<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] This can  also be  done with a  simple generating function.   Call the
desired quantity $T_n.$  We first choose the value  $a_1$ and then the
gaps between successive  values among the $a_q$ which  are either zero
or one.  We  will have $1\le k\le  n$ but for $k=1$ there  is just one
possibility so we may assume $2\le k\le n.$ We get for $a_1,$ which is
positive,
$$\frac{z}{1-z}$$
but it contributes to all $k$ terms so we have in fact
$$\frac{z^k}{1-z^k}.$$
We combine this  with $k-1$ gaps of either zero or  one. The first gap
contributes to $k-1$ terms, the next one to $k-2$ and so on. Using the
variable $u$ to mark gaps we thus obtain
$$\bbox[5px,border:2px solid #00A000]{
1 + \sum_{k=2}^n 
\frac{z^k}{1-z^k} \prod_{q=1}^{k-1} \left(1+uz^{k-q}\right).}$$
Now  from the  constraint  that $a_k-a_1$  be  either zero  or one  it
follows we need  the coefficient on $[u^0]$ and  $[u^1].$ We set $u=0$
for the first one and get
$$\sum_{k=2}^n \frac{z^k}{1-z^k}.$$
We differentiate and set $u=0$ for the second one and obtain
$$\left.\sum_{k=2}^n 
\frac{z^k}{1-z^k} \prod_{q=1}^{k-1} \left(1+uz^{k-q}\right)
\sum_{q=1}^{k-1} \frac{z^{k-q}}{1+uz^{k-q}}\right|_{u=0}
\\ = \sum_{k=2}^n 
\frac{z^k}{1-z^k}\sum_{q=1}^{k-1} z^{k-q}.$$
It follows that the desired result is given by
$$\bbox[5px,border:2px solid #00A000]{
1 + [z^n] \sum_{k=2}^n 
\frac{z^k}{1-z^k}\sum_{q=1}^{k} z^{k-q}.}$$
Evaluating this we initially obtain
$$1 + [z^n] \sum_{k=2}^n 
\frac{z^{2k}}{1-z^k}\sum_{q=1}^{k} z^{-q}
= 1 + [z^n] \sum_{k=2}^n 
\frac{z^{2k}}{1-z^k} \frac{1}{z} \sum_{q=0}^{k-1} z^{-q}
\\ = 1 + [z^n] \sum_{k=2}^n 
\frac{z^{2k}}{1-z^k} \frac{1}{z} \frac{1-1/z^k}{1-1/z}
= 1 + [z^n] \sum_{k=2}^n 
\frac{z^{2k}}{1-z^k} \frac{1-1/z^k}{z-1}
\\ = 1 + [z^n] \sum_{k=2}^n 
\frac{z^k}{1-z^k} \frac{z^k-1}{z-1}
= 1 - [z^n] \sum_{k=2}^n 
\frac{z^k}{z-1}.$$
Now we  may extend $k$ to infinity  because the terms for  $k\gt n$ do
not contribute to the coefficient on $[z^n],$ getting
$$1 + [z^n] \frac{1}{1-z} \sum_{k\ge 2} z^k
= 1 + [z^n] \frac{z^2}{(1-z)^2}
\\ = 1 + [z^{n-2}] \frac{1}{(1-z)^2}
= 1 + {n-2+1\choose 1}.$$
This yields the end result
$$\bbox[5px,border:2px solid #00A000]{T_n = n.}$$
I do think this computation  is interesting and perhaps different from
what one might have expected.
</s>