A364516 - OEIS

%I #14 Oct 05 2023 08:37:56

%S 1,28,3900,685216,133501500,27583083528,5919115212192,

%T 1304298034300800,293086491979934268,66857471357130883000,

%U 15434267149448839091400,3597756971630997935635200,845406187463509505329860000,200002748013094535687584437696

%N a(n) = (2/3) * (8*n)!*(3*n)!^2/((6*n)!*(4*n)!*(2*n)!*n!^2) for n >= 1, with a(0) = 1.

%C Row 6 of A364513.

%H Paolo Xausa, <a href="/A364516/b364516.txt">Table of n, a(n) for n = 0..400</a>

%F a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(6*n-1, (1 + x)/(1 - x)) for n >= 1.

%F a(n) = Sum_{k = 0..n} binomial(6*n - 1, n - k)^2 * binomial(4*n + k - 2, k).

%F a(n) = (6*n-1)!*(4*n-1/2)!*(2*n-1/2)!/((4*n-1)! * (3*n-1/2)!^2 * n!^2) for n >= 1 (fractional factorials are defined in terms of the gamma function, for example, (4*n - 1/2)! = gamma(4*n + 1/2)).

%F a(n) ~ 2^(8*n) * sqrt(6)/(6*Pi*n).

%F P-recursive: a(0) = 1; for n >= 1, a(n) = (8*n-1)*(8*n-3)*(8*n-5)*(8*n-7)*(3*n-1)*(3*n-2)/((6*n-1)*(6*n-5)*(2*n-1)^2*n^2) * a(n-1) with a(1) = 28.

%F 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.

%e Examples of supercongruences:

%e a(7) - a(1) = 1304298034300800 - 28 = (2^2)*(7^4)*103553*1311481 == 0 (mod 7^4).

%e a(11) - a(1) = 3597756971630997935635200 - 28 = (2^2)*(11^3)*22567*7702811* 3887502719 == 0 (mod 11^3).

%p seq( (2/3) * (8*n)!*(3*n)!^2/((6*n)!*(4*n)!*(2*n)!*n!^2), n = 0..15);

%t A364516[n_]:=If[n==0,1,(2/3)(8n)!(3n)!^2/((6n)!(4n)!(2n)!n!^2)];Array[A364516,15,0] (* _Paolo Xausa_, Oct 05 2023 *)

%Y Cf. A364513.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Aug 02 2023