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Closed form for $\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}$

This is equivalent to $\prod_{n=1}^\infty n^{1/2^n}$.

Putting this into Wolframalpha gives that it is approximately 1.661687, and failed to find a closed form.

(1) Is this irrational and transcendental, irrational and algebraic, or rational?

(2) Is there a name for this constant or does there exist a possible closed form?

(3) How does one calculate its partial sum formula? Wolfram Partialsum

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  • $\begingroup$ Putting that number into a search engine gives this or this. $\endgroup$ – mvw Dec 9 '14 at 2:57
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    $\begingroup$ The second link is merely the same question as this one, just with a log taken to make the product a sum. The first link does not match with the value given. $\endgroup$ – Teoc Dec 9 '14 at 3:03
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Yes, there is a name for it. It is called the Somos' Quadratic Recurrence Constant. It has a weird closed-form in terms of the polylogarithm and Lerch transcendent.

P.S. You can search for constants in the OEIS using its decimal expansion. If you're lucky, then it might be there like in this case: 1,6,6,1,6,8, (A112302)

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    $\begingroup$ Sigh, did you look at the definition of oeis.org/A112302? It is precisely $\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}$ $\endgroup$ – Tito Piezas III Dec 9 '14 at 3:06
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    $\begingroup$ Wolfram may be wrong. Check out the Mathworld Link. $\endgroup$ – Cheerful Parsnip Dec 9 '14 at 3:07
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    $\begingroup$ Wolfram alpha and mathematica are not infallible. $\endgroup$ – Cheerful Parsnip Dec 9 '14 at 3:08
  • $\begingroup$ Is it irrational and transcendental? Mathworld does not have a notice on that. $\endgroup$ – Teoc Dec 9 '14 at 3:11
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    $\begingroup$ @MathNoob: We are pretty sure it is transcendental, but probably don't have a proof. Unless a number has a reason to be algebraic, it is transcendental-there are only countably many algebraic numbers. This doesn't seem to have a reason to be algebraic. $\endgroup$ – Ross Millikan Dec 9 '14 at 4:57
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Let $$A=\prod_{n=1}^\infty n^{1/2^n}$$ $$\log(A)=\sum_{n=1}^\infty\frac{1}{2^n}\log(n)$$ and, from a CAS, the result is $$\log(A)=-\text{PolyLog}^{(1,0)}\left(0,\frac{1}{2}\right)$$ $$A=e^{-\text{PolyLog}^{(1,0)}\left(0,\frac{1}{2}\right)}$$ where appears a derivative of the polylogarithm function.

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