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.

Showing 1-9 of 9 results.

A309528 The number of non-equivalent distinguishing colorings of the cycle on n vertices with at most k colors (k>=1). The cycle graph is defined for n>=3; extended to n=1,2 using the closed form. Square array read by descending antidiagonals: the rows are indexed by n, the number of vertices of the cycle and the columns are indexed by k, the number of permissible colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 4, 3, 0, 0, 0, 0, 10, 15, 12, 1, 0, 0, 0, 20, 45, 72, 37, 2, 0, 0, 0, 35, 105, 252, 266, 117, 6, 0, 0, 0, 56, 210, 672, 1120, 1044, 333, 14, 0, 0, 0, 84, 378, 1512, 3515, 5270, 3788, 975, 30, 0, 0, 0, 120, 630, 3024, 9121, 19350, 23475, 14056, 2712, 62, 0
Offset: 1

Views

Author

Bahman Ahmadi, Aug 06 2019

Keywords

Comments

A vertex-coloring of a graph G is called distinguishing if it is only preserved by the identity automorphism of G. This notion is considered in the subject of symmetry breaking of simple (finite or infinite) graphs. Two vertex-colorings of a graph are called equivalent if there is an automorphism of the graph which preserves the colors of the vertices. Given a graph G, we use the notation Phi_k(G) to denote the number of non-equivalent distinguishing colorings of G with at most k colors. The sequence here, displays A(n,k)=Phi_k(C_n), i.e., the number of non-equivalent distinguishing colorings of the cycle C_n on n vertices with at most k colors.

Examples

			The table begins:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
0, 0, 1, 4, 10, 20, 35, 56, 84, 120, ...
0, 0, 3, 15, 45, 105, 210, 378, 630, 990, ...
0, 0, 12, 72, 252, 672, 1512, 3024, 5544, 9504, ...
0, 1, 37, 266, 1120, 3515, 9121, 20692, 42456, 80565, ...
0, 2, 117, 1044, 5270, 19350, 57627, 147752, 338364, 709290, ...
0, 6, 333, 3788, 23475, 102690, 355446, 1039248, 2673810, 6222150, ...
0, 14, 975, 14056, 106950, 555990, 2233469, 7440160, 21493836, 55505550, ...
0, 30, 2712, 51132, 483504, 3009426, 14089488, 53611992, 174189024, 499720518, ...
------
For n=4, we can color the vertices of the cycle C_4 with at most 3 colors, in 3 ways, such that all the colorings distinguish the graph (i.e., no non-identity automorphism of C_4 preserves the coloring) and that all the three colorings are non-equivalent. The color classes are as follows:
{ { 1 }, { 2 }, { 3, 4 } }
{ { 1 }, { 2, 3 }, { 4 } }
{ { 1, 2 }, { 3 }, { 4 } }
		

Crossrefs

Columns k=2..5 for n >= 3 are A032239, A032240, A032241, A032242.
Different from A293496.

Programs

  • PARI
    A(n,k)={sumdiv(n, d, moebius(n/d)*(k^d/n - if(d%2, k^((d+1)/2), (k+1)*k^(d/2)/2)))/2} \\ Andrew Howroyd, Aug 11 2019

Formula

A(n,k) = (A074650(n,k) - A284856(n,k))/2. - Andrew Howroyd, Aug 11 2019

A309651 T(n,k) is the number of non-equivalent distinguishing colorings of the cycle on n vertices with exactly k colors (k>=1). Regular triangle read by rows, n >= 1, 1 <= k <= n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 3, 3, 0, 0, 12, 24, 12, 0, 1, 34, 124, 150, 60, 0, 2, 111, 588, 1200, 1080, 360, 0, 6, 315, 2484, 7845, 11970, 8820, 2520, 0, 14, 933, 10240, 46280, 105840, 129360, 80640, 20160, 0, 30, 2622, 40464, 254664, 821592, 1481760, 1512000, 816480, 181440
Offset: 1

