## 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. 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 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 $n$“).