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For any m = 3 ( 2n + 1 ) with \ n ∈ N^* , the prime counting function π(m) = 4 + A₄(m) + 2 A₆(m) where A₆(m) and A₄(m) are the sets of Twin Primes and "Isolated" Primes, below m, respectively. T(m) = 1 + A₂(m) + A₄(m) + A₆(m) is the number of consecutive odd composite numbers (COCONs) below $ m.$ A₂(m) is the set of COCONs, below m and distant by 2. With m odd, $π(m)$, $ T(m)$ and A₂(m) = m - 9/2 - 3 A₆(m) - 2 A₄(m) lead to 4 A₆(m) + 7 = m - 2 ( T(m) + A₄(m) ). \ Hence \ 0 < 1 - 2 /m ( T(m) + A₄(m) ) < 1. Thus the non-empty unique set $S ≡ 1 - 2 /l ( T(l) + A_4(l) ) \ such that \ l = 3 ( 2k + 1 ) with \ k ∈ N^* ⊂ R $ is bounded. Therefore $ Inf ( S ) $ exists, is unique and finite. By definition, $ Inf(S) > 0.$ $Q$ dense in $R$ also guarantees the latter. S and $ Inf(S)$ are independent of $m.$ We then introduce $α= Inf ( S ):$ $α$ is independent of $m.$ We then have $ 0 < α≤ 1 - 2 /m ( T(m) + A_4(m) )$ for any m as above. Therefore, $ 4 A_6(m) + 7 ≥ αm,$ with $α> 0$ and independent of $m.$ Hence $ limm + ∞ 2 A_6(m) = + ∞.$ Similarly, $ limm + ∞ A_4(m) = + ∞ .$
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Okouma et al. (2022) studied this question.