Views

Author

Bahman Ahmadi, Aug 11 2019

Keywords

Comments

The cycle graph is defined for n>=3; extended to n=1,2 using the closed form.
A vertex-coloring of a graph G is called distinguishing if it is only preserved by the identity automorphism of G. This notion is considered in the subject of symmetry breaking of simple (finite or infinite) graphs. Two vertex-colorings of a graph are called equivalent if there is an automorphism of the graph which preserves the colors of the vertices. Given a graph G, we use the notation phi_k(G) to denote the number of non-equivalent distinguishing colorings of G with exactly k colors. The sequence here, displays T(n,k)=phi_k(C_n), i.e., the number of non-equivalent distinguishing colorings of the cycle C_n on n vertices with exactly k colors.
From Andrew Howroyd, Aug 15 2019: (Start)
First differs from A305541 at n = 6.
Also the number of n-bead asymmetric bracelets with exactly k different colored beads. More precisely the number of chiral pairs of primitive (aperiodic) color loops of length n with exactly k different colors. For example, for n=4 and k = 3, there are 3 achiral loops (1213, 1232, 1323) and 3 pairs of chiral loops (1123/1132, 1223/1322, 1233/1332).
(End)

Examples

			The triangle begins:
  0
  0,  0;
  0,  0,    1;
  0,  0,    3,     3;
  0,  0,   12,    24,     12;
  0,  1,   34,   124,    150,     60;
  0,  2,  111,   588,   1200,   1080,     360;
  0,  6,  315,  2484,   7845,  11970,    8820,    2520;
  0, 14,  933, 10240,  46280, 105840,  129360,   80640,  20160;
  0, 30, 2622, 40464, 254664, 821592, 1481760, 1512000, 816480, 181440;
  ...
For n=4, we can color the vertices of the cycle C_4 with exactly 3 colors, in 3 ways, such that all the colorings distinguish the graph (i.e., no non-identity automorphism of C_4 preserves the coloring) and that all the three colorings are non-equivalent. The color classes are as follows:
{ { 1 }, { 2 }, { 3, 4 } }
{ { 1 }, { 2, 3 }, { 4 } }
{ { 1, 2 }, { 3 }, { 4 } }
		

Crossrefs

Columns k=2..4 are A032239(n>=3), A326660, A326789.
Row sums are A326888.

Programs

  • PARI
    \\ U(n,k) is A309528
    U(n,k)={sumdiv(n, d, moebius(n/d)*(k^d/n - if(d%2, k^((d+1)/2), (k+1)*k^(d/2)/2)))/2}
    T(n,k)={sum(i=2, k, (-1)^(k-i)*binomial(k,i)*U(n,i))} \\ Andrew Howroyd, Aug 12 2019

Formula

Let n>2. For any k >= floor(n/2) we have phi_k(C_n)=k! * Stirling2(n,k)/2n.
T(n, k) = Sum_{i=2..k} (-1)^(k-i)*binomial(k,i)*A309528(n, i). - Andrew Howroyd, Aug 12 2019
Column k is the Moebius transform of column k of A305541. - Andrew Howroyd, Sep 13 2019

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

Views

Author

Geoffrey Critzer, Jan 01 2012

Keywords

Comments

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

Examples

			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)
		

Crossrefs

Cf. A000029 (row sums), A032239 (T(n, 2n) for n >= 3), A056493, A203398.

Programs

  • Mathematica
    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}]]

Formula

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)

A326660 Number of n-bead asymmetric bracelets with exactly 3 different colored beads.

Original entry on oeis.org

0, 0, 1, 3, 12, 34, 111, 315, 933, 2622, 7503, 21033, 59472, 167118, 472120, 1332945, 3777600, 10720869, 30516447, 87032994, 248820704, 712743768, 2045784183, 5882367570, 16942974048, 48876558318, 141204944529, 408494941773, 1183247473872, 3431450670601
Offset: 1

Views

Author

Andrew Howroyd, Sep 12 2019

Keywords

Comments

Asymmetric bracelets are those primitive (period n) bracelets that when turned over are different from themselves.

Examples

			Case n = 4: There are 3 distinct asymmetric bracelets with exactly 3 colors which are aabc, abbc, abcc.
		

Crossrefs

Formula

a(n) = A032240(n) - 3*A032239(n) for n >= 3.
Moebius transform of A305542.

A326789 Number of n-bead asymmetric bracelets with exactly 4 different colored beads.

Original entry on oeis.org

0, 0, 0, 3, 24, 124, 588, 2484, 10240, 40464, 158220, 608951, 2333520, 8894616, 33864340, 128791140, 490027200, 1865614976, 7110959340, 27138170397, 103717719412, 396965536224, 1521562700988, 5840509149020, 22450188684264, 86412086034120, 333035003533660
Offset: 1

Views

Author

Andrew Howroyd, Sep 12 2019

Keywords

Comments

Asymmetric bracelets are those primitive (period n) bracelets that when turned over are different from themselves.

