Explicit and recurrence formulas for generalized Euler numbers (Q1343469)

The generalized Euler numbers \(E_{2n}^{(r)}\) for a complex number \(r\) and positive integer \(n\) are defined by \[ \text{sec}^r z=1/ \cos^r z=1+ \sum_{n=1}^\infty E_{2n}^{(r)} z^{2n}/ (2n)! , \qquad |z|< \pi/2. \tag{1} \] For \(r=1\) we get the ordinary Euler numbers \(E_{2n}\). The author proves the following explicit formula for the generalized Euler numbers: \[ E_{2n}^{(r)}= \sum_{k=1}^n \sum_{j=1}^k (-1)^{n-j} {\textstyle {{r+k-1} \choose k} {{2k} \choose {k-j}}} j^{2n}/ 2^{k-1}, \tag{3} \] which generalizes the known formula for the Euler numbers [\textit{L. Comtet}, Advanced combinatorics (1974; Zbl 0283.05001)]. Furthermore the author gives the recursive formula: \[ E_{2n}^{(r)}= \sum_{m=0}^{n-1} (-1)^{n-1-m} \left[ {\textstyle {{2n-1} \choose {2m}} r+ {{2n-1} \choose {2m-1}}} \right] E_{2m}^{(r)} \tag{8} \] for positive integers \(n\), where \({{2n-1} \choose {-1}} =0\) and \(E_0^{(r)} =1\). We get from the formula (3) that \(E_{2n}^{(r)}\) are polynomials of degree \(n\) and from the formula (8) that the coefficients of these polynomials are integers. Finally it is proved that these coefficients are positive.