<s>[INST] Find the coefficient of $x^{13}$ in the convolution of two generating functions 
Four thiefs have stolen a collection of 13 identical diamonds. After the
  theft, they decided how to distribute them.
3 of them have special requests: 
  
  
*
  
*One of them doesn't want more than 2 diamonds ($\leq2$).
  
*The other one only wants a number of diamonds that's a multiple of 3.
  
*And the other one wants an odd number of diamonds greater or equal than 3.
Find in how many ways they can distribute the diamonds.

My first thought was to use generating functions to find the coefficient of $x^{13}$, for this problem it would be:
$f(x)=(1+x+x^2+x^3+...)(1+x+x^2)(1+x^3+x^6+...)(x^3+x^5+x^7+...)=\frac{1}{1-x} \frac{1-x^3}{1-x} \frac{1}{1-x^3}\frac{x^3}{1-x^3}=\frac{1}{(1-x)^{2}}\frac{x^3}{(1-x^2)}$ 
and that would be equivalent to finding the coefficient of $x^{10}$ in $\frac{1}{(1-x)^{2}}\frac{1}{(1-x^2)}$.
I know that I could use the binomial theorem, but the solution I have says I should be using convolution of these two generating functions, but I have no idea how to use it to find the coefficient of $x^{10}$.
 [/INST] It   is  convenient to   use the  coefficient of operator  $[x^n]$ to  denote   the coefficient  of $x^n$  of  a  series.

We obtain
  \begin{align*}
\color{blue}{[x^{13}]\frac{x^3}{(1-x)^2(1-x^2)}}
&=[x^{10}]\frac{1}{(1-x)^2(1-x^2)}\tag{1}\\
&=[x^{10}]\sum_{j=0}^5 x^{2j}\sum_{k=0}^\infty \binom{-2}{k}(-x)^k\tag{2}\\
&=\sum_{j=0}^5[x^{10-2j}]\sum_{k=0}^\infty\binom{k+1}{k}x^k\tag{3}\\
&=\sum_{j=0}^5[x^{2j}]\sum_{k=0}^\infty(k+1)x^k\tag{4}\\
&=\sum_{j=0}^5(2j+1)\tag{5}\\
&=2\left(\frac{5\cdot6}{2}\right)+6\\
&\,\,\color{blue}{=36}
\end{align*}

Comment:


*

*In (1) we apply the   rule  $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

*In (2) we do a geometric and a binomial series expansion.

*In (3) we select the coefficient of $x^{2j}$ and restrict the upper limit of the outer sum to $5$ since other values do not contribute to $[x^{10}]$. We also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (4) we change the order of summation $j\to 5-j$ and use $\binom{k+1}{k}=\binom{k+1}{1}=k+1$.

*In (5) we select the coefficient of $x^{2j}$.

*In (6) we apply the formula $\sum_{j=0}^n j=\frac{n(n+1)}{2}$.
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