From,
\(p+q=2n\)
\(p_1+q_1=2n+2\)
\(p+q-(p_1+q_1)=2\)
This is true for all consecutive \(p_i\) and \(q_i\) pair,
\(p_i+q_i-(p_{i+1}+q_{i+1})=2\)
This is a simple predictor for \(p_{i+1}\) and \(q_{i+1}\), given \(p_{i}\) and \(q_{i}\) and a list of primes \(p\lt 2(n+i+1)\).
Alternatively, if we use the formulation,
\(p=2n-2n_o+1\)
\(q=2n-2n_g+1\)
\(\cfrac{p-q}{2}=n_g-n_o\) --- (1)
but
\(p+q=2n=4n+2-2(n_o+n_g)\)
\(n_g+n_o=n+1\) --- (2)
We add (1) and (2),
\(2n_g=\cfrac{p-q}{2}+n+1\)
\(n_g=\cfrac{p-q}{4}+\cfrac{n+1}{2}\)
Similarly,
\(n_o=\cfrac{q-p}{4}+\cfrac{n+1}{2}\)
We can find \(n_g\) and \(n_o\) exactly.
For the next pair of primes,
\(p_1+q_1=2(n+1)\)
from (2),
\(p_1+q_1=2(n+1)=2(n_g+n_o)\)
Graphically,
The constrain seems to be on \(n_g\) and \(n_o\) growing in opposite directions about \(\cfrac{n+1}{2}\). \(n_o\) is confined to about \([1,\cfrac{n+1}{2})\).
When \(n_o=1\)
\(p=2n-2(1)+1=2n-1\)
\(q=1\)
that means the last number introduce to the set of primes available to construct \(p+q\) is \(2n-1\). \(2n-1\) must be a prime.
\(n_g\) is confined to about \((\cfrac{n+1}{2},n]\),
When \(n_o=n\)
\(q=2n-2(n)+1=1\)
\(p=2n-1\)
Equivalently, \(2n-1\) must be a prime.
We can squeeze \(n\) such that \(n_o\) and \(n_g\) must be on the boundary that insist on \(2n-1\) being prime that contradicts with not all odd numbers are prime. That happens for \(n\) small, but those values has been enumerated. Squeezing \(n\) will not disprove the conjecture.
However if \(p\) and \(q\) diverges quickly, where \(\cfrac{p-q}{2}\) forces \(n_g\) and \(n_o\) to their boundary values that insist on \(2n-1\) being prime and since not all odd number (\(2n-1\)) admitted as \(2(n-1)\) moves to \(2n\) is prime, then we have a contradiction.
Does \(p\) and \(q\) diverges quickly? Where,
\(\cfrac{p-q}{4}\longrightarrow\cfrac{n+1}{2}\) as primes\(\longrightarrow\)large and \(n\longrightarrow\infty\)
and the Prime Number Theorem says? The average gap between consecutive prime numbers among the first \(n\) integers is roughly \(log(n)\).
No. PNT does not serve to contradict the conjecture. In the worst scenario, when \(p\) and \(q\) are consecutive with increasing gap,
\(\cfrac{p-q}{2}\) does not squeeze close to \(\cfrac{n+1}{2}\) and insist that the last odd number just admitted into \(2n\) is prime.
PNT does not guarantee the existent of \(n_g\) and \(n_o\) either.
