Recall that $(a;q)\_0:=1,\,(a;q)\_n=(1-a)(1-aq)(1-aq^2)\cdots(1-aq^{n-1})$ and
$(a;q)\_{\infty}=(1-a)(1-aq)(1-aq^2)\cdots$. Let's introduce the following (generalized) concept.


A *colored overpartition* (COP) is a partition where the last occurrence of each distinct number may receive any one of $c$ colors. The number of such partitions of $n$ we denote by $\overline{p}\_c(n)$. For example, the $8$ COP partitions of $n=3$ with $c=2$ colors are:
$$3,\, \overline{3}, \, 2+1, \, \overline{2}+1, \, 2+\overline{1}, \, \overline{2}+\overline{1}, \, 1+1+1, \, 1+1+\overline{1}.$$
**Note.** In the literature, (i) ordinary partitions $p(n)=\overline{p}\_1(n)$; (ii) overpartitions $\overline{p}(n)=\overline{p}\_2(n)$. 



> 
> **Question.** Does this generating function hold true?
>  $$\sum\_{n\geq0}\overline{p}\_c(n)q^n=\frac{((1-c)q;q)\_{\infty}}{(q;q)\_{\infty}}.$$
> 
> 
> 



partition theory: meet the COP
Rather than coloring the last occurrence of each distinct number, we
can equivalently for each part $i$ color all the $i$'s with the same
color. Thus
 $$ \sum\_{n\geq 0}\bar{p}\_c(n)q^n = \prod\_{i\geq
 1}(1+c(q^i+q^{2i}+q^{3i}+\cdots)) $$
 $$ = \prod\_{i\geq 1}\frac{1+(c-1)q^i}{1-q^i}. $$


