The Apery Limits for all summands of the form, binomial(n, k) binomial(a1 n + b1 k, k) binomial(a2 n + b2 k, k) binomial(a3 n + b3 k, k) k x , for a1, a2 and a3from 1 to, 1, and b1,b2 and b3 from 0 to, 1, and x from to, 2 By Shalosh B. Ekhad ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 4 ) binomial(n, k) / ----- k By Shalosh B. Ekhad Let ----- \ 4 b(n) = ) binomial(n, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 2 2 (2 n - 1) (3 n - 3 n + 1) b(n - 1) b(n) = ------------------------------------- 3 n 4 (4 n - 3) (4 n - 5) (n - 1) b(n - 2) + -------------------------------------- 3 n and in Maple notation b(n) = 2*(2*n-1)*(3*n^2-3*n+1)/n^3*b(n-1)+4*(4*n-3)*(4*n-5)*(n-1)/n^3*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 2 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 2 2 (2 n - 1) (3 n - 3 n + 1) a(n - 1) a(n) = ------------------------------------- 3 n 4 (4 n - 3) (4 n - 5) (n - 1) a(n - 2) + -------------------------------------- 3 n but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.32898681336964528729448303332920503784378998024135968754711164587400149408\ 06401747667257801239517411 2 Pi This constant is identified as, --- 30 The implied delta is, -0.6698041273376934610299754528407174906437950258342366\ 039746430064656845446898606769598067408655640543 2 Pi Since this is negative, there is no Apery-style irrationality proof of, ---, 30 but still a very fast way to compute it to many digits ----------------------- This took, 5.591, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 4 k ) binomial(n, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 4 k b(n) = ) binomial(n, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 3 2 b(n) = 3 (3456 n - 34560 n + 138012 n - 282164 n + 316147 n - 194631 n / 3 8 7 + 63195 n - 8550) b(n - 1) / (%1 n ) + (188352 n - 2354400 n / 6 5 4 3 2 + 12518814 n - 36978528 n + 66359449 n - 74056272 n + 50166510 n / 3 - 18860355 n + 3019770) b(n - 2) / (n (n - 1) %1) + 15 (n - 2) ( / 7 6 5 4 3 2 44928 n - 539136 n + 2672844 n - 7053704 n + 10623437 n - 9057312 n / 3 + 4025082 n - 727191) b(n - 3) / (n (n - 1) %1) / 4 3 2 3 (n - 2) (864 n - 3888 n + 6099 n - 3915 n + 905) (n - 3) b(n - 4) - --------------------------------------------------------------------- 3 n (n - 1) %1 4 3 2 %1 := 864 n - 7344 n + 22947 n - 31233 n + 15671 and in Maple notation b(n) = 3*(3456*n^7-34560*n^6+138012*n^5-282164*n^4+316147*n^3-194631*n^2+63195* n-8550)/(864*n^4-7344*n^3+22947*n^2-31233*n+15671)/n^3*b(n-1)+(188352*n^8-\ 2354400*n^7+12518814*n^6-36978528*n^5+66359449*n^4-74056272*n^3+50166510*n^2-\ 18860355*n+3019770)/n^3/(n-1)/(864*n^4-7344*n^3+22947*n^2-31233*n+15671)*b(n-2) +15*(n-2)*(44928*n^7-539136*n^6+2672844*n^5-7053704*n^4+10623437*n^3-9057312*n^ 2+4025082*n-727191)/n^3/(n-1)/(864*n^4-7344*n^3+22947*n^2-31233*n+15671)*b(n-3) -(n-2)*(864*n^4-3888*n^3+6099*n^2-3915*n+905)*(n-3)^3/n^3/(n-1)/(864*n^4-7344*n ^3+22947*n^2-31233*n+15671)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 3, b(2) = 37, b(3) = 495 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 3 2 a(n) = 3 (3456 n - 34560 n + 138012 n - 282164 n + 316147 n - 194631 n / 3 8 7 + 63195 n - 8550) a(n - 1) / (%1 n ) + (188352 n - 2354400 n / 6 5 4 3 2 + 12518814 n - 36978528 n + 66359449 n - 74056272 n + 50166510 n / 3 - 18860355 n + 3019770) a(n - 2) / (n (n - 1) %1) + 15 (n - 2) ( / 7 6 5 4 3 2 44928 n - 539136 n + 2672844 n - 7053704 n + 10623437 n - 9057312 n / 3 + 4025082 n - 727191) a(n - 3) / (n (n - 1) %1) / 4 3 2 3 (n - 2) (864 n - 3888 n + 6099 n - 3915 n + 905) (n - 3) a(n - 4) - --------------------------------------------------------------------- 3 n (n - 1) %1 4 3 2 %1 := 864 n - 7344 n + 22947 n - 31233 n + 15671 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very 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, -0.6657637326270667931453364902937010606643610907415391\ 317343840577867519183205089002208257940233323985 Since this is negative, there is no Apery-style irrationality proof of, 1/4 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 1.490, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 ) binomial(n, k) binomial(n + k, k) / ----- k By Shalosh B. Ekhad Let ----- \ 3 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 3 b(n) = 2 (235296 n - 2322864 n + 9245766 n - 19022421 n + 21621181 n 2 / 3 8 - 13561627 n + 4459053 n - 605664) b(n - 1) / (%1 n ) + 2 (1647072 n / 7 6 5 4 3 - 20377728 n + 107506956 n - 315721020 n + 564159163 n - 627527310 n 2 / 3 + 423779896 n - 158592459 n + 25128864) b(n - 2) / (n (n - 1) %1) + 2 / 7 6 5 4 (n - 2) (4176504 n - 49583844 n + 243933522 n - 641841009 n 3 2 / 3 + 971188553 n - 841622632 n + 385567572 n - 72023040) b(n - 3) / (n / (n - 1) %1) - 4 (n - 2) 4 3 2 3 (29412 n - 128592 n + 202011 n - 134886 n + 32480) (n - 3) b(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 29412 n - 246240 n + 764259 n - 1042332 n + 527381 and in Maple notation b(n) = 2*(235296*n^7-2322864*n^6+9245766*n^5-19022421*n^4+21621181*n^3-13561627 *n^2+4459053*n-605664)/(29412*n^4-246240*n^3+764259*n^2-1042332*n+527381)/n^3*b (n-1)+2*(1647072*n^8-20377728*n^7+107506956*n^6-315721020*n^5+564159163*n^4-\ 627527310*n^3+423779896*n^2-158592459*n+25128864)/n^3/(n-1)/(29412*n^4-246240*n ^3+764259*n^2-1042332*n+527381)*b(n-2)+2*(n-2)*(4176504*n^7-49583844*n^6+ 243933522*n^5-641841009*n^4+971188553*n^3-841622632*n^2+385567572*n-72023040)/n ^3/(n-1)/(29412*n^4-246240*n^3+764259*n^2-1042332*n+527381)*b(n-3)-4*(n-2)*( 29412*n^4-128592*n^3+202011*n^2-134886*n+32480)*(n-3)^3/n^3/(n-1)/(29412*n^4-\ 246240*n^3+764259*n^2-1042332*n+527381)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 3, b(2) = 31, b(3) = 399 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 3 a(n) = 2 (235296 n - 2322864 n + 9245766 n - 19022421 n + 21621181 n 2 / 3 8 - 13561627 n + 4459053 n - 605664) a(n - 1) / (%1 n ) + 2 (1647072 n / 7 6 5 4 3 - 20377728 n + 107506956 n - 315721020 n + 564159163 n - 627527310 n 2 / 3 + 423779896 n - 158592459 n + 25128864) a(n - 2) / (n (n - 1) %1) + 2 / 7 6 5 4 (n - 2) (4176504 n - 49583844 n + 243933522 n - 641841009 n 3 2 / 3 + 971188553 n - 841622632 n + 385567572 n - 72023040) a(n - 3) / (n / (n - 1) %1) - 4 (n - 2) 4 3 2 3 (29412 n - 128592 n + 202011 n - 134886 n + 32480) (n - 3) a(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 29412 n - 246240 n + 764259 n - 1042332 n + 527381 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately -0.6502970870924282429866787126026380022680387979254258112293590655115821634\ -1556 809955349844543801287394189 10 This constant is identified as, 0 The implied delta is, -0.5588076599813275570944292142529162276597153955730331\ 783260511590805789608457644252169072563156042125 Since this is negative, there is no Apery-style irrationality proof of, 0, but still a very fast way to compute it to many digits ----------------------- This took, 1.323, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 k ) binomial(n, k) binomial(n + k, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 3 k b(n) = ) binomial(n, k) binomial(n + k, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 3 b(n) = 4 (324380 n - 3211362 n + 12801362 n - 26374163 n + 30084029 n 2 / 3 8 - 19023670 n + 6336372 n - 875448) b(n - 1) / (%1 n ) + 2 (3104780 n / 7 6 5 4 - 38499272 n + 203433983 n - 598443475 n + 1072313479 n 3 2 / 3 - 1198564101 n + 815726644 n - 308596842 n + 49492368) b(n - 2) / (n / 7 6 5 (n - 1) %1) + 4 (n - 2) (14133700 n - 168191030 n + 828727582 n 4 3 2 - 2180471213 n + 3289613941 n - 2828876009 n + 1277859117 n - 234425844 / 3 ) b(n - 3) / (n (n - 1) %1) - 9 (n - 2) / 4 3 2 3 (46340 n - 203896 n + 319075 n - 209138 n + 49200) (n - 3) b(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 389256 n + 1208803 n - 1644336 n + 827649 and in Maple notation b(n) = 4*(324380*n^7-3211362*n^6+12801362*n^5-26374163*n^4+30084029*n^3-\ 19023670*n^2+6336372*n-875448)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+ 827649)/n^3*b(n-1)+2*(3104780*n^8-38499272*n^7+203433983*n^6-598443475*n^5+ 1072313479*n^4-1198564101*n^3+815726644*n^2-308596842*n+49492368)/n^3/(n-1)/( 46340*n^4-389256*n^3+1208803*n^2-1644336*n+827649)*b(n-2)+4*(n-2)*(14133700*n^7 -168191030*n^6+828727582*n^5-2180471213*n^4+3289613941*n^3-2828876009*n^2+ 1277859117*n-234425844)/n^3/(n-1)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+ 827649)*b(n-3)-9*(n-2)*(46340*n^4-203896*n^3+319075*n^2-209138*n+49200)*(n-3)^3 /n^3/(n-1)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+827649)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 5, b(2) = 73, b(3) = 1457 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 3 a(n) = 4 (324380 n - 3211362 n + 12801362 n - 26374163 n + 30084029 n 2 / 3 8 - 19023670 n + 6336372 n - 875448) a(n - 1) / (%1 n ) + 2 (3104780 n / 7 6 5 4 - 38499272 n + 203433983 n - 598443475 n + 1072313479 n 3 2 / 3 - 1198564101 n + 815726644 n - 308596842 n + 49492368) a(n - 2) / (n / 7 6 5 (n - 1) %1) + 4 (n - 2) (14133700 n - 168191030 n + 828727582 n 4 3 2 - 2180471213 n + 3289613941 n - 2828876009 n + 1277859117 n - 234425844 / 3 ) a(n - 3) / (n (n - 1) %1) - 9 (n - 2) / 4 3 2 3 (46340 n - 203896 n + 319075 n - 209138 n + 49200) (n - 3) a(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 389256 n + 1208803 n - 1644336 n + 827649 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very 