A273891 -id:A273891 - cp's OEIS frontend

A152176 Triangle read by rows: T(n,k) is the number of k-block partitions of an n-set up to rotations and reflections.

1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 3, 5, 2, 1, 1, 7, 14, 11, 3, 1, 1, 8, 31, 33, 16, 3, 1, 1, 17, 82, 137, 85, 27, 4, 1, 1, 22, 202, 478, 434, 171, 37, 4, 1, 1, 43, 538, 1851, 2271, 1249, 338, 54, 5, 1, 1, 62, 1401, 6845, 11530, 8389, 3056, 590, 70, 5, 1, 1, 121, 3838, 26148

Offset: 1

Vladeta Jovovic, Nov 27 2008

Number of bracelet structures of length n using exactly k different colored beads. Turning over will not create a new bracelet. Permuting the colors of the beads will not change the structure. - Andrew Howroyd, Apr 06 2017
The number of achiral structures (A) is given in A140735 (odd n) and A293181 (even n).  The number of achiral structures plus twice the number of chiral pairs (A+2C) is given in A152175.  These can be used to determine A+C by taking half their average, as is done in the Mathematica program. - Robert A. Russell, Feb 24 2018
T(n,k)=pi_k(C_n) which is the number of non-equivalent partitions of the cycle on n vertices, with exactly k parts. Two partitions P1 and P2 of a graph G are said to be equivalent if there is a nontrivial automorphism of G which maps P1 onto P2. - Mohammad Hadi Shekarriz, Aug 21 2019

			Triangle begins:
  1;
  1,  1;
  1,  1,   1;
  1,  3,   2,    1;
  1,  3,   5,    2,    1;
  1,  7,  14,   11,    3,    1;
  1,  8,  31,   33,   16,    3,   1;
  1, 17,  82,  137,   85,   27,   4,  1;
  1, 22, 202,  478,  434,  171,  37,  4, 1;
  1, 43, 538, 1851, 2271, 1249, 338, 54, 5, 1;
  ...

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 = 1..1275
B. Ahmadi, F. Alinaghipour and M. H. Shekarriz, Number of Distinguishing Colorings and Partitions, arXiv:1910.12102 [math.CO], 2019.
E. N. Gilbert and J. Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.
Tilman Piesk, Partition related number triangles
Marko Riedel, Bracelets with swappable colors classified by the distribution of colors, Power Group Enumeration algorithm
Marko Riedel, Maple code for the number of bracelets with some number of swappable colors by Power Group Enumeration
Mohammad Hadi Shekarriz, GAP Program

Columns 2-6 are A056357, A056358, A056359, A056360, A056361.
Row sums are A084708.
Partial row sums include A000011, A056353, A056354, A056355, A056356.
Cf. A081720, A273891, A008277 (set partitions), A284949 (up to reflection), A152175 (up to rotation).

Adn[d_, n_] := Adn[d, n] = Which[0==n, 1, 1==n, DivisorSum[d, x^# &],
  1==d, Sum[StirlingS2[n, k] x^k, {k, 0, n}],
  True, Expand[Adn[d, 1] Adn[d, n-1] + D[Adn[d, n - 1], x] x]];
Ach[n_, k_] := Ach[n, k] = Switch[k, 0, If[0==n, 1, 0], 1, If[n>0, 1, 0],
  (* else *) _, If[OddQ[n], Sum[Binomial[(n-1)/2, i] Ach[n-1-2i, k-1],
  {i, 0, (n-1)/2}], Sum[Binomial[n/2-1, i] (Ach[n-2-2i, k-1]
  + 2^i Ach[n-2-2i, k-2]), {i, 0, n/2-1}]]] (* achiral loops of length n, k colors *)
Table[(CoefficientList[DivisorSum[n, EulerPhi[#] Adn[#, n/#] &]/(x n), x]
+ Table[Ach[n, k],{k,1,n}])/2, {n, 1, 20}] // Flatten (* Robert A. Russell, Feb 24 2018 *)

\\ see A056391 for Polya enumeration functions
T(n,k) = NonequivalentStructsExactly(DihedralPerms(n), k); \\ Andrew Howroyd, Oct 14 2017

\\ Ach is A304972 and R is A152175 as square matrices.
Ach(n)={my(M=matrix(n, n, i, k, i>=k)); for(i=3, n, for(k=2, n, M[i, k]=k*M[i-2, k] + M[i-2, k-1] + if(k>2, M[i-2, k-2]))); M}
R(n)={Mat(Col([Vecrev(p/y, n) | p<-Vec(intformal(sum(m=1, n, eulerphi(m) * subst(serlaplace(-1 + exp(sumdiv(m, d, y^d*(exp(d*x + O(x*x^(n\m)))-1)/d))), x, x^m))/x))]))}
T(n)={(R(n) + Ach(n))/2}
{ my(A=T(12)); for(n=1, #A, print(A[n, 1..n])) } \\ Andrew Howroyd, Sep 20 2019

