<s>[INST] Find number of nonnegative integer solutions to x+3y+3z=n, given n, using generating functions 
For every $n,x,y,z\in \mathbb N$, where $x\ge{0}$ and $y,z \ge1$ 
Find the number of nonnegative integer solutions for
  $x+3y+3z = n$

I created a generating function for the problem:
$(1+x+x^2+...)(x^3+x^6+x^9+...)^2$
$(1+x+x^2+...)x^6(1+x^3+x^6+...)^2$
$\frac{1}{1-x} \frac{x^6}{(1-x^3)^2}$
$= \frac{x^6}{(1-x)(1-x^3)^2}$
I can try breaking this expression into partial functions and try to find the coefficient of $x^n$ from there, but I get stuck in that stage.
can you please help me complete this problem?
 [/INST] Call your variables $a_i$, so that you want $a_1 + 3 a_2 + 3 a_3 = n$ with $a_i \ge 0$ integers.
The values of $a_1$ are represented by:
$$
1 + z + z^2 + \dotsb = \frac{1}{1 - z}
$$
The values added by $a_2$ and $a_3$ are:
$$
z^3 + z^6 + \dotsb = \frac{z^3}{1 - z^3}
$$
In all, you want the coeficient of $z^n$ in:
$\begin{align}
[z^n] \frac{z^6}{(1 - z) (1 - z^3)^2}
  &= [z^{n - 6}]
       \left(
          \frac{1}{9 (1 - z)^3}
            + \frac{2}{9 (1 - z)^2}
            + \frac{7}{27 (1 - z)}
            + \frac{1 + 2 z}{(1 + z + z^2)^2}
            + \frac{8 + 7 z}{1 + z + z^2}
       \right)
\end{align}$
The last two terms are troublesome. We can factor more using complex numbers, using the cube root $\omega = \exp(2 \pi \mathrm{i} / 3)$, but it gets quite ugly.
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