# Number of Divisors of most numbers

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.

## 1 Answer

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.

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