John,
I'll take the armchair challenge 
It looks as if 2, 6 & 8 aren't referred to at all.
…but it would seem reasonable to assume that the separations between 9 & 1 are the same as between 1 & 5 and 5 & 9. Likewise 1&3, 3&5, 5&7, 7&9, 9&11 and 11&1 and 1&4, 4&7, 7&10 and 10&1 wouldn't it?
This makes me think that the only number missing is 12 and that otherwise there are 12 equal divisions.
If instead of starting at 1 you number them from zero, the first row is 0, 4, 8 (& 12 i.e. 0)
the second is 0, 3, 6, 9 (& 12 i.e. 0)
and the last row is 0, 2, 4, 6, 8, 10 (& 12 i.e. 0)
Edit: I think this is precisely what Brian said in effect. So +1 for that. I was still compiling my post.
HTH
Jon
Edited By Jon Gibbs on 22/02/2017 09:32:31