cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A110710 Number of ternary necklaces with n beads of each color and no adjacent beads of the same color (i.e., no substrings 00, 11, 22).

Original entry on oeis.org

1, 2, 5, 16, 70, 348, 1948, 11444, 70380, 445944, 2896590, 19186740, 129186596, 881808728, 6089851874, 42482906040, 298976142764, 2120377458900, 15141289233972, 108784152585236, 785869931659980, 5705406374249272
Offset: 0

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Author

Max Alekseyev, Aug 05 2005

Keywords

Comments

The number of circular arrangements (counted up to rotations) of n blue, n red and n green items such that there are no adjacent items of the same color. The number of various linear arrangements is given by A110706, A110707 and A110711.

Examples

			For n=2 there are 5 necklaces: 010212, 012012, 012021, 012102, 021021.
		

Crossrefs

Programs

  • Mathematica
    b = Binomial; A110707[n_] := 2*Sum[b[n - 1, k]*(b[n - 1, k]*(b[2*n + 1 - 2*k, n + 1] - 3*b[2*n - 1 - 2*k, n + 1]) + b[n - 1, k + 1]*(b[2*n - 2*k, n + 1] - 3*b[2*n - 2*k - 2, n + 1])), {k,0, n/2}]; a[n_] := DivisorSum[n, A110707[n/#]*EulerPhi[#]&]/(3n); a[0]=1; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Dec 04 2015, adapted from PARI *)
  • PARI
    { A110707(n) = 2 * sum(k=0,n\2, binomial(n-1,k) * (binomial(n-1,k)*(binomial(2*n+1-2*k,n+1)-3*binomial(2*n-1-2*k,n+1)) + binomial(n-1,k+1)*(binomial(2*n-2*k,n+1)-3*binomial(2*n-2*k-2,n+1)) )); A110710(n) = sumdiv(n,d,A110707(n\d)*eulerphi(d))\(3*n); }

Formula

a(n) = Sum_{d|n} A110707(n/d)*eulerphi(d) / (3n) for n>0, a(0)=1.
a(n) ~ sqrt(3) * 2^(3*n - 1) / (Pi * n^2). - Vaclav Kotesovec, Mar 20 2023

Extensions

a(0)=1 prepended by Alois P. Heinz, Dec 04 2015

A005514 Number of n-bead bracelets (turnover necklaces) with 8 red beads and n-8 black beads.

Original entry on oeis.org

1, 1, 5, 10, 29, 57, 126, 232, 440, 750, 1282, 2052, 3260, 4950, 7440, 10824, 15581, 21879, 30415, 41470, 56021, 74503, 98254, 127920, 165288, 211276, 268228, 337416, 421856, 523260, 645456, 790704, 963793, 1167645, 1408185
Offset: 8

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Author

Keywords

Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 8 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=8 (see our comment at A032279). (End)
From Petros Hadjicostas, Jul 14 2018: (Start)
Let (c(n): n >= 1) be a sequence of nonnegative integers and let C(x) = Sum_{n>=1} c(n)*x^n be its g.f. Let k be a positive integer. Let a_k = (a_k(n): n >= 1) be the output sequence of the DIK[k] transform of sequence (c(n): n >= 1), and let A_k(x) = Sum_{n>=1} a_k(n)*x^n be its g.f. See Bower's web link below. It can be proved that, when k is even, A_k(x) = ((1/k)*Sum_{d|k} phi(d)*C(x^d)^(k/d) + (1/2)*C(x^2)^((k/2)-1)*(C(x)^2 + C(x^2)))/2.
For this sequence, k=8, c(n) = 1 for all n >= 1, and C(x) = x/(1-x). Thus, a(n) = a_8(n) for all n >= 1. Since a_k(n) = 0 for 1 <= n <= k-1, the offset of this sequence is n = k = 8. Applying the formula for the g.f. of DIK[8] of (c(n): n >= 1) with C(x) = x/(1-x) and k=8, we get Herbert Kociemba's formula below.
Here, a(n) is defined to be the number of n-bead bracelets of two colors with 8 red beads and n-8 black beads. But it is also the number of dihedral compositions of n with 8 parts. (This statement is equivalent to Vladimir Shevelev's statement above that a(n) is the "number of non-equivalent necklaces of 8 beads each of them painted by one of n colors." By "necklaces", he means "turnover necklaces". See the second paragraph of Section 2 in his 2004 paper in the Indian Journal of Pure and Applied Mathematics.)
Two cyclic compositions of n (with k = 8 parts) belong to the same equivalence class corresponding to a dihedral composition of n if and only if one can be obtained from the other by a rotation or reversal of order. (End)

