Is there an infinite product for $\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^2$ analogous to the Rogers-Ramanujan identity?

Question

Given

$$ \left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}\;\; =\;\; \frac{r^5}{1-11r^5-r^{10}},\;\;\;\;\;\text{with}\;\;r\; =\; q^{1/5} \prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$

where $\eta(\tau)$ is the Dedekind eta function,

$$\eta(\tau) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$$

and $q = \exp(2\pi i\tau)$, is there an analogous identity for,

$$ \left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2}\;\; =\;\; ???$$

Answer 1

Yes. Michael Somos just today found the identity,

$$ \left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} = \frac{s}{s^2-3s-1}$$

where,

$$s=\frac{1}{q}\; \prod_{n=1}^\infty \frac{ (1-q^{13n-2})(1-q^{13n-5})(1-q^{13n-6})(1-q^{13n-7})(1-q^{13n-8})(1-q^{13n-11}) }{(1-q^{13n-1})(1-q^{13n-3})(1-q^{13n-4})(1-q^{13n-9})(1-q^{13n-10})(1-q^{13n-12})} $$

thus completing the family for $N = 2,3,5,7,13$.

Kindly see this MSE post for the other N. Does address Matt E and Loeffler's comments in that post? Is "s" a modular function?