Crossrefs

Formula

a(n) = A032241(n) - 4*A032240(n) + 6*A032239(n) for n >= 3.
Moebius transform of A305543.

A032253 "DHK" (bracelet, identity, unlabeled) transform of 3,3,3,3,...

Original entry on oeis.org

1, 3, 6, 13, 27, 78, 278, 1011, 3753, 13843, 50934, 187629, 692891, 2568882, 9562074, 35742329, 134117829, 505093740, 1908474674, 7232842785, 27486193251, 104712247296, 399816026490, 1529742725403, 5864036504705, 22517947805343, 86607583200294, 333599771067256
Offset: 0

Views

Author

Keywords

Comments

From Petros Hadjicostas, Jun 17 2019: (Start)
An unmarked cyclic composition of n >= 1 is an equivalence of ordered partitions of n such that two ordered partitions are equivalent iff one can be obtained from the other by rotation.
A dihedral composition of n >= 1 is an equivalence class of ordered partitions of n such that two such ordered partitions are equivalent iff one can be obtained from the other by rotation or reflection. See, for example, Knopfmacher and Robbins (2013).
A cyclic composition (ordered partition) of n >= 1 is called achiral iff it has a reflection symmetry, and it is called chiral otherwise. (This terminology comes from Chemistry in the study of molecules.)
A symmetric (achiral) cyclic composition of n >= 1 is also a symmetric (achiral) dihedral composition of n > = 1 (and vice versa).
Many mathematicians consider a cyclic composition of n >= 1 with one part or with two parts as achiral by default because the axis of symmetry may pass through the parts. When he defines the DHK transform, Bowers (in the link below) does not accept this convention except possibly for a cyclic composition with two identical (in value and color) parts.
Given n >= 1, a(n) here is the number of aperiodic chiral dihedral compositions of n such that the parts may be colored by any one of three colors (say, A, B, C).
Notice that a(1) = 3 because 1_A, 1_B, 1_C are the three colored aperiodic dihedral compositions of n = 1 that (according to Bowers) are considered chiral (= with no reflection symmetry).
In addition, a(2) = 6 because 2_A, 2_B, 2_C, 1_A + 1_B, 1_B + 1_C, 1_C + 1_A are the six colored aperiodic dihedral compositions of n = 2 that (according to Bowers) are considered chiral (= with no reflection symmetry).
In general, for n >= 1, if one disagrees with Bower's conventions about colored aperiodic dihedral compositions with one or two parts, then a(n) - 3*A001651(n) = a(n) - 3 * floor((3*n - 1 )/2) is the actual number of aperiodic chiral dihedral compositions of n such that the parts may be colored by any one of three colors.
Let c = (c(n): n >= 1) be the input sequence and b = (b(n): n >= 1) be the output sequence under Bower's DHK transform; i.e., b = (DHK c). Let C(x) = Sum_{n >= 1} c(n)*x^n; i.e., C(x) is the g.f. of c. Then the g.f. of b is Sum_{n >= 1} b(n)*x^n = -(1/2) * Sum_{d >= 1} (mu(d)/d) * log(1 - C(x^d)) - (1/2) * Sum_{d >= 1} mu(d) * ((C(x^d) + 1)^2/(2 * (1 - C(x^(2*d))) - (1/2)) + C(x) + (1/2) * (C(x)^2 - C(x^2)). Here, c(n) = 3 for all n >= 1 and C(x) = 3*x/(1 - x).
The part of the g.f. that gives the extra aperiodic dihedral compositions due to Bower is C(x) + (1/2) * (C(x)^2 - C(x^2)) = 3*x*(1 + x + x^2)/((1 + x)*(1 - x)^2). This is the g.f. of (3*A001651(n): n >= 1).
Here, D(x) = (C(x) + 1)^2/(2*(1 - C(x^2))) - (1/2) = 3*x/((1 - 2*x)*(1 - x)) = 3*(x + 3*x^2 + 7*x^3 + 15*x^4 + ...) is the g.f. of (3*A000225(n): n >= 1) = (3*(2^n - 1): n >= 1), which counts the symmetric (= achiral) unmarked cyclic compositions of n where up to 3 colors can be used.
Thus, the sequence (3*A038199(n): n >= 1) = (3*Sum_{d|n} mu(d)*A000225(n/d): n >= 1) = (3*Sum_{d|n} mu(d)*(2^(n/d) - 1): n >= 1) counts the aperiodic symmetric (unmarked) cyclic compositions of n where up to three colors can be used (without Bower's conventions for compositions with one or two parts). This latter sequence has g.f. Sum_{d >= 1} mu(d)*D(x^d) = Sum_{d >= 1} mu(d) * ((C(x^d) + 1)^2/(2 * (1 - C(x^(2*d))) - (1/2)).
Finally, -Sum_{d >= 1} (mu(d)/d)*log(1 - C(x^d)), where C(x) = 3*x/(1 - x), is the g.f. of sequence (A185172(n): n >= 1), which counts the aperiodic (unmarked) cyclic compositions of n where up to three colors can be used. See Eqs. (94) and (95) in Novelli and Thibon (2008) or Eqs. (99) and (100) in Novelli and Thibon (2010).
(End)

