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Saturday, December 24, 2022

Silly Me, Goldbach

 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=2n2no+1

q=2n2ng+1

pq2=ngno --- (1)

but

p+q=2n=4n+22(no+ng)

ng+no=n+1   --- (2)

We add (1) and (2),

2ng=pq2+n+1

ng=pq4+n+12

Similarly,

no=qp4+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=2n2(1)+1=2n1

q=1

that means the last number introduce to the set of primes available to construct p+q is 2n12n1 must be a prime.

ng is confined to about (n+12,n]

When no=n

q=2n2(n)+1=1

p=2n1

Equivalently,  2n1 must be a prime.

We can squeeze n such that no and ng must be on the boundary that insist on 2n1 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 pq2 forces ng and no to their boundary values that insist on 2n1 being prime and since not all odd number (2n1) admitted as 2(n1) moves to 2n is prime, then we have a contradiction.

Does p and q diverges quickly?  Where,

pq4n+12 as primeslarge 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,

pq2 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.