Prouhet-Thue-Morse sequence and Arithmetic Progressions This question is a fragment from a question posted by @Mathphile for which the link will be provided below.
Let $(x_i)$ be an arithmetic progression of length $M$. Let $(t_i)$ be the $i$th element in the Prouhet-Thue-Morse sequence where we have:
$$t_0=0$$
$$t_{2n}=t_n \quad (n \in \mathbb{N}_0)$$
$$t_{2n+1}=1-t_n \quad (n \in \mathbb{N}_0)$$
If all the elements of $(t_{x_i})$ are to be $0$'s, can the value of $M$ be arbitrarily large?
Link for Prouhet-Thue-Morse Sequence
Link for @Mathphile problem
 A: Yes. For $r=0,1$ let $T_r=\{n\in\Bbb N_0: t_n=r\}$. The famous van der Waerden's theorem states that for any partition of the set of natural numbers into finitely many parts there exists a part containing an arbitrary long arithmetic progression. So for each natural number $M$ there exists an arithmetic progression $P$ of length $M$ contained in $T_0$ or $T_1$. Moreover, if $P\subset T_1$ then a set $P’=\{2p+1:p\in P\}$ is an arithmetic progression of length $M$ contained in $T_0$.
Similarly we can show that there is no uniform bound $M$ for the claim from Overview in your answer. Namely, for 
$0\le r<b-1$ let $T_r=\{n\in\Bbb N_0: t_n=r\}$. By van der Waerden’s theorem, for each number $M$ there exist  $0\le r<b-1$ and an arithmetic progression $P$ of length $M$ contained in $T_r$. Pick an arbitrary number $0\le d<b$ coprime to $b-1$ and define a map $f:\Bbb N_0\to \Bbb N_0$ by putting $f(n)=bn+d$ for each $n\in\Bbb N_0$. Then a set $f(P)$ is an arithmetic progression of length $M$ contained in $T_{r’}$, where $r’\equiv r+C_d\pmod {b-1}$. Since $C_d=d^d$ is coprime to $b-1$ too, there exists a natural number $l$ such that $r+C_dl\equiv 0\pmod {b-1}$. Iterating $T$, we obtain that $T^r(P)$ is an arithmetic progression of length $M$ contained in $T_0$.