Examples

			From _Petros Hadjicostas_, Jun 17 2019: (Start)
For n = 3, the Bower's extra 3*A001651(3) = 12 aperiodic dihedral compositions of 3 (using three colors) with one or two parts are as follows: 3_A, 3_B, 3_C, 1_A + 2_A, 1_B + 2_B, 1_C + 2_C, 1_A + 2_B, 1_A + 2_C, 1_B + 2_A, 1_B + 2_C, 1_C + 2_A, 1_C + 2_B. Since a(3) - 3*A001651(3) = 13 - 12 = 1, we have only one aperiodic chiral dihedral composition of 3 (with more than two parts): 1_A + 1_B + 1_C.
For n = 4, the Bower's extra 3*A001651(4) = 15 aperiodic dihedral compositions of n = 4 (using three colors) with one or two parts are as follows: 4_X, where X \in {A, B, C}; 2_X + 2_Y,  where (X,Y) \in {(A, B), (B, C), (C, A)}; and 1_X + 3_Y, where (X, Y) \in {(A, A), (A, B), (A, C), (B, A), (B, B), (B, C), (C, A), (C, B), (C, C)}.
The remaining (i.e., the genuine) a(4) - 15 = 27 - 15 = 12 aperiodic chiral dihedral compositions of n = 4 of 3 colors are as follows: 1_X + 2_X + 1_Y, where (X, Y) \in {(A, B), (A, C), (B, A), (B, C), (C, A), (C, B)}; 1_X + 2_Y + 1_Z and 1_X + 1_X + 1_Y + 1_Z, where (X, Y, Z) \in \{(A, B, C), (B, C, A), (C, A, B)}.
(End)
		

Crossrefs

Programs

  • Mathematica
    A001651[n_] := n - 1 + Ceiling[n/2];
    A185172[n_] := If[n==1, 3, Sum[MoebiusMu[d] 4^(n/d), {d, Divisors[n]}]/n];
    A038199[n_] := Sum[((2^d-1) MoebiusMu[n/d]), {d, Divisors[n]}];
    a[n_] := Switch[n, 0, 1, 1, 3, _, 3 A001651[n] + (1/2)(A185172[n] - 3 * A038199[n])];
    a /@ Range[0, 30] (* Jean-François Alcover, Sep 17 2019 *)
  • PARI
    DHK(p,n)={my(q=((1+p)^2/(1-subst(p, x, x^2))-1)/2); p + (p^2-subst(p, x, x^2))/2 + sum(d=1, n, moebius(d)*(log(subst(1/(1+O(x*x^(n\d))-p), x, x^d))/d - subst(q + O(x*x^(n\d)), x, x^d)))/2}
    seq(n)={Vec(1 + DHK(3*x/(1-x) + O(x*x^n), n))} \\ Andrew Howroyd, Sep 21 2018

Formula

From Petros Hadjicostas, Jun 18 2019: (Start)
a(n) = 3*A001651(n) + (1/2)*(A185172(n) - 3*A038199(n)) for n >= 1. Here, A001651(n) = floor((3*n - 1)/2) and A038199(n) = Sum_{d|n} mu(d)*(2^(n/d) - 1) for n >= 1. Also, A185172(1) = 3 and A185172(n) = (1/n)*Sum_{d|n} mu(d) * 4^(n/d) for n >= 2.
G.f.: 1 - (1/2)*Sum_{d >= 1} (mu(d)/d)*log(1 - 3*x^d/(1 - x^d)) - (1/2)*Sum_{d >= 1} mu(d)*3*x^d/((1 - 2*x^d)*(1 - x^d)) + 3*x*(1 + x + x^2)/((1 + x)*(1 - x)^2).
G.f.: 1 - x/2 - (1/2)*Sum_{d >= 1} (mu(d)/d)*log(1 - 4*x^d) - (1/2)*Sum_{d >= 1} mu(d)*3*x^d/((1 - 2*x^d)*(1 - x^d)) + 3*x*(1 + x + x^2)/((1 + x)*(1 - x)^2). (End)

