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

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,

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

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## closed as not a real question by Thomas Andrews, Chris Eagle, Arturo Magidin, J.D., Alex BeckerJul 5 '12 at 17:35

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Your title makes it seem like you know the answer. Is that the case? –  Alex Becker Jul 5 '12 at 17:16
This site is not a site for news announcements, it is a site for posting questions. It is inappropriate to use it to post news. –  Thomas Andrews Jul 5 '12 at 17:20
There's a difference between asking and answering a question (perhaps because you want feedback, perhaps because you found the answer, or other similar circumstances) and making a misleading post for the purpose of advertising something (even if that something is a piece of math). Your post amounts to mathematical spam. If you want to make an announcement, call a press conference or post it in an appropriate forum. This ain't it. –  Arturo Magidin Jul 5 '12 at 17:23
The clear intent of your post was not to present an answer to a question that you wanted asked, but to announce news. I did not object to the title, although the title made it clear your intent - news announcement. –  Thomas Andrews Jul 5 '12 at 17:34
@TitoPiezasIII: I don't know about Thomas. But I would like you to actually read the criticisms you have been given and try to understand them, instead of pretending that it was only the title that people found troublesome. It wasn't the title is objectionable, it is the intent of the post and your attempt to use this site as a way to make a news announcement. That is not what this site is for. It's not the formatting, it's not the title. Read the comments, for crying out loud. –  Arturo Magidin Jul 5 '12 at 18:45

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?

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