Find most stable combination of a vector with itself at different periods
Made up rule nr 1: Every number has to be part of the sum. Therefore the sum cannot be x, if there is a number b, such that no combination of numbers can add up to x-b.

So for [0, 0, 2, 3, 2]: 


the sum cannot be 2 (because 3 cannot be part of that)
it cannot be 3 (because 3 - 2 = 1 and there is no combination of numbers that add
up to 1 (rule 1))  
it cannot be 4 (because 4-3 = 1 again, (rule 1))
it cannot be 5 (because there are twice as many 2-s as 3-s)
it cannot be 6 (because in case of 3-s, only two rows in exactly the
same place can make up that number, and the 2-s would then form 4-s,
which cannot sum up to 6)


Therefore for the specific case of [0, 0, 2, 3, 2], the simplest solution is to have the value of 7

Perhaps this complicated reasoning helps to consider the future vectors as well :)

[0, 0, 2, 0, 3, 0, 4, 0, 2] is more difficult though. While not considering too thoroughly, it looks like for that as well, there isn't anything better than the trivial solution.

Trivial solution:

[a, b, c ....]
   [a, b, c ....]
      [a, b, c ....]
      ...