Conjectured continued fraction for the Generalized Rogers-Ramanujan continued fraction Given the following Generalized Rogers-Ramanujan continued fraction,with $|q|\lt1$,which is equation (38) in mathworld
$F(a,q)=1-\cfrac{aq}{1-\cfrac{aq^2}{1-\cfrac{aq^3}{1-\cfrac{aq^4}{1-\cfrac{aq^{5}}{1-\cfrac{aq^{6}}{1-\dots}}}}}}\tag1$
where $$F(a,q)=\frac{\sum_{n=0}^{\infty}\frac{(-a)^n q^{n^2}}{(q)_{n}}}{\sum_{n=0}^{\infty}\frac{(-a)^n q^{n(n+1)}}{(q)_{n}}}\tag2$$
it is conjectured that it is equivalent to the following continued fraction
$H(a,q)= \cfrac{1}{1+\cfrac{aq}{1-\cfrac{aq}{1-\cfrac{q}{1+\cfrac{q}{1+\cfrac{aq^2}{1-\cfrac{aq^{2}}{1-\cfrac{q^{2}}{1+\cfrac{q^{2}}{1+\cfrac{aq^3}{1-\cfrac{aq^3}{1-\dots}}}}}}}}}}}\tag3$

How do we prove $F(a,q)\overset{\color{red}?}=H(a,q)$

 A: Indeed as suggested by Nemo in his comment,the usual Generalized Rogers-Ramanujan continued fraction $F(a,q)$ is the odd part of $H(a,q)$ according to the formulas from wikipedia,thus $H(a,q)$ is an extension of $F(a,q)$
Moreover the even part of $\frac{1}{H(-a,q)}$ can be determined 
$G(a,q)=\cfrac{1}{1-aq+}\cfrac{(aq)^{2}}{1+aq-q+}\cfrac{q^{2}}{1+q-aq^2+}\cfrac{(aq^2)^{2}}{1+aq^2-q^2+}\cfrac{q^{4}}{1+q^2-aq^3+}\cfrac{(aq^3)^{2}}{1+aq^3-q^3+}\cfrac{q^{6}}{1+q^3-aq^4+}\cdots$
which converges to the same value as $\frac{1}{F(-a,q)}$ and $\frac{1}{H(-a,q)}$,when $|q|\lt1$
Thus $$G(a,q)=\frac{\sum_{n=0}^{\infty}\frac{(a)^n q^{n(n+1)}}{(q)_{n}}}{\sum_{n=0}^{\infty}\frac{(a)^n q^{n^2}}{(q)_{n}}}$$
which leads us to another continued fraction equivalent to the RRCF
$$\cfrac{q^{1/5}}{1-q+}\cfrac{q^{2}}{1+}\cfrac{q^{2}}{1+q-q^2+}\cfrac{q^{4}}{1+}\cfrac{q^{4}}{1+q^2-q^3+}\cfrac{q^{6}}{1+}\cfrac{q^{6}}{1+q^3-q^4+}\cdots=q^{1/5}\frac{(q;q^5)_{\infty}(q^4;q^5)_{\infty}}{(q^2;q^5)_{\infty}(q^3;q^5)_{\infty}}$$
with the use of the notation $$\cfrac{a_{1}}{b_{1}+}\cfrac{a_{2}}{b_{2}+}\cfrac{a_{3}}{b_{3}+}\cdots$$ 
in place of $$\cfrac{a_{1}}{b_{1} + \cfrac{a_{2}}{b_{2} + \cfrac{a_{3}}{b_{3} + \cdots}}}$$
which takes up much more space than necessary 
