# partition theory: meet the COP

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? $$\sum_{n\geq0}\overline{p}_c(n)q^n=\frac{((1-c)q;q)_{\infty}}{(q;q)_{\infty}}.$$

• Just a typographical suggestion, $COP$ uses mathemarics spacing for what is an acronym, so looks wrong. I think the small-caps \textsc{COP}, italic \textit{COP} or just plain COP would look better. – J.J. Green Mar 7 '17 at 12:46
• @J.J.Green: Thanks, I edited as such. How about now? – T. Amdeberhan Mar 7 '17 at 12:52
• Google search returned only 6 hits regarding "colored overpartition". One of them is your paper arxiv.org/abs/1207.4045 where this question goes under Theorem 12.1. This shows that you already have a proof. – Nemo Mar 7 '17 at 13:31
• @Nemo: No, I don't have a proof. In the preprint you saw it should read "conjectures" instead of "theorem". Sorry, it appeared misleading. – T. Amdeberhan Mar 7 '17 at 13:33
• For overpartitions overlined parts form a partition into distinct parts and the non-overlined parts form an ordinary partition, therefore the generating function is $\prod_{n=1}^\infty\frac{1+q^n}{1-q^n}$. In analogy with this well known case, for $\textrm{COP}$ with $c$ colors, the non-overlined parts form an ordinary partition and the overlined parts form a partition into distinct parts such that each part can take one of $c-1$ colors. So it is not surprising that generating function is $\prod_{n=1}^\infty\frac{1+(c-1)q^n}{1-q^n}$. – Nemo Mar 7 '17 at 14:33

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 $$\sum_{n\geq 0}\bar{p}_c(n)q^n = \prod_{i\geq 1}(1+c(q^i+q^{2i}+q^{3i}+\cdots))$$ $$= \prod_{i\geq 1}\frac{1+(c-1)q^i}{1-q^i}.$$