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%I #16 Sep 08 2022 08:45:48
%S 1,3,90,7560,1247400,340540200,138940401600,79196028912000,
%T 60109785944208000,58607041295602800000,71383376298044210400000,
%U 106218463931489785075200000,189599958117709266359232000000
%N a(n) = (3*n)!/(2^n*n!).
%C Integral representation as n-th moment of a positive function on a positive halfaxis (solution of the Stieltjes moment problem), in Maple notation: a(n) = int(x^n*(1/3)*sqrt(2)*BesselK(1/3,(2/9)*sqrt(6*x))/(sqrt(x)*Pi), x=0..infinity), n=0,1... .
%C This solution is unique.
%H G. C. Greubel, <a href="/A166334/b166334.txt">Table of n, a(n) for n = 0..100</a>
%F G.f.: sum(a(n)*x^(n)/(n!)^2,n=0..infinity)=hypergeom([1/3, 2/3], [1], (27/2)*x).
%F Asymptotics: a(n)=(sqrt(3)-(1/18)*sqrt(3)/n+(1/648)*sqrt(3)/n^2 +(463/174960)*sqrt(3)/n^3+O(1/n^4))*(3^n)^3/(((1/n)^n)^2*(exp(n))^2*2^n), n->infinity.
%F E.g.f.: (of aerated sequence) 2*sqrt(2)*cos(arcsin((3*sqrt(6)x/4)/3))/sqrt(8-27x^2). - _Paul Barry_, Jul 27 2010
%F 2*a(n) = 3*(3*n-1)*(3*n-2)*a(n-1). - _R. J. Mathar_, Jul 24 2012
%t Table[(3*n)!/(2^n*n!), {n, 0, 10}] (* _G. C. Greubel_, May 09 2016 *)
%o (Magma) [Factorial(3*n)/(2^n*Factorial(n)): n in [0..20]]; // _Vincenzo Librandi_, May 10 2016
%K nonn
%O 0,2
%A _Karol A. Penson_, Oct 12 2009
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