A376466 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n-1, k).

1, 3, 127, 9435, 866751, 89591753, 9988439203, 1173951006987, 143456999185855, 18063466831218981, 2329136571942011877, 306174745758226208537, 40896708938016175140963, 5536767359542664588001285, 758259747093486125157272779, 104880152366856305370319427435, 14632959744552362547801104612799

Offset: 0

Author

Peter Bala, Sep 25 2024

Keywords

• nonn
• easy

Comments

Cf. A005258(n) = Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(n+k, k)^2.

The sequence of Apéry numbers A005258 satisfies the pair of supercongruences

1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r

and

2) A005258(n*p^r - 1) == A005258(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.

We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Crossrefs

Cf. A005258, A376458 - A376465.

Programs

Maple

A108625(n, k) := add( binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n-1, k), k = 0..n):
seq(a(n), n = 0..25);

Formula

Examples of supercongruences:

a(11) - a(1) = 306174745758226208537 - 3 = 2*(11^3)*17*79367*85245689663 == 0 (mod 11^3).

a(10) - a(0) = 2329136571942011877 - 1 = (2^2)*(11^3)*17011*25717400209 == 0 (mod 11^3).