smc

Plugging in $n=3$ gives $F(4)=\frac{F(3)\cdot F(2)+1}{F(1)}=\frac{1\cdot 1+1}{1}=2$. Plugging in $n=4$ gives $F(5)=\frac{F(4)\cdot F(3)+1}{F(2)}=\frac{2\cdot 1+1}{1}=3$. Plugging in $n=5$ gives $F(6)=\frac{F(5)\cdot F(4)+1}{F(3)}=\frac{3\cdot 2+1}{1}=7$. Thus, the answer is $\boxed{7}$.