Nice graphics, Patrick O'Neil!

> in general, it takes \\(2^n\\) terms to go from 1/n to n/1.

It looks that way. Can anyone here prove this, who hasn't already read all about this stuff? I think attempting to prove this could lead to a lot of extra insights.

> 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}\\)?

Of course you mean not \\(\mathbb{Q}\\) but the nonnegative rationals. This is a very interesting question.

> From numerical evidence, I conjecture "no".

What's the simplest nonnegative rational you can find that appears not to be of the form \\(g(n)\\)?

(Again, I think answering this would lead to a lot of extra insights.)

> in general, it takes \\(2^n\\) terms to go from 1/n to n/1.

It looks that way. Can anyone here prove this, who hasn't already read all about this stuff? I think attempting to prove this could lead to a lot of extra insights.

> 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}\\)?

Of course you mean not \\(\mathbb{Q}\\) but the nonnegative rationals. This is a very interesting question.

> From numerical evidence, I conjecture "no".

What's the simplest nonnegative rational you can find that appears not to be of the form \\(g(n)\\)?

(Again, I think answering this would lead to a lot of extra insights.)