A228514 a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(2k,k+1).

0, 1, 8, 60, 456, 3535, 27888, 223209, 1807760, 14784759, 121909800, 1012208340, 8454274920, 70975888425, 598536562848, 5067375370380, 43052078886048, 366911053809199, 3135773892098520, 26867522192372988, 230731788606093720

Offset

0,3

Comments

Conjecture: Let p > 3 be a prime.

(i) Let A(p) be the p X p determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,p-1. Then A(p) == 0 (mod p^2).

(ii) Let B(p) be the (p-1) X (p-1) determinant with (i,j)-entry equal to a(i+j) for all i,j = 1,...,p-1. If p == 1 (mod 3) then B(p) == (-1)^{(p-1)/2} (mod p); if p == 2 (mod 3), then B(p) == 0 (mod p).

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..100

Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Formula

By Zeilberger's algorithm, we have the following recurrence:

(n+2)*(n+4)^2*a(n+3) = (n+3)*(11*n^2+64*n+89)*a(n+2)

-(n+2)*(19*n^2+94*n+108)*a(n+1)+9*(n+1)^2*(n+3)*a(n).

Recurrence (order 2): (n-1)*(n+1)^2*a(n) = 2*n*(5*n^2-2)*a(n-1) - 9*(n-1)^2*(n+1)*a(n-2). - Vaclav Kotesovec, Aug 25 2013

a(n) ~ 3^(2*n+3/2)/(4*Pi*n). - Vaclav Kotesovec, Aug 25 2013

G.f.: ((1-3*x)*hypergeom([1/3, 1/3],[1],27*(x-1)*x^2/(1-9*x))/(1-9*x)^(1/3)-1)/(6*x). - Mark van Hoeij, Nov 12 2023

Mathematica

a[n_]:=Sum[Binomial[n, k]^2*Binomial[2k, k+1], {k, 0, n}]
Table[a[n], {n, 0, 20}]

Crossrefs

Cf. A086618, A228456.

Sequence in context: A099156 A245391 A254658 * A233666 A199526 A129331

Adjacent sequences: A228511 A228512 A228513 * A228515 A228516 A228517

Keyword

nonn

Author

Zhi-Wei Sun, Aug 24 2013

Status

approved