A recurrence formula for \(D\) numbers \(D_{2n}^{(2n-1)}\) (Q1040142)

Let the \(D\) numbers \(D^{(k)}_{2n}\) be defined by \[ (t\, \text{csc}\, t)^k = \sum_{n=0}^\infty (-1)^n D^{(k)}_{2n}\frac{t^{2n}}{(2n)!},\qquad | t|<\pi\,. \] The author finds a recurrence formula for \(D\) numbers \(D^{(2n-1)}_{2n}\) and a generating function. The following two theorems are proved. Theorem 1. Let \(n\in\mathbb N\). Then \[ \sum_{j=1}^n \binom{2n}{2j}(-1)^{j-1}4^{j-1}((j-1)!)^2 D^{(2n-1-2j)}_{2n-2j}=\frac{(-1)^{n-1}2(2n)!}{4^n}\binom {2n-2}{n-1} \] which gives the values \(D^{(1)}_{2}=-1/3\), \(D^{(3)}_{4}=17/5\), \(D^{(5)}_{6}=-1835/21\), \(D^{(71)}_{8}=195013/45\), \(D^{(9)}_{10}=-3887409/11\), \(\ldots\). Theorem 2. Let \(t\) be a complex number with \(| t|<1\). Then \[ \sum_{n=0}^\infty D^{(2n-1)}_{2n} \frac{t^{2n}}{(2n)!}=\frac 1{\sqrt{1+t^2}}\left(\frac 1{\log(t+\sqrt{1+t^2}}\right)^2. \]