cp's OEIS Frontend

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.

A363871 a(n) = A108625(2*n, 3*n).

Original entry on oeis.org

1, 37, 5321, 980407, 201186025, 43859522037, 9939874413899, 2314357836947571, 549694303511409641, 132569070434503802605, 32360243622138480889321, 7977001183875449759759807, 1982402220908671654519130731, 496031095735572731850517509727
Offset: 0

Views

Author

Peter Bala, Jun 27 2023

Keywords

Comments

a(n) = B(2*n, 3*n, 2*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 A363870 (r = 1, s = 3).

Crossrefs

Programs

  • Magma
    A363871:= func< n | (&+[Binomial(2*n,j)^2*Binomial(5*n-j,2*n): j in [0..2*n]]) >;
    [A363871(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
    seq(simplify(A108625(2*n, 3*n)), n = 0..20);
  • Mathematica
    Table[HypergeometricPFQ[{-2*n,-3*n,2*n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)
  • SageMath
    def A363871(n): return sum(binomial(2*n,j)^2*binomial(5*n-j,2*n) for j in range(2*n+1))
    [A363871(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

Formula

a(n) = Sum_{k = 0..2*n} binomial(2*n, k)^2 * binomial(5*n-k, 2*n).
a(n) = Sum_{k = 0..2*n} (-1)^k * binomial(2*n, k)*binomial(5*n-k, 2*n)^2.
a(n) = hypergeometric3F2([-2*n, -3*n, 2*n+1], [1, 1], 1).
a(n) = [x^(3*n)] 1/(1 - x)*Legendre_P(2*n, (1 + x)/(1 - x)).
a(n) ~ sqrt(1700 + 530*sqrt(10)) * (98729 + 31220*sqrt(10))^n / (120 * Pi * n * 3^(6*n)). - Vaclav Kotesovec, Feb 17 2024
a(n) = Sum_{k = 0..2*n} binomial(2*n, k) * binomial(3*n, k) * binomial(2*n+k, k). - Peter Bala, Feb 26 2024