I encounter the following recurrence relations when trying to find a general formula for the derivative of $f(x) = (1 + e^x)^{-1}$ ,
for $n\geq 0$,
\begin{align*} a(0,1) &= 1,\\ a(n,0) &= 0, \quad \\ a(n+1,k) &= (k - 1)a(n, k - 1) - a(n,k),\quad 1 \leq k \leq n+1, \\ a(n + 1,n+2) &= (n+1)a(n,n+1), \end{align*} For $n \geq 0$, according to that post, the solution seems to be $$a(n,k) = (-1)^n \sum_{j = 0}^{k - 1}(-1)^j\binom{k - 1}{j}(j+1)^n,\quad 0 \leq k \leq n+1$$ but I don't know how to get there. I never encounter recurrence relation of this form before.
So here are my questions.
- Is there any method to solve this type of recurrence? If any, would you mind to tell me the book/reference on this topic might be.
- I'm not sure about this, but is the solution to the problem unique? I think it is, by tring to apply the relations.
Most importantly, how to find the solution above? Any help would be appreciated.
If I had to solve this (I don't have to since you are asking for a method, not a solution), I would use a two-variable generating function In the form
$A(x, y) =\sum_{i=0}^{\infty}\sum_{j=0}^{\infty} a(i,j)x^iy^j $
and see what properties of A follow from the recurrence.