Borders, Palindrome Prefixes, and Square Prefixes

TL;DR

This paper establishes bijections between words with no even palindromic prefixes and unbordered words, providing asymptotic counts and extending results to words with no square prefixes, including a conjecture resolution.

Contribution

It introduces explicit bijections linking palindromic prefix restrictions to unbordered words and solves a conjecture on words with no square prefix.

Findings

Number of words with no even palindromic prefix equals unbordered words.

Asymptotic enumeration for words with no palindromic prefixes.

Resolution of a 2013 conjecture on words with no square prefix.

Abstract

We show that the number of length-n words over a k-letter alphabet having no even palindromic prefix is the same as the number of length-n unbordered words, by constructing an explicit bijection between the two sets. A slightly different but analogous result holds for those words having no odd palindromic prefix. Using known results on borders, we get an asymptotic enumeration for the number of words having no even (resp., odd) palindromic prefix . We obtain an analogous result for words having no nontrivial palindromic prefix. Finally, we obtain similar results for words having no square prefix, thus proving a 2013 conjecture of Chaffin, Linderman, Sloane, and Wilks.

Tables

$n$	1	2	3	4	5	6	7	8	9	10	11	12
$t_2\left(n\right)$	2	4	4	8	12	24	40	80	148	296	568	1136
$v_2\left(n\right)=u_2\left(n\right)$	2	2	4	6	12	20	40	74	148	284	568	1116

$n$	1	2	3	4	5	6	7	8	9	10	11	12
$a_3\left(n\right)$	3	6	12	30	78	222	636	1878	5556	16590	49548	148422

$n$	1	2	3	4	5	6	7	8	9	10	11	12
$v_2\left(n\right)$	2	2	4	6	12	20	40	74	148	284	568	1116
$s_2\left(n\right)$	2	2	4	6	12	20	40	74	148	286	572	1124

$n$	1	2	3	4	5	6	7	8	9	10	11	12
$c_2\left(n\right)$	2	2	4	6	10	20	36	72	142	280	560	1114
$d_2\left(n\right)$	0	2	4	10	20	44	88	182	364	738	1476	2972

Equations

$$x\text{mathcyr}SHy=a_1{b}_1{a}_2{b}_2\cdots a_n{b}_n.$$

$$x=t{t}^R=(t_1\text{mathcyr}SH{t}_2)aa(t_1\text{mathcyr}SH{t}_2{)}^R=(t_1\text{mathcyr}SH{t}_2)aa(t_2^R\text{mathcyr}SH{t}_1^R)=(t_{1}a{t}_2^R)\text{mathcyr}SH(t_{2}a{t}_1^R).$$

$$x=(y_{1}a{y}_2)\text{mathcyr}SH(y_2^{R}a{y}_1^R)=(y_1\text{mathcyr}SH{y}_2^R)aa(y_2\text{mathcyr}SH{y}_1^R).$$

$$f(x):=(y\text{mathcyr}SH{z}^R)a$$

$$f(w)=f(u{v}_{1}a{v}_{2}u)=((u{v}_1)\text{mathcyr}SH(v_{2}u{)}^R)a=(u\text{mathcyr}SH{u}^R)(v_1\text{mathcyr}SH{v}_2^R)a,$$

$$f(w)=((y_1{y}_2)\text{mathcyr}SH(z_2^R{z}_1^R))a=(y_1\text{mathcyr}SH{z}_2^R)(y_2\text{mathcyr}SH{z}_1^R)a.$$

$$\{01000010,01001010,01010010,01011010,10100101,10101101,10110101,10111101\}.$$

$$\{00110000,00110001,00110010,00110011,11001100,11001101,11001110,11001111\}.$$

$$g(a_1{a}_2\cdots a_{2i}{a}_{2i+1})$$

$$=(a_1+a_2)(a_2+a_3)\cdots (a_{2i}+a_{2i+1})$$

$$=(a_1+a_2)(a_2+a_3)\cdots (a_{i-1}+a_i)(a_i+a_{i+1})(a_{i+1}+a_i)(a_i+a_{i-1})\cdots (a_3+a_2)(a_2+a_1),$$

