Smallest and Largest Block Palindrome Factorizations

TL;DR

This paper investigates the properties of block palindrome factorizations in words, providing recurrence relations, expected value analysis, and extremal case studies, revealing structural insights and connections to borders in words.

Contribution

It introduces new recurrence formulas, analyzes the expected width of largest BP-factorizations, and explores the relationship between smallest and largest BP-factorizations and word borders.

Findings

Recurrence for the number of words with a given largest BP-factorization width.

Expected width of largest BP-factorization tends to a constant.

Connection between words with a unique border and coinciding smallest and largest BP-factorizations.

Abstract

A \emph{palindrome} is a word that reads the same forwards and backwards. A \emph{block palindrome factorization} (or \emph{BP-factorization}) is a factorization of a word into blocks that becomes palindrome if each identical block is replaced by a distinct symbol. We call the number of blocks in a BP-factorization the \emph{width} of the BP-factorization. The \emph{largest BP-factorization} of a word $w$ is the BP-factorization of $w$ with the maximum width. We study words with certain BP-factorizations. First, we give a recurrence for the number of length-$n$ words with largest BP-factorization of width $t$. Second, we show that the expected width of the largest BP-factorization of a word tends to a constant. Third, we give some results on another extremal variation of BP-factorization, the \emph{smallest BP-factorization}. A \emph{border} of a word $w$ is a non-empty word that is both a proper prefix and suffix of $w$. Finally, we conclude by showing a connection between words with a unique border and words whose smallest and largest BP-factorizations coincide.

Tables

Table 1. Table 1: Some values of LBP 2 ⁡ ( n , t ) subscript LBP 2 𝑛 𝑡 \operatorname{LBP}_{2}(n,t) for n 𝑛 n , t 𝑡 t where 10 ≤ n ≤ 20 10 𝑛 20 10\leq n\leq 20 and 1 ≤ t ≤ 10 1 𝑡 10 1\leq t\leq 10 .

	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
10	284	12	224	40	168	72	96	64	32	32
11	568	0	472	0	416	0	336	0	192	0
12	1116	20	856	88	656	176	448	224	224	160
13	2232	0	1752	0	1488	0	1248	0	896	0
14	4424	40	3328	176	2544	432	1856	640	1152	640
15	8848	0	6736	0	5440	0	4576	0	3584	0
16	17622	74	13100	372	9896	984	7408	1744	5088	2080
17	35244	0	26348	0	20536	0	16784	0	13664	0
18	70340	148	51936	760	38824	2248	29152	4416	21088	6240
19	140680	0	104168	0	79168	0	62800	0	51008	0
20	281076	284	206744	1592	153344	4992	114688	10912	84704	17312

Table 2. Table 2: Asymptotic expected width of a word’s largest BP-factorization.

$k$	$\approx E_k$
2	6.4686
3	2.5908
4	1.9080
5	1.6314
6	1.4827
7	1.3902
8	1.3272
9	1.2817
10	1.2472
⋮	⋮
100	1.0204

Table 3. Table 3: Probability that a word has a unique border.

$k$	$\approx P_k$
2	0.5155
3	0.3910
4	0.2922
5	0.2302
6	0.1890
7	0.1599
8	0.1384
9	0.1219
10	0.1089
⋮	⋮
100	0.0101

Equations

$$u_n=\{\begin{array}{ll}1,& \text{if }n=0;\\ k{u}_{n-1}-u_{n/2},& \text{if n>0 is even;}\\ k{u}_{n-1},& \text{if n is odd.}\end{array}$$

$$\mathtt{abracadabra},$$

$$\mathtt{abra}\cdot \mathtt{cad}\cdot \mathtt{abra},$$

$$\mathtt{a}\cdot \mathtt{bracadabr}\cdot \mathtt{a},$$

$$\mathtt{a}\cdot \mathtt{br}\cdot \mathtt{acada}\cdot \mathtt{br}\cdot \mathtt{a},$$

