Stirling Number of the Second Kind of n with n-2

Theorem
Let $n \in \Z_{\ge 2}$ be an integer greater than or equal to $2$.

Then:
 * $\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$

where:
 * $\ds {n \brace n - 2}$ denotes an Stirling number of the second kind
 * $\dbinom n 4$ denotes a binomial coefficient.

Proof
The proof proceeds by induction.

Basis for the Induction
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
 * $\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$

$\map P 2$ is the case:

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:
 * $\ds {k \brace k - 2} = \binom {k + 1} 4 + 2 \binom k 4$

from which it is to be shown that:
 * $\ds {k + 1 \brace k - 1} = \binom {k + 2} 4 + 2 \binom {k + 1} 4$

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \forall n \in \Z_{\ge 2}: {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$

Also see

 * Unsigned Stirling Number of the First Kind of n with n-2


 * Particular Values of Stirling Numbers of the Second Kind