My task connecting number spectral analysis and the GC

Consider f \forall_{n \in N, n \ge 1}{f(2n)=p_1} for 2n = p_1 + p_2, for p_1,p_2 prime,  p_1<p_2. Consider F \forall_{n \in N, n \ge 1}{F(2n)=p_2} for 2n = p_1 + p_2, for p_1,p_2 prime,  p_1<=p_2. Due to errors in the statement, a couple of examples:  16 = 3 + 13, f(16)=3, F(16)=13, we take both the smaller of the two primes as well as the smallest prime of all such pairs. 16=11+5, 3<5, 13>11, so we take 3 and 13.

Find the limit lim_{n \rightarrow \infty}{{{f(n)}\over{F(n)}}} and lim_{n \rightarrow \infty}{{{f(n)}\over{n}}}.

Anwer from JoeBlow (to the first one, slightly changed due to Latex syntax rendering problem):

So you’re asking for \lim_{n\to \infty}\frac{p(n)}{2n-p(n)} where p(n) is the smallest prime such that 2n-p(n) is prime (it goes without saying we are assuming the Goldbach conjecture — or at least some weaker asymptotic version of it — throughout this discussion).

If this limit were not 0, it would mean we can find arbitrarily large N such that N-p is not prime for all primes p<aN for some a>0.  If we pretend that primes are randomly distributed in N with a probability distribution consistent with PNT, the probability of such an event for a fixed N is bounded roughly by e^{-\frac{N}{a^2(\log N)^2}}.  But Borel-Cantelli then implies that the event occurs for only finitely many N with probability 1.

Of course the probabilistic argument doesn’t quite hold water, but it suggests heuristically that it would be extremely surprising if this limit weren’t 0 — as surprising as Goldbach failing infinitely many times (because in this probabilstic framework the event \frac{2n-p(n)}{2n}\geq an has the same order of magnitude as the event “the Goldbach conjecture fails for n“).

Advertisements

About misha

Imagine a story that one can't believe. Hi. Life changes here. Small things only.
This entry was posted in Mathematics. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s