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How can I find limit points of $\{2^n \sqrt 2\}$, where $\{\cdot\}$ denotes the fractional part function? This is a subsequence of the sequence $\{\sqrt n\}$ for which we know the set of limit points. However, it is not clear to me if $\{2^n \sqrt 2\}$ converge or has several limit points.

LSpice
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Rabat
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  • Is this a homework question? It is not at the appropriate level for mathoverflow. – Anthony Quas Feb 03 '23 at 17:32
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    I do not know whether the answer is known. For example, a similar question, the equirepartition modulo 1 of $((3/2)^n)_{n \ge 0}$ is an open question (to my knowledge). Indicating where the questions come from would be useful: is it related to a question in research or does it come from an exercise? – Christophe Leuridan Feb 03 '23 at 18:39
  • It is related to a research question. I was reading a survey on some Ramanujan's diophantine equations (see e.g. pp. 12-14 in "The problems submitted by Ramanujan to the indian mathematical society", and I though that some of them can be related to the posted question! – Rabat Feb 03 '23 at 22:07
  • @Anthony Quas: Your comment indicates that you know the answer. Would you please share your knowledge with the community. Thank you. – tj_ Feb 04 '23 at 00:31
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    It is suspected (but not proved) that $\sqrt{2}$ is normal in all bases. This would imply that ${2^n \sqrt{2}}$ has every element of $[0,1]$ as limit point. The reason is that the binary expansion for ${2a}$ is obtained as the left shift of the binary expansion of ${a}$. – Gerald Edgar Feb 04 '23 at 01:28
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    It is certainly known that there is more than one limit point. Write out the base 2 expansion of $\sqrt2$. As @geraldedgar says, Doubling and taking the fractional part is equivalent to shifting the binary expansion left and truncating the first term. Since $\sqrt 2$ is irrational, the binary expansion is not periodic. It follows that the set of limit points is infinite. – Anthony Quas Feb 04 '23 at 04:21
  • Here is the related question on SE: https://math.stackexchange.com/questions/3573684/at-least-one-even-number-among-lfloor-2n-sqrt2-rfloor-lfloor-2n1. Not an answer to the question but there are some estimates for number of ones and zeroes in the binary expression of $\sqrt{2}$. – Pavel Gubkin Feb 07 '23 at 08:16

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