Does there exist a meromorphic function all of whose Taylor coefficients are prime? More precisely, does there exist an unbounded sequence $a_0, a_1, ... \in \mathbb{N}$ of primes such that the function
$\displaystyle O(z) = \sum_{n \ge 0} a_n z^n$
is meromorphic on $\mathbb{C}$?  
[A previous version of the question also asked about the exponential generating function of $(a_n)$.  However, such a function can trivially be entire. - GJK]
 A: Here is the case when you take the sequence of all primes, and you can probably adapt it to handle the general case.
The sequence of primes grows like $p_n \sim n \log n$ hence your series has radius 1 and integer coefficients.
By a theorem of Carlson (a result which was conjectured by Polya) an powerseries
with radius 1 and integer coefficients is either a rational function or has a 
natural boundary at $|z| = 1$. The second is impossible if your function is to
be meromorphic. The first is impossible because if your powerseries is a rational
function then its coefficients satisfy a linear recurrence relation which is not
the case here (a solution to a linear recurrence relation cannot grow like $n \log n$).
A: Borel proved the following much stronger result: if a power series with integer coefficients
represents a function f(z) that is meromorphic in a disk of radius >1, then f(z) extends to a rational function on all of C. I found this result without a reference on page 3 of www.mathematik.uni-bielefeld.de/~anugadre/Adeles.pdf.
A: If we have a function of radius 1 then by Carlson's theorem as noted above the function is either a rational function or has a natural boundary. For it to be meromorphic it must not have a natural boundary so it must be rational but that means that the sequence must satisfy a linear recurrence relation. But a the sequence generated by a linear recurrence relation must have an infinite number of composite values. See page 94 of Recurrence sequences by Graham Everest available here:
[http://books.google.com/books?id=LmfonVHe7MMC&source=gbs_navlinks_s1
Now if the function can be represented by a function whose radius of convergence is greater than one which decays plus a rational function then if the rational function is required to eventually be a sequence of prime numbers then the sequence generated a rational function is a linear recurrence relation it must contain an infinite number of composite numbers which gives a contradiction.
