Just because I don't see the phrase mentioned here: your $c_{k, k-i}$ are precisely the Stirling Numbers and I'd start by looking in that direction.

Is there an approach through geometric group theory here? Specifically the Freudenthal-Hopf theorems that any finitely generated group either has $\leq 2$ ends or infinitely many.

Just to check my understanding: the $F$ in your last-paragraph is well-defined, but not computable itself, correct?)

Setting $W_n$ to be the number of sets with a first player win, then $W_n-2W_{n-1}$ goes $1, -1, 2, 1, -3, 6, -10, 2, -6, -32, 19, -136, 58, -380, -13$, which is distressingly random-looking.

Doesn't the statement for positive $x$ immediately imply it for all $x$? One can just reverse the signs of $a, b, c$...

I'm still curious as to whether there's an answer with $q()$ irreducible but this absolutely answers the question I had. Thank you!

I should have clarified degree $\lt n$; thank you! I was thinking that any polynomials of degree higher than $q$ would yield a polynomial quotient that integrated to a rational value, but forgot that that rational value can be zero; more broadly, if $\int_0^1\frac{p(x)}{q(x)}\in\mathbb{Q}\pi$ then so is $\int_0^1\frac{p(x)+z(x)q(x)}{q(x)}$ for any $z(x)$ with $\int_0^1z(x)=0$.

'Unique' here needs a little bit of clarification; all of the associates of the golden ratio — that is, any irrationals whose continued fraction expansion ends in an infinite streak of ones — have the same approximability properties. Beyond the golden ratio, when you exclude $\varphi$ and its associates there's a second-most-difficult number to approximate, etc. See mathworld.wolfram.com/HurwitzsIrrationalNumberTheorem.html for more thorough details.

@JoelDavidHamkins The difference-of-primes set seems really unlikely to me since there are pretty strong arguments to be made that it's just $\{2n: n\in\mathbb{N}\}\cup\{p-2: p\ \mbox{prime}\}$.

Curious side question: is it possible that the answer depends on the dimension? (e.g., that such an icosahedron exists in $\mathbb{R}^4$ but not $\mathbb{R}^3$)

We can readily count them exactly! If you want you can even turn that into a 'closed form' using inclusion-exclusion, in the usual way of such things. But that formula isn't particularly useful. It sounds like what you're expecting is a 'constant length' formula ($O(1)$ or at worst $O(\log^k n)$ to evaluate, for some small $k$) for $\pi(p_n^2)-\pi(p_n)$ and the simple observations about size of divisors just turn out to not be much help with that.

The latter denominator ($x^4-2x^3+4x-4$) factors as $(x^2-2)(x^2-2x+2)$; the original denominator is irreducible over $\mathbb{Q}$, though, which I think makes it a much more interesting case.