I am trying to solve the following partial difference equation:

$$A_k^{n+1}=(k+1)A_{k+1}^n+(n+2-k)A_{k-1}^n $$

with initial condition:

$$\begin{cases} A_0^0&=1\\ A_1^0&=1 \end{cases}$$

I have tried using generating function method and the detail is given by the following article in Voofie:

The generating function I found is:

$$A(x,y)=\left(\sec \left(y\sqrt{1-x^2}+\sin ^{-1}x\right)+\tan \left(y\sqrt{1-x^2}+\sin ^{-1}x\right)\right)\sqrt{1-x^2}$$

if A(x,y) is defined by:

$$A(x,y)=\sum _{n=0}^{\infty } \sum _{k=0}^{\infty } \frac{A_k^n x^k y^n}{n!}$$

Though I have solved the generating function, I still can't solve explicitly the form of $A_k^n$.

Can anyone help me with it? And is there any other approach other than the generating function method?

P.S. If you would like to know where the partial difference equation comes from, please refer to this article:

Finding nth derivative of the function sec x + tan x and partial difference equation