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Does anyone know of any lower bounds on the number of nonzero digits that appear in powers of 2 when written to base 3? (Other than the easy "If it's more than 8 it has to have at least 3.") I know there's been some stuff done with powers of 2 written to base 3, but I can't seem to find anything that quite answers this question.

(Are there more general such bounds? Powers of a written to base b? (To avoid triviality, suppose that no power of a is a power of b... or are stricter conditions needed?) From what I can tell, it looks like even specific instances of these are hard, so I suppose I should stick to the specific version.)

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  • $\begingroup$ This should be a pretty open problem. I can't find a related problem of can the square be written by two digits only. $\endgroup$ – Wadim Zudilin Jul 2 '10 at 23:46
  • $\begingroup$ arxiv.org/abs/math/0512006 $\endgroup$ – Junkie Jul 3 '10 at 0:07
  • $\begingroup$ @Junkie: Yes, I saw that. It didn't appear to address what I'm asking, though, or at least not that I noticed.... did I miss something? $\endgroup$ – Harry Altman Jul 3 '10 at 0:10
  • $\begingroup$ "Rather than easy" (at least three nonzero digits) is Levi ben Gerson's mathoverflow.net/questions/29926/…. The problem has to do something with simultaneous (special) rational approximations to $\log 2$ and $\log 3$. $\endgroup$ – Wadim Zudilin Jul 3 '10 at 0:16
  • $\begingroup$ So, the insolvability of $a^m-b^n=1$ for $a,b,n,n\ge2$ (Catalan's equation), except $3^2-2^3=1$, implies "at least 3 nonzero digits" for any $a$ and $b$. $\endgroup$ – Wadim Zudilin Jul 3 '10 at 0:25
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A nontrivial lower bound can be found in a paper of Cam Stewart (see http://www.math.uwaterloo.ca/PM_Dept/Homepages/Stewart/Jour_Books/J-reine-ange-Math-1980.pdf). He proves, more generally, for fixed bases a and b for which $\log a/\log b$ is irrational, that the sum of the number of nonzero digits in the base a and base b digits of an integer n exceeds (essentially) $$\log \log n/ \log \log \log n.$$

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  • $\begingroup$ Ooh, thank you! Making use of this might not be easy, but it certainly does answer the question, and in generality, too! $\endgroup$ – Harry Altman Jul 3 '10 at 19:48

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