Timeline for Closed form of $ \sum_{i=1}^{n-k} {n-1-i\choose k-1}i^a + \sum_{i=1}^k {n-1-i\choose n-1-k}$
Current License: CC BY-SA 4.0
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| Jun 15, 2021 at 19:15 | comment | added | Ira Gessel | If you multiply the first sum by $\frac{x^a}{a!}y^nz^k$ and sum on $a$, $n$, and $k$ (starting the sum with $i=0$ to make things a little simpler), you get $$\frac{1-y}{(1-y-yz)(1-ye^x)}$$ from which you can derive whatever explicit formulas exist. When you expand in powers of $x$ you will get Stirling numbers of the second kind. | |
| Jun 15, 2021 at 3:28 | comment | added | Benjamin L. Warren | What about if k was fixed and a was free variable? | |
| Jun 15, 2021 at 2:05 | history | answered | Ira Gessel | CC BY-SA 4.0 |