Circularly squarefree words and unbordered conjugates: a new approach

TL;DR

This paper introduces a new approach using automatic sequences, logic, and decision procedures to analyze circularly squarefree words and unbordered conjugates, providing simpler proofs and new classifications.

Contribution

It offers a novel method for studying unbordered conjugates, completes the classification of binary words with maximum unbordered conjugates, and determines their expected number in random words.

Findings

Complete classification of binary words with maximum unbordered conjugates

Existence of words with any number of unbordered conjugates up to maximum

Calculated the expected number of unbordered conjugates in random words

Abstract

Using a new approach based on automatic sequences, logic, and a decision procedure, we reprove some old theorems about circularly squarefree words and unbordered conjugates in a new and simpler way. Furthermore, we prove three new results about unbordered conjugates: we complete the classification, due to Harju and Nowotka, of binary words with the maximum number of unbordered conjugates; we prove that for every possible number, up to the maximum, there exists a word having that number of unbordered conjugates, and finally, we determine the expected number of unbordered conjugates in a random word.

Figures

Equations

$$\mathbf{c}=210201\cdots ,$$

$$\phi (0)$$

$$\phi (1)$$

$$\phi (2)$$

$$\phi (3)$$

$$\mathrm{crep}(i,m,n,p,s)$$

$$(\forall j((j\ge i)\wedge (j<s+n)\wedge (j+p\ge s+n)\wedge (j+p<i+m))⟹C[j]=C[j+p-n])\wedge$$

$$(\forall j((j\ge i)\wedge (j\ge s+n)\wedge (j+p<i+m))⟹C[j-n]=C[j+p-n])$$

$$facge2(n,s)$$

$$\mathrm{circsf}(n)$$

$$w[j]=\{\begin{array}{ll}\mathbf{c}[j],& \text{if }j<s+n-3;\\ 0,& \text{if }j\in \{s+n-3,s+2n-3\};\\ 2,& \text{if }j\in \{s+n-2,s+2n-2\};\\ 1,& \text{if }j=s+n-1;\\ \mathbf{c}[j-n],& \text{if }s+n\le j<s+2n-3.\end{array}$$

$$\forall j((i\le j)\wedge (j<i+p))⟹w[j]=w[j+p].$$

$$\forall j((i\le j)\wedge (j<i+p)\wedge (w[j]=\alpha )\wedge (w[j+p]=\beta ))⟹\alpha =\beta$$

$$\begin{gathered} sqfree021(i,n,p,s):=(n>3)\wedge (\forall i,p((1\le p)\wedge (2p\le n)\wedge (s\le i)\wedge (i<s+n)) \\ ⟹\neg (sq021(i,n,p,s))). \end{gathered}$$

$$test021(n):=\exists ssqfree021(n,s)$$

$$test2120(n):=\exists ssqfree2120(n,s)$$

$$\mathrm{currie}(n):=test021(n)\vee test2120(n)$$

$$a_i=\{\begin{array}{ll}u,& \text{if }{\sigma }^i(w)\text{ is unbordered;}\\ b,& \text{if }{\sigma }^i(w)\text{ is bordered.}\end{array}$$

$$isBorderC1(k,l,m,n)$$

$$\wedge (\forall i(i<k+l-n)\Rightarrow T[m+i]=T[m+n-k+i])))$$

$$isBorderC2(k,l,m,n)$$

$$T[m+l+i]=T[m+l-k+i]))$$

$$isBorderC3(k,l,m,n)$$

$$=T[m+l+i])\wedge (\forall i(i<l)\Rightarrow T[m+i]=T[m+k+i])))$$

$$\mathrm{isBorder}(k,l,m,n)$$

$$\mathrm{isBordered}(l,m,n):=\exists i(2i\le n\wedge i\ge 1\wedge \mathrm{isBorder}(i,l,m,n)).$$

$$isAlternating0(l,m,n):=$$

$$\forall i(((i\ne l\wedge i<n-1)\Rightarrow (\mathrm{isBordered}(i,m,n)=\neg \mathrm{isBordered}(i+1,m,n))))\wedge$$

