What rational numbers are expressible as "multiple-of-$12$" continued fractions? I need to use a special subclass of continued fraction expansions for a problem I'm studying. Namely, expansions of the form
$$[a_0;a_1;a_2;\dotsc; a_{n}] = a_0+\displaystyle\frac{1}{a_1+\displaystyle\frac{1}{a_2+\displaystyle\frac{1}{\ddots\displaystyle\frac{1}{a_n}}}}$$
where $a_0,\dotsc, a_{n-1}\in 12\mathbb{Z}$ and $a_n\in S\subset\mathbb{Q}\cup\{\infty\}$, a finite subset of the extended rationals.
This problem is in analogue to the "even continued fraction expansions", which have $a_0,\dotsc, a_{n-1}\in 2\mathbb{Z}$ and $a_n\in\{0,1,\infty\}$. It turns out that the set of even continued fraction expansions is just all of $\mathbb{Q}\cup\{\infty\}$. The proof uses the projectivized congruence level $2$ subgroup ${\rm P}\Gamma(2)$, which is generated by the Möbius transformations
$$A: z\mapsto \frac{z}{-2z+1}\text{ and }B: z\mapsto z+2.$$
I've seen it claimed in various places (but never with proof) that the (left) action of ${\rm P\Gamma}(2) = \langle A, B\rangle$ on $\mathbb{Q}\cup\{\infty\}$ has orbit transversal $\{0,1,\infty\}$, meaning that for each $r/s\in\mathbb{Q}\cup\{\infty\}$ there is a word $w\in \langle A,B\rangle$ such that $r/s = w.x$ for some $x\in\{0,1,\infty\}$. Determining an even continued fraction expansion for $r/s$ from there is then simply a matter of decoding the corresponding word $w$.
Now, my questions: is there any hope of a similar result holding in the "multiple-of-$12$" expansion case? Namely, could there be some finite set $S$ such that expansions of that form describe all of $\mathbb{Q}\cup\{\infty\}$? Equivalently, if $G$ is the group generated by the transformations
$$a: z\mapsto z+12, \quad b:z\mapsto -z,\quad c:z\mapsto 1/z,$$
does the action of $G$ on $\mathbb{Q}\cup\{\infty\}$ have a finite orbit transversal? And how would such an orbit transversal be calculated, finite or not?
This seems like a pretty tough/niche problem, so even just pointing me in the direction of some references would be greatly appreciated.
Edit: I gave both $b$ and $c$ as valid transformations since I happen to have those symmetries in the context I'm working in, but if one needs to restrict this to just $z\mapsto -1/z$ so that $G$ is a subset of ${\rm PSL}(2,\mathbb{Z})$, then that is fine.
 A: Let me first indicate the proof in the case of $P\Gamma(2)$, and then I will go on to explain what goes wrong in the case of your "multiple of $12$" question.
The action of $\text{PSL}(2,\mathbb Z)$ on the upper half plane has a fundamental domain $D$ which is a triangle with three hyperbolic geodesic sides, two finite vertices, and with one vertex "at infinity. Two of the sides are on the respective lines $x=-1/2$ and $x=+1/2$ (dotted orange in the linked diagram); the third side is on the circle $z^2=1$ (solid blue in the linked diagram).  This description uniquely determines $D$. The upper half plane is tiled by other fundamental domains, each a translate of $D$ by some element of $\text{PSL}(2,\mathbb Z)$, any two of which meet along a common side, a common vertex, or not at all. The ideal vertex of the fundamental domain $D$ is the point $\infty$, and from this fact the proof of the standard continued fraction theorem can be produced.
Generally speaking, for any subgroup $\Gamma < \text{PSL}(2,\mathbb Z)$, one can produce a fundamental domain of $\Gamma$ that is a union of some translates of $D$. The set of translates corresponds 1-to-1 with the left cosets of $\Gamma$ in $\text{PSL}(2,\mathbb Z)$, and in particular the number of translates is equal to the index of $\Gamma$ in $\text{PSL}(2,\mathbb Z)$.
