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Thread: Combinatorial Proof

  1. #1
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    Combinatorial Proof

    Thanks in advance.
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  2. #2
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    From the way you put the question, I assume that part (a) is no problem.
    Any string in $\displaystyle a_n$ is a string in $\displaystyle a_{n-1}$ with one of 1,2, or 3 added. To avoid repetition we can use only two of three. Therefore $\displaystyle a_n = 2 a_n $, because $\displaystyle a_1 = 3$ the given result easily follows.

    EDIT:
    Clearly any string in $\displaystyle b_n$ with the last number removed is a string in $\displaystyle a_{n-1}$. Therefore, taking any string in $\displaystyle a_{n-1}$ that is not in $\displaystyle b_{n-1}$ and adding the same number as it begins with gives a string in $\displaystyle b_n$. Thus $\displaystyle b_n=a_{n-1}-b_{n-1}$.
    .
    Last edited by Plato; Mar 16th 2008 at 04:20 AM.
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  3. #3
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    Plato, how would I go about proving part a?
    The basis would be when n=2, so their should be 6 A-sequences. How would I show that this is true? just list them?

    Then what would I do for the inductive step?

    Thanks!
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