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.
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$\begingroup$ Is this a homework question? It is not at the appropriate level for mathoverflow. $\endgroup$– Anthony QuasCommented Feb 3, 2023 at 17:32
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2$\begingroup$ 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? $\endgroup$– Christophe LeuridanCommented Feb 3, 2023 at 18:39
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$\begingroup$ 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! $\endgroup$– RabatCommented Feb 3, 2023 at 22:07
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1$\begingroup$ 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 $\{a\}$. $\endgroup$– Gerald EdgarCommented Feb 4, 2023 at 1:28
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2$\begingroup$ 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. $\endgroup$– Anthony QuasCommented Feb 4, 2023 at 4:21
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