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It is well-known that the Stirling numbers of the second kind satisfy the following (vertical) recurrence relation:

$$\sum\limits_{r=k}^n \binom{n}{r}S\left( r,k\right) =S\left( n+1,k+1\right) $$

I was wondering if the following sum satisfies a similar relation:

$$\sum\limits_{r=k}^n a^{r} \binom{n}{r}S\left( r,k\right)=f(a,r,k) S\left( n+1,k+1\right),\ \ \ \ 0<a\leq1 $$

I came across this sum when trying to implement the Stirling numbers in an economics model (I'm not a mathematician). I have already had a look at several papers regarding Stirling numbers and Chapter 8 ("Stirling numbers") in "Enumerative Combinatorics" (Charalambides), but without success.

Any help/useful references would be greatly appreciated!

Abe Doe
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  • Such equations are easier to read as displayed equations, which you get by enclosing them in double dollar signs instead of single dollar signs. – joriki Mar 25 '13 at 11:15
  • Thanks for the advice. Reading through the chapter in Enumerative Combinatorics again, I was wondering if the answer to my question might be related to "generalized factorial coefficients"? – Abe Doe Mar 25 '13 at 11:33
  • I asked a question about the same sum here. There are lots of different results about this sum in the question and answers there. – Mike Spivey Oct 30 '13 at 02:51

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