Let $w=x_0 x_1 x_2 \ldots$ be an infinite word, where each $x_i\in \{0,1\}$. For each positive integer $k$ (thought of as the jump size of an arithmetic progression) and each residue $0\leq a \leq k-1$ we can form the new "arithmetic progression" word $w_{a\ {\rm mod}\ k}=x_a x_{a+k} x_{a+2k}\ldots$.

Question 1: Does there exist a word $w$, as above, so that each of the derived words $w_{a\ {\rm mod}\ k}$ has no arbitrarily long constant subwords?

I tried finding the answer, but failed. I did find in the literature that for non-binary languages it is possible to do this (and more) without any repetitions at all!

I believe that the answer to Question 1 is probably well-known to experts, and it is positive. If so, I'm interested in the following extension.

Question 2: Can the upper bound on the length of the constant subwords of $w_{a\ {\rm mod}\ k}$ be made independent of $a$ and $k$? (If so, what is the smallest such bound?)

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    $\begingroup$ Van der Waerden's theorem says that the answer to question 2 is no, even for larger alphabets. $\endgroup$ – zeb Oct 29 at 23:42
  • $\begingroup$ @zeb I should have seen that. Good point! $\endgroup$ – Pace Nielsen Oct 30 at 0:50

Any Sturmian word will work for Question 1. Before I prove this, I can't resist giving the standard example: the Fibonacci word. The Fibonacci word is defined as the fixed point of the iterative procedure which replaces every $0$ with the string $01$, and replaces every $1$ with the string $0$:


In general, a Sturmian word $w$ is characterized by a pair of real numbers $0 \le \alpha, \beta < 1$ such that $\alpha$ is irrational. The $i$ letter of the Sturmian word corresponding to the pair $\alpha, \beta$ is given by

$x_i = \lfloor\alpha (i+1) + \beta \rfloor - \lfloor\alpha i + \beta \rfloor$.

To see that any Sturmian word will work, note that for any $k \in \mathbb{N}^+$ there is some $\ell \in \mathbb{N}^+$ such that the fractional part $\{\alpha k \ell \}$ of $\alpha k \ell$ is less than $\min(\alpha, 1-\alpha)$. If $m \in \mathbb{N}^+$ is large enough that $m\{\alpha k \ell\} > 1$, then no subword of $w_{a \text{ mod } k\ell}$ of length $m$ can be constant (for any $a$), so no subword of $w_{a \text{ mod } k}$ of length $\ell m$ can be constant.

Question: does the Thue-Morse word also work for Question 1? (The Thue-Morse word isn't a Sturmian word, but it has a similar flavor.)

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    $\begingroup$ I think Olga Parshina has looked at Question 1 for the Thue-Morse word here. $\endgroup$ – Narad Rampersad Oct 30 at 2:15
  • $\begingroup$ Excellent answer. This gives a bound, $m$, in terms of a rather complicated function of $k$. Do you know if a "nicer" bound can be given for $m$ (in terms of a relatively slow growing function of $k$)? $\endgroup$ – Pace Nielsen Oct 30 at 16:06
  • $\begingroup$ I haven't worked out the explicit bound in terms of $k$ - I suspect that without too much effort you may be able to get an explicit quadratic or linear bound in $k$ for the Fibonacci word (since the golden ratio has nice rational approximations). Perhaps you can do better with other words - this seems like a natural followup question to ask! $\endgroup$ – zeb Oct 31 at 14:20

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