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Let $q=e^{2\pi i \tau}$. The Heine continued fraction is $$H_2(\tau)=\frac1{q^{1/24}}\frac{\eta(2\tau)}{\eta(\tau)} =1+\cfrac{q}{1-q+\cfrac{q^3-q^2}{1+\cfrac{q^5-q^3}{1+\cfrac{q^7-q^4}{1+\ddots}}}}$$

Some nice values are,

$$H_2\Big(\tfrac{1+\sqrt{-5}}{2}\Big)=\frac1{q^{1/24}}\frac{\zeta_{48}}{(2^{1/4}\,\phi^{1/4})}=0.9991112\dots$$

$$H_2\Big(\tfrac{1+\sqrt{-11}}{2}\Big)=\frac1{q^{1/24}}\frac{\zeta_{48}}{(1+T^{-1})}=0.9999701\dots$$

$$H_2\Big(\tfrac{1+\sqrt{-23}}{2}\Big)=\frac1{q^{1/24}}\frac{\zeta_{48}}{(\sqrt{2}\,P)}=0.99999971\dots$$

where $\zeta_{48}=e^{2\pi i/48}$, with golden ratio $\phi$, tribonacci constant $T$, and plastic constant $P$.

Q: Does the next step as $\displaystyle H_3(\tau)=\frac1{q^{1/12}}\frac{\eta(3\tau)}{\eta(\tau)}$ have a similar continued fraction as well?

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