A228885 Determinant of the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j is coprime to n or not.

1, -1, -2, 0, 4, -4, -6, 0, 0, -16, -10, 0, 12, -36, -2048, 0, 16, 0, -18, 0, 27648, -100, -22, 0, 0, -144, 0, 0, 28, -4194304, -30, 0, 2048000, -256, -127401984, 0, 36, -324, -14155776, 0, 40, -764411904, -42, 0, 0, -484, -46, 0, 0, 0, -536870912, 0, 52, 0, -419430400000, 0, 3057647616, -784, -58, 0

Offset

1,3

Comments

Conjecture: If n is squarefree, then (-1)^(n*(n-1)/2)*a(n) > 0.

When p^2 divides n with p prime, (i + n/p) + j is coprime to n if and only if i + j is coprime to n. So a(n) = 0 if n is not squarefree.

It is easy to show that Phi(n) divides a(n) for any n > 0, where Phi(n) is Euler's totient function. Also, a(p) = (-1)^((p-1)/2)*(p-1) for any odd prime p.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..500

Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2019.

Example

a(1) = 1 since 1 + 1 = 2 is relatively prime to 1.

Mathematica

a[n_]:=Det[Table[If[GCD[i+j, n]==1, 1, 0], {i, 1, n}, {j, 1, n}]]
Table[a[n], {n, 1, 60}]

Prog

(PARI) a(n) = matdet(matrix(n, n, i, j, gcd(n, i+j)==1)); \\ Michel Marcus, Aug 25 2021

Crossrefs

Cf. A228860, A228884.

Sequence in context: A244340 A363062 A316987 * A323910 A166085 A129760

Adjacent sequences: A228882 A228883 A228884 * A228886 A228887 A228888

Keyword

sign

Author

Zhi-Wei Sun, Sep 06 2013

Status

approved