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 $$x^n(1+x^2)...(1+x^r) $$ let function $$f(n,r) = x^n(1+x^2)...(1+x^r) $$ 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.

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 $3$.