Does there exist a meromorphic function all of whose Taylor coefficients are prime? - MathOverflow

Does there exist a meromorphic function all of whose Taylor coefficients are prime?

2009-11-19T21:50:51Z

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]

Answer by Maksym for Does there exist a meromorphic function all of whose Taylor coefficients are prime?

2009-11-20T03:33:12Z

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 $n \log n$).

Answer by Kristal Cantwell for Does there exist a meromorphic function all of whose Taylor coefficients are prime?

2009-11-21T00:28:14Z

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_s[1]

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.

Answer by Richard Stanley for Does there exist a meromorphic function all of whose Taylor coefficients are prime?

2009-11-21T03:40:39Z

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.