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A302770 a(n) = (4*n-2)*((n-1)*a(n-1) + ((n-2)!)^2), with a(1) = 0, n > 1. 1

%I #17 Apr 18 2018 04:51:31

%S 0,6,130,5516,397800,43770672,6828599232,1434021390720,

%T 390054681930240,133398762996833280,56027485989309542400,

%U 28349908516327342694400,17009945189463951728640000,11940981535393590884843520000,9696077008988591505023631360000

%N a(n) = (4*n-2)*((n-1)*a(n-1) + ((n-2)!)^2), with a(1) = 0, n > 1.

%H Travis Sherman, <a href="http://math.arizona.edu/~rta/001/sherman.travis/series.pdf">Summation of Glaisher- and Apery-like Series</a>, University of Arizona, May 23 2000, p. 14, (3.79) - (3.83).

%F a(n) = f3(n)*((n-1)!)^2, where f3(n) corresponds to the z values such that Sum_{k>=0} 1/(binomial(2*k,k)*(k+n))) = x*Pi*sqrt(3) - y*Pi^2 - z. (See examples for connection with a(n) in terms of material at Links section.)

%F f2(n) corresponds to the y values, so f2(n) = (1/9)*((2*n-1)!/((n-1)!)^2).

%e Examples ((3.79) - (3.83)) at page 14 in Links section as follows, respectively.

%e For n=1, f3(1) = 0, so a(1) = 0.

%e For n=2, f3(2) = 6, so a(2) = 6.

%e For n=3, f3(3) = 65/2, so a(3) = 130.

%e For n=4, f3(4) = 1379/9, so a(4) = 5516.

%e For n=5, f3(5) = 5525/8, so a(5) = 397800.

%t RecurrenceTable[{a[n] == (4*n-2) * ((n-1)*a[n-1] + (n-2)!^2), a[1] == 0}, a, {n, 1, 15}] (* _Vaclav Kotesovec_, Apr 13 2018 *)

%t Table[FullSimplify[Pi^2*Gamma[2*n]/9 - Gamma[n]^2 * HypergeometricPFQ[{1, n, n}, {n + 1/2, n + 1}, 1/4]/n], {n, 1, 15}] (* _Vaclav Kotesovec_, Apr 13 2018 *)

%t nmax = 15; Table[CoefficientList[Expand[FunctionExpand[ Table[-Sum[1/(Binomial[2*j, j]*(j + m)), {j, 0, Infinity}], {m, 1, nmax}]]], Pi][[n, 1]]*(n - 1)!^2, {n, 1, nmax}] (* _Vaclav Kotesovec_, Apr 13 2018 *)

%Y Cf. A302769.

%K nonn

%O 1,2

%A _Detlef Meya_, Apr 13 2018

%E More terms from _Altug Alkan_, Apr 13 2018

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