Given $n \in \mathbb{N}$, what numbers are representable as $a_1a_2 + \cdots + a_{k-1}a_k$, $k \geq 2$, $\sum a_k = n$, $a_i \in \mathbb{N}_+$? Could someone give me a little hint how to solve this ?
Given a number $n\in\mathbb{N},\ n\geq 3$ and the set
$S_n = \{ \left. m \in \mathbb{N}\setminus\left\{0\right\} \ \right| \quad
\exists k \in \mathbb{N},\ k \geq 2,\ \exists a_1, \ldots, a_k \in \mathbb{N}\setminus\left\{0\right\}\ \text{such that}\ a_1 + \cdots + a_k = n$
$\text{and} \ a_1a_2 + a_2a_3 + \cdots + a_{k-1}a_k = m\}
$
one has to find an explicit description of this set.
I thought about it a bit and it seems to me, that $S$ should be of the form $S_n=\left\{ 1,2,\ldots,n \right\} $ but I have no idea how to attempt a proof.
 A: This is not a complete answer, only a prove of the bounds $\min S_n$ and $\max S_n$.
Let $\mathbb N = \lbrace 1,2,3,\dots\rbrace$ denote the positive integers.
We first prove $\min S_n = n-1$ by induction:
Let $a_1,a_2 \in \mathbb N$, then it is easily verified, that $a_1a_2 \geq a_1 + a_2 - 1$.
Now assume, that for $k \geq 2$ and for all $a_1, \dots, a_k \in \mathbb N$ we know that
$$a_1a_2 + \dots a_{k-1}a_k \geq a_1 + \dots + a_k - 1$$
is true, then we conclude
$$a_1a_2 + \dots + a_{k-1}a_k + a_ka_{k+1} \geq$$
$$(a_1 + \dots + a_{k-1} + a_k - 1) + a_ka_{k+1} =$$
$$a_1 + \dots + a_{k-1} + a_k(1 + a_{k+1}) - 1 \geq$$
$$a_1 + \dots + a_k + a_{k+1} - 1.$$
This inequality assures $\min S_n \geq n-1$. If we choose $a_1 = \dots = a_n = 1$, then $a_1a_2 + \dots + a_{n-1}a_n = n-1$.
Now we will show $\max S_{2n} = n^2$ and $\max S_{2n+1} = n(n+1)$:
Let $a_1,a_2 \in \mathbb N$ with $a_1 + a_2 = 2n$ (resp. $a_1 + a_2 = 2n+1$), then it is obvious, that $a_1a_2 \leq n^2$ (resp. $a_1a_2 \leq n(n+1)$) and the bound is actually taken.
We will prove the rest of this part only for the even case (the odd case being completely analogous).
If we have $a_1, a_2, a_3 \in \mathbb N$ satisfying $a_1 + a_2 + a_3 = 2n$, then 
$a_1a_2 + a_2a_3 = a_2(a_1 + a_3) \leq n^2$.
Now assume we know, that for some $k\geq3$ and for all $a_1, \dots ,a_k \in \mathbb N$ with $a_1 + \dots + a_k = 2n$ we have $a_1a_2 + \dots + a_{k-1}a_{k} \leq n^2$.
Let $a_1, \dots, a_{k+1} \in \mathbb N$ with $a_1 + \dots + a_{k+1} = 2n$, then
$$a_1a_2 + \dots + a_{k-2}a_{k-1} + a_{k-1}a_k + a_ka_{k+1} =$$
$$a_1a_2 + \dots + a_{k-2}(a_{k-1} + a_{k+1}) + (a_{k-1} + a_{k+1})a_k - a_{k-2}a_{k+1} \leq$$
$$n^2 - a_{k-2}a_{k+1} < n^2.$$