A087854 Triangle read by rows: T(n,k) is the number of n-bead necklaces with exactly k different colored beads.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 4, 9, 6, 1, 6, 30, 48, 24, 1, 12, 91, 260, 300, 120, 1, 18, 258, 1200, 2400, 2160, 720, 1, 34, 729, 5106, 15750, 23940, 17640, 5040, 1, 58, 2018, 20720, 92680, 211680, 258720, 161280, 40320, 1, 106, 5613, 81876, 510312, 1643544, 2963520, 3024000, 1632960, 362880

Offset: 1

Philippe Deléham

Equivalently, T(n,k) is the number of sequences (words) of length n on an alphabet of k letters where each letter of the alphabet occurs at least once in the sequence. Two sequences are considered equivalent if one can be obtained from the other by a cyclic shift of the letters. Cf. A054631 where the surjective restriction is removed. - Geoffrey Critzer, Jun 18 2013

Robert A. Russell's g.f. for column k >= 1 (in the Formula section below) can be proved by integrating both sides of the formula Sum_{n>=1} S2(n, k)*x^(n-1) = x^(k-1)/((1 - x)* (1 - 2*x) * (1 - 3*x) * ... * (1 - k*x)) w.r.t. x. A variation of this identity (valid for |x| < 1/k) can be found in the Formula section of A008277. - Petros Hadjicostas, Aug 20 2019

			The triangle begins with T(1,1):
  1;
  1,   1;
  1,   2,    2;
  1,   4,    9,     6;
  1,   6,   30,    48,     24;
  1,  12,   91,   260,    300,     120;
  1,  18,  258,  1200,   2400,    2160,     720;
  1,  34,  729,  5106,  15750,   23940,   17640,    5040;
  1,  58, 2018, 20720,  92680,  211680,  258720,  161280,   40320;
  1, 106, 5613, 81876, 510312, 1643544, 2963520, 3024000, 1632960, 362880;
  ...
For T(4,2) = 4, the necklaces are AAAB, AABB, ABAB, and ABBB.
For T(4,4) = 6, the necklaces are ABCD, ABDC, ACBD, ACDB, ADBC, and ADCB.

Alois P. Heinz, Rows n = 1..141, flattened

Columns 1-6: A057427, A052823, A056283, A056284, A056285, A056286.
Diagonals: A000142 and A074143.
Row sums: A019536.
Cf. A000010 (Euler totient phi function), A008277 (Stirling2 numbers), A075195 (table of Jablonski).
Cf. A254040, A273891.

with(numtheory):
T:= (n, k)-> (k!/n) *add(phi(d) *Stirling2(n/d, k), d=divisors(n)):
seq(seq(T(n,k), k=1..n), n=1..12);  # Alois P. Heinz, Jun 19 2013

Table[Table[Sum[EulerPhi[d]*StirlingS2[n/d,k]k!,{d,Divisors[n]}]/n,{k,1,n}],{n,1,10}]//Grid (* Geoffrey Critzer, Jun 18 2013 *)

T(n, k) = (k!/n) * sumdiv(n, d, eulerphi(d) * stirling(n/d, k, 2)); \\ Joerg Arndt, Sep 25 2020

T(n,k) = Sum_{i=0..k-1} (-1)^i * C(k,i) * A075195(n,k-i); A075195 = Jablonski's table.
T(n,k) = (k!/n) * Sum_{d|n} phi(d) * S2(n/d, k), where S2(n,k) = Stirling numbers of 2nd kind A008277.
G.f. for column k: -Sum_{d>0} (phi(d)/d) * Sum_{j = 1..k} (-1)^(k-j) * C(k,j) * log(1 - j * x^d). - Robert A. Russell, Sep 26 2018
T(n,k) = Sum_{d|n} A254040(d, k) for n, k >= 1. - Petros Hadjicostas, Aug 19 2019

Formula section edited by Petros Hadjicostas, Aug 20 2019

A305540 Triangle read by rows: T(n,k) is the number of achiral loops (necklaces or bracelets) of length n using exactly k different colors.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 4, 3, 1, 6, 6, 1, 10, 21, 12, 1, 14, 36, 24, 1, 22, 93, 132, 60, 1, 30, 150, 240, 120, 1, 46, 345, 900, 960, 360, 1, 62, 540, 1560, 1800, 720, 1, 94, 1173, 4980, 9300, 7920, 2520, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 190, 3801, 24612, 71400, 103320, 73080, 20160, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320

Offset: 1

Robert A. Russell, Jun 04 2018

The number of achiral necklaces is equivalent to the number of achiral bracelets.

