A278639 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.

0, 0, 0, 1, 3, 12, 38, 117, 336, 976, 2724, 7689, 21455, 60228, 168714, 475037, 1338861, 3788400, 10742588, 30556305, 87112059, 248967564, 713032782, 2046325125, 5883428618, 16944975048, 48880471500, 141212377489, 408509453511, 1183275193908, 3431504760514

Offset

0,5

Comments

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

Links

Table of n, a(n) for n=0..30.

Formula

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.

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

Example

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.
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

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, 3], {x, 0, mx}], x]
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 *)

Crossrefs

Column 3 of A293496.

Cf. A059076 (2 colors).

a(n) = (A001867(n) - A182751(n-1)) / 2.

Equals A001867 - A027671.

a(n) = A027671(n) - A182751(n-1).

Sequence in context: A048246 A320203 A217093 * A350787 A222643 A129014

Adjacent sequences: A278636 A278637 A278638 * A278640 A278641 A278642

Keyword

nonn

Author

Herbert Kociemba, Nov 24 2016

Status

approved