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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 $\omega(n)$?

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+1 interesting. Would the Riemann Hypothesis make a difference? –  draks ... Aug 6 '12 at 6:08
    
"Robin proves..." - where is this, if I may ask? –  J. M. Aug 6 '12 at 6:16
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@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 6 '12 at 7:17
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NB: According to Hardy and Ramanujan, the normal value of both $\omega(n)$ and $\Omega(n)$ is $\log \log n$. –  user02138 Aug 7 '12 at 1:30
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$\omega(n)\le n-1$ 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 7 '12 at 3:23
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1 Answer 1

up vote 4 down vote accepted

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.

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