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%I #58 Apr 11 2018 12:04:22
%S 0,0,16,1504,361392,161889216,116561117520,123088454370720,
%T 179216801666043120,344096256861334857600,842347637388219301894800,
%U 2560736817470194362347292000,9464483277445284311686053822000,41795158153162079160478354766472000
%N a(n) = 16*(n-1)*((2*n-3)*a(n-1) + (((-1)^n)/9)*Product_{k=0..n-1} (2*k-3)^2) with a(0) = 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. 11, (3.53) - (3.57).
%F a(n) = (-1)^n*f2(n)*Product_{k=0..n-1} (2*k-1)^2 where f2(n) corresponds to the y values such that Sum_{k>=0} (-1)^k/(binomial(2*k,k)*2^k*(2*k+(2*n-1))) = x*log(2) + y. (See examples for connection with a(n) in terms of material at Links section).
%e Examples ((3.53) - (3.57)) at page 11 in Links section as follows, respectively.
%e For n=1, f2(1) = 0, so a(1) = 0.
%e For n=2, f2(2) = 16, so a(2) = 16.
%e For n=3, f2(3) = -1504/9, so a(3) = 1504.
%e For n=4, f2(4) = 120464/75, so a(4) = 361392.
%e For n=5, f2(5) = -53963072/3675, so a(5) = 161889216.
%t RecurrenceTable[{a[n] == 16*(n-1)*((2*n-3)*a[n-1] + (-1)^n / 9 * Product[(2*k-3)^2, {k,0,n-1}]), a[0] == 0}, a, {n, 0, 15}] (* _Vaclav Kotesovec_, Apr 10 2018 *)
%t nmax = 15; Flatten[{0, Table[CoefficientList[TrigToExp[Expand[FunctionExpand[ Table[FullSimplify[Sum[(-1)^j/(Binomial[2*j, j]*2^j*(2*j + (2*m - 1))), {j, 0, Infinity}]]*(-1)^m*Product[(2*k - 1)^2, {k, 0, m - 1}], {m, 1, nmax}]]]], Log[2]][[n, 1]], {n, 1, nmax}]}] (* _Vaclav Kotesovec_, Apr 10 2018 *)
%o (PARI) a=vector(20); a[1]=0; for(n=2, #a, a[n]=16*(n-1)*((2*n-3)*a[n-1]+(((-1)^n)/9)*prod(k=0, n-1, (2*k-3)^2))); concat(0, a) \\ _Altug Alkan_, Apr 09 2018
%Y Cf. A302115.
%K nonn
%O 0,3
%A _Detlef Meya_, Apr 01 2018
%E More terms from _Altug Alkan_, Apr 09 2018
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