Stirling Number of the Second Kind of Number with Self

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Theorem

$\displaystyle {n \brace n} = 1$

where $\displaystyle {n \brace n}$ denotes a Stirling number of the second kind.


Proof

The proof proceeds by induction.


For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\displaystyle {n \brace n} = 1$


Basis for the Induction

$\map P 0$ is the case:

\(\displaystyle {0 \brace 0}\) \(=\) \(\displaystyle \delta_{0 0}\) Stirling Number of the Second Kind of 0
\(\displaystyle \) \(=\) \(\displaystyle 1\) 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:

$\displaystyle {k \brace k} = 1$


from which it is to be shown that:

$\displaystyle {k + 1 \brace k + 1} = 1$


Induction Step

This is the induction step:


\(\displaystyle {k + 1 \brace k + 1}\) \(=\) \(\displaystyle \paren {k + 1} {k \brace k + 1} + {k \brace k}\) Definition of Stirling Numbers of the Second Kind
\(\displaystyle \) \(=\) \(\displaystyle \paren {k + 1} \times 0 + {k \brace k}\) Stirling Number of Number with Greater
\(\displaystyle \) \(=\) \(\displaystyle {k \brace k}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\) Induction Hypothesis


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: {n \brace n} = 1$

$\blacksquare$


Also see


Sources