Note that Thurston says

Mostow has rigorously enumerated examples by hand, so this table can be regarded as just a check.

That is effectively a claim that it suffices to check up to denominator $42$.

This answer is a sketch as to how Mostow might have done it.

The denominators make things unnecessarily long-winded. Normalise by dividing the curvatures by $2\pi$: we have $\mu_1, \ldots, \mu_n$ each in interval $(0, 1)$ and summing to $2$. Let $r_{i,j} = (1 - \mu_i - \mu_j)^{-1}$. We further require that for all $1 \le i < j \le n$ such that $\mu_i + \mu_j < 1$ either $r_{i,j} \in \mathbb{N}$ or $(\mu_i = \mu_j) \wedge 2r_{i,j} \in \mathbb{N}$.

Wlog $\mu_1 \ge \cdots \ge \mu_n$, and then obviously $\mu_i \le \frac2i$ with equality only if $i = n$ and all the parts are equal.

Observe that for all $3 \le i < j$ we must have $\mu_i + \mu_j \le 1$ with equality only if $n = 4$ and $\mu_1 = \mu_2 = \mu_3 = \mu_4 = \frac12$. Proof is by contradiction: since $\mu_1 \ge \mu_i$ and $\mu_2 \ge \mu_j$ the total sum would otherwise exceed $2$.

By inverting the definition of $r_{i,j}$ we get $\mu_i + \mu_j = \frac{r_{i,j} - 1}{r_{i,j}}$.

Case: $n = 5$. From the assumption of descending size of $\mu_i$ we get

• $\mu_5 \le \mu_4$ so $\frac{\mu_4 + \mu_5}2 \le \mu_4$
• $\mu_4 \le \mu_3$
• $\mu_3 \le \frac13(2 - \mu_4 - \mu_5)$

which can be combined to give $\mu_4 + \mu_5 \le \frac45$. Then $1 - \mu_4 - \mu_5 \ge \frac15$ so $r_{4,5} \le 5$.

By substitution, we get $\mu_3 \le \frac{r_{4,5} + 1}{3r_{4,5}}$ and can follow through to $r_{3,5} \le \frac{6r_{4,5}}{r_{4,5} + 1}$.

If $r_{4,5} > 2$ the constraint $\mu_4 \le \mu_3$ allows us to bound $\mu_3 + \mu_4$ away from $1$ and get $r_{3,4} \le \frac{3r_{4,5}}{r_{4,5} - 2}$. We have a finite number of values for $r_{4,5}, r_{3,5}, r_{3,4}$ and each element of the Cartesian product gives three simultaneous linear equations. From $\mu_2 \le \frac12(2 - \mu_3 - \mu_4 - \mu_5)$ we can filter out some of them because they violate the assumption $\mu_2 \ge \mu_3$. This leaves the following possibilities:

$$\begin{array}{ccc} \mu_3 & \mu_4 & \mu_5 \\ 3/10 & 3/10 & 3/10 \\ 11/30 & 3/10 & 3/10 \\ 9/20 & 3/10 & 3/10 \\ 1/3 & 1/3 & 1/3 \\ 5/14 & 5/14 & 13/42 \\ 3/8 & 3/8 & 7/24 \\ 7/18 & 7/18 & 5/18 \\ 2/5 & 2/5 & 4/15 \\ 9/22 & 9/22 & 17/66 \\ 5/12 & 5/12 & 1/4 \\ 11/26 & 11/26 & 19/78 \\ 3/7 & 3/7 & 5/21 \\ 13/30 & 13/30 & 7/30 \\ 7/16 & 7/16 & 11/48 \\ 15/34 & 15/34 & 23/102 \\ 4/9 & 4/9 & 2/9 \\ 5/12 & 1/3 & 1/3 \\ 53/120 & 43/120 & 37/120 \\ 5/14 & 5/14 & 5/14 \\ 11/28 & 5/14 & 5/14 \\ 3/8 & 3/8 & 3/8 \\ 7/18 & 7/18 & 13/36 \\ 2/5 & 2/5 & 7/20 \\ 9/22 & 9/22 & 15/44 \\ 5/12 & 5/12 & 1/3 \\ 7/18 & 7/18 & 7/18 \\ 2/5 & 2/5 & 2/5 \\ \end{array}$$

For each of these, the bound on $\mu_2$ turns out to give $\mu_3 + \mu_2 < 1$ and we could follow through to bound $r_{2,3}$ and enumerate and check all possible cases.

