Lower bound related to the number of distinct prime numbers

Question

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!

Answer 1

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 $k$.

Answer 2

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.