# Any odd > 1 is the average of three primes

I think that any odd integer is the average of three primes. My first question is if this is equivalent to some other conjecture/theorem in number theory (I suspect it is).

But more importantly, I want some insight about the patterns that appear while verifying for a few values. I verified it by hand for enough values for it to interest me (up to 100 or so) but what was interesting was the appearance of cycles within the process. The three primes are not unique, but I found myself following this pattern for quick verification.

For nice example, if we take a prime like 53, then the three primes that it is an average of is just itself 3 times.

53 = avg(53,53,53)

The average for the next odd number, 55, is the result of replacing one 53 with the next highest prime.

55 = avg(53,53,59)

And then 57 comes from doing that again.

57 = avg(53,59,59)

And then obviously, the next odd is 59 itself, so the pattern ends with all 59s.

59 = avg(59,59,59)

61, the next odd, is also a prime, so the pattern ends here. However, sometimes the pattern will continue past the next prime and end at the next prime after that.

These little patterns are not always occurring, but for it not to occur is the oddity. I tried to notice something about where it did not occur but couldn't say anything meaningful about these numbers.

Thank you for any insight.

• See Goldbach's conjecture – Oria Gruber Apr 5 '14 at 0:30

If $n$ is odd, then $3n$ is odd, so writing $n$ as average of three primes is the same as writing the odd number $3n$ as the sum of three primes. This is possible (a theorem of Vinagradov for large $n$, and recently tightened to all $n$ by Helfgott). (See here.)
If $n$ is even, say $n = 2m$, then $3n -2 = 2 (3m -1)$ is even, and Goldbach's conjecture says that we may write $2 (3m -1) = p + q$ as a sum of two primes, so $n = (2 + p + q)/3.$