And I guess that in base $b$, the mean would be $(b+1)/2b$.

Is there a form of the conjecture that depends on $a$? ie, as $a\rightarrow n$?

Chapter 3 of H. Davenport, Multiplicative Number Theory, Third Edition, has some relevant material on this, I think.

Hope you don't mind me switching notation from $PF(n)$ to $\Omega(n)$. :) $\Omega(n) \leq \tau(n)$, the number of divisors of $n$, and $\sum_{n \leq x}\tau(n) \sim x \log x$ by a result of Dirichlet, so this gives what you want.

Let $A_n$ be the roughly $n/6$ integers in $(n/3,n/2)$. Let $B_n$ be the integers in $(n/2 + 1,n]$ that are of the form $2m$, $m \in A_n$. Let $C_n$ be the integers in $(n/2+1,n]$. By considering $C_n \setminus B_n$ we get $2^{n/2-n/6}$. By considering $A_n$ and $B_n$ we get $3^{n/6}$. So a lower bound is $2^{(0.5+r/6)n}$ where $2^r=1.5$.

I was kind of thinking that $\sum_{n \leq x}\mu_k(n)$ would be a way to "measure" the frequencies of $\omega(n) \bmod k$.