$$\mathtt{a}\cdot \mathtt{br}\cdot \mathtt{a}\cdot \mathtt{cad}\cdot \mathtt{a}\cdot \mathtt{br}\cdot \mathtt{a}.$$

$${\mathrm{LBP}}_k(n,t)=\{\begin{array}{ll}{\sum }_{i=1}^{(n-t)/2+1}{u}_i{\mathrm{LBP}}_k(n-2i,t-2),& \text{if n, t even;}\\ {\sum }_{i=1}^{(n-t+1)/2}{u}_{2i}{\mathrm{LBP}}_k(n-2i,t-1),& \text{if n even, t odd;}\\ 0,& \text{if n odd, t even;}\\ {\sum }_{i=1}^{(n-t)/2+1}{u}_{2i-1}{\mathrm{LBP}}_k(n-2i+1,t-1),& \text{if n, t odd.}\end{array}$$

$${\mathrm{LBP}}_k(0,0)$$

$${\mathrm{LBP}}_k(2n,2)$$

$${\mathrm{LBP}}_k(n,1)$$

$${\mathrm{LBP}}_k(n,t)=\sum _{i=1}^{(n-t)/2+1}{u}_i{\mathrm{LBP}}_k(n-2i,t-2).$$

$${\mathrm{LBP}}_k(n,t)=\sum _{i=1}^{(n-t+1)/2}{u}_{2i}{\mathrm{LBP}}_k(n-2i,t-1).$$

$$\sum _{i=1}^{(n-t)/2+1}{u}_{2i-1}{\mathrm{LBP}}_k(n-2i+1,t-1).$$

$$E_{n,k}=\frac{1}{k^n}\sum _{i=1}^{n}i\cdot {\mathrm{LBP}}_k(n,i).$$

$$\frac{{\mathrm{LBP}}_k(n,t)}{k^n}\le \frac{1}{k^{t/2-1}}.$$

$$\frac{{\mathrm{LBP}}_k(n,t)}{k^n}\le \frac{1}{k^{t/2-1}}$$

$$1+\frac{2}{k}+\frac{4}{k^2}+\frac{6}{k^3}+\frac{10}{k^4}+\frac{16}{k^5}+\frac{24}{k^6}+\frac{38}{k^7}+\frac{58}{k^8}+\frac{88}{k^9}+\cdots .$$

$$E_k=1+\sum _{i=1}^{\infty }{a}_i{k}^{-i}$$

$$(0101{)}^l(1001{)}^l$$

$$B_k(n,t)=\{\begin{array}{ll}0,& \text{if n<2t;}\\ u_t{k}^{n-2t}-{\sum }_{i=2t}^{\lfloor n/2\rfloor }{B}_k(i,t)k^{n-2i},& \text{if n≥2t and n+t odd;}\\ u_t{k}^{n-2t}-B_k((n+t)/2,t)-{\sum }_{i=2t}^{\lfloor n/2\rfloor }{B}_k(i,t)k^{n-2i},& \text{if n≥2t and n+t even.}\end{array}$$

$$\overline{B_k}(n,t)=\sum _{i=2t}^{\lfloor n/2\rfloor }{B}_k(i,t)k^{n-2i}.$$

$$\overline{B_k}(n,t)=B_k((n+t)/2,t)+\sum _{i=2t}^{\lfloor n/2\rfloor }{B}_k(i,t)k^{n-2i}.$$

$$B_k(n)=\sum _{t=1}^{\lfloor n/2\rfloor }{B}_k(n,t).$$

$$P_{n,k}=\frac{B_k(n)}{k^n}=\frac{1}{k^n}\sum _{i=1}^{\lfloor n/2\rfloor }{B}_k(n,i).$$

$$\frac{B_k(n,t)}{k^n}\le \frac{1}{k^t}.$$

Full text

1 Introduction

Let $\Sigma_{k}$ denote the alphabet $\{0,1,\ldots,k-1\}$. The length of a word $w$ is denoted by $|w|$. A border of a word $w$ is a non-empty word that is both a proper prefix and suffix of $w$. A word is said to be bordered if it has a border. Otherwise, the word is said to be unbordered. For example, the French word entente is bordered, and has two borders, namely ente and e.

