

A227609


Determinant of the (p_n1)/2 X (p_n1)/2 matrix with (i,j)entry being the Legendre symbol((i^2+j^2)/p_n), where p_n is the nth prime.


9



1, 1, 4, 16, 27, 441, 1024, 1024, 34445, 13778944, 82719025, 48841786125, 67649929216, 564926611456, 153908556861703, 25481517249593344, 2456184022341328125, 399780402627654713344, 14448269983744, 214168150727821285287075
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OFFSET

2,3


COMMENTS

Conjecture: p_n never divides a(n), and moreover a(n) is a quadratic residue mod p_n.
ZhiWei 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 (p1)/2 X (p1)/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 nonresidue 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^((p3)/2), i.e., S(1,p) = (2^((p3)/4)*m)^2 for some positive integer m.  ZhiWei Sun, Sep 09 2018


LINKS



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)))):


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((p1)/2, (p1)/2, i, j, kronecker(i^2+j^2, p))); \\ Michel Marcus, Aug 25 2021


CROSSREFS



KEYWORD

sign


AUTHOR



STATUS

approved



