Let $\omega(n) = \sum_{p \mid n} 1$. Robin proves for $n > 2$, \begin{align} \omega(n) < \frac{\log n}{\log \log n} + 1.4573 \frac{\log n}{(\log \log n)^{2}}. \end{align} Is there a similar tight effective upper bound for $\Omega(n) = \sum_{p \mid n} \text{ord}_{p}(n)$ or at least an upper bound in terms of $\omega(n)$?

  • $\begingroup$ +1 interesting. Would the Riemann Hypothesis make a difference? $\endgroup$ – draks ... Aug 6 '12 at 6:08
  • $\begingroup$ "Robin proves..." - where is this, if I may ask? $\endgroup$ – J. M. is a poor mathematician Aug 6 '12 at 6:16
  • 2
    $\begingroup$ @J.M., Robin proves something like this in Estimation de la fonction de Tchebychef $\theta$ sur le $k$-ieme nombre premier et grandes valeurs de la fonction $\omega(n)$ nombre de diviseurs premiers de $n$, Acta Arith 42 (1983) 367-389, MR0736719 (85j:11109). $\endgroup$ – Gerry Myerson Aug 6 '12 at 7:17
  • 1
    $\begingroup$ NB: According to Hardy and Ramanujan, the normal value of both $\omega(n)$ and $\Omega(n)$ is $\log \log n$. $\endgroup$ – user02138 Aug 7 '12 at 1:30
  • 1
    $\begingroup$ $\omega(n)\le n-1$ is exact precisely twice, $n=1$ and $n=2$. $\Omega(n)\le\log n/\log 2$ is exact infinitely often. Robin's bound isn't very close to the actual value if $n$ is a large prime. So I'm having trouble grasping what you're getting at. $\endgroup$ – Gerry Myerson Aug 7 '12 at 3:23

The number of prime divisors counted with multiplicity is maximized for powers of $2$ and so

$$\Omega(n)\le\frac{\log n}{\log 2}=\log_2 n$$

and since it is exactly equal for infinitely many $n$ it is also the tighest possible bound.


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.