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%I #36 Feb 03 2022 00:51:09
%S 1,30,3150,420420,62355150,9816086280,1605660228900,269764879032000,
%T 46225898052627150,8042050347997165500,1415997888807961859400,
%U 251762943910387780962000,45125969443194371927422500,8143514687130622653091029120,1478138194032735032800001630400
%N a(n) = (5*n)!/((2*n)!*(2*n)!*n!).
%H James Spahlinger, <a href="/A001459/b001459.txt">Table of n, a(n) for n = 0..400</a>
%H Romeo Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.
%F From _Ilya Gutkovskiy_, Feb 07 2017: (Start)
%F O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1/2,1/2,1; 3125*x/16).
%F E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1/2,1/2,1,1; 3125*x/16).
%F a(n) ~ 5^(5*n+1/2)/(4*Pi*16^n*n). (End)
%F From _Peter Bala_, Sep 20 2021: (Start)
%F a(n) = 5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/(4*n^2*(2*n - 1)^2)*a(n-1).
%F a(n) = Sum_{k = n..3*n} binomial(3*n,k)^2*binomial(k,n). Cf. A006480.
%F Congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(5*n,2*n)*C(3*n,n) and apply Mestrovic, equation 39, p. 12). (End)
%p f := n->(5*n)!/((2*n)!*(2*n)!*n!);
%t Table[(5 n)!/((2 n)! (2 n)!*n!), {n, 0, 12}] (* _Michael De Vlieger_, Feb 07 2017 *)
%Y Cf. A006480.
%K nonn
%O 0,2
%A _N. J. A. Sloane_
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