  Let $p_{-r}(n)$ denote the $r$-coloured partition function, and
$\sigma(n)=\sum_{d|n}d$ denote the sum of positive divisors of $n$. The aim of
this note is to prove the following $$
p_{-r}(n)=\theta(n)+\,\sum_{k=1}^{n-1}\frac{r^{k+1}}{(k+1)!} \sum_{\alpha_1\,=
k}^{n-1} \, \sum_{\alpha_2\,= k-1}^{\alpha_1-1} \cdots \sum_{\alpha_k\, =
1}^{\alpha_{k-1}-1}\theta(n-\alpha_1) \theta(\alpha_1 -\alpha_2) \cdots
\theta(\alpha_{k-1}-\alpha_k) \theta(\alpha_k) $$ where $\theta(n)=n^{-1}\,
\sigma(n)$, and its inverse $$\sigma(n) = n\,\sum_{r=1}^n
\frac{(-1)^{r-1}}{r}\, \binom{n}{r}\, p_{-r}(n). $$
