Fix $n \in \Bbb{N}$.
Define $C_2(x) = 2x, \ C_3(x) = 3x$, $C_5(x) = 2 [\frac{x + 2}{3}] + [\frac{ x + 1}{3}] + 2 [ \frac{x}{3}]$, more generally for prime $p_n \geq q \geq 5$ define $C_q(x) = 2[\frac{ x + q - 3}{q - 2}] + [\frac{x + q - 2}{q - 2}] + \dots + [\frac{ x + 1}{q-2}] + 2[\frac{x}{q-2}]$.
If the least non-negative $x \in \Bbb{Z}$ such that $x = C_2(x_2) = C_3(x_3) = \dots =C_q(x_q)$ for some $x_i \in \Bbb{Z}$; is less than $p_{n+1}^2 - 1$, then $x$ is a twin prime average.
So the goal might be to show that for each $n \geq 1$ such a small enough $x \in \Bbb{Z}$ exists as a common image point of all the $C_i$.
Have you seen this form of the twin prime conjecture before?
The opposite of the twin prime conjecture would be that for none but finitely many $n \geq 1$ can you find such a solution.
According to Wikipedia (Quotients with Floor) we actually have an even more simpler (algebraic) form, because for $q \geq 5$ we have that each
$$ C_q(x) = x + [\frac{ x + q - 3}{q - 2}] + [\frac{x}{q - 2}] $$
The majority of the summation collapses down to $x$ according to the Wikipedia article.
Question.
Can we write down a formula for the simultaneous images of the $C_q$? Note that this of course allows you to temporarily ignore the $\lt p_{n+1}^2 - 1$ bound.
Thanks.
Let's first assume we're at step (in some hypothetical proof) $n = 3$ so that $q = 5$ and we want:
$$ 6x = C_5(y) = y + [\frac{y + q - 3}{q-2}] + [\frac{y}{q-2}] $$
well after some thought, let's break this up into cases modulo $y \pmod{q-2}$:
$$ C_5(3y) = 5y \\ C_5(3y + 1) = 5y + 2 \\ C_5(3y + 2) = 5y + 3 \\ $$
I'll do the last one:
$$ C_5(3y + 2) = 3y + 2 + [\frac{3y + 2 + q - 3}{3}] + [\frac{3y + 2}{3}] = \\ 3y + 2 + y + 1 + y = 5y + 3 \ \blacksquare $$
So working the case $y' = 3y + i$ for $0 \leq i \lt 3$ we have that:
$$ 6 x = C_5(y') \iff 6x -5y = i + [i\gt 0] = c $$
where $[i \gt 0] = \begin{cases} 1 \text{ if } i \gt 0 \\ 0 \text{ otherwise}\end{cases}$.
The general solutions to the linear diophantine equation for a given $i$ are: $$ x = c +5k,\ y=c+6k $$
Thus the twin prime average solutions would be $6x = 6(c + 5k) \lt p_{4}^2 - 1$:
Or whenever:
$$ \frac{\frac{p_4^2 - 1}{6} - c}{5} \gt 1 $$