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\begin{center}
\vskip 1cm{\LARGE\bf Concerning Kurosaki's Squarefree Word
}
\vskip 1cm
\large
Serina Camungol and Narad Rampersad\\
Department of Mathematics and Statistics \\
University of Winnipeg\\
515 Portage Ave. \\
Winnipeg, Manitoba R3B 2E9 \\
Canada\\
\href{mailto:serina.camungol@gmailcom}{\tt serina.camungol@gmail.com} \\
\href{mailto:narad.rampersad@gmail.com}{\tt narad.rampersad@gmail.com} 
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\begin{abstract}
In 2008, Kurosaki gave a new construction of a (bi-)infinite
squarefree word over three letters.  We show that in fact Kurosaki's
word avoids $(7/4)^+$-powers, which, as shown by Dejean, is optimal over
a $3$-letter alphabet.
\end{abstract}

\section{Introduction}
In 1906, Thue \cite{Thu06} constructed an infinite word over a
$3$-letter alphabet that avoided \emph{squares}, i.e., repetitions of
the form $xx$.  In 1935, Arshon \cite{Ars35,Ars37} independently
rediscovered this and gave another construction of an infinite
squarefree word.  In 1972, Dejean \cite{Dej72} generalized Thue's
notion of repetition to \emph{fractional powers}.  She constructed an
infinite ternary word avoiding $(7/4)^+$-powers, i.e., repetitions of
the form $x^\alpha$, where $\alpha &gt; 7/4$.  Furthermore, she showed
that this was optimal over a ternary alphabet.  In 2001, Klepinin and
Sukhanov \cite{KS01} examined Arshon's word in more depth and showed
that it also avoided $(7/4)^+$-powers (Krieger \cite{Kri10} extended
this result by analyzing Arshon words over larger alphabets.)  In 2008,
Kurosaki \cite{Kur08} gave a new construction of a squarefree word.  In
this short note, we show that Kurosaki's word also avoids
$(7/4)^+$-powers.

Let us now recall some basic definitions.  Let $\Sigma$ be a finite
alphabet and let $x$ be a word over $\Sigma$.  We denote the length of
$x$ by $|x|$.  Write $x = x_1x_2\cdots x_n$, where each $x_i \in
\Sigma$.  The word $x$ has a \emph{period} $p$ if $x_i = x_{i+p}$ for
all $i$.  The \emph{exponent} of $x$ is the quantity $|x|/p$, where
$p$ is the least period of $x$.  If $x$ has exponent $\alpha$, we say
that $x$ is an \emph{$\alpha$-power}.  A $2$-power is also called a
\emph{square}.  A word $y$ (finite or infinite) \emph{avoids
  $\alpha$-powers} (resp.~\emph{avoids $\alpha^+$-powers}) if every
factor of $y$ has exponent less than (resp.\ at most) $\alpha$.

\section{Kurosaki's word avoids $(7/4)^+$-powers}

We now give the definition of Kurosaki's word.  We will use the notation from his paper.
We define functions $\sigma, \rho : \{1,2,3\}^* \to \{1,2,3\}^*$ as
follows:  If $a$ is a word over $\{1,2,3\}$, then
\begin{itemize}
  \item $\sigma(a)$ is the word obtained by exchanging 1's and 2's in $a$;
  \item $\rho(a)$ is the word obtained by exchanging 2's and 3's in $a$.
\end{itemize}
We define the function $\varphi : \{ 1, 2 ,3 \}^{*} \rightarrow \{ 1,2,3 \}^{*}$ as follows:
$$ \varphi(a) = \sigma(a)a\rho(a) $$

Consider the finite words $\varphi^n(2)$, for $n =1,2,\ldots$.  These
words do not converge to a one-sided infinite word as $n$ tends to
infinity; however, since $\varphi^n(2)$ appears as the middle third of
$\varphi^{n+1}(2)$ for all $n$, we see that this sequence of words
defines, in the limit, a two-sided, or bi-infinite, word.  It is this
bi-infinite word
\[
\cdots 123213231\,213123132\,312132123 \cdots
\]
that we refer to as \emph{Kurosaki's word} (see
\seqnum{A217522} in Sloane's {\it Encyclopedia}).

