4
$\begingroup$

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 $\Omega(n)$.

$\endgroup$
5
  • 2
    $\begingroup$ arxiv.org/pdf/1906.02847.pdf $\endgroup$ – Peter Humphries Jun 21 '20 at 21:58
  • 1
    $\begingroup$ 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. $\endgroup$ – Steven Clark Jun 22 '20 at 1:12
  • $\begingroup$ @StevenClark, thanks for the input but Peter Humphries link gives us our full expected answer. $\endgroup$ – Milo Moses Jun 22 '20 at 3:35
  • $\begingroup$ Perhaps, Milo, you could summarize that link and post your summary as an answer. $\endgroup$ – Gerry Myerson Jun 22 '20 at 7:51
  • $\begingroup$ That's a great idea! I'll do that. $\endgroup$ – Milo Moses Jun 22 '20 at 15:44
7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ should there be an upper bound on $|f(n)|$ in your second paragraph? $\endgroup$ – kodlu Jun 27 '20 at 4:29
  • $\begingroup$ @kodlu yes! Sorry, I forgot to but it. $\endgroup$ – Milo Moses Jun 27 '20 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.