A027671 -id:A027671 - cp's OEIS frontend

A081720 Triangle T(n,k) read by rows, giving number of bracelets (turnover necklaces) with n beads of k colors (n >= 1, 1 <= k <= n).

1, 1, 3, 1, 4, 10, 1, 6, 21, 55, 1, 8, 39, 136, 377, 1, 13, 92, 430, 1505, 4291, 1, 18, 198, 1300, 5895, 20646, 60028, 1, 30, 498, 4435, 25395, 107331, 365260, 1058058, 1, 46, 1219, 15084, 110085, 563786, 2250311, 7472984, 21552969, 1, 78, 3210, 53764, 493131, 3037314

Offset: 1

N. J. A. Sloane, based on information supplied by Gary W. Adamson, Apr 05 2003

From Petros Hadjicostas, Nov 29 2017: (Start)
The formula given below is clear from the programs given in the Maple and Mathematica sections, while the g.f. for column k can be obtained using standard techniques.
If we differentiate the column k g.f. m times, then we can get a formula for row m. (For this sequence, we only need to use this row m formula for 1 <= k <= m, but it is valid even for k>m.) For example, to get the formula for row 8, we have T(n=8,k) = d^8/dx^8 (column k g.f.)/8! evaluated at x=0. Here, "d^8/dx^8" means "8th derivative w.r.t. x" of the column k g.f. Doing so, we get T(n=8, k) = (k^6 - k^5 + k^4 + 3*k^3 + 2*k^2 - 2*k + 4)*(k + 1)*k/16, which is the formula given for sequence A060560. (Here, we use this formula only for 1 <= k <= 8.)
(End)

			1;                                                (A000027)
1,  3;                                            (A000217)
1,  4,  10;                                       (A000292)
1,  6,  21,   55;                                 (A002817)
1,  8,  39,  136,   377;                          (A060446)
1, 13,  92,  430,  1505,   4291;                  (A027670)
1, 18, 198, 1300,  5895,  20646,  60028;          (A060532)
1, 30, 498, 4435, 25395, 107331, 365260, 1058058; (A060560)
...
For example, when n=k=3, we have the following T(3,3)=10 bracelets of 3 beads using up to 3 colors: 000, 001, 002, 011, 012, 022, 111, 112, 122, and 222. (Note that 012 = 120 = 201 = 210 = 102 = 021.) _Petros Hadjicostas_, Nov 29 2017

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

Andrew Howroyd, Table of n, a(n) for n = 1..1275
Yi Hu, Numerical Transfer Matrix Method of Next-nearest-neighbor Ising Models, Master's Thesis, Duke Univ. (2021).
Yi Hu and Patrick Charbonneau, Numerical transfer matrix study of frustrated next-nearest-neighbor Ising models on square lattices, arXiv:2106.08442 [cond-mat.stat-mech], 2021, cites the 4th column.

Cf. A321791 (extension to n >= 0, k >= 0).

Cf. A081721 (diagonal), A081722 (row sums), column sequences k=2..6: A000029, A027671, A032275, A032276, A056341.

A081720 := proc(n, k)
    local d, t1;
    t1 := 0;
    if n mod 2 = 0 then
        for d from 1 to n do
            if n mod d = 0 then
                t1 := t1+numtheory[phi](d)*k^(n/d);
            end if;
        end do:
        (t1+(n/2)*(1+k)*k^(n/2)) /(2*n) ;
    else
        for d from 1 to n do
            if n mod d = 0 then
                t1 := t1+numtheory[phi](d)*k^(n/d);
            end if;
        end do;
        (t1+n*k^((n+1)/2)) /(2*n) ;
    end if;
end proc:
seq(seq(A081720(n,k),k=1..n),n=1..10) ;

t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]); Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 13 2012, after Maple, updated Nov 02 2017 *)
Needs["Combinatorica`"]; Table[Table[NumberOfNecklaces[n,k,Dihedral],{k,1,n}],{n,1,8}]//Grid  (* Geoffrey Critzer, Oct 07 2012, after code by T. D. Noe in A027671 *)

