This question is a follow-up to [About Goldbach's conjecture](https://mathoverflow.net/questions/61842/about-goldbachs-conjecture?r=SearchResults) and as such deals with the notion of primality radius of a composite integer $n$, that is, a positive integer $r$ such that both $n-r$ and $n+r$ are prime.


So, considering the quantity $ord\_{C}(n):=\pi(\sqrt{2n-3})$ and the associated $ord\_{C}(n)$-tuple $C(n)$ made of the finite sequence of residues of $n$ modulo the first $ord\_{C}(n)$ primes, let for such primes $p$ the quantity $a\_{p}(n):=1$ if $p\mid n$ and
$a\_{p}(n)=2$ otherwise.


Let $N\_{2}^{eq}(n):=n\prod\_{p\leq\sqrt{2n-3}}\left(\frac{\pi(n)}{n}\right)^{\frac{a\_{p}(n)}{ord\_{C}(n)}}$. This quantity would equal $\pi(n)$ if we could see primes less than $n$ as primality radii of $0$ (in that case all the $a\_{p}(n)$ would equal $1$), which of course is unrigorous as primes are greater than $0$. The general idea is that the number $N\_{2}(n)$ of primality radii of $n$ should be greater than $\frac{n}{\log^{2}n}$ and less than $\pi(n)\sim\frac{n}{\log n}$, the exponent of the log being a barycenter of $1$ and $2$ whose weight coefficients should be the $a\_{p}(n)$, where $\log n$ is just a rough approximation of the supposedly "exact" quantity $\frac{n}{\pi(n)}$ and that is inspired by Mertens' theorem.


My question is: can one prove that $N\_{2}(n)=N\_{2}^{eq}(n)+O(1)$? If yes can the involved constant be made explicit?



Is this Goldbach conjecture related quantity equal to the number of Goldbach decompositions up to a bounded quantity?
This identity is false. To see that, notice first that
$$
\sum\_{p\leq \sqrt{2n-3}}a\_p(n)=2\sum\_{p\leq \sqrt{2n-3}}1-\sum\_{\substack{p\leq \sqrt{2n-3}\\ p\mid n}}1=2\mathrm{ord}\_C(n)-O(\ln n),
$$
so
$$
N\_2^{eq}(n)=n\prod\_{p\leq \sqrt{2n-3}}\left(\frac{\pi(n)}{n}\right)^{a\_p(n)/\mathrm{ord}\_C(n)}=n\left(\frac{1+o(1)}{\ln n}\right)^{2+O(\ln^2 n/\sqrt n)}=\frac{n}{\ln^2 n}(1+o(1)).
$$
On the other hand, for composite $n$, $N\_2(n)$ is $\frac{1}{2}\#\{p+q=2n\}$. The asymptotic formula for the last quantity is known to be true for almost all $n$:
$$
\#\{p+q=2n\}=2C\_2g(n)\frac{n}{\ln^2 n}(1+o(1))
$$
for almost all $n$, where
$$
g(n)=\prod\_{\substack{p\mid n\\ p>2}}\frac{p-1}{p-2}
$$
and
$$
C\_2=2\prod\_{p>2}\left(1-\frac{1}{(p-1)^2}\right).
$$
This result follows from Theorem 19.1, H. Iwaniec, E. Kowalski, "Analytic number theory" (p. 444).


If we had $N\_2^{eq}(n)=N\_2(n)+O(1)$ even for almost all $n$, the equality
$$
g(n)=\frac{1}{C\_2}+o(1)
$$
would also hold for almost all $n$. But if $n$ is not divisible by $3$ (for example), then
$$
g(3n)=2g(n),
$$
so if you take the numbers $3m+1,3m+4,\ldots,6m-2$ ($m$ of them below $6m$) and $3(3m+1),\ldots,3(6m-2)$ ($m$ below $18m$) for at least one half of them the equality
$$
g(n)=\frac{1}{C\_2}+o(1)
$$
cannot be true, so you get at least $m$ counterexamples below $18m$.


