27 (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 0.04668143933490230516020453466737448799569465023608797352320348835984491608\ 935168757019913138915232614 The implied delta is, 0.15898823937860467861445754692728583404385283305841140\ 8538792381840429827526644985555978547393357429 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0466814393349023051602045346673744879956946502360879735232034883598449\ 1608935168757019913138915232614 ----------------------- This took, 3.627, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 21 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 21 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) 63 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)+63*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (2 n - 1) a(n - 1) 63 (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 0.04552797818341964448821589867244213147742348094198356917936907198856349362\ 680182783945566559168532497 The implied delta is, 0.26777964817509272096295917061879860566963611085557654\ 1141044853441927398674100451480252456054554957 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0455279781834196444882158986724421314774234809419835691793690719885634\ 9362680182783945566559168532497 ----------------------- This took, 2.114, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (13 x + 9)\n |------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (13 x + 9)\n |------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 16 (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 0.04596559751566466915787462096503339943878585363933852661870710542355951011\ 089903070654289422662816390 This constant is identified as, 1/8 ln(13) - 1/4 ln(3) The implied delta is, 0.00816837802705123978114562821130588587828143647283726\ 6412903593053576990351402733733717497630060606 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(13) - 1/4 ln(3) ----------------------- This took, 1.098, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 22 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 22 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 20 (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 0.04484352888786998944070963521251547985643569820521711762243084916499771848\ 169704411768773960615650350 The implied delta is, 0.04345228710234823610804653089872902925199841868863775\ 6911169567427881177512877642957319168077290959 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0448435288878699894407096352125154798564356982052171176224308491649977\ 1848169704411768773960615650350 ----------------------- This took, 3.602, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 23 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 23 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 11 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)+11*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 11 (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 0.04318060420506946996695508989112974642538934363218783432261890072916620644\ 069688938742683754227846868 The implied delta is, 0.09424908812473310387972776499766723099521381573971991\ 5157830531015081719707215612869907653723893316 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0431806042050694699669550898911297464253893436321878343226189007291662\ 0644069688938742683754227846868 ----------------------- This took, 3.389, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 24 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 24 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 72 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-72*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 72 (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 0.04354587144837332331550821966983992764453854901695518810494032672363070806\ 427831377765892343632527161 The implied delta is, 0.00951124588810055322066934407254334040943213456616871\ 2183173665783512395879721797945611171417473241 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0435458714483733233155082196698399276445385490169551881049403267236307\ 0806427831377765892343632527161 ----------------------- This took, 4.939, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 24 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 24 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 36 (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 0.04082977718781069236201374686854596848569068306353067785456265244455918659\ 794034942908902627811699856 This constant is identified as, arctan( 6 5 4 3 2 RootOf(_Z + 24 _Z - 15 _Z - 80 _Z + 15 _Z + 24 _Z - 1, index = 1)) The implied delta is, 0.11466028147835372514261194743378671436646056975831559\ 8851683790120834998642999686323543930617052221 Since this is positive, this suggests an Apery-style irrationality proof of, arctan( 6 5 4 3 2 RootOf(_Z + 24 _Z - 15 _Z - 80 _Z + 15 _Z + 24 _Z - 1, index = 1)) ----------------------- This took, 3.666, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 25 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 25 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 13 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)-13*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 13 (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 0.04028084672199820131161551403333008064340834186599131608273477844423955082\ 170787271275723379569285938 The implied delta is, 0.07659933283810454726267459321566900789168737574000482\ 5849104622699240276925484970252580628739061844 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0402808467219982013116155140333300806434083418659913160827347784442395\ 5082170787271275723379569285938 ----------------------- This took, 5.156, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 26 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 26 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)-28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 28 (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 0.03900616621219270031235406676281317733458068887816479548306424170916916862\ 094441445724560992903497071 The implied delta is, -0.0021907515132339270980877411521375595648536769275683\ 707725341174918791698726231998390580736950490735 Since this is negative, there is no