See Maple code.
From Petros Hadjicostas, Nov 29 2017: (Start)
T(n,k) = ((1+k)*k^{n/2}/2 + (1/n)*Sum_{d|n} phi(n/d)*k^d)/2, if n is even, and = (k^{(n+1)/2} + (1/n)*Sum_{d|n} phi(n/d)*k^d)/2, if n is odd.
G.f. for column k: (1/2)*((k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)) provided we chop off the Taylor expansion starting at x^k (and ignore all the terms x^n with n(End)
2*n*T(n,k) = A054618(n,k)+n*(1+k)^(n/2)/2 if n even, = A054618(n,k)+n*k^((n+1)/2) if n odd. - R. J. Mathar, Jan 23 2022

Name edited by Petros Hadjicostas, Nov 29 2017

A056353 Number of bracelet structures using a maximum of three different colored beads.

Original entry on oeis.org

1, 1, 2, 3, 6, 9, 22, 40, 100, 225, 582, 1464, 3960, 10585, 29252, 80819, 226530, 636321, 1800562, 5107480, 14548946, 41538916, 118929384, 341187048, 980842804, 2824561089, 8147557742, 23536592235, 68087343148, 197216119545, 571924754778, 1660419530056, 4825588205920

Offset: 0

Marks R. Nester

nonn

Turning over will not create a new bracelet. Permuting the colors of the beads will not change the structure.

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]

Andrew Howroyd, Table of n, a(n) for n = 0..200
R. M. Thompson and R. T. Downs, Systematic generation of all nonequivalent closest packed stacking sequences of length N using group theory, Acta Cryst. B57 (2001), 766-771; B58 (2002), 153.

Cf. A002076, A000011, A027671, A114438, A152176.

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.

a(n) = Sum_{k=1..3} A152176(n, k) for n > 0. - Andrew Howroyd, Oct 25 2019

a(0)=1 prepended and terms a(28) and beyond from Andrew Howroyd, Oct 25 2019

A054631 Triangle read by rows: row n (n >= 1) contains the numbers T(n,k) = Sum_{d|n} phi(d)*k^(n/d)/n, for k=1..n.

Original entry on oeis.org

1, 1, 3, 1, 4, 11, 1, 6, 24, 70, 1, 8, 51, 208, 629, 1, 14, 130, 700, 2635, 7826, 1, 20, 315, 2344, 11165, 39996, 117655, 1, 36, 834, 8230, 48915, 210126, 720916, 2097684, 1, 60, 2195, 29144, 217045, 1119796, 4483815, 14913200, 43046889

Offset: 1

N. J. A. Sloane, Apr 16 2000, revised Mar 21 2007

T(n,k) is the number of n-bead necklaces with up to k different colored beads. - Yves-Loic Martin, Sep 29 2020

			1;
1,  3;                                   (A000217)
1,  4,  11;                              (A006527)
1,  6,  24,   70;                        (A006528)
1,  8,  51,  208,   629;                 (A054620)
1, 14, 130,  700,  2635,  7826;          (A006565)
1, 20, 315, 2344, 11165, 39996, 117655;  (A054621)

Seiichi Manyama, Rows n = 1..140, flattened
Wikipedia, Necklace (combinatorics)
Index entries for sequences related to necklaces

Cf. A054630, A054618, A054619, A087854. Lower triangle of A075195.

