A032276 Number of bracelets (turnover necklaces) with n beads of 5 colors.
5, 15, 35, 120, 377, 1505, 5895, 25395, 110085, 493131, 2227275, 10196680, 46989185, 218102685, 1017448143, 4768969770, 22440372245, 105966797755, 501938733555, 2384200683816, 11353290089305
Offset: 1
Keywords
Examples
For n=2, the 15 bracelets are AA, AB, AC, AD, AE, BB, BC, BD, BE, CC, CD, CE, DD, DE, and EE. - _Robert A. Russell_, Sep 24 2018
Links
- 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]
- M. Taniguchi, H. Du, and J. S. Lindsey, Enumeration of virtual libraries of combinatorial modular macrocyclic (bracelet, necklace) architectures and their linear counterparts, Journal of Chemical Information and Modeling, 53 (2013), 2203-2216.
- Index entries for sequences related to bracelets
Crossrefs
Programs
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Mathematica
mx=40;CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-5*x^n]/n,{n,mx}]+(1+5 x+10 x^2)/(1-5 x^2))/2,{x,0,mx}],x] (* Herbert Kociemba, Nov 02 2016 *) k=5; Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}] (* Robert A. Russell, Sep 24 2018 *)
Formula
"DIK" (bracelet, indistinct, unlabeled) transform of 5, 0, 0, 0, ...
a(n) = A081720(n,5), n >= 1. - Wolfdieter Lang, Jun 03 2012
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 5*x^n)/n + (1+5*x+10*x^2)/(1-5*x^2))/2. - Herbert Kociemba, Nov 02 2016
a(n) = (3/2)*5^(n/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is even, and = (1/2)*5^((n+1)/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is odd. - Petros Hadjicostas, Sep 01 2018
a(n) = (A001869(n) + A056487(n+1)) / 2 = A278641(n) + A056487(n+1) = A001869(n) - A278641(n). - Robert A. Russell, Oct 13 2018
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{d divides n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Richard L. Ollerton, May 04 2021
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i), where k=5 is the maximum number of colors. (See A051137.) - Richard L. Ollerton, May 04 2021
Comments