Let $ω(n)$ be the number of distinct prime factors of $n$.
Is the inequality $ω(n)\leq C\log\log(n)$ true and if so what is the value of the constant $C$ ?
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2$\begingroup$ $\log\log(n)$ is the normal order of $\omega(n)$, see, for example, en.wikipedia.org/wiki/…. $\endgroup$– StoppleCommented Aug 19, 2022 at 18:14
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$\begingroup$ @Stopple I want to prove an inequality contains $ω(n)$ like $a^$ω(n)$$$\leq f(n)$ , can I use the above inequality for all values $n$, or it will be changed by changing the value of $n$ ? $\endgroup$– OmegaCommented Aug 20, 2022 at 10:39
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2 Answers
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On numbers $n$ which are primorials, $$\omega(n)\sim \frac{\log n}{\log \log n},$$ so there is no such general constant. This is because for such $n,$ we have $$n=\prod_{p<x} p = e^{x(1+o(1))}.$$
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There is another proof that this equality fails and therefore there is no such value for the constant $C$.
The value of $\omega(2^n-1)$ is at least $d(n)-2$, where $d(n)$ is the number of divisors of $n$. This follows from the Bang's theorem as the special case of the Zsigmondy's theorem – see https://en.m.wikipedia.org/wiki/Zsigmondy%27s_theorem.
The inequality
$$d(n)\gt e^{c\frac{\ln(n)}{\ln(\ln(n))}}$$
holds for infinitely many $n$ and any constant $c\le\ln(2)$.
This gives
$$n\ge e^{c\frac{\ln(\ln(n))}{\ln(\ln(\ln(n)))}}$$
for any constant $c\le\ln(2)$. This completes the proof.