Define. $G_n =\{ x \in \Bbb{Z} : x^2 = 1 \mod n\}$ for any $n \geq 1$. Clearly, moduloing everything by $n$, it is a subgroup of $\Bbb{Z}/n^{\times}$ and has size $2^{\omega(d) - [2\mid n]}$, where $[2\mid n] = \begin{cases}1 \text{ if } 2 \mid n \\ 0 \text{ else }. \end{cases}$, whenever $n$ is square-free.
Clearly, $a \in [p_m + 2, \dots, x]$ is a twin prime average if and only if $a \notin G_n$ for any $n \mid p_m\#$, where $m = \pi(\sqrt{x + 1})$. By the sieve of Eratosthenes.
We can count this explicity using some inequality-solving and inclusion-exclusion principle:
$$ \rho(x) = -1 + \sum_{n \mid p_m\#} \mu(n) \sum_{r^2 = 1 \mod n } \left\lfloor \frac{x - r}{n}\right\rfloor $$
The $-1$ can be brought back in via, Dirac-delta:
$$ \rho(x) = \sum_{n \mid p_m\#} \mu(n) \sum_{r^2 = 1 \mod n} \left( \lfloor\dots \rfloor - \delta_1(n)\right) $$
In other words. If the twin prime conjecture is false, the number of twin primes in $[p_m + 2, x]$ must eventually vanish as $m = \pi(\sqrt{x + 1})$ and $x \to \infty$.
To state this in usual mathy terms we say:
$$ T = \sum_{n \mid p_m\#}\mu(n)\sum_{r^2 = 1\mod n} \bullet $$
to be a transformation from $T: \{\Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z}\} \to \{\Bbb{R}\to \Bbb{Z}\}$.
And so if we define $K = \lim \ker T := \{ f : \Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z} : \lim_{x\to \infty} T(f)(x) = 0 \}$, where it is understood that it's the set of all $f$ such that the limit exists and is $0$.
Then we can say that $f(n,r,x) = \left\lfloor \frac{x - r}{n}\right\rfloor - \delta_1(n) \in K$ the limit kernel of $T$.
But, I came up with this more general identity that is not found in the floor-ceiling article on Wikipedia.
It's closely related to one under "Quotient" (Properties).
Lemma 1.
$$ \sum_{i = 0}^{kn - 1} \left\lfloor\frac{x - i}{n} \right\rfloor = k(x - \frac{(k + 1)}{2} n + 1) $$
But this is totally different from what they have on Wikipedia, and also I think their formula might be off, from experimenting in SymPy. If ound that it's not true that this sum of floors equals just $x$.. Anyway, I've code-checked the above formula.
Question 1.
How can we prove the above formula?
Motivation.
Well, if the limit kernel behaves like $\Bbb{Z}$-submodule of the domain function space, then there's no reason why we can't add in other related floor-functions such as subtracting $i$ and summing up, all the way to a multiple of the denominator, or in our case $p_m\# - 1$. That means in the formula $k = \frac{p_m\#}{n}$.
However, it's not that simple since we don't yet know what happens at negative $x$.
Question 2. What can we prove the bounds to be on $\rho(-x)$ for $x \gt 0$?
From code experiments we have:
$$ -(\rho(x) + 2) \leq \rho(-x) \leq \rho(x) $$
but can we prove this and how?
Alternative
Question 3. Is Question 2 even necessary since we're taking a limit as $x \to \infty$??? I think it still matters because the sum over $x-i$ will always delve into going past $-x$ and so on...
Further Derivation & Question 4.
$$ x \in \Bbb{Z}, x \geq 4, n = \pi(\sqrt{x + 1}) \implies x \in [p_n + 2, x] \text{ is a twin prime average } \\ \iff x^2 \neq 1 \mod m, \forall m \mid p_n \#. $$
Therefore, to discount the sets $X_{m} = \{ x \in I_x := [0, x] : x^2 = 1 \mod m\}$ for $m\mid p_n\#$, we can first solve:
$$ 0 \leq z = mz' + r\leq x $$ for any given $r \in X_m \mod m$, and then add the individual counts.
