# 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]

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The second one is trivial; you can even make the function entire, just by making sure the $a\_n$ don't grow too fast. –  Harald Hanche-Olsen Nov 19 '09 at 22:11
Oh, of course you're right; brain fart. –  Qiaochu Yuan Nov 20 '09 at 0:10
Editing the question like that after a comment makes for the comment to become impossible to understand... –  Mariano Suárez-Alvarez Nov 20 '09 at 3:29
I am tempted to agree with Mariano: use of the strikethrough tag would be preferable, although perhaps not always possible in reasonable fashion –  Yemon Choi Nov 20 '09 at 8:22

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.

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Very nice! So now Kristal's argument works. –  David Speyer Nov 21 '09 at 4:07
I found the precise reference: E. Borel, Bull. Sci. Math. 18 (1894), 22-25. –  Richard Stanley Dec 23 '09 at 21:14
To precise more: Sur une application d'un theoreme de M. Hadamard, E. Borel, Bull. Sci. Math. 18 (1894), 22-25 archive.org/stream/s2bulletindessci18fran#page/n27/mode/2up –  Junkie Apr 26 '10 at 2:28

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:

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.

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This isn't enough. As Greg Kuperberg noted, all we know is that the coefficients of the rational function are required to eventually ROUND to a sequence of prime numbers. –  Qiaochu Yuan Nov 21 '09 at 0:36
If there is a sum of a decaying series and a rational function then I think the decaying series will go to zero which it means that the rational function will have to go arbitrarily close to the primes numbers so it looks like to me they will have to be increasingly close to integers which is more than rounding. –  Kristal Cantwell Nov 21 '09 at 1:26
That's true, Kristal. You could try to use to your advantage that for every $\epsilon > 0$, there is a series-rational sequence whose rounding error is less than $\epsilon^n$. (But not the same sequence each time, unless the function is rational plus entire.) That is much more stringent than merely rounding. –  Greg Kuperberg Nov 21 '09 at 1:36
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 $n \log n$).