What is the lowest sum for K=2 in this infinite sequence with specific criteria?

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SUMMARY

The discussion focuses on determining the lowest sum of an infinite sequence of natural numbers A for K=2, where no two adjacent subsequences can have the same average. The proposed sequence {1, 2, 1, 3, 1, 2, 1, 4, ...} is suggested as a candidate for achieving this condition. The sequence is constructed by minimizing elements while ensuring that adjacent subsequences do not yield equal averages. The challenge lies in proving that this sequence indeed represents the lowest sum for K=2.

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Homework Statement



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.


Homework Equations


None that I know of.


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

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