Sum over k of m-n choose m+k by m+n choose n+k by Stirling Number of the Second Kind of m+k with k

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Theorem

Let $m, n \in \Z_{\ge 0}$.

$\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$

where:

$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
$\ds {m + k \brace k}$ denotes a Stirling number of the second kind
$\ds {n \brack n - m}$ denotes an unsigned Stirling number of the first kind.


Proof

The proof proceeds by induction on $m$.


For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$


$\map P 0$ is the case:

\(\ds \) \(\) \(\ds \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brace k}\)
\(\ds \) \(=\) \(\ds \sum_k \binom {- n} k \binom n {n + k}\) Stirling Number of the Second Kind of Number with Self
\(\ds \) \(=\) \(\ds \sum_k \binom {- n} k \delta_{0 k}\) as $\dbinom n {n + k} = 0$ for $k = 0$, and Binomial Coefficient with Self
\(\ds \) \(=\) \(\ds \binom {- n} 0\) All terms but where $k = 0$ vanish
\(\ds \) \(=\) \(\ds 1\) Binomial Coefficient with Zero
\(\ds \) \(=\) \(\ds {n \brack n - 0}\) Unsigned Stirling Number of the First Kind of Number with Self

So $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:


This needs considerable tedious hard slog to complete it.
In particular: Needs to be established whether the case $\map P 1$ actually needs to be proved separately, or whether we can get away with using $\map P 0$ as our base case.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


Thus $\map P 1$ is seen to hold for all $m$.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.


So this is the induction hypothesis:

$\ds \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brace k} = {n \brack n - r}$


from which it is to be shown that:

$\ds \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}$


Induction Step

This is the induction step:

\(\ds \) \(\) \(\ds \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}\)
\(\ds \) \(=\) \(\ds \sum_k \paren {\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} } \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}\) Pascal's Rule


This needs considerable tedious hard slog to complete it.
In particular: Lots to do, not sure if this is a workable approach.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


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


Therefore:

$\ds \forall m, n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$

$\blacksquare$


Sources

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