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, -0.6205959196707023009532447198191543234176716142685418\ 629622493872084042057588134232000771353760531484 Since this is negative, there is no Apery-style irrationality proof of, 1/6 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 1.574, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 ) binomial(n, k) binomial(n + k, k) / ----- k By Shalosh B. Ekhad Let ----- \ 3 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 3 b(n) = 2 (235296 n - 2322864 n + 9245766 n - 19022421 n + 21621181 n 2 / 3 8 - 13561627 n + 4459053 n - 605664) b(n - 1) / (%1 n ) + 2 (1647072 n / 7 6 5 4 3 - 20377728 n + 107506956 n - 315721020 n + 564159163 n - 627527310 n 2 / 3 + 423779896 n - 158592459 n + 25128864) b(n - 2) / (n (n - 1) %1) + 2 / 7 6 5 4 (n - 2) (4176504 n - 49583844 n + 243933522 n - 641841009 n 3 2 / 3 + 971188553 n - 841622632 n + 385567572 n - 72023040) b(n - 3) / (n / (n - 1) %1) - 4 (n - 2) 4 3 2 3 (29412 n - 128592 n + 202011 n - 134886 n + 32480) (n - 3) b(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 29412 n - 246240 n + 764259 n - 1042332 n + 527381 and in Maple notation b(n) = 2*(235296*n^7-2322864*n^6+9245766*n^5-19022421*n^4+21621181*n^3-13561627 *n^2+4459053*n-605664)/(29412*n^4-246240*n^3+764259*n^2-1042332*n+527381)/n^3*b (n-1)+2*(1647072*n^8-20377728*n^7+107506956*n^6-315721020*n^5+564159163*n^4-\ 627527310*n^3+423779896*n^2-158592459*n+25128864)/n^3/(n-1)/(29412*n^4-246240*n ^3+764259*n^2-1042332*n+527381)*b(n-2)+2*(n-2)*(4176504*n^7-49583844*n^6+ 243933522*n^5-641841009*n^4+971188553*n^3-841622632*n^2+385567572*n-72023040)/n ^3/(n-1)/(29412*n^4-246240*n^3+764259*n^2-1042332*n+527381)*b(n-3)-4*(n-2)*( 29412*n^4-128592*n^3+202011*n^2-134886*n+32480)*(n-3)^3/n^3/(n-1)/(29412*n^4-\ 246240*n^3+764259*n^2-1042332*n+527381)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 3, b(2) = 31, b(3) = 399 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 3 a(n) = 2 (235296 n - 2322864 n + 9245766 n - 19022421 n + 21621181 n 2 / 3 8 - 13561627 n + 4459053 n - 605664) a(n - 1) / (%1 n ) + 2 (1647072 n / 7 6 5 4 3 - 20377728 n + 107506956 n - 315721020 n + 564159163 n - 627527310 n 2 / 3 + 423779896 n - 158592459 n + 25128864) a(n - 2) / (n (n - 1) %1) + 2 / 7 6 5 4 (n - 2) (4176504 n - 49583844 n + 243933522 n - 641841009 n 3 2 / 3 + 971188553 n - 841622632 n + 385567572 n - 72023040) a(n - 3) / (n / (n - 1) %1) - 4 (n - 2) 4 3 2 3 (29412 n - 128592 n + 202011 n - 134886 n + 32480) (n - 3) a(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 29412 n - 246240 n + 764259 n - 1042332 n + 527381 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately -0.6502970870924282429866787126026380022680387979254258112293590655115821634\ -1556 809955349844543801287394189 10 This constant is identified as, 0 The implied delta is, -0.5588076599813275570944292142529162276597153955730331\ 783260511590805789608457644252169072563156042125 Since this is negative, there is no Apery-style irrationality proof of, 0, but still a very fast way to compute it to many digits ----------------------- This took, 0.678, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 k ) binomial(n, k) binomial(n + k, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 3 k b(n) = ) binomial(n, k) binomial(n + k, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 3 b(n) = 4 (324380 n - 3211362 n + 12801362 n - 26374163 n + 30084029 n 2 / 3 8 - 19023670 n + 6336372 n - 875448) b(n - 1) / (%1 n ) + 2 (3104780 n / 7 6 5 4 - 38499272 n + 203433983 n - 598443475 n + 1072313479 n 3 2 / 3 - 1198564101 n + 815726644 n - 308596842 n + 49492368) b(n - 2) / (n / 7 6 5 (n - 1) %1) + 4 (n - 2) (14133700 n - 168191030 n + 828727582 n 4 3 2 - 2180471213 n + 3289613941 n - 2828876009 n + 1277859117 n - 234425844 / 3 ) b(n - 3) / (n (n - 1) %1) - 9 (n - 2) / 4 3 2 3 (46340 n - 203896 n + 319075 n - 209138 n + 49200) (n - 3) b(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 389256 n + 1208803 n - 1644336 n + 827649 and in Maple notation b(n) = 4*(324380*n^7-3211362*n^6+12801362*n^5-26374163*n^4+30084029*n^3-\ 19023670*n^2+6336372*n-875448)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+ 827649)/n^3*b(n-1)+2*(3104780*n^8-38499272*n^7+203433983*n^6-598443475*n^5+ 1072313479*n^4-1198564101*n^3+815726644*n^2-308596842*n+49492368)/n^3/(n-1)/( 46340*n^4-389256*n^3+1208803*n^2-1644336*n+827649)*b(n-2)+4*(n-2)*(14133700*n^7 -168191030*n^6+828727582*n^5-2180471213*n^4+3289613941*n^3-2828876009*n^2+ 1277859117*n-234425844)/n^3/(n-1)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+ 827649)*b(n-3)-9*(n-2)*(46340*n^4-203896*n^3+319075*n^2-209138*n+49200)*(n-3)^3 /n^3/(n-1)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+827649)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 