Stirling Number of the Second Kind of n with n-3

Theorem
Let $n \in \Z_{\ge 3}$ be an integer greater than or equal to $3$.

Then:
 * $\displaystyle \left\{ {n \atop n - 3}\right\} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$

where:
 * $\displaystyle \left\{ {n \atop n - 3}\right\}$ denotes an Stirling number of the second kind
 * $\dbinom n 6$ denotes a binomial coefficient.

Proof
The proof proceeds by induction.

Basis for the Induction
For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \left\{ {n \atop n - 3}\right\} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$

$P \left({3}\right)$ is the case:

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \left\{ {k \atop k - 3}\right\} = \binom {k + 2} 6 + 8 \binom {k + 1} 6 + 6 \binom k 6$

from which it is to be shown that:
 * $\displaystyle \left\{ {k + 1 \atop k - 2}\right\} = \binom {k + 3} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 1} 6$

Induction Step
This is the induction step:

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

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 3}: \left\{ {n \atop n - 3}\right\} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$

Also see

 * Unsigned Stirling Number of the First Kind of n with n-3


 * Particular Values of Stirling Numbers of the Second Kind