Let $s$ be an integer greater than 1. For each natural number $n$, let $\omega(n)$ be the number of prime divisors (counted with multiplicities) of $n$ modulo $s$. For $i \in \{0,1,\dots, s1\}$ and a positive integer $k$, set $$c_i(k) = \{x \in \{1, 2, \dots, k\} : \omega(x) = i\}.$$ Is it true that $\frac{c_i(k)}{k} \rightarrow \frac{1}{s}$ as $k \rightarrow \infty$ for all $i \in \{0,1, \dots, s1\}$?
Yes, this is true. More precisely, it is known that, for fixed $s$ and $k \to \infty$, one has
See 'On the residue class distribution of the number of prime divisors of an integer' by Coons and Dahmen; preprint at http://arxiv.org/abs/0906.1029 They also discuss the nature of the errorterm in more detail, and in particular show that for $s>2$ it cannot be Unrelated side remark: since you define $\omega$ precisely this is not a big deal, but using the standard notation for the number of distinct prime divisors for something else is somewhat 'dangerous' (at least, it almost lead me to give a wrong answer). 

