Is the parity of $\omega(n)$ equally distributed? I recently learned that the prime omega function $\Omega(n)=\Omega\left(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}\right)=\alpha_1+\alpha_2...+\alpha_k$ is very well studied. In particular, we know that $\Omega(n)$ is equally often even and odd. This statement is, in fact, equivalent to the prime number theorem.

My question is, do we know anything about the distribution of parities of $\omega(n)=\omega\left(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}\right)=k$?

It is natural to assume that $\omega(n)$ is equally often even and odd, but perhaps it is much harder to show. From what I understand the reason that the distribution of $\Omega(n)$ is so much easier to analyze is that the Liouville lambda function $\lambda(n)=(-1)^{\Omega(n)}$ is very well understood and it's summary function $L(x)=\sum_{n<x}\lambda(n)$ can be related to the Mobius/Mertens function by
$$L(x)=\sum_{d^2<x}M\left(\frac{x}{d^2}\right)$$
The Mertens function is obviously very well studied, but no such inversion formulas are possible for $\omega(n)$ so we cannot use methods like this. I am curious about not only whether or not the result I ask for is known but whether or not the result is easier/harder to prove than the equivalent result for $\Omega(n)$.
 A: In Peter Humphries link he answers the question very well, but by looking at the results cited I learned that this is in fact a special case of a more general phenomenon.
If $f(n)$ is a (real valued) multiplicative function with $\left|f(n)\right|\leq1$, then it's mean value $M=\lim_{x\to\infty}\frac{1}{x}\sum_{n<x}f(n)$ exists. Moreover, if the series
$$\sum_{p}\frac{1-f(p)}{p}$$
diverges then $M=0$. This is theorem 6.4 In Elliot's "Probabilistic Number Theory", attributed to Wirsing. Both $(-1)^{\Omega(n)}$ and $(-1)^{\omega(n)}$ are multiplicative since $\Omega(n)$ and $\omega(n)$ are additive. They both only take values in $\pm1$ and so their mean values must exist. By definition of $\omega$ and $\Omega$ we have
$$\sum_{p}\frac{1-(-1)^{\Omega(p)}}{p}=\sum_{p}\frac{1-(-1)^{\omega(p)}}{p}=\sum_{p}\frac{1-(-1)}{p}=+\infty$$
and thus they both must have average order $0$, meaning equidistribution of parities.
It is true though that the investigation of the parity of $\omega(n)$ is more complicated though. As I mentioned in the question, the equidistribution of parities of $\Omega(n)$ was known before the proof of the PNT to be equivalent to it, and so when the PNT was proved in 1896 the equidistribution of parities of $\Omega(n)$ was settled. The equidistribution of parities of $\omega(n)$, however, was only settled in 1975 by van de Lune and Dressler.
The "general result" of the mean values of multiplicative functions that can be used to settle the equidistribution of $\omega(n)$ is new, namely, Elliot's book was only published in 1979. It is interesting to think that this is so close to the result of van de Lune and Dressler.
