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A372314
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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).
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3
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1, 3, 0, 125, -1215, 0, 0, 9126441, 0, -187590821, 0, 0, 20686753425, 0, 0, 0, 9224101117395305225, 0, 881852208012283730302080, 624391710361368134976, 0, -3428714319207136609529065, 0, 0, 3878246452353765171209988566241, 0, 0, 4308304210666498856284267223158421
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OFFSET
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2,2
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COMMENTS
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Conjecture 1: Let n be any positive integer.
(i) If a(2*n) is nonzero, then 4*n + 1 is a sum of two squares.
(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.
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}.
The following conjecture is similar to Conjecture 1.
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).
(2) If n == 5 (mod 8), then {4, 2}_n = 0. If n == 5 (mod 12), then {3, 3}_n = 0.
(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).
(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.
We have verified Conjecture 2 for p or n smaller than 2000.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}]];
tab={}; Do[tab=Append[tab, a[n]], {n, 2, 29}]; Print[tab]
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PROG
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(PARI) f(i, j) = i^2 + 3*i*j + 2*j^2;
a(n) = matdet(matrix(2*n-2, 2*n-2, i, j, kronecker(f(i+1, j+1), 2*n+1)));
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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