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
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Sep 06 2013
STATUS
approved