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In this MSE post, 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,

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

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    $\begingroup$ Presumably algebraic numbers only at special values of $q$ (namely $e^{2\pi i z}$ where $z$ is a quadratic irrationality in the upper half-plane; e.g. $z=i$ gives your $e^{-2\pi}$). $\endgroup$ – Noam D. Elkies Jul 7 '15 at 3:54
  • $\begingroup$ @NoamD.Elkies: Yes, I was not being precise. :) $\endgroup$ – Tito Piezas III Jul 7 '15 at 4:43
  • $\begingroup$ Related answer together with the posts linked therein. You may want to flip the signs of $a$ and $q$ (here) for easier comparison. The condition $ab=q$ is preserved under that substitution. To summarize: Yes, and Nicco's cfrac is an instance of entry 11 (not 12) that for $ab=q$ can be simplified to a theta quotient like entry 12. I suppose Ramanujan was aware of that, but sought a cfrac that was not restricted to $ab=q$, and entry 12 was the result. I hesitate to make an answer out of this, as I have already written four related posts on MSE. $\endgroup$ – ccorn Oct 3 '15 at 23:09
  • $\begingroup$ @NoamD.Elkies: Can you kindly look at this answer and see if my Part 2 is correct? (Re $\tau$ as quadratic vs quartic roots.) $\endgroup$ – Tito Piezas III Dec 14 '15 at 4:31

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