We are given a partition of a positive integer $x$. Each step, we make a new partition of $x$ by decreasing each term in the partition by 1, removing all 0 terms, and adding a new term equal to the current partition's size. For example, if our current partition is $(1, 1, 2, 3)$, our next will be $(1, 2, 4)$.
Prove that is $x$ is a triangular number, then the sequence of partitions will eventually converge to $(1, 2, ..., n)$.
My progress: I drew out some state diagrams for small $x$ but can't find any patterns. It's easy to prove that $(1, 2, ..., n)$ is the only possible fixed point, so it boils down to proving that there are no other possible cycles for $x=\frac{n(n+1)}{2}$. Looking at the specific case where we start with $(n)$, we essentially make the prefix $(1, 2, ..., k)$ on the $\frac{k(k+1)}{2}$th, and since each prefix sequence essentially repeats the previous but with one extra step, it should be fairly simple to prove via induction that we'll reach $(1, ..., n)$. So one idea I have is proving the partition will eventually have each prefix $(1)$, $(1, 2)$, ..., $(1, 2, ..., n)$, which possibly can be done with some kind of induction. However I have no idea how one would go about doing this.
Bonus/follow-up (I'm not sure if this is solvable in a nice way): For non-triangular $x$, prove (or provide a counterexample) that starting from any partition of $x$, the sequence will end up in the same cycle. If this is true, characterize the cycle that the sequence will end up in.