Why is $n= \left\lfloor \frac{2p^2+2pq+2pr+q^2+2qr+r^2}{p+q+r} \right\rfloor$, where $n$ is $\text{prime}$, of form such that $p\pm a,p\pm b,$ are $ \text{prime}, 1<a<b<n$?
Consider this:
Let $p,q,r$ be $\mathbb{prime}$ where $r<q<p$.
Let $x\pm y$ denote $(x+y\quad\mathbb{and}\quad x-y)$
I conjecture that if the following conditions are true; $$\mathbb{isprime}( n),\quad n = \left\lfloor \frac{2 p^2+2 p q+2 p r+q^2+2 q r+r^2}{p+q+r} \right\rfloor$$ Then $n$ is a prime of such form that $n \pm a$, and $n \pm b$ are prime numbers where $a$ and $b$ are distinct positive integers with $a < b < n$.$\quad$ (A137669(n) where $n>2$)
However, I have no clue as to why this is the case.
Example:
$x = 29, y = 17,z = 11, a = 12, b = 18$
$$71 = \left\lfloor \frac{2 \cdot 29^2+2 \cdot 29 \cdot 17+2 \cdot 29 \cdot 11+17^2+2 \cdot 17 \cdot 11+11^2}{29+17+11} \right\rfloor \,\text{is prime}$$
$$71\pm 12 = \{83,59\} = \{71+12,71-12\},\,\text{are both prime}$$
$$71\pm 18 = \{89,53\} = \{71+18,71-18\},\,\text{are both prime}$$
Series representation:
$$ \sum_{n = -\infty}^{\infty}=\left\lfloor\left ( \begin{cases} 1, & n = 1 \\ y+z, & n = 0 \\ {y+z\over(-y-z)^n}, & n\geq2 \end{cases} \right ) x^n \right\rfloor = \left\lfloor \frac{2 p^2+2 p q+2 p r+q^2+2 q r+r^2}{p+q+r} \right\rfloor $$
If anyone can explain why this is the case, it would be very much appreciated.
