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Determinant of the (p_n-1)/2-by-(p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol ((i^2+j^2+1)/p_n), where p_n is the n-th prime.
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%I #17 Aug 07 2013 10:08:25

%S 0,-2,1,-2,-72,-672,136,4352,-265600,602048,2658941440,128532940800,

%T 7138246144,-277070036992,-5928696847474688,140393397382594560,

%U -476899996446720000,73073105939334987776,-197109670210161672192

%N Determinant of the (p_n-1)/2-by-(p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol ((i^2+j^2+1)/p_n), where p_n is the n-th prime.

%C Conjecture: We have (a(n)/p_n) = ((2(-1)^{(p_n-1)/2}-4)/p_n).

%C Zhi-Wei Sun also made the following general conjectures for any odd prime p:

%C (i) Let c and d be integers not divisible by p, and let S be the determinant of the (p-1)/2-by-(p-1)/2 matrix whose (i,j)-entry is the Legendre symbol ((i^2+d*j^2+c)/p).

%C Then (S/p) = 1 if (c/p) = 1 and (d/p) = -1; (S/p) = (-1/p) if (c/p) = (d/p) = -1; (S/p) = (-2/p) if (-c/p) = (d/p) = 1; and (S/p) = (-6/p) if (-c/p)= -1 and (d/p) = 1.

%C (ii) Let d be any integer not divisible by p. For k = 1,2 let D_k(c) be the determinant of the (p+1)/2-by-(p+1)/2 matrix with (i,j)-entry (i,j = 0,...,(p-1)/2)) being the Legendre symbol ((i^k+d*j^k+c)/p). Then D_k(c) == D_k(0) (mod p) for every integer c.

%H Zhi-Wei Sun, <a href="/A228005/b228005.txt">Table of n, a(n) for n = 2..70</a>

%e a(2) = 0 since the Legendre symbol ((1^2+1^2+1)/3) is zero.

%t a[n_]:=Det[Table[JacobiSymbol[i^2+j^2+1,Prime[n]],{i,1,(Prime[n]-1)/2},{j,1,(Prime[n]-1)/2}]]

%t Table[a[n],{n,2,20}]

%Y Cf. A227609, A227968, A227971, A226163.

%K sign

%O 2,2

%A _Zhi-Wei Sun_, Aug 07 2013