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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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    Your title makes it seem like you know the answer. Is that the case?2012-07-05
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    Yes, MSE permits question-and-answer if it is in Jeopardy form. Took me some time to type the answer.2012-07-05
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    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.2012-07-05
  • 0
    What about this then: http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/2012-07-05
  • 5
    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.2012-07-05
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    @TitoPiezasIII I believe that Thomas is objecting to the way you presented the question, not that you answer it yourself. Specifically, I think he's objecting to the title, which is reminiscent of "Breaking News! New Differential Equation Solved! Page Five, Read All About It!".2012-07-05
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    Sorry, but the number of new mathematical results would flood this site and make it useless if used this way. The stackoverflow.com situation is different: "I was seeking an answer for question X, did not find it, so I figured it out myself and am posting it on SO.com" is different from posting news.2012-07-05
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    Very well, I will change the title to correspond to the Jeopardy format.2012-07-05
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    I do want feedback. Kindly read the answer.2012-07-05
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    @AlexBecker I am not just objecting to the title. Posting news is not an appropriate use of answering your own question. Look at the blog and read it - it gives specific instances where this is a reasonable thing to do, and new results is not one of them.2012-07-05
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    Thomas: The title has been changed. Is that more acceptable?2012-07-05
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    It seems to me this would be more fitting as an answer to [your previous question](http://math.stackexchange.com/questions/165297/an-infinite-product-for-left-frac-eta13-tau-eta-tau-right2-revis) anyway.2012-07-05
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    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.2012-07-05
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    Thomas: This is the first time I did a question-and-answer post. Is it expected one is to do it perfectly the first time? I have already corrected the title to reflect the requirements of the jeopardy format. What more do you want?2012-07-05
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    @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.2012-07-05
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    @ArturoMagidin: I conflated StackExchange (SE) with StackOverflow (SO), and assumed the policies were the same. My apologies.2012-07-05

1 Answers 1

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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?