A054631 := proc(n,k) add( numtheory[phi](d)*k^(n/d),d=numtheory[divisors](n) ) ;  %/n ; end proc: # R. J. Mathar, Aug 30 2011

Needs["Combinatorica`"]; Table[Table[NumberOfNecklaces[n, k, Cyclic], {k, 1, n}], {n, 1, 8}] //Grid (* Geoffrey Critzer, Oct 07 2012, after code by T. D. Noe in A027671 *)
t[n_, k_] := Sum[EulerPhi[d]*k^(n/d)/n, {d, Divisors[n]}]; Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)

T(n, k) = sumdiv(n, d, eulerphi(d)*k^(n/d))/n; \\ Seiichi Manyama, Mar 10 2021

T(n, k) = sum(j=1, n, k^gcd(j, n))/n; \\ Seiichi Manyama, Mar 10 2021

T(n,k) = Sum_{j=1..k} binomial(k,j) * A087854(n, j). - Yves-Loic Martin, Sep 29 2020

T(n,k) = (1/n) * Sum_{j=1..n} k^gcd(j, n). - Seiichi Manyama, Mar 10 2021

A051137 Table T(n,k) read by antidiagonals: number of necklaces allowing turnovers (bracelets) with n beads of k colors.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 6, 10, 10, 5, 1, 1, 8, 21, 20, 15, 6, 1, 1, 13, 39, 55, 35, 21, 7, 1, 1, 18, 92, 136, 120, 56, 28, 8, 1, 1, 30, 198, 430, 377, 231, 84, 36, 9, 1, 1, 46, 498, 1300, 1505, 888, 406, 120, 45, 10, 1

Offset: 0

Alford Arnold

Unlike A075195 and A284855, antidiagonals go from bottom-left to top-right.

			Table begins with T[0,1]:
1  1    1     1      1       1        1        1         1         1
1  2    3     4      5       6        7        8         9        10
1  3    6    10     15      21       28       36        45        55
1  4   10    20     35      56       84      120       165       220
1  6   21    55    120     231      406      666      1035      1540
1  8   39   136    377     888     1855     3536      6273     10504
1 13   92   430   1505    4291    10528    23052     46185     86185
1 18  198  1300   5895   20646    60028   151848    344925    719290
1 30  498  4435  25395  107331   365260  1058058   2707245   6278140
1 46 1219 15084 110085  563786  2250311  7472984  21552969  55605670
1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022

C. G. Bower, Transforms (2)

Columns 2-6 are A000029, A027671, A032275, A032276, and A056341.
Rows 2-7 are A000217, A000292, A002817, A060446, A027670, and A060532.
Cf. A000031.
Cf. A081720, A081721.
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.

b[n_, k_] := DivisorSum[n, EulerPhi[#]*k^(n/#) &] / n;
c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2)];
T[0, ] = 1; T[n, k_] := (b[n, k] + c[n, k])/2;
Table[T[n, k-n], {k, 1, 11}, {n, k-1, 0, -1}] // Flatten
(* Robert A. Russell, Sep 21 2018 after Jean-François Alcover *)

T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d divides n} phi(d) * k^(n/d). - Robert A. Russell, Sep 21 2018
G.f. for column k: (kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. - Robert A. Russell, Sep 28 2018
T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i). (See A075195 formulas.) - Richard L. Ollerton, May 04 2021

A056343 Number of bracelets of length n using exactly three different colored beads.

Original entry on oeis.org

0, 0, 1, 6, 18, 56, 147, 411, 1084, 2979, 8043, 22244, 61278, 171030, 477929, 1345236, 3795750, 10758902, 30572427, 87149124, 248991822, 713096352, 2046303339, 5883433409, 16944543810, 48879769575

Offset: 1

Marks R. Nester

nonn

Turning over will not create a new bracelet.

			For a(4)=6, the arrangements are ABAC, ABCB, ACBC, AABC, ABBC, and ABCC. Only the last three are chiral, their reverses being AACB, ACBB, and ACCB respectively. - _Robert A. Russell_, Sep 26 2018

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]

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Column 3 of A273891.

