4
$\begingroup$

In the book A Comprehensive Course in Number Theory by Alan Baker. The author mentions that even though the average order of $\tau(n)$ is $\log n$, almost all numbers have about $(\log n)^{\log 2}$ divisors.

$\tau(n)$=number of divisors of n.

I was wondering how one would prove this result. If anyone has any ideas it would be great. Thanks.

$\endgroup$
4
$\begingroup$

We have $$ \sum_{n\le x}\tau(n)\sim \sum_{n\le x}\log(n), $$ so that the average value of $\tau(n)$ is indeed $\log(n)$, but from Hardy and Ramanujan we know, since $2^{\omega(n)}\le \tau(n)\le 2­^{\Omega(n)}$, that for most numbers $n$, $$ \tau(n)=\log(n)^{\log(2)+o(1)}, $$ where $\log (2)$ is around $0.693$. The normal value of $\omega (n)$ resp. $\Omega(n)$ is $\log(\log(n))$, by Hardy and Ramanujan.

$\endgroup$
  • $\begingroup$ can you please tell what are $\omega(n),\Omega(n)$ $\endgroup$ – happymath Apr 23 '15 at 15:37
  • $\begingroup$ $\omega(n)$ denotes the number of distinct prime factors of $n$, and $\Omega(n)$ its multiplicities, see here, $\endgroup$ – Dietrich Burde Apr 23 '15 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.