The key feature of the subgroup $P\Gamma(2) < \text{PSL}(2,\mathbb Z)$ is that it has finite index, in fact the index is equal to $6$. One can demonstrate this by exhibiting a fundamental domain $E$ for $P\Gamma(2)$ that is a union of $6$ translates of $D=D_1$. A picture would be worth a thousand words here, but I'll go with somewhat fewer words and no picture (but with references to the linked diagram)
Starting with $D=D_1$ itself, let $D_2$ be the image of $D$ under the transformation $z \mapsto z+1$. The union $D_1 \cup D_2$ is a quadrilateral with one ideal vertex at $\infty$ and with the opposite vertex being a finite vertex at the point $p = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ (in the linked diagram, $p$ is one of the points where three blue segments and three dotted orange segments meet). The stabilizer of the point $p$ in the group $PSL(2,\mathbb Z)$ is cyclic of order $3$, generated by the transformation $\gamma : z \mapsto \frac{1}{-z+1}$. The other four fundamental domains comprising $E$ are $D_3 = \gamma(D_1)$, $D_4 = \gamma(D_2)$, $D_5=\gamma(D_3)$, $D_6 = \gamma(D_5)$; these are the six fundamental domains which fit around the point $p$. In summary,
$$E = D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5 \cup D_6
$$
is a fundamental domain for $P\Gamma(2)$. It is bounded by six geodesics lying on the following six hyperbolic lines: the lines $x=-\frac{1}{2}$ and $x=\frac{3}{2}$; the circle $(z+1)^2=1$; the circle $(z-2)^2=1$; the circle $(z-1/3)^2 = 1/9$; and the circle $(z-2/3)^2 = 1/9$. The six vertices of $E$ alternate between three ideal vertices at $0$, $1$ and $\infty$, and three finite vertices.  The edge identifications of $E$ are such that for each of the three ideal vertices $\xi \in \{0,1,\infty\}$, the two sides of $E$ incident to $\xi$ are identified by a parabolic transformation that fixes $\xi$. For example, the transformation $z \mapsto z + 2$ that fixes $\infty$ identifies the side of $E$ on the line $x=0$ with the side on the line $x=2$. The key feature of $E$ that follows from this description is that the quotient is a three-cusped sphere, those three cusps being in one-to-one correpsondence with the three orbits of the action of $P\Gamma(2)$ on $\mathbb Q \cup \{\infty\}$, with orbit transversal equal to the three ideal vertices $0$, $1$, $\infty$ of $E$. From this fact the proof of the "even continued fraction" theorem follows.

So, what happens with your "multiple of 12" continued fractions? As suggested at the end of your post, I'll stay in the group $\text{PSL}(2,\mathbb Z)$ and I'll consider its subgroup generated by $z \mapsto z + 12$ and $z \mapsto - \frac{1}{z}$ which I'll denote $\Gamma_{12}$ for lack of a better notation.
The trouble is, the subgroup $\Gamma_{12}$ has infinite index in $\text{PSL}(2,\mathbb Z)$, and what's worse its action on $\mathbb Q \cup \{\infty\}$ has infinitely many orbits. So every orbit transversal is infinite.
In fact it's even worse than that. The action of $\text{PSL}(2,\mathbb Z)$ on $\mathbb R \cup \{\infty\}$ has the property that the orbit $\{\gamma \cdot x \mid \gamma \in \text{PSL}(2,\mathbb Z)\}$ of every $x \in \mathbb R \cup \{\infty\}$ is dense; for example, the orbit $\mathbb Q \cup \{\infty\}$ is dense. This property is called topological minimality. But the topological behavior of action of $\Gamma_{12}$ on $\mathbb R \cup \{\infty\}$ is at the far extreme from topological minimality: there exists an open interval $U \subset \mathbb R \cup \{\infty\}$ such that $\gamma(U) \cap U = \emptyset$ for every nontrivial $\gamma \in \Gamma_{12}$. So any two numbers $x \ne y \in U \cap (\mathbb Q \cup \{\infty\})$ have disjoint orbits. And, of course, $U \cap (\mathbb Q \cup \{\infty\})$ is an infinite set; we could construct an orbit tranversal that contains the set $U \cap (\mathbb Q \cup \{\infty\})$ as a subset, and so that orbit transversal contains a dense subset of $U$.
