Prove that the asymptotic density of $w(n)|n$ is 0

Question

Note $w(n)$ is number of primes dividing $n$. I know the definition of asymptotic density, but I'm not sure how to start with this problem. I can prove that the sets $w(n)|n$ and $w(n)\nmid n$ are infinite. Any hint would be appreciated.

Answer 1

Partitioning according to the value of $\omega(n)$ we obtain $$\sum_{n\leq x\atop \omega(n) \mid n} 1 \leq \sum_{k=1}^\infty \sum_{n\leq x, k \mid n \atop \omega(n) =k } 1 = \sum_{k \leq x } \sum_{m\leq x/k \atop \omega(mk) =k } 1 .$$ By the Hardy-Ramanujan theorem we have that the number of integers $n\leq x $ with $$|\omega(n) - \log \log x| > (\log \log x)^{3/4}$$ is $o(x)$. Therefore, the contribution of $k$ with $|k - \log \log x| > (\log \log x)^{3/4}$ is $o(x)$, which you can get just by ignoring the condition $k \mid n$. Hence,$$\sum_{n\leq x\atop \omega(n) \mid n} 1\leq o(x)+ \sum_{|k - \log \log x| \leq (\log \log x)^{3/4} } \sum_{m\leq x/k \atop \omega(mk) =k } 1 .$$ Now one might get asymptotics for the double sum on the right side by less wasteful arguments but let's only see what happens if you throw away the information $\omega(mk)=k$. We get $$\sum_{n\leq x\atop \omega(n) \mid n} 1\leq o(x)+ \sum_{|k - \log \log x| \leq (\log \log x)^{3/4} } \frac{x}{k} \ll o(x)+ O(x/\log \log x)+x \int_{|u - \log \log x| \leq (\log \log x)^{3/4}} \frac{\mathrm{d}u}{u} .$$ It remains to show that the integral is $o(1)$. Bounding the integrant by its maximum value we get that the integral is $$ \ll \frac{1}{\log \log x } \int_{|u - \log \log x| \leq (\log \log x)^{3/4}} 1\mathrm{d}u\ll (\log \log x)^{-1/4} .$$ This shows that $$\sum_{n\leq x\atop \omega(n) \mid n} 1= o(x) +O(x(\log \log x )^{-1/4})=o(x).$$

It would be interesting to find the true the asymptotic size of $$\sum_{n\leq x\atop \omega(n) \mid n} 1 .$$