A306714 Permanent of the circulant matrix whose first row is given by the binary expansion of n.

0, 1, 1, 2, 1, 2, 2, 6, 1, 2, 4, 9, 2, 9, 9, 24, 1, 2, 2, 13, 2, 13, 13, 44, 2, 13, 13, 44, 13, 44, 44, 120, 1, 2, 4, 20, 8, 17, 17, 80, 4, 17, 36, 82, 17, 80, 82, 265, 2, 20, 17, 80, 17, 82, 80, 265, 20, 80, 82, 265, 80, 265, 265, 720, 1, 2, 2, 31, 2, 24, 24

Offset: 0

Author

Alois P. Heinz, Mar 05 2019

Keywords

• nonn
• base

Examples

			The circulant matrix for n = 23 = 10111_2 is
  [1 0 1 1 1]
  [1 1 0 1 1]
  [1 1 1 0 1]
  [1 1 1 1 0]
  [0 1 1 1 1] and has permanent 44, thus a(23) = 44.
a(10) = 4 != a(12) = 2 although 10 = 1010_2 and 12 = 1100_2 have the same number of 0's and 1's.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..8191
• Wikipedia, Circulant matrix
• Wikipedia, Permanent (mathematics)
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000051, A000079, A000142, A000225, A007283, A007395, A008305, A040000, A113473, A126646, A306595.

Programs

Maple

a:= n-> (l-> LinearAlgebra[Permanent](Matrix(nops(l),
         shape=Circulant[l])))(convert(n, base, 2)):
seq(a(n), n=0..100);

Formula

a(n) = 1 <=> n in { A000079 }.

a(n) = floor(log_2(2n))! for n in { A126646 }.

a(A000225(n)) = A000142(n) for n >= 1.

a(A000051(n)) = A040000(n).

a(A007283(n)) = A007395(n+1).