			The triangle begins with T(1,1):
1;
1,   1;
1,   2;
1,   4,     3;
1,   6,     6;
1,  10,    21,     12;
1,  14,    36,     24;
1,  22,    93,    132,     60;
1,  30,   150,    240,    120;
1,  46,   345,    900,    960,     360;
1,  62,   540,   1560,   1800,     720;
1,  94,  1173,   4980,   9300,    7920,    2520;
1, 126,  1806,   8400,  16800,   15120,    5040;
1, 190,  3801,  24612,  71400,  103320,   73080,   20160;
1, 254,  5796,  40824, 126000,  191520,  141120,   40320;
1, 382, 11973, 113652, 480060, 1048320, 1234800,  745920, 181440;
1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880;
For a(4,2)=4, the achiral loops are AAAB, AABB, ABAB, and ABBB.

Odd rows are A019538.
Even rows are A172106.
Columns 1-6 are A057427, A027383, A056489, A056490, A056491, and A056492.

Table[(k!/2) (StirlingS2[Floor[(n + 1)/2], k] + StirlingS2[Ceiling[(n + 1)/2], k]), {n, 1, 15}, {k, 1, Ceiling[(n + 1)/2]}] // Flatten

T(n, k) = (k!/2)*(stirling(floor((n+1)/2), k, 2)+stirling(ceil((n+1)/2), k, 2));
tabf(nn) = for(n=1, nn, for (k=1, ceil((n+1)/2), print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 02 2018

T(n,k) = (k!/2) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)), where S2(n,k) is the Stirling subset number A008277.
T(n,k) = 2*A273891(n,k) - A087854(n,k).
G.f. for column k>1: (k!/2) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2). - Robert A. Russell, Sep 26 2018

A305541 Triangle read by rows: T(n,k) is the number of chiral pairs of color loops of length n with exactly k different colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 3, 3, 0, 0, 12, 24, 12, 0, 1, 35, 124, 150, 60, 0, 2, 111, 588, 1200, 1080, 360, 0, 6, 318, 2487, 7845, 11970, 8820, 2520, 0, 14, 934, 10240, 46280, 105840, 129360, 80640, 20160, 0, 30, 2634, 40488, 254676, 821592, 1481760, 1512000, 816480, 181440, 0, 62, 7503, 158220, 1344900, 5873760, 14658840, 21772800, 19051200, 9072000, 1814400

Offset: 1

Robert A. Russell, Jun 04 2018

In other words, the number of n-bead bracelets with beads of exactly k different colors that when turned over are different from themselves. - Andrew Howroyd, Sep 13 2019

			Triangle T(n,k) begins:
  0;
  0,  0;
  0,  0,    1;
  0,  0,    3,     3;
  0,  0,   12,    24,     12;
  0,  1,   35,   124,    150,     60;
  0,  2,  111,   588,   1200,   1080,     360;
  0,  6,  318,  2487,   7845,  11970,    8820,    2520;
  0, 14,  934, 10240,  46280, 105840,  129360,   80640,  20160;
  0, 30, 2634, 40488, 254676, 821592, 1481760, 1512000, 816480, 181440;
  ...
For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
For T(4,4)=3, the chiral pairs are ABCD-ADCB, ABDC-ACDB, and ACBD-ADBC.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)

Columns 2-6 are A059076, A305542, A305543, A305544, and A305545.
Row sums are A326895.
Cf. A087854, A273891, A293496, A305540, A309651.

Table[(k!/(2n)) DivisorSum[n, EulerPhi[#] StirlingS2[n/#, k] &] - (k!/4) (StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k]), {n, 1, 15}, {k, 1, n}] // Flatten

T(n,k) = {-k!*(stirling((n+1)\2,k,2) + stirling(n\2+1,k,2))/4 + k!*sumdiv(n,d, eulerphi(d)*stirling(n/d,k,2))/(2*n)} \\ Andrew Howroyd, Sep 13 2019

T(n,k) = -(k!/4)*(S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/(2 n))*Sum_{d|n} phi(d)*S2(n/d,k), where S2(n,k) is the Stirling subset number A008277.
T(n,k) = A087854(n,k) - A273891(n,k).
T(n,k) = (A087854(n,k) - A305540(n,k)) / 2.
T(n, k) = Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*A293496(n, i). - Andrew Howroyd, Sep 13 2019

A056342 Number of bracelets of length n using exactly two different colored beads.

