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Hello,

The Stirling numbers of the second kind count how many ways can a set of $k$ elements be partitioned into $n$ non-empty classes, with $k=n,n+1,\dots$.

My question is: Is there a generalization of these numbers such that the classes are not merely non-empty, but instead occupied to a minimum level with each of the $n$ classes having a minimum of, say, $r$ elements in it? Of course, in this case, $k=rn,rn+1,\dots$.

flag	asked Sep 8 2010 at 11:03 Eduardo Lopez 11●1

Of course one can generalize then in this fashion. Whether such a generalization has been studied is a different question. I think they have but don't have a reference to hand. You might extract some sequences from say the $r=2$ example (say for $n=2$, $n=3$ etc.) and search for them in the OEIS. – Robin Chapman Sep 8 2010 at 11:10

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You can still write down a recurrence relation $$S(n,k,r)=kS(n-1,k,r)+\binom{n-1}{r-1}S(n-r,k-1,r)$$ by observing where the $n$th term can be inserted. – Gjergji Zaimi Sep 8 2010 at 11:36

The exponential formula (en.wikipedia.org/wiki/Exponential_formula) lets you write down a generating function for the numbers $S_X(k,n), the number of partitions of a set of $k$ elements into $n$ classes, where the number of elements of each class belongs to the set $X$ (a subset of the positive integers). Namely $$ \sum_{n,k)S_X(k,n)t^n\frac{x^k}{k!} =\exp t\sum_{i\in X}\frac{x^i}{i!}. $$ – Richard Stanley Sep 8 2010 at 16:18

My previous comment obviously has a bug, such as a missing dollar sign. How does one edit or preview comments? – Richard Stanley Sep 8 2010 at 16:22

Not sure how the person responding edits. I know that when posting I see an edit button. Perhaps you can just copy your response again as a new answer? Thank you – Eduardo Lopez Sep 8 2010 at 16:42

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About generalization of stirling numbers of the second kind

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