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%I #39 Apr 11 2018 11:06:15
%S 0,4,68,2172,104268,6673092,533847780,51249383100,5739930948780,
%T 734711160903300,105798407178183300,16927745148371490300,
%U 2979283146116001209100,572022364054217234904900,118980651723278449796792100,26651665986014341131067107900
%N a(n) = 16*(n-1)*a(n-1) + ((-1)^n)*(4/3)*Product_{k=0..n-1} (2*k-3) 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*f1(n)*3*Product_{k=0..n-1} (2*k-1) where f1(n) corresponds to the x 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, f1(1) = 4/3, so a(1) = 4.
%e For n=2, f1(2) = -68/3, so a(2) = 68.
%e For n=3, f1(3) = 724/3, so a(3) = 2172.
%e For n=4, f1(4) = -34756/15, so a(4) = 104268.
%e For n=5, f1(5) = 2224364/105, so a(5) = 6673092.
%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 * 3 * Product[(2*k - 1), {k, 0, m - 1}], {m, 1, nmax}]]]], Log[2]][[n, 2]], {n, 1, nmax}]}] (* _Vaclav Kotesovec_, Apr 10 2018 *)
%t RecurrenceTable[{a[n] == 16*(n - 1)*a[n - 1] + (-1)^n*(4/3) * Product[(2*k - 3), {k, 0, n - 1}], a[0] == 0}, a, {n, 0, 15}] (* _Vaclav Kotesovec_, Apr 11 2018 *)
%o (PARI) a=vector(20); a[1]=4; for(n=2, #a, a[n]=16*(n-1)*a[n-1]+((-1)^n)*(4/3)*prod(k=0, n-1, (2*k-3))); concat(0, a) \\ _Altug Alkan_, Apr 10 2018
%Y Cf. A302116.
%K nonn
%O 0,2
%A _Detlef Meya_, Apr 01 2018
%E More terms from _Vaclav Kotesovec_, Apr 10 2018
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