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Lowest sum of a sequence

  1. Jan 23, 2013 #1

    000

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    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. jcsd
  3. Jan 23, 2013 #2

    haruspex

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    I don't understand what you mean by an infinite sequence having a lowest sum.
     
  4. Jan 24, 2013 #3

    mfb

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    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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