$${\mathrm{opp}}_{k,S}(n)=k\cdot {\mathrm{epp}}_{k,S}(n-1)={\mathrm{epp}}_{k,S}(n),$$

$${\mathrm{opp}}_{k,S}(n)$$

$$=k\cdot {\mathrm{epp}}_{k,S}(n-1)\text{ (by Eq. eqrefopp1)},$$

$$\left(\begin{array}[]{ccccccc}1&2&3&4&5&6&7\\ 1&7&2&6&3&5&4\end{array}\right).$$

$$A_k(2n)$$

$$A_k(2n+1)$$

$$T_k(2n)$$

$$T_k(2n+1)$$

$$T_k(2n)=T_k(2n-2)-(k+1)T_k(n)k^{-n}.$$

$$T_k(2n)=\frac{k-1}{k}-(k+1)\sum _{i=2}^n{T}_k(i)k^{-i},$$

$$T_k(2n)=2-(k+1)\sum _{i=1}^n{T}_k(i)k^{-i}.$$

$${\rho }_k:=\underset{n\to \infty }{\lim }{T}_k(n)=2-(k+1)h(1/k),$$

$$h(X)(1-X)$$

$$=T_k(1)X+\left(\sum _{n\ge 1}(T_k(2n)-T_k(2n-1))X^{2n}\right)+\left(\sum _{n\ge 1}(T_k(2n+1)-T_k(2n))X^{2n+1}\right)$$

$$=X-\left(\sum _{n\ge 1}{T}_k(n)k^{-n}{X}^{2n}\right)-\left(\sum _{n\ge 1}{T}_k(n+1)k^{-n}{X}^{2n+1}\right)$$

$$=X-h(X^2/k)-\frac{k}{X}\left(f(X^2/k)-\frac{X^2}{k}\right)$$

$$=2X-\left(1+\frac{k}{X}\right)h(X^2/k),$$

$$h(X)=\frac{2X}{1-X}+\frac{X+k}{X(X-1)}h(X^2/k).$$

$$\left(\underset{n\to \infty }{\lim }\frac{{\prod }_{i=1}^n(k^{2^i}+1)}{k^{n+1}{\prod }_{i=1}^n(k^{2^i-1}-1)}\right)-2\sum _{n\ge 1}{k}^{2^{2n-1}-2n}(k^{2^{2n-1}-1}+1)\frac{{\prod }_{i=1}^{2n-2}(k^{2^i}+1)}{{\prod }_{i=1}^{2n}(k^{2^i-1}-1)}.$$

$$h(1/3)$$

$${\rho }_3$$

$$\sum _{i=1}^n{c}_k(i)k^{-2i}\le {\beta }_k\le \left(\sum _{i=1}^n{c}_k(i)k^{-2i}\right)+\sum _{i=n+1}^{\infty }{k}^{-i}=\left(\sum _{i=1}^n{c}_k(i)k^{-2i}\right)+(k^{-n})/(k-1).$$

Full text

1 Introduction

In this note, we work with finite words over a finite alphabet $\Sigma$. For reasons that will be clear later, we assume without loss of generality that $\Sigma=\Sigma_{k}=\{0,1,\ldots,k-1\}$ for some integer $k\geq 1$.

We index words starting at position $1$, so that $w[1]$ denotes the first symbol of $w$, and $w[i..j]$ is the factor beginning at position $i$ and ending at position $j$.

We let $w^{R}$ denote the reverse of a word; thus, for example, $({\tt drawer})^{R}=\tt reward$. A word $w$ is a palindrome if $w=w^{R}$; an example in English is the word radar. A palindrome is even if it is of even length, and odd otherwise. If a palindrome is of length $n$, then its order is defined to be $\lfloor n/2\rfloor$. A palindrome is trivial if it is of length $\leq 1$, and nontrivial otherwise. We are not interested in trivial palindromic prefixes in this paper, since every nonempty word has such prefixes of length [math] and $1$.