Examples

			From _Petros Hadjicostas_, Jul 14 2018: (Start)
Every n-bead bracelet of two colors such that 8 beads are red and n-8 are black can be transformed into a dihedral composition of n with 8 parts in the following way. Start with one R bead and go in one direction (say clockwise) until you reach the next R bead. Continue this process until you come back to the original R bead.
Let b_i be the number of beads from R bead i until you reach the last B bead before R bead i+1 (or R bead 1). Here, b_i = 1 iff there are no B beads between R bead i and R bead i+1 (or R bead 8 and R bead 1). Then b_1 + b_2 + ... + b_8 = n, and we get a dihedral composition of n. (Of course, b_2 + b_3 + ... + b_8 + b_1 and b_8 + b_7 + ... + b_1 belong to the same equivalence class of the dihedral composition b_1 + ... + b_8.)
For example, a(10) = 5, and we have the following bracelets with 8 R beads and 2 B beads. Next to the bracelets we list the corresponding dihedral compositions of n with k=8 parts (they must be viewed on a circle):
RRRRRRRRBB <-> 1+1+1+1+1+1+1+3
RRRRRRRBRB <-> 1+1+1+1+1+1+2+2
RRRRRRBRRB <-> 1+1+1+1+1+2+1+2
RRRRRBRRRB <-> 1+1+1+1+2+1+1+2
RRRRBRRRRB <-> 1+1+1+2+1+1+1+2
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Programs

  • Mathematica
    k = 8; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    k=8;CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)

Formula

S. J. Cyvin et al. (1997) give a g.f. (See equation (18) on p. 870 of their paper. Their g.f. is the same as the one given by V. Jovovic below except for the extra x^8.) - Petros Hadjicostas, Jul 14 2018
G.f.: (x^8/16)*(1/(1 - x)^8 + 4/(1 - x^8) + 5/(1 - x^2)^4 + 2/(1 - x^4)^2 + 4/(1 - x)^2/(1 - x^2)^3) = x^8*(2*x^10 - 3*x^9 + 7*x^8 - 6*x^7 + 7*x^6 - 2*x^5 + 2*x^4 - 2*x^3 + 5*x^2 - 3*x + 1)/(1 - x)^8/(1 + x)^4/(1 + x^2)^2/(1 + x^4). - Vladeta Jovovic, Jul 17 2002
a(n) = ((n+4)/32)*s(n,0,8) + ((n-4)/32)*s(n,4,8) + (48*C(n-1,7) + (n+1)*(n-2)*(n-4)*(n-6))/768, if n is even >= 8; a(n) = (48*C(n-1,7) + (n-1)*(n-3)*(n-5)*(n-7))/768, if n odd >= 8, where s(n,k,d)=1, if n == k (mod d), and 0 otherwise. - Vladimir Shevelev, Apr 23 2011
G.f.: k=8, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor((k+2)/2))/2. - Herbert Kociemba, Nov 05 2016 [edited by Petros Hadjicostas, Jul 18 2018]
From Petros Hadjicostas, Jul 14 2018: (Start)
a(n) = (A032193(n) + A119963(n, 8))/2 = (A032193(n) + C(floor(n/2), 4))/2 for n >= 8.
The sequence (a(n): n >= 8) is the output sequence of Bower's "DIK[ 8 ]" (bracelet, indistinct, unlabeled, 8 parts) transform of 1, 1, 1, 1, ...
(End)

Extensions

Sequence extended and description corrected by Christian G. Bower
Name edited by Petros Hadjicostas, Jul 20 2018

A032276 Number of bracelets (turnover necklaces) with n beads of 5 colors.