$$(((i\ne l)\wedge (i=n-1))\Rightarrow (\mathrm{isBordered}(n-1,m,n)=\neg \mathrm{isBordered}(0,m,n))).$$

$$\mathrm{hasMNUCO}(m,n):=\exists i(((i<n-1\wedge \mathrm{isBordered}(i,m,n)\wedge \mathrm{isBordered}(i+1,m,n))\vee$$

$$(i=n-1\wedge \mathrm{isBordered}(n-1,m,n)\wedge \mathrm{isBordered}(0,m,n)))\wedge isAlternating0(i,m,n)).$$

$$\mathrm{isAlternatingE}(e,l,m,n)$$

$$\neg \mathrm{isBordered}(i+1,m,n))))\wedge (((i\ne l)\wedge (i\ne e)\wedge (i=n-1))\Rightarrow$$

$$(\mathrm{isBordered}(n-1,m,n)\iff \neg \mathrm{isBordered}(0,m,n))))$$

$$\mathrm{hasMNUCE}(m,n)$$

$$((j=n-1\wedge \mathrm{isBordered}(n-1,m,n)\wedge \mathrm{isBordered}(0,m,n))\vee ((j<n-1)\wedge$$

$$\mathrm{isBordered}(j,m,n)\wedge \mathrm{isBordered}(j+1,m,n)))\wedge \mathrm{isAlternatingE}(i,j,m,n)))\vee$$

$$\mathrm{isAlternatingE}(n,n,m,n).$$

$$\forall n((n\ge 2)$$

$${\mathrm{mnuc}}_2(n)=\{\begin{array}{ll}1,& \text{if }n=1;\\ 2,& \text{if }n=2\text{ or }n=3;\\ n/2,& \text{if }n\in \{2^i,3\cdot 2^i:i\ge 1\};\\ n/2-1,& \text{if }n>3\text{ even and }n\notin \{2^i,3\cdot 2^i:i\ge 1\};\\ \lfloor n/2\rfloor ,& \text{if }n>3\text{ odd}.\end{array}$$

$$\begin{gathered} \forall n((n\ge 2)⟹(\exists i\mathrm{hasMNUCO}(i,2n+1))\wedge ((T[i]=0\wedge T[i+1]=0\wedge T[2n+i]=0) \\ \vee (T[i]=0\wedge T[2n-1+i]=0\wedge T[2n+i]=0))), \end{gathered}$$

Full text

1 Introduction

Throughout this paper, $\Sigma_{k}$ denotes the alphabet $\{0,1,\ldots,k-1\}$.

Two words are said to be conjugate if one is a cyclic shift of the other, as in the English words enlist and listen.

A word $w$ has a border $x$ if $x\not\in\{\epsilon,w\}$ and $x$ is both a prefix and suffix of $w$; the two occurrences of $x$ are allowed to overlap each other. For example, alfa is a border of alfalfa. A word $w$ is said to be bordered if it has a border, and otherwise, it is unbordered. It follows immediately from the Lyndon-Schützenberger theorem [8] that a word $w$ if bordered iff it has a border of length $\leq|w|/2$; then the two shorter borders cannot overlap each other. For example, alfalfa is also bordered by a.

A word $w$ is said to be a square if $w=xx$ for some nonempty word $x$. An example in French is the word couscous. A word is squarefree if no nonempty factor is a square. Let $\mu$ be the Thue-Morse morphism, defined by $\mu(0)=01$ and $\mu(1)=10$. The Thue-Morse word $\bf t={\tt 01101001}\cdots$ is the fixed point, starting with [math], of $\mu$. Thue [12, 13, 3] proved that there exist infinite squarefree words over a three-letter alphabet; also see [1]. A famous example of such a word can be obtained from the Thue-Morse word as follows: count the number of $1$’s between two consecutive [math]’s in $\bf t$. This gives the so-called ternary Thue-Morse word

[TABLE]

and is squarefree. An alternative description of $\bf c$ is as follows: it is the image, under $\tau$ of the fixed point of the morphism $\varphi$ defined below:

[TABLE]

A word $w$ is circularly squarefree if every one of its conjugates is squarefree. For example, outshout is squarefree, but not circularly squarefree. Clearly we have

Proposition 1**.**

A word is circularly squarefree iff all its conjugates are unbordered.

