The weak point in the proof of Goldbach's Conjecture is the insistent that all skipped indexes are reachable by selecting the next prime number (backward or forward) and reach out with known primes.
Skipped index, in \(S_{n}\) is the direct consequence of gap between consecutive primes. Such gaps are small compared to the primes numbers that mark the gap. We know from Bertrand–Chebyshev Theorem that there is a prime between \(n\) and \(2n\) (inclusive). So the widest prime gap in this interval is \(n\), in which case the last prime is \(2n\). On average the gap is \(\cfrac{n}{2}\).
Given a prime \(n\), the next prime is at most \(n\) away at \(2n\), from Bertrand–Chebyshev Theorem. But since the average prime gap over \(n\) ratio, tends towards \(\cfrac{1}{log(n)}\)(and is asymptotically zero) by the Prime Number Theorem. The gap is smaller than the last prime \(n\), in the worst case scenario. Please refer to the post "Squeeze Goldbach" dated 24 Dec 2022, also.
The last prime (at \(2n\)) after a widest gap of at most \(n\) remains reachable from the last but one prime \(p=n\).
ie., with \(p=n\) and \(q=n\)
\(p+q=2n\)
which is \(S_{2n}\). In cases when the last prime is \(p\lt 2n\) we will use this prime, and all of any one of the previous prime to reach beyond \(2(n+p)\) and \(2n\).
\(q=3,5,...q_i\t p\)