Apery-style irrationality proof of, 0.039\ 006166212192700312354066762813177334580688878164795483064241709169168620\ 94441445724560992903497071, but still a very fast way to compute it to many digits ----------------------- This took, 2.908, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 26 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 26 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 8 (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 0.03831088470899431649907071985716184712025182099493404867213991725176491748\ 925566336369184568307782589 The implied delta is, 0.18181804224841047541881619713079362512314394115593202\ 3500729277921544402323768366934568703363741070 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0383108847089943164990707198571618471202518209949340486721399172517649\ 1748925566336369184568307782589 ----------------------- This took, 5.038, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 27 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 27 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) 45 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)-45*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (2 n - 1) a(n - 1) 45 (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 0.03782864856335163860425608908910991407865183076911765505981395900960544024\ 684684763116706720747216476 The implied delta is, 0.07399954440446325845593857732847145218556826767628048\ 3179963726192069817217018564011090209682542479 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0378286485633516386042560890891099140786518307691176550598139590096054\ 4024684684763116706720747216476 ----------------------- This took, 3.849, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(4 x + 3) (5 x + 3)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(4 x + 3) (5 x + 3)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (2 n - 1) a(n - 1) 9 (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 0.03719059188570162596104918171830575056243351425800120227854797874789862396\ 137805556973453735005703929 This constant is identified as, 1/6 ln(5) - 1/3 ln(2) The implied delta is, 0.49194606025670296841864250800783184410820041320874671\ 3696518854044035400099816889993501451563849698 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(5) - 1/3 ln(2) ----------------------- This took, 0.693, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 28 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 28 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 28 (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 0.03614880563811591080602061503973709952991715464460921841704403334450803805\ 599532754739165331561488091 The implied delta is, 0.17130762618968464513307172464965950195365570520173940\ 0992292071258006948364516725923146333895250951 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0361488056381159108060206150397370995299171546446092184170440333445080\ 3805599532754739165331561488091 ----------------------- This took, 2.344, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |22 x + 28 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |22 x + 28 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 8 (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 0.03559354691155796956650329350893868963748443204689061046975407457752793177\ 778559120823652682756902032 The implied delta is, 0.28478407519319374641314313172249249454235461548147188\ 0586062101872986657283244476518940240774694673 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0355935469115579695665032935089386896374844320468906104697540745775279\ 3177778559120823652682756902032 ----------------------- This took, 3.834, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 29 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 29 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) 13 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-13*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (2 n - 1) a(n - 1) 13 (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 0.03466210064939951693316961189380272729561624745508721254447046664557311509\ 358120751054274707388679220 The implied delta is, 0.10306793124185817458961779516772481117800645263960086\ 5986871310627955244358369807431225552837115781 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0346621006493995169331696118938027272956162474550872125444704666455731\ 1509358120751054274707388679220 ----------------------- This took, 3.080, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |20 x + 30 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |20 x + 30 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 180 (n - 1) b(n - 2) b(n) = --------------------- - -------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-180*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 180 (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 0.03586741174700033657411943607921711878538089021169080456308384199938010128\ 842766344679085539143499628 The implied delta is, 0.19238621785744988050220396899838232200111210149782834\ 5311204149545642978901878280332709502606786448 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0358674117470003365741194360792171187853808902116908045630838419993801\ 0128842766344679085539143499628 ----------------------- This took, 3.382, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |26 x + 30 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |26 x + 30 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 36 (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 0.03289925997498012639500829419913171557459751729797535025294815670683899499\ 637396428878297133954800735 This constant is identified as, arctan( 6 5 4 3 2 RootOf(_Z + 30 _Z - 15 _Z - 100 _Z + 15 _Z + 30 _Z - 1, index = 1)) The implied delta is, 0.27270782245946457223806389318872976716053439548587568\ 6699451470121308965437156670785535597890142178 Since this is positive, this suggests an Apery-style irrationality proof of, arctan( 6 5 4 3 2 RootOf(_Z + 30 _Z - 15 _Z - 100 _Z + 15 _Z + 30 _Z - 1, index = 1)) ----------------------- This took, 2.368, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |28 x + 30 x + 9| |----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |28 