We now turn to a description of what we do in this paper. Using a complicated case-based argument, Currie [5] proved that there exist circularly squarefree ternary words of every length $n$, except for $\{5,7,9,10,14,17\}$. The first of our main results is a new proof of Currie’s theorem, based on the following result:

Theorem 2**.**

For all natural numbers $n>3$, except $5,7,9,10,14,17,21$, and $28$, there exists a factor $x=x(n)$ of the ternary Thue-Morse word $\bf c$ that is either

• (a)

of length $n-3$, and $x021$ is circularly squarefree;

• (b)

of length $n-4$, and $x2120$ is circularly squarefree.

We now turn to unbordered conjugates. In two fundamental papers, Harju and Nowotka [6, 7] studied the unbordered conjugates of a word. In particular, letting $\operatorname{nuc}(w)$ denote the number of unbordered conjugates of $w$, and $\operatorname{mnuc}_{k}(n)$ denote the maximum number of unbordered conjugates of a length-$n$ word over a $k$-letter alphabet, they proved that

• (a)

for binary words $w$ of length $n\geq 4$ we have $\operatorname{nuc}(w)\leq n/2$;

• (b)

for $n>2$ even, there exists a binary word of length $n$ having $n/2$ unbordered conjugates iff $n=2^{k}$ or $n=3\cdot 2^{k}$ for some $k\geq 1$.

In other words, they explicitly computed $\operatorname{mnuc}_{2}(n)$ for all even $n$ and bounded it above for odd $n$. We complete the understanding of $\operatorname{mnuc}_{2}(n)$ by proving that $\operatorname{mnuc}_{2}(n)=\lfloor n/2\rfloor$ for all odd $n>3$. Our strategy is to show that the maximum of $\operatorname{nuc}(w)$, over all words of length $n$, is actually achieved by a factor of the Thue-Morse word.

More precisely, we prove

Theorem 3**.**

For all $n\geq 1$, there exists a length-$n$ factor $w$ of the Thue-Morse word $\bf t$ with $\operatorname{nuc}(w)=\operatorname{mnuc}_{2}(n)$. Furthermore, such a factor is guaranteed to occur starting at a position $\leq n$ in $\bf t$.

2 Circularly squarefree ternary words via Walnut

Since the ternary Thue-Morse word $\bf c$ is squarefree, it is reasonable to hope its factors might be a good source of circularly squarefree words. Unfortunately, $\bf c$ contains circularly squarefree words of length $n$ for only about $1/8$ of all natural numbers $n$, as the following result shows.

Theorem 4**.**

There is a length-$n$ factor of $\bf c$ that is circularly squarefree iff $(n)_{2}$ is accepted by the automaton in Figure 1.

To prove this result, we make use of the fact that many first-order statements concerning claims about $k$-automatic sequences are decidable [4]. Furthermore, there is free software called Walnut available to decide these claims [9].

Let $(n)_{k}$ denote the canonical base-$k$ representation of $n$, starting with the most significant digit, having no leading zeroes. A sequence $(a_{n})_{n\geq 0}$ is $k$-automatic if there is a deterministic finite automaton with output (DFAO) taking $(n)_{k}$ as input, and reaching a state with $a_{n}$ as output. For example, Figure 2 illustrates the DFAO generating the sequence $\bf c$. The notation $q/a$ in a state means the name of the state is $q$ and the output is $a$.

For more about automatic sequences, see [2].

Proof.

