Does the Prouhet-Thue-Morse sequence eventually contain its bitwise boolean complement in whole?

Question

My intuition says that it does not, but I am struggling to prove it. It definitely contains any finite consecutive subsequence ("substring") of the latter, by its construction. I also suspect that, even under best alignment, the alignment breaks at the $2^n$th term for some nonnegative integer $n$. But I do not have a strong enough grasp of the sequence in orded to prove the result.

Answer 1

The Thue Morse Sequence obey the recurrence relation : $$\begin{cases} t_0 = 0\\ t_{2n} = t_n\\ t_{2n+1} = 1-t_{n}\end{cases}$$

Suppose there exist a subsequence equal to the bitwise boolean complement Thue Morse sequence. If the subsequence start at the $k$-st term, we would have the additional requirement : $$t_n = 1 - t_{n-k}$$ And so, we should have $$t_{2n} = 1 - t_{2n-k}$$ It fails when $n = 2n-k$ i.e. $n = k$, so the subsequence starting at $k$ will be no longer than $k$