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


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 $\Omega(n)$.

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    $\begingroup$ arxiv.org/pdf/1906.02847.pdf $\endgroup$ Commented Jun 21, 2020 at 21:58
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    $\begingroup$ 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. $\endgroup$ Commented Jun 22, 2020 at 1:12
  • $\begingroup$ @StevenClark, thanks for the input but Peter Humphries link gives us our full expected answer. $\endgroup$
    – Milo Moses
    Commented Jun 22, 2020 at 3:35
  • $\begingroup$ Perhaps, Milo, you could summarize that link and post your summary as an answer. $\endgroup$ Commented Jun 22, 2020 at 7:51
  • $\begingroup$ That's a great idea! I'll do that. $\endgroup$
    – Milo Moses
    Commented Jun 22, 2020 at 15:44

1 Answer 1


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


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


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.

  • $\begingroup$ should there be an upper bound on $|f(n)|$ in your second paragraph? $\endgroup$
    – kodlu
    Commented Jun 27, 2020 at 4:29
  • $\begingroup$ @kodlu yes! Sorry, I forgot to but it. $\endgroup$
    – Milo Moses
    Commented Jun 27, 2020 at 21:11

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