A363869 - OEIS

A363869

a(n) = A108625(3*n, 2*n).

1, 55, 12559, 3685123, 1205189519, 418856591055, 151353475289275, 56193989426243199, 21283943385478109071, 8185785098679048061837, 3186604888590691870779559, 1252744279186835597251089055, 496508748101370063304243706939, 198134918989716743103591120933103

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OFFSET

0,2

COMMENTS

a(n) = B(3*n, 2*n, 3*n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.

More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.

For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363868 (r = 3, s = 1), A363870 (r = 1, s = 3) and A363871 (r = 2, s = 3).

LINKS

Paolo Xausa, Table of n, a(n) for n = 0..350

Peter Bala, A recurrence for A363869

FORMULA

a(n) = Sum_{k = 0..2*n} binomial(3*n, k)^2 * binomial(5*n-k, 3*n).

a(n) = Sum_{k = 0..2*n} (-1)^k * binomial(3*n, k)*binomial(5*n-k, 3*n)^2.

a(n) = hypergeom( [-2*n, -3*n, 3*n+1], [1, 1], 1).

a(n) = [x^(2*n)] 1/(1 - x)*Legendre_P(3*n, (1 + x)/(1 - x)).

a(n) ~ 2^(4*n) * 3^(3*n) / (sqrt(5)*Pi*n). - Vaclav Kotesovec, Apr 27 2024

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