

A032196


Number of necklaces with 11 black beads and n11 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 11variate cycle index polynomial for the cyclic group C_11, with substitution x[i]>1/(1x^i), i=1..11. By Polya enumeration, a(n+11) is the number of cyclically inequivalent 11necklaces 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.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
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) / ((1x)^10 * (1x^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



