1,2

Asymmetric rhythm cycles (A115114): binary necklaces of length 2n subject to the restriction that for any k if the k-th bead is of color 1 then the (k+n)-th bead (modulo 2n) is of color 0. - Valery A. Liskovets (liskov(AT)im.bas-net.by), Jan 17 2006

This sequence differs from the Moebius transform of A115114 (for even n). Coincides with the second row (q=3) of array A098691. - Valery A. Liskovets (liskov(AT)im.bas-net.by), Jan 17 2006

This sequence is the number of Lyndon words on {1, 2, 3} with an odd number of 1's. Also, for even n, this sequence represents the differences between the number of Lyndon words on {1, 2, 3} with an odd number of 1's and the number of Lyndon words on {1, 2, 3} with an even number of 1's. - Jennifer Woodcock (jennifer.woodcock(AT)ugdsb.on.ca), Jan 03 2008

D. Shanks and M. Lal, Bateman's constants reconsidered and and the distribution of cubic residues, Math. Comp., 26 (1972), 265-285.

R. W. Hall and P. Klingsberg, Asymmetric Rhythms, Tiling Canons, and Burnside's Lemma,Bridges Proceedings, pp. 189-194, 2004 (Winfield, Kansas).

R. W. Hall and P. Klingsberg, Asymmetric Rhythms and Tiling Canons, Preprint, 2004.

a(n)=(Sum_{d|n, d odd}mu(d)(3^(n/d)-1))/(2n). Also a(n)=(3^n-1)/(2n) for n=2^k, and a(n)=(Sum_{d|n, d odd}mu(d)3^(n/d))/(2n) otherwise. - Valery A. Liskovets (liskov(AT)im.bas-net.by), Jan 17 2006

Example. For n=3, out of 6=A115114(3) admissible rhythm cycles (necklaces) 000000, 100000, 110000, 101000, 111000 and 101010, only the first and the last ones are imprimitive. Thus a(3)=4.

Cf. A133267.

Adjacent sequences: A006572 A006573 A006574 this_sequence A006576 A006577 A006578

Sequence in context: A100087 A088354 A055919 this_sequence A138175 A121691 A124499

nonn

njas

Edited and extended by Valery A. Liskovets (liskov(AT)im.bas-net.by), Jan 17 2006