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A few days ago I wrote the following post Two new conjectures related to Lemoine's and Goldbach's in which conjecture #2 was:

For every odd positive integer O there exists two distinct odd primes (from primes and 1) p0 and p1 such that p1 is the average of O and p0. I verified that the conjecture seems valid with both added constraints O < p0 < p1 or O > p1 > p0. Let's call the first case conjecture 2a and the second case conjecture 2B.

Which leads me to the following derived conjectures about triplets of odd primes...

Assume triplets [p0,p1,p2] where p0 < p1 < p2 and p1 = (p0 + p2)/2 (p1 is the average of p0 and p2) with: p0 odd prime or 1 , p1 and p2 odd primes

We have:

1) For any p0 (1 or odd prime) we can find a p1 and a p2 satisfying the above conditions. (This follows from my conjecture 2A).

2) For any p1 (odd prime) we can find a p0 and a p2 satisfying the above conditions. (This follows from Goldbach's conjecture!)

3) For any p2 (odd prime) we can find a p0 and a p2 satisfying the above conditions. (This follows from conjecture 2B).

These conjectures are of course weaker than the conjectures they derive from but I find the set of 3 pretty cool. (The link with the Goldbach conjecture is also interesting.)

Feedback? Have you seen this anywhere before?

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  • $\begingroup$ Not sure I am following. If $n$ is your odd number (or, indeed, any natural number) then, assuming Goldbach, $2n=p+q$ is the sum of two primes. Thus $n=\frac {p+q}2$ is the average of those two primes. Are you saying something different from that? $\endgroup$ – lulu Sep 20 '17 at 16:21
  • $\begingroup$ I assume you refer to point 2). Yes number 2 is just a special case of the Goldbach conjecture. It is included in my trio because it nicely completes the set of conjectures on (p0,p1,p2) all odd primes. $\endgroup$ – Raymond Duchesne Sep 20 '17 at 16:52
  • $\begingroup$ Well, as for the first point, it is conjectured that there is a prime progression of length $p$ starting with each odd prime $p$...thus $\{3,5,7\}$ or $\{5,11,17,23,29\}$. This is, of course, much stronger than your statement $\endgroup$ – lulu Sep 20 '17 at 16:58
  • $\begingroup$ I did not know about that conjecture... Nice & Thanks! I guess the merits left to my trio are: 1) Point 3 is still possibly original 2) All together they mean that if you build all the possible triplets of odd primes with the expressed conditions, you will find all odd primes including 1 in the head of the triplet, all odd primes in the middle of the triplet and all odd primes greater than 3 in the tail. Still a nice looking symmetry. $\endgroup$ – Raymond Duchesne Sep 20 '17 at 17:24
  • $\begingroup$ I should add that we have the same type of symmetry in this case: Assume triplets [p0,p1,p2] where p0 < p1 < p2 and p1 = (p0 + p2)/2 (p1 is the average of p0 and p2) with: p0 odd prime or 1 , p1 and p2 odd primes $\endgroup$ – Raymond Duchesne Sep 21 '17 at 13:52

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