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?