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

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The discussion centers on finding the lowest sum for the infinite sequence of natural numbers A under the condition that no two adjacent subsequences of length K=2 can have the same average. An example sequence proposed is {1, 2, 1, 3, 1, 2, 1, 4, ...}, which aims to minimize the sum while adhering to the criteria. The sequence must be constructed carefully to avoid violating the average condition, particularly with the placement of specific numbers like 3 and 8. The conversation highlights the challenge of proving the sequence's minimality and the complexities involved in maintaining the required properties. Overall, the exploration seeks to understand the behavior of such sequences and their sums.
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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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