I believe this question is due to Erdős and Graham, and I think it is still open: does the base 3 expansion of $2^n$ avoid the digit 2 for infinitely many $n$?


If we concatenate the digits of $2^i$, $i \geq 0$, we produce the number $0.110100100010000...$. This number is not simply normal in base 2, so it is not normal. Is it simply normal in base 3? I think even that result would not imply that for sufficiently large $n$, 2 doesn't appear in the base 3 expansion of $2^n$.


The number 20 here is not special:


$2^{20} = 1222021101011\_3, \;\;\;\; 2^{21} = 10221112202022\_3, \;\;\;
2^{22} = 21220002111121\_3$


Statistically, we seem to be flipping a fair 3-sided coin, and statistical analysis for larger $n$ bears this out (in the past, I did a p-test on the digits, but don't have the data available here). If we actually produced these digits by flipping this 3-sided coin, for fixed $n$ we would have probability about
$$(2/3)^{n\ln2/\ln3}$$
of having no 2s in the base-3 digit expansion.


What is the state of the art for this problem? Is there a good number-theoretic reason why this problem should be very difficult (e.g. an analogy with other supposed-hard problems)? Are there related problems that have been solved?



Do the base 3 digits of $2^n$ avoid the digit 2 infinitely often -- what is the status of this problem?
As of a few months ago, the status of the problem was: still unsolved. See the slides [Jeff Lagarias](http://www.math.lsa.umich.edu/~lagarias/) put up from a talk he gave in September 2009: <http://www.math.lsa.umich.edu/~lagarias/TALK-SLIDES/ternary-fields-2009sep.pdf>


An older reference is 
<http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.65.6976&rep=rep1&type=pdf> (Brian Hayes, Third Base, American Scientist) which says the problem was still open in late 2001; also that Ilan Vardi searched up to $2^{6973568802}$ without finding any 2-less powers of 2 (other than $2^2$ and $2^8$). 


