Polar Density of a Set of Primes

Solution 1:

We might as well restrict to $K=\mathbb Q$, so that $\zeta_A(s) = \prod_{p\in A} (1 - p^{-s})^{-1}$. If the polar density of $A$ exists, then the Dirichlet density exists and is equal. In particular, if the natural density and polar density of $A$ both exist, then they must be the same.

Since the definition of polar density is only capable of producing rational numbers, it suffices to name a set of primes $A$ having a natural density that is irrational.

I can think of several natural sets of primes which should have irrational density, but I haven't thought of one that is both aesthetically pleasing and of provably irrational density.

Some incomplete examples:

• The set of primes with $p-1$ squarefree has density equal to Artin's constant $\prod_p (1-\frac{1}{p(p-1)})$, see https://math.stackexchange.com/a/155117/30402. Artin's constant is presumed but not known to be irrational.
• The set of primes for which $p-1$ has a prime factor greater than $\sqrt{p}$ is conjectured to be $\log 2$, which is the exact density of integers $n$ possessing a similar property.

Of course it is possible to describe less succinctly a set of primes having any density in $[0,1]$. I still wonder if there is a simple variation that will give an obviously irrational singular series constant like $6/\pi^2$.

Another approach is to take a sparse set of primes for which $\prod_p(1-1/p)^{-1}$ still diverges slowly (for instance, if the counting function is $x/(\log x \log \log x)$). This will have natural density $0$, but the singularity at $s=1$ prevents $\zeta_A(s)$ from being continued to a holomorphic function in a neighbourhood of $1$ (by a result of Landau).

The disjoint union of such a set with one already having polar density should also lack a polar density, even with a rational natural density.