2-split of $n$ is $\left\{ \lfloor \frac{n}{2} \rfloor,\lceil \frac{n}{2} \rceil \right\}$. What about 3, 4, ...? Clarification: $k$-split of $n$ is an ordered integer sequence $\left\{ a_1,\cdots,a_k \right\}\quad \text{s.t.}$


*

*$0\le a_1\le\cdots\le a_k$

*$a_1+\cdots+a_k=n$

*${\left(a_k-a_1\right)}$ is minimized.



I know that
$$
n = \lfloor \frac{n}{3} \rfloor + \lceil \frac{2n}{3} \rceil,
$$
so I guess 3-split of $n$ is
$$
n = \lfloor \frac{n}{3} \rfloor + \lfloor \frac{\lceil \frac{2n}{3} \rceil}{2} \rfloor+\lceil \frac{\lceil \frac{2n}{3} \rceil}{2} \rceil ?
$$
If so, can this be simplified?

It is clear your question is satisfied by the following partition:
$$x = \sum_{k=0}^{n-1}\bigg\lfloor\frac{x+k}{n}\bigg\rfloor$$
