%I #56 Aug 06 2024 21:15:12

%S 1,1,6,26,91,273,728,1768,3978,8398,16796,32066,58786,104006,178296,

%T 297160,482885,766935,1193010,1820910,2731365,4032015,5864750,8414640,

%U 11920740,16689036,23107896,31666376,42975796

%N Number of necklaces with 11 black beads and n-11 white beads.

%C 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

%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>

%H F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>

%H F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]

%H <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>

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

%F 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)).

%F a(n) = ceiling(binomial(n, 11)/n) (conjecture _Wolfdieter Lang_).

%F From _Herbert Kociemba_, Oct 11 2016: (Start)

%F This conjecture indeed is true.

%F Sketch of proof:

%F 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.

%F 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)

%F G.f.: (10/(1 - x^11) + 1/(1 - x)^11)*x^11/11. - _Herbert Kociemba_, Oct 16 2016

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

%t 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 *)

%Y Column k=11 of A047996.

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

%K nonn

%O 11,3

%A _Christian G. Bower_