It is well-known [1] that the number $u_{n}$ of length-$n$ unbordered words over $\Sigma_{k}$ satisfies

[TABLE]

A palindrome is a word that reads the same forwards as it does backwards. More formally, letting $w^{R}=w_{n}w_{n-1}\cdots w_{1}$ where $w=w_{1}w_{2}\cdots w_{n}$ and all $w_{i}$ are symbols, a palindrome is a word $w$ such that $w=w^{R}$. The definition of a palindrome is quite restrictive. The second half of a palindrome is fully determined by the first half. Thus, compared to all length-$n$ words, the number of length-$n$ palindromes is vanishingly small. But many words exhibit palindrome-like structure. For example, take the English word ${\tt marjoram}$. It is clearly not a palindrome, but it comes close. Replacing the block ${\tt jo}$ with a single letter turns the word into a palindrome. In this paper, we consider a generalization of palindromes that incorporates this kind of palindromic structure.

In the 2015 British Olympiad [2], the concept of a block palindrome factorization was first introduced. Let $w$ be a non-empty word. A block palindrome factorization (or BP-factorization) of $w$ is a factorization $w=w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ of a word such that $w_{0}$ is a possibly empty word, and every other factor $w_{i}$ is non-empty for all $i$ with $1\leq i\leq m$. We say that a BP-factorization $w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ is of width $t$ where $t=2m+1$ if $w_{0}$ is non-empty and $t=2m$ otherwise. In other words, the width of a BP-factorization is the number of non-empty blocks in the factorization. The largest BP-factorization333Largest BP-factorizations also appear in https://www.reddit.com/r/math/comments/ga2iyo/i_just_defined_the_palindromity_function_on/. [3] of a word $w$ is a BP-factorization $w=w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ where $m$ is maximized (i.e., where the width of the BP-factorization is maximized). See [4, 5] for more on the topic of BP-factorizations and block reversals. Kolpakov and Kucherov [6] studied a special case of BP-factorizations, the gapped palindrome. If $w_{0}$ is non-empty and $|w_{i}|=1$ for all $i$ with $1\leq i\leq m$, then $w$ is said to be a gapped palindrome. Régnier [7] studied something similar to BP-factorizations, but in her paper she was concerned with borders of borders. See [8, 9] for results on factoring words into palindromes.

Example 1**.**

We use the centre dot $\cdot$ to denote the separation between blocks in the BP-factorization of a word.

Consider the word abracadabra. It has the following BP-factorizations:

[TABLE]

The last BP-factorization is of width $7$ and has the longest width; thus it is the largest BP-factorization of abracadabra.

Let $w$ be a length-$n$ word. Suppose $w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ is the largest BP-factorization of $w$. Goto et al. [3] showed that $w_{i}$ is the shortest border of $w_{i}\cdots w_{1}w_{0}w_{1}\cdots w_{i}$ where $i\geq 1$. This means that we can compute the largest BP-factorization of $w$ by greedily “peeling off” the shortest borders of central factors until you hit an unbordered word or the empty word.

The rest of the paper is structured as follows. In Section 2 we give a recurrence for the number of length-$n$ words with largest BP-factorization of width $t$. In Section 3 we show that the expected width of the largest BP-factorization of a length-$n$ word tends to a constant. In Section 4 we consider smallest BP-factorizations in the sense that one “peels off” the longest non-overlapping border. We say a border $u$ of a word $w$ is non-overlapping if $|u|\leq|w|/2$; otherwise $u$ is overlapping. Finally, in Section 5 we present some results on words with a unique border and show that they are connected to words whose smallest and largest BP-factorizations are the same.

