A306598 Determinant of the circulant matrix whose first column corresponds to the divisors of n.

1, -3, -8, 49, -24, -960, -48, -3375, 676, -8640, -120, -2247392, -168, -34560, -46080, 923521, -288, -28789488, -360, -54867456, -184320, -216000, -528, -89384770560, 15376, -423360, -512000, -438939648, -840, -558786571200, -960, -992436543, -1152000

Offset: 1

Author

Rémy Sigrist, Feb 27 2019

Keywords

• sign
• look

Comments

From Robert Israel, Mar 06 2019: (Start)

a(n) is divisible by A000203(n).

If n is not a square, a(n) is divisible by A000203(n)*A071324(n).

(End)

Examples

			For n = 12:
- the divisors of 12 are: 1, 2, 3, 4, 6, 12,
- the corresponding circulant matrix is:
    [ 1 12  6  4  3  2]
    [ 2  1 12  6  4  3]
    [ 3  2  1 12  6  4]
    [ 4  3  2  1 12  6]
    [ 6  4  3  2  1 12]
    [12  6  4  3  2  1]
- its determinant is -2247392,
- hence, a(12) = -2247392.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Wikipedia, Circulant matrix

Crossrefs

Cf. A027750, A086459, A177894.

Programs

Maple

f:= proc(n) local F,d; uses numtheory, LinearAlgebra;
  F:= sort(convert(divisors(n),list));
  d:= nops(F);
  Determinant(Matrix(d,d,shape=Circulant[F]))
end proc:
map(f, [$1..100]); # Robert Israel, Mar 06 2019

Mathematica

a[n_] := Module[{dd = Divisors[n], m, r}, m = Length[dd]; r = E^(2 Pi I/m); Product[Sum[dd[[j+1]] r^(j k), {j, 0, m-1}], {k, 0, m-1}] // FullSimplify];
Array[a, 100] (* Jean-François Alcover, Oct 17 2020 *)

PARI

a(n) = my (d=divisors(n)); my (m=matrix(#d, #d, i,j, d[1+(i-j)%#d])); return (matdet(m))

Formula

Apparently, a(n) > 0 iff n is a square.

a(p) = p^2 - 1 for any prime number p.

a(p^2) = p^6 - 2*p^3 + 1 for any prime number p.

a(2^k) = A086459(k+1) for any k >= 0.

If p < q are primes, a(p*q) = -(p^4-1)*(q^2-1)^2. - Robert Israel, Mar 06 2019