Given the *Ramanujan theta function*,

$$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$

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, 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 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* regarding PSL(2,5), PSL(2,7), and PSL(2,11). I've tried various denominators $f(-q^n)$ but can't get $q^{m/660}$.

samedenominators as mine, but the numerator for $p=11,13$ is different. Yours give $q^{199/264}$ and $q^{149/156}$. Give me time to write an addendum. $\endgroup$nota quotient? $\endgroup$onlyif $k=5m$. And believe I know why you chose $55$ since $24/(p-1)$ for $p=11$ is $12/\color{blue}5$. The $33$ version is not pretty? $\endgroup$4more comments