I want to know whether the fractional parts of the sequence $i\mapsto \sqrt{2}^i$ are dense in $[0, 1)$. Obviously this is the same as the sequence $i\mapsto \sqrt{2}\cdot2^i$. I did the obvious thing and recasted the problem as asking whether in the $\textit{binary expansion}$ of $\sqrt{2}$, $\exists$ L such that every finite string of $0$'s and $1$'s of length $\geq L$ appears somewhere in that expansion.

I have no idea how to prove this, or what tools I have. I know that the normality of $\sqrt{2}$ is unproved as of yet, but normality is much stronger than what I want. Am I missing something obvious, or is this problem really nontrivial?

EDIT: Here's a similar question with some more comments: The density --- or otherwise --- of $\{\{2^N\,\alpha\}:N\in\mathbb{N}\}$ for ALL irrational $\alpha$.. Can anyone explain to me why his theorem on $O_n$ is untrue for $n>2$?

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    $\begingroup$ Do you mean \sqrt{2^n} → $\sqrt{2^n}$, or (\sqrt 2)^n → $(\sqrt 2)^n$ . . . ? $\endgroup$ – CiaPan Jul 12 '18 at 19:33
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    $\begingroup$ @CiaPan Whats the difference $\endgroup$ – Coolwater Jul 12 '18 at 19:37
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    $\begingroup$ Look at the $\left\{n \frac{\log{2}}{2}\right\}$. Since $\frac{\log{2}}{2}$ is irrational, then KAT applies ... and $e^x$ is continuous. $\endgroup$ – rtybase Jul 12 '18 at 19:39
  • $\begingroup$ The difference is in the look, both expressions represent the same value (as long as $n$ is a real number). However what you wrote is neither of them and it's hardly obvious what you exactly mean in $\sqrt(2)2^i$ – is $2^i$ under the radical or not? $\endgroup$ – CiaPan Jul 12 '18 at 19:49
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    $\begingroup$ To the best of my knowledge, and as surprising as it may sound, I do think we are "that in the dark" as you put it. Unless there has been some recent progress I haven't heard of, we really know very little as something, we would think, as fundamental as base $b$ expansions. $\endgroup$ – Wojowu Jul 12 '18 at 20:46

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