TITLE: Cube-free infinite binary words
QUESTION [19 upvotes]: A word $y$ is a subword of $w$ if there exist words $x$ and $z$ (possibly empty) such that $w=xyz$.  Thus, $01$ is a subword of $0110$, but $00$ is not a subword of $0110$.  I'm interested in right-infinite words over a two-letter alphabet that do not contain subwords of the form $xxx$, where $x$ is a word of one or more letters.  (For example, the Thue-Morse word, the Kolakoski word, Stewart's choral sequence, and so on.)  In particular, I would like to know if there are any general statements about all such words.  For example, are there an infinite number of them?  Is there any way to classify them?
It's possible that one way to classify cube-free infinite binary words is to group them according to their subwords.  For example, the Kolakoski word has the subword $00100$ whereas the Thue-Morse word does not, so they belong in different classes. The words created in Tony's answer (see below) have the same subwords (the Thue-Morse word is recurrent), so they belong to one class. I suppose there an infinite number of these classes.
Another possible way to classify cfib words is to group them according to their subword complexity. For example, Stewart's choral sequence has a subword complexity of $2n$ (where $n$ is the length of the subword), so we can group it with other cfib words with subword complexity $2n$. Is the subword complexity of the Kolakoski word known?

REPLY [4 votes]: Lemma: Someone only able to speak an infinite cubefree word can convey information at least $1/24$ of the speed of an ordinary person able to speak an arbitrary binary sequence.
Proof: Consider the following period-$12$ sequence over the alphabet $\{ 0, 1, ?\}$:
$S = (001?010?1011)^{\infty}$
Now suppose $T$ is a sequence derived from $S$ by replacing each $?$ with a $0$ or $1$.

$T$ clearly has no words of the form $xxx$ for a single digit $x$.

If we take alternate digits in $T$, we obtain $(010011)^{\infty}$. The only cubes in this sequence have length a multiple of $6$, hence:

If $T$ contains a word of the form $WWW$, where $W$ has even length, then $W$ has length divisible by 12.
Also, $T$ contains a word of the form $WWW$ where $W$ has length $3$ if and only if two question marks in the same dodecad $(001?010?1011)$ are both zero. Assert that this will not happen.
If $T$ contains a word of the form $WWW$, where $W$ has length divisible by $3$, then $W$ has length divisible by $12$.

Hence we are only worried about words of length coprime to or divisible by $12$.

If the length of $W$ is $\pm 1 \mod 12$, then the existence of $WWW$ is ruled out by the proof of the non-existence of $xxx$.
Similarly, it is easy to observe that there are none of length $\pm 5 \mod 12$.

So any word of the form $WWW$ in $T$ requires $W$ to have length divisible by $12$.

Now begin with any tripleless sequence over $\{A, B\}$ (such as the Thue-Morse sequence) and apply the substitution rules:
$A \mapsto (001001011011)$
$B \mapsto (0011010?1011)$
Then we obtain an infinite word over $\{0, 1, ?\}$ such that:

Every replacement of the question marks with digits yields an infinite cubefree word.
Every icositetrad contains one question mark.

The result follows.