A372409 Determinant of the matrix obtained from [Jacobi(i-j, 2*n+1)]_{0

1, -1, 1, -3, 3, -8, -5, -5, -24, 7, 0, 0, 9, 9, -81312, -1341867, 11, -19685120, -13, -13, 0, -15, 0, -180287762432, 17, -1407939911477, 10526233598464, 19, 19, 0, 6040299856799, -21, 29830847001120768, 23, -23, 0, 115407361849089836, -25, 0, 27, 104060523591574200

Offset: 2

Author

Zhi-Wei Sun, Apr 29 2024

Keywords

• sign

Comments

Conjecture 1: If 2*n + 1 (with n > 1) is a prime p, then a(n)/floor((p-2)/3) coincides with (-1)^((p+3)/4) if p == 1 (mod 4), and (-1)^((h(-p)+1)/2) if p == 3 (mod 4), where h(-p) is the class number of the imaginary quadratic field Q(sqrt(-p)).

Conjecture 2: Let p > 3 be a prime, and let S(p) and T(p) denote the matrices obtained from [Jacobi(i+j,p)]{1<=i,j<=(p-3)/2} and [Jacobi(i+j,p)]{0<=i,j<=(p-3)/2} (respectively) by replacing all the entries in the first row by 1. Then det S(p) = -det T(p) = 2^((p-5)/2)*s(p), where s(p) is (-1)^((p+3)/4) if p == 1 (mod 4), and (-1)^((h(-p)-1)/2) if p == 3 (mod 4).

Both conjectures are motivated by Conjecture 4.6 in the author's 2019 FFA paper as well as the conjectures in A372385. They have been verified for primes p < 2000.

Examples

			a(3) = -1 since the determinant of the matrix [1, 1; Jacobi(2-1,2*3+1), Jacobi(2-2,2*3+1)] = [1, 1; 1, 0] has the value -1.
		

References

• L.-Y. Wang and H.-L. Wu, On certain determinants involving Legendre symbols, Ramanujan J. 58 (2022), 43-56.

Links

• R. Chapman, Determinants of Legendre symbol matrices, Acta Arith. 115 (2004), 231-244.
• Z.-W. Sun, On some determinants with Legendre symbol entries, Finite Fields Appl. 56 (2019), 285-307.
• 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.
• L.-Y. Wang and H.-L. Wu, On p-th cyclotomic field and Zhi-Wei Sun's Legendre determinants over F_p, arXiv:2401.05853 [math.NT], 2024.

Crossrefs

Cf. A179071, A179072, A372385.

Programs

Mathematica

a[n_]:=a[n]=Det[Table[If[i==1,1,JacobiSymbol[i-j,2*n+1]],{i,1,n-1},{j,1,n-1}]];
tab={};Do[tab=Append[tab,a[n]],{n,2,42}];Print[tab]

PARI

a(n) = matdet(matrix(n-1, n-1, i, j, if (i==1, 1, kronecker(i-j, 2*n+1)))); \\ Michel Marcus, Apr 30 2024

Python

from sympy import Matrix, jacobi_symbol
def A372409(n): return Matrix(n-1,n-1,[jacobi_symbol(i-j,(n<<1)|1) if i else 1 for i in range(n-1) for j in range(n-1)]).det() # Chai Wah Wu, May 01 2024