Solving functional equation for generating function Find the functional equation for the generating function whose coefficients satisfy $$
a_n = \sum_{i=1}^{n-1}2^ia_{n-i}, \text{ for } n\ge 2, a_0 = a_1 = 1
$$
This is what I've tried so far:
$$
\begin{align}
g(x) -1 -x &= \sum_{n\ge2} \sum_{i=1}^{n-1} 2^i a_{n-i}\\
&=\sum_{n\ge2} \sum_{i=1}^n 2^i a_{n-i}x^n\\
\end{align}
$$
at which point I'm stuck. How do I proceed?

Let
$$g(x) = \sum_{k=0}^{\infty} a_k x^k$$
From the defining recurrence:
$$\begin{align}a_2 x^2 &= 2 a_1 x^2\\ a_3 x^3 &= 2 a_2 x^3 + 4 a_1 x^3\\a_4 x^4 &= 2 a_3 x^4 + 4 a_2 x^4 + 8 a_1 x^4\\a_5 x^5 &= 2 a_4 x^5 + 4 a_3 x^5 + 8 a_2 x^5 + 16 a_1 x^5\\ \ldots \end{align}$$
and so on.  We then form 
$$g(x) - a_0 - a_1 x = \sum_{k=2}^{\infty} a_k x^k$$
by summing the above terms.  By rearranging like coefficients, we get
$$\begin{align}\sum_{k=2}^{\infty} a_k x^k &= \frac{1}{2} a_1 \sum_{k=2}^{\infty}(2 x)^k +  \frac{1}{4} a_2 \sum_{k=3}^{\infty}(2 x)^k + \frac{1}{8} a_3 \sum_{k=4}^{\infty}(2 x)^k+\ldots\\ &= (2 x^2 a_1 + 2 x^3 a_2 + 2 x^4 a_3 + \ldots) \sum_{k=0}^{\infty} (2 x)^k \\ &= \frac{2 x}{1-2 x}\sum_{k=1}^{\infty} a_k x^k \end{align}$$
Using the initial conditions and the definition of $g$, we get the following functional equation:
$$g(x) - (1+x) = \frac{2 x}{(1-2 x)} [g(x) - 1] $$
Doing this algebra, we get Marvis's result:
$$g(x) = \frac{1-3 x - 2 x^2}{1-4 x}$$
