From,
p+q=2n
p1+q1=2n+2
p+q−(p1+q1)=2
This is true for all consecutive pi and qi pair,
pi+qi−(pi+1+qi+1)=2
This is a simple predictor for pi+1 and qi+1, given pi and qi and a list of primes p<2(n+i+1).
Alternatively, if we use the formulation,
p=2n−2no+1
q=2n−2ng+1
p−q2=ng−no --- (1)
but
p+q=2n=4n+2−2(no+ng)
ng+no=n+1 --- (2)
We add (1) and (2),
2ng=p−q2+n+1
ng=p−q4+n+12
Similarly,
no=q−p4+n+12
We can find ng and no exactly.
For the next pair of primes,
p1+q1=2(n+1)
from (2),
p1+q1=2(n+1)=2(ng+no)
Graphically,
The constrain seems to be on ng and no growing in opposite directions about n+12. no is confined to about [1,n+12).
When no=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.
ng is confined to about (n+12,n],
When no=n
q=2n−2(n)+1=1
p=2n−1
Equivalently, 2n−1 must be a prime.
We can squeeze n such that no and ng 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 p−q2 forces ng and no 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,
p−q4⟶n+12 as primes⟶large and n⟶∞
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,
p−q2 does not squeeze close to n+12 and insist that the last odd number just admitted into 2n is prime.
PNT does not guarantee the existent of ng and no either.