Given the following Generalized Rogers-Ramanujan continued fraction, with $|q|\lt1$, which is equation (38) in mathworld


where $F(a,q)= \frac{\displaystyle \sum_{n=0}^{\infty}\frac{(-a)^n q^{n^2}}{(q)_{n}}}{\displaystyle\sum_{n=0}^{\infty}\frac{(-a)^n q^{n(n+1)}}{(q)_{n}}}\tag2$

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


If we let $a=-1$ and take the reciprocal of $(3)$, we have the rogers-ramanujan continued fraction formula first conjectured by R. Blecksmith and J. Brillhart and proved in On the generalized rogers-ramanujan continued fraction by Bruce C. Berndt and AE JA YEE

  • $\begingroup$ $F(a,q)$ is an odd part of $H(a,q)$, en.wikipedia.org/wiki/… $\endgroup$ Oct 17, 2017 at 6:40
  • $\begingroup$ How do you prove $(1)=(2)$ $\endgroup$
    – reuns
    Oct 17, 2017 at 18:00

1 Answer 1


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


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


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


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