A103314 conjecture

[wouter meeussen]

a naive interpretation of the n=p^a q^b r^c case.
Or, how to generate "cancellation by missing parts".

Take prime p. Drop up to Floor[p/2] unit vectors of its 'star'.
For each dropped, imagine a shadow negative (blue?) vector to be neutralised by a complementary
'blue' element of a star of either q or r multiplicity. How many ways?

Note that n must be even in order to realise the 'blackness' of the complementary blue star.
This means that, in my opinion, the least prime factor for the 'difficult' cases must be 2. No
special cases for 3*5*7 for instance.

I trust my instincts on this, but do you?
I've errred before.

W.

>The difficulty is with numbers having more than two prime factors
>because of the possibility of interaction. The smallest such n is
>30=2*3*5. a 5-star and 3-star can be combined to make an "off" 6-star
>(Wouter's example). The "base case" of square free numbers needs
>to be worked out.