most recent 30 from http://mathoverflow.net 2013-06-19T09:01:13Z http://mathoverflow.net/feeds/question/82426 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82426/a-conjecture-on-the-relative-size-of-goldbach-pairs Chuck 2011-12-02T02:46:40Z 2011-12-02T16:16:07Z

<p>On leafing through some papers of John Nash (available online on his <a href="http://www.math.princeton.edu/jfnj/" rel="nofollow">webpage</a>) I found this intriguing little observation:</p> <blockquote> <p>Noticing that with larger even numbers it seemed to become possible to find Goldbach decompositions with the smaller prime very very much smaller than the larger I thought of the simple conjecture that only a finite set of even integers would be such that they could not be expressed in the form of a sum of two primes where the size of the cube of the smaller prime would be less than the size of the larger prime.</p> </blockquote> <p>Has anyone ever tackled this conjecture? Would it be an 'easy' conjecture?</p> <p>Let me make the conjecture precise, for the sake of clarity. Let $N$ be an even number and let $s(N)$ stand for the smallest prime such that $N-s(N)$ is prime (if such a prime exists, i.e. if $N$ is expressible as the sum of two primes.) Define the following set: $$S_m = \lbrace N \vert N \text{ is even and } s(N)^m > N-s(n) \rbrace $$ With this notation, Nash's conjecture asks: Is $S_3$ finite or infinite?</p> <p>Nash calculated the first member of $S_3$, which is 63274 = 293 + 62681. </p> <p>What about other values of $m$? If indeed $S_3$ is infinite, is there an $m$ such that $S_m$ is finite?</p> <p>(I'm tagging this as a reference request too since I know very little about the Goldbach literature and would be intrigued to read any related papers.) </p>