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%I #20 Jul 15 2018 17:44:20
%S 1,15,7,495,2145,69615,33915,245157,2523675,150225075,60480225,
%T 187751655,26397882693,7073942421,221214302595,26306570659215,
%U 362711807574025,315526198564395,154309366568181825,495353332454332275,13575552922770178725
%N Let K(n) = integral(t=-1,1, t^(2n)*(1-t^2)^(2n)/(1+it)^(3n+1) dt) and write K(n) = d(n)*Pi - b(n)/a(n) where a(n), b(n), d(n) are positive integers; sequence gives a(n).
%H Bradley Klee, <a href="/A305998/b305998.txt">Table of n, a(n) for n = 1..1000</a>
%H Frits Beukers, <a href="https://dspace.library.uu.nl/handle/1874/26398">A rational approach to Pi</a>, Nieuw archief voor wiskunde 5/1 No. 4, December 2000, p. 378.
%t HermiteReduceRational[num_, den_, m_] := If[m > 1, Module[{cl = CoefficientList[num, t], deg, u, v, sol},If[Length[cl] == 1, cl = PadRight[cl, 3]]; deg = Length[cl] - 1; u = Total[c[#]*t^(2 #) & /@ Range[0, deg/2 - 1]]; v = Plus[Total[-c[#]*(m - 1)/(2*# + 1) t^(2*# + 1) & /@ Range[0, deg/2 - 1]], c[-1] t]; sol = Solve@ MapThread[Equal, {cl,CoefficientList[Expand[Dot[{1 + t^2, 2 t}, {u, v}]], t]}]; Plus[ ReplaceAll[v/(m - 1)/den^(m - 1), sol[[1]]] /. t -> 1, HermiteReduceRational[ Expand@ReplaceAll[u+1/(m-1)*D[v, t], sol[[1]]], den, m - 1]]],0]
%t Denominator[ HermiteReduceRational[ t^(2*#)*(1-t^2)^(2*#)*((1+I*t)^(3*#+1)+(1-I*t)^(3*#+1)), (1+t^2), 3*#+1]]&/@Range[20] (* _Bradley Klee_, Jun 18 2018 *)
%t Denominator@RecurrenceTable[ {64*(1+n)*(2+n)*(1+2*n)*(3+2*n)*(5+2*n)*(816+755*n+165*n^2)*a[n] -48*(2+n)*(3+2*n)*(5+2*n)*(4+3*n)*(2039+4103*n+2595*n^2+495*n^3)*a[n+1] +6*(5+2*n)*(4+3*n)*(5+3*n)*(893628+2406908*n+2163923*n^2+803750*n^3+106095*n^4)*a[n+2] -9*(3+n)*(4+3*n)*(5+3*n)*(7+3*n)*(8+3*n)*(226+425*n+165*n^2)*a[n+3]==0, a[0]==0, a[1]==44, a[2]==45616/15}, a, {n,1,5000}] (* _Bradley Klee_, Jun 25 2018 *)
%Y Cf. A123178, A305997.
%K nonn,frac
%O 1,2
%A _Bradley Klee_, Jun 16 2018
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