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Number of pairs of orientable necklaces with n beads and up to 5 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.
3

%I #16 Sep 25 2018 17:58:38

%S 0,0,0,10,45,252,1130,5270,23520,106960,483756,2211650,10149805,

%T 46911060,217868310,1017057518,4767797895,22438419120,105960938380,

%U 501928967930,2384171386941,11353241261180,54185968572450,259150507387910,1241763071712930,5960463867187752,28656077411358180,137973711706163210

%N Number of pairs of orientable necklaces with n beads and up to 5 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 five different colors.

%F G.f.: k=5, (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=5 is the maximum number of 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,5],{x,0,mx}],x]

%t k=5; 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 5 of A293496.

%Y Cf. A059076 (2 colors), A278639 (3 colors), A278640 (4 colors).

%Y a(n) = (A001869(n) - A056487(n+1)) / 2 = A032276(n) - A056487(n+1).

%Y Equals A001869 - A032276.

%K nonn

%O 0,4

%A _Herbert Kociemba_, Nov 24 2016