Searching for rock-hard integers.

Question

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! $256 = 2^6*4$.

Answer 1

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 ($117649$).