Signed Stirling Number of the First Kind of n+1 with 0
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Theorem
Let $n \in \Z_{\ge 0}$.
Then:
- $\map s {n + 1, 0} = 0$
where $\map s {n + 1, 0}$ denotes a signed Stirling number of the first kind.
Proof
By definition of signed Stirling number of the first kind:
- $\map s {n, k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus
| \(\ds n\) | \(\ge\) | \(\ds 0\) | by hypothesis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds n + 1\) | \(>\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds n + 1\) | \(\ne\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \delta_{\paren {n + 1} 0}\) | \(=\) | \(\ds 0\) |
Hence the result.
$\blacksquare$