Extensions

a(0)=1 prepended and terms a(24) and beyond from Andrew Howroyd, Sep 21 2018

A308583 Triangle read by rows: T(n,k) = number of aperiodic chiral bracelets (turnover necklaces with no reflection symmetry and period n) with n beads, k of which are white and n - k are black, for n >= 1 and 1 <= k <= n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 3, 4, 4, 3, 0, 0, 0, 0, 0, 4, 6, 10, 6, 4, 0, 0, 0, 0, 0, 5, 10, 16, 16, 10, 5, 0, 0, 0, 0, 0, 7, 14, 28, 29, 28, 14, 7, 0, 0, 0, 0, 0, 8, 20, 42, 56, 56, 42, 20, 8, 0, 0, 0, 0, 0, 10, 26, 64, 90, 113, 90, 64, 26, 10, 0, 0, 0, 0, 0, 12, 35, 90, 150, 197, 197, 150, 90, 35, 12, 0, 0, 0, 0, 0, 14, 44, 126, 222, 340, 368, 340, 222, 126, 44, 14, 0, 0, 0
Offset: 1

Views

Author

Petros Hadjicostas, Jun 08 2019

Keywords

Comments

For k = 1, 4, or a prime, the columns of this triangular array are exactly the same as the corresponding columns for the triangular array A180472. In other words, all chiral bracelets with n beads, k of which are white and n - k are black, are aperiodic if k = 1, 4, or a prime.
Note that, T(n, k) is also the number of aperiodic dihedral compositions of n with k parts and no reflection symmetry. Since T(n, k) = T(n, n - k), T(n, k) is also the number of aperiodic dihedral compositions of n with n - k parts and no reflection symmetry.

Examples

			The triangle begins (with rows for n >= 1 and columns for k >= 1) as follows:
  0;
  0, 0;
  0, 0,  0;
  0, 0,  0,  0;
  0, 0,  0,  0,  0;
  0, 0,  1,  0,  0,  0;
  0, 0,  1,  1,  0,  0,   0;
  0, 0,  2,  2,  2,  0,   0,  0;
  0, 0,  3,  4,  4,  3,   0,  0,  0;
  0, 0,  4,  6, 10,  6,   4,  0,  0,  0;
  0, 0,  5, 10, 16, 16,  10,  5,  0,  0,  0;
  0, 0,  7, 14, 28, 29,  28, 14,  7,  0,  0, 0;
  0, 0,  8, 20, 42, 56,  56, 42, 20,  8,  0, 0, 0;
  0, 0, 10, 26, 64, 90, 113, 90, 64, 26, 10, 0, 0, 0;
  ...
Notice, for example, that T(14, 6) = 90 <> 91 = A180472(14, 6). Out of the 91 chiral bracelets with 6 W and 8 B beads, only WWBWBBBWWBWBBB is periodic.
Using Frank Ruskey's website (listed above) to generate bracelets of fixed content (6, 3) with string length n = 9 and alphabet size 2, we get the following A005513(n = 9) = 7 bracelets: (1) WWWWWWBBB, (2) WWWWWBWBB, (3) WWWWBWWBB, (4) WWWWBWBWB, (5) WWWBWWWBB, (6) WWWBWWBWB, and (7) WWBWWBWWB. From these, bracelets 1, 4, 5, and 7 have reflection symmetry, while bracelets 2, 3 and 6 have no reflection symmetry. Because chiral bracelets 2, 3, and 6 are aperiodic as well, we have T(9, 3) = 3 = T(9, 6).
Starting with a black bead, we count that bead and how many white beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence to get the following dihedral compositions of n = 9 into 3 parts: (1) 1 + 7 + 1, (2) 1 + 2 + 6, (3) 1 + 3 + 5, (4) 2 + 5 + 2, (5) 4 + 1 + 4, (6) 2 + 3 + 4, and (7) 3 + 3 + 3. Again, dihedral compositions 1, 4, 5, and 7 are symmetric (have reflection symmetry), while dihedral compositions 2, 3, and 6 are not symmetric. In addition, chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and so (again) T(9, 3) = 3.
We may also start with a white bead and count that bead and how many black beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence again to get the following (conjugate) dihedral compositions of n = 9 into 6 parts: (1) 1 + 1 + 1 + 1 + 1 + 4, (2) 1 + 1 + 1 + 1 + 2 + 3, (3) 1 + 1 + 1 + 2 + 1 + 3, (4) 1 + 1 + 1 + 2 + 2 + 2, (5) 1 + 1 + 2 + 1 + 1 + 3, (6) 1 + 1 + 2 + 1 + 2 + 2, and (7) 1 + 2 + 1 + 2 + 1 + 2. Again, dihedral compositions 1, 4, 5, and 7 have reflection symmetries, while dihedral compositions 2, 3, and 6 do not have reflection symmetries. Chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and hence T(9, 6) = 3.
		

