A365859 Number of self-dual cyclic n-color compositions.
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
Offset: 1
Keywords
Examples
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.
Links
- A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (11) (2000), 1421-1427.
- Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 33.
- Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color cyclic compositions, Discrete Mathematics 341 (2018), 3209-3226.
- Brian Hopkins, Jesús Sistos Barrón, and Hua Wang, Conjugating cyclic n-color compositions, Utilitas Mathematica (2025) Vol. 122, 53-64. See p. 61.
- Jesus Omar Sistos Barron, Counting Conjugates of Colored Compositions, Honors College Thesis, Georgia Southern Univ. (2024), No. 985. See p. 25.
Programs
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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
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Python
from sympy import totient, lucas, divisors def A365859(n): 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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