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

Question

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

Answer 1

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)

Answer 2

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.