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Yemon Choi
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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.

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.

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.

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Johann Cigler
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Reference request: Vanishing coefficients in power series.

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.