Generating Function for 2-Associated Stirling Numbers of the Second Kind

Question

I am looking for a paper which explicitly defines a power series for 2-associated Stirling Numbers of the Second Kind. The paper defines the generating function as follows: Let $S_2(n,k)=b(n,k)$ be the 2 associated Stirling Numbers of the second kind. Then $$\sum_{k,n\ge1}b(n,k)u^k\frac{t^n}{n!}=\exp{(u\exp{(1-Q(2,t)}})$$ where $Q(2,t)=\frac{\Gamma(2,t)}{\Gamma(2)}$ I found the paper, and never wrote down the author or journal. Any help would be amazing!

EDIT: I am not having much luck with help here. Perhaps, instead of finding the actual article itself, outside of just google searching (which is how i found it, I just can't seem to enter the same search and have the same luck now), what might be some strategies to finding such an article? I don't know the author, I don't know the publication, I only know subject matter. I am a first year grad student, and am not accustomed to finding mathematical journal articles except Googling and praying...

Answer 1

Note: I think @jayakrishnan provides an interesting reference about the 2-associated Stirling Numbers of the second kind from F.T.Howard.

Here's another paper which looks promising. The idea is, that even if it's not exactly the article you have in mind the formulae of $S_2(n,k)$ you will find in the referred papers could be nevertheless valuable and a proper substitute for the formulae you want to find.

• So you might find Some Properties of Associated Stirling Numbers (2008) from Feng-Zhen Zhao interesting.

It presents (besides other numbers) r-associated Stirling numbers of the second kind $S_r(n,k)$: $$\sum_{n=k}^{\infty}S_r(n,k)\frac{t^n}{n!} =\frac{1}{k!}\left(e^t-\sum_{j=0}^{r-1}\frac{t^j}{j!}\right)^k$$ and you will find using the definition $$\sum_{n=k}^{\infty}S_2(n,k)\frac{t^n}{n!} =\frac{(e^t-1-t)^k}{k!}$$ some interesting identities about $S_2(n,k)$. Maybe a recurrence formula (2.2) of $S_2(n,k)$ with the help of Bernoulli numbers is useful:

$$\sum_{j=0}^n\binom{n+k-1}{j}S_2(n-j+k,k)B_j=(n+k-1)S_2(n+k-2,k-1)\qquad k\geq 2$$

or some asymptotic expansions of $S_2(n,k)$ at the end of the paper.

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Added 2014-12-03: With respect to the comment of @Eleven-Eleven I have added another hint which could give some more insights.

Regarding associated Stirling numbers of the second kind you can find in Analytic Combinatorics from Flajolet and Sedgewick interesting concepts and some nice examples. You may have a look at chapter II: Labelled Structures and Exponential Generating Functions.

You can read in Section II.3: Surjections, set partitions, and words:

• Example II.4 (13) (p. 124): Stirling numbers of the second kind

The Stirling numbers of the second kind, $S_n^{(k)}$, denoting the number of ways of partitioning the set $[1..n]$ into $k$ disjoint and non-empty equivalence classes (called blocks) can be modelled by the symbolic structure:

$$S^{(k)}=\text{SET}_k\left(\text{SET}_{\leq 1}(\mathcal{Z})\right)$$

which directly translates into the exponential generating function (EGF) for $S_n^{(k)}$

$$S^{(k)}(z)=\sum_{n\geq k}S_n^{(k)}\frac{z^n}{n!}=\frac{1}{k!}\left(e^z-1\right)^k$$

• Example II.6: Smallest and largest blocks in set partitions.

Let $e_b(z)$ denote the truncated exponential function:

$$e_b(z) := 1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\frac{z^b}{b!}$$

The EGFs $S^{(\leq b)}(z)$ and $S^{(>b)}(z)$ with

$$S^{(\leq b)}(z)=\exp\left(e_b(z)-1\right)\qquad \qquad S^{(>b)}(z)=\exp\left(e^z-e_b(z)\right)$$

correspond to partitions with all blocks of size $\leq b$ and all blocks of size $>b$, respectively.

• Example II.7: The EGF of partitions without singleton parts is

$$e^{e^z-1-z}$$

• Example II.8: Restricted words. The EGF of words over an alphabet of cardinality $k$ containing each letter at most $b$ times, and that of words containing each letter more than $b$ times are

$$\mathcal{W}^{(\leq b)}(z)=e_b(z)^k,\qquad \qquad\mathcal{W}^{(> b)}(z)=\left(e^z-e_b(z)\right)^k$$

... and some more examples

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Note: Observe, that setting $b=1$ results in the 2-associated Stirling numbers of the second kind

$$S_2(n,k)=\frac{1}{k!}\left(e(z)-e_b(z)\right)^k$$

Answer 2

I am not sure if this will be helpful or not. You might want to check out the following link, particularly, the reference list in the end :
http://www.fq.math.ca/Scanned/18-4/howard.pdf