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A306714
Permanent of the circulant matrix whose first row is given by the binary expansion of n.
5
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
0,4
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).
EXAMPLE
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.
MAPLE
a:= n-> (l-> LinearAlgebra[Permanent](Matrix(nops(l),
shape=Circulant[l])))(convert(n, base, 2)):
seq(a(n), n=0..100);
KEYWORD
nonn,base
AUTHOR
Alois P. Heinz, Mar 05 2019
STATUS
approved