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a(n) = 2*(n*(2*n+1)*b(n-1) + (-1)^(n-1)*(2*n+1)*((n-1)!)^2), with a(0) = 0, n > 1.
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%I #13 Apr 18 2018 04:51:01

%S 0,6,110,4676,336024,36975312,5767774272,1211248149120,

%T 329458632906240,112674914230625280,47323458446223052800,

%U 23945670579525578956800,14367402268047801262080000,10085916404559453257318400000,8189764118253280338411847680000,7616480630446754091029242183680000

%N a(n) = 2*(n*(2*n+1)*b(n-1) + (-1)^(n-1)*(2*n+1)*((n-1)!)^2), with a(0) = 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, (3.84) - (3.88).

%F a(n-1) = f3(n)*(-1)^n*((n-1)!)^2, where f3(n) corresponds to the z values such that Sum_{k>=0} (-1)^k/(binomial(2*k,k)*(k+n)) = x*sqrt(5)*log(phi) + y*(log(phi))^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*4*((2*n-1)!/((n-1)!)^2).

%e Examples ((3.83) - (3.88)) at page 14 in Links section as follows, respectively.

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

%e For n=2, f3(2) = 6, so a(1) = 6.

%e For n=3, f3(3) = -55/2, so a(2) = 110.

%e For n=4, f3(4) = 1169/9, so a(3) = 4676.

%e For n=5, f3(5) = -4667/8, so a(4) = 336024.

%t Fold[Append[#1, 2 (#2 (2 #2 + 1) #1[[#2]] + (-1)^(#2 - 1) (2 #2 + 1) ((#2 - 1)!)^2)] &, {0}, Range@ 19] (* _Michael De Vlieger_, Apr 14 2018 *)

%t nmax = 15; Table[CoefficientList[TrigToExp[Expand[FunctionExpand[ Table[Sum[(-1)^j/(Binomial[2*j, j]*(j + m)), {j, 0, Infinity}], {m, 1, nmax}]]]], Log[1/2 + Sqrt[5]/2]][[n, 1]]*(-1)^n*(n - 1)!^2, {n, 1, nmax}] (* _Vaclav Kotesovec_, Apr 15 2018 *)

%o (PARI) a=vector(20); a[1]=0; for(n=2, #a, a[n]=2*((n-1)*(2*n-1)*a[n-1] + (-1)^(n)*(2*n-1)*((n-2)!)^2)); a \\

%Y Cf. A302865.

%K nonn

%O 0,2

%A _Detlef Meya_, Apr 14 2018