Let $k = p_1^{a_1}p_2^{a_2}\cdots p_{\omega(k)}^{a_{\omega(k)}}$ be the prime factorization of $k$ where $p_1, p_2, \ldots p_{\omega(k)}$ are the prime factors of $k$. Define
$$ f(k) = \frac{1}{p_1^{a_1}} + \frac{1}{p_2^{a_2}} + \cdots + \frac{1}{p_{\omega(k)}^{a_{\omega(k)}}} $$
What is the limiting value of $\lim_{n \to \infty}\frac{1}{n} \sum_{k = 1}^n f(k)$?
Experimental data: For $n = 17954000000$, the value is $0.330229887335653$ while for $n = 24178000000$, it is $0.330229866905716$ which show that the value is decreasing at a very slow rate.
$$F(s)=\sum_{n=1}^\infty f(n)n^{-s} = \sum_{n\ge 1} n^{-s} \sum_{p^r\| n} p^{-r}=\sum_{n\ge 1} \sum_{p^r\| n} p^{-r-rs} (n/p^r)^{-s}$$ $$= \sum_{p^r} p^{-r-rs} \sum_{n\ge 1,p\ \nmid\ n} n^{-s}= \sum_{p^r} p^{-rs-r}(1-p^{-s})\zeta(s)= G(s)\zeta(s)$$
$G(s)=\sum_{p^r} p^{-rs-r}(1-p^{-s})= \sum_{m\ge 1} g_m m^{-s}$ converges absolutely for $s> 0$ and $G(1)\ne 0$. Not hard to see that $\sum_{m\le N} g_m=o(N)$. Therefore $$\sum_{n\le N} f(n) = \sum_{m\le n} g_m \lfloor N/m \rfloor = \sum_{m\le N} g_m\frac{N}m + O(\sum_{m\le N} g_m) = N (G(1)+ o(1))+o(N)$$ whence $$C=\lim_{N\to \infty}\frac1N \sum_{n\le N} f(n)=G(1)=\sum_p p^{-2}\frac{1-p^{-1}}{1-p^{-2}}$$ This series doesn't have a closed-form. You can compute it efficiently as $$C= \sum_{r\ge 0} (-1)^r \sum_p p^{-2-r}= \sum_{r\ge 0} (-1)^r \sum_{k\ge 1} \frac{\mu(k)}{k} \log \zeta(k(2+r))$$ using that $\log \zeta(k(2+r))=O(2^{-k(2+r)})$ to obtain some rapidly convergent approximations to $C$.