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The prime number theorem is equivalent to the statement that $$\sum_{n\le x}\mu(n)=o(x),$$ which in turn is equivalent to the statement that the total number of prime factors $\Omega(n)$ is even $1/2$ of the time, and odd $1/2$ of the time. I am interested in the natural generalisation, where we consider the probability of $\Omega(n)$ lying in any of the residue classes $\{0,1,2,\cdots,m-1\}$ modulo $m$; by the Pillai-Selberg theorem, $\Omega(n)$ equidistributes amongst these residue classes. But I want to know what the Möbius function analogue of this is?

Is there some function like $\mu(n)$ but which may be adapted to this setting?

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    $\begingroup$ You can count the number of squarefrees up to $x$ with a given value of $\Omega(n)\bmod m$ and prove that it is $\frac 6{\pi^2}\cdot\frac xm+o(x)$ by standard methods. $\endgroup$ Commented Sep 19 at 12:10
  • $\begingroup$ @AlexeiEntin what is the analogue of $\mu(n)$ here though? $\endgroup$ Commented Sep 19 at 13:06

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The statement that $\Omega(n)$ is even half of the time, and odd half of the time, is better captured by the statement $$\sum_{n\leq x}(-1)^{\Omega(n)}=o(x),$$ where $n\mapsto(-1)^{\Omega(n)}$ is the Liouville function. This bound is known of course. You can generalize the Liouville function as $n\mapsto\xi^{\Omega(n)}$, where $\xi$ is any $m$-th root of unity. Such a function is completely multiplicative, because $\Omega(n_1n_2)=\Omega(n_1)+\Omega(n_2)$. By the Selberg-Delange method (see Theorem 2 in Chapter II.6 of Tenenbaum: Introduction to analytic and probabilistic number theory), we have $$\sum_{n\leq x}\xi^{\Omega(n)}=o_m(x)$$ as long as $\xi$ is an $m$-th root of unity different from $1$. In contrast, for $\xi=1$, the left-hand side is asymptotically $x$. Therefore, \begin{align*} \frac{1}{x}\#\{n\leq x:\ \Omega(n)\equiv r\!\!\!\pmod{m}\} &=\frac{1}{x}\sum_{n\leq x}\frac{1}{m}\sum_{\xi^m=1}\xi^{\Omega(n)-r}\\ &=\frac{1}{m}\sum_{\xi^m=1}\xi^{-r}\frac{1}{x}\sum_{n\leq x}\xi^{\Omega(n)}\\ &=\frac{1}{m}+o_m(1). \end{align*} Hence $\Omega(n)$ hits each residue class modulo $m$ asymptotically $1/m$ of the time.

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    $\begingroup$ One addendum is that if you ask for more precise estimates on the behavior of these error terms, you will get very different answers. The reason is that the Dirichlet series $\sum_{n=1}^\infty \xi^{\Omega(n)} n^{-s}$ has a singularity at $s=1$ for $\xi \neq -1$. For $m=-1$ this shows $\sum_{n\leq x} \xi^{\Omega(n)}$ has size comparable to $x$ over some power of $\log x$. The error term in the prime number theorem, on the other hand, decays much faster than this. $\endgroup$ Commented Sep 19 at 17:30
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    $\begingroup$ @WillSawin Absolutely, and I thank you for this remark. In fact the theorem that I quote from Tenenbaum's book gives an asymptotic expansion with precise lower order terms. For $\xi=-1$, all these lower order terms vanish (by the poles of the $\Gamma$ function), but for $\xi\neq -1$ this is no longer the case. $\endgroup$ Commented Sep 19 at 17:33
  • $\begingroup$ See this question. $\endgroup$ Commented Sep 20 at 20:58

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