Counting Colorful Necklaces and Bracelets in Three Colors

TL;DR

This paper derives formulas for counting non-equivalent colorful necklaces and bracelets with n beads in up to three colors, extending known sequences and providing new results for necklace counts.

Contribution

It introduces new formulas for counting colorful necklaces and bracelets under symmetry, with novel expressions for necklaces not previously documented.

Findings

Derived formulas for K(n) and K'(n)

Simplified expressions for bracelets matching OEIS sequence A114438

New formulas for counting necklaces not in OEIS

Abstract

A necklace or bracelet is \textit{colorful} if no pair of adjacent beads are the same color. In addition, two necklaces are \textit{equivalent} if one results from the other by permuting its colors, and two bracelets are \textit{equivalent} if one results from the other by either permuting its colors or reversing the order of the beads; a bracelet is thus a necklace that can be turned over. This note counts the number $K(n)$ of non-equivalent colorful necklaces and the number $K'(n)$ of colorful bracelets formed with $n$-beads in at most three colors. Expressions obtained for $K'(n)$ simplify expressions given by OEIS sequence A114438, while the expressions given for $K(n)$ appear to be new and are not included in OEIS.

Tables

Table 1. Table 1. List of K ​ ( 1 ) , … , K ​ ( 40 ) 𝐾 1 … 𝐾 40 K(1),\ldots,K(40) , which counts colorful necklaces.

$n$	$K\left(n\right)$	$n$	$K\left(n\right)$	$n$	$K\left(n\right)$	$n$	$K\left(n\right)$
1	0	11	31	21	16651	31	11545611
2	1	12	64	22	31838	32	22371000
3	1	13	105	23	60787	33	43383571
4	2	14	202	24	116640	34	84217616
5	1	15	367	25	223697	35	163617805
6	4	16	696	26	430396	36	318150720
7	3	17	1285	27	828525	37	619094385
8	8	18	2452	28	1598228	38	1205614054
9	11	19	4599	29	3085465	39	2349384031
10	20	20	8776	30	5966000	40	4581315968

Table 2. Table 2. List of K ′ ​ ( 1 ) , … , K ′ ​ ( 40 ) superscript 𝐾 ′ 1 … superscript 𝐾 ′ 40 K^{\prime}(1),\ldots,K^{\prime}(40) , which counts colorful bracelets.

$n$	$K^{\prime }\left(n\right)$	$n$	$K^{\prime }\left(n\right)$	$n$	$K^{\prime }\left(n\right)$	$n$	$K^{\prime }\left(n\right)$
1	0	11	21	21	8496	31	5778267
2	1	12	48	22	16431	32	11201884
3	1	13	63	23	30735	33	21702708
4	2	14	133	24	59344	34	42141576
5	1	15	205	25	112531	35	81830748
6	4	16	412	26	217246	36	159140896
7	3	17	685	27	415628	37	309590883
8	8	18	1354	28	803210	38	602938099
9	8	19	2385	29	1545463	39	1174779397
10	18	20	4644	30	2991192	40	2290920128

Equations

$$N(n,c)=\frac{1}{n}\sum _{d|n}\phi (d)c^{n/d},$$

$$N^{\prime }(n,c)=\frac{N(n,c)+R(n,c)}{2},$$

$$R(n,c)=\{\begin{array}{ll}{c}^{(n+1)/2}& \text{if n is odd},\\ \frac{1+c}{2}{c}^{n/2}& \text{if n is even}.\end{array}$$

$$⟨s⟩=\{id,s,s^2,\dots ,s^{n-1}\},$$

$$⟨s,r⟩=\{id,s,s^2,\dots ,s^{n-1},r,rs,r{s}^2,\dots ,r{s}^{n-1}\},$$

$$f{\sim }_{G}g⟺\text{mbox}thereexists\gamma \in G\text{mbox}suchthat\gamma f=g.$$

$$\left|X/G\right|=\frac{1}{\left|G\right|}\sum _{\gamma \in G}\left|\mathfrak{F}(\gamma )\right|,$$

$${\mathfrak{f}}^{(n)}(\sigma ,\epsilon ,j)=\left|{\mathfrak{F}}^{(n)}(\sigma ,\epsilon ,j)\right|$$

$${\mathfrak{F}}^{(n)}(\sigma ,\epsilon ,j)=\left\{f\in {\mathcal{A}}_n:f=\sigma \circ f\circ r^{\epsilon }\circ s^j\right\}.$$

$${\alpha }_n=\left|{\mathcal{A}}_n\right|=2^n+2(-1{)}^n.$$

$${\mathcal{A}}_n^{\prime }$$

$${\mathcal{A}}_n^{\prime \prime }$$

$${\alpha }_n=\left|{\mathcal{A}}_n\right|=\left|{\mathcal{A}}_n^{\prime }\right|+\left|{\mathcal{A}}_n^{\prime \prime }\right|={\alpha }_{n-1}+2{\alpha }_{n-2},$$

$${\mathfrak{f}}^{(n)}(id,0,0)={\alpha }_n.$$

$$\left|{\mathcal{A}}_n^{i\cdot \cdot j}\right|=\frac{{\alpha }_n}{6}.$$

$$\left|{\mathcal{A}}_n^{i\cdot \cdot j}\right|=\left|{\mathcal{A}}_n^{1\cdot \cdot 2}\right|.$$

$${\mathfrak{f}}^{(n)}({\tau }_{12},\epsilon ,\ell )={\mathfrak{f}}^{(n)}({\tau }_{13},\epsilon ,\ell )={\mathfrak{f}}^{(n)}({\tau }_{23},\epsilon ,\ell ).$$

