How to prove the relationship between Stirling numbers first kind and second kind with negative integral values

Question

I am studying Stirling Numbers recently based on the wikipedia information and I am currently stuck on how to prove this relationship between Stirling Number first kind and second kind with negative value.Based on the table, it is easy to see. But how they can make this conclusion. I have checked a lot of material but still could not find out the solution. Here is the link:https://en.wikipedia.org/wiki/Stirling_number#Stirling_numbers_with_negative_integral_values

Answer 1

This is explained in Concrete Mathematics 2nd ed. Graham, Knuth, Patashnik, pp. 266 & 267: the definition of $x\brace k$ and $x\brack k$ are extended to negative value of $x$ and $k$ by computing them recursively with the same recursions as for positive values of $x$ and $k$:

\begin{align*}{x-1\brace k-1}&= {x \brace k} - k{x-1\brace k}\\{x-1\brack k-1}&= {x \brack k} - (x-1){x-1\brack k}\end{align*} with the initial values (additional stipulations in the above reference) ${0\brace x}={0\brack x}={x\brace 0}={x\brack 0}=0$ for all non-zero values of $x$ and, as usual for $x=0$, ${0\brace 0}={0\brack 0}=1$.

Now, if we replace $x$ (resp. $k$) by $-x+1$ (resp. $-k+1$) in the above recursions, we have: \begin{align*}{-x\brace -k}&= {-x+1 \brace -k+1} - (-k+1){-x\brace -k+1}\\{-x\brack -k}&= {-x+1 \brack -k+1} + x{-x\brack -k+1}\end{align*} that is \begin{align*}{-x\brace - k}+(-k+1){-x\brace -k+1}&= {-x+1 \brace -k+1} \\{-x\brack -k}+ (- x){-x\brack -k+1}&= {-x+1 \brack -k+1} \end{align*}

We then see that ${-x\brace - k}$ (resp. ${-x\brack -k}$) obeys the same recursion as ${k\brack x}$ (resp. ${k\brace x}$). Since they have the same initial values, they must be equal.