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

$\displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} \left\{ {m + k \atop k}\right\} = \left[{n \atop n - m}\right]$

where:

$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
$\displaystyle \left\{ {m + k \atop k}\right\}$ denotes a Stirling number of the second kind
$\displaystyle \left[{n \atop n - m}\right]$ 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 $P \left({n}\right)$ be the proposition:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} \left\{ {m + k \atop k}\right\} = \left[{n \atop n - m}\right]$


$P \left({0}\right)$ is the case:

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

So $P \left({0}\right)$ is seen to hold.


Basis for the Induction

$P \left({1}\right)$ is the case:



Thus $P \left({1}\right)$ is seen to hold for all $m$.


This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

$\displaystyle \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} \left\{ {r + k \atop k}\right\} = \left[{n \atop n - r}\right]$


from which it is to be shown that:

$\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} \left\{ {r + 1 + k \atop k}\right\} = \left[{n \atop n - r + 1}\right]$


Induction Step

This is the induction step:

\(\displaystyle \) \(\) \(\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} \left\{ {r + 1 + k \atop k}\right\} = \left[{n \atop n - r + 1}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \left({\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} }\right) \binom {r + 1 + n} {n + k} \left\{ {r + 1 + k \atop k}\right\} = \left[{n \atop n - r + 1}\right]\) Pascal's Rule



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


Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} \left\{ {m + k \atop k}\right\} = \left[{n \atop n - m}\right]$

$\blacksquare$


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