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 $www$.

• 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.

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.