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Let $q=e^{2\pi i\tau}$. If $u(\tau)$ is Ramanujan's octic continued fraction,

$$u(\tau)=\cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$

is it true that the generator of the octahedral group is the continued fraction,

$$\big(u(2\tau)\big)^2=\cfrac{2\,q^{1/2}}{1-q+\cfrac{q(1+q)^2}{1-q^3+\cfrac{q^2(1+q^2)^2}{1-q^5+\cfrac{q^3(1+q^3)^2}{1-q^7+\ddots}}}}$$

for $|q|\lt 1$?

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  • $\begingroup$ I improved your post. Beautiful cfrac for $(u(2\tau))^2$! Where did you find this? $\endgroup$ Jul 23, 2015 at 14:02

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