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Suppose $n\in\mathbb{N}$. I am trying to find an explicit upper bound for the number of prime factors of $n$ as a function of $n$. I know similar problems have been asked before (here, for example) but I am trying to do another approach.

Here is my proof so far. Every $n\in\mathbb{N}$ can be written as a product of primes: \begin{equation} n = p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}. \end{equation} Trivially, we can see that $n$ has $k$ prime factors. Let $\omega(n)=k$. $\omega(n)$ is maximized if $n$ is a product of consecutive primes, or $n=2\cdot3\cdot5\cdot\dots\cdot p$, where $\omega(n)=\pi(p)$. According to the Prime Number Theorem, we have that \begin{equation} \omega(n)=\mathcal{O}\bigg(\frac{p}{\ln p}\bigg). \end{equation}

I don't know where to go from here. I know that the result is supposed to be $$\frac{\ln n}{\ln(\ln n)}$$ but this means that $p\approx\ln n$ and I am unsure of how to show that.

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Matt G
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1 Answers1

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This is a good start! (One should rigorously prove that it suffices to look at the product of the first several primes.) The other piece of information you need is that $$ \log n = \log\bigg( \prod_{q\le p} q \bigg) = \sum_{q\le p} \log q = \theta(p)$$ (where $q$ denotes a prime variable); and the prime number theorem tells us then that $\log n \sim p$. That's how $\frac p{\log p}$ gets transformed into $\frac{\log n}{\log\log n}$.

Greg Martin
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