Original entry on oeis.org

5, 15, 35, 120, 377, 1505, 5895, 25395, 110085, 493131, 2227275, 10196680, 46989185, 218102685, 1017448143, 4768969770, 22440372245, 105966797755, 501938733555, 2384200683816, 11353290089305
Offset: 1

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Keywords

Comments

From Petros Hadjicostas, Sep 01 2018: (Start)
The DIK transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has g.f. -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-C(x^m)) + (1 + C(x))^2/(4*(1-C(x^2))) - 1/4.
Here, c(1) = 5 and c(n) = 0 for n >= 2, and thus, C(x) = 5*x. Substituting this to the above g.f., we get that the g.f. of the current sequence is A(x) = Sum_{n >= 1} a(n)*x^n = -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-5*x^m) + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4. This agrees with Herbert Kociemba's g.f. below except for an extra 1 because (1 + (1+5*x+10*x^2)/(1-5*x^2))/2 = 1 + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4.
(End)

Examples

			For n=2, the 15 bracelets are AA, AB, AC, AD, AE, BB, BC, BD, BE, CC, CD, CE, DD, DE, and EE. - _Robert A. Russell_, Sep 24 2018
		

Crossrefs

Cf. A081720.
Column 5 of A051137.
Cf. A001869 (oriented), A056487 (achiral), A278641 (chiral).

Programs

  • Mathematica
    mx=40;CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-5*x^n]/n,{n,mx}]+(1+5 x+10 x^2)/(1-5 x^2))/2,{x,0,mx}],x] (* Herbert Kociemba, Nov 02 2016 *)
    k=5; Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}] (* Robert A. Russell, Sep 24 2018 *)

Formula

"DIK" (bracelet, indistinct, unlabeled) transform of 5, 0, 0, 0, ...
a(n) = A081720(n,5), n >= 1. - Wolfdieter Lang, Jun 03 2012
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 5*x^n)/n + (1+5*x+10*x^2)/(1-5*x^2))/2. - Herbert Kociemba, Nov 02 2016
a(n) = (3/2)*5^(n/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is even, and = (1/2)*5^((n+1)/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is odd. - Petros Hadjicostas, Sep 01 2018
a(n) = (A001869(n) + A056487(n+1)) / 2 = A278641(n) + A056487(n+1) = A001869(n) - A278641(n). - Robert A. Russell, Oct 13 2018
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{d divides n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Richard L. Ollerton, May 04 2021
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i), where k=5 is the maximum number of colors. (See A051137.) - Richard L. Ollerton, May 04 2021

A056357 Number of bracelet structures using exactly two different colored beads.

Original entry on oeis.org

0, 1, 1, 3, 3, 7, 8, 17, 22, 43, 62, 121, 189, 361, 611, 1161, 2055, 3913, 7154, 13647, 25481, 48733, 92204, 176905, 337593, 649531, 1246862, 2405235, 4636389, 8964799, 17334800, 33588233, 65108061, 126390031, 245492243, 477353375, 928772649, 1808676325
Offset: 1

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Keywords

Comments

Turning over will not create a new bracelet. Permuting the colors of the beads will not change the structure.
Also the number of distinct twills of period n. [Grünbaum and Shephard]

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Column 2 of A152176.
Cf. A056295.

Programs

  • Maple
    with(numtheory);
    rho:=n->(3+(-1)^n)/2;
    f:=n->2^((n+rho(n))/2-2) + (1/(4*n))*(add(phi(d)*rho(d)*2^(n/d), d in divisors(n))) - 1;
    # N. J. A. Sloane, Jul 13 2011
  • PARI
    a(n) = {if(n<1, 0, 2^(n\2-1) - 1 + sumdiv(n, k, eulerphi(2*k) * 2^(n/k)) / (4*n))}; \\ Andrew Howroyd, Oct 24 2019

Formula

a(n) = A000011(n) - 1.
For an explicit formula see the Maple program.