Equals (A056283 + A056489) / 2 = A056283 - A305542 = A305542 + A056489.

t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 3];
Array[a, 26] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=3; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 26 2018 *)

a(n) = A027671(n) - 3*A000029(n) + 3.
From Robert A. Russell, Sep 26 2018: (Start)
a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=3 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=3 is the number of colors. (End)

A056344 Number of bracelets of length n using exactly four different colored beads.

Original entry on oeis.org

0, 0, 0, 3, 24, 136, 612, 2619, 10480, 41388, 159780, 614058, 2341920, 8919816, 33905188, 128907279, 490213680, 1866127840, 7111777860, 27140369148, 103721218000, 396974781456, 1521577377012, 5840547488954

Offset: 1

Marks R. Nester

nonn

Turning over will not create a new bracelet.

			For a(4)=3, the arrangements are ABCD, ABDC, and ACBD, all chiral, their reverses being ADCB, ACDB, and ADBC respectively.

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]

Column 4 of A273891.

Cf. A056284 (oriented), A056490 (achiral), A305543 (chiral).

t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 4];
Array[a, 24] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=4; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 27 2018 *)

a(n) = A032275(n) - 4*A027671(n) + 6*A000029(n) - 4.
From Robert A. Russell, Sep 27 2018: (Start)
a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=4 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=4 is the number of colors.
a(n) = (A056284(n) + A056490(n)) / 2 = A056284(n) - A305543(n) = A305543(n) + A056490(n). (End)

A114438 Number of Barlow packings that repeat after n (or a divisor of n) layers.

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 3, 8, 8, 18, 21, 48, 63, 133, 205, 412, 685, 1354, 2385, 4644, 8496, 16431, 30735, 59344, 112531, 217246, 415628, 803210, 1545463, 2991192, 5778267, 11201884, 21702708, 42141576, 81830748, 159140896, 309590883, 602938099, 1174779397, 2290920128

Offset: 1

N. J. A. Sloane, Feb 28 2006; more terms, Aug 10 2006

nonn

See A011768 for the number of Barlow packings that repeat after exactly n layers.

Like A056353 but with additional restriction that adjacent beads must have different colors.

N. J. A. Sloane, Table of n, a(n) for n = 1..500
Dennis S. Bernstein, Omran Kouba, Counting Colorful Necklaces and Bracelets in Three Colors, arXiv:1901.10703 [math.CO], 2019.
J. H. Conway and N. J. A. Sloane, What are all the best sphere packings in low dimensions?, Discr. Comp. Geom., 13 (1995), 383-403.
E. Estevez-Rams, C. Azanza-Ricardo, J. Martínez García and B. Aragón-Fernández, On the algebra of binary codes representing closed-packed staking sequences, Acta Cryst. A61 (2005), 201-208.
T. J. McLarnan, The numbers of polytypes in close packings and related structures, Zeits. Krist. 155, 269-291 (1981). [See P'(N) on page 272.]
R. M. Thompson and R. T. Downs, Systematic generation of all nonequivalent closest-packed stacking sequences of length N using group theory, Acta Cryst. B57 (2001), 766-771; B58 (2002), 153.

Cf. A027671, A056353.

with(numtheory); read transforms; M:=500;
A:=proc(N,d) if d mod 3 = 0 then 2^(N/d) else (1/3)*(2^(N/d)+2*cos(Pi*N/d)); fi; end;
E:=proc(N) if N mod 2 = 0 then N*2^(N/2) + add( did(N/2,d)*phi(2*d)*2^(N/(2*d)),d=1..N/2) else (N/3)*(2^((N+1)/2)+2*cos(Pi*(N+1)/2)); fi; end;
PP:=proc(N) (1/(4*N))*(add(did(N,d)*phi(d)*A(N,d), d=1..N)+E(N)); end; for N from 1 to M do lprint(N,PP(N)); od: # N. J. A. Sloane, Aug 10 2006