A word $w$ has an even palindromic prefix (resp., odd palindromic prefix) if there is some nonempty prefix $p$ (possibly equal to $w$) that is a palindrome of even (resp., odd) length. Thus, for example, the English word diffident has the even palindromic prefix diffid of order $3$, and the English word selfless has an odd palindromic prefix selfles of order $3$.

A border of a word $w$ is a word $u$, $0<|u|<|w|$, that is both a prefix and suffix of $w$. If a word has a border, we say it is bordered, and otherwise it is unbordered. For example, alfalfa is bordered, but chickpea is unbordered. Bordered and unbordered words have been studied for almost fifty years; see, for example, [12, 4].

Call a border $u$ of a word $w$ long if $|u|>|w|/2$ and short otherwise. If a word has a long border $u$, then by considering the overlap of the two occurrences of $u$, one as prefix and one as suffix, we see that $w$ also has a short border. Given a word $w$, its set of short border lengths is $\{1\leq i\leq|w|/2\ :\ w[1..i]\text{ is a border of }w\}$.

By explicit counting for small $n$, one quickly arrives at the conjecture that $u_{k}(n)$, the number of length-$n$ words over $\Sigma_{k}$ that are unbordered, equals $v_{k}(n)$, the number of length-$n$ words over $\Sigma_{k}$ having no even palindromic prefix. This seems to be true, despite the fact that the individual words being counted differ in the two cases. As an example consider $0011$, which has an even palindromic prefix $00$ but is unbordered. Similarly, if $t_{k}(n)$ denotes the number of length-$n$ words over $\Sigma_{k}$ having no nontrivial odd palindromic prefix, it is natural to conjecture that $t_{k}(n)=u_{k}(n)$ for $n$ odd, and $t_{k}(n)=ku_{k}(n-1)$ for $n$ even. The first few terms of the sequences $(t_{2}(n))$ and $(v_{2}(n))$ are given in the following table.

The sequence $t_{2}(n)$ is sequence A308528 in the On-Line Encyclopedia of Integer Sequences (OEIS) [9], and the sequence $u_{2}(n)$ is sequence A003000.

In fact, even more seems to be true: if $S$ is any set of positive integers, then the number of length-$n$ words $w$ for which $S$ specifies the lengths of all the short borders of $w$ is exactly the same as the number of length-$n$ words having even palindromic prefixes with orders given by $S$. A similar, but slightly different claim seems to hold for the odd palindromic prefixes. How can we explain this?

The obvious attempts at a bijection (e.g., map $uvu$ to $uu^{R}v$) don’t work, because (for example) $00100$ and $00010$ both map to $00001$. Nevertheless, there is a bijection, as we will see below, and this bijection provides even more information.

2 A bijection on $\Sigma^{n}$

The perfect shuffle of two words $x$ and $y$, both of length $n$, is written $x\,\mathcyr{SH}\,y$, and is defined as follows: if $x=a_{1}a_{2}\cdots a_{n}$ and $y=b_{1}b_{2}\cdots b_{n}$, then

[TABLE]

Thus, for example, ${\tt clip}\,\mathcyr{SH}\,{\tt aloe}$ = calliope.

Clearly $(x\,\mathcyr{SH}\,y)^{R}=y^{R}\,\mathcyr{SH}\,x^{R}$, a fact we use below.

Lemma 1**.**

Let $x$ be an even-length word, and (uniquely) write $x=y\,\mathcyr{SH}\,z$ for words $y,z$ with $|y|=|z|$. Then $x$ is a palindrome iff $z=y^{R}$.

Proof.

Suppose $x$ is palindrome. Then $x=tt^{R}$ for some word $t$.

By “unshuffling”, write $t$ as $(t_{1}\,\mathcyr{SH}\,t_{2})a$, for words $t_{1}$ and $t_{2}$, where $a$ is either empty or a single letter, depending on whether $|t|$ is even or odd. Then

[TABLE]

It follows that $y=t_{1}at_{2}^{R}$ and $z=t_{2}at_{1}^{R}$, and hence $z=y^{R}$.

