Closed form solution of Fibonacci-like sequence Could someone please tell me the closed form solution of the equation below.
$$F(n) = 2F(n-1) + 2F(n-2)$$
$$F(1) = 1$$
$$F(2) = 3$$
Is there any way it can be easily deduced if the closed form solution of Fibonacci is known?

Define $g(z) = \sum_{n \ge 0} F(n + 1) z^n$, write the recurrence as:
$$
F(n + 3) = 2 F(n + 2) + 2 F(n + 1) \qquad F(1) = 1, F(2) = 3
$$
Multiply the recurrence by $z$, sum over $n \ge 0$ and get:
$$
\frac{g(z) - F(1) - F(2)}{z^2} = 2 \frac{g(z) - F(1)}{z} + 2 g(z)
$$
Solve for $g(z)$:
$$
g(z) = \frac{1 + z}{1 - 2 z - 2 z^2}
     = \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 - (1 + \sqrt{3}) z)}
        + \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 + (1 - \sqrt{3})z}
$$
Two geometric series:
$$
T(n+ 1) = \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot (1 + \sqrt{3})^n
            + \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot (1 - \sqrt{3})^n
$$