Extensions

Terms a(32) and beyond from Andrew Howroyd, Oct 24 2019

A166315 Lexicographically earliest binary de Bruijn sequences, B(2,n).

Original entry on oeis.org

1, 3, 23, 2479, 73743071, 151050438420815295, 1360791906900646753867474206897715071, 228824044090659455778900855050322128002759787305348791014476408721956007679
Offset: 1

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Author

Darse Billings, Oct 11 2009

Keywords

Comments

Term a(n) is a cyclical bit string of length 2^n, with every possible substring of length n occurring exactly once.
Mathworld says: "Every de Bruijn sequence corresponds to an Eulerian cycle on a de Bruijn graph. Surprisingly, it turns out that the lexicographic sequence of Lyndon words of lengths divisible by n gives the lexicographically earliest de Bruijn sequence (Ruskey). de Bruijn sequences can be generated by feedback shift registers (Golomb 1967; Ronse 1984; Skiena 1990, p. 196)."
Terms grow like Theta(2^(2^n)). - Darse Billings, Oct 18 2009

Examples

			Example: For n = 3, the first de Bruijn sequence, a(n) = B(2,3), is '00010111' = 23.
		

Crossrefs

Cf. A166316 (Lexicographically largest de Bruijn sequences (binary complements)).

Programs

  • Python
    # See Links.

Extensions

a(6)-a(8) from Darse Billings, Oct 18 2009

A276550 Array read by antidiagonals: T(n,k) = number of primitive (period n) bracelets using a maximum of k different colored beads.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 3, 2, 0, 5, 6, 7, 3, 0, 6, 10, 16, 15, 6, 0, 7, 15, 30, 45, 36, 8, 0, 8, 21, 50, 105, 132, 79, 16, 0, 9, 28, 77, 210, 372, 404, 195, 24, 0, 10, 36, 112, 378, 882, 1460, 1296, 477, 42, 0, 11, 45, 156, 630, 1848, 4220, 5890, 4380, 1209, 69, 0
Offset: 1

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Author

Andrew Howroyd, Apr 09 2017

Keywords

Comments

Turning over will not create a new bracelet.

Examples

			Table starts:
  1  2   3    4     5      6      7       8 ...
  0  1   3    6    10     15     21      28 ...
  0  2   7   16    30     50     77     112 ...
  0  3  15   45   105    210    378     630 ...
  0  6  36  132   372    882   1848    3528 ...
  0  8  79  404  1460   4220  10423   22904 ...
  0 16 195 1296  5890  20640  60021  151840 ...
  0 24 477 4380 25275 107100 364854 1057392 ...
  ...
		

Crossrefs

Programs

  • Maple
    A276550 := proc(n,k)
        local d ;
        add( numtheory[mobius](n/d)*A081720(d,k),d=numtheory[divisors](n)) ;
    end proc:
    seq(seq(A276550(n,d-n),n=1..d-1),d=2..10) ; # R. J. Mathar, Jan 22 2022
  • Mathematica
    t[n_, k_] := Sum[EulerPhi[d] k^(n/d), {d, Divisors[n]}]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4;
    T[n_, k_] := Sum[MoebiusMu[d] t[n/d, k], {d, Divisors[n]}];
    Table[T[n-k+1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Mar 26 2020 *)

Formula

T(n, k) = Sum_{d|n} mu(n/d) * A081720(d,k) for k<=n. Corrected Jan 22 2022

A280218 Number of binary necklaces of length n with no subsequence 0000.