$${\mathfrak{f}}^{(n)}(c,\epsilon ,\ell )={\mathfrak{f}}^{(n)}(c^2,\epsilon ,\ell )$$

$${\mathfrak{f}}^{(n)}(\sigma ,1,2\ell )={\mathfrak{f}}^{(n)}(\sigma ,1,0)$$

$${\mathfrak{f}}^{(n)}(\sigma ,1,2\ell +1)={\mathfrak{f}}^{(n)}(\sigma ,1,1)$$

$$\sum _{d|n}(-1{)}^d\phi \left(\frac{n}{d}\right)=\{\begin{array}{ll}0& \text{if n is even,}\\ -n& \text{if n is odd.}\end{array}$$

$$\sum _{d|n}(-1{)}^d\phi \left(\frac{n}{d}\right)=-\sum _{d|n}\phi \left(\frac{n}{d}\right)=-n.$$

$$\sum _{d|n}(-1{)}^d\phi \left(\frac{n}{d}\right)$$

$$=2\sum _{\genfrac{}{}{0ex}{}{d|n}{d\text{ is even}}}\phi \left(\frac{n}{d}\right)-\sum _{d|n}\phi \left(\frac{n}{d}\right)$$

$$=2\sum _{d^{\prime }|m}\phi \left(\frac{m}{d^{\prime }}\right)-\sum _{d|n}\phi \left(\frac{n}{d}\right)$$

$$=2m-n=0,$$

$${\mathfrak{f}}^{(n)}(id,0,\ell )={\alpha }_{\gcd (\ell ,n)}.$$

$${\mathfrak{f}}^{(n)}(c,0,\ell )=0.$$

$${\mathfrak{f}}^{(n)}(c,0,\ell )=\{\begin{array}{ll}0& \text{if 3∣(ℓ/d),}\\ 2^d-(-1{)}^d& \text{if 3∤(ℓ/d),}\end{array}$$

$${\mathfrak{f}}^{(n)}({\tau }_{12},0,\ell )=0.$$

$${\mathfrak{f}}^{(n)}({\tau }_{12},0,\ell )=\{\begin{array}{ll}0& \text{if 2∣(ℓ/d),}\\ 2^d& \text{if 2∤(ℓ/d),}\end{array}$$

$${\mathfrak{F}}^{(n)}((id,0,\ell ))\subset {\mathcal{A}}_n\cap {\mathcal{A}}_{\ell }={\mathcal{A}}_d.$$

Full text

This is a pre-print of an article published in Aequationes Mathematicæ. The final authenticated version is available online at: https://doi.org/10.1007/s00010-019-00645-w.

1. Introduction

A necklace with $n$ beads and $c$ colors is an $n$-tuple, each of whose components can assume one of $c$ values, where not all $c$ colors need appear. Two necklaces are equivalent if one results from the other by either rotating it cyclically or permuting its colors. The classical necklace problem asks to determine the number of non-equivalent necklaces formed with $n$ beads of $c$ colors. The answer to this problem is given by

[TABLE]

where $\varphi$ is the Euler totient function.

A bracelet with $n$ beads and $c$ colors is a necklace of $n$ beads and $c$ colors that can be turned over, and thus the order of its beads is reversed. The number of non-equivalent bracelets with $n$ beads of $c$ colors is given by

[TABLE]

where

[TABLE]

As expected, $N^{\prime}(n,c)\leq N(n,c)$. These results and further details are given in [9, 10] and the references therein.

In this work we consider a variation on this problem that arose from considering sequences of $n$ coordinate-axis rotations defined by Euler angles, where $n=7$ for aircraft [1]. For this problem, it is of interest to count the number of distinct coordinate-axis rotation sequences of length $n$ that are closed in the sense of transforming the starting frame by a sequence of coordinate-axis rotations that lead back to the starting frame [2]. A pair of successive coordinate-axis rotations around the same axis can be combined into a single rotation, and the labeling of the axes of the starting frame is arbitrary. Counting the number of closed sequences consisting of $n$ coordinate-axis rotations is thus equivalent to counting necklaces in $3$ colors, where each color corresponds to an axis label. Furthermore, reversing a sequence of coordinate-axis rotations is equivalent to replacing each Euler angle in the sequence of coordinate-axis rotations by its negative. Hence, for the purpose of determining all feasible Euler angles for each closed sequence of coordinate-axis rotations, it suffices to count bracelets.

Motivated by the fact that successive coordinate-axis rotations around the same axis can be merged, the present paper considers colorful necklaces and bracelets formed with $n$-beads of three colors, where a necklace or bracelet is colorful if no pair of adjacent beads have the same color. Two colorful necklaces are equivalent if one results from the other by permuting its colors, and two colorful bracelets are equivalent if one results from the other by either permuting its colors or reversing the order of the beads; a colorful bracelet is thus a colorful necklace that can be turned over. In fact, the number of colorful bracelets with $n$ beads in three colors appears in the On-line Encyclopedia of Integer Sequences (OEIS) as sequence A114438, which is the “Number of Barlow packings that repeat after $n$ (or a divisor of $n$) layers.” The provided references indicate that the problem of studying this sequence originates in crystallography [4, 5, 8].

The contribution of the present paper is twofold. First, we provide explicit formulas for the number of color bracelets that simplify those given by $P^{\prime}(n)$ in [4, p. 272]. Furthermore, we provide expressions for the number of colorful necklaces with $n$ beads in three colors; this sequence is currently unknown to OEIS.

2. n-Periodic Sequences

Instead of working with necklaces and bracelets of $n$ beads we work with $n$-periodic sequences. To fix notation, let ${\rm Nature}_{q}$ denote the set of natural numbers from $1$ to $q$. In particular, ${\rm Nature}_{3}$ represents the set of the three colors under consideration.

