[Probabilistic number theory]: Sum of exponents from prime factorization of a natural number.

Question

For a natural number $n$ let

$$n=\prod_{k=1}^{\omega(n)} p_k^{e_k} \tag1$$

be its prime factorization where $p_k\in\Bbb P$ are all distinct primes and $e_k\geqslant 1$.

$\omega(n)$ denotes the number of different prime factors of $n$ where $$\omega(n) \stackrel.\approx \log\log n \tag2$$ and the Hardy-Ramanujan theorem gives precise meaning to $(2)$, namely that $\log\log n$ is a normal order of $\omega(n)$.

Question: Are there similar result for the sum of the exponents $e_k$ from $(1)$

$$g(n) = \sum_{k=1}^{\omega(n)} e_k \tag3$$

i.e. we know some function $f(n)$ such that $g(n)\stackrel.\approx f(n)$?

Answer 1

Following a comment, we have

$$n=\prod_{k=1}^{\Omega(n)} p_k$$

where $\Omega(n)$ is the number of prime factors of $n$ with multiplicity. Then

$$\Omega(n) = \sum_{k=1}^{\Omega(n)} 1 = \sum_{k=1}^{\omega(n)} e_k = g(n)$$

According to [1], $\Omega(n)\stackrel.\approx\log \log n$ so that

$$g(n) \stackrel.\approx\log \log n$$