$7 \cdot 93$ If $F(n)$ is a function, where $F(1)=F(2)=F(3)=1$, and $F(n+1)=\frac{F(n) \cdot F(n-1)+1}{F(n-2)}$, then $F(6)$ equals
(A) 2 .
(B) 3 .
(C) 7 .
(D) 11 .
(E) 26 .
(21st American High School Mathematics Examination, 1970)
[Solution] From the given, we can calculate $F(4), F(5)$, and then obtain $F(6)$.
$$
\begin{array}{l}
F(4)=\frac{F(3) F(2)+1}{F(1)}=2, \\
F(5)=\frac{F(4) F(3)+1}{F(2)}=3, \\
F(6)=\frac{F(5) F(4)+1}{F(3)}=7 .
\end{array}
$$

Therefore, the answer is (C).