# 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.

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I wonder what happens if $S$ is the set of numbers of the form $p+1$, $p$ being prime. –  Gerry Myerson Nov 5 '12 at 11:24

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:

1. 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.

2. 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 $\Re{(s)} = 0$

3. 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.

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What do you mean by a "natural boundary"? Unfortunately I think the second condition is never satisfied by a sequence satisfying my conditions, because it implies that $S$ consists only of prime powers, but the set of prime powers that aren't primes has Dirchlet density zero. –  Will Sawin Nov 5 '12 at 15:18
Natural Boundary: analytic continuation to that line, but no further owing to an accumulation of zeros or poles on the line forcing the function to be identically 0 or infinity under the assumption of an analytic continuation beyond the line - a contradiction. –  Jonah Sinick Nov 5 '12 at 16:37
Yes, you're right that the second condition that I list and yours are mutually exclusive - again, the things that I mention are about the cognate question that I mention, not your original one. –  Jonah Sinick Nov 5 '12 at 16:41