A364511 a(n) = (6*n)!*(5*n)!*(2*n)! / ((4*n)!^2 * (3*n)! * n!^2).

1, 50, 8910, 2011100, 503909070, 133954543800, 36992598142500, 10491379251679040, 3034472729231379150, 891028813210575018980, 264787855164104281785160, 79455812929030151249454000, 24035311050907120054564683300, 7320107028326385998504601648000

Offset

0,2

Comments

Row 4 of A364509.

Links

Paolo Xausa, Table of n, a(n) for n = 0..300

Wikipedia, Dixon's identity

Formula

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(6*n, 3*n+i).

Compare with Dixon's identity: Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.

a(n) = (-1)^n*binomial(6*n,2*n) * hypergeom([-2*n, -2*n, -4*n], [1, 2*n+1], 1).

P-recursive: a(n) = (15/4)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*(6*n-1)*(6*n-5)/((4*n-1)^2*(4*n-3)^2*n^2) * a(n-1) with a(0) = 1.

a(n) ~ c^n * sqrt(5)/(4*Pi*n), where c = (3^3)*(5^5)/(2^8).

a(n) = [x^n] G(x)^(10*n), where the power series G(x) = 1 + 5*x + 208*x^2 + 19960*x^3 + 2580710*x^4 + 390721786*x^5 + 65243160516*x^6 + 11646611942100*x^7 + 2182248792056787*x^8 + ... appears to have integer coefficients.

exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^10, where the power series F(x) = 1 + 5*x + 458*x^2 + 69285*x^3 + 13037740*x^4 + 2773287786*x^5 + 638122182196*x^6 + 155077758079485*x^7 + 39234250338228617*x^8 + ... appears to have integer coefficients.

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.

a(n) = (-1)^n * [x^(4*n)] (1 - x)^(8*n)*Legendre_P(2*n, (1 + x)/(1 - x)). - Peter Bala, Aug 14 2023

Maple

seq((6*n)!*(5*n)!*(2*n)! / ((4*n)!^2 * (3*n)! * n!^2), n = 0..15);

Mathematica

A364511[n_]:=(6n)!(5n)!(2n)!/((4n)!^2(3n)!n!^2); Array[A364511, 15, 0] (* Paolo Xausa, Oct 05 2023 *)

Crossrefs

Cf. A006480, A364509, A364510, A364512.

Sequence in context: A229753 A276102 A152258 * A203097 A028465 A252630

Adjacent sequences: A364508 A364509 A364510 * A364512 A364513 A364514

Keyword

nonn,easy

Author

Peter Bala, Jul 28 2023

Status

approved