Let $$\alpha = \frac{{1 - az + \sqrt {(1 - az)^2 - 4bz^2 } }}{2}.$$

In the power series expansion of $\alpha ^n $ the coefficients of $z^k$ vanish for $n + 1 \le k \le 2n - 1.$

Similar results hold for $$\alpha (m) = \frac{{1 - az + \sqrt {(1 - az)^2 - 4bz^m } }}{2}.$$
Since the proofs are very simple such results must be well known. I would be very grateful for references.