A306595 Determinant of the circulant matrix whose first column corresponds to the binary digits of n.

0, 1, 1, 0, 1, 2, 2, 0, 1, 0, 0, 3, 0, -3, 3, 0, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 0, 1, 0, 4, 0, 0, -9, 9, 0, 4, 9, 0, 8, 9, 0, 8, 5, 0, 0, 9, 0, -9, -8, 0, -5, 0, 0, 8, 5, 0, -5, 5, 0, 1, 2, 2, 3, 2, 24, 24, 4, 2, 3, 3, 32, 3, 4, 32, 5, 2, 24, 3

Offset

0,6

Comments

This sequence is the binary variant of A177894.

From Robert Israel, Mar 05 2019: (Start)

a(n) is divisible by A000120(n).

If A070939(n) is even then n is divisible by A000120(n)*A065359(n). (End)

Links

Robert Israel, Table of n, a(n) for n = 0..10000

Wikipedia, Circulant matrix

Index entries for sequences related to binary expansion of n

Formula

a(A121016(n)) = 0 for any n > 0.

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

a(A219325(n)) = A219325(n) for any n > 0.

Example

For n = 13:
- the binary representation of 13 is "1101",
- the corresponding circulant matrix is:
    [1 1 0 1]
    [1 1 1 0]
    [0 1 1 1]
    [1 0 1 1]
- its determinant is -3,
- hence a(13) = -3.

Maple

a:= n-> `if`(n=1, 1, (l-> LinearAlgebra[Determinant](Matrix(nops(l),
       shape=Circulant[l[-i]$i=1..nops(l)])))(convert(n, base, 2))):
seq(a(n), n=0..100);  # Alois P. Heinz, Mar 05 2019

Prog

(PARI) a(n) = my (d=if (n, binary(n), [0])); my (m=matrix(#d, #d, i, j, d[1+(i-j)%#d])); return (matdet(m))

Crossrefs

Cf. A000120, A065359, A070939, A121016, A177894, A219325, A306714.

Sequence in context: A049502 A242284 A333624 * A332996 A292592 A292274

Adjacent sequences: A306592 A306593 A306594 * A306596 A306597 A306598

Keyword

sign,base,look

Author

Rémy Sigrist, Feb 27 2019

Status

approved