5, b(2) = 73, b(3) = 1457 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 3 a(n) = 4 (324380 n - 3211362 n + 12801362 n - 26374163 n + 30084029 n 2 / 3 8 - 19023670 n + 6336372 n - 875448) a(n - 1) / (%1 n ) + 2 (3104780 n / 7 6 5 4 - 38499272 n + 203433983 n - 598443475 n + 1072313479 n 3 2 / 3 - 1198564101 n + 815726644 n - 308596842 n + 49492368) a(n - 2) / (n / 7 6 5 (n - 1) %1) + 4 (n - 2) (14133700 n - 168191030 n + 828727582 n 4 3 2 - 2180471213 n + 3289613941 n - 2828876009 n + 1277859117 n - 234425844 / 3 ) a(n - 3) / (n (n - 1) %1) - 9 (n - 2) / 4 3 2 3 (46340 n - 203896 n + 319075 n - 209138 n + 49200) (n - 3) a(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 389256 n + 1208803 n - 1644336 n + 827649 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very 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, -0.6205959196707023009532447198191543234176716142685418\ 629622493872084042057588134232000771353760531484 Since this is negative, there is no Apery-style irrationality proof of, 1/6 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 0.756, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 2 2 ) binomial(n, k) binomial(n + k, k) / ----- k By Shalosh B. Ekhad Let ----- \ 2 2 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 2 3 (2 n - 1) (17 n - 17 n + 5) b(n - 1) (n - 1) b(n - 2) b(n) = ------------------------------------- - ----------------- 3 3 n n and in Maple notation b(n) = (2*n-1)*(17*n^2-17*n+5)/n^3*b(n-1)-(n-1)^3/n^3*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 5 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 2 3 (2 n - 1) (17 n - 17 n + 5) a(n - 1) (n - 1) a(n - 2) a(n) = ------------------------------------- - ----------------- 3 3 n n but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.20034281719326571423328969358524166512749771539008314696537859255697303429\ 77188483644093122682225430 This constant is identified as, 1/6 Zeta(3) The implied delta is, 0.08507477688006303521147749746906580809236227288658715\ 3176053247978253317356214127149095473736359157 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 Zeta(3) ----------------------- This took, 5.791, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 2 2 k ) binomial(n, k) binomial(n + k, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 2 2 k b(n) = ) binomial(n, k) binomial(n + k, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 4 3 2 (2 n - 1) (952 n - 4760 n + 7918 n - 5049 n + 1182) b(n - 1) b(n) = --------------------------------------------------------------- - 3 %1 n 6 5 4 3 2 (2 n - 3) (18088 n - 162792 n + 583938 n - 1061748 n + 1023813 n / 3 - 491265 n + 90714) b(n - 2) / (n (2 n - 5) (n - 1) %1) + / 4 3 2 (n - 2) (2 n - 1) (952 n - 6664 n + 16486 n - 16755 n + 5889) b(n - 3) ------------------------------------------------------------------------- 3 (n - 1) %1 n 2 3 (n - 2) (2 n - 1) (28 n - 56 n + 27) (n - 3) b(n - 4) - ------------------------------------------------------- 3 n (2 n - 5) (n - 1) %1 2 %1 := 28 n - 112 n + 111 and in Maple notation b(n) = (2*n-1)*(952*n^4-4760*n^3+7918*n^2-5049*n+1182)/(28*n^2-112*n+111)/n^3*b (n-1)-(2*n-3)*(18088*n^6-162792*n^5+583938*n^4-1061748*n^3+1023813*n^2-491265*n +90714)/n^3/(2*n-5)/(n-1)/(28*n^2-112*n+111)*b(n-2)+(n-2)*(2*n-1)*(952*n^4-6664 *n^3+16486*n^2-16755*n+5889)/(n-1)/(28*n^2-112*n+111)/n^3*b(n-3)-(n-2)*(2*n-1)* (28*n^2-56*n+27)*(n-3)^3/n^3/(2*n-5)/(n-1)/(28*n^2-112*n+111)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 9, b(2) = 217, b(3) = 7089 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 4 3 2 (2 n - 1) (952 n - 4760 n + 7918 n - 5049 n + 1182) a(n - 1) a(n) = --------------------------------------------------------------- - 3 %1 n 6 5 4 3 2 (2 n - 3) (18088 n - 162792 n + 583938 n - 1061748 n + 1023813 n / 3 - 491265 n + 90714) a(n - 2) / (n (2 n - 5) (n - 1) %1) + / 4 3 2 (n - 2) (2 n - 1) (952 n - 6664 n + 16486 n - 16755 n + 5889) a(n - 3) ------------------------------------------------------------------------- 3 (n - 1) %1 n 2 3 (n - 2) (2 n - 1) (28 n - 56 n + 27) (n - 3) a(n - 4) - ------------------------------------------------------- 3 n (2 n - 5) (n - 1) %1 2 %1 := 28 n - 112 n + 111 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very 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, -0.6782475847599052291513165997915666727839775217204705\ 850478429842306869829330883162947906773570689183 Since this is negative, there is no Apery-style irrationality proof of, 1/4 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 1.484, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 ) binomial(n, k) binomial(n + k, k) / ----- k By Shalosh