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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then:

$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$

where:

$\ds {n \brace n - 3}$ 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 $\map P n$ be the proposition:

$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$


$\map P 3$ is the case:

\(\ds {3 \brace 0}\) \(=\) \(\ds \delta_{3 0}\) Definition of Stirling Numbers of the Second 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:

$\ds {k \brace k - 3} = \binom {k + 2} 6 + 8 \binom {k + 1} 6 + 6 \binom k 6$


from which it is to be shown that:

$\ds {k + 1 \brace k - 2} = \binom {k + 3} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 1} 6$


Induction Step

This is the induction step:


\(\ds {k + 1 \brace k - 2}\) \(=\) \(\ds \paren {k - 2} {k \brace k - 2} + {k \brace k - 3}\) Definition of Stirling Numbers of the Second Kind
\(\ds \) \(=\) \(\ds \paren {k - 2} \paren {\binom {n + 1} 4 + 2 \binom n 4} + {k \brace k - 3}\) Stirling Number of the Second Kind of n with n-2
\(\ds \) \(=\) \(\ds \paren {k - 2} \paren {\binom {n + 1} 4 + 2 \binom n 4} + \binom {k + 2} 6 + 8 \binom {k + 1} 6 + 6 \binom k 6\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {k - 2} \paren {\dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} } {4!} + 2 \dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4!} }\) Definition of Binomial Coefficient
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \dfrac {\paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 8 \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 6 \dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} \paren {k - 5} } {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2}^2 \paren {\dfrac {\paren {k + 1} + 2 \paren {k - 3} } {4!} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds k \paren {k - 1} \paren {k - 2} \paren {k - 3} \dfrac {\paren {k + 2} \paren {k + 1} + 8 \paren {k + 1} \paren {k - 4} + 6 \paren {k - 4} \paren {k - 5} } {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2}^2 \paren {\dfrac {3 k - 5} {4!} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds k \paren {k - 1} \paren {k - 2} \paren {k - 3} \dfrac {k^2 + 3 k + 2 + 8 \paren {k^2 - 3 k - 4} + 6 \paren {k^2 - 9 k + 20} } {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2}^2 \paren {\dfrac {\paren {3 k - 5} } {4!} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds k \paren {k - 1} \paren {k - 2} 15 \paren {k - 3} \dfrac {k^2 - 5 k + 6} {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2}^2 15 \paren {\dfrac {6 k - 10} {6!} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds k \paren {k - 1} \paren {k - 2}^2 15 \paren {k - 3} \dfrac {k - 3} {6!}\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2}^2 15 \paren {\dfrac {6 k - 10 + k^2 - 6 k + 9} {6!} }\)
\(\ds \) \(=\) \(\ds k \paren {k - 1} \paren {k - 2}^2 15 \paren {\dfrac {k^2 - 1} {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} 15 \paren {\dfrac {\paren {k - 1} \paren {k - 2} } {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {15 k^2 - 45 k + 30} {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {\paren {6 k^2 - 42 k + 72} + \paren {8 k^2 - 8 k - 48} + \paren {k^2 + 5 k + 6} } {6!} }\)
\(\ds \) \(=\) \(\ds \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {\paren {k + 3} \paren {k + 2} + 8 \paren {k + 2} \paren {k - 3} + 6 \paren {k - 3} \paren {k - 4} } {6!} }\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {k + 3} \paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 8 \dfrac {\paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {6!}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 6 \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} } {6!}\)
\(\ds \) \(=\) \(\ds \binom {k + 3} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 1} 6\) Definition of Binomial Coefficient


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


Therefore:

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

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Stirling_Number_of_the_Second_Kind_of_n_with_n-3&oldid=534717"