# Cantor set intersecting a geometric sequence

I was working on a problem involving finding all points in the intersection of the Cantor set $C$ and the geometric sequence $\{ (2/3)^i \}_{i=1}^\infty$. The only points I have in this intersection at the moment are $2/3$, $(2/3)^3$, and $(2/3)^9$.

With the ternary representation of the Cantor set, the question can be translated to "Which powers of $2$ have no $1$'s in their ternary representations?" I have verified this in Mathematica for powers of $2$ with exponent $i \leq 10000$, and the only results I obtained was $2^1=2=(2)_3$, $2^3=8=(22)_3$, and $2^9=512=(200222)_3$.

Can one actually prove (or disprove) that there is no more points in this intersection? I have worked on this seemingly simple question for a while but could not find a way to go. Thanks a lot!

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Probably this is very hard. There's a well known open problem, showing that every sufficiently large power of 2 has at least one 7 in its decimal expansion. I don't see why the base 3 would be any easier. [ You can, however, show that for almost every $n$, $2^n$ has at least one 1 in its ternary expansion ]. – Anthony Quas Jun 27 '14 at 20:43

This is an old unsolved problem. Erdos conjectured (see the first problem in that paper) that for all $n\ge 9$ the ternary expansion of $2^n$ contains the ternary digit $2$ (this is equivalent to for every $n\ge 10$ the ternary expansion of $2^n$ contains a $1$). For recent work related to this (and references) see these papers of Abram and Lagarias, and Lagarias.