I would like to find the number of ordered triples $(x_1,x_2,x_3)$ such that $x_1 + 2x_2 + 3x_3 = n$. For each $n$ call this number $r(n)$. So after reading some similar questions on this site I'm pretty sure this is just some form of the partition function, and my book states explicitly that it is equivalent to asking for all partitions $x_1 + x_2 + x_3 = n$ where $x_1 \leq x_2 \leq x_3$, unfortunately I don't fully understand the wikipedia article on the partition function, so let's just leave that aside for now..
What I was first asked to do was show that $\sum_0^{\infty}r(n)z^n = \frac{1}{(1-z)(1-z^2)(1-z^3)}$. So I successfully did this using the Cauchy Product Formula, and then I was asked to find the partial fraction decomposition of this rational equation into a particular form, which I did:
$ -\frac{\frac{1}{6}}{(z-1)^3} + \frac{\frac{1}{4}}{(z-1)^2} - \frac{\frac{17}{72}}{z-1} + \frac{\frac{1}{8}}{z+1} - \frac{\frac{1}{9}e^{\frac{2\pi i}{3}}}{z-e^{\frac{2\pi i}{3}}} - \frac{\frac{1}{9}e^{\frac{-2\pi i}{3}}}{z-e^{\frac{-2\pi i}{3}}}$
So now I'm asked to show that $r(n)$ is the integer nearest $\frac{(n+3)^2}{12}$.
My thinking was that I was supposed to use this decomposition and convert each fraction back to series form, combine them all, and then the coefficients of this series would be given by $\frac{(n+3)^2}{12}$. But I can't seem to simplify the series(plural) even close to that, can anyone take a look at this for me? Thanks.