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Discrete: Recurrence relation for sum of integer using only 2's and 3's.

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Find a recurrence relation for Tn, the number of ways to write an integer n as the sum of terms, each of which is 2 or 3, and the order matters. [So 2+3 and 3+2 are different sums for 5.]


    2. Relevant equations

    (if I had one, this would be easier)

    3. The attempt at a solution
    So I tried at first to just map out some of the smaller n's:
    T0=0
    T1=0
    T2=1
    T3=1
    T4=2 (2+2) & (2'+2)
    T5=2 (as above)

    A nice pattern seemed to emerge for a moment, but then it fell apart:
    T6 had (2+2'+2'') and it's permutations (3!=6) and (3+3') and it's 2 permutations. So T6=8.
    But T7 only has (2+2'+3) and it's 6 permutations, so T7=6.

    I'm thinking that f(n) is probably dependent on f(n/3) or f(n/2), or both. I don't think it's going to be a simple f(n) based on f(n-1)... but I could be wrong.

    It seems like when finding recurrence relations you just have to have an aha! moment - and I'm not having one on this. Any help would be awesome!

    FYI: I'm using 2' and 2'' just to keep track of the permutations, since order matters, just to help me remember that (2+2')≠(2'+2).

    EDIT: I misunderstood the problem. The permutations of 2's and 3's, like in the above sums of 5, have order matter. But for 2+2+2...+2, order does not matter. So I think I actually have it now:

    T0=0
    T1=0
    T2=1
    T3=1
    T4=1 (2+2)
    T5=2 (as above)
    etc...

    So for n>3, f(n)=f(n-2)+f(n-3).
     
    Last edited: Apr 23, 2012
  2. jcsd
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