Given $\large q=e^{2\pi i \tau}$. Define, $$\alpha(\tau) = \sqrt2\,q^{1/8}\prod_{n=1}^\infty\frac{ (1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}$$ $$\beta(\tau) = q^{1/5}\prod_{n=1}^\infty\frac{ (1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$ $$\gamma(\tau) = q^{1/4}\prod_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-2})(1-q^{6n-4})}$$ $$\delta(\tau) = q^{1/3}\prod_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}$$ $$\epsilon(\tau) = q^{1/2}\prod_{n=1}^\infty\frac{ (1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}$$ $$\lambda(\tau) = q^{1/1}\prod_{n=1}^\infty\frac{ (1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}$$

each of which has a beautiful $q$-continued fraction. Then we propose the *functional equations*,

$$\begin{aligned} \alpha^8\Big(\frac{-1}{4\tau}\Big)&=1-\alpha^8(\tau)\\[2mm] \beta^5\Big(\frac{-1}{5\tau}\Big)&=\frac{\phi^5\beta^5(\tau)-1}{-\beta^5(\tau)-\phi^5},\quad\phi=\tfrac{1+\sqrt5}2\\[2mm] \gamma^4\Big(\frac{-1}{6\tau}\Big)&=\frac{\gamma^4(\tau)-\tfrac19}{\gamma^4(\tau)-1}\\[2mm] \delta^3\Big(\frac{-1}{6\tau}\Big)&=\frac{\delta^3(\tau)-\tfrac18}{-\delta^3(\tau)-1}\\[2mm] \epsilon^2\Big(\frac{-1}{8\tau}\Big)&=\frac{u^2\,\epsilon^2(\tau)-1}{\epsilon^2(\tau)-u^2},\quad u =1+\sqrt2\\[2mm] \lambda\Big(\frac{-1}{12\tau}\Big)&=\frac{v\,\lambda(\tau)-1}{\lambda(\tau)-v},\quad v = 2+\sqrt3\end{aligned}$$

*Note 1*: Excepting $\gamma^4\big(\tfrac{-1}{6\tau}\big)$ which is in this post and served as the model, I just found the rest empirically, but their consistent forms suggest these are correct.

*Note 2*: Also, the two order $6$ obey the simple $\displaystyle \frac1{\gamma^4(\tau)}-\frac1{\delta^3(\tau)}=1$.

**Questions:**

- How do we derive the easiest one from first principles and generalize it for the rest?
- Is the list complete? Without requiring representation as a q-continued fraction, is there a similar functional equation for, say, a $p=7$ order?