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Let \begin{equation} z := \prod_p p^{1/p^2}, \end{equation}

where the product is over all prime numbers $p$, and we always take the positive real root. Is $z$ transcendental or algebraic, or (as I suspect) is the answer not at all clear (meaning it's "probably" transcendental)?

More generally, let $(p_i)$ be an increasing sequence of prime numbers, and $(e_i)$ a sequence of integers such that the infinite product $\prod_{i=1}^\infty (p_i)^{1/e_i}$ converges.

Are there some numbers of this form where we can say anything, other than those where the convergence is so fast that we can use classical Liouville-type arguments?

Edit: Just to clarify: I'm not sure if there's any natural reason to look at such numbers, except that I had a vague idea (which is too long to get into here) that I might be able to prove something about them. I guess that was the case in the beginning of transcendence theory (a subject I don't usually think about) -- the first numbers to be proved transcendental were not numbers anyone cared about for any other reason.

Another comment about these numbers: they "look" transcendental, because the partial products live in larger and larger extensions of the rationals. However, a power of each partial product is an integer, so in some sense we are not getting too far away from the rationals -- or away from $\mathbb{Q}^\times \otimes \mathbb{Q}$.

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    $\begingroup$ @Hjalmar not that I know of! I had a vague idea about a possible novel way to show that some numbers of this form are transcendental. Before I spent too long figuring out if this idea worked, I wanted to get a sense of whether or not anything could be said already. $\endgroup$
    – Bobby Grizzard
    Commented Jan 23, 2015 at 12:16
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    $\begingroup$ My sense: A proof that this is rational would be interesting. A proof that this is transcendental would be very interesting. A proof that this is algebraic but irrational would be stupendous. $\endgroup$
    – Gerald Edgar
    Commented Jan 23, 2015 at 14:29
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    $\begingroup$ By the way, what about the (presumably) easier case $\prod_{n=1}^\infty n^{1/n^2}$ ?? $\endgroup$
    – Gerald Edgar
    Commented Jan 23, 2015 at 14:30
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    $\begingroup$ $\prod_{n=1}^\infty n^{1/n^2} = \exp(-\zeta'(2))$, so show that is transcendental! $\endgroup$
    – Gerald Edgar
    Commented Jan 23, 2015 at 14:59
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    $\begingroup$ Similar infinite products can even be rational: For example, $\prod_{i=1}^{\infty} \frac{p_{i}^{2}+1}{p_{i}^{2}-1} = \frac{5}{2},$ where $p_{i}$ is the $i$-th prime. $\endgroup$
    – Geoff Robinson
    Commented Jan 23, 2015 at 14:59

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I'm not sure this really constitutes an answer, but I think that it may not be fruitful to study the type of generalized question you discussed at the start of your post; you should be able to prove rather easily that for any real number $r>1$ you can find a sequence of primes $p_i$ with exponents $e_i=p_i$ such that the infinite product $\prod_{i=1}^\infty (p_i)^{1/e_i}$ converges to exactly $r$. Just pick the primes in a greedy fashion. Or maybe you use every prime, and choose the exponents in a greedy fashion. Or mix and match, and end up with many ways to represent your real in this fashion (if it matters, or you are diligent enough, you can show that there are uncountably many such representations for each $r>1$).

Given that there is so much choice and so little effect, it's my opinion that you're going to have to rely on the incredibly specific structure of your product to go anywhere fast, and perhaps this isn't so unreasonable of an opinion given the nature of almost all results we have about transcendence (namely that they exploit the very special definitions, structures, relations, etc., that their subject numbers enjoy, like $e$ and $\pi$). But hey, I'm no expert.

Edit: New account, can't make comments, so this space is for Wojowu's request (and perhaps others in the future).

You're right in questioning me on that point; this is not a proof that I have done myself, but it seems fairly straight forward given that the product $\prod_{i=1}^\infty (p_i)^{1/p_i}$ diverges if you take it over all primes. You can't fail to make it past $r$ since you can always just add more primes in sequence (there is always a "partial tail" that is as large as you want), but by selectively (greedily) deleting certain primes we can lower the partial products below $r$. So you should converge exactly to $r$. If you go about formalizing that and run into problems, let me know.

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    $\begingroup$ Could you elaborate on why you think your first claim is true? I guess that you mean to always choose the least $p_i$ such that adding $p_i^{1/e_i}$ factor will make the product not exceed $r$, but I don't see how it guaranteed the product actually converges to $r$ and not something smaller. $\endgroup$
    – Wojowu
    Commented May 19, 2015 at 16:38
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    $\begingroup$ For reference, Merten's first theorem says $\sum_{p\leq n} \ln(p)/p = \ln(n) + E_n$ where $|E_n| \leq 2$. $\endgroup$
    – user13113
    Commented May 19, 2015 at 17:20

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