# a conjecture of two equivalent q-continued fractions related to the reciprocal of the Göllnitz-Gordon continued fraction A111374-OEIS

Given the square of the nome $q=e^{2i\pi\tau}$ and ramanujan theta function $f(a,b)=\sum_{k=-\infty}^{\infty}a^{k(k+1)/2}b^{k(k-1)/2}$ with $|q|\lt1$, define,

\begin{aligned}M(q)=\cfrac{1-q^3}{1-q^2+\cfrac{q^2(1-q^{-1})(1-q^5)}{1-q^6+\cfrac{q^4(1-q)(1-q^7)}{1-q^{10}+\cfrac{q^6(1-q^3)(1-q^{9})}{1-q^{14}+\ddots}}}}\overset{\color{blue}{?}}=\prod_{n=1}^\infty\frac{(1-q^{8n-3})(1-q^{8n-5})}{(1-q^{8n-1})(1-q^{8n-7})}\\[1.5mm]&\end{aligned} $$=\frac{f(-q^3,-q^5)}{f(-q,-q^7)}$$ and \begin{aligned}N(q)=\cfrac{1-q^3}{1+q^2-\cfrac{q^2(1+q^{-1})(1+q^5)}{1+q^6+\cfrac{q^4(1-q)(1-q^{7})}{1+q^{10}-\cfrac{q^6(1+q^3)(1+q^{9})}{1+q^{14}+\ddots}}}}\overset{\color{blue}{?}}=\prod_{n=1}^\infty\frac{(1-q^{8n-3})(1-q^{8n-5})}{(1-q^{8n-1})(1-q^{8n-7})}\\[1.5mm]&\end{aligned} $$=\frac{f(-q^3,-q^5)}{f(-q,-q^7)}$$

Q: How do we prove that the two q-continued fractions are equal to the q-products such that the continued fractions are equivalent $M(q)=N(q)$?

• I have given an answer, but I suppose that you can do that yourself. There are several cfrac questions of yours which I won't answer because there is nothing new to tell. I'd appreciate it if you answer those remaining questions. That would demonstrate progress and unclutter the Unanswered queue. Nov 8, 2015 at 17:06

For $n\in\mathbb{N}$, let $q_n=\exp\frac{2\pi\mathrm{i}\tau}{n}$, so $q_n^n=q$.
Use formula $(***)$ from that post, but with $q$ replaced with $q^2$, so it reads $$\small\cfrac{1}{1-q^2+\cfrac{(a+bq^2)(aq^2+b)} {1-q^6+\cfrac{q^2(a+bq^4)(aq^4+b)} {1-q^{10}+\cfrac{q^4(a+bq^6)(aq^6+b)}{1-q^{14}+\cdots}}}} = \frac{(-a^2q^6;q^8)_\infty\,(-b^2q^6;q^8)_\infty} {(-a^2q^2;q^8)_\infty\,(-b^2q^2;q^8)_\infty} \qquad(ab=q^2)\tag{1}$$
Now set $a=-\mathrm{i}q_2^5$, $b=\mathrm{i}q_2^{-1}$. This implies $ab=q^2$, $a/b=-q^3$, $-a^2=q^5$, $-b^2=q^{-1}$, hence $$\underbrace{\cfrac{1}{1-q^2+\cfrac{q^2(1-q^{-1})(1-q^5)} {1-q^6+\cfrac{q^4(1-q)(1-q^7)} {1-q^{10}+\cfrac{q^6(1-q^3)(1-q^9)}{1-q^{14}+\cdots}}}}}_{\frac{M(q)}{1-q^3}} = \frac{(q^{11};q^8)_\infty\,(q^5;q^8)_\infty} {(q^7;q^8)_\infty\,(q;q^8)_\infty}$$ which implies your statement for $M(q)$.
In $(1)$, replace $q$ with $\mathrm{i}q$. The formula requires $ab=-q^2$ then, and odd powers of $q^2$ in $(1)$ change sign. Now set $a=-q_2^5$, $b=q_2^{-1}$. This implies $ab=-q^2$, $a/b=-q^3$, $-a^2=-q^5$, $-b^2=-q^{-1}$, hence $$\underbrace{\cfrac{1}{1-q^2-\cfrac{q^2(1+q^{-1})(1+q^5)} {1+q^6+\cfrac{q^4(1-q)(1-q^7)} {1+q^{10}-\cfrac{q^6(1+q^3)(1+q^9)}{1+q^{14}+\cdots}}}}}_{\frac{N(q)}{1-q^3}} = \frac{(q^{11};q^8)_\infty\,(q^5;q^8)_\infty} {(q^7;q^8)_\infty\,(q;q^8)_\infty}$$ which implies your statement for $N(q)$.
From a bird's view, this works because the right-hand side in (1) is preserved if the signs of $q^2$, $a^2$ and $b^2$ are changed simultaneously. The corresponding left-hand sides must then be equivalent too.