A364508 - OEIS

%I #18 Oct 13 2024 07:07:47

%S 1,350,594594,1299170600,3164045050530,8188909171581600,

%T 22035578229399735000,60924423899585957558848,

%U 171839010049825493742617250,492149504510899056782561257748,1426695143534668869395862598229344,4176678405144148418744441910948978000

%N a(n) = (7*n)!*(6*n)!*(2*n)! / ((4*n)!^2 * (3*n)!^2 * n!).

%C Row 3 of A364506.

%H Paolo Xausa, <a href="/A364508/b364508.txt">Table of n, a(n) for n = 0..250</a>

%H 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).

%F a(n) = Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k) * binomial(6*n, 3*n + k)^2 (showing a(n) to be integral). Compare with Dixon's identity  Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k)^3 = (3*n)!/n!^3 = A006480(n).

%F P-recursive: a(n) = (7/4)*(6*n-1)*(6*n-5)*(7*n-1)*(7*n-2)*(7*n-3)*(7*n-4)*(7*n-5)*(7*n-6)/((3*n-1)*(3*n-2)*(4*n-1)^2*(4*n-3)^2*n^2) * a(n-1) with a(0) = 1.

%F a(n) ~ c^n * sqrt(21)/(12*Pi*n), where c = (7^7)/(2^8).

%F a(n) = [x^n] G(x)^(14*n), where the power series G(x) = 1 + 25*x + 12798*x^2 + 13543850*x^3 + 18933663145*x^4 + 30733263922830*x^5 + 54771428143877503*x^6 + 104061045049532102971*x^7 + 207134582792235253663131*x^8 + ... appears to have integer coefficients.

%F exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^14, where the power series F(x) = 1 + 25*x + 21548*x^2 + 31466125*x^3 + 57506245907*x^4 + 119069165444705*x^5 + 266966985031172547*x^6 + 632553825380957995891*x^7 + 1560815989686060202098169*x^8 + ... appears to have integer coefficients.

%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 [added Oct 11 2024: follows from Meštrović, Section 6, equation 39, since a(n) = binomial(7*n, 3*n)*binomial(6*n, 3*n) *binomial(2*n, n)/binomial(4*n, n)].

%e Examples of supercongruences:

%e a(7) - a(1) = 60924423899585957558848 - 350 = 2*(7^4)*12687301936606821649 == 0 (mod 7^4).

%e a(11) - a(1) = 4176678405144148418744441910948978000 - 350 = 2*(5^2)*7*(11^3)* 29*139*1181*1883311859633620981885519 == 0 (mod 11^3).

%p seq( (7*n)!*(6*n)!*(2*n)! / ((4*n)!^2 * (3*n)!^2 * n!), n = 0..15);

%t A364508[n_]:=(7n)!(6n)!(2n)!/((4n)!^2(3n)!^2n!);Array[A364508,15,0] (* _Paolo Xausa_, Oct 06 2023 *)

%Y Cf. A006480, A364506, A364507.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Jul 27 2023