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Math Help - recurrence involving floors and square roots

  1. #1
    schumi316
    Guest

    recurrence involving floors and square roots

    Hey,

    Would appreciate any help on this one.

    Solve the recurrence:

    a(0) = 1;

    a(n) = a(n-1) + floor(sqrt[a(n-1)]) ; for n >0.

    This is exercise number 3.28 from Concrete mathematics. The solution at the back kinda makes a bit of sense....but i still cant grasp entirely how they've reached the solution.

    Thanks
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  2. #2
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    Quote Originally Posted by schumi316 View Post
    Hey,

    Would appreciate any help on this one.

    Solve the recurrence:

    a(0) = 1;

    a(n) = a(n-1) + floor(sqrt[a(n-1)]) ; for n >0.

    This is exercise number 3.28 from Concrete mathematics. The solution at the back kinda makes a bit of sense....but i still cant grasp entirely how they've reached the solution.

    Thanks
    In general the formula is,
    a_n=(1+2+...+n)+([\sqrt(1)]+[sqrt(2)]+...+[\sqrt(n)])
    With some work I think this can be simplified.
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