A363868 a(n) = A108625(3*n, n).

1, 13, 505, 24691, 1337961, 76869013, 4586370139, 280973874215, 17552736006121, 1113134497824901, 71437216036404505, 4629194489296980715, 302391678415222922475, 19886936616891022422159, 1315438146193644502479255, 87445220568000089973356191, 5838332204000163260729138153

Offset: 0

Author

Peter Bala, Jun 27 2023

Keywords

• nonn
• easy

Comments

a(n) = B(3*n, n, 3*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), 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

A363868:= func< n | (&+[Binomial(3*n,n-j)^2*Binomial(3*n+j,j): j in [0..n]]) >;
[A363868(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(3*n, n)), n = 0..16);

Mathematica

Table[HypergeometricPFQ[{-3*n,-n,3*n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)

SageMath

def A363868(n): return sum(binomial(3*n,n-j)^2*binomial(3*n+j, j) for j in range(n+1))
[A363868(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

Formula

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

a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(3*n, n-k) * binomial(3*n+k, k)^2.

a(n) = hypergeometric3F2( [-3*n, -n, 3*n+1], [1, 1], 1).

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

P-recursive: 3*(4797*n^4 - 26076*n^3 + 53055*n^2 - 47886*n + 16178)*(3*n - 1)^2*(3*n - 2)^2*n^2*a(n) = (82935333*n^10 - 699633963*n^9 + 2570641767*n^8 - 5402404662*n^7 + 7171181427*n^6 - 6264762171*n^5 + 3637752517*n^4 - 1382756780*n^3 + 328531700*n^2 - 44004160*n + 2529600)*a(n-1) + 3*(4797*n^4 - 6888*n^3 + 3609*n^2 - 816*n + 68)*(n - 1)^2*(3*n - 4)^2*(3*n - 5)^2*a(n-2) with a(0) = 1 and a(1) = 13.

a(n) ~ sqrt(17 + 61/sqrt(13)) * ((1921 + 533*sqrt(13))/54)^n / (6*Pi*sqrt(2)*n). - Vaclav Kotesovec, Feb 17 2024

a(n) = Sum_{k = 0..n} binomial(n, k) * binomial(3*n, k) * binomial(3*n+k, k). - Peter Bala, Feb 26 2024