For an $n$-tuple $S$ of decreasing positive integers, we can define $f(S)$ as subtracting $1$ from every element of $S$, prepending $n$, and then removing $0$s and re-ordering in decreasing order if neccecary. For example, $f((4,2,1))=(3,3,1)$.
I've proven that repeated iteration of this function always ends at a cycle, by noticing that $f$, $f(f(x))$, etc. is a finite sequence. Now how can I prove that every tuple with the same sum will eventually go to the same cycle? I've verified that this is true with Mathematica.
EDIT: I'm looking to either prove or disprove this statement. I've verified it for a lot of cases, but it may not be true. This wasn't given as an assignment so I have no idea whether it's true.
Let $S$ be your initial $n$-tuple of decreasing integers. If $N := \sum_{x\in S} x$, note that $S$ describes a partition of $N$ (any partition of a positive integer can be written uniquely as a weakly decreasing sequence of numbers). Furthermore, $f$ is defined such that $f(S)$ is also a partition of $N$. Since there are only finitely many partitions of a given positive integer, the sequence $\{f^{k}(S)\}$ must eventually repeat.
Edit: However, the claim that there is only one cycle for each $N$ is false. For example, note that when $N = 17$, the partitions $(6,5,3,2,1)$ and $(6,4,4,2,1)$ are contained in disjoint cycles under the action of $f$.