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.

  • $\begingroup$ Where did you see this? $\endgroup$ – marty cohen Jul 23 at 6:50
  • $\begingroup$ What is $w(n)$? In what context are you investigating this question? $\endgroup$ – Greg Martin Jul 23 at 6:51
  • $\begingroup$ @GregMartin $n\in\mathbb N$. I am considering fixing values of $w(n)$ and find n, but that fails. $\endgroup$ – Kai Jul 23 at 7:13
  • $\begingroup$ 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. $\endgroup$ – Kai Jul 23 at 7:16
  • $\begingroup$ The number of proper powers (of integers even, not just primes) up to $x$ is $\sum_{k=2}^{\lfloor \log_2 x\rfloor} x^{1/k} \le x^{1/2} + \sum_{k=2}^{\lfloor \log_2 x\rfloor} x^{1/3} \le \sqrt x + x^{1/3}\log_2 x$. $\endgroup$ – 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 $$\sum_{n\leq x\atop \omega(n) \mid n} 1 .$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.