Similarly, suppose $z=y^{R}$. Write $y=y_{1}ay_{2}$, where $y_{1},y_{2}$ are words of equal length, and $a$ is either empty or a single letter, depending on whether $|y|$ is even or odd. Then $z=y_{2}^{R}ay_{1}^{R}$. Hence

[TABLE]

Letting $t=(y_{1}\,\mathcyr{SH}\,y_{2}^{R})a$, we see that $x=tt^{R}$, and so $x$ is a palindrome. ∎

For a related result, see [10].

We now define a certain map from $\Sigma^{n}$ to $\Sigma^{n}$, as follows:

[TABLE]

if $x=yaz$ with $|y|=|z|$ and $a$ empty or a single letter (depending on whether $|x|$ is even or odd). Thus, for example, $f({\tt preserve})={\tt perverse}$ and $f({\tt cider})={\tt cried}$. Clearly this map is a bijection.

Theorem 2**.**

Let $w$ be a word and let $1\leq i\leq|w|/2$. Then $w$ has a border of length $i$ iff $f(w)$ has an even palindromic prefix of order $i$.

Roughly speaking, this theorem says that $f$ “maps borders to orders”.

Proof.

Suppose $w$ has a border of length $i$. Then $w=uvu$, where $|u|=i$. Write $v=v_{1}av_{2}$, where $|v_{1}|=|v_{2}|$ and $a$ is either empty, or a single letter, depending on whether $|v|$ is even or odd. Then

[TABLE]

which by Lemma 1 has a palindromic prefix of length $2i$ and order $i$.

Suppose $f(w)$ has an even palindromic prefix of order $i$. Write $w=yaz$, so that $f(w)=(y\,\mathcyr{SH}\,z^{R})a$. Write $y=y_{1}y_{2}$ and $z=z_{1}z_{2}$ such that $|y_{1}|=|z_{2}|=i$. Now

[TABLE]

It follows that $y_{1}\,\mathcyr{SH}\,z_{2}^{R}$ is a palindrome and thus $z_{2}=y_{1}$ by Lemma 1. Hence $w=yaz=y_{1}y_{2}az_{1}z_{2}=y_{1}y_{2}az_{1}y_{1}$ has a length-$i$ border, namely $y_{1}$. ∎

Corollary 3**.**

Let $S\subseteq\{1,\ldots,\lfloor n/2\rfloor\}$. Then the number of length-$n$ words whose short borders are exactly those in $S$ equals the number of length-$n$ words whose even palindromic prefixes are of orders exactly those in $S$.

In particular, this holds for $S=\emptyset$, so the equality $u_{k}(n)=v_{k}(n)$ holds for all $k\geq 1$ and $n\geq 1$: the number of length-$n$ words that are unbordered is the same as the number of length-$n$ words having no even palindromic prefix.

Proof.

As we have seen in Theorem 2, the map $f$ is a bijection from the the first set to the second. ∎

Example 4**.**

As an example, consider the length-$8$ binary words with short borders of length $1$ and $3$ only. There are 8 of them:

[TABLE]

By applying the map $f$ to each word, we get the length-$8$ binary words having even palindromic prefixes of orders $1$ and $3$ only:

[TABLE]

Let $\operatorname{epp}_{k,S}(n)$ denote the number of length-$n$ words over a $k$-letter alphabet having even palindromic prefixes of order $i$ for each $i\in S$, and no other orders.

Proposition 5**.**

We have $\operatorname{epp}_{k,S}(n+1)=k\cdot\operatorname{epp}_{k,S}(n)$ for $n$ even.

Proof.

Let $n$ be even. Let $w$ be a word over a $k$-letter alphabet with even palindromic prefix orders given by $S$, and let $a$ be a single letter. Then clearly $wa$ has exactly the same palindromic prefixes as $w$. Since $a$ is arbitrary, the result follows. ∎

3 Odd palindromic prefixes

Let $S$ be any subset of $\{1,2,\ldots,\lfloor n/2\rfloor\}$. Let $\operatorname{opp}_{k,S}(n)$ denote the number of length-$n$ words over a $k$-letter alphabet having odd palindromic prefixes of order $i$ for each $i\in S$, and no others. Notice that we are not concerned here with those words having the trivial odd palindromic prefix of a single letter.

