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%I #10 Oct 08 2022 10:18:19
%S 0,4,0,94,500,19262,50421,2929583,25197642,2007045752,3634262225,
%T 368738402141,6908530637021,852421484283739,1168833981781025,
%U 56641833705924527,276827636652242789,46345946530867053437,51051733540797155872,9673584199611903429172
%N a(n) = the numerator of ( Sum_{k = 1..n} (-1)^(n+k)*(1/k)*binomial(n,k)* binomial(n+k,k)^2 ).
%C 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 of the two types of Apéry numbers. The present sequence is the case m = 1. See A357561 for the case m = 3.
%C Conjectures:
%C 1) for even m >= 2, S_m(p-1) == 0 (mod p^3) for all primes p > m + 3.
%C 2) for odd m >= 1, S_m(p-1) == 0 (mod p^4) for all primes p > m + 4.
%H A. Straub, <a href="https://arxiv.org/abs/1401.0854">Multivariate Apéry numbers and supercongruences of rational functions</a>, arXiv:1401.0854 [math.NT] (2014).
%F Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 7 (checked up to p = 499).
%F 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).
%e Example of a supercongruence:
%e p = 19: a(19 - 1) = 51051733540797155872 = (2^5)*(19^4)*12241823444801 == 0 (mod 19^4).
%p seq( numer(add( (-1)^(n+k) * (1/k) * binomial(n,k) * binomial(n+k,k)^2, k = 1..n )), n = 0..20 );
%Y Cf. A005258, A357510, A357511, A357512, A357513, A357558, A357559, A357561.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Oct 04 2022
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