Question

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?

Answer 1

As of a few months ago, the status of the problem was: still unsolved. See the slides Jeff Lagarias put up from a talk he gave in September 2009: http://www.math.lsa.umich.edu/~lagarias/talks-files/fields-horz5.pdf

An older reference is http://www.americanscientist.org/issues/id.3268,y.0,no.,content.true,page.2,css.print/issue.aspx (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 $2^8$).

Answer 2

... and $\sum (2/3)^{n\log 2/\log 3} < \infty$ so (if things were random) we would expect only finitely many such occurrences by Borel-Cantelli easy direction.

But of course proving this is anything like random is far too hard for today's tools, I think.

Answer 3

Something way easier that one can prove is that your sequence is disjunctive, which is much weaker than normality. This is true since for every string of digits $k$ you can find a power of two with a base 3 expansion that starts with k.

I don't know much about normality except that there are many conjectures (for example that every irrational algebraic number is normal) and no techniques to answer such questions except in trivial cases. I believe the current state is similar for simply normal numbers.

Another question that is similar in spirit and that was answered recently is Stolarsky's conjecture that says $$\liminf_{n\to \infty}\frac{s_q(n^k)}{s_q(n)}=0$$ where $s_q(\cdot)$ is the sum of digits in base $q$. Intuitively it is hard to come up with examples that $s _3(2^{nk)} < s_3(2^k)$ for most $k$, let alone that the $\liminf$ is zero. However this question is much weaker than the one you ask since the sum of digits puts very few restraints on the digits themselves.