OFFSET
1,3
COMMENTS
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
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..1000
P. Hadjicostas, Cyclic, dihedral and symmetric Carlitz compositions of a positive integer, Journal of Integer Sequences, 20 (2017), Article 17.8.5.
Petros Hadjicostas, Proof of an explicit formula for Bower's CycleBG transform
FORMULA
CycleBG transform of (1, 1, 1, 1, ...).
CycleBG transform T(A) = invMOEBIUS(invEULER(Carlitz(A)) + A(x^2) - A) + A.
Carlitz transform T(A(x)) has g.f. 1/(1 - Sum_{k>0} (-1)^(k+1)*A(x^k)).
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
Conjecture: a(n) ~ A241902^n / n. - Vaclav Kotesovec, Sep 06 2017
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
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
EXAMPLE
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.
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Christian G. Bower, Apr 29 2005
EXTENSIONS
Name clarified by Andrew Howroyd, Oct 12 2017
STATUS
approved