Proposition 6**.**

We have $\operatorname{opp}_{k,S}(n+1)=k\cdot\operatorname{opp}_{k,S}(n)$ for $n$ odd.

Proof.

Exactly like the proof of Proposition 5. ∎

Theorem 7**.**

We have

• (a)

$\operatorname{opp}_{k,S}(n)=\operatorname{epp}_{k,S}(n)$* for $n$ odd; and*

• (b)

$\operatorname{opp}_{k,S}(n)=k\cdot\operatorname{epp}_{k,S}(n-1)$* for $n$ even.*

Proof.

We begin by proving $\operatorname{opp}_{k,S}(n)=k\cdot\operatorname{epp}_{k,S}(n-1)$ for $n$ odd. We do this by creating a $k$ to $1$ map from the length-$n$ words with odd palindromic prefix orders given by $S$ to the length-$(n-1)$ words with even palindromic prefix orders given by $S$.

Here is the map. Let $w=a_{1}a_{2}\cdots a_{n}$ be a word of odd length, and define $g(w)=(a_{1}+a_{2})(a_{2}+a_{3})\cdots(a_{n-1}+a_{n})$, where the addition is performed modulo $k$. We claim that this is a $k$ to $1$ map, and furthermore, it maps words with odd palindromic prefix orders given by $S$ to words with even palindromic prefix orders also given by $S$.

To see the first claim, observe that if both $g(w)$ and $a_{1}$ are given, then we can uniquely reconstruct $w$. Since $a_{1}$ is arbitrary, this gives a $k$ to $1$ map.

To see the second claim, suppose $w=a_{1}a_{2}\cdots a_{n}$ has an odd palindromic prefix of order $i$. Then $a_{2i+2-j}=a_{j}$ for $1\leq j\leq i+1$. Hence, applying the map $g$ to a prefix of $w$ we get

[TABLE]

which is clearly an even palindrome of order $i$.

On the other hand, if $(a_{1}+a_{2})(a_{2}+a_{3})\cdots(a_{2i-1}+a_{2i})(a_{2i}+a_{2i+1})$ is a palindrome, then by examining the two elements in the middle, we get $a_{i}+a_{i+1}\equiv a_{i+1}+a_{i+2}\ \mbox{\rm(mod}\ k\mbox{\rm)}$, which forces $a_{i}=a_{i+2}$. Continuing from the middle out to the end, we successively obtain $a_{i-1}=a_{i+3}$, …, $a_{1}=a_{2i+1}$, which shows that $w$ starts with an odd palindrome of order $i$.

Hence for $n$ odd we get

[TABLE]

where we have used Proposition 5.

For $n$ even we get

[TABLE]

which completes the proof. ∎

Corollary 8**.**

Consider words over a $k$-letter alphabet.

For $n$ odd, we have $t_{k}(n)=u_{k}(n)$; that is, the number of length-$n$ words having no nontrivial odd palindromic prefix is the same as the number of length-$n$ words that are unbordered.

For $n$ even, we have $t_{k}(n)=ku_{k}(n-1)$; that is, the number of length-$n$ words having no nontrivial odd palindromic prefix is $k$ times the number of length-$(n-1)$ unbordered words.

Remark 9*.*

It is seductive, but wrong, to think that the map $g$ also maps even-length palindromic prefixes in a $k$ to $1$ manner to odd-length palindromic prefixes, but this is not true (consider what happens to the center letter).

4 An application

As an application of our results we can (for example) determine the asymptotic fraction of length-$n$ words having no nontrivial even palindromic prefix (resp., having no nontrivial odd palindromic prefix).

Corollary 10**.**

For all $k\geq 2$ there is a constant $\gamma_{k}>0$ such that the number of length-$n$ words having no nontrivial even palindromic prefix (resp., having no nontrivial odd palindromic prefix) is asymptotically equal to $\gamma_{k}\cdot k^{n}$.

