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 $\left\{ a_1,\cdots,a_k \right\}\quad \text{s.t.}$

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?

could you please define what a "2-split" is, and more generally what an "n-split" is? It it supposed to be a sequence of $n$ consecutive integers which add to $n$? — Brevan Ellefsen, Commented May 1, 2020 at 3:11
@BrevanEllefsen Sorry. I proposed it intuitively. In general I think $k$-split of $n$ is $\left\{ a_1,\cdots,a_k \right\}\quad \text{s.t.}\quad$ (1) $a_1\le\cdots\le a_k$ (2) $a_1+\cdots+a_k=n$ (3) $\sum_{i<j}{a_j-a_i}$ is minimized. — SnzFor16Min, Commented May 1, 2020 at 3:17

Brevan Ellefsen · Accepted Answer · 2020-05-19 23:21:00Z

2

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

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

answered May 1, 2020 at 3:37

Brevan Ellefsen

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$\begingroup$ The partition formula is amazing. How should I understand it properly, or if not, prove it? $\endgroup$
– SnzFor16Min
Commented May 1, 2020 at 10:51
$\begingroup$ @SneezeFor16Min You can think of it as splitting up the possible cases mod $n$ $\endgroup$
– Brevan Ellefsen
Commented May 9, 2020 at 2:10
$\begingroup$ It's not that obvious to me that possible cases $\mod n$ make up $x$... $\endgroup$
– SnzFor16Min
Commented May 19, 2020 at 17:22
$\begingroup$ There're some mistakes in the derivations, e.g., $3$-split should be $\big\lfloor \frac{x}{3}\big\rfloor +\big\lfloor \frac{x+1}{3}\big\rfloor +\big\lceil \frac{x}{3}\big\rceil$; $4$-split should be $\big\lfloor \frac{x}{4}\big\rfloor +\big\lfloor \frac{x+1}{4}\big\rfloor +\big\lceil \frac{x-1}{4}\big\rceil +\big\lceil \frac{x}{4}\big\rceil$; $5$-split should be $\big\lfloor \frac{x}{5}\big\rfloor +\big\lfloor \frac{x+1}{5}\big\rfloor +\big\lfloor \frac{x+2}{5}\big\rfloor +\big\lceil \frac{x-1}{5}\big\rceil +\big\lceil \frac{x}{5}\big\rceil$; $\cdots$ $\endgroup$
– SnzFor16Min
Commented May 19, 2020 at 17:28
$\begingroup$ @SneezeFor16Min you are right, I had some mistakes. Not sure what happened... I must have had an error in the code I wrote to check them. I removed them accordingly, though the initial partition should be fine. $\endgroup$
– Brevan Ellefsen
Commented May 19, 2020 at 23:23

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