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Number of bracelets (turnover necklaces) with n beads of 5 colors.
7

%I #55 May 05 2021 00:18:53

%S 5,15,35,120,377,1505,5895,25395,110085,493131,2227275,10196680,

%T 46989185,218102685,1017448143,4768969770,22440372245,105966797755,

%U 501938733555,2384200683816,11353290089305

%N Number of bracelets (turnover necklaces) with n beads of 5 colors.

%C From _Petros Hadjicostas_, Sep 01 2018: (Start)

%C The DIK transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has g.f. -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-C(x^m)) + (1 + C(x))^2/(4*(1-C(x^2))) - 1/4.

%C Here, c(1) = 5 and c(n) = 0 for n >= 2, and thus, C(x) = 5*x. Substituting this to the above g.f., we get that the g.f. of the current sequence is A(x) = Sum_{n >= 1} a(n)*x^n = -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-5*x^m) + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4. This agrees with _Herbert Kociemba_'s g.f. below except for an extra 1 because (1 + (1+5*x+10*x^2)/(1-5*x^2))/2 = 1 + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4.

%C (End)

%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 M. Taniguchi, H. Du, and J. S. Lindsey, <a href="https://doi.org/10.1021/ci400175f">Enumeration of virtual libraries of combinatorial modular macrocyclic (bracelet, necklace) architectures and their linear counterparts</a>, Journal of Chemical Information and Modeling, 53 (2013), 2203-2216.

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

%F "DIK" (bracelet, indistinct, unlabeled) transform of 5, 0, 0, 0, ...

%F a(n) = A081720(n,5), n >= 1. - _Wolfdieter Lang_, Jun 03 2012

%F 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

%F 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

%F a(n) = (A001869(n) + A056487(n+1)) / 2 = A278641(n) + A056487(n+1) = A001869(n) - A278641(n). - _Robert A. Russell_, Oct 13 2018

%F 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

%F 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

%e 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

%t 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 *)

%t 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 *)

%Y Cf. A081720.

%Y Column 5 of A051137.

%Y Cf. A001869 (oriented), A056487 (achiral), A278641 (chiral).

%K nonn

%O 1,1

%A _Christian G. Bower_