A000029 -id:A000029 - cp's OEIS frontend

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

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)

A091696 Number of classes of compositions of n equivalent under reflection or cycling.

Original entry on oeis.org

1, 2, 3, 5, 7, 12, 17, 29, 45, 77, 125, 223, 379, 686, 1223, 2249, 4111, 7684, 14309, 27011, 50963, 96908, 184409, 352697, 675187, 1296857, 2493725, 4806077, 9272779, 17920859, 34669601, 67159049, 130216123, 252745367, 490984487, 954637557, 1857545299

Offset: 1

Neil Fernandez, Jan 29 2004

nonn

Equivalently, the number of radial cutting patterns of a (maximally symmetric) circular cake such that all resulting pieces are a multiple of 1/n of the whole. - Peter Munn, Oct 27 2020

			7 has 15 partitions and 64 compositions. Compositions can be mapped to other compositions by reflection, cycling, or both, e.g., {1,2,4} -> {4,2,1} (reflection), {2,4,1} (cycling), or {1,4,2} (both); this defines the equivalence relation used. The number of equivalence classes so defined is 2 greater than the number of partitions because only {3,1,2,1} and {2,1,2,1,1} (and their equivalents) cannot be mapped to the conventionally stated forms of partitions (here, {3,2,1,1} and {2,2,1,1,1} respectively). So a(7) = 15 + 2 = 17.

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A000029, A056342.

DIK transform of A057427.

with(numtheory):
a:= n-> add(phi(d)*2^(n/d)/(2*n), d=divisors(n))
        +`if`(irem(n, 2)=0, 2^(n/2-1) +2^(n/2-2), 2^((n-1)/2)) -1:
seq(a(n), n=1..40);  # Alois P. Heinz, Oct 20 2012

Needs["Combinatorica`"]
nn=40;Apply[Plus,Table[CoefficientList[Series[CycleIndex[DihedralGroup[n], s]/.Table[s[i]->x^i/(1-x^i),{i,1,nn}],{x,0,nn}],x],{n,1,nn}]]  (* Geoffrey Critzer, Oct 18 2012 *)
mx:=50;CoefficientList[Series[(Sum[(EulerPhi[n] Log[2+1/(-1+x^n)])/n,{n,1,mx}]+(1-1x^2+ x^3)/((x-1) (1-2 x^2)))/(-2),{x,0,mx}],x] (* Herbert Kociemba, Dec 04 2016 *)
a[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#) & ]/(2*n) - 1; Array[a, 37] (* Jean-François Alcover, Nov 05 2017 *)

a(n) = A000029(n) - 1.
a(n) = A056342(n) + 1.
G.f.: ( Sum_{n>=1} phi(n)*log(2+1/(-1+x^n))/n + (1-1x^2+x^3)/((x-1)*(1-2*x^2)) )/(-2). - Herbert Kociemba, Dec 04 2016

More terms from Sean A. Irvine, Feb 09 2012

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

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)

A157371 a(n) = (n+1)(9-9n+5*n^2-n^3).

Original entry on oeis.org

9, 8, 9, 0, -55, -216, -567, -1216, -2295, -3960, -6391, -9792, -14391, -20440, -28215, -38016, -50167, -65016, -82935, -104320, -129591, -159192, -193591, -233280, -278775, -330616, -389367, -455616, -529975, -613080, -705591, -808192, -921591, -1046520, -1183735, -1334016, -1498167

Offset: 0

Paul Curtz, Feb 28 2009

This is the fourth in a family of sequences that appear in columns on pages 36 and 56 of the reference: (i) sequence n+1, A000029, (ii) sequence (n+1)*(1-n), A147998 and (iii) (n+1)*(5-5*n+2*n^2), A152064.

First differences along columns shown on page 56 of the reference are columns of what is shown on page 36 of the reference. Example: the third column of page 56, A152064, has first differences which constitute the third column p page 36, A140811.

Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Note 12, Arcueil (1969).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

[(n+1)*(9-9*n+5*n^2-n^3): n in [0..40] ]; // Vincenzo Librandi, Jul 14 2011

LinearRecurrence[{5,-10,10,-5,1},{9,8,9,0,-55},40] (* or *) Table[(n+1)(9-9n+5n^2-n^3),{n,0,40}] (* or *) CoefficientList[ Series[ (55x^3- 59x^2+ 37x-9)/ (x-1)^5,{x,0,40}],x] (* Harvey P. Dale, Jul 13 2011 *)

a(n)=(n+1)*(9-9*n+5*n^2-n^3) \\ Charles R Greathouse IV, Oct 16 2015

First differences: a(n+1)-a(n) = -A141530(n).
Fourth differences: a(n+4)-4*a(n+3)+6*a(n+2)-4*a(n+1)+a(n) = -24 = -A010863(n).
From Harvey P. Dale, Jul 13 2011: (Start)
a(0)=9, a(1)=8, a(2)=9, a(3)=0, a(4)=-55, a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5).
G.f.: (9-37*x+59*x^2-55*x^3)/(1-x)^5. (End)
E.g.f.: (9 - x + x^2 - 2*x^3 - x^4)*exp(x). - G. C. Greubel, Feb 02 2018

Edited, extended by R. J. Mathar, Sep 25 2009

A227910 The number of necklaces with n beads of white and red colors, including at least three white ones.

Original entry on oeis.org

1, 2, 4, 8, 13, 24, 40, 71, 119, 216, 372, 678, 1215, 2240, 4102, 7674, 14299, 27000, 50952, 96896, 184397, 352684, 675174, 1296843, 2493711, 4806062, 9272764, 17920843, 34669585, 67159032, 130216106, 252745349, 490984469, 954637538, 1857545280, 3617214660, 7048675939, 13744694906

Offset: 3

Vladimir Letsko, Oct 13 2013

nonn

a(n) is the number of classes (not necessarily convex) n-gons. Two n-gons are assigned to the same class, if (under suitable start and direction) their vertex at the same time belong or do not belong to the convex hull of the corresponding polygon.
a(n) is also the number of classes (not necessarily convex) n-gons at the other classification. Two n-gons belong to one class, if (under suitable start and direction) their angles simultaneously have a measure either less or greater than 180 degrees.
Note that the equation quantities of classes of these classifications does not mean equality classes themselves.
a(n) is also the number of non-isomorphic n-vertex undirected graphs in the form of a simple cycle with any number of degree-1 vertices attached to each cycle vertex. To transform a necklace into a graph of this type, create a cycle vertex for each white bead and a pendant vertex for each red bead, with the pendant vertex for a red bead attached to the cycle vertex for the clockwise next white bead. - David Eppstein, Oct 25 2015

			a(4)=2 because there are exactly 2 necklaces with 4 beads of white and red colors, including at least three white ones:
1) all beads are white;
2) one bead is red.

Andrew Howroyd, Table of n, a(n) for n = 3..200
Peter Kagey, Examples of classes of n-gons for a(4), a(5), and a(6)
Vladimir Letsko, Mathematical Marathon, Problem 150 (in Russian)

Cf. A000029, A262232.

g:=(n,m)->add(phi(d)*binomial(n/d,m/d),d in divisors(igcd(n,m)))/2/n+1/2*binomial(floor(m/2)+floor((n-m)/2),floor(m/2));
for n from 3 do s:=0:for m from 3 to n dos:=s+g(n,m) od: print(n,s) od:

g[n_, m_] := Sum[EulerPhi[d]*Binomial[n/d, m/d], {d , Divisors[GCD[n, m]]}]/2/n + 1/2*Binomial[Floor[m/2] + Floor[(n - m)/2], Floor[m/2]];
A227910 = Table[Sum[g[n, m], {m, 3, n}], {n, 3, 40}] (* Jean-François Alcover, Jul 02 2018, from Maple *)

a(n) = (sum(m=3, n, (1/n*sumdiv(gcd(m,n), d, eulerphi(d)*binomial(n/d,m/d))) + binomial(m\2+(n-m)\2, m\2) ))/2; \\ Andrew Howroyd, Oct 11 2017

a(n) = (Sum_{m=3..n} (1/n*sum_{d|gcd(m,n)} phi(d)*binomial(n/d,m/d)) + binomial(floor(m/2)+floor((n-m)/2), floor(m/2)))/2.

A056346 Number of bracelets of length n using exactly six different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 0, 60, 1080, 11970, 105840, 821952, 5874480, 39713550, 258136200, 1631273220, 10096734312, 61536377700, 370710950400, 2213749658880, 13132080672480, 77509456944318, 455754569692680

Offset: 1

Marks R. Nester

nonn

Turning over will not create a new bracelet.

			For a(6)=60, pair up the 120 permutations of BCDEF, each with its reverse, such as BCDEF-FEDCB.  Precede the first of each pair with an A, such as ABCDEF.  These are the 60 arrangements, all chiral.  If we precede the second of each pair with an A, such as AFEDCB, 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 6 of A273891.
Equals (A056286 + A056492) / 2 = A056286 - A305545 = A305545 + A056492.
Cf. A008277.

