A357561 a(n) = the numerator of ( Sum_{k = 1..n} (-1)^(n+k)*(1/k^3)*binomial(n,k)* binomial(n+k,k)^2 ).

0, 4, -27, 1367, -15625, 3129353, -14749, 308477847, 14343020119, 80826490175689, 618729030402659, 6526775794564145231, 52975460244520902439, 965428117884339747694757, 8161435689582967449592663, 70159702295938799645630801, 4897311439674525483507166097, 212741477113936719632186271679919

Offset: 0

Author

Peter Bala, Oct 04 2022

Keywords

• sign
• easy

Comments

Define S_m(n) = the numerator of Sum_{k = 1..n} (-1)^(n+k)*(1/k^m)*binomial(n,k)* binomial(n+k,k)^2, so that S_0(n) = -1 + A005258(n), one type of Apéry numbers. The present sequence is the case m = 3. See A357560 for the case m = 1.

Conjecture: for odd m >= 1, S_m(p-1) == 0 (mod p^4) for all primes p > m+4.

Examples

			Example of a supercongruence:
p = 17: a(17 - 1) = 212741477113936719632186271679919 = (17^4)*4871421029* 12036670481533 == 0 (mod 17^4).
		

Links

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

Crossrefs

Cf. A005258, A357510, A357511, A357512, A357513, A357558, A357559, A357560.

Programs

Maple

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

Formula

Conjecture: a(p-1) == 0 (mod p^4) for p = 5 and all primes p >= 11 (checked up to p = 499).

Note: the Apéry numbers B(n) = A005258(n) = Sum_{k = 0..n} (-1)^(n+k)* binomial(n,k)*binomial(n+k,k)^2 satisfy the supercongruences B(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Example 3.4).