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

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Theorem

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

Then:

$\ds {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$

where:

$\ds {n \brack n - 2}$ denotes an unsigned Stirling number of the first 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:

$\ds {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$


$\map P 2$ is the case:

\(\ds {2 \brack 0}\) \(=\) \(\ds \delta_{2 0}\) Definition of Unsigned Stirling Numbers of the First Kind
\(\ds \) \(=\) \(\ds 0\) 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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds {k \brack k - 2} = \binom k 4 + 2 \binom {k + 1} 4$


from which it is to be shown that:

$\ds {k + 1 \brack k - 1} = \binom {k + 1} 4 + 2 \binom {k + 2} 4$


Induction Step

This is the induction step:


\(\ds {k + 1 \brack k - 1}\) \(=\) \(\ds k {k \brack k - 1} + {k \brack k - 2}\) Definition of Unsigned Stirling Numbers of the First Kind
\(\ds \) \(=\) \(\ds k \binom k 2 + {k \brack k - 2}\) Unsigned Stirling Number of the First Kind of n with n-1
\(\ds \) \(=\) \(\ds k \binom k 2 + \binom k 4 + 2 \binom {k + 1} 4\) Induction Hypothesis
\(\ds \) \(=\) \(\ds k \frac {k \paren {k - 1} } 2 + \dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4!} + \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} } {4!} + \binom {k + 1} 4\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds \binom {k + 1} 4 + k \paren {k - 1} \frac {12 k + \paren {k - 2} \paren {k - 3} + \paren {k + 1} \paren {k - 2} } {4!}\)
\(\ds \) \(=\) \(\ds \binom {k + 1} 4 + k \paren {k - 1} \frac {12 k + k^2 - 5 k + 6 + k^2 - k - 2} {4!}\)
\(\ds \) \(=\) \(\ds \binom {k + 1} 4 + k \paren {k - 1} \frac {2 \paren {k^2 + 3 k + 2} } {4!}\)
\(\ds \) \(=\) \(\ds \binom {k + 1} 4 + \frac {2 \paren {k + 2} \paren {k + 1} k \paren {k - 1} } {4!}\)
\(\ds \) \(=\) \(\ds \binom {k + 1} 4 + 2 \binom {k + 2} 4\) Definition of Binomial Coefficient


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


Therefore:

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

$\blacksquare$


Also see


Sources

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