Proof.

Follows immediately from the same result for unbordered words; see [8, 1, 5]. For related results, see [11]. ∎

5 Interlude: the permutation defined by $f$

The map $f$ defined in Section 2 can be considered as a permutation on $a_{1}a_{2}\cdots a_{n}$. In this case, we write it as $f_{n}$. For example, if $n=7$, the resulting permutation $f_{n}$ is

[TABLE]

This is an interesting permutation that has been well-studied in the context of card-shuffling, where it is called the milk shuffle. A classic result about the milk shuffle is the following [6]:

Theorem 11**.**

The order of the permutation $f_{n}$ is the least $m$ such that $2^{m}=\pm 1$ (mod $2n-1$).

This is sequence A003558 in the OEIS.

6 No palindromic prefix

In this section we consider the words having no nontrivial palindromic prefix. (Recall that a palindrome is trivial if it is of length $\leq 1$.) This is only of interest for alphabet size $k\geq 3$, for if $k=2$, the only such words are of the form $01^{i}$ and $10^{i}$.

Let $A_{k}(n)$ denote the number of such words over a $k$-letter alphabet. We use the technique of [1, 5] to show that $A_{k}(n)\sim\rho_{k}k^{n}$ for a constant $\rho_{k}$ and large $n$. First we need a lemma, which can essentially be found in (for example) [2, Prop. 6].

Lemma 12**.**

Let $w$ be a palindrome and let $p$ be a proper palindromic prefix of $w$. If $|p|>|w|/2$, then $w$ also has a nonempty palindromic prefix of length $<|w|/2$.

Proof.

If $p$ is a prefix of $w$, then $p^{R}$ is a suffix of $w^{R}$. Since both $p$ and $w$ are palindromes, this means $p$ is a suffix of $w$. Hence there exist nonempty words $y,z$ such that $w=py=zp$. By the Lyndon-Schützenberger theorem [7] there exist $u,v$ with $u$ nonempty, and an integer $e\geq 0$ such that $z=uv$, $p=(uv)^{e}u$, and $y=vu$. Since $|p|>|w|/2$, it follows that $|u|\leq|y|<|w|/2<|p|$. Since $p$ is a palindrome, we have $w=zp=zp^{R}$. Since $w$ is a palindrome, we have $py=w=w^{R}=pz^{R}$. So $vu=y=z^{R}=v^{R}u^{R}$, and so $u=u^{R}$. Thus $u$ is a nonempty palindromic prefix of $z$, which is a prefix of $w$, and $|u|<|w|/2$. ∎

Lemma 13**.**

Let $w,a$ be words with $|a|\leq 1$. Then $wa$ has a nontrivial palindromic prefix iff $waw^{R}$ has a nontrivial proper palindromic prefix.

Proof.

One direction is trivial.

For the other direction, let the shortest nontrivial proper palindromic prefix of $s:=waw^{R}$ be $q$. If $|q|\leq|wa|$, then $q$ is a prefix of $wa$ as desired. Otherwise we have $|q|>|wa|\geq|s|/2$. Then by Lemma 12, the word $s$ also has a nontrivial palindromic prefix of length $<|s|/2$, contradicting the definition of $q$. ∎

Proposition 14**.**

For $n\geq 1$ we have

[TABLE]

Proof.

Consider the words of length $2n-1$ having no nontrivial palindromic prefix. By appending a new letter, we get $kA_{k}(2n-1)$ words. However, some of these words can be palindromes of length $2n$, and we do not want to count these. By Lemma 13, the number of length-$2n$ palindromes having no proper palindromic prefix is $A_{k}(n)$. This gives (2).

