A376460 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n, k).

1, 11, 559, 42923, 3996751, 416148761, 46607742091, 5494953548459, 673006663951183, 84887862132737261, 10960384987436995309, 1442356991928315362009, 192832695330486199076011, 26126080107002322789528197, 3580244300364122205875141059, 495479786625880934937914634923, 69162655828624756613259127413583

Offset: 0

Author

Peter Bala, Sep 24 2024

Keywords

• nonn
• easy

Comments

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

Examples

			Examples of supercongruences:
(1a) a(13) - a(1) = 26126080107002322789528197 - 11 = 2*3*(7^3)*11*(13^3)*101* 5200975019533651  == 0 (mod 13^3);
(1b) a(13 - 1) - a(1 - 1) = 192832695330486199076011 - 1 = 2*3*5*(13^3)*107*457* 1366649*43779661 == 0 (mod 13^3).
(2a) a(5^2) - a(5) = 1806832316206138774124773283611717478292644188023761 - 416148761 = (2^3)*(3^3)*(5^8)*953*22470418607359387809254496551076506667197 = 0 (mod 5^8);
(2b) a(5^2 - 1) - a(5 - 1) = 12360782465873293068286987426696670402529985934251 - 3996751 = (2^2)*3*(5^6)*11779*1268093*9320371*634752259*746015256807592939 == 0 (mod 5^6).
		

Crossrefs

Cf. A005258, A376458 - A376466.

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