Sum over k of Stirling Numbers of Second Kind by x^k
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Theorem
| \(\ds \sum_k {k \brace n} z^k\) | \(=\) | \(\ds \dfrac {z^n} {\prod \limits_{k \mathop = 1}^n \paren {1 - k n} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {z^n} {\paren {1 - z} \paren {1 - 2 z} \cdots \paren {1 - n z} }\) |
where:
- $\ds {k \brace n}$ denotes a Stirling number of the second kind.
Proof
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Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.9$: Generating Functions: $(28)$