@BenjaminSteinberg The dihedral group in the last example ($p=7, q=5$) is $D_{2\cdot10}$ which doesn't have $\mathbb{Q}$ as a splitting field but the eigenvalues are shown as integral, so either there's something off in the calculation or something else is going on?

Do you have any pointers to more info on these order matrices? I'd like to know more about them but all I'm finding for Dedekind group matrices seem to be ones that treat the elements of the group as variables and take the determinant of the multiplication table as a polynomial in the group members.

Is the statement about the matrix of orders having integer spectra specific to the dihedral groups being generated by inversion permutations in this way, or does it seem to be the case for all dihedral groups?

This is a wonderful answer, and I'm sorely tempted to try and go do some exploring (though I have so many projects on my plate right now that coding this up is likely to fall by the wayside). One note on the tractable search space, though — it certainly seems to me that there should be infinitely many paths satisfying those constraints, since e.g. after any 121 sequence of walls we still have both 3 or 4 as valid options to continue the sequence. Whether all of these are realizable is of course a different matter, but I don't see any a priori reasons why one wouldn't be.

@DavidESpeyer It seems like that should at the worst be solvable as a linear programming problem; the condition on internality can be written as a set of inequalities on the bounce point and given the sequence of reflections each bounce point is a linear expression in the coefficients of the initial point and vector.

@DavidESpeyer Oh, that makes perfect sense — $1$ as a multiple eigenvalue means that there's an at-least linear set of solutions to the equation $M_rv=v$ where $M_r$ is the product of all the reflection matrices for the sequence of reflecting faces. Very neat!

Another related question (or family): if you take as your tetrahedron the alternating vertices of a unit cube, then any billiard path that starts with two rational points on different faces of the tetrahedron will only bounce at rational points. Then you can ask about realizability of periodic paths as rational ones, etc...

What's the origin of this question? What prompted you to look at these particular continued fractions, these particular generating functions? Without some sense of the background here it's hard to even suggest good directions to look in for a proof.

It very much feels to me like the small-$n$ values are 'accidental' and that the behavior should be exactly $n$ roots for all $n$ greater than some single-digit value. I'm not sure there's really a 'deeper meaning' here other than that the values of $\tan z$ and $z^n$ are just comparable enough for small $n$ to lead to this sort of accident.

I wasn't one of the downvotes, but I feel like the point in my initial comment still holds strongly.

Maybe I'm confused, but aren't all of the values inside your example integrals 'constants'? I would expect, for instance, that $\int_0^1\omega\ dt=\omega$ for any 'reasonable' definition of integration...

The particular arrangement in the second one doesn't have any such decomposition, but I haven't checked to see whether there's possibly an arrangement that does (e.g., whether the pieces other than the $\frac14+\frac16+\frac1{12}$ ones can be shaped into a $\frac12\times1$ rectangle; I don't think there's anywhere for the $\frac16\times\frac1{30}$ sliver to go, but I haven't wholly convinced myself). It might be interesting to know whether indecomposable tilings exist for all sufficiently large $n$, how many there are compared to the decomposable ones, etc...

Another possibly-interesting question prompted by the $n=10$ examples: The first one is decomposable; we can form two rectangles of pieces each summing to $\frac12$ (though note that one of the two rectangles isn't really Egyptian in its own right, since if we scale to give it unit area then there are e.g. pieces with area $\frac29$). (cont)

The Gauss-Kuzmin distribution would seem to suggest that these numbers would have positive density (in some suitable sense) in e.g. $\mathbb{Q}\cap[0,1]$ but I'm not entirely sure that the density is even well-defined in this case — in particular, whether the ratio of the number of rationals with denominator less than $n$ with last coefficient 2 to the total number of rationals with denominator less than $n$ even converges.

Think about it this way: start with your initial grid and choose one of the dimensions. 'Reduce by two' along this dimension: pair each $\langle 2i, j, k, \ldots\rangle$ with $\langle 2i+1, j, k, \ldots\rangle$ to get a collection of $(\frac N2\times N^{d-1})$ $2\times1\times\ldots\times1$ rectangles and now treat these rectangles as your new points. Pair them up along a different dimension to get $(\frac N2\times\frac N2\times N^{d-2})$ $1\times2\times1\ldots\times1$ rectangles of these new points (which is to say, that many $2\times2\times1\ldots$ rectangles of the original points); repeat.