%I #23 Aug 21 2020 14:33:38
%S 2,4,8,24,70,232,782,2744,9710,34990,127102,466152,1720742,6391714,
%T 23860936,89479864,336860182,1272587758,4822419422,18325211326,
%U 69810262088,266548336954,1019836872142,3909374909672,15011998757958
%N 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.
%H R. W. Hall and P. Klingsberg, <a href="https://archive.bridgesmathart.org/2004/bridges2004-189.html">Asymmetric Rhythms, Tiling Canons and Burnside's Lemma</a>, Bridges Proceedings, pp. 189-194, 2004 (Winfield, Kansas).
%H R. W. Hall and P. Klingsberg, <a href="https://doi.org/10.1080/00029890.2006.11920376">Asymmetric Rhythms and Tiling Canons</a>, Preprint, 2004; The American Mathematical Monthly, Volume 113, 2006 - Issue 10, [<a href="https://www.jstor.org/stable/27642087">alternative link</a>].
%F 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.
%F a(n) ~ 4^n / (3*n). - _Vaclav Kotesovec_, Aug 28 2019
%t a[n_] := (Sum[EulerPhi[3d], {d, Divisors[n]}] + Sum[Boole[CoprimeQ[3, d]] EulerPhi[d] 4^(n/d), {d, Divisors[n]}])/(3n);
%t Array[a, 25] (* _Jean-François Alcover_, Aug 28 2019 *)
%o (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
%Y Cf. A115114.
%K easy,nonn
%O 1,1
%A _Valery A. Liskovets_, Jan 17 2006