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I would like to find informations regarding this way of doing Integer Partitions or this conjecture,

Suppose you have all the ordered partitions of 5:
5
4 1
3 2
2 2 1
3 1 1
2 1 1 1
1 1 1 1 1

Then extend those partitions to 5 digits by adding 0:
5 0 0 0 0
4 1 0 0 0
3 2 0 0 0
2 2 1 0 0
3 1 1 0 0
2 1 1 1 0
1 1 1 1 1

And finaly for each extended partitions, count the digit repetitions, like this:
5 0 0 0 0 -> 1 4
4 1 0 0 0 -> 1 1 3
3 2 0 0 0 -> 1 1 3
2 2 1 0 0 -> 2 1 2
3 1 1 0 0 -> 1 2 2
2 1 1 1 0 -> 1 3 1
1 1 1 1 1 -> 5

And you can repeat !
5 0 0 0 0 -> 1 4 -> 1 4 0 0 0 -> 1 1 3 -> 1 1 3 0 0 -> 2 1 2 -> 2 1 2 0 0 -> 1 1 1 2 -> ..
4 1 0 0 0 -> 1 1 3 -> 1 1 3 0 0 -> 2 1 2 -> 2 1 2 0 0 -> 1 1 1 2 -> 1 1 1 2 0 -> 3 1 1
3 2 0 0 0 -> 1 1 3 -> 1 1 3 0 0 -> 2 1 2 -> 2 1 2 0 0 -> 1 1 1 2 -> 1 1 1 2 0 -> 3 1 1
2 2 1 0 0 -> 2 1 2 -> 2 1 2 0 0 -> 1 1 1 2 -> 1 1 1 2 0 -> 3 1 1 -> 3 1 1 0 0 -> 1 2 2
3 1 1 0 0 -> 1 2 2 -> 1 2 2 0 0 -> 1 2 2 -> 1 2 2 0 0
2 1 1 1 0 -> 1 3 1 -> 1 3 1 0 0 -> 1 1 1 2 -> .... -> 1 2 2
1 1 1 1 1 -> 5 -> 5 0 0 0 0 -> 1 4 -> 1 4 0 0 0 -> 1 1 3 -> ... -> 1 2 2

All of them finish in 1 2 2!

I want informations, documentations

Thanks a lot!

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What's your conjecture? That they sum to five can hardly be surprising. –  M.B. Aug 9 '12 at 10:16
    
What's the question here? Couldn't see it... –  DonAntonio Aug 9 '12 at 10:19
    
Following my add, I would like to know why all of them finish with the partion 1 2 2? Is there documentations, articles?? –  Yvan Aug 9 '12 at 11:08
    
For all the partitions of 6, they end with partition 1 1 1 3 !!! –  Yvan Aug 9 '12 at 11:45

1 Answer 1

up vote 2 down vote accepted

It’s not clear that there’s a general phenomenon to be explained.

Note that $n=2,3$, and $4$ behave somewhat differently: none of them ends up in a single one-element loop irrespective of the starting point. For $n=2$ we have the loop

$$2,0\to1,1\to2,0\;,$$

a single two-element loop that does not depend on the starting point. For $n=3$ each starting point leads to the loop

$$3,0,0\to1,2,0\to1,1,1\to3,0,0\;.$$

For $n=4$ all starting points save $2,2,0,0$ lead to the loop

$$4,0,0,0\to 1,3,0,0\to 1,1,2,0\to2,1,1,0\to1,2,1,0\to1,1,1,1\to4,0,0,0\;,$$

while $2,2,0,0$ is a fixed point (a one-element loop).

That’s already several different behaviors in the first few values of $n$. The only thing that’s clear is that for each $n$, each starting point must eventually fall into a cycle, since there are only finitely many $n$-tuples that sum to $n$.

share|improve this answer
    
Thank you Brian! For 4 the perfect symetry of 2, 2, 0, 0 (one-element loop) seems to be the only one possible for all possible n. Did I am wrong? –  Yvan Aug 9 '12 at 12:23
    
@Yvan: I’m not sure what you mean: there’s $1\to 1$ for $n=1$, and there’s $1,2,2,0,0\to1,2,2,0,0$ for $n=5$. –  Brian M. Scott Aug 9 '12 at 12:41
    
Brian: I mean : –  Yvan Aug 9 '12 at 13:02
    
Brian : For an even number n; is there 2 sets of same size, one containing (n/2) * 0, and the other one (n/2) * 2, such that the enumeration of the n digits, leads to 2 digits having value (n/2) forming a one-element loop .E.g. 2 2 2 2 0 0 0 0 -> 4 4 0 0 0 0 0 0 -> 2 6 -> not possible. –  Yvan Aug 9 '12 at 13:37
    
To generalise: Is there a number n divisible in m sets of equal size, such that each set contains (n/m) times the same digit and where the digits take value from 0 to m - 1. The enumeration of the n digits, leads to m digits having value (n/m). Can we form a one element loop as 2 2 0 0. If there is more loops, the loop is symetric (centered)? –  Yvan Aug 9 '12 at 13:38

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