What a weird function! Here are a few scattered observations and curiosities that came to mind as I was playing around with it:

If you plot the first few thousand terms, \\(f^n(0)\\) appears to be \\(O(\log(n))\\):



This is consilient with the arrangement of the terms John (#2) posted: in general, it takes 2^n terms to go from 1/n to n/1. The linearly increasing subsequence \\(\frac{1}{1}, \frac{2}{1}, \frac{3}{1}\ldots\\) occurs at times that are coming exponentially further apart, so you'd expect growth to be logarithmic in general.

For convenience, let's define \\(g(n) = f^n(0)\\). So \\(f\\) is a function from \\(\mathbb{Q}\\) to \\(\mathbb{Q}\\), and \\(g: \mathbb{N} \rightarrow \mathbb{Q}\\) is a sequence of iterates starting at 0.

One natural question to ask is: "is \\(g\\) surjective onto \\(\mathbb{Q}\\)?" I.e., for every \\(\frac{p}{q} \in \mathbb{Q}\\), is there an \\(n \in \mathbb{N}\\) such that \\(g(n) = \frac{p}{q}\\)? From numerical evidence, I conjecture "no". For example, here is a plot of the first ten thousand iterates, with the numerator along the x axis and the denominator along the y:



So not only do the iterates not appear to be filling in \\(\mathbb{N}^2\\) in any systematic way, but the plot appears to show some kind of fractal bifurcation structure. What's up with that? Considering each rational number \\frac{p}{q}\\) as a point \\((p, q) \in \mathbb{N}^2\\), what is the image of \\(g\\)? (NB we haven't shown yet that \\(g(n) \geq 0\\) for all \\(n\\), so it's slightly unsporting to even talk about \\(\mathbb{N}^2\\) at this point, but one gets the idea.)