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%I #32 Jan 23 2024 17:39:53
%S 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,
%T 44,44,120,1,2,4,20,8,17,17,80,4,17,36,82,17,80,82,265,2,20,17,80,17,
%U 82,80,265,20,80,82,265,80,265,265,720,1,2,2,31,2,24,24
%N Permanent of the circulant matrix whose first row is given by the binary expansion of n.
%H Alois P. Heinz, <a href="/A306714/b306714.txt">Table of n, a(n) for n = 0..8191</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Circulant_matrix">Circulant matrix</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Permanent_(mathematics)">Permanent (mathematics)</a>
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%F a(n) = 1 <=> n in { A000079 }.
%F a(n) = floor(log_2(2n))! for n in { A126646 }.
%F a(A000225(n)) = A000142(n) for n >= 1.
%F a(A000051(n)) = A040000(n).
%F a(A007283(n)) = A007395(n+1).
%e The circulant matrix for n = 23 = 10111_2 is
%e [1 0 1 1 1]
%e [1 1 0 1 1]
%e [1 1 1 0 1]
%e [1 1 1 1 0]
%e [0 1 1 1 1] and has permanent 44, thus a(23) = 44.
%e a(10) = 4 != a(12) = 2 although 10 = 1010_2 and 12 = 1100_2 have the same number of 0's and 1's.
%p a:= n-> (l-> LinearAlgebra[Permanent](Matrix(nops(l),
%p shape=Circulant[l])))(convert(n, base, 2)):
%p seq(a(n), n=0..100);
%Y Cf. A000051, A000079, A000142, A000225, A007283, A007395, A008305, A040000, A113473, A126646, A306595.
%K nonn,base
%O 0,4
%A _Alois P. Heinz_, Mar 05 2019
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