A363867 a(n) = A108625(n,2*n).

1, 5, 61, 923, 15421, 272755, 5006275, 94307855, 1811113021, 35301145037, 696227550811, 13863654392945, 278264498108611, 5622746346645953, 114268249446672151, 2333733620675302423, 47868774493665731645, 985608360056821004233, 20362035153323824192645

Offset: 0

Author

Peter Bala, Jun 27 2023

Keywords

• nonn
• easy

Comments

a(n) = B(n,2*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), A363868 (r = 3, s = 1), A363869 (r = 3, s = 2), A363870 (r = 1, s = 3) and A363871 (r = 2, s = 3).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..500

Crossrefs

Cf. A005258, A099601, A108625, A363864 - A363871.

Programs

Magma

A363867:= func< n | (&+[Binomial(n,j)^2*Binomial(2*n+j,n): j in [0..n]]) >;
[A363867(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(n, 2*n)), n = 0..18);

Mathematica

Table[HypergeometricPFQ[{-n,-2*n,n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)

SageMath

def A363867(n): return sum(binomial(n,j)^2*binomial(2*n+j,n) for j in range(n+1))
[A363867(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

Formula

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

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

a(n) = hypergeom( [-n, -2*n, n+1], [1, 1], 1).

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

P-recursive: 4*(2*n - 1)^2*n^2*(85*n^2 - 235*n + 163)*a(n) = (29665*n^6 - 141345*n^5 + 264772*n^4 - 249181*n^3 + 124975*n^2 - 31902*n + 3276)*a(n-1) + 4*(2*n - 3)^2*(n-1)^2*(85*n^2 - 65*n + 13)*a(n-2) with a(0) = 1 and a(1) = 5.

a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*binomial(2*n, k). - Peter Bala, Feb 25 2024

a(n) ~ sqrt(13 + 53/sqrt(17)) * (349 + 85*sqrt(17))^n / (Pi * n * 2^(5*n + 5/2)). - Vaclav Kotesovec, Apr 26 2024