TITLE: Gaps between Bogotá numbers
QUESTION [5 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).
Bogotá numbers are known to have natural density 0 (On the density of a certain sequence of integers).
Also many instances of two consecutive integers being both Bogotá numbers (such as 24 and 25) are known (https://puzzling.stackexchange.com/questions/98998/pairs-of-bogot%c3%a1-numbers?noredirect=1#comment281441_98998).
Question: Can the gap between two consecutive Bogotá numbers be arbitrarily large?
Bogotá numbers less than or equal to 1000 are as follows: 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.
The initial sequence of gaps is: 1, 3, 5, 2, 5, 8, 1, 11, 3, 3, 7, 7, 8, 11, 6, 7, 5, 3, 15, 8, 19, 6, 20, 7, 21, 32, 18, 8, 5, 42, 15, 24, 3, 27, 12, 15, 15, 14, 26, 8, 32, 9, 19, 4, 2, 3, 39, 11, 73, 48, 8, 34, 6, 11, 22, 15, 3, 24, 29, 36, 16, 12, 21, 11, 22, 26, 16, 45, 18, 9, 24, 7, 23, 2, 61, 53, 28, 34, 4, 59, 11, 62, 64, 8, 13, 3, 72, 36, 12, 24, 32, 4, 3, 45, 12, 2, 94, 54, 2...
Among the first 10,000 Bogota numbers, the largest gap Freddy Barrera found was one of size 5712.

REPLY [5 votes]: Any set of natural numbers having a natural density of $0$ has no upper limit on the size of the gaps between consecutive elements of the set, i.e., they become arbitrarily large.
To show this, suppose instead there is a maximum gap of some integer $m \ge 1$. Also, let the set be $B$ and $b(n)$ be the number of elements of $B$ less than or equal to $n$. Note there must be at least one element of $B$ in each group of $m$ consecutive elements. In particular, there must be at least one between $im + 1$ and $(i+1)m$, inclusive, for all $i \ge 0$. Adding up these elements means that, for all $j \ge 0$, there must be at least $j$ elements between $1$ and $jm$, inclusive. Thus, if $n = km + r$, for some integers $k \ge 0$ and $0 \le r \lt m$, you have
$$b(n) \ge k = \left\lfloor \frac{n}{m} \right\rfloor \tag{1}\label{eq1A}$$
As Wikipedia's Natural density article states, the natural density $\alpha$ is
$$\alpha = \lim_{n\to \infty}\frac{b(n)}{n} \ge \lim_{n\to \infty}\frac{\left\lfloor\frac{n}{m}\right\rfloor}{n} = \frac{1}{m} \tag{2}\label{eq2A}$$
However, since $\alpha = 0$ for the Bogotá numbers, as was proven in the answer to your other question, this means the original assumption of a maximum gap cannot be true, i.e., the gaps become arbitrarily large.