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%I #11 Oct 08 2022 10:12:35
%S 0,4,81,14651,956875,1335793103,697621869,3929170277787,
%T 573290332967211,8235727724024089939,172296487023049395523,
%U 5032311952710217004416313,114828404520381550476341513,5947240175728534283432460589661,144126887537331651710781931325261
%N a(n) = numerator of Sum_{k = 1..n} (1/k^3) * binomial(n,k)^2 * binomial(n+k,k)^2 for n >= 1 with a(0) = 0
%C Let m be a nonnegative integer and set u(n) = the numerator of Sum_{k = 0..n} 1/k^(2*m+1) * binomial(n,k)^2 * binomial(n+k,k)^2. We conjecture that u(p-1) == 0 (mod p^4) for all primes p, with a finite number of exceptions that depend on m.
%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 >= 3 except p = 7 (checked up to p = 499).
%F Note: the Apery numbers A(n) = A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2 satisfy the supercongruence A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction).
%e a(11 - 1) = 172296487023049395523 = (11^4)*(43^2)*6163*1032705769 == 0 (mod 11^4).
%p seq(numer(add( (1/k^3) * binomial(n,k)^2 * binomial(n+k,k)^2, k = 1..n )), n = 0..20);
%o (PARI) a(n) = if (n, numerator(sum(k=1, n, binomial(n,k)^2*binomial(n+k,k)^2/k^3)), 0); \\ _Michel Marcus_, Oct 04 2022
%Y Cf. A005259, A357510, A357511, A357512.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Oct 02 2022
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