# 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 $$14$$.

• This might be an interesting question if you edit to include a short table showing the sequence for the first few values of $n$. Do that for several bases. Binary might be the cleanest - then you're just counting the number of $1$s. – Ethan Bolker Jun 23 at 14:00
• According to my program, $$n=20619661$$ has order $14$ – Peter Jun 23 at 14:13
• Yeap, but there is also the case that none of them is true. – Βασίλης Μάρκος Jun 23 at 14:56
• Upto $n=10^9$ , there is no number with an order greater than $14$ – Peter Jun 23 at 17:00
• @Peter I think there is a global maximum. – Grešnik Jun 23 at 17:12

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 $$W$$.
• But, you assume that "assuming any slight independence between the events $S_d(n^m)$ (for n fixed as m ranges), which is reasonable,", can you explain why do you think that is reasonable? – Grešnik Jun 27 at 9:13
• @Grešnik I see no reason why the fact that the sum of the digits of one power of $n$ being prime should influence the fact that the sum of digits of another power of $n$ is prime. Of course, everything here is deterministic, so it is a bit silly to speak in the language of probability and hard for me to explain myself, but that is the best I can do. – mathworker21 Jun 27 at 9:32