Elementary counting formula for the number of $2k$-separated prime pair averages in an interval.

Let all intervals be of integers. Fix a natural number $k \geq 1$. Primorial $p_n\#$ is defined as usual to be $p_1 p_2 \cdots p_n$. Let an undeclared $p$ or $q$ just mean any prime number.

Lemma 1. The number of $x \in [0,y]$ such that $x^2 = k^2 \pmod {p_i}$ for no $i = 1..n$ is given by:
$$f(y) = \sum_{d \mid p_n\#, \ \\ d \leq y } (-1)^{\omega(d)}\sum_{c \mid d, \ \\ \gcd(c,2k) = 1} \lfloor \dfrac{y - (z_{c,d})_{(d)}}{d}\rfloor \tag{0}$$

Where $(z_{c,d})_{(d)}$ is the unique least non-negative residue in $\Bbb{Z}$ modulo $d$ such that $z_{c,d} = k \pmod c, \ z_{c,d} = -k \pmod {\frac{d}{c}}$. Then condition $(z_{c,d})_{(d)} \lt d \leq y$ ensures that there exists at least one such $x = (z_{c,d})_{(d)}$ in the interval $[0, y]$. Otherwise this term needs to equal zero as it counts the number of such solutions in the interval.

Proof. Define $A_i = \{ x \in [0, y] : x^2 = k^2 \pmod {p_i}\}$ where $p_i$ is the $i$th prime number.
Then the expression we’re after is $y - |A_1 \cup A_2 \cup \dots \cup A_n|$. We can compute this via inclusion-exclusion to be:

$$y +\sum_{1 \lt d \mid p_n\#, \ \\ d \leq y} (-1)^{\omega(d)}|\bigcap_{p \mid d} A_{\pi(p)}| \tag{1}$$

The size of each intersection term counts the number $x \in [0, y]$ such that $x^2 = k^2 \pmod {p}$ for all prime $p \mid d$. But modulo each $p$, we have that $x^2 = k^2 \pmod p \iff x = k$ or $x = -k \pmod p$. So what we can do is split up $d$ into $c, \dfrac{d}{c}$ and sum up the count for each special case $x = k \pmod c$ (equivalently modulo each $p \mid c$) and $x = -k \pmod {\frac{d}{c}}$ (equivalently modulo each $p \mid \frac{d}{c}$).

Except if $\gcd(c, 2k) \neq 1$ in which case we have that there exist a prime $p \mid c$ such that $p \mid (k - (-k)) = 2k$. Which means the two possibilities $x = k$ or $x = -k$ modulo $p$ are equivalent. What happens is a double counting of $|\bigcap_{p \mid d} A_{\pi(p)}|$ in the case that $\gcd(c, 2k) \neq 1$. Thus we only consider when $x = -k \pmod 2$ if we remove that case from the final summation.

Now focusing on a single special case $x = k \pmod c, x = -k \pmod d, \gcd(c, 2k) = 1$. Let $(z_{c,d})_{(d)} = x$. We know that we want to count all solutions to:
$$0 \leq dz +(z_{c,d})_{(d)} \leq y \tag{2}$$

So simply rearrange this:

$$\iff 0 - (z_{c,d})_{(d)} \leq dz \leq y - (z_{c,d})_{(d)} \\ \ \\ \iff \dfrac{-(z_{c,d})_{(d)}}{d} \leq z \leq \dfrac{y - (z_{c,d})_{(d)}}{d}, \text{ but } z \in \Bbb{Z}, \\ \iff \lceil \dfrac{-(z_{c,d})_{(d)}}{d} \rceil \leq z \leq \lfloor\dfrac{y - (z_{c,d})_{(d)}}{d} \rfloor$$

The expression on the left is just zero, by definition of $(\cdot)_{(d)}$ taking things to the least non-negative residue $\pmod d$ in $\Bbb{Z}$. Thus for this special case, the count is:

$$\lfloor \dfrac{ y - (z_{c,d})_{(d)}}{d} \rfloor \tag{3}$$

The reason we can (and may have to, in summation bound) take $d \leq y$ is that if $d \gt y \geq x$, then $d z + (z_{c,d})_{(d)} \leq y \lt d \implies dz \leq y - (z_{c,d})_{(d)} \lt d$ so that $z \leq \dfrac{y - (z_{c,d})_{(d)}}{d} \lt \dfrac{d}{d} = 1$, which means $z = 0$

All together and referring back to formula (2) we have that we must sum up these special case counts over all possible choices of $c \mid d, \gcd(c, 2k) = 1$:

$$f(y) = y + \sum_{1 \lt d \mid p_n\#, \ \\ d \leq y } (-1)^{\omega(d)}\sum_{c \mid d, \\ \gcd(c,2k) = 1} \lfloor \dfrac{y - (z_{c,d})_{(d)}}{d}\rfloor$$

Now notice that we rewrite it in our goal form (0) by noticing that $z_{1, 1} = 0$ and so we can allow in $d = 1$ and when you evaluate at $d = 1$ the inner sum, you get $y$ itself. $\blacksquare$

Corollary 1. With $y = p_{n+1}^2 - 2, n \in \Bbb{N}$. The $2k$-separated prime pair conjecture is proven if $f(y)$ is not eventually $0$. For example if $f(y)$ is always growing as $n \to \infty$.

Proof. If $x \in [0, y]$ is such that $x^2 \neq k^2$ modulo any $p_i, i = 1..n$ (that’s what $f(y)$ counts !), then by Sieve-of-Eratosthenes, we have that $x^2 -k^2$ is a product of primes or that $x$ is a $2k$-separated prime average. So $f(y)$ essentially counts all $2k$-separated prime pair averages in the interval $[0, y]$ whenever $y \leq p_{n+1}^2 - 2$. That is kind of a strict maximum because if you let in $y = p_{n+1}^2 - 1$ then $y + 1 = p_{n+1}^2$ is also coprime to all $p_i, i=1..n$ and yet it’s not a prime number, and so it would get falsely counted. $\blacksquare$

Corollary 2. The formula is true for any $y \leq p_{n+1}^2 -2$, that is $f(y)$ counts the number of $2k$-separated prime pair averages in that interval, as long as of course $y \gt p_n + 1$.

Proof. It could turn out that $p_n$ is a twin prime first and so $p_n + 1$ is a twin prime average, but it would not be “picked up by the count” because $p_n$ is part of the set $p_i, i=1..n$ which has canceled out that average. The first statement follows from the discussion in the proof of Corollary 1. $\blacksquare$