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A363865

a(n) = A143007(3*n,n).

1, 25, 2773, 430081, 77620661, 15276834025, 3180268712125, 688612022804773, 153504724110658741, 34994264014955310181, 8120680975872203708773, 1911897036160037674700065, 455553725980571500127902109

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OFFSET

0,2

COMMENTS

a(n) = A(3*n,n,3*n,n) in the notation of Straub, equation 8. It follows from Straub, Theorem 1.2, that the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and all positive integers n and k.

More generally, for positive integers r and s the sequence {A143007(r*n, s*n) : n >= 0} satisfies the above supercongruences. For other cases, see A005259 (r = s = 1), A363864 (r = 2, s = 1) and A363866 (r = 3, s = 2).

LINKS

Table of n, a(n) for n=0..12.

Peter Bala, A recurrence for A363865

Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

FORMULA

a(n) = Sum_{k = 0..n} binomial(3*n,n-k)^2*binomial(3*n+k,k)^2.

a(n) = Sum_{k = 0..n} binomial(3*n+k,2*k)*binomial(2*k,k)^2*binomial(n+k,2*k).

a(n) = hypergeom([3*n+1, -3*n, n+1, -n], [1, 1, 1], 1)

a(n) = [x^n] 1/(1 - x)*( Legendre_P(3*n,(1 + x)/(1 - x)) )^2 = [x^(3*n)] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^2.

MAPLE

A143007 := proc(n, k); add(binomial(n+j, 2*j)*binomial(2*j, j)^2*binomial(k+j, 2*j), j = 0..n) end:

seq(A143007(3*n, n), n = 0..20);