A228095 Determinant of the p_n X p_n matrix with (i,j)-entry equal to the Legendre symbol ((i^2 + 3*i*j + 3*j^2)/p_n) for all i,j = 0, 1, ..., p_n-1, where p_n is the n-th prime.

0, 0, -72, 0, 9600, 0, 23970816, 0, 0, -8814759178752000000, -1217765613534782800527360, 0, 2555625991208076641833058304, 0, 0, 0, 164525463228624478317575381527120287356682240, -33094833021317386202938131485140597289779200, 0

Offset: 2

Author

Zhi-Wei Sun, Aug 12 2013

Keywords

• sign

Comments

Conjecture: a(n) = 0 if p_n == 5 (mod 6).

For an odd prime p and integers c and d, Zhi-Wei Sun defined [c,d]_p to be the determinant of the p X p matrix with (i,j)-entry equal to the Legendre symbol ((i^2+c*i*j+d*j^2)/p). It is easy to see that [-c,d]_p = (-1/p)*[c,d]_p. Sun conjectured that for any integer c and nonzero integer d there are infinitely many odd primes p with [c,d]_p = 0. He also conjectured that if (c,d)_p (defined in the comments in A225611) is nonzero with d not divisible by p, then [c,d]_p/(c,d)_p equals (1-p)/(p-2) or (p-1)/2 according as p divides c^2-4*d or not. He also made some concrete conjectures, for example, [3,2]_p = 0 if p == 3 (mod 4); and [4,2]_p = 0 if p == 5 or 7 (mod 8).

On Aug 12 2013, Zhi-Wei Sun conjectured that for any odd prime p and integers c and d with d not divisible by p, if [c,d]_p is nonzero then its p-adic valuation (i.e., p-adic order) must be even.

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 2..50
• Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Crossrefs

Cf. A225611, A226163, A227609, A227968, A227971, A228005, A228077.

Programs

Mathematica

a[n_]:=Det[Table[JacobiSymbol[i^2+3*i*j+3*j^2,Prime[n]],{i,0,(Prime[n]-1)},{j,0,(Prime[n]-1)}]]
Table[a[n],{n,2,20}]

PARI

a(n) = my(p=prime(n)); matdet(matrix(p, p, i, j, i--; j--; kronecker(i^2+3*i*j+3*j^2, p))); \\ Michel Marcus, Aug 25 2021