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Hello,

This question is a follow-up from About Goldbach's conjecture.

As $N_{2}(n)=\sum_{r\leq n}1_{\mathbb{P}}(n-r)1_{\mathbb{P}}(n+r)$, Chebotarev's theorem allows to write:

$$\dfrac{N_{2}(n)}{\pi(n)}\sim \dfrac{N_{1}(n)}{\varphi(P_{ord_C(n)})}$$

So that

$$N_2(n)\sim \pi(n) \dfrac{N_{1}(n)}{\prod_{2\lt p\le \sqrt{2n-3}}(p-1)}$$.

One can easily show that $$N_{1}(n)={\prod_{2\lt p\le \sqrt{2n-3}}(p-2)}\prod_{p\vert n \\ p>2}\dfrac{p-1}{p-2}$$.

Thus one gets

$$N_{2}(n)\sim \pi(n)\dfrac{C}{\log\sqrt{2n-3}}\prod_{p\vert n \\ p>2}\dfrac{p-1}{p-2}$$ and, through the prime number theorem:

$$N_{2}(n)\sim \dfrac{Kn}{\log^{2} n}\prod_{p\vert n \\ p>2}\dfrac{p-1}{p-2}$$ with $K>0$.

As $N_{2}(n)$ is the number $G(2n)$ of couples $(p,q)$ such that $p+q=2n$, $p\leq q$, $p$ and $q$ primes, this shows that

$$G(2n)\sim \dfrac{Kn}{\log^{2} n}\prod_{p\vert n \\ p>2}\dfrac{p-1}{p-2}$$.

I have been told on a French maths forum that Hardy and Littlewood rigorously proved that if such a $K$ exists, then its value is such that their conjecture sometimes knowns as "extended Goldbach's conjecture" is true. Would you have a reference where this proof is given?
Thanks in advance.

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This is from Section 4 of Hardy and Littlewood's "On some problems of partitio numerorum III: On the expression of a number as a sum of primes" 1923 Acta Math. 44: 1–70.

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  • $\begingroup$ Thank you but I can't access. $\endgroup$ Aug 4, 2012 at 21:56
  • $\begingroup$ Is there no library that would do an interlibrary loan for you? $\endgroup$ Aug 4, 2012 at 22:57
  • $\begingroup$ Finally, I could get the whole article. Thank you again. $\endgroup$ Aug 5, 2012 at 11:58
  • $\begingroup$ I removed the link given in the answer, as it (inadvertently I suppose) was the link through some proxy of some specific institution, which as such is useful for hardly anybody and might rather cause confusion. $\endgroup$
    – user9072
    Aug 5, 2012 at 12:38
  • $\begingroup$ That was indeed inadvertent. I replaced the link with a generic link to the Springerlink access point (paywalled).This, however, should work on many institutions networks. $\endgroup$
    – Mark Lewko
    Aug 5, 2012 at 17:20

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