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Let $\omega(n)$ be the number of distinct prime factors of $n$ (without multiplicity, of course). I know some results about average of $\omega(n)$. But I didn't find any result about the following: Let $k\geq 2$ be an integer, then there exists some good lower bound for $$ \# \{n\leq x : \omega(n)\geq k\}? $$

Good means we can ensure that the set $\{n\geq 1 : \omega(n)\geq k\}$ has positive upper density, i.e., $$ \lim_{x\to \infty}\sup\displaystyle\frac{ \# \{n\leq x : \omega(n)\geq k\} }{x} $$ is positive? If so, some lower bound?

Thanks a lot!

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2 Answers 2

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In fact, for any fixed $k$, the set $\{n\colon \omega(n)\ge k\}$ has density $1$. This follows from the Hardy–Ramanujan theorem, which is the even stronger statement that for any fixed $\varepsilon>0$, the set $$ \{ n\colon |\omega(n) - \log\log n| < (\log\log n)^{1/2+\varepsilon} \} $$ has density $1$.

A related result, often called the Selberg–Sathe theorem, actually gives the asymptotic formula $$ \#\{n\le x\colon \omega(n)=k\} \sim \frac x{\log x}\frac{(\log\log x)^{k-1}}{(k-1)!}. $$ In particular, this set has density $0$ for every fixed $k$.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – Jean
    Commented Mar 3, 2020 at 19:56
  • $\begingroup$ Dear @Greg Martin, where can I find the asymptotic formula $$ \#\{n\le x\colon \omega(n)=k\} \sim \frac x{\log x}\frac{(\log\log x)^{k-1}}{(k-1)!}? $$ I found here: arxiv.org/pdf/1709.01963.pdf, but with some factor $G((k-1)/\log \log x)$, where $G$ is a function depending on an infinite product and the Gamma function. Thanks $\endgroup$
    – Jean
    Commented Mar 5, 2020 at 17:39
  • $\begingroup$ That version is also correct, indeed even better! That version works even if $k$ is growing with $x$. If $k$ is fixed, then the argument of $G$ tends to $0$ and the factor $G(0)=1$ disappears, leaving us with the version I put in this answer.—Two books one can find this theorem (indeed the uniform version you linked to, which implies the version in the answer) are Tenenbaum's Introduction to Probabilistic and Analytic Number Theory and Montgomery & Vaughan's Multiplicative Number Theory I. $\endgroup$
    – Greg Martin
    Commented Mar 6, 2020 at 0:59
  • $\begingroup$ I would like to know if this function $G(z)=\Gamma(1+z)^{-1}\prod_{p} (1+z/p)(1-1/p)^z$ is well defined for all $z\in (0,2)$. Any suggestion? $\endgroup$
    – Jean
    Commented May 19, 2020 at 19:51
  • $\begingroup$ Yes, one can show that the infinite product converges uniformly and absolutely for $z$ in any bounded set. $\endgroup$
    – Greg Martin
    Commented May 19, 2020 at 22:05
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It follows from the Erdős–Kac theorem that this natural density is $1$ for all $k$. Informally speaking, the distribution of $\omega(n)$ approaches a normal distribution with mean and variance $\log\log n$, so the proportion of values below any fixed finite bound goes to zero.

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