Original entry on oeis.org

1, 2, 3, 5, 6, 11, 15, 27, 43, 75, 125, 228, 391, 707, 1262, 2285, 4119, 7525, 13691, 25111, 46033, 84740, 156123, 288529, 533670, 989305, 1835983, 3412885, 6351031, 11834623, 22074821, 41222028, 77048131, 144148859, 269913278, 505826391, 948652695, 1780473001, 3343960175, 6284560319, 11818395345
Offset: 1

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Author

Petros Hadjicostas, Dec 29 2016

Keywords

Comments

a(n) is the number of cyclic sequences of length n consisting of zeros and ones that do not contain four consecutive zeros provided we consider as equivalent those sequences that are cyclic shifts of each other.

Examples

			a(5)=6 because we have six binary cyclic sequences of length 5 that avoid four consecutive zeros: 00011, 00101, 00111, 01101, 01111, 11111.
		

Crossrefs

Programs

  • Mathematica
    Table[(1/n) Sum[EulerPhi[n/d] SeriesCoefficient[(4 - 3 x - 2 x^2 - x^3)/(1 - x - x^2 - x^3 - x^4), {x, 0, d}], {d, Divisors@ n}], {n, 41}] (* Michael De Vlieger, Dec 30 2016 *)
  • PARI
    N=44; x='x+O('x^N);
    B(x)=x*(1+x+x^2+x^3);
    Vec(sum(k=1, N, eulerphi(k)/k * log(1/(1-B(x^k))))) \\ Joerg Arndt, Dec 29 2016

Formula

a(n) = (1/n) * Sum_{d divides n} totient(n/d) * A073817(d).
G.f.: Sum_{k>=1} (phi(k)/k) * log(1/(1-B(x^k))) where B(x) = x*(1+x+x^2+x^3).

A032240 Number of identity bracelets of n beads of 3 colors.

Original entry on oeis.org

3, 3, 1, 3, 12, 37, 117, 333, 975, 2712, 7689, 21414, 60228, 168597, 475024, 1338525, 3788400, 10741575, 30556305, 87109332, 248967446, 713025093, 2046325125, 5883406830, 16944975036, 48880411272, 141212376513
Offset: 1

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Keywords

Comments

For n>2 also number of asymmetric bracelets with n beads of three colors. - Herbert Kociemba, Nov 29 2016

Crossrefs

Column k=3 of A309528 for n >= 3.

Programs

  • Mathematica
    m = 3; (* asymmetric bracelets of n beads of m colors *) Table[Sum[MoebiusMu[d] (m^(n/d)/n - If[OddQ[n/d], m^((n/d + 1)/2), ((m + 1) m^(n/(2 d))/2)]), {d, Divisors[n]}]/2, {n, 3, 20}] (* Robert A. Russell, Mar 18 2013 *)
    mx=40;gf[x_,k_]:=Sum[MoebiusMu[n]*(-Log[1-k*x^n]/n-Sum[Binomial[k,i]x^(n i),{i,0,2}]/(1-k x^(2n)))/2,{n,mx}];ReplacePart[Rest[CoefficientList[Series[gf[x,3],{x,0,mx}],x]],{1->3,2->3}] (* Herbert Kociemba, Nov 29 2016 *)
  • PARI
    a(n)={if(n<3, binomial(3,n), sumdiv(n, d, moebius(n/d)*(3^d/n - if(d%2, 3^((d+1)/2), 2*3^(d/2))))/2)} \\ Andrew Howroyd, Sep 12 2019

Formula

"DHK" (bracelet, identity, unlabeled) transform of 3, 0, 0, 0...
From Herbert Kociemba, Nov 29 2016: (Start)
More generally, gf(k) is the g.f. for the number of asymmetric bracelets with n beads of k colors.
gf(k): Sum_{n>=1} mu(n)*( -log(1-k*x^n)/n - Sum_{i=0..2} binomial(k,i)x^(n*i)/(1-k*x^(2*n)) )/2 (End)

A032280 Number of bracelets (turnover necklaces) of n beads of 2 colors, 7 of them black.

