Plain integer partitions of $n$ using $r$ parts Division of number $n$ on parts $a_1,...,a_r$ where $a_1 \le ... \le a_r$ we call a plain if $a_1 = 1$ and $a_i - a_{i-1} \le 1$ for $2 \le i \le r$. Find enumerator (generating function) for plain divisions.
my try
The hint was to use bijection between plain divisions and some commonly know enumerator. I tried to use enumerator of divisions on different parts:
$$ (1+x)(1+x^2)...(1+x^r)$$
where number of plain divisions is
$$[x^n](1+x)(1+x^2)...(1+x^r) $$
let function 
$$f(n,r) = [x^n](1+x)(1+x^2)...(1+x^r) $$
For some first divisions it works. For example:
$$f(4,3) = 1 $$
$$f(6,3) = 1 $$
$$f(11,5) = 2$$
But when I tried to find bijection, I failed. I found that this function isn't correct because
$f(15,6) = 4$ but should be equal to $3$ because:
$$15 = 1,1, 2, 3, 4, 4 \\
15 = 1, 2, 2, 3, 3, 4\\
15 = 1, 2, 3, 3, 3, 3 $$. There I stucked.
 A: Are you familiar with Ferrer's Diagrams?  Make a row of dots for each part, so the plain partition $(4,4,3,2,1,1,1)$ is $$\begin{matrix} \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet &  \\ \bullet & \bullet &  &  \\ \bullet & & & \\ \bullet & & & \\ \bullet & & & \\ \bullet & & & \\ \end{matrix}$$
Now look at the columns.  You get the partition into distinct parts $(8,4,3,2)$.  Try to prove that this bijection works.
Now, $$\sum_{n \geq 1} d(n) x^n = \prod_{n \geq 1} \left(1+x^n \right),$$ where $d(n)$ is the number of partitions of $n$ into distinct parts.  But by the bijection, $d(n)$ equals the number of plain partitions of $n$, so the above product is the required generating function.
A: The set of plain division of $n$ into $r$ parts is in bijection with the set of divisions of $n$ into distinct parts whose largest part is equal to $r$. The bijection is conjugation, i.e. reflecting the Ferrer's diagram. Since there must be a part of size $r$, the factor must be $x^r$ instead of $(1+x^r)$, while all other parts are the same as what you had. Therefore, the generating function is 
$$
(1+x)(1+x^2)\dots(1+x^{r-1})x^r.
$$
Note that the coefficient of $x^{15}$ in $(1+x)\cdots(1+x^5)x^6$ is indeed $3$. 
A: Let $0\leq d_{i-1}:=a_i - a_{i-1} \le 1$  be the $i$-th increment for $2\leq i\le r$ then
$$d_1+\dots+d_{i-1}=a_i-1$$
and 
$$(r-1)d_1+(r-2)d_2+\dots+1\cdot d_{r-1}=a_2+a_3+\dots +a_r-(r-1)=n-r$$
that is
$$r+1\cdot d_{r-1}+\dots+(r-2)d_2+(r-1)d_1=n.$$
with $d_i\in\{0,1\}$.
It follows that the generating function for a given $r\geq 2$ is
$$x^{r}\prod_{k=1}^{r-1}(1+x^k).$$
