%I #8 May 12 2016 12:46:43
%S 1,60,151200,1816214400,60339831552000,4274473667143680000,
%T 553761949463615692800000,118528911546113729396736000000,
%U 38996486014317601426443730944000000
%N a(n) = (5*n)!/(2*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:
%C a(n) = int(x^n*((1/3125)*sqrt(5)*((625/4)*Pi*csc((2/5)*Pi)*csc((1/5)*Pi)*sin((3/10)*Pi)*GAMMA(7/10)*2^(2/5)*hypergeom([7/10], [2/5, 3/5, 4/5], -(4/3125)*x)/(GAMMA(4/5)*x^(4/5)) - (625/4)*Pi*sec((3/10)*Pi) *csc((2/5)*Pi)*sin((1/10)*Pi)*GAMMA(9/10)*2^(4/5)*hypergeom([9/10], [3/5, 4/5, 6/5], -(4/3125)*x)/(GAMMA(3/5)*x^(3/5)) - (125/8)*Pi*sec((3/10)*Pi)*GAMMA(3/5)*2^(1/5)*hypergeom([11/10], [4/5, 6/5, 7/5], -(4/3125)*x)/(GAMMA(9/10)*x^(2/5)) + (125/8)*Pi*csc((2/5)*Pi)*GAMMA(4/5)*2^(3/5)*hypergeom([13/10], [6/5, 7/5,8/5], -(4/3125)*x)/(GAMMA(7/10)*x^(1/5)))/Pi^(3/2)), x=0..infinity),n=0,1... .
%C This solution is not unique.
%C From _G. C. Greubel_, May 12 2016: (Start)
%C 2^n * 10^(floor[n/2]) | a(n), for n>=0.
%C 2^(2*n) * 10^(floor[n/2]) | a(n), for n>=2. (End)
%H G. C. Greubel, <a href="/A166384/b166384.txt">Table of n, a(n) for n = 0..100</a>
%F G.f.: sum(a(n)*x^n/((n!)^4),n=0..infinity) = hypergeom([1/5, 2/5, 3/5, 4/5],[1/2, 1, 1, 1], (3125/4)*x).
%F asymptotics: a(n) = sqrt(10)*(1/2-1/(80*n)+1/(6400*n^2)+(619/3840000)*sqrt(2)/n^3+O(1/n^4))*(5^n)^5/(((1/n)^n)^3*(exp(n))^3*(2^n)^2),n->infinity.
%t Table[ (5*n)!/(2*n)!, {n, 0, 25}](* _G. C. Greubel_, May 12 2016 *)
%K nonn
%O 0,2
%A _Karol A. Penson_, Oct 13 2009