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 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. 5
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Number of chiral bracelets of n beads using up to three different colors. LINKS 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 * A222643 A129014 A335412 Adjacent sequences:  A278636 A278637 A278638 * A278640 A278641 A278642 KEYWORD nonn AUTHOR Herbert Kociemba, Nov 24 2016 STATUS approved

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Last modified August 11 00:32 EDT 2020. Contains 336403 sequences. (Running on oeis4.)