I know that a large number $n$ has probability $\frac{1}{\ln (n)}$ of having exactly 1 prime factor (i.e. it's prime). But is there any statement on the exact distribution for the number of prime factors $n$ will have in general? Feel free to choose whether $12 = 2\cdot 2\cdot 3$ counts as 3 separate factors, or 2 unique factors; whichever makes answering the question easier.

If there's no definitive answer, any approximate results are also appreciated. (e.g. what is the average of the distribution, the variance of the distribution, etc.).

  • 1
    $\begingroup$ if $3$ has two factors then $12$ has six factors $(1,2,3,4,6,12)$. If we call this $d(n)$ thenWikipedia says Dirichlet showed $\sum\limits_{n\leq x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt{x})$ $\endgroup$
    – Henry
    Aug 15 at 0:54
  • $\begingroup$ @Henry Thanks so much! But I realized that I was more concerned with prime factors (not simply divisors), so I updated the question. $\endgroup$
    – chausies
    Aug 15 at 1:00
  • 6
    $\begingroup$ In that case, it is Erdos-Kac. en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem see Kac's little bookhttps://bookstore.ams.org/car-12/ $\endgroup$
    – Will Jagy
    Aug 15 at 1:02


You must log in to answer this question.