This question and its answer incidentally show that every non-zero rational number $q$ can be written as a finite generalised continued fraction of the form:
$$ \dfrac{2^{n_0}}{1- \dfrac{2^{n_1}}{1 - \dfrac{2^{n_2}}{1- \dfrac{2^{n_3}}{\ddots\dfrac{\ddots}{1-2^{n_{r(q)}}}}}}}$$
with $n_i \in \Bbb N$ and $r(q) \ge 1$ depending on $q$. For example: $$ \frac{14}{9} = \dfrac2{1- \dfrac2{1 - 8}} $$ and
$$\frac{9}{5} = \dfrac1{1- \dfrac4{1 - \dfrac8{1- 2}}} $$ and $$-17 = \dfrac1{1- \dfrac2{1 - \dfrac8{1- \dfrac2{1-\dfrac4{1-\dfrac4{1-2}}}}}}$$
(How to get the powers of $2$: Let $v_2$ be the $2$-adic valuation, then if $v_2(q) \ge 0$, set $n_i = v_2(f^i(q))$; if $v_2(q) <0$, set $n_0=0, n_{i+1}= v_2(f^i\circ g(q))$ where $f(x)= 1-\frac{1}{x|x|_2}$ and $g(x) = 1-\frac{1}{x}$ are the functions in that question resp. answer.)
Now, if we generalise from $2$ to an arbitrary natural $k > 1$ (in particular prime $k$, but why restrict to that), I wondered if it is also true that:
For every $ q \in \Bbb Q^*$ there is a finite tuple $(n_0, ..., n_{r(q)})$ such that $$ q = \dfrac{k^{n_0}}{1- \dfrac{k^{n_1}}{1 - \dfrac{k^{n_2}}{1- \dfrac{k^{n_3}}{\ddots\dfrac{\ddots}{1-k^{n_{r(q)}}}}}}}$$
or something similar/more general (what if the "$1$"'s are replaced by another natural number etc).
If not, which rationals can be written that way?
Also, as I basically know nothing about continued fractions and made up an ad hoc approach for the other question, so:
Are there standard tools for handling this kind of question from the theory of continued fractions?
Edit: After playing around a bit, it seems more natural to me (albeit I've neither proven sufficiency nor necessity) that for general $k$, one would allow a set of representatives modulo $k$ instead of the leading "$1$"'s, i.e.
$$ q = \dfrac{k^{n_0}}{b_1- \dfrac{k^{n_1}}{b_2 - \dfrac{k^{n_2}}{b_3- \dfrac{k^{n_3}}{\ddots\dfrac{\ddots}{b_{n_q}-k^{n_{r(q)}}}}}}}$$
with the $b_i \in \{1,2, ..., k-1\}$. This would make a characterisation of the ones where all $b_i=1$ more subtle.
