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

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.

• Where did you see this? – marty cohen Jul 23 at 6:50
• What is $w(n)$? In what context are you investigating this question? – Greg Martin Jul 23 at 6:51
• @GregMartin $n\in\mathbb N$. I am considering fixing values of $w(n)$ and find n, but that fails. – Kai Jul 23 at 7:13
• Oh wait nvm I proved density is 0 for all $w(n)\ne 1$, but not sure about how to deal with perfect prime powers. – Kai Jul 23 at 7:16
• The number of proper powers (of integers even, not just primes) up to $x$ is $\sum_{k=2}^{\lfloor \log_2 x\rfloor} x^{1/k} \le x^{1/2} + \sum_{k=2}^{\lfloor \log_2 x\rfloor} x^{1/3} \le \sqrt x + x^{1/3}\log_2 x$. – Greg Martin Jul 23 at 16:26

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 $$\sum_{n\leq x\atop \omega(n) \mid n} 1 .$$