Definition 2.1**.**

For a positive integer $n$, a colorful $n$-periodic sequence is a function $f:\mathbb{Z}/n\mathbb{Z}\to{\rm Nature}_{3}$ defined on the set of integers modulo $n$, such that $f(i)\neq f(i+1)$ for all $i\in\mathbb{Z}/n\mathbb{Z}$. The set of colorful $n$-periodic sequences is denoted by $\mathcal{A}_{n}$.

The set $\mathcal{A}_{n}$ represents the colorful $n$-bead necklaces or bracelets of three colors.

Next, we consider the permutations $r$ and $s$ on $\mathbb{Z}/n\mathbb{Z}$ defined by $s(i)=i+1$ and $r(i)=-i$ for all $i\in\mathbb{Z}/n\mathbb{Z}$. The group generated by $s$ is

[TABLE]

which is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. The group generated by $r$ is $\langle r\rangle=\{id,r\}\simeq\mathbb{Z}/2\mathbb{Z}$. Furthermore, since $rsr=s^{-1}$, the group generated by $r$ and $s$ is

[TABLE]

which is isomorphic to the dihedral group $D_{n}$ of order $2n$.

We also consider the symmetric group $\mathfrak{S}_{3}$ of ${\rm Nature}_{3}$, where $\mathfrak{S}_{3}$ consists of the identity $id$, the three substitutions $\tau_{12},$ $\tau_{13},$ and $\tau_{23}$ (with $\tau_{ij}$ representing the $2$-cycle $(i,j)$), and the two $3$-cycles $c=(1,2,3)$ and $c^{2}=c^{-1}=(1,3,2)$. We recall that two permutations $\sigma$ and $\sigma^{\prime}$ are conjugates if there exists a permutation $\tau$ such that $\sigma^{\prime}=\tau^{-1}\circ\sigma\circ\tau$; this is equivalent to the fact that $\sigma$ and $\sigma^{\prime}$ have the same cyclic decomposition. Hence, for $\mathfrak{S}_{3}$ all transpositions are conjugates and all $3$-cycles are conjugates, which can be checked directly.

The group acting on the elements of $\mathcal{A}_{n}$, considered as colorful $n$-bead necklaces, is $G=\mathfrak{S}_{3}\times\langle s\rangle$, with group action defined by $\gamma f=\sigma\circ f\circ t^{-1}$ for $\gamma=(\sigma,t)\in G$ and $f\in\mathcal{A}_{n}$. Similarly, the group acting on the elements of $\mathcal{A}_{n}$, considered as colorful $n$-bead bracelets, is $G^{\prime}=\mathfrak{S}_{3}\times\langle s,r\rangle$, with group action defined again by $\gamma f=\sigma\circ f\circ t^{-1}$ for $\gamma=(\sigma,t)\in G^{\prime}$ and $f\in\mathcal{A}_{n}$.

The action of $G$ on $\mathcal{A}_{n}$ defines an equivalence relation $\sim_{G}$ on colorful $n$-bead necklaces given by

[TABLE]

The set of equivalence classes of this relation is denoted by $\mathcal{A}_{n}/G$, and the desired number of non-equivalent colorful $n$-bead necklaces is exactly $K(n)=\left|\mathcal{A}_{n}/G\right|$. Similarly, $K^{\prime}(n)=\left|\mathcal{A}_{n}/G^{\prime}\right|$, where $\mathcal{A}_{n}/G^{\prime}$ is the set of equivalence classes of colorful $n$-bead bracelets defined by the action of $G^{\prime}$ on $\mathcal{A}_{n}$.

The basic tool in this investigation is the classical Burnside’s Lemma [7, Theorem 3.22]. (While this lemma bears the name of Burnside, it seems that it was well-known to Frobenius (1887) and before him to Cauchy (1845). An account on the history of this lemma can be found in [9], see also [11]).

Theorem 2.2** (Burnside’s Lemma).**

Let $G$ be a finite group acting on a finite set $X$. Then

[TABLE]

where $\mathfrak{F}(\gamma)$ is the set of elements $x\in X$ fixed by $\gamma$ (i.e. $\gamma x=x$.)

Noting that $G$ is a subgroup of $G^{\prime}$, our task is to determine the numbers

[TABLE]

with $\sigma\in\mathfrak{S}_{3}$, $\varepsilon\in\{0,1\},$ and $j\in\{0,\ldots,n-1\}$, where

[TABLE]

This paper is organized as follows: In Section 3, we gather some useful properties and lemmas. In Section 4 the case of necklaces is considered. Finally, in Section 5 we consider the case of bracelets.

3. Useful Properties and Lemmas

Lemma 3.1**.**

If $n$ and $m$ are positive integers, then $\mathcal{A}_{n}\cap\mathcal{A}_{m}=\mathcal{A}_{\gcd(n,m)}.$

Proof.

This result follows from the fact that $n\mathbb{Z}+m\mathbb{Z}=\gcd(n,m)\mathbb{Z}$. ∎

Our first step is to determine the number of $n$-periodic colorful sequences, that is $\alpha_{n}{\,{\buildrel\rm def\over{=}}\,}\left|\mathcal{A}_{n}\right|$. This is the object of the next proposition.

Proposition 3.2**.**

For all $n\geq 1$,

[TABLE]

Proof.