Following Kurosaki, we write $\varphi^n(2) = \{ a^{n}_{i}
\}_{i=0}^{3^{n}-1}$.  Proofs of Theorem~\ref{squarefree},
Lemma~\ref{permutation} and Lemma~\ref{preimage} can be found in
\cite{Kur08}.

\begin{theorem}
\label{squarefree}
$\{ a^{n}_{i} \}_{i=0}^{3^{n}-1}$ is squarefree for all $n \geq 1$.
\end{theorem}

\begin{lemma}
\label{permutation}
Each triple of the form $a^{n}_{3k}a^{n}_{3k+1}a^{n}_{3k+2}$, $k = 0,
1, \ldots, 3^{n-1}-1$, is some permutation of $123$.
\end{lemma}

From now on, when we use the term \emph{triple} we mean a permutation
of $123$ occurring at a position congruent to $0$ modulo $3$ in
$\varphi^n(2)$.  Next we define the operator $f$, which extracts the
middle term of each triple: $f(\varphi^{n}(2)) = \{
a^{n}_{3i+1}\}^{3^{n-1}-1}_{i=0}$.

\begin{lemma}
\label{preimage}
The sequence $f(\varphi^{n}(2)) = \{ a^{n}_{3i+1} \}^{3^{n-1}-1}_{i=0}$  is equal to $\varphi^{n-1}(2) = \{ a^{n-1}_{i} \}^{3^{n-1}-1}_{i=0}$.
\end{lemma}

\begin{lemma}
\label{4tuple}
$\{ a^{n}_{i} \}_{i=0}^{3^{n}-1}$ has no factors of the form
$a_{3q}a_{3q+1} \cdots a_{3q+11}$ such that $a_{3q}=a_{3q+3}=a_{3q+6}=a_{3q+9}$.
\end{lemma}

\begin{proof}
For convenience we omit the superscript $n$ from the terms
$a^n_i$.  Let $a_{3q}a_{3q+1}\cdots a_{3q+11}$ be a factor of $\{
a^{n}_{i} \}_{i=0}^{3^{n}-1}$.  As $a_{3q}a_{3q+1}a_{3q+2}$,
$a_{3q+3}a_{3q+4}a_{3q+5}$, $a_{3q+6}a_{3q+7}a_{3q+8}$ and
$a_{3q+9}a_{3q+10}a_{3q+11}$ are permutations of 123
(Lemma~\ref{permutation}), it is not difficult to see that having
$a_{3q}=a_{3q+3}=a_{3q+6}=a_{3q+9}$ creates a square of length 6
or 12, in contradiction to Theorem~\ref{squarefree}.
\end{proof}

We now establish the main result.

\begin{theorem}
\label{7/4powerfree}
$\{ a^n_{i} \}_{i=0}^{3^{n}-1}$ is $(7/4)^{+}$-power-free for all $n
\geq 1$.
\end{theorem}

\begin{proof}
We will prove that $\varphi^{n}(2) = \{ a^n_{i} \}_{i=0}^{3^{n}-1}$
does not contain a $(7/4)^{+}$-power of the form $xyx$, where $|x| +
|y| = s$, $|x| = r$ and $r/s&gt;3/4$.  We prove this by induction on
$n$, considering various cases for the possible lengths of $s$.  We
first check the base cases $n = 1, 2, 3$.  Clearly
$\varphi^{1}(2)=123$ and $\varphi^{2}(2)=213 123 132$ avoid
$(7/4)^{+}$-powers, and by observation $\varphi^{3}(2)=123 213 231 213
123 132 312 132 123$ avoids $(7/4)^{+}$-powers.  So we assume
$\varphi^{n}(2)$ avoids $(7/4)^{+}$-powers and we show that
$\varphi^{n+1}(2)$ is $(7/4)^{+}$-power-free.

