A228005 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.

0, -2, 1, -2, -72, -672, 136, 4352, -265600, 602048, 2658941440, 128532940800, 7138246144, -277070036992, -5928696847474688, 140393397382594560, -476899996446720000, 73073105939334987776, -197109670210161672192

Offset: 2

Author

Zhi-Wei Sun, Aug 07 2013

Keywords

• sign

Comments

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

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

(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).

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.

(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.

Examples

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

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 2..70

Crossrefs

Cf. A227609, A227968, A227971, A226163.

Programs

Mathematica

a[n_]:=Det[Table[JacobiSymbol[i^2+j^2+1,Prime[n]],{i,1,(Prime[n]-1)/2},{j,1,(Prime[n]-1)/2}]]
Table[a[n],{n,2,20}]