A357510 a(n) = Sum_{k = 0..n} k * binomial(n,k)^2 * binomial(n+k,k)^2.

0, 4, 108, 3144, 95000, 2935020, 92054340, 2918972560, 93330811440, 3003683380020, 97177865060540, 3157623679795992, 102973952434618824, 3368460743291372092, 110480459392323735540, 3631941224582026770720, 119637879389041977365600, 3947968300820696313987780

Offset

0,2

Links

Table of n, a(n) for n=0..17.

Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Formula

Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 5 (checked up to p = 499).

Note: the Apery numbers A(n) = A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2 satisfy the supercongruence A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction).

Recurrence: a(0) = 0, a(1) = 4, a(2) = 108 and thereafter n^2*(n-1)*(2*n-3)*(3*n^2-9*n+7)*a(n) = (2*n-1)*(105*n^5 - 576*n^4 + 1208*n^3 - 1195*n^2 + 556*n - 104)*a(n-1) - (2*n-3)*(105*n^5 - 474*n^4 + 800*n^3 - 629*n^2 + 240*n - 36)*a(n-2) + (n-1)*(2*n-1)*(3*n^2-3*n+1)*(n-2)^2*a(n-3).

a(n) ~ (1 + sqrt(2))^(4*n + 2) / (2^(11/4) * Pi^(3/2) * sqrt(n)). - Vaclav Kotesovec, Oct 04 2022

Example

a(11 - 1) = 97177865060540 = (2^2)*5*(11^4)*37*239*37529 == 0 (mod 11^4).

Maple

seq( add( k*binomial(n, k)^2 * binomial(n+k, k)^2, k = 0..n ), n = 0..20 );

Prog

(PARI) a(n) = sum(k = 0, n, k * binomial(n, k)^2 * binomial(n+k, k)^2); \\ Michel Marcus, Oct 04 2022

Crossrefs

Cf. A005259, A357506, A357507, A357511, A357512, A357513.

Sequence in context: A128865 A269270 A302113 * A336028 A131092 A360092

Adjacent sequences: A357507 A357508 A357509 * A357511 A357512 A357513

Keyword

nonn,easy

Author

Peter Bala, Oct 01 2022

Status

approved