# 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 can be related to the Mobius/Mertens function by

$$L(x)=\sum_{d^2

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 $$\Omega(n)$$.

• arxiv.org/pdf/1906.02847.pdf – Peter Humphries Jun 21 '20 at 21:58
• Perhaps the conjectured relationship $H(x)=\sum\limits_{n\le x}(-1)^{\omega(n)}=\sum\limits_{n\le x}a(n)\ M\left(\frac{x}{n}\right)$ where $a(n)=\sum\limits_{d|n}\mu(rad(d))$ and $rad(d)$ is the square-free kernel of $d$ might be of some use. – Steven Clark Jun 22 '20 at 1:12
• @StevenClark, thanks for the input but Peter Humphries link gives us our full expected answer. – Milo Moses Jun 22 '20 at 3:35
• Perhaps, Milo, you could summarize that link and post your summary as an answer. – Gerry Myerson Jun 22 '20 at 7:51
• That's a great idea! I'll do that. – Milo Moses Jun 22 '20 at 15:44

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 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.

• should there be an upper bound on $|f(n)|$ in your second paragraph? – kodlu Jun 27 '20 at 4:29
• @kodlu yes! Sorry, I forgot to but it. – Milo Moses Jun 27 '20 at 21:11