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%I #17 Sep 25 2018 14:56:52
%S 0,0,0,4,15,72,270,1044,3795,14060,51204,188604,694130,2572920,
%T 9567090,35758704,134137875,505159200,1908554190,7233104844,
%U 27486506268,104713296760,399817262550,1529746919604,5864041395730,22517964582504,86607602546220,333599838189804,1286742419927070,4969488707124120,19215357085867800
%N Number of pairs of orientable necklaces with n beads and up to 4 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 four different colors.
%F G.f.: k=4, (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=4 is the maximum number of colors. - _Robert A. Russell_, Sep 24 2018
%e Example: For 3 beads and the colors A, B, C and D the 4 orientable necklaces are ABC, ABD, ACD and BCD. The turned-over necklaces ACB, ADB, ADC and BDC are not included in the count.
%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,4],{x,0,mx}],x]
%t k=4; 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 4 of A293496.
%Y Cf. A059076 (2 colors), A278639 (3 colors).
%Y Equals (A001868 - A056486) / 2 = A001868 - A032275 = A032275 - A056486.
%K nonn
%O 0,4
%A _Herbert Kociemba_, Nov 24 2016