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.)
Version 2.1.8p2