Original entry on oeis.org

0, 1, 2, 4, 6, 11, 16, 28, 44, 76, 124, 222, 378, 685, 1222, 2248, 4110, 7683, 14308, 27010, 50962, 96907, 184408, 352696, 675186, 1296856, 2493724, 4806076, 9272778, 17920858, 34669600, 67159048, 130216122, 252745366, 490984486, 954637556, 1857545298, 3617214679, 7048675958, 13744694926, 26818405350

Offset: 1

Marks R. Nester

nonn

Turning over will not create a new bracelet.

			For a(6)=11, the arrangements are AAAAAB, AAAABB, AAABAB, AAABBB, AABAAB, AABBBB, ABABAB, ABABBB, ABBABB, ABBBBB, and AABABB, the last being chiral. Its reverse is AABBAB. - _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]

G. C. Greubel, Table of n, a(n) for n = 1..3000

Column 2 of A273891.
Equals A052823 - A059076.
Cf. A008277, A027383.

a[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2*n) - 2; Array[a, 41] (* Jean-François Alcover, Nov 05 2017 *)
k=2; 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) = my(k=2); (k!/4)*(stirling(floor((n+1)/2),k,2) + stirling(ceil((n+1)/2),k,2)) + (k!/(2*n))*sumdiv(n,d,eulerphi(d)*stirling(n/d,k,2)); \\ Michel Marcus, Sep 28 2018

a(n) = A000029(n) - 2.
From Robert A. Russell, Sep 26 2018: (Start)
a(n) = (A052823(n) + A027383(n-2)) / 2 = A059076(n) + A027383(n-2).
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=2 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=2 is the number of colors. (End)

More terms from Joerg Arndt, Jun 10 2016

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

Andrew Howroyd, Apr 09 2017

Turning over will not create a new bracelet.

			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 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..1275
G. Melançon, C. Reutenauer, On a Class of Lyndon Words Extending Christoffel Words and Related to a Multidimensional Continued Fraction Algorithm, J. Int. Seq. 16 (2013) #13.9.7, Corollary 6.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]

Columns 2-6 are A001371, A032294, A032295, A032296, A056347.

Cf. A081720, A273891, A276543, A152176.

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

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 *)

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

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)

A214306 Triangle with entry a(n,m) giving the total number of bracelets of n beads (D_n symmetry) with n colors available for each bead, but only m distinct colors present, with m from {1, 2, ..., n} and n >= 1.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 24, 24, 3, 5, 60, 180, 120, 12, 6, 165, 1120, 2040, 900, 60, 7, 336, 5145, 21420, 25200, 7560, 360, 8, 784, 23016, 183330, 442680, 335160, 70560, 2520, 9, 1584, 91056, 1320480, 5846400, 8890560, 4656960, 725760, 20160, 10

Offset: 1

Wolfdieter Lang, Jul 20 2012

This triangle is obtained from the array A213941 by summing in row n, for n >= 1, all entries related to partitions of n with the same number of parts m.
a(n,m) is the total number of necklaces of n beads (dihedral D_n symmetry) corresponding to all the color multinomials obtained from all p(n,m) = A008284(n,m) partitions of n with m parts, written in nonincreasing form, by 'exponentiation'. Therefore only m from the available n colors are present, and a(n,m) gives the number of bracelets with n beads with only m of the n available colors present, for m from 1,2,...,n, and n >= 1. All of the possible color assignments are counted.
See the comments on A212359 for the Abramowitz-Stegun (A-St) order of partitions, and the 'exponentiation' to obtain multisets, used to encode color multinomials, from partitions. See a link in A213938 for representative multisets for given signature used to define color multinomials.
The row sums of this triangle coincide with the ones of array A213941, and they are given by A081721.

