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!



Cantor set intersecting a geometric sequence
This is an old unsolved problem. [Erdos](https://www.renyi.hu/~p_erdos/1979-22.pdf) 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](http://arxiv.org/pdf/1308.3133v1.pdf), and [Lagarias](http://arxiv.org/pdf/math/0512006.pdf). 


