%I #19 Sep 25 2018 14:57:12
%S 0,0,0,1,3,12,38,117,336,976,2724,7689,21455,60228,168714,475037,
%T 1338861,3788400,10742588,30556305,87112059,248967564,713032782,
%U 2046325125,5883428618,16944975048,48880471500,141212377489,408509453511,1183275193908,3431504760514
%N Number of pairs of orientable necklaces with n beads and up to 3 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.
%C Number of chiral bracelets of n beads using up to three different colors.
%F G.f.: k=3, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} binomial(k,i)*x^i / (1 - k*x^2))/2.
%F For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=3 is the maximum number of colors. - _Robert A. Russell_, Sep 24 2018
%e Example: The 3 orientable necklaces with 4 beads and the colors A, B and C are AABC, BBAC and CCAB. The turned-over necklaces AACB, BBCA and CCBA are not included in the count.
%e For n=6, the three chiral pairs using just two colors are AABABB-AABBAB, AACACC-AACCAC, and BBCBCC-BBCCBC. The other 35 use three colors. - _Robert A. Russell_, Sep 24 2018
%t mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,3],{x,0,mx}],x]
%t k=3; 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 3 of A293496.
%Y Cf. A059076 (2 colors).
%Y a(n) = (A001867(n) - A182751(n-1)) / 2.
%Y Equals A001867 - A027671.
%Y a(n) = A027671(n) - A182751(n-1).
%K nonn
%O 0,5
%A _Herbert Kociemba_, Nov 24 2016