A228511 a(n) = sum_{k=0}^n binomial(n,k)^2*4^k*A000108(k).

1, 5, 49, 645, 9921, 167909, 3030705, 57284901, 1120905985, 22531796805, 462793508529, 9674942743365, 205261950829761, 4409503432713765, 95746612458475569, 2098428359692863717, 46366172896708865025, 1031886636204630031493, 23112239140054942651185, 520644236358436868354565, 11789139538117859937032385

Offset: 0

Author

Zhi-Wei Sun, Aug 23 2013

Keywords

• nonn

Comments

Conjecture: Let p be any odd 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 we have A(p) == (-1)^{(p-1)/2} (mod p).

(ii) Let B(p) be the p X p determinant with (i,j)-entry equal to b(i+j) for all i,j = 0,...,p-1, where b(n) denotes sum_{k=0}^n binomial(n,k)^2*binomial(2k,k)*4^k or sum_{k=0}^n binomial(n,k)^2*binomial(2k,k)*(-2)^(n-k). Then B(p) is congruent to the Legendre symbol (p/3) modulo 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.

Crossrefs

Cf. A000108, A086618.

Programs

Mathematica

a[n_]:=Sum[Binomial[n,k]^2*4^k*CatalanNumber[k],{k,0,n}]
Table[a[n],{n,0,20}]

Formula

By Zeilberger's algorithm, we have the following recurrence: 225*(12*n+43)*(n+1)^2*(n+2)^2*a(n)

- (n+2)^2*(3108*n^3+20869*n^2+42172*n+26271)*a(n+1)

+ (n+3)*(420*n^4+4037*n^3+13835*n^2+19872*n+9840)*a(n+2)

= (n+1)*(n+3)*(12*n+31)*(n+4)^2*a(n+3).

a(n) ~ 5^(2*n+5/2)/(32*Pi*n^2). - Vaclav Kotesovec, Aug 25 2013