<s>[INST] Solve using generating functions the following recurrence $F_0 = F_1 = 0$
$F_2 = 1$

$F_n = F_{n-1} + F_{n-2} + F_{n-3} , n \ge 3 $

Hint: You will have to write the recurrence in relation to the Fibonacci sequence.  I seem to be stuck in this method since I am at a point where $G(z) = \frac{z}{1-z-z^2-z^3}$ and I do not know how to proceed
 [/INST] The sequence 
\begin{align*}
\left(F_n\right)_{n\geq 0}=(0,0,1,1,2,4,7,13,24,44,81,\ldots)
\end{align*}
has the corresponding generating function
\begin{align*}
G(z)=z^2+z^3+2z^4+4z^5+7z^6+\cdots
\end{align*}
Since the series expansion of $G(z)$ at $z=0$ starts with $z^2$ and the geometric series $\frac{1}{1-y}=1+y+y^2+\cdots$ starts with the constant term $1$ we use $z^2$ in the numerator of OPs generating function 

\begin{align*}
G(z)&=\frac{z^2}{1-\left(z+z^2+z^3\right)}\tag{1}
\end{align*}

In the following it is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series.

We apply the geometric series expansion of (1) and obtain for $n\geq 2$
  \begin{align*}
[z^n]G(z)&=[z^n]z^2\sum_{k=0}^\infty \left(z+z^2+z^3\right)^k\\
&=[z^{n-2}]\sum_{k=0}^\infty z^k\left(1+z+z^2\right)^k\tag{2}\\
&=\sum_{k=0}^{n-2}[z^{n-2-k}]\left(1+z+z^2\right)^k\tag{3}\\
&=\sum_{k=0}^{n-2}[z^{n-2-k}]\sum_{l=0}^k\binom{k}{l}z^l\left(1+z\right)^l\tag{4}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}[z^{n-2-k-l}]\binom{k}{l}\sum_{j=0}^l\binom{l}{j}z^j\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}\binom{k}{l}\binom{l}{n-2-k-l}\tag{5}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}\binom{k}{n-2-k-l}\binom{t-2-k-l}{l}\tag{6}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{k}\binom{n-2-k}{k-l}\binom{k-l}{l}\tag{7}\\
\end{align*}

Comment:


*

*In (2) we use the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$.

*In (3) we use the linearity of the coefficient of operator and use the rule from (2) again. Since the powers of $z$ are non-negative we restrict the upper limit of the left-most sum with $n-2$.

*In (4) we apply the binomial theorem and continue here and in the next line similarly as before.

*In (5) we select the coefficient of $z^{n-2-k-l}$. Next we transform the double sum to simplify the representation somewhat.

*In (6) we change the summation order of the inner sum by replacing the index $l$ with $n-2-k-l$.

*In (7) we change the summation order of the outer sum by replacing the index $k$ with $n-2-k$.

We conclude, the coefficients $F_n$ are
  \begin{align*}
&F_0=F_1=0\\
&F_2=1\\
&F_n=\sum_{k=0}^{n-2}\sum_{l=0}^{k}\binom{n-2-k}{k-l}\binom{k-l}{l}\qquad\qquad n\geq 3
\end{align*}

Note: The coefficients $F_n$ are called Tribonacci numbers and are archived as OEIS/A000073.
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