A372314 - OEIS

%I #34 Apr 27 2024 12:30:18

%S 1,3,0,125,-1215,0,0,9126441,0,-187590821,0,0,20686753425,0,0,0,

%T 9224101117395305225,0,881852208012283730302080,624391710361368134976,

%U 0,-3428714319207136609529065,0,0,3878246452353765171209988566241,0,0,4308304210666498856284267223158421

%N Determinant of the matrix [Jacobi(i^2 + 3*i*j + 2*j^2, 2*n + 1)]_{1 < i, j < 2*n}, where Jacobi(a, m) denotes the Jacobi symbol (a / m).

%C Conjecture 1: Let n be any positive integer.

%C (i) If a(2*n) is nonzero, then 4*n + 1 is a sum of two squares.

%C (ii) a(2*n + 1) is divisible by phi(4*n + 3)/2, where phi is Euler's totient function. If n is even, then a(2*n + 1)/(phi(4*n + 3)/2) is a square. This has been verified for n = 2..1000.

%C For any odd integer n > 3 and integers c and d, we introduce the notation: {c,d}_n = det[Jacobi(i^2 + c*i*j + d*j^2, n)]_{1 < i, j < n-1}.

%C The following conjecture is similar to Conjecture 1.

%C Conjecture 2: (1) {2, 2}_p = 0 for any prime p == 13,19 (mod 24), and {2, 2}_p == 0 (mod p) for any prime p == 17,23 (mod 24).

%C (2) If n == 5 (mod 8), then {4, 2}_n = 0. If n == 5 (mod 12), then {3, 3}_n = 0.

%C (3) If n == 5 (mod 12) and n is a sum of two squares, then {10, 9}_n = 0. Also, {10, 9}_p == 0 (mod p) for any prime p == 11 (mod 12).

%C (4) {8, 18}_p == 0 (mod p^2) for any prime p == 19 (mod 24), and {8,18}_p == 0 (mod p) for any prime p == 23 (mod 24). If n == 13,17 (mod 24) and n is a sum of two squares, then {8, 18}_n = 0.

%C We have verified Conjecture 2 for p or n smaller than 2000.

%H Zhi-Wei Sun, <a href="/A372314/b372314.txt">Table of n, a(n) for n = 2..73</a>

%H D. Krachun, F. Petrov, Z.-W. Sun and M. Vsemirnov, <a href="https://doi.org/10.1016/j.ffa.2020.101672">On some determinants involving Jacobi symbols</a>, Finite Fields Appl. 64 (2010), Article 101672.

%H Z.-W. Sun, <a href="https://doi.org/10.1016/j.ffa.2018.12.004">On some determinants with Legendre symbol entries</a>, Finite Fields Appl. 56 (2019), 285-307.

%e a(2) = 1 since the determinant of the matrix [Jacobi(i^2 + 3*i*j + 2*j^2, 5)]_{1 < i, j < 2*2} = [1,0; 0,1] is 1.

%t a[n_]:=a[n]=Det[Table[JacobiSymbol[i^2+3*i*j+2*j^2,2n+1],{i,2,2n-1},{j,2,2n-1}]];

%t tab={};Do[tab=Append[tab,a[n]],{n,2,29}];Print[tab]

%o (PARI) f(i,j) = i^2 + 3*i*j + 2*j^2;

%o a(n) = matdet(matrix(2*n-2, 2*n-2, i, j, kronecker(f(i+1,j+1), 2*n+1)));

%o vector(25, n, a(n+1)) \\ _Michel Marcus_, Apr 27 2024

%Y Cf. A000010, A000040, A001481.

%K sign

%O 2,2

%A _Zhi-Wei Sun_, Apr 27 2024