B. Ekhad Let ----- \ 3 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 3 b(n) = 2 (235296 n - 2322864 n + 9245766 n - 19022421 n + 21621181 n 2 / 3 8 - 13561627 n + 4459053 n - 605664) b(n - 1) / (%1 n ) + 2 (1647072 n / 7 6 5 4 3 - 20377728 n + 107506956 n - 315721020 n + 564159163 n - 627527310 n 2 / 3 + 423779896 n - 158592459 n + 25128864) b(n - 2) / (n (n - 1) %1) + 2 / 7 6 5 4 (n - 2) (4176504 n - 49583844 n + 243933522 n - 641841009 n 3 2 / 3 + 971188553 n - 841622632 n + 385567572 n - 72023040) b(n - 3) / (n / (n - 1) %1) - 4 (n - 2) 4 3 2 3 (29412 n - 128592 n + 202011 n - 134886 n + 32480) (n - 3) b(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 29412 n - 246240 n + 764259 n - 1042332 n + 527381 and in Maple notation b(n) = 2*(235296*n^7-2322864*n^6+9245766*n^5-19022421*n^4+21621181*n^3-13561627 *n^2+4459053*n-605664)/(29412*n^4-246240*n^3+764259*n^2-1042332*n+527381)/n^3*b (n-1)+2*(1647072*n^8-20377728*n^7+107506956*n^6-315721020*n^5+564159163*n^4-\ 627527310*n^3+423779896*n^2-158592459*n+25128864)/n^3/(n-1)/(29412*n^4-246240*n ^3+764259*n^2-1042332*n+527381)*b(n-2)+2*(n-2)*(4176504*n^7-49583844*n^6+ 243933522*n^5-641841009*n^4+971188553*n^3-841622632*n^2+385567572*n-72023040)/n ^3/(n-1)/(29412*n^4-246240*n^3+764259*n^2-1042332*n+527381)*b(n-3)-4*(n-2)*( 29412*n^4-128592*n^3+202011*n^2-134886*n+32480)*(n-3)^3/n^3/(n-1)/(29412*n^4-\ 246240*n^3+764259*n^2-1042332*n+527381)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 3, b(2) = 31, b(3) = 399 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 3 a(n) = 2 (235296 n - 2322864 n + 9245766 n - 19022421 n + 21621181 n 2 / 3 8 - 13561627 n + 4459053 n - 605664) a(n - 1) / (%1 n ) + 2 (1647072 n / 7 6 5 4 3 - 20377728 n + 107506956 n - 315721020 n + 564159163 n - 627527310 n 2 / 3 + 423779896 n - 158592459 n + 25128864) a(n - 2) / (n (n - 1) %1) + 2 / 7 6 5 4 (n - 2) (4176504 n - 49583844 n + 243933522 n - 641841009 n 3 2 / 3 + 971188553 n - 841622632 n + 385567572 n - 72023040) a(n - 3) / (n / (n - 1) %1) - 4 (n - 2) 4 3 2 3 (29412 n - 128592 n + 202011 n - 134886 n + 32480) (n - 3) a(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 29412 n - 246240 n + 764259 n - 1042332 n + 527381 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately -0.6502970870924282429866787126026380022680387979254258112293590655115821634\ -1556 809955349844543801287394189 10 This constant is identified as, 0 The implied delta is, -0.5588076599813275570944292142529162276597153955730331\ 783260511590805789608457644252169072563156042125 Since this is negative, there is no Apery-style irrationality proof of, 0, but still a very fast way to compute it to many digits ----------------------- This took, 0.894, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 k ) binomial(n, k) binomial(n + k, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 3 k b(n) = ) binomial(n, k) binomial(n + k, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 3 b(n) = 4 (324380 n - 3211362 n + 12801362 n - 26374163 n + 30084029 n 2 / 3 8 - 19023670 n + 6336372 n - 875448) b(n - 1) / (%1 n ) + 2 (3104780 n / 7 6 5 4 - 38499272 n + 203433983 n - 598443475 n + 1072313479 n 3 2 / 3 - 1198564101 n + 815726644 n - 308596842 n + 49492368) b(n - 2) / (n / 7 6 5 (n - 1) %1) + 4 (n - 2) (14133700 n - 168191030 n + 828727582 n 4 3 2 - 2180471213 n + 3289613941 n - 2828876009 n + 1277859117 n - 234425844 / 3 ) b(n - 3) / (n (n - 1) %1) - 9 (n - 2) / 4 3 2 3 (46340 n - 203896 n + 319075 n - 209138 n + 49200) (n - 3) b(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 389256 n + 1208803 n - 1644336 n + 827649 and in Maple notation b(n) = 4*(324380*n^7-3211362*n^6+12801362*n^5-26374163*n^4+30084029*n^3-\ 19023670*n^2+6336372*n-875448)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+ 827649)/n^3*b(n-1)+2*(3104780*n^8-38499272*n^7+203433983*n^6-598443475*n^5+ 1072313479*n^4-1198564101*n^3+815726644*n^2-308596842*n+49492368)/n^3/(n-1)/( 46340*n^4-389256*n^3+1208803*n^2-1644336*n+827649)*b(n-2)+4*(n-2)*(14133700*n^7 -168191030*n^6+828727582*n^5-2180471213*n^4+3289613941*n^3-2828876009*n^2+ 1277859117*n-234425844)/n^3/(n-1)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+ 827649)*b(n-3)-9*(n-2)*(46340*n^4-203896*n^3+319075*n^2-209138*n+49200)*(n-3)^3 /n^3/(n-1)/(46340*n^4-389256*n^3+1208803*n^2-1644336*n+827649)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 5, b(2) = 73, b(3) = 1457 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 3 a(n) = 4 (324380 n - 3211362 n + 12801362 n - 26374163 n + 30084029 n 2 / 3 8 - 19023670 n + 6336372 n - 875448) a(n - 1) / (%1 n ) + 2 (3104780 n / 7 6 5 4 - 38499272 n + 203433983 n - 598443475 n + 1072313479 n 3 2 / 3 - 1198564101 n + 815726644 