# Lowest sum of a sequence

1. Jan 23, 2013

### 000

1. The problem statement, all variables and given/known data

Say we have an infinite sequence of natural numbers A such that no K subsequences can be found adjacent such that the average of the elements in any subsequence is equal for all K subsequences. Sorry about my poor description, an example would be that {2, 3, 4, 1} wouldn't work for K=2 because {2, 3} and {4, 1} are adjacent and both their averages are 5/2. {2, 3, 10, 4, 1} would work however because {2, 3} and {4, 1} are no longer adjacent. Anyway, my question is: which sequence that follows this has the lowest sum for K? If that's too general, then which for K=2? Honestly, any information on the behavior of this sequence would be great.

2. Relevant equations
None that I know of.

3. The attempt at a solution

It looks like for k=2 the lowest A is {1, 2, 1, 3, 1, 2, 1, 4 ...} but I have no idea how to prove that it is.

2. Jan 23, 2013

### haruspex

I don't understand what you mean by an infinite sequence having a lowest sum.

3. Jan 24, 2013

### Staff: Mentor

You could find some value which might stay finite - for example, the sum of the first n elements divided by n, and try to minimize this.

Minimizing every single element before adding a new one, your series continues
1,2,1,3,1,2,1,4,2,1,2,5,2,1,3,1,2,1,3,4,1,2,1,7,2,3,1,2,1,5,1,2,1,8,2,4,2,3,2,1,5,4,3,3,2,5,4,2,5,3,1,2,9,1,7,2

The 8 is required, as a 3 would violate 1,4,2,1,2,5,2,1,3,1,2,1,3,4 next to 1,2,1,7,2,3,1,2,1,5,1,2,1,3 (groups of 14 elements each) and the other numbers would give problems easier to spot.