This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A357512 #18 Jun 02 2025 15:26:20
%S A357512 0,4,1188,126144,10040000,682492500,41503541940,2325305113600,
%T A357512 122429236976640,6140504039242500,296222848665342500,
%U A357512 13841644170257145792,629814531655430506944,28019919084086921883892,1222770835880665252492500,52476371578141941012480000,2219374467089388085650636800
%N A357512 a(n) = Sum_{k = 0..n} k^5 * binomial(n,k)^2 * binomial(n+k,k)^2.
%C A357512 Let m >= 2 and set u(n) = Sum_{k = 0..n} k^(2*m+1) * binomial(n,k)^2 * binomial(n+k,k)^2. We conjecture that there is a finite set of primes P(m) such that u(n-1) == 0 (mod n^4) for all odd numbers n not divisible by an element of P(m). For example, calculation suggests that P(2) = {3} (the present sequence), while conjecturally P(3) = {3, 5}, P(4) = {3, 7}, P(5) = {3, 5}, P(6) = {3, 11}, P(7) = {3, 5, 7, 13}, P(8) = {3}, P(9) = {3} and P(10) = {3, 7, 19}.
%H A357512 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 A357512 Conjecture: a(n-1) == 0 (mod n^4) for all n congruent to 1 or 5 mod 6 (checked up to n = 499).
%F A357512 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 A357512 Recurrence: a(1) = 4, a(2) = 1188, a(3) = 126144 and thereafter
%F A357512 (n - 3)*(2*n - 3)*(3*n^4 - 18*n^3 + 40*n^2 - 39*n + 13)*(n - 2)^3*(n - 1)^4*a(n) = (n - 3)*(2*n - 1)*(n + 1)*(105*n^7 - 879*n^6 + 2981*n^5 - 5124*n^4 + 4376*n^3 - 1236*n^2 - 442*n + 195)*(n - 2)^3*a(n-1) - n^3*(n - 3)*(2*n - 3)*(n + 1)*(105*n^7 - 591*n^6 + 1253*n^5 - 1346*n^4 + 784*n^3 - 124*n^2 - 90*n + 33)*a(n-2) + n^3*(2*n - 1)*(n + 1)*(3*n^4 - 6*n^3 + 4*n^2 - n - 1)*(n - 1)^4*a(n-3).
%F A357512 a(n) ~ n^(7/2) * (1 + sqrt(2))^(4*n + 2) / (2^(19/4) * Pi^(3/2)). - _Vaclav Kotesovec_, Oct 04 2022
%p A357512 seq( add( k^5 * binomial(n,k)^2 * binomial(n+k,k)^2, k = 0..n ), n = 0..20 );
%o A357512 (PARI) a(n) = sum(k = 0, n, k^5 * binomial(n,k)^2 * binomial(n+k,k)^2); \\ _Michel Marcus_, Oct 04 2022
%Y A357512 Cf. A005259, A007310, A357510, A357511, A357513.
%K A357512 nonn,easy
%O A357512 0,2
%A A357512 _Peter Bala_, Oct 02 2022