Note that $\alpha_{1}=0$ and $\alpha_{2}=6$. Suppose that $n\geq 3$, and define

[TABLE]

The mapping $\mathcal{A}_{n}^{\prime}\to\mathcal{A}_{n-1}:f\mapsto\tilde{f}$ with $\tilde{f}$ defined by $\tilde{f}_{|{\rm Nature}_{n-1}}=f_{|{\rm Nature}_{n-1}}$ is bijective because $f(n)$ is uniquely defined by the knowledge of $f(n-1)$ and $f(1)$, indeed $\{f(1),f(n-1),f(n)\}={\rm Nature}_{3}$. Hence $\left|\mathcal{A}_{n}^{\prime}\right|=\left|\mathcal{A}_{n-1}\right|=\alpha_{n-1}$.

Also, the mapping $\mathcal{A}_{n}^{\prime\prime}\to\mathcal{A}_{n-2}:f\mapsto\hat{f}$ with $\hat{f}$ defined by $\hat{f}_{|{\rm Nature}_{n-2}}=f_{|{\rm Nature}_{n-2}}$ is surjective and the pre-image of each $g\in\mathcal{A}_{n-2}$ consists of exactly two elements, namely $f_{1}$ and $f_{2}$ defined by $f_{1|{\rm Nature}_{n-2}}=f_{2|{\rm Nature}_{n-2}}=g_{|{\rm Nature}_{n-2}}$, $f_{1}(n)=\min({\rm Nature}_{3}\setminus\{f(1)\})$ and $f_{2}(n)=\max({\rm Nature}_{3}\setminus\{f(1)\})$. Hence $\left|\mathcal{A}_{n}^{\prime\prime}\right|=2\left|\mathcal{A}_{n-2}\right|=2\alpha_{n-2}$. But $\left\{\mathcal{A}_{n}^{\prime},\mathcal{A}_{n}^{\prime\prime}\right\}$ is a partition of $\mathcal{A}_{n}$, so

[TABLE]

and the desired conclusion follows by induction. ∎

In particular, since the neutral element of $G$ (or $G^{\prime}$) fixes the whole set $\mathcal{A}_{n}$, the next corollary is immediate.

Corollary 3.3**.**

[TABLE]

Corollary 3.4**.**

For distinct $i,j\in{\rm Nature}_{3}$, let $\mathcal{A}_{n}^{i\cdot\cdot j}$ denote the subset of $\mathcal{A}_{n}$ consisting of functions $f$ satisfying $f(1)=i$ and $f(n)=j$. Then

[TABLE]

Proof.

Given $i$ and $j$, there is a unique permutation $\sigma\in\mathfrak{S}_{3}$ such that $\sigma(i)=1$ and $\sigma(j)=2$, and with this $\sigma$ the mapping $f\mapsto\sigma\circ f$ defines a bijection between $\mathcal{A}_{n}^{i\cdot\cdot j}$ and $\mathcal{A}_{n}^{1\cdot\cdot 2}$. Thus

[TABLE]

The conclusion follows since $\{\mathcal{A}_{n}^{1\cdot\cdot 2},\mathcal{A}_{n}^{1\cdot\cdot 3},\mathcal{A}_{n}^{2\cdot\cdot 1},\mathcal{A}_{n}^{2\cdot\cdot 3},\mathcal{A}_{n}^{3\cdot\cdot 1},\mathcal{A}_{n}^{3\cdot\cdot 2}\}$ constitutes a partition of $\mathcal{A}_{n}$. ∎

The next lemma helps to reduce the number of cases to be considered. The proof is immediate and left to the reader.

Lemma 3.5** (Reduction).**

Suppose that a group $G$ acts on a set $X$, and consider two elements $g$ and $g^{\prime}$ from $G$. If there is $h\in G$ such that $g^{\prime}=h^{-1}gh$, then the mapping $x\mapsto hx$ defines a bijection from $\mathfrak{F}(g^{\prime})$ onto $\mathfrak{F}(g)$. In particular, if $X$ and $G$ are finite and if $g$ and $g^{\prime}$ are conjugate elements from $G$ then $\left|\mathfrak{F}(g)\right|=\left|\mathfrak{F}(g^{\prime})\right|$.

Remark 3.6**.**

Another simple remark from group theory is that if $G=A\times B$ is the direct product of two groups $A$ and $B$, and if $a$ and $a^{\prime}$ are conjugate elements from $A$, then $(a,e_{B})$ and $(a^{\prime},e_{B})$, (with $e_{B}$ denoting the neutral element of $B$), are also conjugate elements in $G$.

With Lemma 3.5 and Remark 3.6 at hand, the next corollary is immediate:

Corollary 3.7**.**

**

1. (a)

For all $\varepsilon\in\{0,1\}$ and all $\ell\in\{0,1,\ldots,n-1\}$ we have

[TABLE] 2. (b)

For all $\varepsilon\in\{0,1\}$ and all $\ell\in\{0,1,\ldots,n-1\}$ we have

[TABLE] 3. (c)

For all $\sigma\in\mathfrak{S}_{3}$ and all $\ell\in\{0,1,\ldots,\left\lfloor{(n-1)/2}\right\rfloor\}$ we have

[TABLE] 4. (d)

For all $\sigma\in\mathfrak{S}_{3}$ and all $\ell\in\{0,1,\ldots,\left\lfloor{n/2-1}\right\rfloor\}$ we have

[TABLE]

Proof.

Both (a) and (b) follow from the fact that all permutations of the same cycle structure are conjugate. On the other hand, since $rs^{2\ell}=s^{-\ell}rs^{\ell}$ and $rs^{2\ell+1}=s^{-\ell}(rs)s^{\ell}$ for all $\ell$, both (c) and (d) follow from Corollary 3.7. ∎

The final result in this preliminary section is a simple formula concerning sums involving Euler’s totient function $\varphi$ (see [3, Chapter V, Section 5.5]), recall that $\varphi(n)$ is the number of integers in ${\rm Nature}_{n}$ coprime to $n$.

