I4.2 Let $F(n)$ be a function with $F(1)=F(2)=F(3)=a$ and $F(n+1)=\frac{F(n) \cdot F(n-1)+1}{F(n-2)}$ for
positive integer $n \geq 3$, find the value of $b=F(6)$.
$$
\begin{array}{l}
F(4)=F(3+1)=\frac{F(3) \cdot F(3-1)+1}{F(3-2)}=\frac{F(3) \cdot F(2)+1}{F(1)}=\frac{1 \cdot 1+1}{1}=2 \\
F(5)=F(4+1)=\frac{F(4) \cdot F(4-1)+1}{F(4-2)}=\frac{F(4) \cdot F(3)+1}{F(2)}=\frac{2 \cdot 1+1}{1}=3 \\
F(6)=\frac{F(5) \cdot F(4)+1}{F(3)}=\frac{3 \cdot 2+1}{1}=7
\end{array}
$$