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%I #14 Jun 02 2023 14:40:14
%S 0,4,270,8448,192000,3669300,62952162,1003770880,15182515584,
%T 220700443500,3110529630450,42769154678976,576313309494000,
%U 7636526099508852,99765264496070250,1287663145631539200,16446680778536421888,208154776511034178380,2613380452317012835386
%N a(n) = Sum_{k = 0..n} (-1)^(n+k)*k^3*binomial(n,k)*binomial(n+k,k)^2.
%C Define S_m(n) = Sum_{k = 0..n} (-1)^(n+k)*k^m*binomial(n,k)*binomial(n+k,k)^2, so that S_0(n) = A005258(n), one type of Apéry numbers. The present sequence is the case m = 3. See A357558 for the case m = 1.
%C It is known that S_0(2*n) - 1 is divisible by (2*n + 1)^3 provided 2*n + 1 is a prime greater than 3 (see, for example, Straub, Example 3.4).
%C Let C_m = Product_{odd primes p <= m} p.
%C Conjectures:
%C 1) S_2(2*n) is divisible by n^2*(2*n + 1)^2.
%C 2) for even m >= 4, C_m * S_m(2*n) is divisible by n^3*(2*n + 1)^2.
%C 3) for odd m >= 5, C_m * S_m(2*n) is divisible by n^2*(2*n + 1)^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: 3*a(2*n) == 0 ( mod n^2*(2*n + 1)^3 ).
%F Recurrence a(0) = 0, a(1) = 4, and for n >= 2, n*(n + 3)*(5*n - 6)*(n - 1)^4*a(n) = (n - 1)*(55*n^4 - 66*n^3 - 22*n^2 + 15*n + 6)*(n + 1)^2*a(n-1) + n^4*(5*n - 1)*(n + 1)^2*a(n-2).
%F a(n) ~ n^2 * phi^(5*n + 11/2) / (2*Pi*5^(1/4)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Oct 07 2022
%p seq( add( (-1)^(n+k) * k^3 * binomial(n, k) * binomial(n+k,k)^2, k = 0..n ), n = 0..20 );
%Y Cf. A005258, A357510, A357512, A357558, A357560, A357561.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Oct 04 2022
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