Lemma 3.8**.**

For every positive integer $n$ we have

[TABLE]

Proof.

If $n$ is odd then all its divisors are odd and using [3, Theorem 63], we get

[TABLE]

Now, if $n=2m$ for some positive integer $m$, then

[TABLE]

where we used again [3, Theorem 63]. ∎

4. Counting Colorful Necklaces

In this section we consider $G=\mathfrak{S}_{3}\times\langle s\rangle$. According to Corollary 3.7 we need to determine $\mathfrak{f}^{(n)}(\sigma,0,\ell)$, for $\sigma\in\{id,\tau_{12},c\}$ and $\ell\in\mathbb{Z}/n\mathbb{Z}$. The next proposition gives the answer.

Proposition 4.1**.**

**

1. (a)

If $\gcd(\ell,n)=d$ then $\mathfrak{F}^{(n)}(id,0,\ell)=\mathfrak{F}^{(n)}(id,0,d)=\mathcal{A}_{d}$. In particular,

[TABLE]

Thus, $\gcd(n,\ell)=1$ implies $\mathfrak{F}^{(n)}(id,0,\ell)=\emptyset$. 2. (b)

1. i.

Suppose that $3\nmid n$, then for all $\ell\in\mathbb{Z}/n\mathbb{Z}$ we have

[TABLE] 2. ii.

Suppose that $n=3m$ for some positive integer $m$, then for all $\ell\in\mathbb{Z}/n\mathbb{Z}$ we have

[TABLE]

where $d=\gcd(m,\ell)$. 3. (c)

1. i.

Suppose that $2\nmid n$, then for all $\ell\in\mathbb{Z}/n\mathbb{Z}$ we have

[TABLE] 2. ii.

Suppose that $n=2m$ for some positive integer $m$, then for all $\ell\in\mathbb{Z}/n\mathbb{Z}$ we have

[TABLE]

where $d=\gcd(m,\ell)$.

Proof.

(a) A sequence $f\in\mathfrak{F}^{(n)}((id,0,\ell))$ satisfies $f\circ s^{\ell}=f$, so it belongs to $\mathcal{A}_{\ell}$. Thus, by Lemma 3.1, we have

[TABLE]

The converse inclusion: $\mathcal{A}_{d}\subset\mathfrak{F}^{(n)}((id,0,\ell))$ is trivial, because both $\ell$ and $n$ are multiples of $d$.

(b,c) i. Let $\sigma$ be any permutation from $\mathfrak{S}_{3}$, and suppose that $\mathfrak{F}^{(n)}(\sigma,0,\ell)\neq\emptyset$ so there is $f\in\mathfrak{F}^{(n)}(\sigma,0,\ell)$. From

[TABLE]

we conclude by an easy induction that for all integers $p$ we have

[TABLE]

• •

If $\sigma=c$ and $3\nmid n$ we have $c^{3}=id$, so (4.7) implies that $f\circ s^{3p\ell}=f$ for all integers $p$. But, because $3\nmid n$ there is $r\in\{1,2\}$ such that $n-r=3p$ for some $p$. Consequently, $f\circ s^{(n-r)\ell}=f$, or equivalently $f=f\circ s^{r\ell}=c^{-r}\circ f$ because $f$ is $n$-periodic. This is a contradiction because neither $c$ nor $c^{2}$ has fixed points. Thus $\mathfrak{F}^{(n)}(c,0,\ell)=\emptyset$. This proves (b) $i$.

• •

If $\sigma=\tau_{12}$ and $n$ is odd, we have $\tau_{12}^{2}=id$, so (4.7) implies that $f\circ s^{2p\ell}=f$ for all integers $p$. But, because $n=2p+1$ for some $p$ we conclude that $f\circ s^{(n-1)\ell}=f$, or equivalently $f=f\circ s^{\ell}=\tau_{12}\circ f$. This is a contradiction because $f$ takes two different values, and $\tau_{12}$ has only one fixed point. Thus $\mathfrak{F}^{(n)}(\tau_{12},0,\ell)=\emptyset$. This proves (c) $i$.

(b) $ii$. Assume that $\mathfrak{F}^{(n)}(c,0,\ell)\neq\emptyset$ and consider $f\in\mathfrak{F}^{(n)}(c,0,\ell)$. From $c\circ f\circ s^{\ell}=f$ we conclude that $f\circ s^{3\ell}=f$. Thus $f\in\mathcal{A}_{3j}\cap\mathcal{A}_{3m}=\mathcal{A}_{3d}$, with $d=\gcd(m,\ell)$. Further, if $\ell/d=3q+r$ with $r\in\{0,1,2\}$ then $\ell=3dq+dr$ and consequently

[TABLE]

• •

If $r=0$ then (4.8) implies $f=c^{2}\circ f$ which is impossible since $f$ is not constant. Thus $\mathfrak{F}^{(n)}(c,0,\ell)=\emptyset$ in this case.

• •

If $r=1$ then (4.8) shows that $f\in\mathfrak{F}^{(3d)}(c,0,d)$. Conversely, it is easy to check that any $f\in\mathfrak{F}^{(3d)}(c,0,d)$ belongs to $\mathfrak{F}^{(n)}(c,0,\ell)$. Thus, we have shown that

[TABLE]

Now, when $f$ belongs to $\mathfrak{F}^{(3d)}(c,0,d)$ it is completely determined by its restriction to ${\rm Nature}_{d}$, and the mapping (see Figure 1):

[TABLE]

where $\tilde{f}$ is the unique sequence from $\mathfrak{F}^{(3d)}(c,0,d)$ which coincides with $f$ on ${\rm Nature}_{d}$, is a bijection.