A similar argument works to prove (3). ∎

For $k=3$ the corresponding sequence is given below and is sequence A252696 in the OEIS:

Now define $T_{k}(n)$ by $T_{k}(n)=A_{k}(n)k^{-n}$. From (2) and (3) we get

[TABLE]

It now follows that

[TABLE]

By telescoping cancellation applied to (4), we now get

[TABLE]

or

[TABLE]

Next, define $h(X)=\sum_{n\geq 1}T_{k}(n)X^{n}$. Since $0\leq T_{k}(n)\leq 1$ for all $n$, it follows that the series defining $h(X)$ is convergent for $|X|<1$. Then

[TABLE]

where $\rho_{k}$ is the limiting frequency of words having no nontrivial palindromic prefix.

Using (4), we now get

[TABLE]

and so we get a functional equation for $h(X)$:

[TABLE]

By iterating this functional equation, and using the fact that $h(\epsilon)\sim\epsilon$ for small real $\epsilon$, we get an expression for $h(1/k)$:

[TABLE]

This is very rapidly converging; for $k=3$ only $6$ terms are enough to get 60 decimal places of $h(1/k)$:

[TABLE]

7 Square prefixes

It is natural to conjecture that our bijections connecting words with no border and no even palindromic prefix might also apply to words having no square prefix. However, this is not the case. Let $s_{k}(n)$ denote the number of length-$n$ words over $\Sigma_{k}$ having no square prefix. When $k=2$, for example, the two sequences $s_{k}(n)$ and $v_{k}(n)$ differ for the first time at $n=10$, as the following table indicates.

The sequence $s_{2}(n)$ is sequence A122536 in the OEIS.

Chaffin, Linderman, Sloane, and Wilks [3, §3.7] conjectured that $s_{2}(n)\sim\alpha_{2}\cdot 2^{n}$ for a constant $\alpha_{2}\doteq 0.27$. In this section we prove this conjecture in more generality.

Theorem 15**.**

The limit $\lim_{n\rightarrow\infty}s_{k}(n)/k^{n}$ exists and equals a constant $\alpha_{k}$ with $\alpha_{k}>1-1/(k-1)$.

Proof.

Let $d_{k}(n)=k^{n}-s_{k}(n)$ be the number of length-$n$ words over $\Sigma_{k}$ having a nonempty square prefix, and let $c_{k}(n)$ be the number of squares of length $2n$ over $\Sigma_{k}$ having no nonempty proper square prefix. Hence $c_{k}(1)=k$ and $c_{k}(2)=k(k-1)$.

The first few values of $c_{2}(n)$ and $d_{2}(n)$ are given in the following table.

The sequence $c_{2}(n)$ is sequence A216958 in the OEIS, and the sequence $d_{2}(n)$ is sequence A121880.

Let $w$ be a word of length $n$. Either its shortest square prefix is of length 2 (and there are $c_{k}(1)k^{n-2}$ such words), or of length 4 (and there are $c_{k}(2)k^{n-4}$ such words), and so forth.

So $d_{k}(n)$, the number of words of length $n$ having a nonempty square prefix, is exactly $\sum_{2i\leq n}c_{k}(i)k^{n-2i}$. Hence $d_{k}(n)/k^{n}=\sum_{2i\leq n}c_{k}(i)k^{-2i}$. Thus $\lim_{n\rightarrow\infty}d_{k}(n)/k^{n}$ exists iff the infinite sum $\sum_{i=1}^{\infty}c_{k}(i)k^{-2i}$ converges. But, since $c_{k}(i)\leq k^{i}$, this sum converges to some constant $\beta_{k}<1/(k-1)$, by comparison with the sum $\sum_{i=1}^{\infty}k^{-i}=1/(k-1)$. It follows that $s_{k}(n)=k^{n}-d_{k}(n)\sim(1-\beta_{k})k^{n}$. Letting $\alpha_{k}=1-\beta_{k}$, the result follows. ∎

To estimate the value of $\beta_{k}$ (and hence $\alpha_{k}$) we use the inequalities

[TABLE]

For example, if we take $n=20$, we get $\beta_{2}\in(0.7299563,0.7299574)$ and hence $\alpha_{2}\in(0.2700426,0.2700437)$. This can be compared to the analogous constant $\gamma_{2}\doteq 0.2677868$ for even palindromes.

Acknowledgments

We are grateful to the referees for their helpful comments.