There is a somewhat forgotten sieve-theoretic approach to the Goldbach conjecture, due to Buchstab et al, see e.g. pp.247-248 of R.D. James.


On p.247, James defines some function $F$ such that for any fixed $a \in \mathbb{N}$ and even $x \geq 6$:


1. $F(x ; 2, a, 1) = F(x; 2)$ with $a=1$ is the number of positive integers $n \leq x$ such that $n \equiv a\pmod{2}$. Thus $F(x; 2) = x/2$.
2. $F(x; 2, √x, a)=F(x; 2, √x)$ with $a=1$ is the number of odd positive integers $n<x$ (without double counting $n$ and $x-n$), such that each of $n$ and $x-n$ is either a prime or equal to 1. Thus if it could be shown that $F(x; 2, √x) \geq 2$, it would follow that there exists at least one representation $x= n+(x-n)$ whereby each of $n$ and $x-n$ is either a prime or equal to 1. Thus if $x-1$ is composite, it would suffice to show that $F(x; 2, √x) = F(x; 2, √x) \geq 1$.


On the bottom of p.248, James states that
$$
F(x; 2, √x) = F(x; 2) - 2\sum\_{r=1}^{k} F(x; 2p\_r, p\_{r-1}) = x/2 - 2\sum\_{r=1}^{k} F(x; 2p\_{r}, p\_{r-1})
,$$
where $p\_i$ denotes the $i$-th odd prime $\leq √x$. T. Kubalalika, in his preprint [2], lets $6 \leq x \equiv 2\pmod{4}$ where $x-1$ is composite. Now suppose that $x$ is a counterexample to Goldbach, so that $F(x; 2, √x)=0$. Putting this into the above equality gives
$$
x/2 = 2\sum\_{r=1}^{k} F(x; 2p\_r, p\_{r-1}),
$$
contradicting the fact that $x/2$ is odd. One therefore deduces that if $x\equiv 2\pmod{4}$ and $x-1$ is composite, then $x$ is a sum of two primes.


My question is, given the strength of Buchstab et al's sieve (as evidenced by how easily it leads to a proof of the above result), are there any modern improvements to it, such that it could possibly lead to even more powerful results ? A quick Google search seems to suggest that the sieve became forgotten as soon as the Hardy-Littlewood circle method lead to Vinogradov's 3-primes theorem.


**References**


[1] R. D. James, "[Recent progress in the Goldbach problem](https://www.ams.org/journals/bull/1949-55-03/S0002-9904-1949-09180-2/S0002-9904-1949-09180-2.pdf)" Bulletin of the American Mathematical Society 55, 246-260 (1949), [MR0028893](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0028893), [Zbl 0034.02301](https://zbmath.org/?q=an%3A0034.02301).


[2] T. Kubalalika, "[On the binary Goldbach conjecture for certain even integers](https://figshare.com/articles/preprint/On_the_binary_Goldbach_conjecture/21342042)", figSHARE preprint.



On Buchstab et al's "forgotten" sieve and the Goldbach conjecture for certain integers
The result (aka Buchstab's identity) you mentioned is not forgotten. In modern sieve theory texts such as Halberstam & Richert's *Sieve Methods* and Friedlander & Iwaniec's *Opera de Cribro*, the identity is written as


$$
S(\mathcal A,z)=S(\mathcal A,w)-\sum\_{w\le p<z}S(\mathcal A\_p,p)
$$


where $S(\mathcal A,z)$ counts the integers in $\mathcal A$ that are free of prime divisors $<z$.


One of its generalization (Kuhn's weighted sieve) is used to prove Chen's theorem. When $N$ is a positive even integer, $\mathcal A=\{N-p:p\le N\}$, we have


$$
\begin{aligned}
r\_{1,2}(N)&=\#\{p\le N:n-p\text{ prime or product of two primes}\} \\
&>S(\mathcal A,N^{1/10})-
\frac12\sum\_{N^{1/10}\le p<N^{1/3}}S(\mathcal A\_p,N^{1/10})-\frac\Omega2+O(N^{9/10})
\end{aligned}\tag1
$$


in which


$$
\Omega=\#\{p\le N:N-p=p\_1p\_2p\_3,N^{1/10}\le p\_1<N^{1/3}\le p\_2<(N/p\_1)^{1/2}\}.
$$


By evaluating the right hand side using Jurkat-Richert's theorem and Selberg's sieve, Chen found out that for large $N$ there is


$$
r\_{1,2}(N)>{0.67N\over\log^2N}\prod\_{2<p|N}{p-1\over p-2}\prod\_{p>2}\left(1-{1\over(p-1)^2}\right).
$$