We conclude, according to Corollary 3.4 that

[TABLE]

• •

If $r=2$, then a similar argument to the previous one (with $c$ replaced by $c^{2}$), yields the desired conclusion. This completes the proof of (b) $ii$.

(c) $ii$. For simplicity we write $\tau$ for $\tau_{12}$. Suppose that $\mathfrak{F}^{(n)}(\tau,0,\ell)\neq\emptyset$ and consider $f\in\mathfrak{F}^{(n)}(\tau,0,\ell)$. We have

[TABLE]

Hence $f\circ s^{2\ell}=f$ and consequently $f\in\mathcal{A}_{2\ell}$, this implies that $f\in\mathcal{A}_{2m}\cap\mathcal{A}_{2\ell}=\mathcal{A}_{2d}$ where $d=\gcd(m,\ell)$. Now write $\ell/d=2q+r$ with $r\in\{0,1\}$, then $\ell=2dq+dr$ and consequently

[TABLE]

• •

If $r=0$, then (4.11) implies $f=\tau\circ f$ which is impossible since $f$ is not constant. Thus $\mathfrak{F}^{(n)}(\tau,0,\ell)=\emptyset$ in this case.

• •

If $r=1$, then (4.11) implies $f\circ s^{d}=\tau\circ f$, that is $f\in\mathfrak{F}^{(2d)}(\tau,0,d)$. Conversely, it is easy to check that any $f\in\mathfrak{F}^{(2d)}(\tau,0,d)$ belongs to $\mathfrak{F}^{(n)}(\tau,0,\ell)$. Thus, we have shown that in this case

[TABLE]

Clearly, if $d=1$ then $\mathfrak{F}^{(2d)}(\tau,0,d)$ consists exactly of two elements: namely $f_{1}$, defined by $f_{1}(1)=1,f_{1}(2)=2$, and $f_{2}=\tau\circ f_{1}$. So,

[TABLE]

Now suppose that $d>1$. Any $f\in\mathfrak{F}^{(2d)}((\tau,d))$ is completely determined its restriction to ${\rm Nature}_{d}$ (note that $f(d)$ should be different from $\tau(f(1))$ and $f(d-1)$,) so considering the different possibilities for $f(1)$ we see that the mapping $\Psi$, (see Figure 2):

[TABLE]

where $\hat{f}$ is the unique sequence from $\mathfrak{F}^{(2d)}(\tau,0,d)$ that coincides with $f$ on ${\rm Nature}_{d}$, is a bijection.

Thus, according to Corollary 3.4, we have

[TABLE]

This concludes the proof of (c) $ii$, in view of (4.12). ∎

The final step is to put all the pieces together to get the expression of $K_{n}$ in terms of $n$ using Burnside’s Lemma.

Theorem 4.2**.**

The number of non-equivalent colorful $n$-bead necklaces with three colors is given by

[TABLE]

where $\mathbb{I}_{X}$ is the indicator function of the set $X$, ( i.e. $\mathbb{I}_{X}(k)=1$ if $k\in X$ and $\mathbb{I}_{X}(k)=0$ if $k\notin X$,) and $3^{\nu_{3}(n)}$ is the largest power of $3$ dividing $n$.

Proof.

According to Corollary 3.7 and Burnside’s Lemma 2.2 we have

[TABLE]

with

[TABLE]

Using part (a) of Proposition 4.1 we have

[TABLE]

Thus, using the expression of $\alpha_{n}$ from Proposition 3.2, we get

[TABLE]

Similarly, according part (c) of Proposition 4.1 we know that $B_{n}=0$ if $n$ is odd, while we have the following when $n=2m$:

[TABLE]

Finally we get

[TABLE]

Now, we come to $C_{n}$. According to part (b) of Proposition 4.1 we know that $C_{n}=0$ if $n$ is not a multiple of $3$ while if $n=3m$ we have

[TABLE]

Thus,

[TABLE]

Replacing (4.19), (4.20) and (4.21) in (4.15) we get

[TABLE]

with

[TABLE]

and

[TABLE]

In order to reduce a little bit the expression of $K(n)n$ we use Lemma 3.8. Indeed, Suppose that $n=3^{\nu}m$ where $\nu=\nu_{3}(n)$ is the exponent of $3$ in the prime factorization of $n$, thus $3\nmid m$. Clearly if $\nu=0$ then using Lemma 3.8 we get

[TABLE]

Now if $\nu>0$ then

[TABLE]

Finally, noting that $n=m\mod 2$, we obtain the following formula for $\varepsilon_{n}$ which is also valid when $\nu=0$ according to (4.25):

[TABLE]

Now note that $b_{n}$ can be written ans follows

[TABLE]

with

[TABLE]

where

[TABLE]

equivalently

[TABLE]

Thus

[TABLE]

So, we may write $b_{n}$ in the following form

[TABLE]

with $\chi:\mathbb{Z}\to{\rm Nature}_{6}$ defined by

[TABLE]

This can also be written in the form $\chi(k)=(1+\mathbb{I}_{2\mathbb{Z}\setminus 3\mathbb{Z}}(k))\gcd(k,6)$, and the announced expression (4.13) for $K(n)$ is obtained. Finally, the formula $K(n)=\left\lfloor{b_{n}}\right\rfloor$, follows from the fact that. $-\frac{1}{3}\leq\varepsilon_{n}\leq 0$. ∎

We conclude our discussion of the case of necklaces by noting that there are some simple cases where the formula for $K(n)$ is particularly appealing, for example, if $n=p>3$ is prime, then

[TABLE]

and if $\gcd(n,6)=1$, then

[TABLE]

Remark 4.3**.**

If $6$ and $n$ are coprime, then $K(n)$ is related to the number $N(n,2)$ of $n$-bead necklaces of two colors (1.1) by the formula

[TABLE]

Remark 4.4**.**

An equivalent formula for $K(n)$ that does not use the indicator function of the set $2\mathbb{Z}\setminus 3\mathbb{Z}$ is the following

[TABLE]

Table 1 lists the first 40 terms of the sequence $(K(n))_{n\geq 1}$.

