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Thread: Sequence

  1. #1

    Thread Starter
    New Member
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    May 2007
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    2

    Sequence

    For every positive integer n, let an denote the number of ways n is obtained as a sum of terms that are all 1, 3 or 4. The order of the terms also matters. Prove that a2006a2007a2008 is a perfect cube.

  2. #2
    Frenzied Member
    Join Date
    Jun 2006
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    1,098

    Re: Sequence

    While I am unable to prove this, I have found the following to be true:

    a2006 and a2008 are squares of consecutive Fibonacci numbers
    a2007 is the product of those numbers

    As such, a2006a2008 = a20072
    and a2006a2007a2008 = a20073

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