2 Counting largest BP-factorizations

In this section, we prove a recurrence for the number $\operatorname{LBP}_{k}(n,t)$ of length-$n$ words over $\Sigma_{k}$ with largest BP-factorization of width $t$. See Table 1 for sample values of $\operatorname{LBP}_{2}(n,t)$ for small $n$, $t$. For the following theorem, recall the definition of $u_{n}$ from Equation 1.

Theorem 2**.**

Let $n,t\geq 0$, and $k\geq 2$ be integers. Then

[TABLE]

where

[TABLE]

Proof.

Let $w$ be a length-$n$ word whose largest BP-factorization $w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ is of width $t$. Clearly $\operatorname{LBP}_{k}(0,0)=1$. We know that each block in a largest BP-factorization is unbordered, since each block is a shortest border of some central factor. This immediately implies $\operatorname{LBP}_{k}(n,1)=u_{n}$ and $\operatorname{LBP}_{k}(2n,2)=u_{n}$.

Now we take care of the other cases.

• •

Suppose $n$, $t$ are even. Then by removing both instances of $w_{1}$ from $w$, we get $w^{\prime}=w_{m}\cdots w_{2}w_{2}\cdots w_{m}$, which is a length-$(n-2|w_{1}|)$ word whose largest BP-factorization is of width $t-2$. This mapping is clearly reversible, since all blocks in a largest BP-factorization are unbordered, including $w_{1}$. Thus summing over all possible $w_{1}$ and all length-$(n-2|w_{1}|)$ words with largest BP-factorization of width $t-2$ we have

[TABLE]

• •

Suppose $n$ is even and $t$ is odd. Then by removing $w_{0}$ from $w$, we get $w^{\prime}=w_{m}\cdots w_{1}w_{1}\cdots w_{m}$, which is a length-$(n-|w_{0}|)$ word whose largest BP-factorization is of width $t-1$. This mapping is reversible for the same reason as in the previous case. The word $w^{\prime}$ is of even length since $|w^{\prime}|=2|w_{1}\cdots w_{m}|$. Since $n$ is even and $|w^{\prime}|$ is even, we must have that $|w_{0}|$ is even as well. Thus summing over all possible $w_{0}$ and all length-$(n-|w_{0}|)$ words with largest BP-factorization of width $t-1$, we have

[TABLE]

• •

Suppose $n$ is odd and $t$ is even. Then the length of $w$ is $2|w_{1}\cdots w_{m}|$, which is even, a contradiction. Thus $\operatorname{LBP}_{k}(n,t)=0$.

• •

Suppose $n$, $t$ are odd. Then by removing $w_{0}$ from $w$, we get $w^{\prime}=w_{m}\cdots w_{1}w_{1}\cdots w_{m}$, which is a length-$(n-|w_{0}|)$ word whose largest BP-factorization is of width $t-1$. This mapping is reversible for the same reasons as in the previous cases. Since $n$ is odd and $|w^{\prime}|$ is even (proved in the previous case), we must have that $|w_{0}|$ is odd. Thus summing over all possible $w_{0}$ and all length-$(n-|w_{0}|)$ words with largest BP-factorization of width $t-1$, we have

[TABLE]

∎

3 Expected width of largest BP-factorization

In this section, we show that the expected width $E_{n,k}$ of the largest BP-factorization of a length-$n$ word over $\Sigma_{k}$ is bounded by a constant. From the definition of expected value, it follows that

[TABLE]

Table 2 shows the behaviour of $\lim\limits_{n\to\infty}E_{n,k}$ as $k$ increases.

Lemma 3**.**

Let $k\geq 2$ and $n\geq t\geq 1$ be integers. Then

[TABLE]

Proof.

Let $w$ be a length-$n$ word whose largest BP-factorization $w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ is of width $t$. Since $w_{i}$ is non-empty for every $1\leq i\leq m$, we have that $\operatorname{LBP}_{k}(n,t)\leq k^{n-m}\leq k^{n-t/2+1}$. So

[TABLE]

for all $n\geq t\geq 1$. ∎

Theorem 4**.**

The limit $E_{k}=\lim\limits_{n\to\infty}E_{n,k}$ exists for all $k\geq 2$.

