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%I #21 Aug 05 2019 02:09:14
%S 1,5,49,645,9921,167909,3030705,57284901,1120905985,22531796805,
%T 462793508529,9674942743365,205261950829761,4409503432713765,
%U 95746612458475569,2098428359692863717,46366172896708865025,1031886636204630031493,23112239140054942651185,520644236358436868354565,11789139538117859937032385
%N a(n) = sum_{k=0}^n binomial(n,k)^2*4^k*A000108(k).
%C Conjecture: Let p be any odd 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 we have A(p) == (-1)^{(p-1)/2} (mod p).
%C (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.
%H Zhi-Wei Sun, <a href="/A228511/b228511.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: 225*(12*n+43)*(n+1)^2*(n+2)^2*a(n)
%F - (n+2)^2*(3108*n^3+20869*n^2+42172*n+26271)*a(n+1)
%F + (n+3)*(420*n^4+4037*n^3+13835*n^2+19872*n+9840)*a(n+2)
%F = (n+1)*(n+3)*(12*n+31)*(n+4)^2*a(n+3).
%F a(n) ~ 5^(2*n+5/2)/(32*Pi*n^2). - _Vaclav Kotesovec_, Aug 25 2013
%t a[n_]:=Sum[Binomial[n,k]^2*4^k*CatalanNumber[k],{k,0,n}]
%t Table[a[n],{n,0,20}]
%Y Cf. A000108, A086618.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Aug 23 2013