From the previous post "More Prime Spiral Goldbach" dated 24 Dec 2022,
\(n_{gi+1}+n_{oi+1}=n_{g}+n_{o}+i+1\) \(i\) is odd, 1, 3, 5...
predicts the next pair \(n_{gi+1}\) and \(n_{oi+1}\) from an odd position \(i\) odd.
\(n_{gi+1}+n_{oi+1}=(2n+(i+1)-2)-n_{g}-n_{o}\) \(i\) is even, 2, 4, 6
predicts the next pair \(n_{gi+1}\) and \(n_{oi+1}\) from a even position \(i\) even.
Since \(n_{gi+1}\) and \(n_{oi+1}\) is not guaranteed to ensure,
\(2(n+i+1)-2n_{oi+1}+1=p_{i+1}\) and \(2(n+i+1)-2n_{gi+1}+1=q_{i+1}\)
where both \(p_{i+1}\) and \(q_{i+1}\) are prime and,
\(p_{i+1} +q_{i+1}=2(n+i+1)\)
Goldbach's Conjecture not proven here.
But take any pair of primes, \(p\) and \(q\) and calculate,
\(n=\cfrac{p+q}{2}\)
\(n_o=\cfrac{2n-p+1}{2}\) and
\(n_p=\cfrac{2n-q+1}{2}\)
starting with the odd position prediction find candidates for \(n_{g1}\) and \(n_{o1}\) under the constrain,
\(p_{1} +q_{1}=2(n+1)\)
then use the even position prediction to find candidates for \(n_{g2}\) and \(n_{o2}\) under the constrain,
\(p_{2} +q_{2}=2(n+2)\)
repeat till no possible candidates for \(n_{gi}\) and \(n_{oi}\) can be found. The series ends.
Does Goldbach has long hairs? Spirally long hairs?
