conjectured new generating function of fibonacci numbers I conjecture a new generating function for the fibonacci numbers  $F_{n}$. Given,the following conjectured q-continued fraction
$$\chi(q)=\cfrac{1}{1+q-\cfrac{(1+q^2)}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}$$
how do we show that,
$\chi\Big(\frac{1}{q}\Big)= \sum_{n=1}^\infty (-1)^{n-1} F_{n}q^n$
is true?
 A: (A partial answer.)
This is a special case of a conjectured equality discussed in this MO post. Let $|q|<1$, then,
$$\begin{aligned}U(q)
&= \prod_{n=0}^\infty \frac{\big(1-a^2q^3(q^4)^n\big)\big(1-b^2q^3(q^4)^n\big)}{\big(1-a^2q(q^4)^n\big)\big(1-b^2q(q^4)^n\big)}\\ 
&= \dfrac{1} {1+ab-\dfrac{(a+bq)(b+aq)} {1+(ab)^3+\dfrac{(a-bq^2)(b-aq^2)q} {1+(ab)^5-\dfrac{(a+bq^3)(b+aq^3)q^2} {1+(ab)^7+\dfrac{(a-bq^4)(b-aq^4)q^3} {(1+(ab)^9-\ddots }}}}}
\end{aligned}$$
If $a=q,\;b=1$, and $|q|<1$ then,
$$\begin{aligned}U(q)
&=\prod_{n=0}^\infty \frac{\big(1-q^5(q^4)^n\big)\big(1-q^3(q^4)^n\big)}{\big(1-q^3(q^4)^n\big)\big(1-q(q^4)^n\big)} =\prod_{n=0}^\infty \frac{\big(1-q^5(q^4)^n\big)}{\big(1-q(q^4)^n\big)} = \frac{1}{1-q}\\
&=\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}
\end{aligned}$$
and it only takes a little algebraic manipulation to get the brown part of this cfrac to the form in the post. I get,
If $|q|<1$:
$$\chi(q) = \frac{1}{q}\tag1$$
If $|q|>1$:
$$\chi(q) = \sum_{n=1}^\infty (-1)^{n-1} \frac{F_{n}}{q^n} = \frac{q}{q^2+q-1}\tag2$$
where $(2)$ is a variant of the identity in this post. Thus, it should be specified that the generating function is only valid when $\color{blue}{|q|>1}$. 
P.S. Such behavior is present in other cfracs. For example, for the Rogers-Ramanujan cfrac $R(q)$, if $|q|<1$, then $R(q) = R(q)$, but if $\color{blue}{|q|>1}$, then,
$$R(q) \to R(1/q^4),\quad \text{(even convergents)}$$
$$R(q) \to -1/R(-1/q),\quad \text{(odd convergents)}$$
See Section 2 of Berndt's paper.
