Given the [*Ramanujan theta function*][1], $$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$ Let $q = e^{2\pi i \tau}$ and assume $\tau = \sqrt{-d}.$ Then the following functions for levels $p=5, 7, 11, 13$ are radicals, $$q^{11/60} \frac{f(-q,-q^4)}{f(-q)}$$ $$q^{61/168} \frac{f(-q,-q^6)}{f(-q^2)}$$ $$q^{199/264} \frac{f(-q,-q^{10})}{f(-q^4)}$$ $$q^{149/156} \frac{f(-q,-q^{12})}{f(-q^5)}$$ (Note: The numerators for levels $p=11,13$ are corrected courtesy of Somos.) Without the $q$-factor, they are analogues of the [Rogers-Ramanujan identities][2], the first being the namesake, and which are sum-products, $$\sum_{n=0}F_1(q) = \prod_{n=1}F_2(q)$$ Three of the integers involved, namely $60, 168, 156,$ are [orders of transitive groups][3] which in Magma notation are $5T4, 7T5, 13T6$. The one for $p=11$ is the odd one out since $264$ is not a group order. But we can use alternative quotients, such as the one for level $7$, $$q^{17/42} \frac{f(-q,-q^6)}{\color{blue}{f(-q)}}$$ which is also a radical and $42$ is the order of $7T4$ (and the answer to the Hitchhiker's universe). It is not necessary that there be a sum-product associated with it. **Question:** For level $p=11$, can we find an *alternative* quotient (both numerator and denominator) such that, together with the particular factor $q^{m/660},$ then it is also a radical? **P.S.** The desired integer $660$ is the order of $11T5$ so this would be another manifestation of [*V. Arnold's trinities*][4] regarding PSL(2,5), PSL(2,7), and PSL(2,11). I've tried various denominators $f(-q^n)$ but can't get $q^{m/660}$. [1]: https://en.wikipedia.org/wiki/Ramanujan_theta_function [2]: https://mathworld.wolfram.com/Rogers-RamanujanIdentities.html [3]: http://galoisdb.math.upb.de/groups [4]: https://en.wikipedia.org/wiki/ADE_classification#Trinities