M = 40;
did[m_, n_] := If[Mod[m, n] == 0, 1, 0];
A[n_, d_] := If[Mod[d, 3] == 0, 2^(n/d), (1/3)(2^(n/d) + 2 Cos[Pi n/d])];
EE[n_] := If[Mod[n, 2] == 0, n 2^(n/2) + Sum[did[n/2, d] EulerPhi[2d] 2^(n/(2d)), {d, 1, n/2}], (n/3)(2^((n+1)/2) + 2 Cos[Pi(n+1)/2])];
a[n_] := (1/(4n))(Sum[did[n, d] EulerPhi[d] A[n, d], {d, 1, n}] + EE[n]);
Array[a, M] (* Jean-François Alcover, Apr 20 2020, from Maple *)

A214307 a(n) is the number of representative three-color bracelets (necklaces with turn over allowed) with n beads for n >= 3.

Original entry on oeis.org

1, 2, 6, 20, 40, 106, 304, 731, 1936, 5769, 14343, 39583, 117957, 305576, 855474, 2565922, 6793516, 19242857, 57827068, 155681341, 444461623, 1337436721, 3645877447, 10471728930, 31534868169, 86818242806, 250543852080, 754851821246, 2094887707000

Offset: 3

Wolfdieter Lang, Jul 31 2012

nonn

This is the third column (m=3) of triangle A213940.
The relevant p(n,3)= A008284(n,3) representative color multinomials have exponents (signatures) from the 3 part partitions of n, written with nonincreasing parts. E.g., n=6: [4,1,1], [3,2,1] and [2,2,2] (p(6,3)=3). The corresponding representative bracelets have the three-color multinomials c[1]^4*c[2]*c[3], c[1]^3*c[2]^2*c[3] and c[1]^2*c[2]^2*c[3]^2. Therefore, color c[1] is dominant, except for the last case.
Compare this with A027671 where also bracelets with less than three colors are included and not only three-color representatives are counted.
Number of n-length bracelets w over ternary alphabet {a,b,c} such that #(w,a) >= #(w,b) >= #(w,c) >= 1, where #(w,x) counts the letters x in word w (bracelet analog of A226882). The number of 3 color bracelets up to permutations of colors is given by A056358. - Andrew Howroyd, Sep 26 2017

			a(5) = A213939(5,4) + A213939(5,5) = 2 + 4 = 6 from the representative bracelets (with colors j for c[j], j=1,2,3) 11123, 11213, 11223, 11232, 12123 and 12213 , all taken cyclically. The first two have color signature (exponents) [3,1,1] and the other four ones have signature [2,2,1].
a(6) = A213939(6,5) + A213939(6,6) + A213939(6,7) = 3 + 6 + 11 = 20. The first three representative bracelets have color signature [4,1,1], the next six have signature [3,2,1] and the remaining 11 ones have signature [2,2,2]. The corresponding representative color multinomials are c[1]^4*c[2]*c[3], c[1]^3*c[2]^2*c[3] and c[1]^2*c[2]^2*c[3]^2.

Andrew Howroyd, Table of n, a(n) for n = 3..200

Cf. A213939, A213940, A214309 (m=4), A214310 (m=3, all bracelets).

Cf. A056358, A226882.

Cyc(v)={my(g=fold(gcd,v), s=vecsum(v)); sumdiv(g, d, eulerphi(d)*(s/d)!/prod(i=1, #v, (v[i]/d)!))/s}
CPal(v)={my(odds=#select(t->t%2,v), s=vecsum(v));  if(odds>2, 0, ((s-odds)/2)!/prod(i=1, #v, (v[i]\2)!))}
a(n)={my(t=0); forpart(p=n, t+=Cyc(Vec(p))+CPal(Vec(p)), [1,n], [3,3]); t/2} \\ Andrew Howroyd, Sep 26 2017

a(n) = A213940(n,3), n >= 3.

a(n) = sum(A213939(n,k),k=(2+floor(n/2))..p(n,3)+1+floor(n/2)), n >= 3, with p(n,3) = A008284(n,3) the number of partitions of n with three parts.

