Let ${\cal L}\_n$ be the set of all subwords of length $n$ of a biinfinite Sturmian sequence, induced by a rotation coding with irrational angle $\theta$.

Take a word $w \in {\cal L}\_{2^n}$ and write it as the concatenation of its two halves: $w=w\_1w\_2$.
Say that $w$ is *bad* if $w\_1 \neq w\_2$ and $w\_2w\_1 \in {\cal L}\_{2^n}$. 

For example there is always a bad word for $n=1$: the word $ab$, because $ab, ba \in {\cal L}\_2$.

I want to know whether there is a value of $\theta$ for which there is no bad word for $n \geq 2$. 

For example, using a computer, I find:

* for a small $\theta \approx 0.015$ there are some bad words for $n=2$, such as $abbb$;
* for the "golden" case $\theta=1/\phi^2 \approx 0.381966$, I found the first bad word for $n=3$, namely $abbabbab$;
* for $\theta=\sqrt{2}-1 \approx 0.4142136$, the first bad word occurs for $n=7$;
* for $\theta \approx 0.4548312$, I didn't find any bad word until $n=10$.

The smallest $n\geq 2$ for which there is a bad word seems to increase when $\theta$ increases from $0$ to $1/2$. 

I have no reason to expect that there is no bad word for some value of $\theta$, but I ask this question because a positive answer would help me to solve a certain problem: I try to prove a property $P\_n$ on ${\cal L}\_{2^n}$ which satisfies $P\_n \implies P\_{n+1}$ if there is no bad word.

Update
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I just think to a better question. Let $n(\theta)$ the smallest $n \geq 2$ for which there is a bad word in ${\cal L}\_{2^n}(\theta)$. Does $n(\theta) \to \infty$ as $\theta \to 1/2$?