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a(n) = 4*((2*n-1)*a(n-1) + (-1)^n*(n-2)!), with a(1) = 8, n > 1.
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%I #14 Apr 18 2018 04:52:48

%S 8,100,1996,55896,2012232,88538304,4603991328,276239482560,

%T 18784284793920,1427605644499200,119918874136481280,

%U 11032536420570792960,1103253642056919628800,119151393342149235916800,13821561627689286458265600,1713873641833471869538099200,226231320722018281548331622400

%N a(n) = 4*((2*n-1)*a(n-1) + (-1)^n*(n-2)!), with a(1) = 8, 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, (3.89) - (3.93).

%F a(n) = f1(n)*(-1)^(n-1)*3*(n-1)!, where f1(n) corresponds to the x values such that Sum_{k>=0} (-1)^k/(binomial(2*k,k)*2^k*(k+n)) = x*log(2) + y*(log(2))^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)^n*2^n*((2*n-1)!/((n-1)!)^2).

%e Examples ((3.89) - (3.93)) at page 14 in Links section as follows, respectively.

%e For n=1, f1(1) = 8/3, so a(1) = 8.

%e For n=2, f1(2) = -100/3, so a(2) = 100.

%e For n=3, f1(3) = 998/3, so a(3) = 1996.

%e For n=4, f1(4) = -9316/3, so a(4) = 55896.

%e For n=5, f1(5) = 83843/3, so a(5) = 2012232.

%t RecurrenceTable[{a[n] == 4*((2*n-1)*a[n-1] + (-1)^n*(n-2)!), a[1] == 8}, a, {n, 1, 20}] (* _Altug Alkan_, Apr 16 2018 *)

%t nmax = 15; Table[CoefficientList[TrigToExp[Expand[FunctionExpand[Table[Sum[(-1)^j/(Binomial[2*j, j]*2^j*(j + m)), {j, 0, Infinity}], {m, 1, nmax}]]]], Log[2]][[n, 2]]*(-1)^(n - 1)*3*(n - 1)!, {n, 1, nmax}] (* _Vaclav Kotesovec_, Apr 18 2018 *)

%o (PARI) a=vector(20); a[1]=8; for(n=2, #a, a[n]=4*((2*n-1)*a[n-1] + (-1)^n*((n-2)!))); a \\

%Y Cf. A302945.

%K nonn

%O 1,1

%A _Detlef Meya_, Apr 16 2018