A357558 - OEIS

%I #11 Oct 08 2022 10:15:01

%S 0,4,54,648,7500,85440,965202,10849552,121566744,1359160020,

%T 15172321890,169175039616,1884704860116,20982512553912,

%U 233474575117770,2596777575029280,28872014164369968,320917108809011868,3566175414049854306,39620770883613043240,440115513924937822020

%N a(n) = Sum_{k = 0..n} (-1)^(n+k)*k*binomial(n,k)*binomial(n+k,k)^2.

%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 - 1)*p^4) for all primes p >= 5 (checked up to p = 499).

%F Note: Let B(n) = A005258(n). It is known that B(n) = Sum_{k = 0..n} (-1)^(n+k)* binomial(n,k)*binomial(n+k,k)^2 and the supercongruences B(p-1) == 1 (mod p^3) hold for all primes p >= 5 (see, for example, Straub, Example 3.4).

%F Recurrence: a(0) = 0, a(1) = 4 and for n >= 2, (5*n - 2)*(n^2 - 1)*a(n) = (55*n^3 - 22*n^2 - 19*n + 10)*a(n-1) + n*(5*n + 3)*(n-1)*a(n-2).

%F a(n) ~ phi^(5*n + 7/2) / (2*Pi*5^(1/4)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Oct 05 2022

%e Example of a supercongruence:

%e p = 17: a(17 - 1) = 28872014164369968 = (2^4)*3*(17^4)*107*251*268153 == 0 (mod 16*7^4)

%p seq( add( (-1)^(n+k)*k*binomial(n, k)*binomial(n+k, k)^2, k = 0..n ), n = 0..20 );

%Y Cf. A005258, A357510, A357511, A357512, A357513, A357559, A357560, A357561.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Oct 03 2022