Numbers that don't start with (p-1) in base p for any p Say that an integer $n$ is $p$-leading if its expansion in base $p$ starts with the digit $p-1$. My postdoc, Lifan Guan, asks: are there infinitely many positive integers $n$ that are not $p$-leading for any odd prime $p$? That is: is the set
$$S = \mathbb{N}\setminus \bigcup_{p\geq 3} \bigcup_{d\geq 1} [p^d - p^{d-1},p^d)$$
infinite?
[My estimation is this is interesting, doable and non-trivial. I also suspect it must have been studied before!]
 A: Following some discussion with Harald (offline)... It appears that the answer is YES, the set is indeed infinite.
First ingredient:
We take the idea of adding upper densities, but use logarithmic densities instead of natural densities (logarithmic density meaning one sums reciprocals to $x$, divides by $\log{x}$, and sends $x$ to infinity). A short computation bounds the upper logarithmic density of integers which start with $p-1$ in some base $p > C$ by $\sum_{p > C} \frac{1}{(p-1)\log{p}}$. Unfortunately, this sum is larger than $1$ when $C=2$, so this does not immediately imply a positive answer to the question. So we have to work harder.
2nd component: One shows that if $g_1,\dots,g_m \ge 2$ are integers with $1/\log{g_1}, \dots, 1/\log{g_m}$ being $\mathbb{Q}$-linearly independent, then the fractional parts $\{\log_{g_1} n\}, \dots, \{\log_{g_m} n\}$ are distributed like independent uniform random variables on $[0,1)$, with respect again to logarithmic density. This uses a Weyl-type criterion; the key point is that if $\gamma \ne 0$, then $e^{2\pi i \gamma \log{n}} = n^{2\pi i \gamma}$ has logarithmic mean value $0$. (The condition that $\gamma \ne 0$ explains the linear independence condition required on the numbers $1/\log g_i$.)
Now $1/\log{3}$ and $1/\log{5}$ are easily seen to be $\mathbb{Q}$-linearly independent, and so one gets from the second component that the integers which do not begin with $2$ in base $3$, and do not begin with $4$ in base $5$, comprise a set with logarithmic density
$$ (1-\log_3(3/2)) (1-\log_5(4/5)) \approx 0.543.$$
Then one uses the first component with $C=5$. The sum of $\frac{1}{(p-1)\log{p}}$ for $p>5$ can be proved to be $< 0.53$ (say). It follows that the set in question is indeed infinite, with lower logarithmic density $> 1/100$.
If $\{1/\log{p}\}_{p\ge 3}$ is a linearly independent set (over $\mathbb{Q}$), modifying the above argument would show that the logarithmic density is exactly $\prod_{p \ge 3} (1-\log_p(p/(p-1))$. But I think this linear independence is open.
