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%I #23 Oct 08 2022 10:09:02
%S 0,4,108,3144,95000,2935020,92054340,2918972560,93330811440,
%T 3003683380020,97177865060540,3157623679795992,102973952434618824,
%U 3368460743291372092,110480459392323735540,3631941224582026770720,119637879389041977365600,3947968300820696313987780
%N a(n) = Sum_{k = 0..n} k * binomial(n,k)^2 * binomial(n+k,k)^2.
%H Armin 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 >= 5 (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).
%F Recurrence: a(0) = 0, a(1) = 4, a(2) = 108 and thereafter n^2*(n-1)*(2*n-3)*(3*n^2-9*n+7)*a(n) = (2*n-1)*(105*n^5 - 576*n^4 + 1208*n^3 - 1195*n^2 + 556*n - 104)*a(n-1) - (2*n-3)*(105*n^5 - 474*n^4 + 800*n^3 - 629*n^2 + 240*n - 36)*a(n-2) + (n-1)*(2*n-1)*(3*n^2-3*n+1)*(n-2)^2*a(n-3).
%F a(n) ~ (1 + sqrt(2))^(4*n + 2) / (2^(11/4) * Pi^(3/2) * sqrt(n)). - _Vaclav Kotesovec_, Oct 04 2022
%e a(11 - 1) = 97177865060540 = (2^2)*5*(11^4)*37*239*37529 == 0 (mod 11^4).
%p seq( add( k*binomial(n,k)^2 * binomial(n+k,k)^2, k = 0..n ), n = 0..20 );
%o (PARI) a(n) = sum(k = 0, n, k * binomial(n,k)^2 * binomial(n+k,k)^2); \\ _Michel Marcus_, Oct 04 2022
%Y Cf. A005259, A357506, A357507, A357511, A357512, A357513.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Oct 01 2022
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