x + 30 x + 9| |----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 108 (n - 1) b(n - 2) b(n) = --------------------- + -------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+108*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 108 (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 0.03208847096118561879394595787719031883309564257503570881510202180837592172\ 478624937219715439876890303 The implied delta is, 0.07949868979512726044250781940130086600501239294960041\ 4219459547376722514919070474707205412604297561 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0320884709611856187939459578771903188330956425750357088151020218083759\ 2172478624937219715439876890303 ----------------------- This took, 3.577, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 20 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 20 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 20 (2 n - 1) b(n - 1) 40 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 20*(2*n-1)/n*b(n-1)+40*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 20 (2 n - 1) a(n - 1) 40 (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 0.04842670411701946246901902500592326026755587116454064533465773417374428298\ 822154940811526578349125121 The implied delta is, 0.05773499370554175330731823631097035020848250640477256\ 1852740643560102740298910950674333307825105535 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0484267041170194624690190250059232602675558711645406453346577341737442\ 8298822154940811526578349125121 ----------------------- This took, 2.277, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (11 x + 10)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (11 x + 10)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 21 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 21*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 21 (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 0.04765508990216243002197606164038254611030268265432209959261990408150050711\ 794211641952875145651824654 This constant is identified as, 1/2 ln(11) - 1/2 ln(2) - 1/2 ln(5) The implied delta is, 0.58448163333107330960348282540393061902711881756838552\ 1535141988810736997507168010457014961147704158 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(11) - 1/2 ln(2) - 1/2 ln(5) ----------------------- This took, 1.727, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |11 x + 22 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |11 x + 22 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 44 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)-44*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 44 (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 0.04691233882745658088469094517599776535573223426848000037555554869189588873\ 318733018462329129940292162 The implied delta is, 0.17365920458969177782584436581682405206744689026287059\ 9984227227928722096106320149129986935460878405 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0469123388274565808846909451759977653557322342684800003755555486918958\ 8873318733018462329129940292162 ----------------------- This took, 2.172, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (13 x + 10)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (13 x + 10)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (2 n - 1) a(n - 1) 9 (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 0.04372737741124850867258266448015906620069440935523902339761960161598200960\ 784294986059624305848186507 This constant is identified as, 1/6 ln(13) - 1/6 ln(2) - 1/6 ln(5) The implied delta is, 0.13483860817151004634809823925040754975159289838216717\ 2217996162065164183767152605569956017751349947 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(13) - 1/6 ln(2) - 1/6 ln(5) ----------------------- This took, 1.077, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 24 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 24 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 24 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+24*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 24 (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 0.04110201346700524332646682755080302988958301481227112460848308056441555358\ 699858757241274207689709603 The implied delta is, 0.15872249030202671343471562268121857169174306272504912\ 2716760813338163056076735287433594279115223953 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0411020134670052433264668275508030298895830148122711246084830805644155\ 5358699858757241274207689709603 ----------------------- This took, 3.329, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 25 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 25 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 15 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)+15*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 15 (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 0.03968453045189532147432668567759031502719236015744145360172697912997143110\ 035955414660503628935818368 The implied delta is, 0.25649525891783396468389747770110005966362539540892215\ 3712282901928783672291550522568259165330188929 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0396845304518953214743266856775903150271923601574414536017269791299714\ 3110035955414660503628935818368 ----------------------- This took, 2.388, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 26 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 26 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)+4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 4 (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 0.03838594563488901961466106657369275405330559858823606215271357234470055890\ 908422389093713234096969154 1/2 1/2 (14450 - 1105 170 ) This constant is identified as, arcsin(------------------------) 170 The implied delta is, 0.42299041940762591479476249471398482856165028885904779\ 8664755127441808854624178396980003016345605063 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 1/2 (14450 - 1105 170 ) arcsin(------------------------) 170 ----------------------- This took, 3.843, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(6 x + 5) (3 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(6 