\begin{case}
$s \equiv 0\pmod{3}$
\end{case}
Suppose towards a contradiction that $\varphi^{n+1}(2)$ contains a
$(7/4)^{+}$-power of the form $xyx$, where $|x| + |y| = s$, $|x| = r$,
$r/s&gt;3/4$ and $s \equiv 0\pmod 3$. Let us write $$xyx = a_{p}a_{p+1}
\cdots a_{p+s-1}a_{p+s}a_{p+s+1} \cdots a_{p+s+r-1},$$ where $a_{p+i}
= a_{p+s+i}$ for $0 \leq i \leq r-1$.  Again, for convenience we omit
the superscript $n$ from the terms $a^n_i$.  If $p \equiv r \equiv 0
\pmod 3$, then $f(\varphi^{n+1}(2))$ contains a $(7/4)^{+}$-power and by
Lemma~\ref{preimage} the word $\varphi^{n}(2)$ contains a
$(7/4)^{+}$-power, a contradiction to our assumption.

We will show that the repetition $xyx$ can always be extended to the
left to give a $(7/4)^+$-power that has the same period $s$, but now starts
at a position congruent to $0$ modulo $3$.  A similar argument shows
that the repetition can also be extended to the right so that
its ending position is congruent to $2$ modulo $3$, and therefore that
the argument of the previous paragraph can be applied to obtain a
contradiction.

If $p \equiv 1 \pmod3$, let $p=3q+1$. The first two terms of $x$
form a suffix of the triple $a_{3q}a_{3q+1}a_{3q+2}$.
Lemma~\ref{permutation} allows us to uniquely determine $a_{3q}$ from
$a_{3q+1}$ and $a_{3q+2}$.  Similarly, the values of $a_{3q+s+1}$ and
$a_{3q+s+2}$ uniquely determine the value of $a_{3q+s}$.  Since
$a_{3q+1}a_{3q+2} = a_{3q+s+1}a_{3q+s+2}$, we have $a_{3q} =
a_{3q+s}$.  Thus $a_{p-1}a_p \cdots a_{p+s+r-1}$ is a $(7/4)^+$-power
with period $s$ starting at position $3q$.

If $p \equiv 2 \pmod3$, let $p=3q+2$. In order to determine the two
elements preceding $a_{3q+2}$, we look at the triple succeeding
$a_{3q+2}$, namely $a_{3q+3}a_{3q+4}a_{3q+5}$.  By
Lemma~\ref{permutation}, $a_{3q}a_{3q+1}a_{3q+2}$ is some permutation
of $a_{3q+3}a_{3q+4}a_{3q+5}.$ We must have either $a_{3q+2} =
a_{3q+4}$ or $a_{3q+2} = a_{3q+5}$ (for otherwise we would have a
square, contrary to Theorem~\ref{squarefree}).  Without loss of
generality, suppose $a_{3q+2} = a_{3q+4}$; we will determine the
values of $a_{3q}$ and $a_{3q+1}$.  If $a_{3q+1} = a_{3q+3}$, then
$a_{3q+1}a_{3q+2}a_{3q+3}a_{3q+4}$ is a square, contrary to
Theorem~\ref{squarefree}, so it must be that $a_{3q+1} = a_{3q+5}$,
which implies $a_{3q} = a_{3q+3}$.  In other words, $a_{3q}a_{3q+1}$
is uniquely determined by $a_{3q+2}a_{3q+3}a_{3q+4}a_{3q+5}$.
Similarly, $a_{3q+s}a_{3q+s+1}$ is uniquely determined by
$a_{3q+s+2}a_{3q+s+3}a_{3q+s+4}a_{3q+s+5}$.  We deduce that
$a_{p-2}a_p \cdots a_{p+s+r-1}$ is a $(7/4)^+$-power with period $s$
starting at position $3q$.

