A059076 Number of pairs of orientable necklaces with n beads and two colors; i.e., turning the necklace over does not leave it unchanged.
0, 0, 0, 0, 0, 0, 1, 2, 6, 14, 30, 62, 128, 252, 495, 968, 1866, 3600, 6917, 13286, 25476, 48916, 93837, 180314, 346554, 666996, 1284570, 2477342, 4781502, 9240012, 17871708, 34604066, 67060746, 130085052, 252548760, 490722344
Offset: 0
Keywords
Examples
For n=6, the only chiral pair is AABABB-AABBAB. For n=7, the two chiral pairs are AAABABB-AAABBAB and AABABBB-AABBBAB. - _Robert A. Russell_, Sep 24 2018
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Daniel Gabric and Joe Sawada, Efficient Construction of Long Orientable Sequences, arXiv:2401.14341 [cs.DS], 2024.
- Petros Hadjicostas, Formulas for chiral bracelets, 2019; see Section 5.
- John P. McSorley and Alan H. Schoen, Rhombic tilings of (n,k)-ovals, (n, k, lambda)-cyclic difference sets, and related topics, Discrete Math., 313 (2013), 129-154. - From _N. J. A. Sloane_, Nov 26 2012
Crossrefs
Programs
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Mathematica
nn=35;Table[CoefficientList[Series[CycleIndex[CyclicGroup[n],s]-CycleIndex[DihedralGroup[n],s]/.Table[s[i]->2,{i,1,n}],{x,0,nn}],x],{n,1,nn}]//Flatten (* Geoffrey Critzer, Mar 26 2013 *) mx=40; CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-2*x^n]/n, {n, mx}]-(1+x)^2/(1-2*x^2))/2, {x, 0, mx}], x] (* Herbert Kociemba, Nov 02 2016 *) terms = 36; a29[0] = 1; a29[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#) & ]/(2*n); Array[a29, 36, 0] - LinearRecurrence[{0, 2}, {1, 2, 3}, 36] (* Jean-François Alcover, Nov 05 2017 *) k = 2; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n)(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)
Formula
a(n) = A000031(n) - A000029(n) = A000029(n) - A029744(n) = (A000031(n) - A029744(n))/2 = A008965(n) - A091696(n)
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 2*x^n)/n - (1 + x)^2/(1 - 2*x^2))/2. - Herbert Kociemba, Nov 02 2016
For n > 0, a(n) = -(k^floor((n + 1)/2) + k^ceiling((n + 1)/2))/4 + (1/(2*n))* Sum_{d|n} phi(d)*k^(n/d), where k = 2 is the maximum number of colors. - Robert A. Russell, Sep 24 2018
Comments