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, 6];
Array[a, 21] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=6; 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) = my(k=6); (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 29 2018

a(n) = A056341(n) - 6*A032276(n) + 15*A032275(n) - 20*A027671(n) + 15*A000029(n) - 6.
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=6 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=6 is the number of colors. (End)

A092481 Number of different sets of n-gons labeled 1...n such that all members of each set contain equivalent paths with increasing labels; i.e., the number of isotemporal classes of n-gons.

Original entry on oeis.org

1, 3, 3, 8, 9, 20, 29, 60, 93, 189, 315, 618, 1095, 2114, 3855, 7414, 13797, 26478, 49939, 95838, 182361, 350572, 671091, 1292604, 2485533, 4797616, 9256395, 17903928, 34636833, 67125304, 130150587, 252677904, 490853415, 954502948

Offset: 3

Benjamin de Bivort (bivort(AT)fas.harvard.edu), Apr 03 2004

nonn

B. de Bivort, Isotemporal classes of n-gons, preprint, 2004.
B. de Bivort, An introduction to temporal networks, preprint, 2004.

B. de Bivort, Isotemporal classes of n-gons

Cf. A000031, A000029, A027671.

f[n_] := Block[{d = Divisors[n], c = Divisors[n/2]}, Switch[ Mod[n, 4], 0, (Plus @@ (2^(n/d - 1)EulerPhi[d]) - Plus @@ (2^(n/(2c) - 1)EulerPhi[2c]))/n + 2^((n - 4)/2) + 2^((n - 8)/4) - 2^(Ceiling[(n - 4)/8] - 1), 1, (Plus @@ ((2^(n/d - 1) - 1)EulerPhi[d]))/n, 2, (Plus @@ (2^(n/d - 1)EulerPhi[d]) - Plus @@ (2^(n/(2c) - 1)EulerPhi[2c]))/n + 2^((n - 4)/2), 3, (Plus @@ ((2^(n/d - 1) - 1)EulerPhi[d]))/n]]; Table[ f[n], {n, 3, 36}]

If n odd: (1/n) sum_{d|n} (2^(n/d-1)-1) phi(d).
If n = 4k + 2: (1/n) {sum_{d|n} (2^(n/d-1) phi(d)) - sum_{c|n/2} (2^(n/2c-1) phi(2c)} + 2^(n-4)/2
If n = 4k: (1/n) {sum_{d|n} (2^(n/d-1) phi(d)) - sum_{c|n/2} (2^(n/2c-1) phi(2c))} + 2^(n-4)/2 + 2^(n-8)/4 - 2^(ceiling[(n-4)/8]-1).

Edited by Robert G. Wilson v, Apr 09 2004

A203399 T(n, k), a triangular array read by rows, is the number of classes of equivalent 2-color n-bead necklaces (turning over is allowed) that contain k necklaces.

Original entry on oeis.org

2, 0, 2, 1, 0, 0, 2, 0, 2, 0, 0, 0, 2, 1, 0, 3, 0, 0, 0, 0, 2, 0, 0, 0, 6, 0, 0, 0, 0, 0, 2, 1, 2, 0, 0, 7, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 14, 0, 0, 0, 0, 0, 0, 2, 2, 1, 0, 3, 0, 0, 0, 18, 0, 0, 0, 0, 0, 0, 0, 6, 2, 0, 2, 0, 0, 0, 0, 0, 28, 0, 0, 0, 0, 0, 0, 0, 0, 14, 2, 1, 0, 0, 6, 0, 0, 0, 0, 39, 0, 0, 0, 0, 0, 0, 0, 0, 0, 30

Offset: 1

Geoffrey Critzer, Jan 01 2012

Equivalently, the dihedral group of order n acts on the set of length n binary sequences. T(n,k) is the number of orbits that contain k elements.