\begin{case}
$s &lt; 14$ and $s\not\equiv 0\pmod3$
\end{case}
As $\varphi^{n}(2)$ is $(7/4)^{+}$-power-free by the induction
hypothesis, it is not hard to see that $\sigma(\varphi^{n}(2))$ and
$\rho(\varphi^{n}(2))$ are also $(7/4)^{+}$-power-free.  Therefore, if
$\varphi^{n+1}(2)$ contains a $(7/4)^{+}$-power, it must be a factor of
$\sigma(\varphi^{n}(2))\varphi^{n}(2)$ or
$\varphi^{n}(2)\rho(\varphi^{n}(2))$.  Note that here we only consider
$n \geq 3$, as the smaller values of $n$ are dealt with in our base
cases, and $|\varphi^{n}(2)| \geq 27$, so if $\varphi^{n+1}(2)$
contains a $(7/4)^{+}$-power it cannot extend from
$\sigma(\varphi^{n}(2))$ over into $\rho(\varphi^{n}(2))$.
Furthermore, if $s&lt;14$, then we may assume that the entire repetition
$xyx$ has length at most $\lceil 13(7/4) \rceil = 23$.

Observe that for even $n$, $\varphi^{n}(2)$ has the form
$$213 123 132 123 213 231 321 231 \cdots 312 321 312 132 123 213 123
132,$$
so 
$$\sigma(\varphi^{n}(2))\varphi^{n}(2) = \cdots 321 312 321 231 213
123 213 231 \cdot 213 123 132 123 213 231 321 231 \cdots$$
and
$$\varphi^{n}(2)\rho(\varphi^{n}(2)) = \cdots 312 321 312 132 123 213
123 132 \cdot 312 132 123 132 312 321 231 321\cdots,$$
 where the single dots represent the boundaries between
 $\sigma(\varphi^{n}(2))$, $\varphi^n(2)$ and $\rho(\varphi^{n}(2))$.

Similarly, for odd $n$, $\varphi^{n}(2)$ has the form
$$123 213 231 213 123 132 312 132 \cdots 213 231 213 123 132 312 132
123,$$
so 
$$\sigma(\varphi^{n}(2))\varphi^{n}(2) =  \cdots 123 132 123 213 231
321 231 213 \cdot 123 213 231 213 123 132 312 132 \cdots$$
and
$$\varphi^{n}(2)\rho(\varphi^{n}(2)) = \cdots 213 231 213 123 132 312 132 123 \cdot 132 312 321 312 132 123 213 123 \cdots.$$

It suffices to check the factors of length at most $23$ crossing the
indicated boundaries; by inspection we are able to conclude that
$\varphi^{n+1}(2)$ is $(7/4)^{+}$-power-free for $s &lt; 14$.

\begin{case}
$s \geq 14$ and $s\not\equiv 0\pmod3$
\end{case}
This case is further divided into six subcases.  We will prove this
case for $p \equiv 1\pmod 3$, where $p$ is the starting index of our
$(7/4)^{+}$-power and $s \equiv 2\pmod 3$.  The proofs for the other five
cases are similar, all involving use of Lemma~\ref{4tuple}.  Suppose
towards a contradiction that $\varphi^{n+1}(2)$ contains a
$(7/4)^{+}$-power of the form $xyx$, where $|x| + |y| = s$, $|x| = r$
and $s \equiv 2\pmod 3$. We can write this as $a_{3q+1}a_{3q+2} \cdots
a_{3q+s}a_{3q+s+1}a_{3q+s+2} \cdots a_{3q+s+r}$, where $a_{3q+i} =
a_{3q+s+i}$ for $1 \leq i \leq r$.  Now consider the triples
$a_{3q}a_{3q+1}a_{3q+2}$ and $a_{3q+s+1}a_{3q+s+2}a_{3q+s+3}$.  As
$a_{3q+1} = a_{3q+s+1}$ and $a_{3q+2} = a_{3q+s+2}$ and both
$a_{3q}a_{3q+1}a_{3q+2}$ and $a_{3q+s+1}a_{3q+s+2}a_{3q+s+3}$ are
permutations of $123$, it must be that $a_{3q} = a_{3q+s+3}$.
Continuing in this manner we see that $a_{3q} = a_{3(q+1)} = \cdots =
a_{3(q+j)}$, where $j \leq r$ is maximal and $j \equiv 0\pmod 3$.
This is easy to see using the diagram below, where the lines signify
that the terms are equivalent.
\begin{equation*}
\xymatrix@R20pt@C0pt{ a_{3q} \ar@{-}[drrr] &amp; a_{3q+1} \ar@{-}[d] &amp; a_{3q+2} \ar@{-}[d] &amp; a_{3q+3}\ar@{-}[d]\ar@{-}[drrr] &amp; a_{3q+4} \ar@{-}[d] &amp; a_{3q+5} \ar@{-}[d] &amp; a_{3q+6}\ar@{-}[d] &amp; \cdots &amp; a_{3q+r-2}\ar@{-}[d]&amp; a_{3q+r-1}\ar@{-}[d] &amp; a_{3q+r}\ar@{-}[d]\\
&amp; a_{3q+s+1} &amp;  a_{3q+s+2} &amp; a_{3q+s+3} &amp; a_{3q+s+4} &amp; a_{3q+s+5} &amp; a_{3q+s+6} &amp; \cdots &amp; a_{3q+s+r-2} &amp; a_{3q+s+r-1} &amp; a_{3q+s+r}}
\end{equation*}
However, as $s \geq 14$ it follows that $r \geq 11$ and that $j \geq
3$, so we always have $a_{3q} = a_{3(q+1)} = a_{3(q+2)} = a_{3(q+3)}$,
in contradiction to Lemma~\ref{4tuple}.

