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
Contents
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:
Needs to be established whether the case $P \left({1}\right)$ actually needs to be proved separately, or whether we can get away with using $P \left({0}\right)$ as our base case.
To discuss it in more detail, feel free to use the talk page.
When this has been done, {{Finish}} should be removed 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 $P \left({1}\right)$ 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 |
Lots to do, not sure if this is a workable approach.
To discuss it in more detail, feel free to use the talk page.
When this has been done, {{Finish}} should be removed 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 $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} \left\{ {m + k \atop k}\right\} = \left[{n \atop n - m}\right]$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(54)$
