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A032196 Number of necklaces with 11 black beads and n-11 white beads. 1
1, 1, 6, 26, 91, 273, 728, 1768, 3978, 8398, 16796, 32066, 58786, 104006, 178296, 297160, 482885, 766935, 1193010, 1820910, 2731365, 4032015, 5864750, 8414640, 11920740, 16689036, 23107896, 31666376, 42975796 (list; graph; refs; listen; history; text; internal format)
OFFSET

11,3

COMMENTS

The g.f. is Z(C_11,x)/x^11, the 11-variate cycle index polynomial for the cyclic group C_11, with substitution x[i]->1/(1-x^i), i=1..11. By Polya enumeration, a(n+11) is the number of cyclically inequivalent 11-necklaces whose 11 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_11,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005

LINKS

Table of n, a(n) for n=11..39.

C. G. Bower, Transforms (2)

F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.

Index entries for sequences related to necklaces

FORMULA

"CIK[ 11 ]" (necklace, indistinct, unlabeled, 11 parts) transform of 1, 1, 1, 1...

G.f.: (x^11) * (1 - 9*x + 41*x^2 - 109*x^3 + 191*x^4 - 229*x^5 + 191*x^6 - 109*x^7 + 41*x^8 - 9*x^9 + x^10) / ((1-x)^10 * (1-x^11)).

a(n) = ceiling(binomial(n, 11)/n)) (conjecture Wolfdieter Lang).

From Herbert Kociemba, Oct 11 2016: (Start)

This conjecture indeed is true.

Sketch of proof:

There are binomial(n,11) ways to place the 11 black beads in the necklace with n beads. If n is not divisible by 11 there are no necklaces with a rotational symmetry. So exactly n necklaces are equivalent up to rotation and there are binomial(n,11)/n = ceiling(binomial(n,11)/n) equivalence classes.

If n is divisible by 11 the only way to get a necklace with rotational symmetry is to space out the 11 black beads evenly. There are n/11 ways to do this and all ways are equivalent up to rotation. So there are  binomial(n,11) - n/11 unsymmetric necklaces which give binomial(n,11)/n - 1/11 equivalence classes. If we add the single symmetric equivalence class we get Binomial(n,11)/n - 1/11 + 1 which also is ceiling(binomial(n,11)/n). (End)

G.f.: (10/(1 - x^11) + 1/(1 - x)^11)*x^11/11. - Herbert Kociemba, Oct 16 2016

MATHEMATICA

k = 11; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)

DeleteCases[CoefficientList[Series[(x^11) (1 - 9 x + 41 x^2 - 109 x^3 + 191 x^4 - 229 x^5 + 191 x^6 - 109 x^7 + 41 x^8 - 9 x^9 + x^10)/((1 - x)^10 (1 - x^11)), {x, 0, 39}], x], 0] (* Michael De Vlieger, Oct 10 2016 *)

CROSSREFS

Column k=11 of A047996.

Cf. A004526, A007997, A008610, A008646, A032191, A032192, A032193, A032194, A032195.

Sequence in context: A036422 A166214 A032169 * A011780 A036631 A224288

Adjacent sequences:  A032193 A032194 A032195 * A032197 A032198 A032199

KEYWORD

nonn

AUTHOR

Christian G. Bower

STATUS

approved

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Last modified February 22 19:36 EST 2018. Contains 299469 sequences. (Running on oeis4.)