Proof.

Follows from the definition of $E_{n,k}$, Lemma 3, and the direct comparison test for convergence. ∎

Interpreting $E_{k}$ as a power series in $k^{-1}$, we empirically observe that $E_{k}$ is approximately equal to

[TABLE]

We conjecture the following about $E_{k}$.

Conjecture 5**.**

Let $k\geq 2$. Then

[TABLE]

where the sequence $(a_{n}/2)_{n\geq 1}$ is A274199 in the On-Line Encyclopedia of Integer Sequences (OEIS) [10].

Cording et al. [11] proved that the expected length of the longest unbordered factor in a word is $\Theta(n)$. Taking this into account, it is not surprising that the expected length of the largest BP-factorization of a word tends to a constant.

4 Smallest BP-factorization

A word $w$, seen as a block, clearly satisfies the definition of a BP-factorization. Thus, taken literally, the smallest BP-factorization for all words is of width $1$. But this is not very interesting, so we consider a different definition instead. A border $u$ of a word $w$ is non-overlapping if $|u|\leq|w|/2$; otherwise $u$ is overlapping. We say that the smallest BP-factorization of a word $w$ is a BP-factorization $w=w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ where each $w_{i}$ is the longest non-overlapping border of $w_{i}\cdots w_{1}w_{0}w_{1}\cdots w_{i}$, except $w_{0}$, which is either empty or unbordered. For example, going back to Example 1, the smallest BP-factorization of ${\tt abracadabra}$ is ${\tt abra}\cdot{\tt cad}\cdot{\tt abra}$ and the smallest BP-factorization of ${\tt reappear}$ is ${\tt r}\cdot{\tt ea}\cdot{\tt p}\cdot{\tt p}\cdot{\tt ea}\cdot{\tt r}$.

A natural question to ask is: what is the maximum possible width $f_{k}(n)$ of the smallest BP-factorization of a length-$n$ word? Through empirical observation, we arrive at the following conjectures:

• •

We have $f_{2}(8n+i)=6n+i$ for $i$ with $0\leq i\leq 5$ and $f_{2}(8n+6)=f_{2}(8n+7)=6n+5$.

• •

We have $f_{k}(n)=n$ for $k\geq 3$.

To calculate $f_{k}(n)$, two things are needed: an upper bound on $f_{k}(n)$, and words that witness the upper bound.

Theorem 6**.**

Let $l\geq 0$ be an integer. Then $f_{2}(8l+i)=6l+i$ for $i$ with $0\leq i\leq 5$ and $f_{2}(8l+6)=f_{2}(8l+7)=6l+5$.

Proof.

Let $n\geq 0$ be an integer. We start by proving lower bounds on $f_{2}(n)$. Suppose $n=8l$ for some $l\geq 0$. Then the width of the smallest BP-factorization of

[TABLE]

is $6l$, so $f_{2}(8l)\geq 6l$. To see this, notice that the smallest BP-factorization of $01011001$ is $01\cdot 0\cdot 1\cdot 1\cdot 0\cdot 01$, and therefore is of width $6$. Suppose $n=8l+i$ for some $i$ with $1\leq i\leq 7$. Then one can take $(0101)^{l}(1001)^{l}$ and insert either [math], $00$, $010$, $0110$, $01010$, $010110$, or $0110110$ to the middle of the word to get the desired length.

