Problem

If $F(n)$ is a function such that $F(1)=F(2)=F(3)=1$ , and such that $F(n+1)= \frac{F(n)\cdot F(n-1)+1}{F(n-2)}$ for $n\ge 3,$ then $F(6)=$

$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 7\quad \text{(D) } 11\quad \text{(E) } 26$

Solution

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 $\fbox{C}$ .

All the numbers in the sequence $F(n)$ are integers. In fact, the function $F$ satisfies $F(n)=4F(n-2)-F(n-4)$ . (Prove it!).

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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