A 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
$$
(1, 0, -1, 1, -1, 0, 1, 0, -1, 0, 1, -1, 1, 0, -1, \ldots )
$$
is not three-halves-free: e.g., $(-1,1,-1)$ and $(-1,0,1,0,-1,0)$
are three-halves patterns.
Best Answer
(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
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.
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.
$$w=abcdbadcbdacdbcadcbacdabdcadbcdacbadcabdacdbadcbdabcadbacdab...$$
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).