We can use the ideas in [11], adapted for our case. We create first-order logical predicates $\operatorname{crep}$, $\operatorname{facge2}$, and $\operatorname{circsf}$ as follows:

• •

$\operatorname{crep}(i,m,p,n,s)$ evaluates to true iff in the length-$n$ word (considered circularly) starting at position $s$ of the word $\bf c$, there is a factor $w$ of length $m$ and (not necessarily least) period $p\geq 1$ starting at position $i$;

• •

$\operatorname{facge2}(n,s)$ evaluates to true iff in the length-$n$ word (considered circularly) starting at position $s$ of the word $\bf c$ there is a square or higher power;

• •

$\operatorname{circsf}(n)$ evaluates to true iff some length-$n$ factor (considered circularly) of the word $\bf c$ has no squares.

[TABLE]

When we evaluate these predicates in Walnut, we get the automaton depicted in Figure 1. It accepts those $(n)_{2}$ for which circsf evaluates to true. ∎

Remark 5*.*

All the Walnut code for the theorems in this paper is available at

https://cs.uwaterloo.ca/~shallit/papers.html  .

The reader can therefore verify our results.

Corollary 6**.**

The number of lengths $\ell$, with $2^{n}\leq\ell<2^{n+1}$ and $n\geq 4$, such that $\bf c$ contains a circularly squarefree factor of length $\ell$, is $2^{n-3}-F_{n-3}+2$, where $F_{n}$ is the $n$’th Fibonacci number.

Proof.

By standard techniques, by determining the roots of the characteristic polynomial of the $15\times 15$ matrix encoding transitions of the automaton in Fig. 1. ∎

So while the factors of the ternary Thue-Morse word alone do not suffice for our purpose, it turns out that a small modification of them do. We now give the proof of our first main result, Theorem 2.

Proof.

(of Theorem 2) Let $n\geq 4$ and $w\in\{x021,y2120\}$, where $x,y$ are factors of the ternary Thue-Morse word $\bf c$ of lengths $n-3$ and $n-4$, respectively.

First, we create a predicate $\operatorname{sq021}(i,n,p,s)$ which evaluates to true if $w^{\prime}:=x021x02$ contains a square of order $p$ with $p\geq 1$ and $2p\leq n$ beginning at index $i-s$, where $x={\bf c}[s..s+n-4]$. We do this by defining $w[j]$ for all $j$ such that $i\leq j<i+p$ as follows:

[TABLE]

The goal is that $\operatorname{sq021}$ should represent the implication

[TABLE]

It is formed by constructing the conjunction of the predicates

[TABLE]

for each possible combination $j$ and $j+p$, and simplifying.

Next, we create a second predicate $\operatorname{sqfree021}(n,s)$, which evaluates to true if there exists $x$ where $w=x021$ is circularly squarefree, for the given values of $n$ and $s$:

[TABLE]

Similarly, we create the analogous predicates $\operatorname{sq2120}(i,n,p,s)$ and $\operatorname{sqfree2120}(n,s)$ for the word $w^{\prime}:=y2120y212$.

Finally, the predicates

[TABLE]

return true if there exists a length-$n$ squarefree word formed by concatenating some factor of $\bf c$ with $021$ (respectively, $2120$). The automaton for $\operatorname{test021}(n)$ is depicted in Figure 3 and the automaton for $\operatorname{test2120}(n)$ is depicted in Figure 4.

When we now evaluate the predicate

[TABLE]

with Walnut, we get the automaton depicted in Figure 5.

By inspection we easily see that the automaton in Figure 5 accepts the base-$2$ representation of all $n$ except $0,1,2,3,5,7,9,10,14,17,21,28$. ∎

As a consequence we now get Currie’s theorem:

Corollary 7**.**

There exist circularly squarefree ternary words of every length $n$, except for $n\in\{5,7,9,10,14,17\}$.

Proof.

Theorem 2 gives the result for all but finitely many $n$. It is easy to verify by a short computation that there are cyclically squarefree words of lengths $0,1,2,3,21,28$, and none for lengths $5,7,9,10,14,17$. ∎

Remark 8*.*

These calculations were done in Walnut on a Linux machine (2 CPU — Intel E5-2697 v3 Xeon, 256 GB of RAM). Computing the automaton for $\operatorname{sq021}$ took 115.505 seconds, and the automaton for $\operatorname{sq2120}$ took 124.908 seconds.

