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Math Help - formula for recursive definition

  1. #1
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    Post formula for recursive definition

    Hello,

    What is the formula for the following recursive definition...

     f(0)=1, f(1)=0, f(2)=2, f(n)=2f(n-3), n>=3

    The sequence is 2, 0, 4, 4, 0, 8, 8, 0 ...
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  2. #2
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    Quote Originally Posted by robocop_911 View Post
    Hello,

    What is the formula for the following recursive definition...

     f(0)=1, f(1)=0, f(2)=2, f(n)=2f(n-3), n>=3

    The sequence is 2, 0, 4, 4, 0, 8, 8, 0 ...
    Actually its a tricky way of combining three separate sequences.

    Let n = 3k and f(3m) = a_m then:
    f(n)=2f(n-3), n \geq 3 \Rightarrow a_k = f(3k)=2f(3k-3) = 2f(3(k-1)) = 2a_{k-1}, k \geq 1 with a_0 = 1

    Now, for n = 3k+1 and f(3m + 1) = b_m then:
    b_k = f(3k + 1)=2f(3k+1 - 3) = 2f(3(k-1) + 1) = 2b_{k-1}, k \geq 1 with b_0 = 0

    Finally n = 3k+2 and f(3m + 2) = c_m then:
    c_k = f(3k + 2)=2f(3k+2 - 3) = 2f(3(k-1) + 2) = 2c_{k-1}, k \geq 1with c_0 = 2

    So a_k = 2^k , b_k = 0, c_k = 2^{k+1}
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    Actually its a tricky way of combining three separate sequences.

    Let n = 3k and f(3m) = a_m then:
    f(n)=2f(n-3), n \geq 3 \Rightarrow a_k = f(3k)=2f(3k-3) = 2f(3(k-1)) = 2a_{k-1}, k \geq 1 with a_0 = 1

    Now, for n = 3k+1 and f(3m + 1) = b_m then:
    b_k = f(3k + 1)=2f(3k+1 - 3) = 2f(3(k-1) + 1) = 2b_{k-1}, k \geq 1 with b_0 = 0

    Finally n = 3k+2 and f(3m + 2) = c_m then:
    c_k = f(3k + 2)=2f(3k+2 - 3) = 2f(3(k-1) + 2) = 2c_{k-1}, k \geq 1with c_0 = 2

    So a_k = 2^k , b_k = 0, c_k = 2^{k+1}
    Hi, I just need a simple formula what's all this?

    I have to first figure out a sequence from the recursive definition e.g. here it is 1 0 2 2 0 4 4 0 8 8 0... then find a formula which gives such a sequence and then prove it using mathematical induction!
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