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Math Help - sequence, not easy but not hard

  1. #1
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    sequence, not easy but not hard

    A prime number p is given. A sequence of positive integers a_{1}, a_{2}, a_{3},... is determined by this conditon:
    a_{n+1}=a_{n}+p \lfloor \sqrt[p]{a_{n}} \rfloor
    Prove that there is a term in this sequence, which is a p-power of an integer number.
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  2. #2
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    Anybody can help me ?? Is this so hard?
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  3. #3
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    a_1 is a 1-power of an integer .

    or have I misunderstood?
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  4. #4
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    1 is not prime number. It is harder question.

    Please help me with this question, I cannot do it anyway..... Please
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  5. #5
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    Quote Originally Posted by Ununuquantium View Post
    A prime number p is given. A sequence of positive integers a_{1}, a_{2}, a_{3},... is determined by this conditon:
    a_{n+1}=a_{n}+p \lfloor \sqrt[p]{a_{n}} \rfloor
    Prove that there is a term in this sequence, which is a p-power of an integer number.
    What do you mean by "p-power?" For example, choose p = 2 (to keep things simple.) Then given an a_1 = 2 (for instance) we get
    a_2 = a_1 + 2\sqrt{a_1} = 2 + 2\sqrt{2}

    a_3 = a_2 + 2\sqrt{a_2} = 2 + \sqrt{2} + 2\sqrt{2 + \sqrt{2}}

    etc.

    Are you saying that for some n of this sequence that a_n = k^p where k is some integer?

    -Dan
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  6. #6
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    Yes exactly!!
    p power of an integer means
    a_{n}=k^{p} k integer, p prime
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