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Let's define an integer $n$ as zero-avoiding in some base $b$ when its representation in base $b$ has no zero digits (excluding leading zeros of course)

I'd like to prove the following claim which empirically looks true:

an integer $n > 7$ is zero avoiding in base $2$ and $3$ if and only if $n = 2^{15}-1 = 32767$

To make things simpler, we can remove base $2$ and replace $n$ by a certain $2^m - 1$, making my initial claim equivalent to: $$2^m-1 \text{ is zero avoiding in base $3$ if and only if } m = 15$$

I've checked this claim for $m$ up to $100000$, and so far the only candidate is indeed at $m=15$

A few notes:

  • We can suppose $m$ odd: when $m$ is even, $2^m-1$ is divisible by $3$, so it has a digit $0$ in base $3$
  • For a given ternary length $d$, only a unique $m$ can make $2^m - 1$ zero-avoiding in base $3$

That second point seems to me as the most promising if we can reduce the claim to a finite number of lengths $d$. Here's a short proof of the uniqueness given a length:

Let $d$ be the length wanted, and $m$ such that $2^m-1$ is zero-avoiding in base $3$

Therefore, we have $(a_0, \ldots, a_d)$ all in the set $\{ 1,2\}$ such that $2^m-1 = \displaystyle \sum_{k=0}^{d-1} a_k3^k$

So that gives $\displaystyle \sum_{k=0}^{d-1} 3^k \leq 2^m-1 \leq 2\sum_{k=0}^{d-1} 3^k \iff N \leq 2^m-1 \leq 2N$ for $N = \displaystyle \sum_{k=0}^{d-1} 3^k = \frac{1}{2} (3^d - 1)$

Solving for $m$ gives $\log_2(N+1) \leq m \leq \log_2(2N+1)$, but the interval $\log_2(N+1), \log_2(2N+1)$ is of length less than $1$

This means $m = \lfloor d\log_2(3) \rfloor$, which shows uniqueness of $m$ given the length $d$

I'm not well versed enough regarding this, but maybe Baker's method can help find an upper bound for $d$?

I am aware that a similar question has been asked before, but here we restrict ourselves to base $3$, and the addition of uniqueness along with Baker's theorem might bring something new

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    $\begingroup$ Note that even the question, are there only finitely many zeroless powers of two in base ten, is unsolved. oeis.org/wiki/Zeroless_powers $\endgroup$ Commented Nov 12 at 11:48
  • $\begingroup$ We can also skip $m = 6k+1$ for $m>1$, as in that case $2^m-1$ has a zero in the threes position. Specifically, $2^m-1 = 2\times 64^k-1 = 2\times (64^k-1)+1 = 2\times 63 \times (64^{k-1}+64^{k-2}...+64+1) + 1$, i.e. it's a multiple of 9 plus one. You can do similar analysis for each symbol in the trinary expansion of $2^m-1$... perhaps eventually ruling out all but finitely many answers? $\endgroup$ Commented Nov 12 at 13:05
  • $\begingroup$ Let $f\left(m\right)$ denote the number of nonzero digits at the end of the ternary expansion of $2^{m}-1$. Then $f\left(9\right)=f\left(11\right)=3$, $f\left(15\right)=10$, $f\left(101\right)=16$, $f\left(143\right)=25$. Are there numbers of the form $2^{m}-1$ (for positive integer $m$) whose ternary expansion ends with more than 25 nonzero digits? If yes, can the ``nonzero tail'' be arbitrarily long? $\endgroup$ Commented Nov 12 at 14:11
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    $\begingroup$ Looks like $f(21527) = 28$, can't find better with my wonky python code $\endgroup$ Commented Nov 12 at 16:34
  • $\begingroup$ (cont.) I'm not sure if this line can help with answering the question. In case it is useful: I think $f\left(21527\right)=f\left(238877\right)=28$, $f\left(380843\right)=29$, $f\left(511773\right)=31$, $f\left(1377105\right)=f\left(1496483\right)=32$, $f\left(1572525\right)=33$, $f\left(3483159\right)=37$, $f\left(3917963\right)=40$, $f\left(9827775\right)=f\left(12409983\right)=44$, $f\left(43622201\right)=f\left(48082239\right)=47$, $f\left(75765431\right)=50$. It is likely possible to push the sequence further with something faster than a Python script. $\endgroup$ Commented Nov 18 at 23:05

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