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

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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:

\(\ds {3 \brack 0}\) \(=\) \(\ds \delta_{3 0}\) 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 $k \ge 3$, 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:

\(\ds {k + 1 \brack k - 2}\) \(=\) \(\ds k {k \brack k - 2} + {k \brack k - 3}\) Definition of Unsigned Stirling Numbers of the First Kind
\(\ds \) \(=\) \(\ds k \paren {\binom n 4 + 2 \binom {n + 1} 4} + {k \brack k - 3}\) Unsigned Stirling Number of the First Kind of n with n-2
\(\ds \) \(=\) \(\ds k \paren {\binom n 4 + 2 \binom {n + 1} 4} + \binom k 6 + 8 \binom {k + 1} 6 + 6 \binom {k + 2} 6\) Induction Hypothesis
\(\ds \) \(=\) \(\ds k \paren {\dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4!} + 2 \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} } {4!} }\) Definition of Binomial Coefficient
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} \paren {k - 5} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 8 \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 6 \dfrac {\paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2} \paren {\dfrac {k \paren {k - 3} + 2 k \paren {k + 1} } {4!} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds k \paren {k - 1} \paren {k - 2} \paren {k - 3} \dfrac {\paren {k - 4} \paren {k - 5} + 8 \paren {k + 1} \paren {k - 4} + 6 \paren {k + 2} \paren {k + 1} } {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2} \paren {\dfrac {k^2 - 3 k + 2 k^2 + 2 k} {4!} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds k \paren {k - 1} \paren {k - 2} \paren {k - 3} \dfrac {k^2 - 9 k + 20 + 8 \paren {k^2 - 3 k - 4} + 6 \paren {k^2 + 3 k + 2} } {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2} \paren {\dfrac {3 k^2 - k} {4!} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds k \paren {k - 1} \paren {k - 2} 15 \paren {k - 3} \dfrac {k^2 - k} {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2} 15 \paren {\dfrac {6 k^2 - 2 k + k^3 - 4 k^2 + 3 k} {6!} }\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2} 15 \paren {\dfrac {k^3 + 2 k^2 + k} {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} 15 \paren {\dfrac {k^2 + k} {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {15 k^2 + 15 k} {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {\paren {k^2 - 7 k + 12} + \paren {8 k^2 - 8 k - 48} + \paren {6 k^2 + 30 k + 36} } {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {\paren {k - 3} \paren {k - 4} + 8 \paren {k + 2} \paren {k - 3} + 6 \paren {k + 3} \paren {k + 2} } {6!} }\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 8 \dfrac {\paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 6 \dfrac {\paren {k + 3} \paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} } {6!}\)
\(\ds \) \(=\) \(\ds \binom {k + 1} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 3} 6\) Definition of Binomial Coefficient


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


Therefore:

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

$\blacksquare$


Also see


Sources

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