Let $p_i$ be the $i^{th}$ prime number. For Goldbach sums of $2n$, let $p_i$ be the largest prime less than $\sqrt{2n}$, define
$$
P(p_i,n,x)=\sum_{p\le{p_i}}\frac{c_p}{p}\left(1+2\sum_{k=1}^{p-1}(1-\frac{k}{p})\cos\frac{2k\pi n}{p}\cos\frac{2k\pi x}{p}\right),
c_p =
\begin{cases}
1, \text{ when $p \mid 2n$} \\
2, \text{ when $p \nmid 2n$}
\end{cases}
$$
which zeros show
$$
\begin{cases}
\text{When $n=0$: prime sieve (primes when $x<p_i^2$)} \\
\text{When $n=1$: twin prime sieve as $(x-1,x+1)$ (twin primes when $x<p_i^2-1$)} \\
\text{When $n>1$ and $0\le{x}<n$: sieve of Goldbach sums as $(n-x, n+x)$ (prime sum when $x<\frac{p_i^2}{2}$)} \\
\text{When $n\ge{1}$ and $x>n$: sieve of prime pairs of distance of $2n$ as $(x-n, x+n)$}
\end{cases}
$$
And there's an unified formula to calculate number of zeros $L$ on its period $p_i\#$. For all $3\le{p}\le{p_i}$,
\begin{equation}
L=\prod_{p|n}(p-1)\prod_{p \nmid n}(p-2)
\end{equation}
For prime function, $n=0$ so $L=\prod(p-1)$; For twin prime function, $n=1$ so $L=\prod(p-2)$.
Here show some examples for $p_4=7$.
$P(7,0,x)$ zeros show prime sieve bigger than $7$, with $1 \cdot 2 \cdot 4 \cdot 6 = 48$ zeros on period $210$.
$P(7,1,x)$ zeros show twin prime sieve bigger than $7$ as $(x-1,x+1):(11,13),(17,19),(29,31),(41,43),(59,61),(71,73),(101,103),(107,109)$
$P(7,50,x-50)$ zeros show Goldbach sums of $100$ as $(n-x,n+x)$. Graphic is shifted $n=50$ to show prime sum of $2n=100:(11+89),(17+83),(29+71),(41+59),(47+53)$
Under the hood, this function sieves $\{x: x \neq \pm 1 \text{( mod $p$)}, x \in \mathbb{Z}, p \le p_i\}$ for twin primes, and sieves $\{x: x \neq \pm n \text{( mod $p$)}, x \in \mathbb{Z}, p \le p_i\}$ for Goldbach sums of $2n$, instead of directly sieve prime numbers.
Question: Is this an existing sieve method? Can you direct me to some articles that study or apply such sieve method?