Let $A$ be a finite set. Let $A^k$ be the set of words in the alphabet $A$ of length $k$ and $A^\*$ be the set of infinite words. I was looking for an element $a = \lbrace a\_n \rbrace\_{n \in \mathbb{N}}$ of $A^\*$ so that, $\exists k\_0 \in \mathbb{N}$ so that $\forall k>k\_0$, any finite subsequence (of length $k \cdot m$) of $a$ is different from the sequences obtained by repeating some word at least $k$ times. 


In other words, I was looking for an infinite sequence so that all subwords are not element of the diagonal in $X^k$ where $X= A^n$ and $n$ runs over all integers, $k$ runs over all integers larger than some $k\_0$. 


There is an [I suppose] absurdly complicated way to answer this by looking at a geodesic ray in the Burnside groups (this requires $|A| \geq 4$ and gives a $k\_0$ around 665/2). Curiosity pushes me to ask


$\mathbf{Question:}$ How to produce (elementarily) such a sequence?



Totally aperiodic sequence
[Square free words](http://en.wikipedia.org/wiki/Square-free_word) exist over all alphabet sizes greater than 2, and cube free words exist over all alphabet sizes greater than or equal to 2 (same article).


