What I want to ask is about the structure of the Goldbach function that defined by $$ R(x)=\#\{ p \mid x-p \in \mathbb{P} , \ p\leq x/2\}$$ for $x\in 2\mathbb{N}$, where $\mathbb{P}$ is the set of prime numbers. As I understand, the methods which are developed to prove the Goldbach type problems have difficulties in the case of the binary Goldbach problem and it seems that there have to be more essential principle to prove the problem that $R(x)$ follows (though it is still unsolved that $R(x)>0$) $$ c\frac{x}{\log^2 x}\prod_{p|x, \ \ p>2} \frac{p-1}{p-2} $$ with a positive constant $c$.

If I make an explanation for my question, we would consider a correspondence between the unary problems and binary problems (or a natural extension of the unary problem under given specific condition), for example, the prime counting function $\pi(x)$ can be expressed as $$ -\sum_{p\in \mathbb{P}}\mu(p), $$ and $R(x)$ can be expressed as $$ \sum_{p,q\in \mathbb{P} \atop p+q=x}\mu(p)\mu(q).$$ Except the difference of signs in the point of view that the importance is to estimate these moduluses, we would say that there is a correspondence under the condition $a+b=x$ which seems as one of the most simple conditions we can add. And then a question rises, whilst there is a logical equivalence between the estimates of $\pi(x)$ and $\sum_{n\leq x}\mu(n)$, are there any known relation between $R(x)$ and the sum $$\sum_{ns+mt=x \text{ for some } s,t\in\mathbb{2N}-1}\mu(n)\mu(m)? $$ (where $n,m\in 2\mathbb{N}-1$ such that $(x,n)=1=(x,m)$) On the function above, the odd numbers $s,t$ be put as a condition via some deductions and because it seems more meaningful than $a+b=x$ at least as I thought.