x + 5) (3 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (2 n - 1) a(n - 1) 9 (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 0.03719059188570162596104918171830575056243351425800120227854797874789862396\ 137805556973453735005703929 This constant is identified as, 1/6 ln(5) - 1/3 ln(2) The implied delta is, 0.49194606025670296841864250800783184410820041320874671\ 3696518854044035400099816889993501451563849698 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(5) - 1/3 ln(2) ----------------------- This took, 0.830, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 28 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 28 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 24 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-24*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 24 (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 0.03608556077752524585121785586387858459931301750783264106339228311467710361\ 832741731220394269060379775 The implied delta is, 0.04733140523208429384806242188618913865185827390046612\ 0467225526249069510586502787861949355866459324 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0360855607775252458512178558638785845993130175078326410633922831146771\ 0361832741731220394269060379775 ----------------------- This took, 5.177, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 28 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 28 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 56 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+56*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 56 (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 0.03489862698628779901503432896696407856305315525689241589229888341007549137\ 445117500163539028108535097 The implied delta is, 0.09853845222615003892283687190987722507020990657857969\ 9491746469521925964992532836471108774208542425 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0348986269862877990150343289669640785630531552568924158922988834100754\ 9137445117500163539028108535097 ----------------------- This took, 2.952, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(7 x + 5) (3 x + 2)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(7 x + 5) (3 x + 2)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (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 0.03449643574347572573670985262367852323025352407457536553363603284034989810\ 225137405511390366131074253 This constant is identified as, 1/2 ln(3) + 1/2 ln(5) - 1/2 ln(2) - 1/2 ln(7) The implied delta is, 0.61160364913601387555756460059385440034977790821303047\ 2575466750919779724547038506469358114765993755 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(3) + 1/2 ln(5) - 1/2 ln(2) - 1/2 ln(7) ----------------------- This took, 1.808, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 30 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 30 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 60 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-60*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 60 (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 0.03410519202747599285289283866324677857230240163234248312183066455215426010\ 342716356232240718469065848 The implied delta is, 0.21449668881929892212985124471994174435050344873837277\ 3112054043963511093357556763599800523359493280 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0341051920274759928528928386632467785723024016323424831218306645521542\ 6010342716356232240718469065848 ----------------------- This took, 4.481, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |23 x + 30 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |23 x + 30 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 20 (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 0.03308966056439973332320641668126660804476898793168568746943276811334298734\ 316153430255377745843493225 The implied delta is, 0.32996963410349629380836122555785715348988491958500822\ 1902871862113051783210509049185316032672913616 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0330896605643997333232064166812666080447689879316856874694327681133429\ 8734316153430255377745843493225 ----------------------- This took, 4.525, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |27 x + 30 x + 10| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |27 x + 30 x + 10| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 180 (n - 1) b(n - 2) b(n) = --------------------- + -------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+180*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 180 (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 0.03134477868558781172722866759134927502643099103764911139181795286848225732\ 793693101084231671168322876 The implied delta is, 0.07717227807642883477565094274521825147668078016164325\ 4196336740077504899488680117899951085238282781 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0313447786855878117272286675913492750264309910376491113918179528684822\ 5732793693101084231671168322876 ----------------------- This took, 4.130, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |12 x + 22 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |12 x + 22 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 22 (2 n - 1) b(n - 1) 44 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 22*(2*n-1)/n*b(n-1)+44*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 22 (2 n - 1) a(n - 1) 44 (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 0.04414770892688094839122755472000589003314247279112032720951919800110029585\ 545739363254014668320056548 The implied delta is, 0.31904683270275864997035803794180299564090441379877852\ 9952642273025375214391323269795803271821149175 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0441477089268809483912275547200058900331424727911203272095191980011002\ 9585545739363254014668320056548 ----------------------- This took, 2.897, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (12 x + 11)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (12 x + 11)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 23 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 23*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 23 (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 0.04350568849481488308388295093687477048839198630292139237874332174685395662\ 238084250514385040196733623 This constant is identified as, ln(2) + 1/2 ln(3) - 1/2 ln(11) The implied delta is, 0.59061022211924122523032694961230388993126871412053070\ 1772420840244819687524482681229208270558273191 Since this is positive, this suggests an Apery-style irrationality proof of, ln(2) + 1/2 ln(3) - 1/2 ln(11) ----------------------- This took, 1.840, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |12 x + 24 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |12 x + 24 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 48 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-48*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 48 (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 0.04288562875437406118966634228769673892137278485454824723322714044890793084\ 562942232575487758897964246 The implied delta is, 0.07048937710408519875204268691709063762316391085982990\ 2271699474377253841902578520155197927263391537 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0428856287543740611896663422876967389213727848545482472332271404489079\ 3084562942232575487758897964246 ----------------------- This took, 2.894, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (13 x + 11)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (13 x + 11)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 4 (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 0.04176352116579154799788596590004732624589027270569748530011945038322276085\ 279336658112998885946367434 This constant is identified as, 1/4 ln(13) - 1/4 ln(11) The implied delta is, 0.31080800740956958081817834162879740399338741068828733\ 2542153849421700061408577747519159378919201074 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(13) - 1/4 ln(11) ----------------------- This took, 1.103, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (14 x + 11)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (14 x + 11)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 9 (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 0.04019367613614801174344021448937116631514633500078321122691707505006664346\ 808787838641118626360923719 This constant is identified as, 1/6 ln(2) + 1/6 ln(7) - 1/6 ln(11) The implied delta is, 0.15123323781287010337951270302675968596573262243711156\ 7255500385633944528650135517398966804449329108 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 ln(2) + 1/6 ln(7) - 1/6 ln(11) ----------------------- This took, 1.850, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 25 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 25 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) 35 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)+35*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (2 n - 1) a(n - 1) 35 (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 0.03927745961004989326089986825622190595231731105387262980829393139219234312\ 960108634534598708617937180 The implied delta is, 0.04915424834806473128556544143064023199238562129935150\ 4650545688042729156503697687462713613386380244 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0392774596100498932608998682562219059523173110538726298082939313921923\ 4312960108634534598708617937180 ----------------------- This took, 2.908, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (15 x + 11)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (15 x + 11)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 16 (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 0.03876936603797994024175762402294800554392313226923124980359681449763745712\ 662875230358686561303461352 This constant is identified as, 1/8 ln(3) + 1/8 ln(5) - 1/8 ln(11) The implied delta is, 0.03983017924688135875912247778507833943026664990467426\ 5767787554012210646603744396427949259161198093 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(3) + 1/8 ln(5) - 1/8 ln(11) ----------------------- This took, 1.185, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 26 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 26 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 28 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)+28*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 28 (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 0.03794333014712563097333896478995175977317776031455459818147007170416288639\ 786763645409614661496772456 The implied delta is, 0.08340378383065119697303860113949391987709550310928589\ 8320999289313543357103642581980700404368667872 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0379433301471256309733389647899517597731777603145545981814700717041628\ 8639786763645409614661496772456 ----------------------- This took, 3.958, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 27 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 27 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) 19 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)+19*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (2 n - 1) a(n - 1) 19 (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 0.03672021029665081672301215468855480489632763517956081441463021613029875019\ 837407880873091686149621604 The implied delta is, 0.01531187258072094630964278113423925791988565342169706\ 5194986815635226631820672448142358747767664333 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0367202102966508167230121546885548048963276351795608144146302161302987\ 5019837407880873091686149621604 ----------------------- This took, 3.960, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 28 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 28 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 80 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-80*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 80 (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 0.03700932148478349164457639353289065183422895220921315812272228905079891310\ 317458716004213224264228231 The implied delta is, 0.00956118891211741119858615233544006783765490209990440\ 3725939659517625294668216129688522012053434211 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0370093214847834916445763935328906518342289522092131581227222890507989\ 