OFFSET
2,3
COMMENTS
In a 2003 preprint, R. Chapman conjectured that if 2*n + 1 (with n > 1) is a prime congruent to 3 modulo 4, then a(n) = 1.
Conjecture 1: If p = 2*n + 1 is a prime congruent to 1 modulo 4, then det[x + Jacobi(i - j, p)]_{0 < i, j < n} = r_p - 2*s_p + (-1)^{n/2}*x*(2*r_p - p*s_p), and det[x + Jacobi(i - j, p)]_{0 <= i, j <= n} = -r_p + (-1)^{n/2}*p*s_p*x, where r_p and s_p are rational numbers such that e_p^{(2 - Jacobi(2, p))*h(p)} = r_p + s_p*sqrt(p), e_p and h(p) are the fundamental unit and the class number of the real quadratic field Q(sqrt(p)), respectively.
Conjecture 2: Suppose that p = 2*n + 1 is a prime greater than 3, and write e_p^{h(p)} = a_p + b_p*sqrt(p) with 2*a_p and 2*b_p integral.
(1) det[x + Jacobi(i + j, p)]_{0 < i, j < n} is (-1)^{n/2}*2^{n - 1}*(b_p - a_p*x) if p == 1 (mod 4), and 2^{n-1}*x if p == 3 (mod 4).
(2) det[x + Jacobi(i + j, p)]_{0 <= i, j < n} is (-1)^{n/2}*2^{n-1}*(2*b_p - a_p + x*(p*b_p - 2*a_p)) if p == 1 (mod 4), and -2^{n - 1}*(2*x + 1) if p == 3 (mod 4).
We have checked both conjectures for n up to 1000.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 2..144
R. Chapman, Determinants of Legendre symbol matrices, preprint, 2003. (Cf. Conjecture 2 of this paper.)
R. Chapman, Determinants of Legendre symbol matrices, Acta Arith. 115 (2004), 231-244.
Z.-W. Sun, Conjectural values of some determinants involving Legendre symbols (I), Question 470324 at MathOverflow, April 30, 2024.
M. Vsemirnov, On the evaluation of R. Chapman's "evil determinant", Linear Algebra Appl. 436 (2012), 4101-4106.
M. Vsemirnov, On R. Chapman's "evil determinant": case p == 1(mod 4), Acta Arith. 159 (2013), 331-344; see also the arXiv version, arXiv:1108.4031 [math.NT], 2011-2012.
EXAMPLE
a(3) = 1 since the determinant of the matrix [Jacobi(i-j, 2*3+1)]_{0 < i, j < 3} = [0, -1; 1, 0] has the value 1.
MATHEMATICA
a[n_]:=a[n]=Det[Table[JacobiSymbol[i-j, 2n+1], {i, 1, n-1}, {j, 1, n-1}]];
tab={}; Do[tab=Append[tab, a[n]], {n, 2, 44}]; Print[tab]
PROG
(PARI) a(n) = matdet(matrix(n-1, n-1, i, j, kronecker(i-j, 2*n+1))); \\ Michel Marcus, Apr 29 2024
(Python)
from sympy import Matrix, jacobi_symbol
def A372385(n): return Matrix(n-1, n-1, [jacobi_symbol(i-j, (n<<1)|1) for i in range(n-1) for j in range(n-1)]).det() # Chai Wah Wu, Apr 29 2024
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Apr 29 2024
STATUS
approved