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A302945 a(n) = 4*(2*n+1)(n*a(n-1) + (-1)^(n-1)*((n-1)!)^2), with a(0) = 0, n > 0. 2

%I #21 Dec 02 2020 17:51:15

%S 0,12,460,38752,5578992,1227403584,382949169408,160838682255360,

%T 87496241419607040,59847429254564597760,50271840562772982988800,

%U 50875102650737732213145600,61050123180725943563550720000,85714372945764004556767887360000,139200141663916245408777987686400000

%N a(n) = 4*(2*n+1)(n*a(n-1) + (-1)^(n-1)*((n-1)!)^2), with a(0) = 0, n > 0.

%H Robert Israel, <a href="/A302945/b302945.txt">Table of n, a(n) for n = 0..212</a>

%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-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)*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).

%F (2*n + 1)*a(n + 1) - (14*n^3 + 45*n^2 + 44*n + 12)*a(n) -4*n^3*(4*n^2 + 8*n + 3)*a(n - 1) = 0. - _Robert Israel_, Dec 02 2020

%e Examples ((3.89) - (3.93)) 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) = 12, so a(1) = 12.

%e For n=3, f3(3) = -115, so a(2) = 460.

%e For n=4, f3(4) = 9688/9, so a(3) = 38752.

%e For n=5, f3(5) = -38743/4, so a(4) = 5578992.

%p f:= gfun:-rectoproc({(2*n + 1)*a(n + 1) - (14*n^3 + 45*n^2 + 44*n + 12)*a(n) -4*n^3*(4*n^2 + 8*n + 3)*a(n - 1) = 0, a(0)=0, a(1)=12}, a(n),remember):

%p map(f, [$0..40]); # _Robert Israel_, Dec 02 2020

%t RecurrenceTable[{a[n] == 4*(2*n+1)*(n*a[n-1] + (-1)^(n-1)*((n-1)!)^2), a[0] == 0}, a, {n, 0, 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, 1]]*(-1)^n*(n - 1)!^2, {n, 1, nmax}] (* _Vaclav Kotesovec_, Apr 18 2018 *)

%o (PARI) a=vector(20); a[1]=12; for(n=2, #a, a[n]=4*(2*n+1)*(n*a[n-1] + (-1)^(n-1)*((n-1)!)^2)); concat(0, a) \\

%Y Cf. A302944.

%K nonn

%O 0,2

%A _Detlef Meya_, Apr 16 2018

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