  For any $m = 3 \left( 2n + 1 \right) with \ n \in \mathbb{N^*} ,$ the prime
counting function $\pi(m) = 4 + \left \vert A_4(m) \right \vert + 2 \left \vert
A_6(m) \right \vert $ where $A_6(m) $ and $ A_4(m) $ are the sets of Twin
Primes and "Isolated" Primes, below $m$, respectively. $T(m) = 1 + \left \vert
A_2(m) \right \vert + \left \vert A_4(m) \right \vert + \left \vert A_6(m)
\right \vert$ is the number of consecutive odd composite numbers (COCONs) below
$ m.$ $ A_2(m)$ is the set of COCONs, below $ m$ and distant by 2. With $m$
odd, $\pi(m)$, $ T(m)$ and $ \left \vert A_2(m) \right \vert = \frac{m - 9}{2}
- 3 \left \vert A_6(m) \right \vert - 2 \left \vert A_4(m) \right \vert $ lead
to $ 4 \left \vert A_6(m) \right \vert + 7 = m - 2 \left( T(m) + \left \vert
A_4(m) \right \vert \right). \ Hence \ 0 < 1 - \frac{ 2 }{m} \left( T(m) +
\left \vert A_4(m) \right \vert \right) < 1.$ Thus the non-empty unique set $S
\equiv \left\lbrace 1 - \frac{ 2 }{l} \left( T(l) + \left \vert A_4(l) \right
\vert \right) \ \mbox{such that} \ l = 3 \left( 2k + 1 \right) with \ k \in
\mathbb{N^*} \right\rbrace \subset \mathbb{R} $ is bounded. Therefore $ Inf
\left( S \right) $ exists, is unique and finite. By definition, $ Inf(S) > 0.$
$\mathbb{Q}$ dense in $\mathbb{R}$ also guarantees the latter. S and $ Inf(S)$
are independent of $m.$ We then introduce $\alpha = Inf \left( S \right):$
$\alpha$ is independent of $m.$ We then have $ 0 < \alpha \leq 1 - \frac{ 2
}{m} \left( T(m) + \left \vert A_4(m) \right \vert \right)$ for any m as above.
Therefore, $ 4 \left \vert A_6(m) \right \vert + 7 \geq \alpha m,$ with $\alpha
> 0$ and independent of $m.$ Hence $ \displaystyle \lim_{ m \longrightarrow +
\infty } \left \vert 2 A_6(m) \right \vert = + \infty.$ Similarly, $
\displaystyle \lim_{ m \longrightarrow + \infty } \left \vert A_4(m) \right
\vert = + \infty .$
