Solution: Given that \[ \frac{z}{1 - z - z^2} = \sum_{n=0}^{\infty} a_n z^n, \] we need to find $a_5 + a_6$. Note that \begin{align*} & \frac{z}{1 - z - z^2} = \sum_{n=0}^{\infty} a_n z^n \\ \implies & z = (1- z - z^2) \sum_{n=0}^{\infty} a_n z^n \\ \implies & z = (1 - z - z^2) \left( a_0 + a_1 z + a_2 z^2 + \dots \right) \\ \implies & z = \left( a_0 + a_1 z + a_2 z^2 + a_3 z^3 + \dots \right) \\ & \kern 0.6cm - \left( a_0 z + a_1 z^2 + a_2 z^3 + a_3 z^4 + \dots \right) \\ & \kern 0.6cm - \left( a_0 z^2 + a_1 z^3 + a_2 z^4 + a_3 z^5 + \dots \right). \end{align*}

Now comparing the coefficients, we get \begin{align*} & a_0 = 0 \\ & a_1 - a_0 = 1 \implies a_1 = 1 \\ & a_2 - a_1 - a_0 = 0 \implies a_2 = 1 \\ & a_3 - a_2 - a_1 = 0 \implies a_3 = 2 \\ & a_4 - a_3 - a_2 = 0 \implies a_4 = 3 \\ & a_5 - a_4 - a_3 = 0 \implies a_5 = 5 \\ & a_6 - a_5 - a_4 = 0 \implies a_6 = 8. \end{align*} Therefore, \[ a_5 + a_6 = 13. \]