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%I #42 Apr 12 2018 05:07:54
%S 0,0,1,50,6027,1350948,486396405,256822659990,186967652864895,
%T 179489092842045000,219694686618136235625,333935935534086791456250,
%U 617113613582364168765061875,1362586861058382580086938587500,3542725840051847141662287901708125
%N a(n) = 8*(n-2)*(2*n-5)*a(n-1) + ((n-2)/9)*Product_{k=0..n-2} (2*k-3)^2 with a(1) = 0.
%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. 10, (3.43) - (3.47).
%F a(n) = (f2(n-1)/8)*Product_{k=0..n-2} (2*k-1)^2 where f2(n) corresponds to the y values such that Sum_{k>=0} 1/(binomial(2*k,k)*(2*k+(2*n-1))) = x*Pi*sqrt(3) - y. (See examples for connection with a(n) in terms of material at Links section).
%e Examples ((3.43) - (3.47)) at page 10 in Links section as follows, respectively.
%e For n=1, f2(1) = 0, so a(2) = 0.
%e For n=2, f2(2) = 8, so a(3) = 1.
%e For n=3, f2(3) = 400/9, so a(4) = 50.
%e For n=4, f2(4) = 16072/75, so a(5) = 6027.
%e For n=5, f2(5) = 3602528/3675, so a(6) = 1350948.
%t RecurrenceTable[{b[n + 1] == 8*(n - 1)*(2*n - 3)*b[n] + (n - 1)/9 * Product[(2*k - 3)^2, {k, 0, n - 1}], b[1] == 0}, b, {n, 1, 20}] (* _Vaclav Kotesovec_, Mar 30 2018 *)
%t nmax = 15; Flatten[{0, Table[CoefficientList[Expand[FunctionExpand[Table[ FullSimplify[-Sum[1/(Binomial[2*j, j]*(2*j + (2*m - 1))), {j, 0, Infinity}]]*Product[(2*k - 1)^2, {k, 0, m - 1}]/8, {m, 0, nmax}]]], Pi][[n, 1]], {n, 2, nmax}]}] (* _Vaclav Kotesovec_, Apr 12 2018 *)
%o (PARI) a=vector(20); a[1]=0; for(n=2, #a, a[n]=8*(n-2)*(2*n-5)*a[n-1] + (n-2)*prod(k=0, n-2, (2*k-3)^2)/9); a \\ _Altug Alkan_, Mar 30 2018
%Y Cf. A000984, A301990.
%K nonn
%O 1,4
%A _Detlef Meya_, Mar 30 2018
%E More terms from _Vaclav Kotesovec_, Mar 30 2018
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