Stirling Number of the Second Kind of n+1 with 1
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Theorem
Let $n \in \Z_{\ge 0}$.
Then:
- $\ds {n + 1 \brace 1} = 1$
where $\ds {n + 1 \brace 1}$ denotes a Stirling number of the second kind.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds {n + 1 \brace 1} = 1$
Basis for the Induction
$\map P 0$ is the case:
\(\ds {1 \brace 1}\) | \(=\) | \(\ds \delta_{1 1}\) | Stirling Number of the Second Kind of 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Definition of Kronecker Delta |
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\ds {k + 1 \brace 1} = 1$
from which it is to be shown that:
- $\ds {k + 2 \brace 1} = 1$
Induction Step
This is the induction step:
\(\ds {k + 2 \brace 1}\) | \(=\) | \(\ds 1 \times {k + 1 \brace 1} + {k + 2 \brace 0}\) | Definition of Stirling Numbers of the Second Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds {k + 1 \brace 1} + 0\) | Stirling Number of the Second Kind of n+1 with 0 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Induction Hypothesis |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 0}: {n + 1 \brace 1} = 1$
$\blacksquare$
Also see
- Unsigned Stirling Number of the First Kind of n+1 with 1
- Signed Stirling Number of the First Kind of n+1 with 1
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(50)$