<s>[INST] 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.
 [/INST] Every solution is a tuple in order of increasing integers. Also, $a_k=a_1$ or $a_k=a_1+1$. Therefore, we have all of the $a_1$s at the beginning and $a_1+1$s at the end. We can say there are $l$ instances of $a_1$ and thus $k-l$ instances of $a_1+1$. Since the sum of the tuples is $n$, we find:
$$l*a_1+(k-l)*(a_1+1)=n$$
Simplify the left side:
$$a_1k+k-l=n$$
Now, $l$ is the number of instances of $a_1$ in the sequence. Thus, $l$ is at least $1$ and at most $k$. Therefore,  $k-l$ is at least $0$ and at most $k-1$, so $k-l$ is basically the remainder of $n$ when divided by $k$ and $a_1$ is the quotient. Thus, by the Division Theorem, we know there are unique integer solutions of $a_1$ and $l$ for this equation for any $k$ and $n$.
However, the problem says that $a_1 > 0$, so we need to exclude the solutions where $a_1=0$. When $a_1=0$, we get:
$$0k+k-l=n \implies k-l=n$$
Since $k-l$ is at most $k-1$, this means $n$ is at most $k-1$ and thus $n < k$. Therefore, exclude all solutions where $k > n$.
Also, obviously, $k \geq 1$. This leaves us with the solutions $k \geq 1$ and $k \leq n$, so there are $n$ solutions.
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