Does there exist a power-summer of order $+ \infty$?

We start with some number $n$ and sum its digits (we can denote sum-of-digits function as $S_d$) to obtain number $S_d(n)$.

If $S_d(n)$ is prime then we calculate number $n^2$ and sum its digits to obtain $S_d(n^2)$. If $S_d(n^2)$ is prime then we calculate number $n^3$ and sum its digits to obtain $S_d(n^3)$, and so on...

We can call number $n$ a power-summer of order m if the numbers $S_d(n),...S_d(n^m)$ are all primes.

We can call number $n$ a power-summer of order $+ \infty$ if $n$ is power-summer of order m for every $m \in \mathbb N$

A question is:

Does there exist a power-summer of order $+ \infty$?

Are you of the opinion that there is some global maximum, that is, a natural number $W$ such that order of every $n$ is less than $W$?

An answer is not in my reach, I do not know much about sum-of-digits functions, but maybe someone has some good ideas.

Peter found a number of order $14$, a number $20619661$ and calculated that upto $n=10^9$ there is no number with an order greater than $14$.

Solution 1:

Basic probabilistic heuristics suggest that there is no power-summer of order $+\infty$ nor is there some global maximum $W$.

If we view $n^m$ as a random number between $0$ and $9\log_{10}(n^m) = 9m\log(n)$, then $Pr(S_d(n^m) \text{ is prime}) \approx \frac{1}{S_d(n^m)} \approx \frac{1}{4.5m\log(n)}$. So assuming any slight independence between the events $S_d(n^m)$ (for $n$ fixed as $m$ ranges), which is reasonable, will give that the probability that each $S_d(n^m)$ is prime is $0$.

However, for any given positive integer $W$, $Pr(S_d(n^1),\dots,S_d(n^W) \text{ are prime}) \approx \frac{1}{4.5^W W!}\frac{1}{(\log n)^W}$, so since $\sum_{n \ge 1} \frac{1}{(\log n)^W} = +\infty$, Borel-Cantelli suggests that infinitely many positive integers are power-summer of order $W$.