Set up difference equation for the following recurrence. I have the following recurrence: 
$t=0: 0$
$t=1: 0$ 
$t=2: 1$
$t=3: \beta+\alpha$
$t=4: (\beta+\alpha)\alpha+\beta^2$
$t=5: ((\beta+\alpha)\alpha+\beta^2)\alpha+\beta^3$
...
I was hoping to do something like:
$x_{t+1}=x_{t}f_{t}+g_{t}$ with  $f_{t}=\alpha$ and $g_{t}=\beta^{t-1}$ 
But it does not fit with $t=0$ and $t=1$ both has to be $0$.
Is there anyway to set up this difference equation without using indicator functions or piecewiese functions for $g_{t}$?

Call your value $a_t$, after $t = 3$ it is $a_{t + 1} = \alpha a_t + \beta^{t - 2}$, with $a_3 = \alpha  + \beta$. Set up a generating function $A(z) = \sum_{t \ge 0} a_{t + 3} z^t$, multiply the recurrence by $z^t$ and sum over $t \ge 0$:
$\begin{align}
\sum_{t \ge 0} a_{t + 4} z^t
  &= \alpha \sum_{t \ge 0} a_{t + 3} z^t
       + \sum_{t \ge 0} \beta^{t + 1} z^t \\
\frac{A(z) - a_3}{z}
  &= \alpha A(z) + \frac{\beta}{1 - \beta z}
\end{align}$
Plugging in $a_3 = \alpha + \beta$ and solving for $A(z)$ gives, unless $\alpha = \beta$:
$\begin{align}
A(z)
  &= \frac{(\alpha + \beta) - \beta(\alpha + \beta - 1) z}
          {(1 - \alpha z) (1 - \beta z)} \\
  &= \frac{\beta^2 - \beta - \alpha^2}{\beta - \alpha}
       \cdot \frac{1}{1 - \alpha z}
       + \frac{\beta}{\beta - \alpha} \cdot \frac{1}{1 - \beta z}
\end{align}$
and your solution is:
$$
a_{t + 3}
  = \frac{\beta^2 - \beta - \alpha^2}{\beta - \alpha} \cdot \alpha^t
      + \frac{\beta^{t + 1}}{\beta - \alpha}
$$
If $\alpha = \beta$:
$\begin{align}
A(z) 
  &= \frac{2 \alpha - \alpha(2 \alpha - 1) z}{(1 - \alpha z)^2} \\
  &= \frac{2 \alpha - 1}{1 - \alpha z}
       + \frac{1}{(1 - \alpha z)^2}
\end{align}$
from which you have directly:
$$
a_{t + 3}
  = (2 \alpha - 1) \cdot \alpha^t
     + (n + 1) \cdot \alpha^t
  = (n + 2 \alpha) \cdot \alpha^t
$$
Earlier values don't follow the pattern.
