A228514 - OEIS

%I #18 Nov 12 2023 12:07:22

%S 0,1,8,60,456,3535,27888,223209,1807760,14784759,121909800,1012208340,

%T 8454274920,70975888425,598536562848,5067375370380,43052078886048,

%U 366911053809199,3135773892098520,26867522192372988,230731788606093720

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

%C Conjecture: Let  p > 3 be a prime.

%C (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).

%C (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).

%H Zhi-Wei Sun, <a href="/A228514/b228514.txt">Table of n, a(n) for n = 0..100</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1308.2900">On some determinants with Legendre symbol entries</a>, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

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

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

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

%F 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

%F a(n) ~ 3^(2*n+3/2)/(4*Pi*n). - _Vaclav Kotesovec_, Aug 25 2013

%F 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

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

%t Table[a[n],{n,0,20}]

%Y Cf. A086618, A228456.

%K nonn

%O 0,3

%A _Zhi-Wei Sun_, Aug 24 2013