For $a,b \in \omega$ with $a > 0$, let $f\_{a,b}: \omega\to\omega$ be defined by $n \mapsto an+b$. What is an example of an infinite binary string $s:\omega\to\{0,1\}$ with the following property?



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> Whenever $(a,b), (a\_1,b\_1)\in (\omega\setminus\{0\})\times \omega$ with $(a,b)\neq (a\_1,b\_1)$, then $s\circ f\_{a,b} \neq s\circ f\_{a\_1,b\_1}$.
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"Arithmetically diverse" infinite binary string
Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$:
$$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$
Observe that all but finitely many blocks of 1s of $s\circ f\_{a,b}$ have size of the form $\lfloor2^{2^{k}}/a\rfloor$ or $\lceil 2^{2^k}/a\rceil$.


We claim that we cannot keep having $k>j$ and $2^{2^k}/{a\_1}\le 2^{2^j}/{a\_2}+1$ (and therefore all $s\circ f\_{a,b}$ are distinct).


Indeed, when this happens then $2^{2^k-2^j}\le a\_1/{a\_2}+a\_1 2^{-2^j}$ is bounded. But as $k>j\to\infty$, $2^{2^k-2^j}$ is unbounded.


(Note that $2^k/a\_1=2^{k-1}/a\_2$ with $a\_1=4$ and $a\_2=2$, so a single-exponential $2^k$ in place of a double-exponential $2^{2^k}$ is not enough for this construction.)


