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A364512 a(n) = (6*n)!^2/((5*n)! * (3*n)!^2 * n!).

%I A364512 #19 Oct 13 2024 07:08:59
%S A364512 1,120,60984,39673920,28734361656,22105177305120,17676475936257600,
%T A364512 14521297485225136320,12168600808728479801400,
%U A364512 10353699767677668805341120,8916443122582617618026013984,7754263877699070505609688536320,6798445963232542402250454047721024
%N A364512 a(n) = (6*n)!^2/((5*n)! * (3*n)!^2 * n!).
%C A364512 Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L  we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence.
%C A364512 The present sequence is an integral factorial ratio sequence of height 2.
%C A364512 It is the particular case a = 2, b = 3 of the 2-parameter family of factorial ratio sequences defined by u(n) = (2*n)!/n! * ((a+b+1)*n)! * (2*a*n)! * (2*b*n)! /( ((a+b)*n)! * ((a+1)*n)! * ((b+1)*n)! * (a*n)! * (b*n)! ). These sequences are shown to be integral by the identity u(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i) * binomial(2*a*n, a*n+i) * binomial(2*b*n, b*n+i).
%C A364512 For other cases see A006480 (a = b = 1), A364507 (a = b = 2), A364508 (a = b = 3), A364510 (a = 1, b = 2) and A364511 (a = 1, b = 3).
%H A364512 Paolo Xausa, <a href="/A364512/b364512.txt">Table of n, a(n) for n = 0..300</a>
%H A364512 Romeo Meštrović, <a href="https://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012)</a>, arXiv:1111.3057 [math.NT], (2011).
%H A364512 Wikipedia, <a href="https://en.wikipedia.org/wiki/Dixon%27s_identity">Dixon's identity</a>
%F A364512 a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i) * binomial(4*n, 2*n+i) * binomial(6*n, 3*n+i), showing that the sequence is integral. Compare with Dixon's identity Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.
%F A364512 a(n) = (-1)^n * (6*n)!/((3*n)!*(2*n)!*n!) * hypergeom([-2*n, -3*n, -4*n], [n + 1, 2*n + 1], 1).
%F A364512 P-recursive: a(n) = (576/5)*(2*n-1)^2*(6*n-1)^2*(6*n-5)^2/((5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*n^2) * a(n-1) with a(0) = 1.
%F A364512 a(n) ~ c^n * 1/(sqrt(5)*Pi*n), where c = (2^12)*(3^6)/(5^5).
%F A364512 a(n) = [x^n] G(x)^(24*n), where the power series G(x) = 1 + 5*x + 683*x^2 + 205020*x^3 + 81906321*x^4 + 38109640996*x^5 + 19499018805299*x^6 + 10646310099966919*x^7 + 6093917580539621690*x^8 + ... appears to have integer coefficients.
%F A364512 exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^24, where the power series F(x) = 1 + 5*x + 1283*x^2 + 557400*x^3 + 302894393*x^4 + 186417421346*x^5 + 124214055930695*x^6 + 87454455447781703*x^7 + 64116544959085589954*x^8 + ... appears to have integer coefficients.
%F A364512 Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3^r)) hold for all primes p >= 5 and all positive integers n and r [added Oct 11 2024: follows from Meštrović, Section 6, equation 39, since a(n) = binomial(6*n, 3*n)*binomial(6*n, n)].
%p A364512 seq( (6*n)!^2/((5*n)! * (3*n)!^2 * n!), n = 0..15);
%t A364512 A364512[n_]:=(6n)!^2/((5n)!(3n)!^2n!);Array[A364512,15,0] (* _Paolo Xausa_, Oct 05 2023 *)
%Y A364512 Cf. A006480, A364507, A364508, A364510, A364511.
%K A364512 nonn,easy
%O A364512 0,2
%A A364512 _Peter Bala_, Jul 29 2023