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%I #23 Mar 06 2019 15:42:14
%S 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,
%T 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,
%U 2,3,2,24,24,4,2,3,3,32,3,4,32,5,2,24,3
%N Determinant of the circulant matrix whose first column corresponds to the binary digits of n.
%C This sequence is the binary variant of A177894.
%C From _Robert Israel_, Mar 05 2019: (Start)
%C a(n) is divisible by A000120(n).
%C If A070939(n) is even then n is divisible by A000120(n)*A065359(n). (End)
%H Robert Israel, <a href="/A306595/b306595.txt">Table of n, a(n) for n = 0..10000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Circulant_matrix">Circulant matrix</a>
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%F a(A121016(n)) = 0 for any n > 0.
%F a(2^k) = 1 for any k >= 0.
%F a(A219325(n)) = A219325(n) for any n > 0.
%e For n = 13:
%e - the binary representation of 13 is "1101",
%e - the corresponding circulant matrix is:
%e [1 1 0 1]
%e [1 1 1 0]
%e [0 1 1 1]
%e [1 0 1 1]
%e - its determinant is -3,
%e - hence a(13) = -3.
%p a:= n-> `if`(n=1, 1, (l-> LinearAlgebra[Determinant](Matrix(nops(l),
%p shape=Circulant[l[-i]$i=1..nops(l)])))(convert(n, base, 2))):
%p seq(a(n), n=0..100); # _Alois P. Heinz_, Mar 05 2019
%o (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))
%Y Cf. A000120, A065359, A070939, A121016, A177894, A219325, A306714.
%K sign,base,look
%O 0,6
%A _Rémy Sigrist_, Feb 27 2019
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