A115114 Asymmetric rhythm cycles (patterns): 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.
2, 3, 6, 11, 26, 63, 158, 411, 1098, 2955, 8054, 22151, 61322, 170823, 478318, 1345211, 3798242, 10761723, 30585830, 87169619, 249056138, 713205903, 2046590846, 5883948951, 16945772210, 48882035163, 141214768974
Offset: 1
Examples
For n=3, the 27=3^3 admissible words are separated into 6 shift-equivalence classes (necklaces) containing, resp., the words 000000, 100000, 110000, 101000, 111000 and 101010. Thus a(3)=6.
Links
- 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; The American Mathematical Monthly, Volume 113, 2006 - Issue 10, [alternative link].
Programs
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Mathematica
a[n_] := Sum[EulerPhi[2d] + Boole[OddQ[d]] EulerPhi[d] 3^(n/d), {d, Divisors[n]}]/(2n); Array[a, 27] (* Jean-François Alcover, Aug 29 2019 *)
Formula
a(n) = (Sum_{d|n}phi(2d)+Sum_{d|n, d odd}phi(d)3^(n/d))/(2n), where phi(n) is the Euler function A000010.
a(n) ~ 3^n/(2*n). - Vaclav Kotesovec, Oct 27 2024