Now we prove upper bounds on $f_{2}(n)$. Let $t\leq n$ be a positive integer. Let $w$ be a length-$n$ word whose largest BP-factorization $w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$ is of width $t$. One can readily verify that $f_{2}(0)=0$, $f_{2}(1)=1$, $f_{2}(2)=2$, $f_{2}(3)=3$, $f_{2}(4)=4$, and $f_{2}(5)=f_{2}(6)=f_{2}(7)=5$ through exhaustive search of all binary words of length $<8$. Suppose $m\geq 4$, so $n\geq t\geq 8$. Then we can write $w=w_{m}w_{m-1}w_{m-2}\cdots w_{m-2}w_{m-1}w_{m}$ where $|w_{m-2}|,|w_{m-1}|,|w_{m}|>0$. It is easy to show that $|w_{m-2}w_{m-1}w_{m}|\geq 4$ by checking that all binary words of length $<8$ do not admit a smallest BP-factorization of width $6$. In the worst case, we can peel off prefixes and suffixes of length $4$ while accounting for the $6$ blocks they add to the BP-factorization until we hit the middle core of length $<8$. Thus, we have $f_{2}(8l+i)\leq 6l+j$ where $j$ is the width of the smallest BP-factorization of the middle core, which is of length $i$. We have already computed $f_{2}(i)$ for $0\leq i\leq 7$, so the upper bounds follow. ∎

Theorem 7**.**

Let $n\geq 0$ and $k\geq 3$ be integers. Then $f_{k}(n)=n$.

Proof.

Clearly $f_{k}(n)\leq n$. We prove $f_{k}(n)\geq n$. If $n$ is divisible by $6$, then consider the word $(012)^{n/6}(210)^{n/6}$. If $n$ is not divisible by $6$, then take $(012)^{\lfloor n/6\rfloor}(210)^{\lfloor n/6\rfloor}$ and insert either [math], $00$, $010$, $0110$, or $01010$ in the middle of the word. When calculating the smallest BP-factorization of the resulting words, it is easy to see that at each step we are removing a border of length $1$. Thus, their largest BP-factorization is of width $n$. ∎

5 Equal smallest and largest BP-factorizations

Recall back to Example 1, that ${\tt abracadabra}$ has distinct smallest and largest BP-factorizations, namely ${\tt abra}\cdot{\tt cad}\cdot{\tt abra}$ and ${\tt a}\cdot{\tt br}\cdot{\tt a}\cdot{\tt cad}\cdot{\tt a}\cdot{\tt br}\cdot{\tt a}$. However, the word ${\tt alfalfa}$ has the same smallest and largest BP-factorizations, namely ${\tt a}\cdot{\tt lf}\cdot{\tt a}\cdot{\tt lf}\cdot{\tt a}$. Under what conditions are the smallest and largest BP-factorizations of a word the same? Looking at unique borders seems like a good place to start, since the shortest border and longest non-overlapping border coincide when a word has a unique border. However, the converse is not true—just consider the previous example ${\tt alfalfa}$. The shortest border and longest non-overlapping border are both ${\tt a}$, but ${\tt a}$ is not a unique border of ${\tt alfalfa}$.

In Theorem 8 we characterize all words whose smallest and largest BP-factorization coincide.

Theorem 8**.**

Let $m,m^{\prime}\geq 1$ be integers. Let $w$ be a word with smallest BP-factorization $w_{m^{\prime}}^{\prime}\cdots w_{1}^{\prime}w_{0}^{\prime}w_{1}^{\prime}\cdots w_{m^{\prime}}^{\prime}$ and largest BP-factorization $w_{m}\cdots w_{1}w_{0}w_{1}\cdots w_{m}$. Then $m=m^{\prime}$ and $w_{i}=w_{i}^{\prime}$ for all $i$, $0\leq i\leq m$ if and only if for all $i\neq 2$, $0<i\leq m$, we have that $w_{i}$ is the unique border of $w_{i}\cdots w_{1}w_{0}w_{1}\cdots w_{i}$ and for $i=2$ we have that either

$w_{2}$* is the unique border of $w_{2}w_{1}w_{0}w_{1}w_{2}$, or* 2. 2.

$w_{2}w_{1}w_{0}w_{1}w_{2}=w_{0}w_{1}w_{0}w_{1}w_{0}$* where $w_{0}$ is the unique border of $w_{0}w_{1}w_{0}$.*

Proof.

