A364243 a(n) = A108625(2*n-1, n-1) for n >= 1.

1, 13, 271, 6637, 176251, 4914427, 141573251, 4174790893, 125288929171, 3811804637263, 117248436333601, 3638993432201563, 113790712076898871, 3580847269415337487, 113299135244467189771, 3601766951734150461677, 114973519461796962202051

Offset: 1

Author

Peter Bala, Jul 16 2023

Keywords

• nonn
• easy

Comments

The sequence of Apéry numbers A005258 forms the main diagonal of A108625, i.e., A005258(n) = A108625(n, n). The Apéry numbers satisfy the supercongruences A005258(n*p^r) == A005258(n^p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

More generally, for positive integers r and s, the sequence defined by a(r,s;n) = A108625(r*n - 1, s*n - 1) may also satisfy the same supercongruences. This is the case r = 2, s = 1. Compare with the comments in A363867.

Crossrefs

Cf. A005258, A108625, A363867, A364244.

Programs

Maple

seq(add(binomial(2*n-1,k)^2 * binomial(3*n-2-k,2*n-1), k = 0..n-1), n = 1..20);
# alternative program
seq( simplify(binomial(2*n-1,n-1)^2 * hypergeom([1 - n, 1 - n, 2*n], [1 + n, 1 + n], 1)), n = 1..20);

Mathematica

Table[Binomial[2*n-1, n-1]^2 * HypergeometricPFQ[{1 - n, 1 - n, 2*n}, {1 + n, 1 + n}, 1], {n, 1, 20}] (* Vaclav Kotesovec, Jul 16 2023 *)

Formula

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

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

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

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

P-recursive: (n-1)^2*(2*n-1)^2*(48*n^2-162*n+137)*a(n) = (6528*n^6-48144*n^5+144800*n^4-226714*n^3+194349*n^2-86299*n+15515)*a(n-1) - (2*n-3)^2*(n-2)^2*(48*n^2-66*n+23)*a(n-2) with a(1) = 1 and a(2) = 13.

a(n) ~ (1 + sqrt(2))^(4*n - 3/2) / (2^(9/4)*Pi*n). - Vaclav Kotesovec, Jul 16 2023