A376465 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n-1, k).

1, 5, 289, 32909, 4846001, 824723005, 154077084505, 30725590395197, 6429986799134257, 1396511202452212733, 312375607924873231289, 71567605248444374973205, 16725718218774077760354953, 3974882968204513021199112653, 958241214962583413382405975337, 233879553339714596653197104362909

Offset

0,2

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

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.

Links

Table of n, a(n) for n=0..15.

Formula

a(n) ~ sqrt((41*cos(Pi/7) - 8) / (224*(44*cos(Pi/7) - 39))) * (1 + 14/(10*cos(Pi/7) - 9))^n / (Pi^(5/2) * n^(5/2) * 5^n). - Vaclav Kotesovec, May 08 2026

Example

Examples of supercongruences:
a(11) - a(1) = 71567605248444374973205 - 5 = (2^4)*(5^2)*(11^3)*19*56040893*126246629 == 0 (mod 11^3).
a(10) - a(0) = 312375607924873231289 - 1 = (2^3)*(11^3)*17*1725679541724893 == 0 (mod 11^3).

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-1, k), k = 0..n):
seq(a(n), n = 0..25);

Mathematica

Table[Sum[Binomial[n, k]^2 * Binomial[n+k, k]^2 * HypergeometricPFQ[{-n+1, -k, n}, {1, 1}, 1], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 08 2026 *)

Crossrefs

Cf. A005259, A376458 - A376466.

Sequence in context: A252173 A386657 A265966 * A337540 A354962 A227579

Adjacent sequences: A376462 A376463 A376464 * A376466 A376467 A376468

Keyword

nonn,easy

Author

Peter Bala, Sep 25 2024

Status

approved