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I want to prove that if $\omega (n)$ is the number of distinct prime factors of $n$ for $n>2$ we have $\omega (n) \leq \frac{\ln n}{\ln \ln n} + O(\frac{\ln n}{(\ln \ln n)^2})$.

How can I do this?

sBs
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    This is a standard result - see http://mathoverflow.net/questions/23867/bound-on-the-number-of-prime-factors-of-logarithmically-rough-numbers. Also http://math.stackexchange.com/questions/179353/effective-upper-bound-for-the-number-of-prime-divisors and the proof by Robin. – Dietrich Burde Sep 19 '14 at 19:11
  • http://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem – Snufsan Sep 19 '14 at 19:42
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    Well okay I now know that this is a standard result. Nevertheless I can't find a proof...So maybe someone can help me? – sBs Sep 19 '14 at 19:52
  • Noone? - maybe you can help me out with an english translation of the proof by Robin? – sBs Sep 21 '14 at 00:15

2 Answers2

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$\omega(n)$ reaches a maximum at the primorial numbers $p\#=2\cdot3\cdot5\cdot\cdots\cdot p,$ where $\omega(p\#)=\pi(p).$ The Prime Number Theorem (in one of its forms) says that $\vartheta(x):=\log x\#\sim x,$ and so $p\#=e^{(1+o(1))p}$. If $p$ is the $n$-th prime then $p\sim n\log n$ and so $$ n=\pi(p)=\omega(p\#)=\omega(e^{(1+o(1))p})=\omega(e^{(1+o(1))(1+o(1))n\log n})=\omega(n^{(1+o(1))n}). $$

Setting $x=n^{(1+o(1))n}$ and solving for $n$ you get $$ n\sim\frac{\log x}{W(\log x)}=\frac{\log x}{W(\log x)}=\frac{\log x}{\log\log x-O(\log\log\log x)}\sim\frac{\log x}{\log\log x} $$ which has a slightly weaker error term than the quoted result.

Charles
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One can use the fact that $w(n) \leq \sum _{d|n} \frac{\Lambda (d)}{\log d}$ and it can be shown that for any $x$ we've $\sum _{d|n} \frac{\Lambda (d)}{\log d} \leq \frac{\log n}{\log x}+ O(\frac{x}{(\log x)^2})$ and now we just need to take $x=\log n$

dragoboy
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