TITLE: On the density of a certain sequence of integers
QUESTION [10 upvotes]: A Bogotá number is a positive integer equal to some smaller number, or itself, times its digital product, i.e. the product of its digits. For example, 138 is a Bogotá number because 138 = 23 x (2 x 3).
Here is the list of Bogotá numbers not greater than 1000: 0, 1, 4, 9, 11, 16, 24, 25, 36, 39, 42, 49, 56, 64, 75, 81, 88, 93, 96, 111, 119, 138, 144, 164, 171, 192, 224, 242, 250, 255, 297, 312, 336, 339, 366, 378, 393, 408, 422, 448, 456, 488, 497, 516, 520, 522, 525, 564, 575, 648, 696, 704, 738, 744, 755, 777, 792, 795, 819, 848, 884, 900, 912, 933, 944, 966, 992.
What is the density of these numbers?
Another issue regarding these numbers has been dealt with at https://puzzling.stackexchange.com/questions/98998/pairs-of-bogot%c3%a1-numbers/99006#99006
https://oeis.org/A336826

REPLY [5 votes]: The natural density is $0$. We will show this by showing that the natural density is at most $9!^{-n}$. Fix $n>0$, and consider the Bogotá numbers less than $10^m$ for some large $m$. 
Let $p(k)$ denote the product of the digits of $k$. Firstly, it is clear that the Bogotá numbers that can be written as $kp(k)$ with each digit from $1$ to $9$ occurring at least $n$ times in $k$ must all be multiples of $9!^{-n}$, and thus there are at most
$$\frac{10^m}{9!^n}$$
of them less than $10^m$, and so it suffices to consider the Bogotá numbers that are not of this type. 
Consider the set of numbers less than $10^m$ for some $m$ with $<n$ copies of the digit $i$ for some $1\leq i\leq 9$. For each $i$, there are at most
$$\sum_{k=i}^{n-1}\binom{m}{i}9^{m-i}\leq n9^m\binom{m}{n-1}$$
of these numbers as long as $m\geq 2n$. So, there are in total at most
$$n9^{m+1}\binom{m}{n-1}$$
of these numbers. If $9!^n\nmid K$ and $K$ is a Bogotá number, then we must be able to write it as $kp(k)$ for some $k\leq K$ so that $k$ has $<n$ copies of each digit from $1$ to $9$. As such, there are at most
$$n9^{m+1}\binom{m}{n-1}$$
number of this form (we may have duplicates). Therefore, there are at most
$$\frac{10^m}{9!^n}+n9^{m+1}\binom{m}{n-1}$$
Bogotá numbers less than $10^m$ for $m$ large. 
Now, as $m\to\infty$,
$$\frac{n9^{m+1}\binom{m}{n-1}+\frac{10^m}{9!^n}}{10^m}\to \frac1{9!^n},$$
so the natural density of the Bogotá numbers is at most $\frac1{9!^n}$. Taking $n\to\infty$, this gives that this natural density must be $0$.