Sum over k of m-n choose m+k by m+n choose n+k by Unsigned Stirling Number of the First Kind of m+k with k
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Theorem
Let $m, n \in \Z_{\ge 0}$.
- $\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
where:
- $\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
- $\ds {m + k \brack k}$ denotes an unsigned Stirling number of the first kind
- $\ds {n \brace n - m}$ denotes a Stirling number of the second kind.
Proof
The proof proceeds by induction on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\forall n \in \Z_{\ge 0}: \ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
Basis for the Induction
$\map P 0$ is the case:
\(\ds \) | \(\) | \(\ds \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brack k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \binom {- n} k \binom n {n + k}\) | Unsigned Stirling Number of the First 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds {n \brace n - 0}\) | Stirling Number of the Second Kind of Number with Self |
So $\map P 0$ is seen to hold.
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:
- $\ds \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brack k} = {n \brace n - r}$
from which it is to be shown that:
- $\ds \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_k \paren {\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} } \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}\) | 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:
- $\forall m, n \in \Z_{\ge 0}: \ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(54)$