[Algebra]
Problem: The function $f$ satisfies the relation $f(n)=f(n-1) f(n-2)$ for all integers $n$, and $f(n)>0$ for all positive integers $n$. If $f(1)=\frac{f(2)}{512}$ and $\frac{1}{f(1)}=2 f(2)$, compute $f(f(4))$.

Solution: Substituting yields $\frac{512}{f(2)}=2 f(2) \Rightarrow(f(2))^{2}=256 \Rightarrow f(2)=16$. Therefore $f(1)=\frac{1}{32}$. Using the recursion, $f(3)=\frac{1}{2}$ and $f(4)=8$. So $f(f(4))=f(8)$. Continue to apply the recursion:

$$
f(5)=4, \quad f(6)=32, \quad f(7)=128, \quad f(8)=\mathbf{4 0 9 6} .
$$

Alternate Solution: Let $g(n)=\log _{2} f(n)$. Then $g(n)=g(n-1)+g(n-2)$, with initial conditions $g(1)=g(2)-9$ and $-g(1)=1+g(2)$. From this, $g(1)=-5$ and $g(2)=4$, and from the recursion,

$$
g(3)=-1, \quad g(4)=3
$$

so $f(4)=2^{g(4)}=8$. Continue to apply the recursion:

$$
g(5)=2, \quad g(6)=5, \quad g(7)=7, \quad g(8)=12
$$

Because $g(f(4))=12$, it follows that $f(f(4))=2^{12}=\mathbf{4 0 9 6}$.

Final answer: 4096