5. Counting Colorful Bracelets

As we explained before, bracelets are turnover necklaces. It is the action of the group $G^{\prime}=\mathfrak{S}_{3}\times\langle r,s\rangle$ on the set of $n$-periodic colorful sequences $\mathcal{A}_{n}$ that is considered.

We are interested in the number of orbits $\mathcal{A}_{n}/G^{\prime}$ denoted by $K^{\prime}(n)$. Again Burnside’s Lemma comes to our rescue. We need to determine the numbers $\mathfrak{f}^{(n)}(\sigma,\varepsilon,\ell)$ with $\sigma\in\mathfrak{S}_{3}$, $\varepsilon\in\{0,1\}$ and $\ell\in\mathbb{Z}/n\mathbb{Z}$, but we have already done this in the case $\varepsilon=0$ in the previous section.

Further, based Corollary 3.7, we only need to determine $\mathfrak{f}^{(n)}(\sigma,1,0)$ and $\mathfrak{f}^{(n)}(\sigma,1,1)$ for $\sigma$ in $\{id,\tau_{12},c\}$. This is the object of the next proposition.

Proposition 5.1**.**

**

1. (a)

2. i.

If $n$ is odd then $\mathfrak{f}^{(n)}(id,1,0)=0$, otherwise $\mathfrak{f}^{(n)}(id,1,0)=3\times 2^{n/2}$. 2. ii.

$\mathfrak{f}^{(n)}(id,1,1)=0$. 2. (b)

1. i.

$\mathfrak{f}^{(n)}(\tau_{12},1,0)=\alpha_{\left\lfloor{(n+1)/2}\right\rfloor}/3.$** 2. ii.

$\mathfrak{f}^{(n)}(\tau_{12},1,1)=\alpha_{\left\lfloor{n/2+1}\right\rfloor}/3.$** 3. (c)

$\mathfrak{f}^{(n)}(c,1,0)=\mathfrak{f}^{(n)}(c,1,1)=0.$**

Proof.

(a) Suppose that $\mathfrak{F}^{(n)}(id,1,0)\neq\emptyset$ and consider $f\in\mathfrak{F}^{(n)}(id,1,0)$. Write $n=2m+t$ with $t\in\{0,1\}$. Because $f(k)=f(-k)=f(n-k)$ for every $k$, we conclude by considering $k=m$ that $f(m+t)=f(m)$. But $f(m)\neq f(m+1)$ so we must have $t=0$ and $n=2m$. Now, from the fact that $f(2m-k)=f(k)$ for every $k$ we conclude that

[TABLE]

So, the mapping

[TABLE]

defines a bijection between $\mathfrak{F}^{(n)}(id,1,0)$ and the set

[TABLE]

Now, $x_{0}$ may take any one of three possible values and each other $x_{i}$ has two possible values. So, the cardinality of this set is $3\times 2^{m}$. Thus (a) i. is proved.

Now suppose that $\mathfrak{F}^{(n)}(id,1,1)\neq\emptyset$ and consider $f$ from ${\mathfrak{F}^{(n)}(id,1,1)}$. We have $f(-k-1)=f(k)$ for every $k$, in particular, for $k=0$ we get $f(-1)=f(0)$ which is absurd, and (a) ii. follows.

(b) i. we write $\tau$ for $\tau_{12}$. Suppose that $\mathfrak{F}^{(n)}(\tau,1,0)\neq\emptyset$ and consider $f\in\mathfrak{F}^{(n)}(\tau,1,0)$. We have

[TABLE]

Taking $k=0$ we get $f(0)=\tau(f(0))$, and this implies that $f(0)=3$.

• •

If $n=2m$ then $f(m)=f(m-n)=f(-m)=\tau(f(m))$ and consequently $f(m)=3$. The restriction of $f$ to the period $\{-m+1,\ldots,m-1,m\}$ has the form

[TABLE]

So, $f$ is completely determined by the knowledge of $(f(1),\ldots,f(m-1))$ and consequently there is a bijection between $\mathfrak{F}^{(2m)}(\tau,1,0)$ and $\mathcal{A}_{m}^{3\cdot\cdot 1}\cup\mathcal{A}_{m}^{3\cdot\cdot 2}$. Thus, by Corollary 3.4, we have

[TABLE]

• •

If $n=2m+1$, then $f(m+1)=f(m+1-n)=f(-m)=\tau(f(m))$ and consequently $f(m)\neq 3$. The restriction of $f$ to the period $\{-m,\ldots,m\}$ takes the form

[TABLE]

So, $f$ is completely determined by the knowledge of $(f(1),\ldots,f(m))$ and consequently there is a bijection between $\mathfrak{F}^{(2m+1)}(\tau,1,0)$ and $\mathcal{A}_{m+1}^{3\cdot\cdot 1}\cup\mathcal{A}_{m+1}^{3\cdot\cdot 2}$. Thus

[TABLE]

(b) ii. Now suppose that $\mathfrak{F}^{(n)}(\tau,1,1)\neq\emptyset$ and consider $f\in\mathfrak{F}^{(n)}(\tau,1,1)$. We have

[TABLE]

Taking $k=0$ we get $f(-1)=\tau(f(0))$, but $f(-1)\neq f(0)$ thus $f(0)\in\{1,2\}$.

