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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 $n>2$?

BRSTCohomology
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    Do you mean \sqrt{2^n} → $\sqrt{2^n}$, or (\sqrt 2)^n → $(\sqrt 2)^n$ . . . ? – CiaPan Jul 12 '18 at 19:33
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    @CiaPan Whats the difference – Coolwater Jul 12 '18 at 19:37
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    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. – rtybase Jul 12 '18 at 19:39
  • 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? – CiaPan Jul 12 '18 at 19:49
  • edited. it should be clear now. – BRSTCohomology Jul 12 '18 at 19:58
  • Is it true that $\exp(n\frac{\ln2}{2} \mod 1)\mod1 = \exp(n\frac{\ln2}{2})\mod1$? @rtybase – BRSTCohomology Jul 12 '18 at 20:09
  • @MarcusAurelius not yet, that was just brainstorming ... – rtybase Jul 12 '18 at 20:13
  • I'm afraid we know literally nothing about expansions of natural constants (algebraic irrationals included), apart from some basic facts like they are nonperiodic, and they can't be too "close" to being periodic (irrationality measure). You shouldn't expect this kind of question to be solved, despite how much weaker this is than normality. – Wojowu Jul 12 '18 at 20:43
  • @Wojowu is anyone working on questions like these at all? Are there any first steps towards theory/results like these? Or are we really that in the dark? – BRSTCohomology Jul 12 '18 at 20:44
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    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. – Wojowu Jul 12 '18 at 20:46
  • We do not know whether any given decimal digit appears infinite many times in $\pi$. I do not think the situation is better for the binary digits of $\sqrt{2}$. The only breakthrough I am aware of was a formula calculating specific binary digits of $\pi$ WITHOUT calculating the complete binary expansion, but that still does not help to prove or disprove that every finite binary sequence appears in $\pi$. Seems that we know almost nothing when it comes to such questions. – Peter Jul 13 '18 at 19:35
  • What we know, that the digits $0$ and $1$ both occur infinite many often in the binary expansion of $\sqrt{2}$, but this is trivially the case for every irrational number. – Peter Jul 13 '18 at 19:40

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