Stirling Number of the Second Kind of n+1 with 2

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Theorem

Let $n \in \Z_{\ge 0}$.

Then:

$\ds {n + 1 \brace 2} = 2^n - 1$

where $\ds {n + 1 \brace 2}$ 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:

$\ds {n + 1 \brace 2} = 2^n - 1$


Basis for the Induction

$\map P 0$ is the case:

\(\ds {1 \brace 2}\) \(=\) \(\ds \delta_{1 2}\) Stirling Number of the Second Kind of 1
\(\ds \) \(=\) \(\ds 0\) Definition of Kronecker Delta
\(\ds \) \(=\) \(\ds 2^0 - 1\) Definition of Kronecker Delta

So the result holds for $\map P 0$.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds {k + 1 \brace 2} = 2^k - 1$


from which it is to be shown that:

$\ds {k + 2 \brace 2} = 2^{k + 1} - 1$


Induction Step

This is the induction step:


\(\ds {k + 2 \brace 2}\) \(=\) \(\ds 2 \times {k + 1 \brace 2} + {k + 1 \brace 1}\) Definition of Stirling Numbers of the Second Kind
\(\ds \) \(=\) \(\ds 2 \times {k + 1 \brace 2} + 1\) Stirling Number of the Second Kind of n+1 with 1
\(\ds \) \(=\) \(\ds 2 \times {2^k - 1} + 1\) Induction Hypothesis
\(\ds \) \(=\) \(\ds 2^{k + 1} - 2 + 1\)
\(\ds \) \(=\) \(\ds 2^{k + 1} - 1\)


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


Therefore:

$\ds \forall n \in \Z_{\ge 0}: {n + 1 \brace 2} = 2^n - 1$

$\blacksquare$


Also see


Sources

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