1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – J.J. Green Mar 7 '17 at 12:46
  • $\begingroup$ @J.J.Green: Thanks, I edited as such. How about now? $\endgroup$ – T. Amdeberhan Mar 7 '17 at 12:52
  • $\begingroup$ 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. $\endgroup$ – Nemo Mar 7 '17 at 13:31
  • $\begingroup$ @Nemo: No, I don't have a proof. In the preprint you saw it should read "conjectures" instead of "theorem". Sorry, it appeared misleading. $\endgroup$ – T. Amdeberhan Mar 7 '17 at 13:33
  • 1
    $\begingroup$ 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 $\prod_{n=1}^\infty\frac{1+(c-1)q^n}{1-q^n}$. $\endgroup$ – Nemo Mar 7 '17 at 14:33
5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.