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 $\omega(n)$?
Effective Upper Bound for the Number of Prime Divisors
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0+1 interesting. Would the Riemann Hypothesis make a difference? – 2012-08-06
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0"Robin proves..." - where is this, if I may ask? – 2012-08-06
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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). – 2012-08-06
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0I 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"? – 2012-08-07
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0Perhaps 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. – 2012-08-07
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1NB: According to Hardy and Ramanujan, the normal value of both $\omega(n)$ and $\Omega(n)$ is $\log \log n$. – 2012-08-07
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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. – 2012-08-07
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0Perhaps $\Omega(n) \leq \log_{2} n$ is the best one can hope for.... Thanks, again, Gerry! – 2012-08-07
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0I deleted my answer because I thought I misunderstood the question but based on the comments I no longer think I did. – 2012-08-07
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0Yes, 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. – 2012-08-07
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0The 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. – 2014-11-25
1 Answers
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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.