In this [MSE post][1], Nicco Mnisi defined a particular $q$-continued fraction of order $12$. More generally, define the cfrac found in *Ramanujan's Notebooks, Vol III*, Chap. 16, page 24, where $|ab|<1$ and $|q|<1$,

$$\begin{aligned}U(q)
&= \prod_{n=0}^\infty \frac{\big(1-a^2q^3(q^4)^n\big)\big(1-b^2q^3(q^4)^n\big)}{\big(1-a^2q(q^4)^n\big)\big(1-b^2q(q^4)^n\big)}\\ 
&= \dfrac{1} {1-ab+\dfrac{(a-bq)(b-aq)} {(1-ab)(1+q^2)+\dfrac{(a-bq^3)(b-aq^3)} {(1-ab)(1+q^4)+\dfrac{(a-bq^5)(b-aq^5)} {(1-ab)(1+q^6)+\ddots }}}}
\end{aligned}$$

I extrapolated that the general form of Nicco's cfrac, without a factor $q^{k_1}(1-q^{k_2})$, apparently is,

$$V(q) = \dfrac{1} {1+ab-\dfrac{(a+bq)(b+aq)} {1+(ab)^3+\dfrac{(a-bq^2)(b-aq^2)q} {1+(ab)^5-\dfrac{(a+bq^3)(b+aq^3)q^2} {1+(ab)^7+\dfrac{(a-bq^4)(b-aq^4)q^3} {(1+(ab)^9-\ddots }}}}} $$

>**Question:** If $ab=q$, and $|q|<1$, is it true that $U(q) = V(q)$?

I tested these numerically for various $a,b,q$ and it seems to be true with $V(q)$ converging faster. However, a rigorous proof is needed.

**P.S.** With the appropriate factor $q^{k_1}(1-q^{k_2})$ affixed, these cfracs are algebraic numbers. For example, in [this post][2],

$$\frac{1}{N(e^{-2\pi})}+N(e^{-2\pi}) = \frac{8}{\sqrt{2}( 3+ \sqrt{2}) - \sqrt[4]{3}(3+\sqrt{3})}=536.4953904\dots$$
 


  [1]: https://math.stackexchange.com/questions/1350614/
  [2]: https://math.stackexchange.com/a/1350892/4781