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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 $H(x,q)+\dfrac{1}{H(x,q)}-1$?

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  • $\begingroup$ 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 $H(x;q)$? $\endgroup$
    – Mark Fischler
    Commented Feb 24, 2016 at 16:31
  • $\begingroup$ 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 $H(x;q)$? $\endgroup$
    – Mark Fischler
    Commented Feb 24, 2016 at 16:34
  • $\begingroup$ Why is it essential to have $-1$ is the last expression? Won't "1+nice formula" be a nice formula itself? $\endgroup$
    – Max Alekseyev
    Commented Feb 24, 2016 at 20:51
  • $\begingroup$ Max, -1 is not essential. $\endgroup$
    – Adam Gyenge
    Commented Feb 25, 2016 at 21:43
  • $\begingroup$ @Adam Gyenge :Perhaps this post is in the spirit of your question $\endgroup$
    – Nicco
    Commented Aug 25, 2017 at 1:56

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