2-split of $n$ is $\left\{ \lfloor \frac{n}{2} \rfloor,\lceil \frac{n}{2} \rceil \right\}$. What about 3, 4, ...?

Question

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.

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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?

Answer 1

It is clear your question is satisfied by the following partition:

$$x = \sum_{k=0}^{n-1}\bigg\lfloor\frac{x+k}{n}\bigg\rfloor$$