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A227609 Determinant of the (p_n-1)/2 X (p_n-1)/2 matrix with (i,j)-entry being the Legendre symbol((i^2+j^2)/p_n), where p_n is the n-th prime. 9
-1, 1, -4, -16, -27, 441, -1024, -1024, 34445, -13778944, 82719025, 48841786125, -67649929216, -564926611456, -153908556861703, -25481517249593344, 2456184022341328125, -399780402627654713344, -14448269983744, -214168150727821285287075 (list; graph; refs; listen; history; text; internal format)
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
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.
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
Sequence in context: A271936 A046358 A046366 * A219338 A275211 A046361
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Jul 17 2013
STATUS
approved

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Last modified May 28 01:34 EDT 2024. Contains 372900 sequences. (Running on oeis4.)