TITLE: Can the expansion of a large integer in all bases consist of almost all zeroes?
QUESTION [18 upvotes]: Let $n$ be a positive integer. Given an integer base $b\ge 2$, let $C_b(n)$ be the number of non-zero digits in the expansion of $N$ in base $b$. Further, let $M(n)=\max\{C_b(n):b\ge 2\}$ be the maximum of all $C_b$. Obviously, for each $n$ only $b\le n$ count, and $M(n)\le 1+\log_2 n$. 
I would like to understand the relation between the $C_b(n)$ for different bases $b$. It seems reasonable to conjecture (and I'm sure stronger conjectures exist) that if $b, b'$ are multiplicatively independent, then $C_b(n)+C_{b'}(n)>\varepsilon \log n$ for some $\varepsilon(b,b')>0$ and all $n$. However, this seems to be out of reach. The best I have found are estimates of the form $C_b(n)+C_{b'}(n)\ge \frac{\log\log n}{\log\log\log n}$, using Baker's Theorem.

My question is:

Can we do better by considering more bases? In particular, is it true that there is $\varepsilon>0$ such that $M(n) \ge \varepsilon \log n$ for all $n$?

The special case in which $n$ is a power would already be interesting, and is perhaps a little easier:

Does there exist $\varepsilon>0$ such that $M(2^k) > \varepsilon k$ for all $k$?


This question came up in studying the absolute continuity of the convolution of certain self-similar measures, a question at the interface of fractal geometry/dynamics/number theory.

REPLY [7 votes]: A full answer to your question seems tough, because for any fixed value of $\epsilon$, only finitely many bases are available to establish the bound.  However, there is an elementary partial answer to your second question:  The digits of $2^{n^2}$ in base $2^n-1$ are exactly the $n$th row of Pascal's triangle.  A variation of this idea yields $M(2^k) = \Omega(\sqrt{k})$.  This suggests the following strategy for general $n$:  Choose a base $b$ so that $n = d+\prod_i (b+c_i)$ with $d \ll n$ and $0 < c_i \ll b$, in such a way that the leading digits of $n$ in base $b$ are the same as the coefficients of the polynomial $p(x) = \prod_i (x+c_i)$.  I have not done a calculation, but maybe this could give you $M(n) = \Omega((\log n)^\alpha)$ for some $\alpha > 0$.