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I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less or equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set $$ \mathcal{J}_T(x,k) = \{ n \in \mathbb{N} : n \le x, \Omega(n)=k , n \mbox{ is square free } \}.$$ Other way to describe this cardinal, using the Möbius function, is $$ |\mathcal{J}_T(x,k) | = \sum_{\Omega(n) = k, n \le x } |\mu(n)|.$$

I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using $$ \sum_{\Omega(n) = k, n \le x } |\mu(n)| \le \sum_{n \le x } |\mu(n)| \le \frac{6x}{\pi^2} + O(\sqrt{x}) \ll x, $$
is not good enought for my purposes.

Thanks in advanced, any reference or idea is helpful

user64494
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Martin Mansilla
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  • Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using \sum_{d^2 | n} \mu(d)) the number of such squarefree integers \leq x with \omega(n) = k as an alternating sum of, for each d\leq x^{1/2}, the number of integers \leq x/d^2 with exactly k - \omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below: – alpoge May 29 '19 at 23:16
  • https://mathoverflow.net/questions/35927/asymptotic-density-of-k-almost-primes – alpoge May 29 '19 at 23:17
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    Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers \leq x with two prime factors seems to grow like x \log\log{x} / \log{x}. Who among us, even with our computers, has seen \log\log{x} grow?! (Canonical quote about \log\log{x} going to infinity with dignity.) What intuition... – alpoge May 29 '19 at 23:21

2 Answers2

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You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.

I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula $$\sum_\text{$n$ square-free}z^{\omega(n)}n^{-s}=\prod_p\left(1+\frac{z}{p^s}\right).$$ Then you will need to "factor out" $\zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.

GH from MO
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It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas \begin{align*} \#\{ n\le x\colon \omega(n) = k \} &\sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!} \\ \#\{ n\le x\colon \Omega(n) = k \} &\sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!}. \end{align*} (Here $\omega$ and $\Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)

Note that any nonsquarefree integer $n$ with $\Omega(n) = k$ satisfies $\omega(n) = j$ for some $1\le j\le k-1$. Therefore the bounds \begin{align*} \#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} &\le \#\{ n\le x\colon \Omega(n) = k \} \\ \#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} &\ge \#\{ n\le x\colon \Omega(n) = k \} - \sum_{j=1}^{k-1} \#\{ n\le x\colon \omega(n) = j \}, \end{align*} together with the Selberg–Sathe asymptotic formulas above, immediately imply that $$ \#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} \sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!}. $$

This argument can be made to work with some uniformity in $k$ as well (I think $k=o(\log\log x)$ is what is needed).

Greg Martin
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