Question

\begin{cases}
f(1)=-2
\\\\
f(2)=5
\\\\
f(n)=f(n-2) \cdot f(n-1)
\end{cases}

Answer 1

Let a = f(1) = −2 and b = f(2) = 5. One checks
f(3) = ab, f(4) = b·ab = ab^2, f(5) = ab·ab^2 = a^2 b^3, …
The exponents satisfy the Fibonacci recurrence, so (with F1 = F2 = 1)
for n ≥ 3
f(n) = a^{F_{n-2}} b^{F_{n-1}}.

Substituting a = −2, b = 5,
for n ≥ 3
f(n) = (−2)^{F_{n-2}} · 5^{F_{n-1}}.
Hence the magnitude is |f(n)| = 2^{F_{n-2}} 5^{F_{n-1}} and the sign is (−1)^{F_{n-2}}.
Because F_k is even iff 3 | k, the sign is + precisely when n ≡ 2 (mod 3) and is − otherwise.

Some initial terms:
f(1) = −2, f(2) = 5, f(3) = −10, f(4) = −50, f(5) = 500, f(6) = −25000, ….

Also ln|f(n)| satisfies the Fibonacci recurrence, so |f(n)| grows extremely fast (roughly like exp(const · φ^n), φ = (1+√5)/2).