%I #53 Oct 26 2017 09:28:51

%S 1,1,2,2,3,6,7,11,18,29,42,73,111,183,299,491,796,1333,2188,3652,6073,

%T 10155,16959,28500,47813,80508,135621,228967,386749,654535,1108353,

%U 1879478,3189495,5418556,9212099,15676275,26694509,45493327,77580915

%N Number of circular compositions of n such that no two adjacent parts are equal.

%C By "circular compositions" here we mean equivalence classes of compositions with parts on a circle such that two compositions are equivalent if one is a cyclic shift of the other. - _Petros Hadjicostas_, Oct 15 2017

%H Vaclav Kotesovec, <a href="/A106369/b106369.txt">Table of n, a(n) for n = 1..1000</a>

%H P. Hadjicostas, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL20/Hadjicostas/hadji5.html">Cyclic, dihedral and symmetric Carlitz compositions of a positive integer</a>, Journal of Integer Sequences, 20 (2017), Article 17.8.5.

%H Petros Hadjicostas, <a href="/A106369/a106369.pdf">Proof of an explicit formula for Bower's CycleBG transform</a>

%H <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>

%F CycleBG transform of (1, 1, 1, 1, ...).

%F CycleBG transform T(A) = invMOEBIUS(invEULER(Carlitz(A)) + A(x^2) - A) + A.

%F Carlitz transform T(A(x)) has g.f. 1/(1 - Sum_{k>0} (-1)^(k+1)*A(x^k)).

%F G.f.: x/(1-x) - Sum_{s>=1} (phi(s)/s)*f(x^s), where f(x) = log(1 - Sum_{n>=1} x^n/(1 + x^n)) + Sum_{n>=1} log(1 + x^n) and phi(s)=A000010 is Euler's totient function. - _Petros Hadjicostas_, Sep 06 2017

%F Conjecture: a(n) ~ A241902^n / n. - _Vaclav Kotesovec_, Sep 06 2017

%F General formula for the CycleBG transform: T(A)(x) = A(x) - Sum_{k>=0} A(x^(2k+1)) + Sum_{k>=1} (phi(k)/k)*log(Carlitz(A)(x^k)). For a proof, see the links above. (For this sequence, A(x) = x/(1-x).) - _Petros Hadjicostas_, Oct 08 2017

%F G.f.: -Sum_{s>=1} x^(2s+1)/(1-x^(2s+1)) - Sum_{s>=1} (phi(s)/s)*g(x^s), where g(x) = log(1 + Sum_{n>=1} (-x)^n/(1 - x^n)). (This formula can be proved from the general formula for the CycleBG transform given above.) - _Petros Hadjicostas_, Oct 10 2017

%e a(6) = 6 because the 6 circular compositions of 6: 6, 5+1, 4+2, 3+2+1, 3+1+2, 2+1+2+1.

%t nmax = 40; Rest[CoefficientList[Series[x/(1-x) - Sum[EulerPhi[s]/s*(Log[1 - Sum[x^(s*n)/(1 + x^(s*n)), {n, 1, nmax}]] + Sum[Log[1 + x^(s*n)], {n, 1, nmax}]), {s, 1, nmax}], {x, 0, nmax}], x]] (* _Vaclav Kotesovec_, Sep 06 2017, after _Petros Hadjicostas_ *)

%Y Cf. A000031, A008965, A212322, A292906.

%K nonn

%O 1,3

%A _Christian G. Bower_, Apr 29 2005

%E Name clarified by _Andrew Howroyd_, Oct 12 2017