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A032191
Number of necklaces with 6 black beads and n-6 white beads.
11
1, 1, 4, 10, 22, 42, 80, 132, 217, 335, 504, 728, 1038, 1428, 1944, 2586, 3399, 4389, 5620, 7084, 8866, 10966, 13468, 16380, 19811, 23751, 28336, 33566, 39576, 46376, 54132, 62832, 72675, 83661, 95988, 109668, 124936, 141778
OFFSET
6,3
COMMENTS
The g.f. is Z(C_6,x)/x^6, the 6-variate cycle index polynomial for the cyclic group C_6, with substitution x[i]->1/(1-x^i), i=1,...,6. Therefore by Polya enumeration a(n+6) is the number of cyclically inequivalent 6-necklaces whose 6 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_6,x). Note the equivalence of this formulation with the one given as this sequence's name: start with a black 6-necklace (all 6 beads have labels 0). Insert after each of the 6 black beads k white ones if the label was k and then disregard the labels. - Wolfdieter Lang, Feb 15 2005
The g.f. of the CIK[k] transform of the sequence (b(n): n>=1), which has g.f. B(x) = Sum_{n>=1} b(n)*x^n, is CIK[k](x) = (1/k)*Sum_{d|k} phi(d)*B(x^d)^{k/d}. Here, k = 6, b(n) = 1 for all n >= 1, and B(x) = x/(1-x), from which we get another proof of the g.f.s given below. - Petros Hadjicostas, Jan 07 2018
LINKS
C. G. Bower, Transforms (2)
Christian Meyer, On conjecture no. 75 arising from the OEIS, preprint, 2004. [cached copy]
Mónica A. Reyes, Cristina Dalfó, Miguel Àngel Fiol, and Arnau Messegué, A general method to find the spectrum and eigenspaces of the k-token of a cycle, and 2-token through continuous fractions, arXiv:2403.20148 [math.CO], 2024. See p. 6.
Ralf Stephan, Prove or disprove: 100 conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,-1,1,4,-3,-3,4,1,-1,-3,1,2,-1).
FORMULA
"CIK[ 6 ]" (necklace, indistinct, unlabeled, 6 parts) transform of 1, 1, 1, 1, ...
G.f.: (1-x+x^2+4*x^3+2*x^4+3*x^6+x^7+x^8)/((1-x)^6*(1+x)^3*(1+x+x^2)^2*(1-x+x^2)) (conjectured). - Ralf Stephan, May 05 2004
G.f.: (x^6)*(1-x+x^2+4*x^3+2*x^4+3*x^6+x^7+x^8)/((1-x)^2*(1-x^2)^2*(1-x^3)*(1-x^6)). (proving the R. Stephan conjecture (with the correct offset) in a different version; see Comments entry above). - Wolfdieter Lang, Feb 15 2005
G.f.: (1/6)*x^6*((1-x)^(-6)+(1-x^2)^(-3)+2*(1-x^3)^(-2)+2*(1-x^6)^(-1)). - Herbert Kociemba, Oct 22 2016
EXAMPLE
From Petros Hadjicostas, Jan 07 2018: (Start)
We explain why a(8) = 4. According to the theory of transforms by C. G. Bower, given in the weblink above, a(8) is the number of ways of arranging 6 indistinct unlabeled boxes (that may differ only in their size) as a necklace, on a circle, such that the total number of balls in all of them is 8. There are 4 ways for doing that on a circle: 311111, 221111, 212111, and 211211.
To translate these configurations of boxes into necklaces with 8 beads, 6 of them black and 2 of them white, we modify an idea given above by W. Lang. We replace each box that has m balls with a black bead followed by m-1 white beads. The four examples above become BWWBBBBB, BWBWBBBB, BWBBWBBB, and BWBBBWBB.
(End)
MATHEMATICA
k = 6; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)
CROSSREFS
Column k=6 of A047996.
Sequence in context: A023609 A055364 A284870 * A065568 A007825 A008256
KEYWORD
nonn
STATUS
approved