# Average order and upper bound of $r_{0}(n)$

Assume Goldbach's conjecture. Then for every integer $n>1$ there exists a non-negative integer $r$ such that $n-r$ and $n+r$ are both primes. For a given $n>1$, the smallest such $r$ will be denoted as $r_{0}(n)$. Let's now define the quantity $k_{0}(n)$ as $k_{0}(n):=\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$. One can easily figure out that the number of integers $x$ below $n$ such that $k_{0}(x)=1$ is equal to $\pi(n)+O(1)$. One can write the following equality:
$$r_{0}(n)=\sum_{k_{0}(m)=1, \vert n-m\vert\leqslant r_{0}(n)}r_{0}(m)$$

One can deduce from the PNT that the average order of $r_{0}(m)$, denoted by $\langle r_{0}(m)\rangle$, verifies $\langle r_{0}(m)\rangle\sim \frac{1}{2}\log m$. If one can prove that $\displaystyle{\dfrac{1}{2r_{0}(n)}\sum_{k_{0}(m)=1, \vert n-m\vert\leqslant r_{0}(n)}1\sim\frac{1}{\log n}}\ \ (*)$, then one gets that $\langle r_{0}(m)\rangle\sim\dfrac{1}{\log n}r_{0}(n)$ and thus the average order of $r_{0}(n)$, denoted by $\langle r_{0}(n)\rangle$, verifies $\langle r_{0}(n)\rangle\sim\frac{1}{2}\log^{2}n$.

Now let $\displaystyle{r_{0,min}(n):=\inf_{x\leqslant 2n}r_{0}(x)}$ and $\displaystyle{r_{0,max}(n):=\sup_{x\leqslant 2n}r_{0}(x)}$. Is it true that $\langle r_{0}(n)\rangle\sim \dfrac{r_{0,min}(n)+r_{0,max}(n)}{2}$?
If so, one could get under Goldbach's conjecture and provided $(*)$ is true that $\displaystyle{\lim\sup_{n\to\infty}\dfrac{r_{0}(n)}{\log^{2}n}=1}$.

EDIT March 24th 2014: the quantity $\sum_{k_{0}(m)=1, |n-m|\leqslant r_{0}(n)}1$ is simply $k_{0}(n)$. As $\dfrac{2r_{0}(n)}{k_{0}(n)}$ is the average gap between two consecutive primes of size $n$, $(*)$ holds true. So that it's very likely that Goldbach's conjecture alone implies Cramer's conjecture.

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I suspect that if it were true that Goldbach implies Cramer then we'd know it by now. Anyway, it should be easy to concoct a sequence of integers that satisfy Goldbach but not Cramer. –  Gerry Myerson Mar 25 at 2:10