A376464 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, n-k).

1, 7, 307, 20167, 1667251, 157682257, 16284029959, 1790285284039, 206214335367859, 24625438825569757, 3026463150129458557, 380777119628689277809, 48850700118326352063559, 6371139828304049908181989, 842708419059535649574061807, 112830751689834347753141396167, 15268616117602266095803172887219

Offset: 0

Author

Peter Bala, Sep 24 2024

Keywords

• nonn
• easy

Comments

The sequence of Apéry numbers A005259 defined by A005259(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2 satisfies the pair of supercongruences

1) A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r

and

2) A005259(n*p^r - 1) == A005259(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. A005259, A376458 - A376466.

Programs

Maple

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

Formula

Examples of supercongruences:

a(11) - a(1) = 380777119628689277809 - 7 = 2*(3^3)*7*(11^3)*(19^2)*83*103*587*417773 == 0 (mod 11^3).

a(10) - a(0) = 3026463150129458557 - 1 = (2^2)*3*(11^3)*17*19*191*251*12236761 == 0 (mod 11^3).