A278640 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.

0, 0, 0, 4, 15, 72, 270, 1044, 3795, 14060, 51204, 188604, 694130, 2572920, 9567090, 35758704, 134137875, 505159200, 1908554190, 7233104844, 27486506268, 104713296760, 399817262550, 1529746919604, 5864041395730, 22517964582504, 86607602546220, 333599838189804, 1286742419927070, 4969488707124120, 19215357085867800

Offset: 0

Author

Herbert Kociemba, Nov 24 2016

Keywords

• nonn

Comments

Number of chiral bracelets of n beads using up to four different colors.

Examples

			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.
		

Crossrefs

Column 4 of A293496.

Cf. A059076 (2 colors), A278639 (3 colors).

Equals (A001868 - A056486) / 2 = A001868 - A032275 = A032275 - A056486.

Programs

Mathematica

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]
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 *)

Formula

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.

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