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

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

Then:
 * $\ds {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$

where:
 * $\ds {n \brack n - 3}$ denotes an unsigned Stirling number of the first 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 $\map P n$ be the proposition:
 * $\ds {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$

$\map P 3$ 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 $k \ge 3$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds {k \brack k - 3} = \binom k 6 + 8 \binom {k + 1} 6 + 6 \binom {k + 2} 6$

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

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:
 * $\ds \forall n \in \Z_{\ge 3}: {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$

Also see

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


 * Particular Values of Unsigned Stirling Numbers of the First Kind