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Mathematics Journal, Proof for

  1. Sep 26, 2005 #1
    Compositions into Relatively Prime Parts

    Hello.

    I was reading a journal and an interesting problem came up. I believe the journal was in the American Mathematics Society publications. Well, here's the statement.

    "For all integers, n greater than or equal to 3, the number of compositions of n into relatively prime parts is a multiple of 3."

    Example : For 4: the compositions of relatively prime parts are:

    (1,3), (3,1), (2,1,1), (1,2,1), (1,1,2), (1,1,1,1).

    This is what I have so far for a "proof":

    Let n be an integer greater than or equal to 3.
    Then the first composition will be given by (n-1, 1), (1, n-1); since for all k, an integer, (k, 1) and (1, k) are always relatively prime.
    Also, (1, 1, ..., 1) where the composition adds to n is also an obtainable composition.

    (So basically, I've gotten the end points of the compositions to be a multiple of 3, then I need to prove that the "in-between" compositions will also be a multiple of 3.)

    Well, obviously I'm stuck there. I've tried to split it into two cases afterwards where the cases involve n - odd and n - even but it has come to no avail. Also I've tried to find a formula where the compositions of relatively prime parts is a multiple of 3 but it fails at "6". Here was the formula I came up with that failed, if it could be potentially be improved upon.

    Formula: Starting at n=1, where i=3, i being the starting point.

    (i)!/2^n

    Like:
    For 3, 3! = 6 divided by 2^1 = 2 will equal 3 compositions- a multiple of 3
    For 4, 4! = 24 divided by 2^2 = 4 will equal 6 compositions - a multiple of 3
    For 5, 5! = 120 divided by 2^3 = 8 will equal 15 compositions - multiple of 3

    Well, hopefully people will post their ideas...
     
    Last edited: Sep 27, 2005
  2. jcsd
  3. Sep 29, 2005 #2
    Any algebraist in here that could help?
     
  4. Oct 8, 2005 #3
    Well, I've still been trying to prove this for the past week. I think there may be some recursion sequence or something. Well, I've tried listing the compositions and here's the results:

    n=3 : 3 compositions
    n=4 : 6 compositions
    n=5 : 15 compositions
    n=6 : 27 compositions
    n=7 : 63 compositions
    n=8 : 129 compositions

    I'm predicting that n=9 : 387 compositions
    n=10 : 687 compositions

    Anyone feel like lending a hand...?
     
    Last edited: Oct 8, 2005
  5. Oct 8, 2005 #4

    VietDao29

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    Homework Helper

    Maybe I don't really get what you mean... But for n = 5, I get 10 compositions, which means it fails for n = 5.
    n = 5:
    (1, 4), (4, 1)
    (1, 1, 3), (1, 3, 1), (3, 1, 1)
    (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), (2, 1, 1, 1)
    (1, 1, 1, 1, 1)
    That's 2 + 3 + 4 + 1 = 10... :confused:
    Viet Dao,
     
  6. Oct 9, 2005 #5
    You're forgetting:
    (3,2), (2,3), (2,2,1), (2,1,2), (1,2,2),

    which adds to 15.
    :tongue2:
     
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