By "n-bead necklaces (turning over is allowed)" the author means "bracelets". By saying that the classes of these n-bead bracelets (turnover necklaces) "contain k necklaces" the author means that the classes "contain k marked necklaces". - Petros Hadjicostas, Jun 06 2019

			Triangle begins (with rows n >= 1 and columns k >= 1):
  2  0
  2  1  0  0
  2  0  2  0  0  0
  2  1  0  3  0  0  0  0
  2  0  0  0  6  0  0  0  0  0
  2  1  2  0  0  7  0  0  0  0  0  1
  2  0  0  0  0  0 14  0  0  0  0  0  0  2
  2  1  0  3  0  0  0 18  0  0  0  0  0  0  0  6
  2  0  2  0  0  0  0  0 28  0  0  0  0  0  0  0  0 14
From _Petros Hadjicostas_, Jun 07 2019: (Start)
Consider all bracelets (turnover necklaces) of two colors (B and W) that can be generated using one of Ruskey's websites above. We keep his numbering, declare whether it has reflexive symmetry or not (achiral or chiral, resp.), and find its value of k (= number of different marked necklaces belonging to its equivalence class).
We have: (1) BBBBBB (k=1, achiral), (2) BBBBBW (k=6, achiral), (3) BBBBWW (k=6, achiral), (4) BBBWBW (k=6, achiral), (5) BBBWWW (k=6, achiral), (6) BBWBBW (k=3, achiral), (7) BBWBWW (k=12, chiral), (8) BBWWWW (k=6, achiral), (9) BWBWBW (k=2, achiral), (10) BWBWWWW (k=6, achiral), (11) BWWBWW (k=3, achiral), (12) BWWWWW (k=6, achiral), (13) WWWWWW (k=1, achiral).
Hence, only bracelet (7) has no reflection symmetry, and thus it is chiral. The k=12 marked necklaces of its equivalence class are as follows:
BBWBWW, WBBWBW, WWBBWB, BWWBBW, WBWWBB, BWBWWB, and their mirror images BWWBWB, BBWWBW, WBBWWB, BWBBWW, WBWBBW, WWBWBB.
We see that T(n=6, k=1) = 2, T(n=6, k=2) = 1, T(n=6, k=3) = 2, T(n=6, k=6) = 7, and T(n=6, k=12) = 1, which agree with line n=6 in the triangle above. (End)

Petros Hadjicostas, Table of n, a(n) for n = 1..210, flattened.
Petros Hadjicostas, Formulas for chiral bracelets, 2019.
S. Karim, J. Sawada, Z. Alamgirz, and S. M. Husnine, Generating bracelets with fixed content, 2011.
S. Karim, J. Sawada, Z. Alamgirz, and S. M. Husnine, Generating bracelets with fixed content, Theoretical Computer Science, Volume 475, 4 March 2013, Pages 103-112.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
Frank Ruskey, Combinatorial Generation Algorithm Algorithm 4.24, p. 95.
Joe Sawada, Generating bracelets in constant amortized time, SIAM J. Comput. 31 (2001), 259-268.
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. A000029 (row sums), A032239 (T(n, 2n) for n >= 3), A056493, A203398.

Needs["Combinatorica`"];
f[list_]:= Sort[NestList[RotateLeft, list, Length[list]-1] ~Join~NestList[RotateLeft, Reverse[list], Length[list]-1]]; Flatten[Table[Distribution[Map[Length, Map[Union, Union[Map[f, Strings[{0, 1}, n]]]]], Range[2 n]], {n, 1, 10}]]

From Petros Hadjicostas, Jun 06 2019: (Start)
Conjectures: For n >= 1, let b(n) be the number of bracelets of two colors with n beads that are either periodic (period >= 2), or have reflection symmetry (achiral), or both. Then b(n) = A000029(n) - A032239(n) for n >= 3 with b(n) = A000029(n) for n = 1, 2. We have A000029(n) = 2^floor(-3 + n/2) * (7 - (-1)^n) + (1/(2*n)) * Sum_{d|n} phi(d) * 2^(n/d) for n >= 1 and A032239(n) = (1/2) * Sum_{d|n} mu(d) * (-2^floor(-2 + (n/(2*d))) * (7 - (-1)^(n/d)) + 2^(n/d)/n) for n >= 3.
For 1 <= k <= n, we conjecture that T(n, k) = Sum_{d|k} mu(d)*b(k/d) for k|n, and = 0 otherwise. Note that, if 3 <= n <= 11, we have T(n, k) = A056493(k) when k|n, but this is not true (for example) for n = 12 and n = 14. We have T(12, 12) = 82 <> 81 = A056493(12) and T(14, 14) = 177 <> 175 = A056493(14).
Apparently, T(n, 2*n) = A032239(n) for all n >= 3, and T(n, k) = 0 for n < k < 2*n. (End)
From Petros Hadjicostas, Jun 16 2019: (Start)
I ran the author's Mathematica program for n = 1..21 and I saw that the conjecture is OK except for n = 18 and n = 21. The program gives T(n=18, k=12) = 1 and T(n=18, k=18) = 742 while my conjecture implies that T(n=18, k=12) = 0 (since k = 12 does not divide n = 18) and T(n=18, k=18) = 743. In addition, the program gives T(n=21, k=14) = 2 and T(n=21, k=21) = 2030, while my conjecture implies that T(n=21, k=14) = 0 (since k = 14 does not divide n = 21) and T(n=21, k=21) = 2032. Apparently, my conjecture needs to be refined.
For n = 18, the single bracelet whose equivalence class has 12 marked necklaces is (BBWBWW)^3 (with period 3).
For n = 21, the two bracelets whose equivalence classes have 14 marked necklaces each are (BBWBWWW)^3 and (WWBWBBB)^3 (each with period 3). (End)

