Lah numbers and Stirling numbers relation

Question

The generalized Stirling number is $S(n,k;\alpha,\beta,r)$ where the special case $S(n,k;-1,1,0)$ is the Lah number, $$S(n,k;-1,1,0)=\frac{n!}{k!}{{n-1}\choose{k-1}}$$ I wonder if there is any well-known form for $S(n,k;-1,-b,0)$. Even if there is no name for it, is there a clean (factorial-based) representation for it similar to Lah number's?

Answer 1

As per here, this set of generalized Stirling numbers may be given, when $ n \ne 0$ by: $$ S(n,k;-1,-b,0) = \frac{1}{(-b)^k k!} \sum_{j = 0}^n (-1)^j \binom{k}{j} ((j-k)b)^\overline{n} $$ Where $ (\cdot)^\overline{n} $ denotes the rising factorial. Due to the binomial coefficient, when $ j > k $ , the summand is zero. Due to the rising factorial, while $ j \le k $ the summand is zero, thus making the entire sum zero.

When $ n = 0$ : $$ S(0,k;a, b, c) = \delta_{0,k} $$

Therefore, the item is very well known and is commonly used, either $ 1 $ while $ n = k = 0 $ or $ 0 $ otherwise.