On the density of a certain sequence of integers

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 $0$.