A368875 - OEIS

%I #43 Jan 11 2024 16:48:08

%S 6,16,105,1008,12012,164736,2494206,40646320,701149020,12655450080,

%T 237026033790,4577828250240,90739095674400,1838979005667840,

%U 37993593597567210,798259862714284080,17022152442879594780,367791659430639444000,8040845154302354844450

%N a(n) = 24*(3*n + 1)!/(n!*((n + 2)!)^2).

%C According to A. Adolphson and S. Sperber (see Links), see page 14, second equation after Eq.(7.4): for any two integers K, L, the ratios (3*K+1)!*(3*L+1)!/(K!*L!*((K+L+1)!)^2) are proven to be integers. Here a(n) results from K = 1 and L = n, n >= 0.

%H A. Adolphson and S. Sperber, <a href="https://arxiv.org/abs/2001.03296">On the integrality of hypergeometric series whose coefficients are factorial ratios</a>, arXiv:2001.03296 [math.NT], 2020.

%H A. Adolphson and S. Sperber, <a href="https://doi.org/10.4064/aa200427-5-4">On the integrality of hypergeometric series whose coefficients are factorial ratios</a>, Acta Arithmetica 200 (2021), no.1, 39-59.

%F G.f.:   6*hypergeometric3F2([2/3, 1, 4/3], [3, 3], 27*z).

%F G.f.:   -(hypergeometric2F1([-4/3, -2/3], [1], 27*z) - 1)/(3*z^2) + 8/z.

%F E.g.f.: 6*hypergeometric3F3([2/3, 1, 4/3], [3, 3, 1], 27*z).

%F a(n) = Integral_{x=0..27} x^n*W(x) dx, n >= 0, where

%F   W(x) = (243*2^(2/3)*Gamma(5/6)*Gamma(2/3)*hypergeometric2F1([-4/3, -4/3], [1/3], x/27)) / (16*Pi^(5/2)*x^(1/3)) - (3*sqrt(3)*2^(1/3)*x^(1/3)* hypergeometric2F1([-2/3, -2/3], [5/3], x/27))/(2*sqrt(Pi)*Gamma(5/6)* Gamma(2/3)).

%F W(x) is a positive function in the interval [0, 27], is singular at x = 0 with the singularity x^(-1/3), and monotonically decreases to zero at x = 27, with W'(x) tending to zero at x = 27. This integral representation as the n-th power moment of the positive function W(x) in the interval [0, 27] is unique, as W(x) is the solution of the Hausdorff moment problem.

%p seq(24*(3*n + 1)!/(n!*((n + 2)!)^2),n=0..17);

%t Table[24*(3*n + 1)!/(n!*((n + 2)!)^2),{n,0,16}] (* _James C. McMahon_, Jan 08 2024 *)

%o (SageMath)

%o def a(n): return (24 * (n + 1) * (n + 2) * gamma(3*n + 2)) / gamma(n + 3)^3

%o print([a(n) for n in range(19)])  # _Peter Luschny_, Jan 09 2024

%Y Cf. A368650, A304126, A368545, A082368, A113424, A368692.

%K nonn

%O 0,1

%A _Karol A. Penson_, Jan 08 2024