$\Longrightarrow:$ Let $i$ be an integer such that $0<i\leq m$. Let $u_{i}=w_{i}\cdots w_{1}w_{0}w_{1}\cdots w_{i}$. Since $w_{i}$ is both the shortest border and longest non-overlapping border of $u_{i}$ (i.e., $w_{i}=w_{i}^{\prime}$), we have that $u_{i}$ has exactly one border of length $\leq|u_{i}|/2$. Thus, either $w_{i}$ is the unique border of $u_{i}$, or $u_{i}$ has a border of length $>|u_{i}|/2$. If $w_{i}$ is the unique border of $u_{i}$, then we are done. So suppose that $u_{i}$ has a border of length $>|u_{i}|/2$. Let $v_{i}$ be the shortest such border. We have that $w_{i}$ is both a prefix and suffix of $v_{i}$. In fact, $w_{i}$ must be the unique border of $v_{i}$. Otherwise we contradict the minimality of $v_{i}$, or the assumption that $w_{i}$ is both the shortest border and longest non-overlapping border of $u_{i}$. Since $w_{i}$ is unbordered, it cannot overlap itself in $v_{i}$ and $w_{i}$. So we can write $v_{i}=w_{i}yw_{i}$ for some word $y$ where $u_{i}=w_{i}yw_{i}yw_{i}$, or $u_{i}=w_{i}xw_{i}x^{\prime}w_{i}x^{\prime\prime}w_{i}$ such that $y=xw_{i}x^{\prime}=x^{\prime}w_{i}x^{\prime\prime}$. If $u_{i}=w_{i}xw_{i}x^{\prime}w_{i}x^{\prime\prime}w_{i}$, then we see that $w_{i}x^{\prime}$ is a suffix of $y$ and $x^{\prime}w_{i}$ is a prefix of $y$, implying that $w_{i}x^{\prime}w_{i}$ is a new smaller border of $u_{i}$. This either contradicts the assumption that $v_{i}$ is the shortest border of length $>|u_{i}|/2$, or the assumption that $u_{i}$ has exactly one border of length $\leq|u_{i}|/2$. Thus, we have that $u_{i}=w_{i}yw_{i}yw_{i}$. The shortest border and longest non-overlapping border of $yw_{i}y$ must be $y$, by assumption. Additionally, $w_{i}$ is unbordered, so $u_{i}$ is of width $5$ and $i=2$. This implies that $w_{i}=w_{2}=w_{0}$ and $y=w_{1}$.

$\Longleftarrow:$ Let $i$ be an integer such that $0<i\leq m$. We omit the case when $i=0$, since proving $w_{i}=w_{i}^{\prime}$ for all other $i$ is sufficient. Since $w_{i}$ is the unique border of $u_{i}=w_{i}\cdots w_{1}w_{0}w_{1}\cdots w_{i}$, we have that the shortest border and longest non-overlapping border of $u_{i}$ is $w_{i}$. In other words, we have that $w_{i}=w_{i}^{\prime}$. Suppose $i=2$ and $u_{2}=w_{2}w_{1}w_{0}w_{1}w_{2}=w_{0}w_{1}w_{0}w_{1}w_{0}$ where $w_{0}$ is the unique border of $w_{0}w_{1}w_{0}$. Since $w_{0}$ is the unique border of $w_{0}w_{1}w_{0}$, it is also the shortest border of $u_{2}$. Additionally, the next longest border of $u_{2}$ is $w_{0}w_{1}w_{0}$, which is overlapping. So $w_{0}$ is also the longest non-overlapping border of $u_{2}$. Thus $w_{2}=w_{2}^{\prime}$. ∎

Just based on this characterization, finding a recurrence for the number of words with a coinciding smallest and largest BP-factorization seems hard. So we turn to a different, related problem: counting the number of words with a unique border.

