A115115 Number of 3-asymmetric rhythm cycles: 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.

2, 4, 8, 24, 70, 232, 782, 2744, 9710, 34990, 127102, 466152, 1720742, 6391714, 23860936, 89479864, 336860182, 1272587758, 4822419422, 18325211326, 69810262088, 266548336954, 1019836872142, 3909374909672, 15011998757958

Offset: 1

Author

Valery A. Liskovets, Jan 17 2006

Keywords

• easy
• nonn

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].

Crossrefs

Cf. A115114.

Programs

Mathematica

a[n_] := (Sum[EulerPhi[3d], {d, Divisors[n]}] + Sum[Boole[CoprimeQ[3, d]] EulerPhi[d] 4^(n/d), {d, Divisors[n]}])/(3n);
Array[a, 25] (* Jean-François Alcover, Aug 28 2019 *)

PARI

a(n) = (sumdiv(n, d, eulerphi(3*d)) + sumdiv(n, d, if (gcd(d, 3)==1, eulerphi(d)*4^(n/d))))/(3*n); \\ Michel Marcus, Aug 28 2019

Formula

a(n) = (Sum_{d|n}phi(3d) + Sum_{d|n, (3, d)=1}phi(d)*4^(n/d))/(3n), where phi(n) is the Euler function A000010.

a(n) ~ 4^n / (3*n). - Vaclav Kotesovec, Aug 28 2019