• •

If $n=2m$, then $f(m-1)=f(m-1-n)=f(-m-1)=\tau(f(m))$ but $f(m-1)\neq f(m)$ thus $f(m-1)\in\{1,2\}$. The restriction of $f$ to the period $\{-m,\ldots,m-1\}$ takes the form

[TABLE]

So, $f$ is completely determined by the knowledge of $(f(0),\ldots,f(m-1))$. We can partition the set $\mathfrak{F}^{(2m)}(\tau,1,1)$ according to the values taken by ${(f(0),f(m-1))}$, and we have obvious bijective mappings:

[TABLE]

Thus

[TABLE]

• •

If $n=2m+1$ then $f(m)=f(m-n)=f(-m-1)=\tau(f(m))$ and consequently $f(m)=3$. The restriction of $f$ to the set $\{-m,\ldots,m\}$ takes the form

[TABLE]

So, $f$ is completely determined by the knowledge of $(f(0),\ldots,f(m-1))$ and there is an obvious bijective mapping between $\mathfrak{F}^{(2m+1)}(\tau,1,1)$ and $\mathcal{A}_{m+1}^{1\cdot\cdot 3}\cup\mathcal{A}_{m+1}^{2\cdot\cdot 3}$. Thus

[TABLE]

This concludes the proof of part (b).

(c) First, suppose that $\mathfrak{F}^{(n)}(c,1,0)\neq\emptyset$ and consider $f\in\mathfrak{F}^{(n)}(c,1,0)$. We have $f(k)=c(f(-k))$ for all $k\in\mathbb{Z}$. In particular, $f(0)=c(f(0))$ which is absurd because $c$ has no fixed points.

Next suppose that $\mathfrak{F}^{(n)}(c,1,1)\neq\emptyset$ and consider $f\in\mathfrak{F}^{(n)}(c,1,1)$. We have $f(k)=c(f(-k-1))$ for all $k\in\mathbb{Z}$.

• •

If $n=2m+1$ then

[TABLE]

which is absurd because $c$ has no fixed points.

• •

If $n=2m$ then

[TABLE]

which is also absurd because $c^{2}$ has no fixed points.

This achieves the proof of the proposition. ∎

Finally we arrive to the main theorem of this section.

Theorem 5.2**.**

The number of non-equivalent colorful $n$-bead Bracelets with three colors is given by

[TABLE]

with

[TABLE]

where $K(n)$ is given by Theorem 4.2.

Proof.

We only need to put things together. We know that

[TABLE]

Thus $K^{\prime}(n)=(K(n)+R(n))/2$ with

[TABLE]

where we used Corollary 3.7. Now using Proposition 5.1 we get

[TABLE]

and

[TABLE]

Replacing in the expression of $R(n)$ we obtain

[TABLE]

and the announced result follows. ∎

Table 2 lists the first 40 terms of the sequence $(K^{\prime}(n))_{n}$.

Remark 5.3**.**

Although $K(n)\leq K^{\prime}(n)$ for all $n\geq 1,$ a surprising fact about $(K(n))_{n\geq 1}$ and $(K^{\prime}(n))_{n\geq 1}$ is that they coincide for the first $8$ values!.

Remark 5.4**.**

The equality $2K^{\prime}(n)=K(n)+R(n)$ and the easy-to-prove fact that $R(n)=n\mod 2$ for $n\geq 3$, allow us to find the parity pattern of the $K(n)$’s. The fact that $K(n)=n\mod 2$ for $n\geq 3$ seems difficult to prove directly.

6. Related Combinatorial Sequences

Colorful necklaces or bracelets with $n$ beads and two colors are easy to determine. There are none when $n$ is odd and just one equivalence class when $n$ is even. Thus the sequences $(K^{*}(n))_{n\geq 1}$ and $(K^{\prime*}(n))_{n\geq 1}$ defined by

[TABLE]

represent the number of non-equivalent colorful necklaces in $n$ beads with exactly $3$ colors and the number of non-equivalent colorful bracelets in $n$ beads with exactly $3$ colors, respectively. Both sequences $(K^{*}(n))_{n\geq 1}$, and $(K^{\prime*}(n))_{n\geq 1}$ are currently not recognized by the OEIS.

Further, if we are interested in periodic colorful sequences of exact period $n$ in at most $3$ colors then the number $\widetilde{K}(n)$ of non-equivalent such sequences assuming that reversing is not allowed is given by

[TABLE]

where $\mu$ is the well known Moebius function. Indeed, this follows from the classical result [6, Theorem 1.5], because clearly $K(n)=\sum_{d|n}\widetilde{K}(d)$.

OEIS recognizes $(\widetilde{K}(n))_{n\geq 1}$ as the “Number of Zn S polytypes that repeat after $n$ layers” A011957.

Similarly, if we are interested in periodic colorful sequences of exact period $n$ in at most $3$ colors then the number $\widetilde{K}^{\prime}(n)$ of non-equivalent such sequences assuming that reversing is allowed is given by

[TABLE]

OEIS recognizes $(\widetilde{K}^{\prime}(n))_{n\geq 1}$ as the “Number of Barlow packings that repeat after exactly $n$ layers” A011768.

7. Future Research

This paper has counted non-equivalent colorful necklaces and non-equivalent colorful bracelets in $n$ beads with $3$ colors. An open problem is to count non-equivalent colorful necklaces and colorful bracelets in $n$ beads with $c\geq 4$ colors.