Order of a function related to divisors Let $f(n)=\max(\{d(ab):\ a,b\le n\})$ where $d(m)$ is the number of divisors of $m.$ What is the order of $f$? In particular I'm looking for an asymptotic upper bound.
 A: Note that the product of all the primes between $y$ and $x$ is approximately $e^{y-x}$. (This follows from the prime number theorem, in the form $\theta(x) := \sum_{p\le x} \log p \sim x$.)
Thus if you take $a$ to be the product of all the primes up to about $\log n$, and $b$ to the the product of all the primes between about $\log n$ and $2\log n$, then both $a$ and $b$ are about $n$ in size. Meanwhile, $d(ab) = 2^{\pi(2\log n)} \approx 2^{2\log n/\log\log n} = \exp(2\log 2\cdot \log n/\log\log n)$.
This construction gives a lower bound for $f(n)$; however, the upper bound that Peter Košinár mentioned in his answer has this same order of magnitude. So that pins down the size of $f(n)$ (more precisely, it gives an asymptotic formula for $\log f(n)$).
A: Observation #1: $$f(n)=\max_{a,b\leq n} \mathrm{d}(ab) \leq \max_{k\leq n^2} \mathrm{d}(k)$$
(and it seems the equality happens quite often)
Observation #2: According to Wikipedia, we have $$ \limsup_{n\rightarrow\infty} \frac{\log d(n)}{\log n / \log \log n}=\log 2$$
This gives us quite a good idea of the asymptotic bounds we can expect for $f(n)$. 
After playing with the values of $\max\limits_{k\leq n^2} \mathrm{d}(k)$ for a while, it also seems that we might have $f(n) = o(n^{1/\log \log n})$ (although I haven't tried proving that yet). Do you need the upper bound to be as tight as possible, or do you "just" need to prove that, for example, it grows slower than any polynomial?