Original entry on oeis.org

1, 1, 4, 8, 20, 38, 76, 133, 232, 375, 600, 912, 1368, 1980, 2829, 3936, 5412, 7293, 9724, 12760, 16588, 21287, 27092, 34112, 42640, 52819, 65008, 79392, 96405, 116280, 139536, 166464, 197676, 233529, 274740, 321741, 375364
Offset: 7

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Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of nonequivalent necklaces of 7 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=7 (see our comment to A032279).
(End)

References

  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Column k=7 of A052307.

Programs

  • Mathematica
    k = 7; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    CoefficientList[Series[-(4 x^6 - 2 x^5 - 2 x^4 + 4 x^3 + x^2 - 2 x + 1)/((x - 1)^7 (x + 1)^3 (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 19 2013 *)
    k=7; CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)

Formula

S. J. Cyvin et al. (1997) give a g.f.
"DIK[ 7 ]" (necklace, indistinct, unlabeled, 7 parts) transform of 1, 1, 1, 1...
From Vladimir Shevelev, Apr 23 2011: (Start)
Put s(n,k,d) = 1, if n == k(mod d); 0, otherwise. Then
a(n) = (3/7)*s(n,0,7) + (48*C(n-1,6) + 7*(n-2)*(n-4)*(n-6))/672, if n is even;
a(n) = (3/7)*s(n,0,7) + (48*C(n-1,6) + 7*(n-1)*(n-3)*(n-5))/672, if n is odd. (End)
G.f.: -x^7*(4*x^6-2*x^5-2*x^4+4*x^3+x^2-2*x+1) / ((x-1)^7*(x+1)^3*(x^6+x^5+x^4+x^3+x^2+x+1)). - Colin Barker, Feb 06 2013
From Herbert Kociemba, Nov 05 2016: (Start)
G.f.: (1/2)*x^7*((1+x)/(1-x^2)^4 + 1/7*(1/(1-x)^7 + 6/(1-x^7))).
G.f.: k=7, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor((k+2)/2))/2. [edited by Petros Hadjicostas, Jul 18 2018] (End)

A032281 Number of bracelets (turnover necklaces) of n beads of 2 colors, 9 of them black.

Original entry on oeis.org

1, 1, 5, 12, 35, 79, 185, 375, 750, 1387, 2494, 4262, 7105, 11410, 17930, 27407, 41107, 60335, 87154, 123695, 173173, 238957, 325845, 438945, 585265, 772252, 1009868, 1308742, 1682660, 2146420, 2718806, 3419924, 4274905, 5310667, 6560225, 8059021, 9849925
Offset: 9

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From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 9 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in the case k=9 (see our comment to A032279). (End)

References

  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Column k=9 of A052307.

Programs

  • Mathematica
    k = 9; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    k=9;CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)

Formula

"DIK[ 9 ]" (necklace, indistinct, unlabeled, 9 parts) transform of 1, 1, 1, 1...
Put s(n,k,d) = 1, if n == k (mod d), and s(n,k,d) = 0, otherwise. Then a(n) =(1/3)*s(n,0,9) + (n-3)*(n-6)*s(n,0,3)/162 + (n-2)(n-4)*(n-6)*(n-8)*(945 + (n-1)*(n-3)*(n-5)*(n-7))/725760, if n is even; a(n) = (1/3)*s(n,0,9) + (n-3)*(n-6)*s(n,0,3)/162 +(n-1)*(n-3)*(n-5)*(n-7)*(945 + (n-2)*(n-4)*(n-6)*(n-8))/725760, if n is odd. - Vladimir Shevelev, Apr 23 2011
From Herbert Kociemba, Nov 05 2016: (Start)
G.f.: (1/2)*x^9*((1+x)/(1-x^2)^5 + 1/9*(1/(1-x)^9 - 2/(-1+x^3)^3 - 6/(-1+x^9))).
G.f.: k=9, x^k*((1/k)*(Sum_{d|k} phi(d)*(1-x^d)^(-k/d)) + (1+x)/(1-x^2)^floor((k+2)/2))/2. [edited by Petros Hadjicostas, Jul 18 2018] (End)
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