@SamHopkins Good catch; yes, assume that I want a lattice here too. Thank you!

I'm confused by this — a finite set of points and a set of edges connecting them is a (finite) graph. What's toy about this model?

I think it's worth asking: why do you want this thing? Is there something in particular you're hoping to learn about the rational points, or the measure itself, or? This feels like one step along the path of a larger question and having a better sense of that larger question can help people with providing a meaningful answer to this one.

Worth noting: if you can tile arbitrarily large regions of the plane in consistent fashion, then you can tile the whole plane by the compactness theorem. I feel like there's a very good chance of constructing a triangle spiral that will accomplish this.

@Gro-Tsen I think at this point the folks looking at it are primarily working with the 'modified' version in which a player has to win if they can — i.e., the only legal move from a prime is the (winning) one to 1; this is much more amenable to analysis for exactly these sorts of reasons.

@Gro-Tsen I believe we can do this by starting with the primes as seeds. Each prime is an $\mathcal{N}$-position (to use the Winning Ways terminology; i.e., the next player to move wins), and a position is a $\mathcal{P}$-position iff all of its options are $\mathcal{N}$-positions. So e.g. we set 2, 3, 5, 7, 11, 13 as $\mathcal{N}$; then 4 is $\mathcal{P}$ since its two legal options are $\mathcal{N}$. Work backwards along the prime gaps: 10 is $\mathcal{P}$ because all three of its options are $\mathcal{N}$; then 9 is $\mathcal{N}$ since it has 10 as an option, etc.

Note that this is a 'mostly' solution to the problem — the question of whether there's such a tile using translations and rotations only (as opposed to translations, rotations and reflections) is still open.

I'm legitimately a bit surprised that Mathematica wasn't able to simplify e.g. $(\sqrt{8x+1}-1)^2(\sqrt{8x+1}+1)^2$ to $64x^2$.

@AlapanDas Even if they do converge to such a solution, that doesn't mean that the sequence 'leads to' periodicity in the sense OP means — they're specifically asking (AFAICT) if 1/3 is a preperiodic point.

This question would likely be a better fit for math.stackexchange.com; I wouldn't put it in the domain of 'research mathematics' per se (with a few rare exceptions, questions about digital representations and digit patterns tend not to be).

@PeterMueller Thank you! That's exactly the bit that I was missing.

This makes rough sense to me, but I think I'm having a little bit of trouble visualizing: what in the argument needs $n\geq7$?

@SylvainJULIEN Wouldn't that immediately imply this specific case of Bunyanakovsky?

@MartinBrandenburg Followed immediately by "More generally, based on what just said, are there other areas that you feel like suggesting that I find out about?", and with a question title about 'domains that may require a categorical background'.

@Wojowu Glah, thank you. I was translating to the sum via logs and blanked on the fact that the multiplicative factor of 2 would actually translate back to an exponential factor on the asymptotics.

At first glance this seems unlikely to me because it doesn't pass a 'smell test'. The main issue from my perspective is that the desnity of twin primes is expected to be less than the density of primes and in fact the sum over the set of twin primes $\sum_{p\in T}\frac1p$ is known to converge (Brun's theorem) but the infinite product $\prod_2^\infty\left(1-\frac2p\right)$ diverges at exactly the same rate as the analogous product $\prod_2^\infty\left(1-\frac1p\right)$ does.