1310317458716004213224264228231 ----------------------- This took, 2.820, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 28 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 28 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 8 (n - 1) b(n - 2) b(n) = --------------------- + ------------------ n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+8*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 8 (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 0.03559354691155796956650329350893868963748443204689061046975407457752793177\ 778559120823652682756902032 The implied delta is, 0.28478407519319374641314313172249249454235461548147188\ 0586062101872986657283244476518940240774694673 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0355935469115579695665032935089386896374844320468906104697540745775279\ 3177778559120823652682756902032 ----------------------- This took, 2.357, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 29 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 29 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) 5 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-5*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (2 n - 1) a(n - 1) 5 (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 0.03455134024907905076553730686423286510096031652949966731144727917437933167\ 215780574666732734815263498 The implied delta is, 0.29214221454036240139497788115121425792485519844299716\ 3405241366757420324933448953233043122140370426 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0345513402490790507655373068642328651009603165294996673114472791743793\ 3167215780574666732734815263498 ----------------------- This took, 3.849, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |20 x + 30 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |20 x + 30 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 20 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-20*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 20 (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 0.03358359227143402643320552117187005268768476621664609744380694789654247765\ 893381602028830168902042423 The implied delta is, 0.44981763382868784816509170128757430661149287710737608\ 3983313222691768516320522462905131939354815609 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0335835922714340264332055211718700526876847662166460974438069478965424\ 7765893381602028830168902042423 ----------------------- This took, 3.721, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |21 x + 30 x + 11| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |21 x + 30 x + 11| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 24 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+24*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 24 (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 0.03304168931087764128548778239898299520056922995433342143962716676342898443\ 844261790116835371694318096 The implied delta is, 0.11258670609930560160343086173120280955318523045954220\ 0319139913660325060881651730283256602411728251 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0330416893108776412854877823989829952005692299543334214396271667634289\ 8443844261790116835371694318096 ----------------------- This took, 2.967, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 24 x + 12| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 24 x + 12| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 24 (2 n - 1) b(n - 1) 48 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 24*(2*n-1)/n*b(n-1)+48*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 24 (2 n - 1) a(n - 1) 48 (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 0.04056389400858234823793319306789228951506510599346487717300925603795477881\ 325097947585585930030492910 The implied delta is, 0.08062003780628902964119366194352728449204052365986734\ 5012548928776943906543232305913013925127070320 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0405638940085823482379331930678922895150651059934648771730092560379547\ 7881325097947585585930030492910 ----------------------- This took, 3.817, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (13 x + 12)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (13 x + 12)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 25 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 25*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 25 (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 0.04002135383676821291188898086321988200338855910847357822149557901959156508\ 320589065711612731696001245 This constant is identified as, 1/2 ln(13) - ln(2) - 1/2 ln(3) The implied delta is, 0.60013843912498329579659016153611624482487887296542625\ 4327249019701506377029369398348550371353894390 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(13) - ln(2) - 1/2 ln(3) ----------------------- This took, 1.524, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |13 x + 26 x + 12| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |13 x + 26 x + 12| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 52 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)-52*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 52 (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 0.03949590898297444826361771077780509713006592963243033333146688359082842118\ 562386693121531132475216387 The implied delta is, 0.33011636597427641707125778927539804180275764233083940\ 1816596861959797166976907290124841208852473049 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0394959089829744482636177107778050971300659296324303333314668835908284\ 2118562386693121531132475216387 ----------------------- This took, 4.123, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |17 x + 28 x + 12| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |17 x + 28 x + 12| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 32 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+32*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 32 (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 0.03523994059392463141853456799633693015344126837961752952854109179432045033\ 404048118530178008193806455 The implied delta is, 