3 Unbordered conjugates

Let $\sigma:\Sigma_{k}^{*}\to\Sigma_{k}^{*}$ denote the cyclic shift function, where $\sigma(\epsilon)=\epsilon$, $\sigma(cw)=wc$ for $w\in\Sigma_{k}^{*}$ and $c\in\Sigma_{k}$. Let $\sigma^{0}(w)=w$ and $\sigma^{i}(w)=\sigma^{i-1}(\sigma(w))$ for $i\geq 1$.

Suppose $w$ is a binary word of length $n$. Let $\beta:\Sigma_{k}^{*}\to\Sigma_{k}^{*}$ be the border correlation function of a word (introduced by Harju and Nowotka [6]), and defined as follows: $\beta(w)=a_{0}a_{1}\cdots a_{n-1}$, where

[TABLE]

For example, $\beta(0001)=ubbu$ since $0001$ is unbordered, while $0010,$ and $0100$ are both bordered, and $1000$ is unbordered. Let $u,v\in\Sigma_{k}^{*}$. We say $u$ is the $i$’th cyclic shift of $v$ if $\sigma^{i}(v)=u$.

A result from Harju and Nowotka [6] shows that a binary word has no two consecutive cyclic shifts that are unbordered. This result immediately tells us that a binary word of length $n$ can have at most $\lfloor n/2\rfloor$ unbordered conjugates. For a binary word $w$ of even length to achieve this bound, every other cyclic shift must be unbordered, or, in other words either $\beta(w)=(ub)^{|w|/2}$ or $\beta(w)=(bu)^{|w|/2}$. Harju and Nowotka [6] showed that the only words of even length that achieve this bound are the circularly overlap-free words, which are of length $3\cdot 2^{i}$ and $2^{i}$ for $i\geq 1$.

Let $w$ be a binary word. Suppose $w$ is of even length and is not circularly overlap-free. Clearly $w$ cannot have $|w|/2$ unbordered conjugates, but it could potentially have $|w|/2-1$ unbordered conjugates. Then $\beta(w)=(ub)^{i}b(ub)^{|w|/2-i-1}b$ for some $i\geq 0$, up to conjugation. Now suppose $w$ is of odd length. No circularly overlap-free words exist of odd length, so it makes sense to think that $w$ could contain a maximum of $\lfloor|w|/2\rfloor$ unbordered conjugates. Then $\beta(w)=(ub)^{\lfloor|w|/2\rfloor}b$, up to conjugation.

Let $w$ be a bordered binary word. Then $w=uvu$ for some words $u$ and $v$. We say $w[1..|u|]$ is the first border of $w$, and $w[|w|-|u|+1..|w|]$ is the second border of $w$.

Now we prove Theorem 3.

Proof.

When $n=1,2,3$ the maxmium number of unbordered conjugates $\operatorname{mnuc}_{2}(n)$ is achieved by the words [math], $01$, and $011$ respectively. Specifically we have that $\operatorname{mnuc}_{2}(1)=1$, $\operatorname{mnuc}_{2}(2)=2$, and $\operatorname{mnuc}_{2}(3)=2$. It is readily verified that each of these words occur as a factor of the Thue-Morse word at position $\leq n$.

Let $w$ be a length-$n$ word at position $m$ of the Thue-Morse word. The first step is to create a first-order predicate $\operatorname{isBorder}(l,m,n)$ that asserts that a cyclic shift of $w$ has a border of a certain length. More specifically, we want to know whether the $l$’th cyclic shift of $w$ has a border of length $k$. There are three cases to consider.

When a prefix of the second border is a suffix of $w$ and a suffix of the second border is a prefix of $w$. In other words, $w=yuvx$ for words $u,v,x,y$ where $xy=u$, $|y|=l$, and $|u|=k$. This predicate is denoted by $\operatorname{isBorderC1}(k,l,m,n)$. 2. 2.