Wwe have $\frac{-r}{m} \leq z' \leq \frac{x-r}{m}$ which means a total possible $z'$ count of:
$$ \left\lfloor\frac{x - r}{m} \right\rfloor-\left\lceil\frac{-r}{m} \right\rceil +1 $$
Then to count:
$$ |[0,x]\setminus \bigcup_{m \mid p_n\#}(m\Bbb{Z} +r)| $$
what we're after (twin prime average count), then it's inclusion-exclusion:
$$ F(x) = -1 +\sum_{m \mid \sqrt{x + 1}\#}\mu(m)\sum_{r^2 =1 \mod m}\left\lfloor\frac{x-r}{m} \right\rfloor $$
Define $f(m,r,x) := \left\lfloor\frac{x - r}{m} \right\rfloor - \delta_1(m)$.
We can say that if the twin prime conjecture is false, then $f$ sits nicely in $f \in K = \lim \ker T = \{ f : \Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z} : \lim_{x \to \infty} T(f)(x) = 0 \}$ where $T$ is the apparent double-summation operator w.r.t. $x$.
Now, by nature of limits, for each fixed $i \in \Bbb{N}$ we may not have:
$$ \lim_{x \to \infty} T(f)(x + i) = 0 $$
as this augments the interval range and allows in twin pseudoprime averages or averages that are not provably anything because they are outside of the Eratosthenes bound $x$. On the other hand, certainly as $x \to \infty$ we have that $x - i \in [p_n + 2, x]$ so that since there are no twin prime averages in the larger interval, there can't possibly be any in the smaller, and the same thing happens for sufficiently large $x$, as $x \to \infty$.
Therefore, we have:
$$ 0 = \sum_{i = 0}^{\sqrt{x + 1}\#-1}\lim_{x\to\infty} T(f)(x-i) = \lim_{x \to \infty}\sum_{i = 0}^{\sqrt{x + 1}\# - 1}\sum_{m \mid \sqrt{x + 1}\#}\mu(m)\sum_{r^2 = 1 \mod m} f(m,r,x-i) \\ $$
Question 4. Is this a valid rearrangement: I commuted a sum of $0$-valued limits with the sum which is bounded in the limiting variable?
Then there is a floor identity:
$$ \sum_{i = 0}^{km - 1} \left\lfloor\frac{x - i}{m} \right\rfloor = k(x - \frac{(k + 1)}{2} m + 1) $$
So since at each $x \in \Bbb{R}$ we have that the limit expression is a finite sum of sums, we can rearrange those sums and move the outer $\sum_{i}$ to be innermost:
$$ g(m,r,x) := \sum_{i = 0}^{\sqrt{x + 1}\# - 1}\left(\left\lfloor\frac{x - r - i}{m} \right\rfloor- \delta_1(m)\right) \in K = \lim \ker T $$
by closure of a $\Bbb{Z}$-submodule (the limit kernel) of $\{\Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z}\}$.
But by the formula:
$$ g(m,r,x) = \frac{\sqrt{x + 1}\#}{m}\left(x - \frac{\frac{\sqrt{x + 1}\#}{m} + 1}{2}m + 1 \right) - \sqrt{ x + 1}\# \delta_1(m) \\ = \frac{\sqrt{x + 1}\#}{m}(x -\frac{\sqrt{x + 1}\#}{2} -\frac{m}{2} +1) - \sqrt{x + 1}\#\delta_1(m) \\ \ \\ \ = \frac{x\sqrt{x + 1}\#}{m} - \frac{(\sqrt{x + 1}\#)^2}{2m} -\frac{\sqrt{x + 1}\#}{2} +\frac{ \sqrt{x + 1}\#}{m} -\sqrt{x + 1}\#\delta_1(m) $$
Under the $T$-transformation, constants such as $\sqrt{x + 1}\#/2$ (i.e. constant with respect to the summation) will vanish, so we have:
$$T(g)(x) =-\sqrt{x+1}\# + \left(\frac{(x + 1)\sqrt{x + 1}\#}{1} - \frac{(\sqrt{ x + 1}\#)^2}{2}\right) \sum_{m \mid \sqrt{x + 1}\#}\frac{\mu(m)}{m}\sum_{r^2 = 1 \mod m}1$$
This is clearly a Merten's factor (which is always a positive number - easy to prove), times a much faster negatively growing number. That means $\lim_{x \to\infty}(g)(x) \neq 0$ which is a contradiction.
