%I #13 Dec 28 2023 11:32:15
%S 1,105,25740,7759752,2574148500,902522205585,328074738591600,
%T 122332313750680800,46485667563689950596,17924037162454524601500,
%U 6991900927489809108938160,2753354160571011216583946400,1092796344333659321191117573200,436609643814534385348768088729640,175431288302508215774213129529432000
%N a(n) = (6*n+1)!/(n!*(2*n)!*(3*n+1)!).
%F G.f.: hypergeometric3F2([1/3, 5/6, 7/6], [1, 4/3], 432*z).
%F E.g.f.: hypergeometric3F3([1/3, 5/6, 7/6], [1, 1, 4/3], 432*z).
%F a(n) = Integral_{x=0..432} x^n*W(x) dx, n>=0, where W(x) = sqrt(3)/(12*Pi*x^(2/3)) - gamma(2/3)*gamma(5/6)*sqrt(3)*hypergeometric3F2([1/2, 5/6, 5/6], [2/3, 3/2], x/432)/(432*Pi^(5/2)*x^(1/6)) + x^(1/6)*hypergeometric3F2([5/6, 7/6, 7/6], [4/3, 11/6], x/432)/(12960*sqrt(Pi)*gamma(2/3)*gamma(5/6)). W(x) is positive on x = [0, 432], it diverges at x=0, and monotonically decreases for x>0. It appears that at x=432, W(x) tends to a constant value close to 0.0007368284. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 432] is unique, as W(x) is the solution of the Hausdorff moment problem.
%F The shape of W(x) in the above integral representation of a(n) resembles very much the shape of the corresponding W(x) in A113424.
%p seq((6*n+1)!/(n!*(2*n)!*(3*n+1)!),n=0..14)
%Y Cf. A113424.
%K nonn
%O 0,2
%A _Karol A. Penson_, Dec 28 2023