When both borders are completely contained inside of $w$. In other words, $w=yuux$ for words $y,u,x$ where $|yu|=l$, and $|u|=k$. This predicate is denoted by $\operatorname{isBorderC2}(k,l,m,n)$. 3. 3.

When a prefix of the first border is a suffix of $w$ and a suffix of the first border is a prefix of $w$. In other words, $w=yvux$ for words $u,v,x,y$ where $xy=u$, $|yvu|=l$, and $|u|=k$. This predicate is denoted by $\operatorname{isBorderC3}(k,l,m,n)$.

[TABLE]

We define the predicate $\operatorname{isBordered}(l,m,n)$ that asserts that the $l$’th cyclic shift of a length $n$ word at position $m$ in the Thue-Morse word is bordered. We can create this predicate by checking whether this word has a border of size $\leq n/2$.

[TABLE]

Recall that when $|w|$ is odd and $w$ has a maximum number of unbordered conjugates, we have that $\beta(w)=(ub)^{\lfloor|w|/2\rfloor}b$, up to conjugation. So we have exactly one pair of adjacent bordered cyclic shifts, and the rest of the cyclic shifts of $w$ alternate between bordered and unbordered. The predicate $\operatorname{isAlternating0}(l,m,n)$ asserts that all of the cyclic shifts of a length $n$ word at position $m$ in the Thue-Morse word alternate between unbordered and bordered, except for the $l$’th and $l+1$’th cyclic shifts, which are both bordered.

[TABLE]

Now we create a predicate $\operatorname{hasMNUCO}(m,n)$ that asserts that a length $n$ word at position $m$ in the Thue-Morse word achieves the maximum number of unbordered conjugates.

[TABLE]

Similarly, recall that when $|w|$ is even and $w$ has a maximum number of unbordered conjugates, we have that $\beta(w)=(ub)^{i}b(ub)^{|w|/2-i-1}b$ for some $i\geq 0$ or $\beta(w)=(ub)^{|w|/2}$, up to conjugation. So we have that either all of the cyclic shifts of $w$ alternate between bordered and unbordered, or there are exactly two pairs of adjacent bordered cyclic shifts, and the rest of the cyclic shifts of $w$ alternate between bordered and unbordered. The predicate $\operatorname{isAlternatingE}(e,l,m,n)$ asserts that all of the cyclic shifts of a length $n$ word at position $m$ in the Thue-Morse word alternate between unbordered and bordered, except for the $l$’th, $l+1$’th, $e$’th, and $e+1$’th cyclic shifts, which are all bordered. Note that $\operatorname{isAlternatingE}(n,n,m,n)$ asserts that all of the cyclic shifts of a length $n$ word at position $m$ in the Thue-Morse word alternate between unbordered and bordered.

[TABLE]

Now we create a predicate $\operatorname{hasMNUCE}(m,n)$ that asserts that a length $n$ word at position $m$ in the Thue-Morse word achieves the maximum number of unbordered conjugates.

[TABLE]

With these predicates we can write a predicate asserting that the Thue-Morse word contains factors of every length $n>3$ that are maximally unbordered and occur at position $\leq n$. We split the computation into cases, one for even length words, and one for odd:

[TABLE]

and Walnut evaluates these predicates to be true. ∎

Thus we have that

[TABLE]

Theorem 9**.**

Let $f(n)=\operatorname{mnuc}_{2}(n)-\lfloor n/2\rfloor$. Then $f$ is a $2$-automatic sequence.

4 More about unbordered conjugates

In this section we show that there exist binary words of length $n$ that have exactly $i$ unbordered conjugates where $1<i\leq\operatorname{mnuc}_{2}(n)$.

