Given a positive number n, how many tuples $(a_1,...,a_k)$ are there such that $a_1+..+a_k=n$ with two extra constraints The problem was: Given a positive integer $n$, how many tuples $(a_1,...,a_k)$ of positive integers are there such that $a_1+a_2+...+a_k=n$. And $0< a_1 \le a_2 \le a_3 \le...\le a_k$. Also, $a_k-a_1$ is either $0$ or $1$. 
Here is what I did:
For $n=1$, there is one way $1=1$.
For $n=2$, there are $2$ ways, $2=1+1,2=2$
For $n=3$, there are $3$ ways, $3=1+1+1,3=1+2,3=3$
For $n=4$, there are $4$ ways, $4=1+1+1+1,4=1+1+2,4=2+2,4=4$
So it seems that there are $n$ tuples that satisfies the three conditions for each $n$. But I'm not sure how to prove it.

The answer is $n$.
Given $n$, you need to proof that for each $k$, where$k\leq n$, there exists exactly one tuple.
First, you can proof that for each $k$, there exists at least one tuple.
$n=kt+r$
where $r<k$.
Make the tuple $(a_1,...,a_k)=(t,...,t)$. Then add to the last $r$ components $1$ unit to get a valid tuple.
Secondly, you need to prove that the constructed tuple is unique, for each $k$ and $n$. If there is another tuple, such that its elements add to $n$, then you can get it from the first constructed tuple, by moving units between components. But, if you move a single unit, then you destroy the increasing order of components.
Finally, you just need to know how many possible $k$ you can have for a number $n$, which is $n$.
