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%I #57 Mar 21 2024 12:40:09
%S 0,0,0,0,0,0,1,2,6,14,30,62,128,252,495,968,1866,3600,6917,13286,
%T 25476,48916,93837,180314,346554,666996,1284570,2477342,4781502,
%U 9240012,17871708,34604066,67060746,130085052,252548760,490722344
%N Number of pairs of orientable necklaces with n beads and two colors; i.e., turning the necklace over does not leave it unchanged.
%C Number of chiral bracelets with n beads and two colors.
%H G. C. Greubel, <a href="/A059076/b059076.txt">Table of n, a(n) for n = 0..1000</a>
%H Daniel Gabric and Joe Sawada, <a href="https://arxiv.org/abs/2401.14341">Efficient Construction of Long Orientable Sequences</a>, arXiv:2401.14341 [cs.DS], 2024.
%H Petros Hadjicostas, <a href="/A059076/a059076.pdf">Formulas for chiral bracelets</a>, 2019; see Section 5.
%H John P. McSorley and Alan H. Schoen, <a href="http://dx.doi.org/10.1016/j.disc.2012.08.021">Rhombic tilings of (n,k)-ovals, (n, k, lambda)-cyclic difference sets, and related topics</a>, Discrete Math., 313 (2013), 129-154. - From _N. J. A. Sloane_, Nov 26 2012
%F a(n) = A000031(n) - A000029(n) = A000029(n) - A029744(n) = (A000031(n) - A029744(n))/2 = A008965(n) - A091696(n)
%F 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
%F 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
%e 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
%t 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 *)
%t 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 *)
%t 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 *)
%t 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 *)
%Y Column 2 of A293496.
%Y Cf. A059053.
%Y Column 2 of A305541.
%Y Equals (A000031 - A164090) / 2.
%Y a(n) = (A052823(n) - A027383(n-2)) / 2.
%K nonn
%O 0,8
%A _Henry Bottomley_, Dec 22 2000
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