On Buchstab et al's "forgotten" sieve and the Goldbach conjecture for certain integers

Question

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" Bulletin of the American Mathematical Society 55, 246-260 (1949), MR0028893, Zbl 0034.02301.

[2] T. Kubalalika, "On the binary Goldbach conjecture for certain even integers", figSHARE preprint.

Answer 1

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

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