Q:

Combinatorics: Distributing $n$ bananas to $k$ people

If I have $n$ bananas that are all looking the same and distribute them to $k$ people, how many bananas does the person with the most bananas have at least, if no two persons can have the same amount of bananas?
The $k$ people can also have no bananas $n$.
Lets make a case analysis:

n > k: If there are $8$ bananas and $3$ persons, we distribute like that: $1,2,5$ -> max = $5$. Or if there are $13$ bananas and $4$ persons, we distribute like that: $1,3,4,5$ or $1,3,4$ -> max = $5$ or $4$. If there are $10$ bananas and $2$ persons, then we distribute: $4,6$ -> max = $6$. I can't find a formula that satisfies altogether.
n < k: If there are $2$ bananas and $5$ persons, then the person with most bananas would have $2$, meaning $n$
n = k: If there are $5$ bananas and $5$ persons, we distribute like that: $2,3$ (because no two persons can have the same amount of bananas). So the person with most bananas has $\lfloor\frac{n}{2}+1\rfloor $ 

So the premise is $$n \geq \frac{k\cdot(k-1)}{2}$$

A:

Let's call the minimum number of bananas needed for $k$ people as $$n_0 = \frac{k\cdot(k-1)}{2}$$
The minimum number of bananas that the person with the most bananas can have is then $$\text{Min} = \lceil{\frac{n-n_0}{k}}\rceil + k - 1$$
An example with $k=4$ (and hence $n_0=6$) and $n$ varying:

For $n=n_0+1$, we need to add a banana to someone. We cannot add it to anyone other than the person with the most bananas, otherwise we would have two people with the same number of bananas. For $n=n_0+2$ we can add the new banana to the person with the most bananas or the person with the next-most. We don't want to give the person with the most any more than absolutely necessary, so we give it to the one with the next-most. This continues until we have added a banana to all $k$ people. The pattern then starts over. 

