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Math Help - Covering integers with arithmetic sequences

  1. #1
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    Covering integers with arithmetic sequences

    Are there 11 infinite arithmetric sequences (consist of positive integers) with differences 10, 11, ... ,20 that cover (starting somewhere) all positive integers?

    Thank you very much in advanve!
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  2. #2
    Super Member PaulRS's Avatar
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    The question is somewhat unclear to me as stated.

    Do you mean that you have one arithmetic sequence with difference 10, one with difference 11, and so on?

    If that is what you meant, then the answer is no :

    Suppose it were possible, then working with the generating functions you have: (*)

    x^{a_1} / (1 - x^{10}) + ... + x^{a_{11}} / (1 - x^{20} ) = x^C /(1-x) + extra terms with positive coefficients

    ( C -> the somewhere you mentioned )

    Then multiplying both sides by (1-x) and letting x -> 1^- we get that

    1/10 + 1/11 + ... + 1/20 = 1 + something non-negative ... which is absurd since the LHS is < 1.

    (*) Note here that x^a / (1-x^b) = x^a ( x^{b 0 } + x^{b 1} + ...) = x^{a + b 0} + x^{a + b 1} ...
    which corresponds to the arithmetic sequence starting at 'a' and with difference 'b'.


    Sorry for the disgusting looking post, but latex appears not to be working properly.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by PaulRS View Post
    Sorry for the disgusting looking post, but latex appears not to be working properly.
    See post 8 of this thread.

    -Dan
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  4. #4
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    Yes, that is what I meant. Thank you very much!
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