A213942 a(n) is the number of representative two-color bracelets (necklaces with turnover allowed) with n beads for n >= 2.

Original entry on oeis.org

1, 1, 3, 3, 7, 8, 18, 22, 46, 62, 136, 189, 409, 611, 1344, 2055, 4535, 7154, 15881, 25481, 56533, 92204, 204759, 337593, 748665, 1246862, 2762111, 4636389, 10253938, 17334800, 38278784, 65108061, 143534770, 245492243, 540353057, 928772649, 2041154125

Offset: 2

Wolfdieter Lang, Jul 31 2012

nonn

This is the second column (m=2) of triangle A213940.
The relevant floor(n/2) representative color multinomials are c[1]^(n-1)*c[2], c[1]^(n-2)*c[2]^2, ..., c[1]^(n-floor(n/2))* c[2]^(floor(n/2)). For such representative bracelets the color c[1] is therefore preferred. Only for even n can c[2] appear as often as c[1], namely, n/2 times.
Note that beads with different colors are always present. This is in contrast to, e.g., A000029, where not only representatives but also one-color bracelets are counted. This sequences gives the number of binary bracelets with at least as many 0's as 1's and at least one 1 (bracelet analog of A226881). The number of two-color bracelets up to permutations of colors is given by A056357. For odd n these two sequences are equal. For a(8), the bracelets 00011011 and 11100100 are equivalent in A056357 but distinct in this sequence. - Andrew Howroyd and Wolfdieter Lang, Sep 25 2017

			a(5) = A213939(5,2) + A213939(5,3) = 1 + 2 = 3 from the representative bracelets (with colors j for c[j], j=1,2) cyclic(11112), cyclic(11122) and cyclic(11212). The first one has color signature (exponents) [4,1] and the two others have signature [3,2]. For the number of all two-color 5-bracelets with beads of five colors available see A214308(5) = 60.
a(8) = 18 =  1 + 4 + 5 + 8 for the partitions of 8 with 2 parts (7,1), (6, 2), (5,3), (4,4), respectively. see A213939(5, k), k = 2..5). The 8 representative bracelets for the exponents (signature) from partition (4,4) are B1 = (11112222), B2 = (11121222), B3 = (11212122), B4 = (11212212), B5 = (11221122), B6 = (12121212), B7 = (11122122) and B8 = (11211222). B1 to B6 are color exchange (1 <-> 2) invariant (modulo D_8 symmetry, i.e., cyclic or anti-cyclic operations). B7 is equivalent to B8 under color exchange.
This explains why A056357(8) = 17. The difference between the present sequence and A056357 is that there, besides D_n symmetry, also color exchange is allowed. Here only color exchange compatible with D_n symmetry is allowed. - _Wolfdieter Lang_, Sep 28 2017

G. C. Greubel, Table of n, a(n) for n = 2..1000

Cf. A213939, A213940, A214307 (m=3), A214308 (m=2, all bracelets).

Cf. A056357, A226881.

a29[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#)&]/(2*n);
a5648[n_] := 1/2*(Binomial[2*Quotient[n, 2], Quotient[n, 2]] + DivisorSum[n, EulerPhi[#]*Binomial[2*n/#, n/#]&]/(2*n));
a[n_] := a29[n]/2 - 1 + If[EvenQ[n], a5648[n/2]/2, 0];
Array[a, 37, 2] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)

a(n) = A213940(n,2), n >= 2.
a(n) = Sum_{k=2..A008284(n,2)+1} A213939(n,k), n >= 2, with A008284(n,2) = floor(n/2).
a(2n) = (A000029(2n) + A005648(n)) / 2 - 1, a(2n+1) = A000029(2n+1) / 2 - 1. - Andrew Howroyd, Sep 25 2017

Terms a(26) and beyond from Andrew Howroyd, Sep 25 2017

cp's OEIS Frontend

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