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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.

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Looks interesting - is there direct relevance to questions about intersections of Cantor sets, Furstenberg's conjecture etc? A numerical test (limit $10^{18}$) gives for the largest integer $n$ with $\max{C_2(n),C_3(n)}=k$ for $k=1\ldots 6$ as $2,513,65664, 272630016,10737942552,20342038921218$ This sequence is rather rapidly growing, but plausibly exponential, consistent with the first conjecture. –  Carl Mar 20 '13 at 11:27
    
Carl - It is related in spirit to Furstenberg's conjectures but not directly. If the conjecture is true, I can use it to show that the arithmetic sum of $A_{1/3}$ and $A_{1/4}$ has positive Lebesgue measure, where $A_r$ is the central Cantor set with rate of dissection $r$. Morally this says something about the intersections, but far too weak to give a dimension bound on all of them. –  Pablo Shmerkin Mar 21 '13 at 1:01
    
For a fixed value of k, do you know anything about numbers which, when expressed in bases 2, 3, and 6, have at most k nonzero digits? Gerhard "Guesses They Are Finitely Many" Paseman, 2013.03.20 –  Gerhard Paseman Mar 21 '13 at 2:11
    
Gerhard - for every $k$ there is $N(k)$ such that for every $n\ge N(k)$, the expansion of $n$ in one of the bases $2$, $3$ has at least $k$ nonzero digits. This follows from the lower bound on $C_2(n)+C_3(n)$ in my post (so ultimately from Baker's Theorem), and gives a lower bound for $N(k)$ which is (roughly) doubly exponential in $k$. It'd be interesting to get better bounds for this as well. –  Pablo Shmerkin Mar 21 '13 at 2:24

1 Answer 1

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 $\alpha > 0$.

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Thanks! It is interesting that an elementary argument can go far beyond what one gets from Baker's Theorem at least when $n=2^k$. I'll check the details of your idea for the general case. –  Pablo Shmerkin Mar 23 '13 at 20:51

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