Terms a(26) and beyond from Andrew Howroyd, Sep 26 2017

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.

Original entry on oeis.org

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

Herbert Kociemba, Nov 24 2016

nonn

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

			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

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).

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 *)

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

A321791 Table read by descending antidiagonals: T(n,k) is the number of unoriented cycles (bracelets) of length n using up to k available colors.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 6, 1, 0, 1, 6, 15, 20, 21, 8, 1, 0, 1, 7, 21, 35, 55, 39, 13, 1, 0, 1, 8, 28, 56, 120, 136, 92, 18, 1, 0, 1, 9, 36, 84, 231, 377, 430, 198, 30, 1, 0

Offset: 0

Robert A. Russell, Dec 18 2018

			Table begins with T(0,0):
  1 1  1    1     1      1       1        1        1         1         1 ...
  0 1  2    3     4      5       6        7        8         9        10 ...
  0 1  3    6    10     15      21       28       36        45        55 ...
  0 1  4   10    20     35      56       84      120       165       220 ...
  0 1  6   21    55    120     231      406      666      1035      1540 ...
  0 1  8   39   136    377     888     1855     3536      6273     10504 ...
  0 1 13   92   430   1505    4291    10528    23052     46185     86185 ...
  0 1 18  198  1300   5895   20646    60028   151848    344925    719290 ...
  0 1 30  498  4435  25395  107331   365260  1058058   2707245   6278140 ...
  0 1 46 1219 15084 110085  563786  2250311  7472984  21552969  55605670 ...
  0 1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022 ...
For T(3,3)=10, the unoriented cycles are 9 achiral (AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, CCC) and 1 chiral pair (ABC-ACB).

Index entries for sequences related to bracelets

Cf. A075195 (oriented), A293496(chiral), A284855 (achiral).
Cf. A051137 (ascending antidiagonals).
Columns 0-6 are A000007, A000012, A000029, A027671, A032275, A032276, and A056341.
Rows 0-7 are A000012, A001477, A000217, A000292, A002817, A060446, A027670, and A060532.
Main diagonal gives A081721.

Table[If[k>0, DivisorSum[k, EulerPhi[#](n-k)^(k/#)&]/(2k) + ((n-k)^Floor[(k+1)/2]+(n-k)^Ceiling[(k+1)/2])/4, 1], {n, 0, 12}, {k, 0, n}] // Flatten

T(n,k) = [n==0] + [n>0] * (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d|n} phi(d) * k^(n/d).
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.
T(n,k) = A075195(n,k) - A293496(n,k) = A293496(n,k) + A284855(n,k).
Linear recurrence for row n: T(n,k) = Sum_{j=0..n} -binomial(j-n-1,j+1) * T(n,k-1-j) for k >= n + 1.
O.g.f. for column k >= 0: Sum_{n>=0} T(n,k)*x^n = 3/4 + (1 + k*x)^2/(4*(1 - k*x^2)) - (1/2) * Sum_{d >= 1} (phi(d)/d) * log(1 - k*x^d). - Petros Hadjicostas, Feb 07 2021

Showing 1-10 of 15 results. Next

cp's OEIS Frontend

A081720 Triangle T(n,k) read by rows, giving number of bracelets (turnover necklaces) with n beads of k colors (n >= 1, 1 <= k <= n).

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A056353 Number of bracelet structures using a maximum of three different colored beads.

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A054631 Triangle read by rows: row n (n >= 1) contains the numbers T(n,k) = Sum_{d|n} phi(d)*k^(n/d)/n, for k=1..n.

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A051137 Table T(n,k) read by antidiagonals: number of necklaces allowing turnovers (bracelets) with n beads of k colors.

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A056343 Number of bracelets of length n using exactly three different colored beads.

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A056344 Number of bracelets of length n using exactly four different colored beads.

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A114438 Number of Barlow packings that repeat after n (or a divisor of n) layers.

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