			n\m 1    2     3       4       5       8       7      8     9
1   1
2   2    1
3   3    6     1
4   4   24    24       3
5   5   60   180     120      12
6   6  165  1120    2040     900     60
7   7  336  5145   21420   25200   7560      360
8   8  784 23016  183330  442680  33516    70560   2520
9   9 1584 91056 1320480 5846400 8890560 4656960 725760 20160
...
Row n=10:  10, 3420, 357480, 8691480, 64420272, 172609920, 177811200, 68040000, 8164800, 181440;
Row n=11:  11, 6820, 1327095, 52727400, 622175400, 2714009760, 4837417200, 3592512000, 1047816000, 99792000, 1814400.
a(2,2) = 1 from the color monomial c[1]^1*c[2]^1 = c[1]*c[2] (from the m=2 partition [1,1] of n=2). The bracelet in question is cyclic(12) (we use j for color c[j] in these examples). The same holds for the necklace case.
a(5,3) = 60 + 120 = 180, from A213941(5,4) + A213941(5,5), because k(5,3,1) = A214314(5,3) = 4 and p(5,3)=2.
a(3,1) = 3 from the color monomials c[1]^3, c[2]^3 and c[3]^3. The three bracelets are cyclic(111), cyclic(222) and cyclic(333). The same holds for the necklace case.
In general a(n,1)=n from the partition [n] providing the color signature (exponent), and the n color choices.
a(3,2) = 6 from the color signature c[.]^2 c[.]^1, (from the m=2 partition [2,1] of n=3), and there are 6 choices for the color indices. The 6 bracelets are cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and cyclic(332). The same holds for the necklace case.
a(3,3) = 1. The color multinomial is c[1]*c[2]*c[3] (from the m=3 partition [1,1,1]). All three available colors are used. There is only one bracelet: cyclic(1,2,3). The necklace cyclic(1,3,2) becomes equivalent under D_3 operation.
a(4,2) = 24 from two color signatures c[.]^3 c[.] and c[.]^2 c[.]^2 (from the two m=2 partitions of n=4: [3,1] and [2,2]). The first one produces 4*3=12 bracelets, namely 1112, 1113, 1114, 2221, 2223, 2224, 3331, 3332, 3334, 4441, 4442 and 4443, all taken cyclically. The second color signature leads to another 6*2=12 bracelets: 1122, 1133, 1144, 2233, 2244, 3344, 1212, 1313, 1414, 2323, 2424 and 3434, all taken cyclically. Together they provide the 24 bracelets counted by a(4,2). The same holds for the necklace case.
a(4,3) = 24 from the color signature c[.]^2 c[.]c[.]. There are 4*3 =12 color choices each with two bracelets: 1123, 1213, 1124, 1214, 1134, 1314, 2213, 2123, 2214, 2124, 2234, 2324, 3312, 3132, 3314, 3134, 3324, 3234, 4412, 4142, 4413, 4143, 4423 and 4243, each taken cyclically.

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Cf. A081721, A212360 (necklaces), A213941, A273891.

(* t = A081720 *) 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_] := Binomial[n, k]*Sum[(-1)^i * Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 08 2017, after Andrew Howroyd *)

a(n,m) = Sum_{j=1..p(n,m)} A213941(n, k(n,m,1)+j-1), with k(n,m,1) = A214314(n,m) the position where in the list of partitions of n in A-St order the first with m parts appears, and p(n,m) is the number of partitions of n with m parts shown in the array A008284. E.g., n=5, m=3: k(5,3,1) = A214314(5,3) = 4, p(5,3) = 2.

a(n,m) = binomial(n,m) * A273891(n,m). - Andrew Howroyd, Mar 25 2017

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)

A056345 Number of bracelets of length n using exactly five different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 12, 150, 1200, 7905, 46400, 255636, 1346700, 6901725, 34663020, 171786450, 843130688, 4110958530, 19951305240, 96528492700, 466073976900, 2247627076731, 10832193571460, 52194109216950

Offset: 1

Marks R. Nester

nonn

Turning over will not create a new bracelet.

			For a(5)=12, pair up the 24 permutations of BCDE, each with its reverse, such as BCDE-EDCB.  Precede the first of each pair with an A, such as ABCDE.  These are the 12 arrangements, all chiral.  If we precede the second of each pair with an A, such as AEDCB, we get the chiral partner of each. - _Robert A. Russell_, Sep 27 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]

Column 5 of A273891.

Equals (A056285 + A056491) / 2 = A056285 - A305544 = A305544 + A056491.

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, 5];
Array[a, 22] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=5; 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) = A032276(n) - 5*A032275(n) + 10*A027671(n) - 10*A000029(n) + 5.
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=5 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=5 is the number of colors. (End)

cp's OEIS Frontend

A152176 Triangle read by rows: T(n,k) is the number of k-block partitions of an n-set up to rotations and reflections.

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A087854 Triangle read by rows: T(n,k) is the number of n-bead necklaces with exactly k different colored beads.

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A305540 Triangle read by rows: T(n,k) is the number of achiral loops (necklaces or bracelets) of length n using exactly k different colors.

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A305541 Triangle read by rows: T(n,k) is the number of chiral pairs of color loops of length n with exactly k different colors.

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A056342 Number of bracelets of length n using exactly two different colored beads.

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

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

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