A364516 a(n) = (2/3) * (8*n)!*(3*n)!^2/((6*n)!*(4*n)!*(2*n)!*n!^2) for n >= 1, with a(0) = 1.

1, 28, 3900, 685216, 133501500, 27583083528, 5919115212192, 1304298034300800, 293086491979934268, 66857471357130883000, 15434267149448839091400, 3597756971630997935635200, 845406187463509505329860000, 200002748013094535687584437696

Offset

0,2

Comments

Row 6 of A364513.

Links

Paolo Xausa, Table of n, a(n) for n = 0..400

Formula

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

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

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)).

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

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.

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.

Example

Examples of supercongruences:
a(7) - a(1) = 1304298034300800 - 28 = (2^2)*(7^4)*103553*1311481 == 0 (mod 7^4).
a(11) - a(1) = 3597756971630997935635200 - 28 = (2^2)*(11^3)*22567*7702811* 3887502719 == 0 (mod 11^3).

Maple

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

Mathematica

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 *)

Crossrefs

Cf. A364513.

Sequence in context: A201099 A290214 A036525 * A355999 A193985 A262018

Adjacent sequences: A364513 A364514 A364515 * A364517 A364518 A364519

Keyword

nonn,easy

Author

Peter Bala, Aug 02 2023

Status

approved