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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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    For all the partitions of 6, they end with partition 1 1 1 3 !!!2012-08-09

1 Answers 1

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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 $n$.

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    Brian: If you want to take a look http://math.stackexchange.com/questions/180704/shifted-young-tableaux-hook-numbers-bulgarian-solitaire2012-08-09