A372394 Determinant of the matrix [Jacobi(i^2+5*i*j+5*j^2,2*n+1)]_{1

0, 0, 0, 33, 0, 0, 0, -77539, 1811939328, -405798912, 0, 0, 649564705105200, -2787119627540625, 86463597248512, 0, 0, 0, 353143905335474188320, -66016543975248459410178048, 0, 23092056382629010556862857216, 0, 0, 0, 0, -5310136941067623723354761986048

Offset: 2

Author

Zhi-Wei Sun, Apr 29 2024

Keywords

• sign

Comments

Conjecture: (i) If n == 6, 8 (mod 10), and 2*n + 1 is a sum of two squares, then a(n) = 0.

(ii) If n == 5, 9 (mod 10), then a(n) is not relatively prime to 2*n + 1.

See also A372314 for other similar conjectures.

Examples

			a(2) = 0 since the determinant of the matrix [Jacobi(i^2+5*i*j+5*j^2,2*2+1)]_{1<i,j<2*2} = [1,1;1,1] has the value 0.
		

Links

• D. Krachun, F. Petrov, Z.-W. Sun, and M. Vsemirnov, On some determinants involving Jacobi symbols, Finite Fields Appl. 64 (2010), Article 101672.
• Z.-W. Sun, On some determinants with Legendre symbol entries, Finite Fields Appl. 56 (2019), 285-307.

Crossrefs

Cf. A000290, A001481, A372314, A372385.

Programs

Mathematica

a[n_]:=a[n]=Det[Table[JacobiSymbol[i^2+5*i*j+5*j^2, 2n+1], {i, 2, 2n-1}, {j, 2, 2n-1}]];
tab={}; Do[tab=Append[tab, a[n]], {n, 2, 28}]; Print[tab]

Python

from sympy import Matrix, jacobi_symbol
def A372394(n): return Matrix(n-1<<1,n-1<<1,[jacobi_symbol(i*(i+5*j+14)+j*(5*j+30)+44,(n<<1)|1) for i in range(n-1<<1) for j in range(n-1<<1)]).det() # Chai Wah Wu, Apr 30 2024