A [square-free word](http://en.wikipedia.org/wiki/Square-free_word)
is a string of symbols (a "word") that avoids the pattern $XX$, where $X$ is any
consecutive sequence of symbols in the string.
For alphabets of two symbols, the longest square-free word has length $3$.
But for alphabets of three symbols, there are infinitely long square-free words,
due to Thue.


Define a *three-halves-free word* as one that avoids the pattern $XYX$, where $|X|=|Y|$.
Another view is that these words avoid $Z(\frac{1}{2}Z)$, i.e., $Z=XY$ with $X$ and $Y$ the
same length. So $Z$ has even length, and we want to avoid immediately
following with the first half of $Z$.


For an alphabet of two symbols, $0$ & $1$, here are the three-halves-free words
of length $5$:
\begin{eqnarray\*}
&(0, 1, 1, 0, 0)\\
&(0, 0, 1, 1, 0)\\
&(1, 1, 0, 0, 1)\\
&(1, 0, 0, 1, 1)
\end{eqnarray\*}
All the $28$ other length-$5$ words fail. 
E.g., $(0,1,0,1,0)$ fails because $(0,1,0)$ matches with $XYX=010$.
But there are no three-halves-free words of length $\ge 6$.
For example, extending $(0, 1, 1, 0, 0)$ with $0$ gives $(0, 1, 1, 0, 0, 0)$, matching
$XYX=000$, and extending with $1$ gives $(0, 1, 1, 0, 0, 1)$, matching with $X=01$.



> 
> ***Q***. Are there infinitely long words in three-letter alphabets that are three-halves-free?
> 
> 
> 


If a word has a square $ZZ$ with $|Z|$ even, then it has a three-halves pattern.
If a word is square-free, it may not be three-halves-free.
For example, both $(1,0,1)$ and $(0,1,2,1,0,1)$ are square-free.
And the square-free infinite word [A029883](http://oeis.org/A029883)
$$
(1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, \ldots )
$$
is not three-halves-free: e.g., $(-1,1,-1)$ and $(-1,0,1,0,-1,0)$
are three-halves patterns.



Three-halves-free words (analogous to square-free)
(Not a real answer, just a conjecture)


Let $w$ be the Pansiot word on $4$ letters defined here :  

[J.J. Pansiot, A propos d'une conjecture de F. Dejean sur les répétitions dans les mots, Discrete Applied Mathematics Volume 7, Issue 3, March 1984, Pages 297–311](https://www.sciencedirect.com/science/article/pii/0166218X84900064)  

in French, or also in  

 [J. Moulin Ollagnier, Proof of Dejean's conjecture for alphabets with 5, 6, 7, 8, 9, 10 and 11 letters, Theoretical Computer Science, Volume 95, Issue 2, 30 March 1992, Pages 187–205](https://www.sciencedirect.com/science/article/pii/030439759290264G).  

This word proves Dejean's Conjecture for $4$ letters, that is:  

it avoids fractional repetitions of exponent greater than $7/5$, and so it is also three-halves-free.


$$w=abcdbadcbdacdbcadcbacdabdcadbcdacbadcabdacdbadcbdabcadbacdab...$$



> 
> **Conjecture** **Theorem**
> 
> 
> Let $h(a)=aabbaccabc$, $h(b)=aacbacbacc$, $h(c)=abbaaccbbc$, $h(d)=abcaacbbcc$. 
> 
> 
> Then $h(w)$ is a three-halves-free word on $3$ letters.
> 
> 
> 


I checked up to size 412030. I do not tried to prove the result, but maybe the proof is simple and "standard" (suppose that the image has a three-halves, then prove that the pre-image has a three-halve).


edit:
The proof is easy. $h$ is $10$-uniform and is *synchronizing*, that is for every $x,y,z\in\{a,b,c,d\}$, $h(x)$ is not a proper infix of $h(yz)$. Thus, if a factor $XYX$ of $h(w)$ is a three-halve, then either $XYX$ is small (of size at most $18\times 3$), or $\vert X \vert = \vert Y \vert$ is a multiple of $10$. There are only a finite number of "small" factors to check.  

Now, if we are in the second case, and if $\vert X \vert$ if big (at least $42$), then $XYX$ has a unique de-substitution in $w$, and $w$ has a factor $X'Y'X'$ with $\vert Y' \vert / \vert X' \vert > 2/5$. Thus $w$ has a repetition of exponent greater that $7/5$, and we have a contradiction with the fact that $w$ is a Dejean word (i.e. it is $7/5+$-free).