n - 308596842 n + 49492368) a(n - 2) / (n / 7 6 5 (n - 1) %1) + 4 (n - 2) (14133700 n - 168191030 n + 828727582 n 4 3 2 - 2180471213 n + 3289613941 n - 2828876009 n + 1277859117 n - 234425844 / 3 ) a(n - 3) / (n (n - 1) %1) - 9 (n - 2) / 4 3 2 3 (46340 n - 203896 n + 319075 n - 209138 n + 49200) (n - 3) a(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 389256 n + 1208803 n - 1644336 n + 827649 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very 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, -0.6205959196707023009532447198191543234176716142685418\ 629622493872084042057588134232000771353760531484 Since this is negative, there is no Apery-style irrationality proof of, 1/6 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 0.794, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 2 2 ) binomial(n, k) binomial(n + k, k) / ----- k By Shalosh B. Ekhad Let ----- \ 2 2 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 2 3 (2 n - 1) (17 n - 17 n + 5) b(n - 1) (n - 1) b(n - 2) b(n) = ------------------------------------- - ----------------- 3 3 n n and in Maple notation b(n) = (2*n-1)*(17*n^2-17*n+5)/n^3*b(n-1)-(n-1)^3/n^3*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 5 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 2 3 (2 n - 1) (17 n - 17 n + 5) a(n - 1) (n - 1) a(n - 2) a(n) = ------------------------------------- - ----------------- 3 3 n n but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.20034281719326571423328969358524166512749771539008314696537859255697303429\ 77188483644093122682225430 This constant is identified as, 1/6 Zeta(3) The implied delta is, 0.08507477688006303521147749746906580809236227288658715\ 3176053247978253317356214127149095473736359157 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 Zeta(3) ----------------------- This took, 2.651, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 2 2 k ) binomial(n, k) binomial(n + k, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 2 2 k b(n) = ) binomial(n, k) binomial(n + k, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 4 3 2 (2 n - 1) (952 n - 4760 n + 7918 n - 5049 n + 1182) b(n - 1) b(n) = --------------------------------------------------------------- - 3 %1 n 6 5 4 3 2 (2 n - 3) (18088 n - 162792 n + 583938 n - 1061748 n + 1023813 n / 3 - 491265 n + 90714) b(n - 2) / (n (2 n - 5) (n - 1) %1) + / 4 3 2 (n - 2) (2 n - 1) (952 n - 6664 n + 16486 n - 16755 n + 5889) b(n - 3) ------------------------------------------------------------------------- 3 (n - 1) %1 n 2 3 (n - 2) (2 n - 1) (28 n - 56 n + 27) (n - 3) b(n - 4) - ------------------------------------------------------- 3 n (2 n - 5) (n - 1) %1 2 %1 := 28 n - 112 n + 111 and in Maple notation b(n) = (2*n-1)*(952*n^4-4760*n^3+7918*n^2-5049*n+1182)/(28*n^2-112*n+111)/n^3*b (n-1)-(2*n-3)*(18088*n^6-162792*n^5+583938*n^4-1061748*n^3+1023813*n^2-491265*n +90714)/n^3/(2*n-5)/(n-1)/(28*n^2-112*n+111)*b(n-2)+(n-2)*(2*n-1)*(952*n^4-6664 *n^3+16486*n^2-16755*n+5889)/(n-1)/(28*n^2-112*n+111)/n^3*b(n-3)-(n-2)*(2*n-1)* (28*n^2-56*n+27)*(n-3)^3/n^3/(2*n-5)/(n-1)/(28*n^2-112*n+111)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 9, b(2) = 217, b(3) = 7089 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 4 3 2 (2 n - 1) (952 n - 4760 n + 7918 n - 5049 n + 1182) a(n - 1) a(n) = --------------------------------------------------------------- - 3 %1 n 6 5 4 3 2 (2 n - 3) (18088 n - 162792 n + 583938 n - 1061748 n + 1023813 n / 3 - 491265 n + 90714) a(n - 2) / (n (2 n - 5) (n - 1) %1) + / 4 3 2 (n - 2) (2 n - 1) (952 n - 6664 n + 16486 n - 16755 n + 5889) a(n - 3) ------------------------------------------------------------------------- 3 (n - 1) %1 n 2 3 (n - 2) (2 n - 1) (28 n - 56 n + 27) (n - 3) a(n - 4) - ------------------------------------------------------- 3 n (2 n - 5) (n - 1) %1 2 %1 := 28 n - 112 n + 111 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very 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, -0.6782475847599052291513165997915666727839775217204705\ 850478429842306869829330883162947906773570689183 Since this is negative, there is no Apery-style irrationality proof of, 1/4 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 1.133, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 2 2 ) binomial(n, k) binomial(n + k, k) / ----- k By Shalosh B. Ekhad Let ----- \ 2 2 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 2 3 (2 n - 1) (17 n - 17 n + 5) b(n - 1) (n - 1) b(n - 2) b(n) = ------------------------------------- - ----------------- 3 3 n n and in Maple notation b(n) = (2*n-1)*(17*n^2-17*n+5)/n^3*b(n-1)-(n-1)^3/n^3*b(n-2) Of course, the initial conditions are b(0) = 1, b(1) = 5 