A115117 Number of primitive (aperiodic, or Lyndon) 3-asymmetric rhythm cycles: ones having no nontrivial shift automorphism. 3-asymmetric rhythm cycles (A115115): binary necklaces of length 3n subject to the restriction that for any k if the k-th bead is of color 1 then the (k+n)-th and (k+2n)-th beads (modulo 3n) are of color 0.
1, 2, 7, 20, 68, 224, 780, 2720, 9709, 34918, 127100, 465920, 1720740, 6390930, 23860928, 89477120, 336860180, 1272578048, 4822419420, 18325176316, 69810262080, 266548209850, 1019836872140, 3909374443520, 15011998757888
Offset: 1
Links
- Jinyuan Wang, Table of n, a(n) for n = 1..1000
- 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.
Programs
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Mathematica
a[n_] := Sum[MoebiusMu[3d] + Boole[GCD[3, d] == 1] MoebiusMu[d] 4^(n/d), {d, Divisors[n]}]/(3n); Array[a, 25] (* Jean-François Alcover, Aug 30 2019 *) -
PARI
a(n) = 1/(3*n) * sumdiv(n,d, moebius(3*d) + if(gcd(3,d)==1, moebius(d)*4^(n/d),0) ); \\ Joerg Arndt, Aug 29 2019
Formula
a(n) = (Sum_{d|n} mu(3d) + Sum_{d|n, (3,d)=1} mu(d) 4^(n/d))/(3n), where mu(n) is the Moebius function A008683.
a(n) ~ 4^n / (3*n). - Vaclav Kotesovec, Oct 27 2024