A357513 a(n) = numerator of Sum_{k = 1..n} (1/k^3) * binomial(n,k)^2 * binomial(n+k,k)^2 for n >= 1 with a(0) = 0.

0, 4, 81, 14651, 956875, 1335793103, 697621869, 3929170277787, 573290332967211, 8235727724024089939, 172296487023049395523, 5032311952710217004416313, 114828404520381550476341513, 5947240175728534283432460589661, 144126887537331651710781931325261

Offset: 0

Author

Peter Bala, Oct 02 2022

Keywords

• nonn
• easy

Comments

Let m be a nonnegative integer and set u(n) = the numerator of Sum_{k = 0..n} 1/k^(2*m+1) * binomial(n,k)^2 * binomial(n+k,k)^2. We conjecture that u(p-1) == 0 (mod p^4) for all primes p, with a finite number of exceptions that depend on m.

Examples

			a(11 - 1) = 172296487023049395523 = (11^4)*(43^2)*6163*1032705769 == 0 (mod 11^4).
		

Links

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

Crossrefs

Cf. A005259, A357510, A357511, A357512.

Programs

Maple

seq(numer(add( (1/k^3) * binomial(n,k)^2 * binomial(n+k,k)^2, k = 1..n )), n = 0..20);

PARI

a(n) = if (n, numerator(sum(k=1, n, binomial(n,k)^2*binomial(n+k,k)^2/k^3)), 0); \\ Michel Marcus, Oct 04 2022

Formula

Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 3 except p = 7 (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).