# Identity with Ramanujan's generalized continued fraction

Let $F(x,q)=\sum_{n\geq 0}x^n\dfrac{q^{n^2}}{(q)_n}$, where $(q)_n=(1-q)(1-q^2)\dots(1-q^n)$. Then: $$H(x,q)=\frac{F(-xq,q)}{F(-x,q)}=\dfrac{1}{1-\dfrac{qx}{1-\dfrac{q^2x}{1-\dots}}}$$ is the generalized continued fraction of Ramanujan.

Does anybody know any nice formula for $H(x,q)+\dfrac{1}{H(x,q)}-1$?

• Excuse my notational ignorance but in the opening statement, what do you mean by $(q)_n$? And, in the definition of $H(x)$, $q$ looks like an ordinary parameter; in that case, why did you not call this $H(x,q)$ or something like $H(x;q)$? – Mark Fischler Feb 24 '16 at 16:31
• Excuse my notational ignorance, but in the opening statement, what is the meaning of $(q)_n$? And in the definition of $H(x)$, it appears that $q$ is a numerical parameter; in that case, why is $H(x)$ not called $H(x,q)$ or possibly $H(x;q)$? – Mark Fischler Feb 24 '16 at 16:34
• Why is it essential to have $-1$ is the last expression? Won't "1+nice formula" be a nice formula itself? – Max Alekseyev Feb 24 '16 at 20:51
• Max, -1 is not essential. – Adam Gyenge Feb 25 '16 at 21:43
• @Adam Gyenge :Perhaps this post is in the spirit of your question – Nicco Aug 25 '17 at 1:56