Mathematics Journal, Proof for

by TimNguyen
Tags: journal, mathematics, proof
 P: 81 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...
 P: 81 Any algebraist in here that could help?
 P: 81 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...?
HW Helper
P: 1,422

Mathematics Journal, Proof for

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...
Viet Dao,
P: 81
 Quote by VietDao29 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... Viet Dao,
You're forgetting:
(3,2), (2,3), (2,2,1), (2,1,2), (1,2,2),

which adds to 15.

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