%I #27 Aug 05 2019 02:09:17

%S 1,0,-14,0,786,0,-61340,0,5562130,0,-549676764,0,57440496036,0,

%T -6242164112184,0,698300344311570,0,-79881547652046140,0,

%U 9301427008157320036,0,-1098786921802152516024,0,131361675994216221116836,0,-15863471168011822803270200,0,1932252897656224864335299400,0,-237114404923760858875375113840

%N a(n) = Sum_{k=0..n} C(n,k)^4*(-1)^k.

%C As (-1)^n*a(n) = a(n), we have a(n) = 0 for n = 1,3,5,... For any odd prime p, the author could show that a(p-1) == 1 + 4*(2^{p-1}-1) + 6*(2^{p-1}-1)^2 (mod p^3).

%C Conjecture: Let p be any odd prime, and 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) == (-1)^{(p-1)/2} (mod p). Similarly, if c(n) = sum_{k=0}^n (-1)^k*C(n,k)^2*C(2k,k)*C(2(n-k),n-k) and C(p) is the p X p determinant with (i,j)-entry equal to c(i+j) for all i,j = 0,...,p-1, then we have C(p) == 1 (mod p).

%H Zhi-Wei Sun, <a href="/A228304/b228304.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 Conjecture: n^3*(n-1)*(12*n^2-63*n+83)*a(n) +(n-2)*(12*n^2-87*n+158)*(n-1)^3*a(n-1) +4*(408*n^6-3774*n^5+13760*n^4-25203*n^3+24465*n^2-11970*n+2340)*a(n-2) +4*(408*n^6-6222*n^5+38750*n^4-126143*n^3+226494*n^2-212867*n+81920)*a(n-3) +16*(n-2)*(12*n^2-15*n+5)*(n-3)^3*a(n-4) +16*(n-3)*(12*n^2-39*n+32)*(n-4)^3*a(n-5)=0. - _R. J. Mathar_, Aug 21 2013

%F a(2n) = A050983(n) * (-1)^n. - _Vaclav Kotesovec_, Feb 01 2014

%t a[n_]:=Sum[Binomial[n,k]^4*(-1)^k,{k,0,n}]

%t Table[a[n],{n,0,30}]

%t Table[HypergeometricPFQ[{-n, -n, -n, -n}, {1, 1, 1}, -1],{n,0,20}] (* _Vaclav Kotesovec_, Feb 01 2014 *)

%Y Cf. A050983, A228289.

%K sign

%O 0,3

%A _Zhi-Wei Sun_, Aug 20 2013