Let $\omega(n)$ the number of distinct prime numbers in the factorisation of $n$.
Many results are known about the average of $\omega(n)$. One of them is that $\omega(n)\sim \log \log n$ for almost all positive integers. However, $|\omega(n)-\log \log n|$ can be arbitrarily large (for taking $n=2^a$, for example, as $a\to \infty$). In these negligible examples, we have $\omega(n)<C\log \log n$, for some computable constant.
My question is: Is it correct to say that $\omega(n)\leq C\log \log n$, for some positive constant $C$? If so, is it possible to give a value of $C$?
Thanks in advance,
Let $n$ be the product of all primes less than $x$. Then: $$n = \prod_{p < x} p \sim e^x,$$ So $x \sim \log(n)$. The estimate above is closely related to the prime number theorem. On the other hand, $\omega(n)$ is the number of primes less than $x$, which is $\pi(x)$. So, again using the prime number theorem,
$$\omega(n) = \pi(x) \sim \frac{x}{\log(x)} \sim \frac{\log n}{\log \log n} \gg \log \log n.$$
In particular, these numbers have many more factors than the "average" number. You don't even need the prime number theorem to prove this, BTW, since the weaker estimates: $$\prod_{p < x} p < 4^x, \qquad \pi(x) > \frac{x}{2 \cdot \log(x)}, \ x \ge 3$$ which are more elementary (and proved by Chebyshev) are enough to give similar estimates.