The general idea behind the proof is to pick some $i>1$ and then pick a word $w$ of odd length such that $\operatorname{nuc}(w)=i$ and $\operatorname{mnuc}_{2}(|w|)=i$. Furthermore we only consider such words $w$ such that one of $w$’s conjugates contain $000$ as a factor. Then we keep adding [math]’s to $w$ precisely where $000$ first occurs. This keeps the number of unbordered conjugates the same. Then we can keep increasing the size of $w$ in this way until we hit the length we want.

Lemma 10**.**

For $n>4$ odd, there exists a word $w\in\Sigma_{2}^{n}$ such that $\operatorname{nuc}(w)=\lfloor n/2\rfloor$ and $000$ is a factor of some conjugate of $w$.

Proof.

By Theorem 3, such a word $w$ exists as a factor of the Thue-Morse word. It is well known that the Thue-Morse word is overlap-free. So $000$ cannot be a factor of such a word $w$. But it is possible that $w=0u00$, or $w=00u0$ for some word $u$. We can check whether this is the case for all odd $n>4$ by modifying our predicate from the proof of Theorem 3:

[TABLE]

which evaluates to true. ∎

Lemma 11**.**

Let $n>4$ be odd and $w$ be a binary word of length $n$ such that a conjugate of $w$ has $000$ as a factor and $\operatorname{nuc}(w)=\lfloor n/2\rfloor$. Then every conjugate of $w$ contains at most one distinct occurrence of $000$ as a factor.

Proof.

Suppose, contrary to what we want to prove that a conjugate of $w$ contains at least two distinct occurrences of $000$ as a factor. Call this conjugate $w^{\prime}$.

If the two occurrences of $000$ overlap, then we can write $w^{\prime}=s0000t$ for some words $s,t$. Then the cyclic shifts $0ts000$, $00ts00$, and $0ts000$ are bordered. This means that only $\lfloor|ts|/2\rfloor+1$ of the remaining cyclic shifts of $w$ can be unbordered since any unbordered cyclic shift must be followed by a bordered one. But $\lfloor|ts|/2\rfloor+1=\lfloor(n-4)/2\rfloor+1<\lfloor n/2\rfloor$, so the two occurrences of $000$ cannot overlap.

If the two occurrences of $000$ do not overlap, then we can write $w^{\prime}=s000t000$ for some words $s,t$ where $s$, and $t$ are non-empty. Then the conjugates $00t000s0$, $0t000s00$, $00s000t0$, and $0s000t00$ are bordered. By the same argument as above, of the remaining cyclic shifts, a maximum of $\lfloor|st|/2\rfloor+2$ of them can be unbordered. But $\lfloor|st|/2\rfloor+2=\lfloor(n-6)/2)\rfloor+2<\lfloor n/2\rfloor$, a contradiction. ∎

Lemma 12**.**

Let $n>4$ be odd and $w$ be a binary word of length $n$ such that a conjugate $w^{\prime}$ of $w$ has $000$ as a prefix and $\operatorname{nuc}(w)=\lfloor n/2\rfloor$. Then $\operatorname{nuc}(w)=\operatorname{nuc}(w^{\prime})=\operatorname{nuc}(0^{i}w^{\prime})$ for all $i\geq 0$.

Proof.

Let $i\geq 0$ be an integer. We can write $w^{\prime}=000u$ for some word $u$. It is clear that $0^{j}u0^{i+3-j}$ is bordered for all $1\leq j\leq i+2$. Therefore, it suffices to prove that $s000t$ is bordered if and only if $s0^{i+3}t$ is bordered where $u=ts$.

First we prove the forward direction. Suppose $s000t$ is bordered. By Lemma 11 we have that $s000t$ contains only one occurrence of $000$ as a factor. So $000$ is neither a prefix of $s00$ nor a suffix of $00t$. Thus, any border of $s000t$ must of length $\leq\min\{|s|,|t|\}+2$. But such a border would also be a border of $s0^{i+3}t$.

A similar argument works for the reverse direction. Therefore $\operatorname{nuc}(w)=\operatorname{nuc}(w^{\prime})=\operatorname{nuc}(0^{i}w^{\prime})$ for all $i\geq 0$. ∎

Theorem 13**.**

For all $1<i\leq\operatorname{mnuc}_{k}(n)$ there exists $w\in\Sigma_{k}^{n}$ such that $\operatorname{nuc}(w)=i$.

