A kind reader (which also has preferred to remain anonymous) has rightly pointed out that the case of three starters less than it first with it is even easier to prove when n is odd.
If n is odd, one of the first three children it is just the number 2 = p.
Then Q = 2n - and R q = 2n - r p can not have in their composition (given by contradiction that none of them are first, see previous post).
In particular,
Q p can not be made, nor by q (q otherwise would be to also be a factor n) nor can it be a power of r (as we recall r> n / 2). It follows the absurd.
In other words, if 2 is a prime not belonging to the factorization of n, then Q = 2n - q (where q is the second first outside the factorization of n) must be necessarily first.
And so the case of 3 primes less than n is completely closed!
