Prime factorization "demoted" leads to function whose fixed points are primes 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.)
 A: Here is a (portion of a) histogram of $|f^{-1}(n)|$ for $n=5,\ldots,10^6$
(newly updated from $10^5$ to $10^6$):

 
 
 
 
 


$266429$ of those numbers $n \le 10^6$ map $f(n)=5$;
$152548$ map $f(n)=7$.
A: 

Was thinking about this old question again, and 
made another image (sorry it is unreadable),
in the style of user @მამუკა ჯიბლაძე, of the numbers $\le 2000$ that ultimately map to $5$ ($5$ and $6$ are in the center, with $8$ right
and $9$ left, each connecting to $6$).
For example:
\begin{eqnarray}
n = 1788 &=& 2^2 \; 3^1 \; 149^1\\
g(1788) &=& 2 \cdot 2 + 3 \cdot 1 + 149 \cdot 1 = 156\\
n = 156 &=& 2^2 \; 3^1 \; 13^1 \\
g(156) &=&  2 \cdot 2 + 3 \cdot 1 + 13 \cdot 1 = 20\\
n = 20 &=& 2^2 \;  5^1\\
g(20) &=& 2 \cdot 2 + 5 \cdot 1 = 9\\
n = 9 &=& 3^2\\
g(9) &=& 3 \cdot 2 = 6\\
n = 6 &=& 2^1 \; 3^1 = 5\\
g(5) &=& 5
\end{eqnarray}
The overall structure of the $5$-sink remains the same as far as I can
take the computations.
A: 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.)
