I apologize for the cryptic title, this issue came up while obtaining a solution to my question here. I was given a power series:
$\sum_{n=0}^{\infty}r(n)z^n = \frac{1}{(1-z)(1-z^2)(1-z^3)}$
I was then asked to perform a partial fraction decomposition on the closed form and obtained:
$ -\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}}}$
Now upon individually converting these fractions back into power series, in short, what I obtained is a power series with coefficients $\frac{(n+3)^2}{12} + x_n$ for $\frac{1}{3} \leq x_n \leq \frac{2}{3}$.
This was expected, since my book states that $r(n)$ is given by the integer closest to $\frac{(n+3)^2}{12}$. What I don't understand is how can we conclude such a close correspondence between $r(n)$ and $\frac{(n+3)^2}{12} + x_n$? Just because two power series converge to the same closed form function, can't their coefficients not still be very different for any particular $n$? I know Taylor series are unique, but I don't think my latter power series is a Taylor series, and even if the power series expansion of a function is always unique, it still has that extra $x_n$ added to or subtracted from each coefficient. What is going on here?
Edit: Ok so $r(n)$ is supposed to be the number of ordered triples $(x_1,x_2,x_3)$ such that $x_1 + 2x_2 + 3x_3 = n$. I formed this by employing the Cauchy product $(\sum z^n)(\sum z^{2n})(\sum z^{3n})$.