Sets with zeta functions that are not the primes Does there exist a set $S \subset \mathbb N$ such that the Dirchlet density of $S$ is well-defined and positive, the Dirchlet density of $S \cap \operatorname{PRIMES}$ is well-defined and zero, and:
$ \prod_{n \in S} \frac{1}{1-n^{-s}}$ 
has a meromorphic continuation to the whole complex plane? Can we construct it?
Motivation: How special is the set of primes, among sets of natural numbers of approximately the same size, for having a meromorphic continuation? I would guess that analytic continuation should be a very rare occurrence, but I don't have an intuition for how rare, or how hard it is to find an example if one exists.
Extra credit for a functional equation.
 A: You're asking about how special the prime numbers are as a subset of the integers. One can equally well ask how special the sequence $a_k = 1$ is when viewed as the sequence of coefficients of a Dirichlet series. I don't have anything to offer on your original question, but have read a few things about the latter question that may be of some interest:


*

*According to Section 8 of David
Farmer's article titled Basic
Analytic Number Theory, if $f(s)
    = \displaystyle \sum_{k = 1}^{\infty} \frac{a_k}{k^s}$ where
the $a_k$ are integers, then a
sufficient condition for $f(s)$ to
admit a meromorphic continuation to $\Re{(s)} = 0$ is that:
(1) $a_k$ is of subpolynomial growth
(2) $a_k$ is multiplicative
(3) If $p$ is prime then $a_{p^m}$ is
independent of $m$
and may or may not have a natural boundary there.

*According to section 9.5 of Titchmarsh's The Theory of the Riemann Zeta-function, if $a_k = 0$ when $k$ is composite and $a_{k} = 1$ when $k$ is prime then $f(s)$ (provably) has a natural boundary at $\Re{(s)} = 0$ 

*I've also heard of results of the type "a Dirichlet series with $a_k$ chosen at random uniformly from $[-1, 1]$ has natural boundary $\Re{(s)} = 0$ with probability $1$," but don't know  a precise statement or a reference.