0.06967615826851729523922404003119868773484750926220299\ 6086933602569373661066409207743046583483351374 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0352399405939246314185345679963369301534412683796175295285410917943204\ 5033404048118530178008193806455 ----------------------- This took, 4.288, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |19 x + 30 x + 12| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |19 x + 30 x + 12| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 12 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+12*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 12 (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 0.03318635919868501822099375740480416430232164338860244747585706777078874237\ 554279168244226399572043139 The implied delta is, 0.48802669867671894536801578996129298051155581766744855\ 2992109897042376091495885966438741426051876815 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0331863591986850182209937574048041643023216433886024474758570677707887\ 4237554279168244226399572043139 ----------------------- This took, 3.050, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |25 x + 30 x + 12| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |25 x + 30 x + 12| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 300 (n - 1) b(n - 2) b(n) = --------------------- + -------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+300*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 300 (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 0.03022998940390363084323463762736926220473443746821234292616474892313538635\ 210589806140208313686766948 1/2 3 Pi This constant is identified as, ------- 180 The implied delta is, 0.14482930678786744528674223345530138443443426372757266\ 0367569595197085015359093368262721084508644519 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 3 Pi ------- 180 ----------------------- This took, 3.658, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 26 x + 13| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 26 x + 13| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 26 (2 n - 1) b(n - 1) 52 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 26*(2*n-1)/n*b(n-1)+52*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 26 (2 n - 1) a(n - 1) 52 (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 0.03751850165322590252698129214550130546753619616645005977428264638936114568\ 751259475342993192572040525 The implied delta is, 0.20194164999205287483007940581543500666596839230011704\ 1541842281546554549064206272877922319564519915 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0375185016532259025269812921455013054675361961664500597742826463893611\ 4568751259475342993192572040525 ----------------------- This took, 2.875, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (14 x + 13)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (14 x + 13)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 27 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 27*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 27 (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 0.03705398607686093923454871166801884645365845959095466308051232438375440869\ 867690199697358107190036288 This constant is identified as, 1/2 ln(2) + 1/2 ln(7) - 1/2 ln(13) The implied delta is, 0.60494236728464687620594805518205321956339755800831773\ 2934297810296642755586658327001457976860588016 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(2) + 1/2 ln(7) - 1/2 ln(13) ----------------------- This took, 2.025, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |14 x + 28 x + 13| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |14 x + 28 x + 13| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 56 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-56*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 56 (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 0.03660303806605900133075137642080336005426892669479481601059981887942434063\ 514341334867673600291314363 The implied delta is, 0.09055938412400127852616928981575939585676901176033805\ 5393153080350367728073750651073382222076708088 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0366030380660590013307513764208033600542689266947948160105998188794243\ 4063514341334867673600291314363 ----------------------- This took, 2.385, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (15 x + 13)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (15 x + 13)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 4 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)-4*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 4 (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 0.03577521091016833248562928214584868484195599183276501430707417861205215340\ 046413802604374236660555270 This constant is identified as, 1/4 ln(3) + 1/4 ln(5) - 1/4 ln(13) The implied delta is, 0.33441180985896824616450203240752203864563918354904647\ 1916995912682711251829924101311728622028277602 Since this is positive, this suggests an Apery-style irrationality proof of, 1/4 ln(3) + 1/4 ln(5) - 1/4 ln(13) ----------------------- This took, 1.502, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 28 x + 13| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 28 x + 13| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 48 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+48*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 48 (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 0.03501107950117672887015340100053086599677098767706598014240261626988368706\ 701376567764934854186424461 The implied delta is, 0.15862543632556278453236911268726879922458069740061803\ 8156104188081462490945303437261615783667759081 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0350110795011767288701534010005308659967709876770659801424026162698836\ 8706701376567764934854186424461 ----------------------- This took, 5.462, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (16 x + 13)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (16 x + 13)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) 9 (n - 1) b(n - 2) b(n) = --------------------- - ------------------ n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-9*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (2 n - 1) a(n - 1) 9 (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 0.03460656079637408360257350737789794458278876544680231667727908788501830342\ 572811383731310742419754357 This constant is identified as, 2/3 ln(2) - 1/6 ln(13) The implied delta is, 0.17522030265818860356785395470584241849907664784203849\ 1426654627989669007376983660394887316878375752 Since this is positive, this suggests an Apery-style irrationality proof of, 2/3 ln(2) - 1/6 ln(13) ----------------------- This took, 1.542, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (17 x + 13)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (17 x + 13)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 16 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-16*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 16 (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 0.03353299832433491802450589703847599134786688347819220885977402840235348855\ 956513768906502557462668234 This constant is identified as, 1/8 ln(17) - 1/8 ln(13) The implied delta is, 0.06934995465479629828851167750946425973667586676147809\ 3686172562420206574430538832801424443004342546 Since this is positive, this suggests an Apery-style irrationality proof of, 1/8 ln(17) - 1/8 ln(13) ----------------------- This took, 1.668, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |18 x + 30 x + 13| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |18 x + 30 x + 13| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 36 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+36*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 36 (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 0.03289925997498012639500829419913171557459751729797535025294815670683899499\ 637396428878297133954800735 This constant is identified as, arctan( 6 5 4 3 2 RootOf(_Z + 30 _Z - 15 _Z - 100 _Z + 15 _Z + 30 _Z - 1, index = 1)) The implied delta is, 0.27270782245946457223806389318872976716053439548587568\ 6699451470121308965437156670785535597890142178 Since this is positive, this suggests an Apery-style irrationality proof of, arctan( 6 5 4 3 2 RootOf(_Z + 30 _Z - 15 _Z - 100 _Z + 15 _Z + 30 _Z - 1, index = 1)) ----------------------- This took, 3.348, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 28 x + 14| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 28 x + 14| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 28 (2 n - 1) b(n - 1) 56 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 28*(2*n-1)/n*b(n-1)+56*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 28 (2 n - 1) a(n - 1) 56 (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 0.03489862698628779901503432896696407856305315525689241589229888341007549137\ 445117500163539028108535097 The implied delta is, 0.09853845222615003892283687190987722507020990657857969\ 9491746469521925964992532836471108774208542425 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0348986269862877990150343289669640785630531552568924158922988834100754\ 9137445117500163539028108535097 ----------------------- This took, 2.263, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, /(x + 1) (15 x + 14)\n |-------------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, /(x + 1) (15 x + 14)\n |-------------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 29 (2 n - 1) b(n - 1) (n - 1) b(n - 2) b(n) = --------------------- - ---------------- n n and in Maple notation b(n) = 29*(2*n-1)/n*b(n-1)-(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 29 (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 0.03449643574347572573670985262367852323025352407457536553363603284034989810\ 225137405511390366131074253 This constant is identified as, 1/2 ln(3) + 1/2 ln(5) - 1/2 ln(2) - 1/2 ln(7) The implied delta is, 0.61160364913601387555756460059385440034977790821303047\ 2575466750919779724547038506469358114765993755 Since this is positive, this suggests an Apery-style irrationality proof of, 1/2 ln(3) + 1/2 ln(5) - 1/2 ln(2) - 1/2 ln(7) ----------------------- This took, 1.210, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |15 x + 30 x + 14| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |15 x + 30 x + 14| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 60 (n - 1) b(n - 2) b(n) = --------------------- - ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)-60*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 60 (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 0.03410519202747599285289283866324677857230240163234248312183066455215426010\ 342716356232240718469065848 The implied delta is, 0.21449668881929892212985124471994174435050344873837277\ 3112054043963511093357556763599800523359493280 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0341051920274759928528928386632467785723024016323424831218306645521542\ 6010342716356232240718469065848 ----------------------- This took, 3.344, seconds. ------------------------------------------------------------- The Apery limit generated by the sequence that are constant terms in x of, / 2 \n |16 x + 30 x + 15| |-----------------| \ x / By Shalosh B. Ekhad Let , b(n), be the coefficient of x^0 (i.e. constant term of, / 2 \n |16 x + 30 x + 15| |-----------------| \ x / The famous Almkvist-Zeilberger algorithm finds (and proves!, but proof omitt\ ed) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 30 (2 n - 1) b(n - 1) 60 (n - 1) b(n - 2) b(n) = --------------------- + ------------------- n n and in Maple notation b(n) = 30*(2*n-1)/n*b(n-1)+60*(n-1)/n*b(n-2) Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 30 (2 n - 1) a(n - 1) 60 (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 0.03262088066935705467466427056746132749411161860024825595766466030099607238\ 908802594372488216150018487 The implied delta is, 0.35555010412699671480676403410379995196689699434768789\ 4692178833768165992540867494430602673175437238 Since this is positive, this suggests an Apery-style irrationality proof of, 0.0326208806693570546746642705674613274941116186002482559576646603009960\ 7238908802594372488216150018487 ----------------------- This took, 2.771, seconds. ------------------------------ The whole thing took, 828.101, seconds.