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Is it known whether $e^{\frac{\pi^2}{12 \log 2}}$ is transcendental or algebraic?

This number showed up in this other question.

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    $\begingroup$ exp(pi) is transcendental by Gelfond-Schneider: en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem $\endgroup$ – Adam P. Goucher Jan 19 '17 at 11:34
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    $\begingroup$ What is the motivation behind this question ? $\endgroup$ – Sylvain JULIEN Jan 19 '17 at 11:49
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    $\begingroup$ @Sylvain, there was a question maybe yesterday about algebraic non-examples of a result of Levy on continued fractions, and this question came up there. (But OP should have linked to it) $\endgroup$ – Gerry Myerson Jan 19 '17 at 11:52
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    $\begingroup$ $e^\pi=(-1)^{-i}$ so it's transcendental. The open one is $\pi^e$ but I don't think that $\pi^e$ has some profound meaning, it's just something that looks as silly as $e^\pi$. $\endgroup$ – Kevin Buzzard Jan 19 '17 at 11:54
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    $\begingroup$ =3.275822918721811159787681882453843863608475525982374149405198924190723215644960355... Clearly trascendental! :) $\endgroup$ – Yaakov Baruch Jan 19 '17 at 13:01
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This is most likely open, since alredy $e^{\pi^2}$ is not known to be transcendental.

As an added difficulty, I don't think that $\frac{\pi^2}{12 \log 2}$ is known to be transcendental either.

There are very few, very limited, tricks to prove this kind of result: things like taking $(-1)^{-i}$ and $i^i$ and applying Gelfond–Schneider, or building the Weierstrass $\wp$-function of $\mathbb{Q}(\sqrt{-d})$ to get the transcendence of $e^{\pi\sqrt{d}}$ from its invariants.

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  • $\begingroup$ If $\pi^2/12\log 2$ were algebraic, its exponential would be transcendental, so I don't see why it's a difficulty :P $\endgroup$ – Wojowu Jan 20 '17 at 6:02

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