Entry 4

These q-continued fractions were "missed(?)" by Ramanujan. Given the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$, then $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-v}+\sqrt{1+v^2}\big)^{1/5}}-\phi = \cfrac{e^{-2\pi\sqrt{5}/5}}{1 + \cfrac{e^{-2\pi\sqrt{5}}}{1 + \cfrac{e^{-4\pi\sqrt{5}}}{1 + \cfrac{e^{-6\pi\sqrt{5}}}{1 + \ddots}}}}\quad\text{where}\; v = \phi^5\tag1$$ $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-v}+\sqrt{1+v^2}\big)^{1/5}}-\phi = \cfrac{e^{-2\pi\sqrt{10}/5}}{1 + \cfrac{e^{-2\pi\sqrt{10}}}{1 + \cfrac{e^{-4\pi\sqrt{10}}}{1 + \cfrac{e^{-6\pi\sqrt{10}}}{1 + \ddots}}}}\quad\text{where}\; v = \small{18-5\sqrt{5}}\tag2$$ $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-v}+\sqrt{1+v^2}\big)^{1/5}}-\phi = \cfrac{e^{-2\pi\sqrt{15}/5}}{1 + \cfrac{e^{-2\pi\sqrt{15}}}{1 + \cfrac{e^{-4\pi\sqrt{15}}}{1 + \cfrac{e^{-6\pi\sqrt{15}}}{1 + \ddots}}}}\quad\text{where}\; v = \tfrac{147-55\sqrt{5}}{4}\tag3$$ where, as $\sqrt{5n}$ increases, then $v$ is an algebraic number of generally increasing high degree. I found this family using Mathematica and the second cfrac in Entry 3. This implies that the Rogers-Ramanujan continued fraction $R(\tau)$ $$R(i \sqrt{n})=\frac{\sqrt{5}}{1+\phi^{-1}R(i/\sqrt{n})}-\phi$$ for some positive real $n$, though I have no proof of this assertion.