\end{proof}

\section{Conclusion}

Kurosaki also gives an alternate definition of his sequence.  In this
other formulation, the $n$-th term of the sequence is computed by a
finite automaton that reads the \emph{balanced ternary} representation
of $n$.  This representation is a base-$3$ representation with digit
set $\{-1,0,1\}$.  Kurosaki says that his sequence can be viewed as an
analogue of the classical Thue--Morse sequence for the balanced ternary
numeration system.  His sequence certainly seems to have many
interesting properties and perhaps should be better known.

We would also like to mention that Daniel Go\v{c} has verified the
main result of this paper by a computer calculation using the
techniques described by Go\v{c}, Henshall, and Shallit \cite{GHS12}.

\begin{thebibliography}{11}
\bibitem{Ars35}
S. E. Arshon, A proof of the existence of infinite asymmetric
sequences on $n$ symbols, \textit{Matematicheskoe Prosveshchenie
  (Mathematical Education)} {\bf 2} (1935), 24--33 (in
Russian). Available electronically at
\url{http://ilib.mccme.ru/djvu/mp1/mp1-2.htm}~.

\bibitem{Ars37}
S. E. Arshon, A proof of the existence of infinite asymmetric
sequences on $n$ symbols, \textit{Mat. Sb.} {\bf 2} (1937), 769--779
(in Russian).

\bibitem{Dej72}
F. Dejean, Sur un th\'eor\`eme de Thue, \textit{J. Combin. Theory Ser. A}
{\bf 13} (1972), 90--99.

\bibitem{GHS12}
D. Go\v{c}, D. Henshall, and J. Shallit, Automatic theorem-proving in
combinatorics on words,  in N. Moreira and
R. Reis eds., \textit{Conference on Implementation and Application of
  Automata: Proc. CIAA 2012}, Lect. Notes in Comp. Sci., Vol. 7381,
Springer, 2012, pp.~180--191.

\bibitem{KS01}
A. V. Klepinin and E. V. Sukhanov, On combinatorial properties of the
Arshon sequence, \textit{Disc. Appl. Math.} {\bf 114} (2001), 155--169.

\bibitem{Kri10}
D. Krieger, The critical exponent of the Arshon words, \textit{RAIRO
  Theor. Inform. Appl.} {\bf 44} (2010), 139--150.

\bibitem{Kur08}
T. Kurosaki, Direct definition of a ternary infinite square-free
sequence, \textit{Inform. Process. Lett.} {\bf 106} (2008), 175--179.

\bibitem{Thu06}
A. Thue, \"Uber unendliche Zeichenreihen, \textit{Kra. Vidensk. Selsk.
Skrifter. I. Mat. Nat. Kl.} {\bf 7} (1906), 1--22.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary
68R15.

\noindent \emph{Keywords:} repetition, squarefree word, $(7/4)^+$-power.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A217522}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  August 9 2013;
revised version received October 3 2013.
Published in {\it Journal of Integer Sequences}, October 12 2013.

\bigskip
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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