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I am looking for a particular ordinary generating function, if it exists for the Associated Stirling Numbers of the second kind $$b(1;n,j)=b(n,j)=\sum_{k=0}^j(-1)^k\binom{n}{k}S(n-k,j-k)$$
Where $S(n,k)$ are the Stirling Numbers of the Second Kind. I am motivated to take up the search again after reading this page. It mentions that "whenever possible" you can convert the function using a Laplace Transform. I am interested in finding out about this technique and was hoping to see papers or books that deal specifically with this kind of problem.

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  • $\begingroup$ On the l.h.s. there is a function of $k$ and $n$, and on the r.h.s. a function of $j$ and $n$. Could you correct this please ? $\endgroup$ – Tom-Tom Dec 18 '14 at 12:16
  • $\begingroup$ oops! will do... $\endgroup$ – Eleven-Eleven Dec 18 '14 at 13:21
  • $\begingroup$ Are you sure you really need an ordinary generating function ? The binomial coefficient calls for an exponential generating function. $\endgroup$ – Tom-Tom Dec 19 '14 at 8:55
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I would use a variant of the binomial transform and then sum the associated Stirling numbers you define as $$b(n,j) = \sum_{k=0}^{j} \binom{n}{n-j+k} \left\lbrace{n-j+k \atop k}\right\rbrace (-1)^{j-k},$$ using the ordinary generating function for the Stirling numbers of the second kind. Alternately, you can define an exponential generating function for the Stirling numbers in the last sum, multiply by $e^{-z}$ and then perform a Laplace transform to the resulting product to account for the multiplier term of $n!$.

Have you seen this article? It gives an exponential generating function for the $r$-associated Stirling numbers of the second kind. You can also apply a Laplace transform to this to obtain the sum you are looking for,

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