Crossrefs

Cf. A032239 (row sums for n >= 3), A180472.
Cf. A001399 (column k = 3 with a different offset), A008804 (column k = 4 with a different offset), A032246 (column k = 5), A032247 (column k = 6), A032248 (column k = 7), A032249 (column k = 8).

Formula

T(n, k) = Sum_{d|gcd(n,k)} mu(d) * A180472(n/d, k/d) for 1 <= k <= n.
T(n, k) = T(n, n - k) for 1 <= k <= n - 1.
T(n, k) = (1/(2*k)) * Sum_{d|gcd(n, k)} mu(d) * (binomial(n/d - 1, k/d - 1) - k * binomial(floor(b(n, k, d)/2), floor(k/(2*d)))) for 1 <= k <= n, where b(n, k, d) = (n/d) + ((-1)^(k/d) - 1)/2.
T(n, k) = (1/(2*n)) * Sum_{d|gcd(n, k)} mu(d) * (binomial(n/d, k/d) - n * binomial(floor(b(n, k, d)/2), floor(k/(2*d)))) for 1 <= k <= n, where b(n, k, d) = (n/d) + ((-1)^(k/d) - 1)/2.
G.f. for column k >= 1: (x^k/(2*k)) * Sum_{d|k} mu(d) * (1/(1 - x^d)^(k/d) - k * (1 + x^d)/(1 - x^(2*d))^floor((k/(2*d)) + 1)).
Bivariate g.f.: Sum_{n,k >= 1} T(n, k)*x^n*y^k = (1/2) * Sum_{d >= 1} mu(d) * (1 - (1 + x^d) * (1 + x^d*y^d) / (1 - x^(2*d) * (1 + y^(2*d)))) - (1/2) * Sum_{d >= 1} (mu(d)/d) * log(1 - x^d * (1 + y^d)).

A032337 Number of identity bracelets with n labeled beads of 2 colors.

Original entry on oeis.org

2, 2, 0, 0, 0, 720, 10080, 241920, 5080320, 108864000, 2474841600, 60833203200, 1569209241600, 42978897561600, 1265828788224000, 38916389191680000, 1280474741145600000, 44189183316934656000
Offset: 1

Views

Author

Keywords

Formula

"DHJ" (bracelets, identity, labeled) transform of 2, 0, 0, 0...
n! * A032239.

A377502 Number of minimum distinguishing labelings in the cycle graph C_n.

Original entry on oeis.org

6, 24, 120, 12, 28, 96, 252, 600, 1364, 3048, 6552, 13804, 29040, 59520, 122400, 248472, 504868, 1017840, 2054388, 4126100, 8294444, 16628160, 33349800, 66784536, 133775712, 267736392, 535920696, 1072242540, 2145452092, 4291768320, 8585609340, 17173070880, 34350563940
Offset: 3

Views

Author

Eric W. Weisstein, Oct 30 2024

Keywords

Comments

The distinguishing number of the n-cycle graph is 3 for n = 3, 4, 5 and 2 for n >= 6.

Crossrefs

Cf. A032239.

Programs

  • PARI
    a(n)={2*n*if(n<6, if(n>2,[1,3,12][n-2]), sumdiv(n, d, moebius(n/d)*(2^d/n - if(d%2, 2^((d+1)/2), 3*2^(d/2)/2)))/2)} \\ Andrew Howroyd, May 27 2025

Formula

a(n) = 2*n*A032239(n) for n >= 6. - Andrew Howroyd, May 27 2025

Extensions

a(27) onwards from Andrew Howroyd, May 27 2025
Showing 1-9 of 9 results.