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A363870
a(n) = A108625(n, 3*n).
6
1, 7, 127, 2869, 71631, 1894007, 51978529, 1464209383, 42050906191, 1225778575021, 36156060825127, 1076772406867549, 32324178587781393, 976893529756053501, 29693248490460447747, 907027175886637081619, 27826656707376811715663, 856949305975908664414097
OFFSET
0,2
COMMENTS
a(n) = B(n, 3*n, n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363868 (r = 3, s = 1), A363869 (r = 3, s = 2) and A363871 (r = 2, s = 3).
FORMULA
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(3*n+k, n).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(3*n+k, n)^2.
a(n) = hypergeometric3F2( [-n, -3*n, n+1], [1, 1], 1).
a(n) = [x^(3*n)] 1/(1 - x)*Legendre_P(n, (1 + x)/(1 - x)).
a(n) ~ sqrt(25 + 151/sqrt(37)) * (11906 + 1961*sqrt(37))^n / (Pi * 2^(3/2) * n * 3^(6*n+1)). - Vaclav Kotesovec, Feb 17 2024
a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*binomial(3*n, k). - Peter Bala, Feb 25 2024
MAPLE
A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
seq(simplify(A108625(n, 3*n)), n = 0..20);
MATHEMATICA
Table[HypergeometricPFQ[{-n, -3*n, n+1}, {1, 1}, 1], {n, 0, 30}] (* G. C. Greubel, Oct 05 2023 *)
PROG
(Magma)
A363870:= func< n | (&+[Binomial(n, j)^2*Binomial(3*n+j, n): j in [0..n]]) >;
[A363870(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023
(SageMath)
def A363870(n): return sum(binomial(n, j)^2*binomial(3*n+j, n) for j in range(n+1))
[A363870(n) for n in range(31)] # G. C. Greubel, Oct 05 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jun 27 2023
STATUS
approved