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 2 3 (2 n - 1) (17 n - 17 n + 5) a(n - 1) (n - 1) a(n - 2) a(n) = ------------------------------------- - ----------------- 3 3 n n but with the following simpler initial conditions a(0) = 0, a(1) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.20034281719326571423328969358524166512749771539008314696537859255697303429\ 77188483644093122682225430 This constant is identified as, 1/6 Zeta(3) The implied delta is, 0.08507477688006303521147749746906580809236227288658715\ 3176053247978253317356214127149095473736359157 Since this is positive, this suggests an Apery-style irrationality proof of, 1/6 Zeta(3) ----------------------- This took, 2.868, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 2 2 k ) binomial(n, k) binomial(n + k, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 2 2 k b(n) = ) binomial(n, k) binomial(n + k, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 4 3 2 (2 n - 1) (952 n - 4760 n + 7918 n - 5049 n + 1182) b(n - 1) b(n) = --------------------------------------------------------------- - 3 %1 n 6 5 4 3 2 (2 n - 3) (18088 n - 162792 n + 583938 n - 1061748 n + 1023813 n / 3 - 491265 n + 90714) b(n - 2) / (n (2 n - 5) (n - 1) %1) + / 4 3 2 (n - 2) (2 n - 1) (952 n - 6664 n + 16486 n - 16755 n + 5889) b(n - 3) ------------------------------------------------------------------------- 3 (n - 1) %1 n 2 3 (n - 2) (2 n - 1) (28 n - 56 n + 27) (n - 3) b(n - 4) - ------------------------------------------------------- 3 n (2 n - 5) (n - 1) %1 2 %1 := 28 n - 112 n + 111 and in Maple notation b(n) = (2*n-1)*(952*n^4-4760*n^3+7918*n^2-5049*n+1182)/(28*n^2-112*n+111)/n^3*b (n-1)-(2*n-3)*(18088*n^6-162792*n^5+583938*n^4-1061748*n^3+1023813*n^2-491265*n +90714)/n^3/(2*n-5)/(n-1)/(28*n^2-112*n+111)*b(n-2)+(n-2)*(2*n-1)*(952*n^4-6664 *n^3+16486*n^2-16755*n+5889)/(n-1)/(28*n^2-112*n+111)/n^3*b(n-3)-(n-2)*(2*n-1)* (28*n^2-56*n+27)*(n-3)^3/n^3/(2*n-5)/(n-1)/(28*n^2-112*n+111)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 9, b(2) = 217, b(3) = 7089 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 4 3 2 (2 n - 1) (952 n - 4760 n + 7918 n - 5049 n + 1182) a(n - 1) a(n) = --------------------------------------------------------------- - 3 %1 n 6 5 4 3 2 (2 n - 3) (18088 n - 162792 n + 583938 n - 1061748 n + 1023813 n / 3 - 491265 n + 90714) a(n - 2) / (n (2 n - 5) (n - 1) %1) + / 4 3 2 (n - 2) (2 n - 1) (952 n - 6664 n + 16486 n - 16755 n + 5889) a(n - 3) ------------------------------------------------------------------------- 3 (n - 1) %1 n 2 3 (n - 2) (2 n - 1) (28 n - 56 n + 27) (n - 3) a(n - 4) - ------------------------------------------------------- 3 n (2 n - 5) (n - 1) %1 2 %1 := 28 n - 112 n + 111 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very 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, -0.6782475847599052291513165997915666727839775217204705\ 850478429842306869829330883162947906773570689183 Since this is negative, there is no Apery-style irrationality proof of, 1/4 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 0.930, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 ) binomial(n, k) binomial(n + k, k) / ----- k By Shalosh B. Ekhad Let ----- \ 3 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 b(n) = 1/2 (4176504 n - 38122740 n + 140783586 n - 270139161 n 3 2 / 3 + 288226505 n - 170040251 n + 51625509 n - 6283008) b(n - 1) / (%1 n ) / 8 7 6 5 + 1/2 (1647072 n - 19152000 n + 94636812 n - 258386460 n 4 3 2 + 423728203 n - 423743982 n + 249392728 n - 77793627 n + 9736704) / 3 7 6 b(n - 2) / (n (n - 1) %1) + 1/2 (n - 2) (235296 n - 2618352 n / 5 4 3 2 + 11905158 n - 28432149 n + 38188669 n - 28610816 n + 10954716 n / 3 - 1618272) b(n - 3) / (n (n - 1) %1) - 1/4 (n - 2) / 4 3 2 3 / (29412 n - 106704 n + 136347 n - 71238 n + 12608) (n - 3) b(n - 4) / / 3 (n (n - 1) %1) 4 3 2 %1 := 29412 n - 224352 n + 632931 n - 781692 n + 356309 and in Maple notation b(n) = 1/2*(4176504*n^7-38122740*n^6+140783586*n^5-270139161*n^4+288226505*n^3-\ 170040251*n^2+51625509*n-6283008)/(29412*n^4-224352*n^3+632931*n^2-781692*n+ 356309)/n^3*b(n-1)+1/2*(1647072*n^8-19152000*n^7+94636812*n^6-258386460*n^5+ 423728203*n^4-423743982*n^3+249392728*n^2-77793627*n+9736704)/n^3/(n-1)/(29412* n^4-224352*n^3+632931*n^2-781692*n+356309)*b(n-2)+1/2*(n-2)*(235296*n^7-2618352 *n^6+11905158*n^5-28432149*n^4+38188669*n^3-28610816*n^2+10954716*n-1618272)/n^ 3/(n-1)/(29412*n^4-224352*n^3+632931*n^2-781692*n+356309)*b(n-3)-1/4*(n-2)*( 29412*n^4-106704*n^3+136347*n^2-71238*n+12608)*(n-3)^3/n^3/(n-1)/(29412*n^4-\ 224352*n^3+632931*n^2-781692*n+356309)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 9, b(2) = 271, b(3) = 11193 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 