Let $n$ be a natural number whose prime factorization is
$$n=\prod\_{i=1}^{k}p\_i^{\alpha\_i} \; .$$
Define a function $g(n)$ as follows
$$g(n)=\sum\_{i=1}^{k}p\_i {\alpha\_i} \;,$$
i.e., exponentiation is "demoted" to multiplication,
and multiplication is demoted to addition.
For example: $n=200=2^3 5^2$, $f(n) = 2 \cdot 3 + 5 \cdot 2 = 16$.


Define $f(n)$ to repeat $g(n)$ until a cycle is reached.
For example: 
$n=154=2^1 7^1 11^1$, 
$g(n)=20$,
$g^2(n)=g(20)=9$,
$g^3(n)=g(9)=6$,
$g^4(n)=g(6)=5$, and now $g^k(n)=5$ for $k \ge 4$. So $f(154)=5$.


It is clear that every prime is a fixed point of $f(\;)$.
I believe that $n=4$ is the only composite fixed point of $f(\;)$.



> 
> **Q1**. Is it the case that $4$ is the only composite fixed point
>  of $f(\;)$, and that there are no cycles of length greater than $1$?
>  (**Yes**: See *EmilJeřábek*'s comment.)
> 
> 
> **Q2**. Does every prime $p$ have an $n \neq p$ such that $f(n) = p$,
>  i.e., is every prime "reached" by $f(\;)$?
>  (**Yes**: See *JeremyRouse*'s answer.)
> 
> 
> 


There appear to be interesting patterns here. For example, it seems
that $f(n)=5$ is common.
(**Indeed**: See მამუკა ჯიბლაძე's graphical display.)



Prime factorization "demoted" leads to function whose fixed points are primes
A way to get a non-trivial solution to $f(n) = p$ is that every odd number $\geq 7$ can be written as a sum of three primes (by Helfgott's recent work), so if $p \geq 7$ is prime, we can write $p = q + r + s$, and we have $g(qrs) = q + r + s = p$. (This is of course a bit overkill, we don't really need such a difficult result to see this.)


