# Searching for rock-hard integers.

For integer $$k\geq 0$$ written $$d_jd_{j-1}\dots d_1$$, where $$d_i$$ are single-digit integers, define $$R(k) := \begin{cases} d_j^{d_1}d_{j-1}^{d_2} \dots d_{j/2}^{{d_{j/2 + 1}}}, & j \text{ is even.}\\ d_j^{d_1}d_{j-1}^{d_2} \dots d_{(j+1)/2 -1}^{d_{(j+1)/2+1}} d_{(j+1)/2}, & j \text{ is odd.} \end{cases}$$ Which is a bit of an eyesore, though some examples may help: $$R(1392) = 1^2 * 3^9$$, $$R(793) = 7^3*9$$, $$R(12345678) = 1^8 * 2^7*3^6 * 4^5$$.

Now, say an integer $$n \geq 0$$ is rock-hard if $$R(n) =n$$. $$0, 1, 2, 3 \dots 9$$ are rock-hard. There are no rock-hard two-digit integers. This can be verified in about five seconds with the question: is there a power of $$n$$ between $$10n$$ and $$10(n+1)$$ for $$0 < n < 10$$? I do not believe there are any three-digit rock hard integers.

My question: is there a more efficient way to search for rock-hard numbers than just checking all numbers? Certainly, we can omit $$111$$, $$121 \dots 191$$, $$11111,$$ $$11211,$$ etc., but can we further narrow down what numbers we check for rock-hardness?

Motivation: John Conway did something similar to this, though I can't remember what he did nor find a reference. I believe N.J.A. Sloane chipped in as well.

Various comments: $$256$$ is almost rock-hard but not quite! $$256 = 2^6*4$$.

• I would just write a program and search through eight digits, which should run quickly. I suspect there are very few. After that there are so many ways to miss I am sure there are no more. Of course, this is just a hand wave. Commented Jan 14, 2020 at 3:17

You can do much better than checking all numbers... note that any rock-hard integer is equal to a product of powers of $$1,2,\ldots,9$$; indeed, all rock-hard integers must be $$7$$-smooth. There aren't that many $$7$$-smooth numbers... below $$N$$, there can't be more than $$(\log_2 N + 1)(\log_3 N + 1)(\log_5 N + 1)(\log_7 N + 1)$$ of them. A simple script verifies that there are only $$2867708$$ candidates below $$10^{48}$$, and checking them all turns up only one rock-hard integer with more than one digit ($$117649$$).