Proof.

Let $C=\{5,7,9,10,14,17\}$. For $k\geq 4$, Harju and Nowotka [7] showed that for all integers $i$ with $1<i\leq n$ there exists a word $w\in\Sigma_{k}^{n}$ such that $\operatorname{nuc}(w)=i$. For $k=3$, Harju and Nowotka [7] showed that if $n\not\in C$ then for all integers $i$ with $1<i\leq n$ there exists a word $w\in\Sigma_{k}^{n}$ such that $\operatorname{nuc}(w)=i$, and if $n\in C$ then for all integers $i$ with $1<i<n$ there exists a word $w\in\Sigma_{k}^{n}$ such that $\operatorname{nuc}(w)=i$.

To the best of the authors’ knowledge, there is no known proof of the existence of such words for $k=2$. Suppose $k=2$. By Theorem 3 there exists a $w\in\Sigma_{2}^{n}$ such that $w$ is a factor of the Thue-Morse word and $\operatorname{mnuc}_{2}(n)=\operatorname{nuc}(w)$. So assume $i<\operatorname{mnuc}_{2}(n)$. By Lemma 10 there exists a binary word $u$ of odd length $m$ such that $\operatorname{nuc}(u)=i=\lfloor m/2\rfloor$ and $000$ is a factor of some conjugate of $u$. Let $u^{\prime}$ be the conjugate of $u$ such that $000$ is a prefix of $u^{\prime}$. Lemma 12 tells us $\operatorname{nuc}(u)=\operatorname{nuc}(u^{\prime})=\operatorname{nuc}(0^{n-m}u^{\prime})$. Since $\operatorname{nuc}(0^{n-m}u^{\prime})=i$ and $|0^{n-m}u^{\prime}|=n$, we have that for all $1<i\leq\operatorname{mnuc}_{2}(n)$, there exists a $w\in\Sigma_{2}^{n}$ such that $\operatorname{nuc}(w)=i$. ∎

5 Expected number of unbordered conjugates

What is the expected number of unbordered conjugates of a randomly-chosen word of length $n$?

Let $u_{k}(n)$ denote the number of unbordered binary words of length $n$. It is known that $\lim_{n\rightarrow\infty}u_{k}(n)/k^{n}=\alpha_{k}$, where $\alpha_{k}$ is a certain real number. We have $\alpha_{2}\doteq 0.2677$; see [10].

Theorem 14**.**

The expected number of unbordered conjugates of a randomly-chosen word of length $n$ over a $k$-letter alphabet is $(u_{k}(n)/k^{n})n$, which is asymptotically equal to $\alpha_{k}n$.

Proof.

To see this, let $x$ be a word of length $n$. There are two cases: $x$ is not primitive, and $x$ is primitive.

If $x$ is not primitive, it equals $y^{e}$ for some $e\geq 2$. Then by a well-known result, every conjugate of $x$ is bordered. So $x$ has [math] unbordered conjugates.

If $x$ is primitive, then all its cyclic shifts are distinct. (For otherwise $x=uv=vu$, and then by Lyndon-Schützenberger we know $x$ is a power.) There are $n$ of these cyclic shifts. So if we consider all conjugates of all primitive words, each primitive word appears exactly $n$ times.

Putting this all together, we get

[TABLE]

as claimed. Here $\operatorname{nuc}(x)$ is the number of unbordered conjugates of $x$, and $\sigma^{i}(x)$ denotes the left shift by $i$ locations of $x$, and $[p]$ is $1$ if $p$ is true and [math] otherwise. ∎

6 Conclusions

We want to emphasize that our experience shows that rephrasing problems in combinatorics on words using the first-order logical theory of automatic sequences can be a useful tool in solving these problems. We encourage others to adopt this approach.

Acknowledgments

We thank Dirk Nowotka for helpful discussions.