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A365859
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Number of self-dual cyclic n-color compositions.
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3
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1, 1, 2, 1, 3, 2, 5, 1, 10, 3, 19, 2, 41, 5, 94, 1, 211, 10, 493, 3, 1170, 19, 2787, 2, 6713, 41, 16274, 5, 39651, 94, 97109, 1, 238838, 211, 589527, 10, 1459961, 493, 3626242, 3, 9030451, 1170, 22542397, 19, 56393862, 2787, 141358275, 2, 354975433, 6713, 892893262, 41, 2249412291, 16274, 5674891017
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OFFSET
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1,3
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COMMENTS
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A cyclic composition is a sum in which the order of the parts is considered up to cyclic permutation. In other words, it is the collection of components remaining in the cycle graph C_n on n vertices when one or more edges are removed, and rotations are considered equivalent. In an n-color composition, each part of size k is assigned one of k "colors" which may be represented graphically by marking one vertex in the part. The dual of a cyclic n-color composition is obtained by switching the roles of edges and vertices in C_n, then removing each edge that came from a previously marked vertex while marking each vertex that came from a previously removed edge. A cyclic n-color composition is self-dual if it is invariant under this process.
a(n) is also the number of cyclic compositions of A000265(n) into odd parts.
This sequence is self-similar; removing all odd-indexed terms results in the same sequence.
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LINKS
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FORMULA
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G.f.: Sum_{k>=1} phi(2*k)/(2*k) * log((1+x^k-x^(2*k))/(1-x^k-x^(2*k))).
a(n) = (1/(b(n)))*[Sum_{k divides A000265(n)} phi(k)*lucas(b(n)/k)], where b(n) = A000265(n) and lucas(n) = A000204(n).
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EXAMPLE
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Every power of 2 has only one self-dual cyclic n-color composition, which has all parts of size 1.
The self-dual cyclic n-color compositions of 5 are 1_1+1_1+1_1+1_1+1_1, 1_1+2_2+2_1, and 5_3, where the subscript indicates the color of the part, or which vertex is marked within the part.
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PROG
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(PARI) my(N=66, x='x+O('x^N)); Vec( sum(k=1, N, eulerphi(2*k)/(2*k) * log((1+x^k-x^(2*k))/(1-x^k-x^(2*k))) ) ) \\ Joerg Arndt, Sep 21 2023
(Python)
from sympy import totient, lucas, divisors
m = n>>(~n&n-1).bit_length()
return sum(totient(k)*lucas(m//k) for k in divisors(m, generator=True))//m # Chai Wah Wu, Sep 23 2023
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CROSSREFS
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KEYWORD
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nonn,changed
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AUTHOR
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STATUS
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approved
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