5.1 Unique borders

Harju and Nowotka [12] counted the number $B_{k}(n)$ of length-$n$ words over $\Sigma_{k}$ with a unique border, and the number $B_{k}(n,t)$ of length-$n$ words over $\Sigma_{k}$ with a length-$t$ unique border. However, through personal communication with the authors, a small error in one of the proofs leading up to their formula for $B_{k}(n,t)$ was discovered. Thus, the formula for $B_{k}(n,t)$ as stated in their paper is incorrect. In this section, we present the correct recurrence for the number of length-$n$ words with a length-$t$ unique border. We also show that the probability a length-$n$ word has a unique border tends to a constant. See A334600 in the OEIS [10] for the sequence $(B_{2}(n))_{n\geq 0}$.

Suppose $w$ is a word with a unique border $u$. Then $u$ must be unbordered, and $|u|$ must not exceed half the length of $w$. If either of these were not true, then $w$ would have more than one border. By combining these ideas, we get Theorem 9 and Theorem 10.

Theorem 9**.**

Let $n>t\geq 1$ be integers. Then the number of length-$n$ words with a unique length-$t$ border satisfies the recurrence

[TABLE]

Proof.

Let $w$ be a length-$n$ word with a unique length-$t$ border $u$. Since $u$ is the unique border of $w$, it is unbordered. Thus, we can write $w=uvu$ for some (possibly empty) word $v$. For $n<2t$, we have that $B_{k}(n,t)=0$ since $u$ is unbordered and thus cannot overlap itself in $w$.

Suppose $n\geq 2t$. Let $\overline{B_{k}}(n,t)$ denote the number of length-$n$ words that have a length-$t$ unbordered border and have a border of length $>t$. Clearly $B_{k}(n,t)=u_{t}k^{n-2t}-\overline{B_{k}}(n,t)$. Suppose $w$ has another border $u^{\prime}$ of length $>t$. Furthermore, suppose that there is no other border $u^{\prime\prime}$ with $|u|<|u^{\prime\prime}|<|u^{\prime}|$. Then $u$ is the unique border of $u^{\prime}$. Since $u$ is the shortest border, we have $|u|\leq n/2$. But we could possibly have $|u^{\prime}|>n/2$. The only possible way for $|u^{\prime}|$ to exceed $n/2$ is if $w=uv^{\prime}uv^{\prime}u$ for some (possibly empty) word $v$. But this is only possible if $n+t$ is even; otherwise we cannot place $u$ in the centre of $w$. When $n+t$ is odd, we compute $\overline{B_{k}}(n,t)$ by summing over all possibilities for $u^{\prime}$ (i.e., $2t\leq|u^{\prime}|\leq\lfloor n/2\rfloor$) and the middle part of $w$ (i.e., $v^{\prime\prime}$ where $w=u^{\prime}v^{\prime\prime}u^{\prime}$). This gives us the recurrence,

[TABLE]

When $n+t$ is even, we compute $\overline{B_{k}}(n,t)$ in the same fashion, except we also include the case where $|u^{\prime}|=(n+t)/2$. This gives us the recurrence,

[TABLE]

∎

Theorem 10**.**

Let $n\geq 2$ be an integer. Then the number of length-$n$ words with a unique border is

[TABLE]

5.2 Limiting values

We show that the probability that a random word of length $n$ has a unique border tends to a constant. Table 3 shows the behaviour of this probability as $k$ increases.

Let $P_{n,k}$ be the probability that a random word of length $n$ has a unique border. Then

[TABLE]

Lemma 11**.**

Let $k\geq 2$ and $n\geq 2t\geq 2$ be integers. Then

[TABLE]

Proof.

Let $w$ be a length-$n$ word. Suppose $w$ has a unique border of length $t$. Since $t\leq n/2$, we can write $w=uvu$ for some words $u$ and $v$ where $|u|=t$. But this means that $B_{k}(n,t)\leq k^{n-t}$, and the lemma follows. ∎

Theorem 12**.**

Let $k\geq 2$ be an integer. Then the limit $P_{k}=\lim\limits_{n\to\infty}P_{n,k}$ exists.

Proof.

Follows from the definition of $P_{n,k}$, Lemma 11, and the direct comparison test for convergence. ∎