Does there exist a meromorphic function all of whose Taylor coefficients are prime? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:28:34Z http://mathoverflow.net/feeds/question/6179 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6179/does-there-exist-a-meromorphic-function-all-of-whose-taylor-coefficients-are-prim Does there exist a meromorphic function all of whose Taylor coefficients are prime? Qiaochu Yuan 2009-11-19T21:50:51Z 2009-11-21T03:40:39Z <p>More precisely, does there exist an unbounded sequence $a_0, a_1, ... \in \mathbb{N}$ of primes such that the function</p> <p>$\displaystyle O(z) = \sum_{n \ge 0} a_n z^n$</p> <p>is meromorphic on $\mathbb{C}$? </p> <p>[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]</p> http://mathoverflow.net/questions/6179/does-there-exist-a-meromorphic-function-all-of-whose-taylor-coefficients-are-prim/6213#6213 Answer by Maksym for Does there exist a meromorphic function all of whose Taylor coefficients are prime? Maksym 2009-11-20T03:33:12Z 2009-11-20T03:33:12Z <p>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 <a href="http://mathworld.wolfram.com/PowerSeries.html" rel="nofollow">theorem</a> 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 $n \log n$).</p> http://mathoverflow.net/questions/6179/does-there-exist-a-meromorphic-function-all-of-whose-taylor-coefficients-are-prim/6334#6334 Answer by Kristal Cantwell for Does there exist a meromorphic function all of whose Taylor coefficients are prime? Kristal Cantwell 2009-11-21T00:28:14Z 2009-11-21T00:28:14Z <p>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 <em>Recurrence sequences</em> by Graham Everest available here:</p> <p>[<a href="http://books.google.com/books?id=LmfonVHe7MMC&amp;source=gbs_navlinks_s" rel="nofollow">http://books.google.com/books?id=LmfonVHe7MMC&amp;source=gbs_navlinks_s</a>[1]</p> <p>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.</p> http://mathoverflow.net/questions/6179/does-there-exist-a-meromorphic-function-all-of-whose-taylor-coefficients-are-prim/6347#6347 Answer by Richard Stanley for Does there exist a meromorphic function all of whose Taylor coefficients are prime? Richard Stanley 2009-11-21T03:40:39Z 2009-11-21T03:40:39Z <p>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.</p>