The remaining subcases of $n=5$ are $r_{4,5} \in \{\frac32, 2\}$.

If $r_{4,5} = \frac32$ then since it's non-integral we have $\mu_4 = \mu_5$, and we can solve to find $\mu_5 = \frac16$, $\mu_3 \in \{\frac16, \frac13, \frac12\}$. $\mu_5 + \mu_2$ is bounded away from $1$, so we can bound $r_{2,5}$ and enumerate and check all possible cases.

$r_{4,5} = 2$ is the tricky one. We have $\mu_4 + \mu_5 = \frac12$ and $\mu_3 \le \frac12$. That gives $\mu_5 \le \frac14$ and $\mu_2 \le \frac58$ so that $r_{2,5} \le 8$, but we need to bound $\mu_3 + \mu_4$ away from $1$ to pin the values down. $r_{3,5} \le 4$ so either $\mu_3 = \mu_5 = \frac14$ or $r_{3,5} \in \{2,3,4\}$. Take the non-equal subcases one by one in order of difficulty:

• $r_{3,5} = 4$. Then $\mu_3 + \mu_5 = \frac34$ so $\mu_5 \ge \frac14$, giving $\mu_5 = \mu_4 = \frac14$, $\mu_3 = \mu_2 = \mu_1 = \frac12$.

• $r_{3,5} = 3$. Then $\mu_3 + \mu_5 = \frac23$ so $\mu_5 \ge \frac16$, whence $\mu_4 \le \frac13$ and $\mu_3 + \mu_4 \le \frac56$.

• $r_{3,5} = 2$. Then $\mu_3 = \mu_4$. We must have $\mu_3 + \mu_4 < 1$ since otherwise $\mu_5 = 0$. Then $2r_{3,4} = \frac{2}{1 - 2\mu_4} \in \mathbb{N}$. Rearranging, $\mu_4 = \frac12 - \frac{1}{2r_{3,4}}$, so $\mu_5 = \frac{1}{2r_{3,4}}$ is an Egyptian fraction. Let $k = 2r_{3,4}$.

  If $r_{2,5} = 2$ then $\mu_2 = \mu_4$. We have a family parameterised by $4 \le k \in \mathbb{N}$: $\frac{k+4}{2k}, \frac{k-2}{2k}, \frac{k-2}{2k}, \frac{k-2}{2k}, \frac{2}{2k}$. When $k > 6$ we need $r_{1,5} = \frac{2k}{k-6} \in \mathbb{N}$, so we only need to consider $k \in \{4,5,6,7,8,9,10,12,18\}$.

  Otherwise we have variables $k$ and $2 < r_{2,5} \le 8$ which between them determine all of the parts. $\mu_2 + \mu_5 = 1 - \frac{1}{r_{2,5}}$. But $\mu_1 = 2 - \mu_2 - \mu_3 - \mu_4 - \mu_5 \ge \mu_2$, so $k \le \frac{3r_{2,5}}{r_{2,5} - 2}$ and we again have a finite number of cases to check.

For $n \ge 6$ the bound $\mu_i \le \frac2i$ suffices to bound $r_{5,6}, r_{4,6}, r_{4,5}$, so it should be easier than for $n=5$. Alternatively, it may be sensible to consider possible refinements of the finite list of solutions for $n=5$, and so on.

Note also that $n \le 12$, since we get the bound $r_{12,13} \le \frac{78}{53} < \frac 32$.