So the summary is that I don't see any hope for a nice theory of "multiple-of-12" continued fraction expansions of rational numbers, that would generalize the theory of "even continued fractions".

A few final words: this answer is informed by the theory of Fuchsian groups, which are discrete subgroups $\Gamma < \text{PSL}(2,\mathbb R)$ acting on the upper half plane. Each such $\Gamma$ has a "limit set" which is the unique minimal closed subset $\Lambda \subset \mathbb R \cup \{\infty\}$ on which the restricted action of $\Gamma$ is topologically minimal. There are many equivalent descriptions of $\Lambda$: one of them says that if there is a fundamental domain with an ideal point (a vertex at $\infty$) that maps to a cusp of the quotient, then $\Lambda$ is the closure of the $\Gamma$ orbit of that ideal point. So, for example, the limit sets of the Fuchsian groups $\text{PSL}(2,\mathbb Z)$ and of $P\Gamma(2)$ are both equal to the whole circle $\mathbb R \cup \{\infty\}$.
If the limit set $\Lambda$ is not the whole of the circle $\mathbb R \cup \{\infty\}$ (as in the group $\Gamma_{12}$), then the action of $\Gamma$ on the open set $(\mathbb R \cup \{\infty\}) - \Lambda$ is properly discontinuous, implying that each point has an open subset $U$ such that $U$ is disjoint from all but finitely many of its translates $\gamma \cdot U$ ($\gamma \in \Gamma$). This open set $(\mathbb R \cup \{\infty\}) - \Lambda$ is called the domain of discontinuity of $\Gamma$.
If the domain of discontinuity of a subgroup $\Gamma < \text{PSL}(2,\mathbb Z)$ is not empty, i.e. if $\Lambda$ is a proper subset of $\mathbb R \cup \{\infty\}$, it follows that for each point $x$ in the domain of discontinuity there exists a neighborhood $U$ of $x$ and an orbit transversal $T$ of the action of $\Gamma$ on $\mathbb Q \cup \{\infty\}$ such that $U \cap T$ is a dense subset of $U$.
When thinking about your subgroup $\Gamma(12)$, I could see that its domain of discontinuity of was nonempty, leading to my answer.

Coda: Regarding the domain of discontinuity. Here's an impressionistic proof that the domain of discontinuity of $\Gamma(12)$ is nonempty.
Consider the invariant tree $T$ of the action of $\text{PSL}(2,\mathbb Z)$, which is the blue tree $T$ in the linked diagram. The transformation $z \mapsto z + 12$ translates along an "axis" in $T$ which is the uppermost blue bouncy "line" $L$ in the tree $T$, comprised of a bi-infinite concatenation of circular arcs; the action of this translation is to move each circular arc to the $12^{\text{th}}$ circular arc along $L$ to the right. Also, the transformation $z \mapsto -\frac{1}{z}$ fixes the midpoint of one of those circular arcs (that fixed point is, of course, $i \in \mathbb C$). As you translate that bouncy line around by all possible products of those two transformations (which generate a group isomorphic to the free product $\mathbb Z * (\mathbb Z / 2 \mathbb Z)$), and then take the union of those translates, you obtain a tree $T(12)$ in which any two translates of $L$ are either disjoint or intersect along one blue circular arc. Furthermore, each component of $T - T(12)$ is an infinite tree whose set of accumulation points in $\mathbb R \cup \{\infty\}$ is an arc. From this description, one sees that $T(12)$ is a proper subtree of $T$ whose set of accumulation points in $\mathbb R \cup \infty$ is a Cantor set, and the complement of that Cantor set is the union of the interiors of the open arcs corresponding to the components of $T \setminus T(12)$. That Cantor set is the limit set of $\Gamma(12)$, and its (nonempty) complement is the domain of discontinuity.