a(n) = 1/2 (4176504 n - 38122740 n + 140783586 n - 270139161 n 3 2 / 3 + 288226505 n - 170040251 n + 51625509 n - 6283008) a(n - 1) / (%1 n ) / 8 7 6 5 + 1/2 (1647072 n - 19152000 n + 94636812 n - 258386460 n 4 3 2 + 423728203 n - 423743982 n + 249392728 n - 77793627 n + 9736704) / 3 7 6 a(n - 2) / (n (n - 1) %1) + 1/2 (n - 2) (235296 n - 2618352 n / 5 4 3 2 + 11905158 n - 28432149 n + 38188669 n - 28610816 n + 10954716 n / 3 - 1618272) a(n - 3) / (n (n - 1) %1) - 1/4 (n - 2) / 4 3 2 3 / (29412 n - 106704 n + 136347 n - 71238 n + 12608) (n - 3) a(n - 4) / / 3 (n (n - 1) %1) 4 3 2 %1 := 29412 n - 224352 n + 632931 n - 781692 n + 356309 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.11090354888959124950675713943330825089208002149764084065930880151894297951\ 51511544969381323194269900 This constant is identified as, 4/25 ln(2) The implied delta is, -0.1594303395173079116519182627166088255502522780855829\ 468090872908932501292394523093769548875346116181 Since this is negative, there is no Apery-style irrationality proof of, 4/25 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 1.598, seconds. ------------------------------------------------------------- The Apery limit generated by the binomial coefficient sum, ----- \ 3 k ) binomial(n, k) binomial(n + k, k) 2 / ----- k By Shalosh B. Ekhad Let ----- \ 3 k b(n) = ) binomial(n, k) binomial(n + k, k) 2 / ----- k The famous Zeilberger algorithm finds (and proves!, but proof omitted) that , b(n), sastisfies the following linear recurrence with polynomial coefficients 7 6 5 4 b(n) = 4/9 (14133700 n - 128616670 n + 472558342 n - 900999967 n 3 2 / + 955325065 n - 560820748 n + 169708536 n - 20619036) b(n - 1) / (%1 / 3 8 7 6 5 n ) + 2/9 (3104780 n - 36015448 n + 177353831 n - 481433171 n 4 3 2 + 781539619 n - 768045837 n + 439113040 n - 130973730 n + 15565608) / 3 7 6 b(n - 2) / (n (n - 1) %1) + 4/9 (n - 2) (324380 n - 3600618 n / 5 4 3 2 + 16304666 n - 38651497 n + 51198473 n - 37401839 n + 13750485 n / 3 - 1949346) b(n - 3) / (n (n - 1) %1) - 1/9 (n - 2) / 4 3 2 3 (46340 n - 166824 n + 207859 n - 103290 n + 17496) (n - 3) b(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 352184 n + 986371 n - 1204840 n + 541809 and in Maple notation b(n) = 4/9*(14133700*n^7-128616670*n^6+472558342*n^5-900999967*n^4+955325065*n^ 3-560820748*n^2+169708536*n-20619036)/(46340*n^4-352184*n^3+986371*n^2-1204840* n+541809)/n^3*b(n-1)+2/9*(3104780*n^8-36015448*n^7+177353831*n^6-481433171*n^5+ 781539619*n^4-768045837*n^3+439113040*n^2-130973730*n+15565608)/n^3/(n-1)/( 46340*n^4-352184*n^3+986371*n^2-1204840*n+541809)*b(n-2)+4/9*(n-2)*(324380*n^7-\ 3600618*n^6+16304666*n^5-38651497*n^4+51198473*n^3-37401839*n^2+13750485*n-\ 1949346)/n^3/(n-1)/(46340*n^4-352184*n^3+986371*n^2-1204840*n+541809)*b(n-3)-1/ 9*(n-2)*(46340*n^4-166824*n^3+207859*n^2-103290*n+17496)*(n-3)^3/n^3/(n-1)/( 46340*n^4-352184*n^3+986371*n^2-1204840*n+541809)*b(n-4) Of course, the initial conditions are b(0) = 1, b(1) = 17, b(2) = 973, b(3) = 76385 Let's consider the related sequence, let's call it, a(n) satisying the same recurrence, i.e. 7 6 5 4 a(n) = 4/9 (14133700 n - 128616670 n + 472558342 n - 900999967 n 3 2 / + 955325065 n - 560820748 n + 169708536 n - 20619036) a(n - 1) / (%1 / 3 8 7 6 5 n ) + 2/9 (3104780 n - 36015448 n + 177353831 n - 481433171 n 4 3 2 + 781539619 n - 768045837 n + 439113040 n - 130973730 n + 15565608) / 3 7 6 a(n - 2) / (n (n - 1) %1) + 4/9 (n - 2) (324380 n - 3600618 n / 5 4 3 2 + 16304666 n - 38651497 n + 51198473 n - 37401839 n + 13750485 n / 3 - 1949346) a(n - 3) / (n (n - 1) %1) - 1/9 (n - 2) / 4 3 2 3 (46340 n - 166824 n + 207859 n - 103290 n + 17496) (n - 3) a(n - 4) / 3 / (n (n - 1) %1) / 4 3 2 %1 := 46340 n - 352184 n + 986371 n - 1204840 n + 541809 but with the following simpler initial conditions a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 1 Using the recurrence, we can compute many terms, how about, 2000, terms for e\ ach sequence and get a very good approximation to the Apery limit, i.e. \ a(n) the limit of, ----, as n goes to infinity b(n) that is approximately 0.05885783827376579738390512966417971337335344856713625449235691802091783937\ 484239131306083629606162276 This constant is identified as, 9/62 ln(3) - 9/62 ln(2) The implied delta is, -0.1684848501305901275496584062921850791009314497191021\ 591201372711112484432829191342629182442077191632 Since this is negative, there is no Apery-style irrationality proof of, 9/62 ln(3) - 9/62 ln(2), but still a very fast way to compute it to many digits ----------------------- This took, 2.136, seconds. ---------------------------------- This ends this book that took altogether, 31.714, seconds to produce.