I'm still trying to find a tight upper bound for the quantity $r_{0}(n):=\inf\{r\geq 0,(n-r,n+r)\in\mathbb{P}^{2}\}$.

My idea is that one should have $\sum_{r=1}^{r_{0}(n)}\Lambda(n-r)\Lambda(n+r)\approx r_{0}(n)$. This boils down to $\log(p)\log(q)\approx r_{0}(n)$ with $p=n-r_{0}(n)$ and $q=n+r_{0}(n)$. So we end up solving the equation $\log(n-x)\log(n+x)-x=0$.

It seems this equation as two roots $a\approx -n$ and $b\approx r_{0}(n)$ between which the above expression is positive.

Now computing $J(n)=\int_{a}^{b}(\log(n-t)\log(n+t)-t)dt$ for $n=28$ I get $J\approx 689$. As this number is not that different from $(n-r_{0}(n))(n+r_{0}(n))$ I solved the equation in $y$ $(28-y)(28+y)=689$ whose roots are $\pm\sqrt{95}$.

Dividing $95$ by the known value of $r_{0}(28)=9$ I get a value close to $b$.

So does one have $J(n)\sim (n-\sqrt{br_{0}(n)})(n+\sqrt{br_{0}(n)})$?

Edit: numerically, the quantity $\dfrac{n^2-J(n)}{b}$ for $n=28$ is very close to $\sqrt{r_{0}(n)^{2}+\frac{\pi}{4}}$.

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    $\begingroup$ An upper bound for $r_0(n)$, you are asking us to prove the Goldbach conjecture ??? $\endgroup$ – reuns Aug 13 at 16:47
  • $\begingroup$ I don't understand if you agree that we need the random model for the primes to say anything. $\endgroup$ – reuns Aug 13 at 16:49
  • $\begingroup$ $n-3$ is required at points like $2n=8,10$ $\endgroup$ – Roddy MacPhee Aug 13 at 18:00
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    $\begingroup$ I told you dozen of times, and it is completely obvious, that assuming the Goldbach conjecture is true doesn't help at all for the asymptotic of those things. If your goal is to discuss of unexpected research level results on GC then at least find the relevant papers... Without such papers and results then the answer is the one I give every time : your question doesn't make sense. To find the relevant papers search for "vinogradov theorem". What they don't prove in there is unprovable at our level. $\endgroup$ – reuns Aug 13 at 20:42
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    $\begingroup$ Also do you have some troubles visualizing that the GC could hold with $r_0(n) = n-3$ for at lot of $n$ ? That $\pi(x) \sim x/\ln x$ implies " a lot" will be a set of density zero, the exact form of this density of integers (for which GC is uncertain) is the point of vinogradov's theorem. $\endgroup$ – reuns Aug 13 at 20:48

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