18

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 $\omega(n)$?

user02138
  • 17,314
  • +1 interesting. Would the Riemann Hypothesis make a difference? – draks ... Aug 06 '12 at 06:08
  • "Robin proves..." - where is this, if I may ask? – J. M. ain't a mathematician Aug 06 '12 at 06:16
  • 2
    @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). – Gerry Myerson Aug 06 '12 at 07:17
  • I don't understand the comment on Rick Sladkey's (deleted) answer. How can you get anything like $\Omega(n)=O(\log n/\log\log n)$, when for $n=2^k$ we have $\Omega(n)=\log n/\log2$? What is the distinction between "tightness" and "sharpness"? – Gerry Myerson Aug 07 '12 at 00:20
  • Perhaps I'm overlooking something obvious, but it seems to me that the primorials govern the behavior of $\omega(n)$ and they aren't encoded in Robin's upper bound. (Sharp meaning equal at certain points.) Robin's bound isn't sharp but it is tight. (Tight meaning close to the actual value.) Rick's bound is sharp but seems rather weak for all other integers. – user02138 Aug 07 '12 at 01:20
  • 1
    NB: According to Hardy and Ramanujan, the normal value of both $\omega(n)$ and $\Omega(n)$ is $\log \log n$. – user02138 Aug 07 '12 at 01:30
  • 1
    $\omega(n)\le n-1$ 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. – Gerry Myerson Aug 07 '12 at 03:23
  • Perhaps $\Omega(n) \leq \log_{2} n$ is the best one can hope for.... Thanks, again, Gerry! – user02138 Aug 07 '12 at 04:10
  • I deleted my answer because I thought I misunderstood the question but based on the comments I no longer think I did. – Rick Sladkey Aug 07 '12 at 07:04
  • Yes, Rick, you deleted your very reasonable answer. I thank Gerry for clarifying things. One always hopes for complicated answers, but sometimes the easy ones are the best. – user02138 Aug 07 '12 at 14:16
  • The original paper uses the constant 1.45743, and the bound applies for $n \geq 3$, with equality for $N_{47}$, which I believe to be the $47$th prime. – void-pointer Nov 25 '14 at 02:22

1 Answers1

17

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.

Rick Sladkey
  • 971