OFFSET
2,3
COMMENTS
Conjecture: p_n never divides a(n), and moreover -a(n) is a quadratic residue mod p_n.
Zhi-Wei Sun also made the following conjecture:
Let p be any odd prime. For each integer d let S(d,p) be the determinant of the (p-1)/2 X (p-1)/2 matrix whose (i,j)-entry is the Legendre symbol ((i^2+d*j^2)/p). If d is a quadratic residue mod p, then so is -S(d,p). If d is a quadratic non-residue mod p, then we have S(d,p) = 0.
These were proved in version 9 of arXiv:1308.2900 (2018). In addition, the author has the following new conjecture.
Conjecture: For any prime p == 3 (mod 4), the number -S(1,p) is a positive square divisible by 2^((p-3)/2), i.e., -S(1,p) = (2^((p-3)/4)*m)^2 for some positive integer m. - Zhi-Wei Sun, Sep 09 2018
LINKS
Alois P. Heinz, Table of n, a(n) for n = 2..100
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.
Zhi-Wei Sun, Is -det[Legendre(i^2+j^2,p)]_{i,j=1,...,(p-1)/2} always a square for each prime p == 3 (mod 4)?, Question 310192 in MathOverflow, Sept. 9, 2018.
EXAMPLE
a(2) = -1 since the Legendre symbol ((1^2 + 1^2)/3) is -1.
MAPLE
with(numtheory): with(LinearAlgebra):
a:= n-> Determinant(Matrix((ithprime(n)-1)/2, (i, j)->
jacobi(i^2+j^2, ithprime(n)))):
seq(a(n), n=2..20); # Alois P. Heinz, Jul 18 2013
MATHEMATICA
a[n_]:=Det[Table[JacobiSymbol[i^2+j^2, Prime[n]], {i, 1, (Prime[n]-1)/2}, {j, 1, (Prime[n]-1)/2}]]
Table[a[n], {n, 2, 20}]
PROG
(PARI) a(n) = my(p=prime(n)); matdet(matrix((p-1)/2, (p-1)/2, i, j, kronecker(i^2+j^2, p))); \\ Michel Marcus, Aug 25 2021
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Jul 17 2013
STATUS
approved