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I am trying to proof some identies from concrete mathmatics p 265. But i cant get nowhere. What i have found out its a recurrence, in the Stirling Numbers Triangle, its a vertical one. I get what it means, e.g. n=4 m=2 the Number of Partitioning 5 Elements into 4 Partitions ${5 \brace 3}$ can be created from from $ {2 \brace 2}*(2+1)^2 + {3 \brace 2}*(2+1)^1 + {4 \brace 2}(2+1)^0 $ but i cant interprete combinatorically the $(2+1)^2,(2+1)^1, (2+1)^0$. Any ideas. It sould also work with Induction but,....well there are 2 pages of identities, i am not sure what to choose.

thetha
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  • What does ${x\brace y}$ mean for some numbers $x$ and $y$? This is not a rhetorical question --- I am just a noob and am curious. Please explain to me. – Mr Pie May 06 '18 at 11:30
  • its a little more complex: https://en.wikipedia.org/wiki/Stirling_number – thetha May 06 '18 at 11:31
  • @thetha With such a broad question, pointing you to Wikipedia was the thing I was about to do. You are basically asking: "How can I solve equations involving Stirling numbers?" without further detail. There is no generic way (well, apart from the techniques introduced in the book A=B perhaps, you may like to have a look at it). – SK19 May 06 '18 at 11:34
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    maybe also see https://math.stackexchange.com/questions/742083/book-on-combinatorial-identities/748850#748850 – SK19 May 06 '18 at 11:36
  • @thetha Oh wait, you want to explicitly show the identity in your title, don't you? – SK19 May 06 '18 at 11:37
  • basically there are two methodes: algebraic and combinatorical(counting two ways or something equivalent),...and something with generating functions, but i have now clue yet on this topic – thetha May 06 '18 at 11:42

2 Answers2

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Starting from the LHS we write with $n\ge m$ and an Iverson bracket

$$\sum_{j=m}^n {j\brace m} (m+1)^{n-j} = \sum_{j\ge m} {j\brace m} (m+1)^{n-j} [[j\le n|j\ge 0]] \\ = \sum_{j\ge m} {j\brace m} (m+1)^{n-j} [z^n] \frac{z^j}{1-z} \\ = [z^n] \frac{1}{1-z} \sum_{j\ge m} {j\brace m} (m+1)^{n-j} z^j \\ = [z^n] \frac{(m+1)^n}{1-z} \sum_{j\ge m} {j\brace m} \frac{z^j}{(m+1)^j}.$$

Recall the OGF

$$\sum_{j\ge m} {j\brace m} x^j = \prod_{r=1}^m \frac{x}{1-rx}$$

which yields for the present case

$$(m+1)^n [z^n] \frac{1}{1-z} \prod_{r=1}^m \frac{z/(m+1)}{1-rz/(m+1)} \\ = [z^n] \frac{1}{1-(m+1)z} \prod_{r=1}^m \frac{z}{1-rz} = [z^{n+1}] \frac{z}{1-(m+1)z} \prod_{r=1}^m \frac{z}{1-rz} \\ = [z^{n+1}] \prod_{r=1}^{m+1} \frac{z}{1-rz} = {n+1\brace m+1}.$$

This is the claim.

Marko Riedel
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  • Highly instructive and extremely elegant transformation. A pleasure to go through this answer. (+1) – Markus Scheuer May 07 '18 at 17:30
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    Thank you very much. I noticed it was special while I was writing it. – Marko Riedel May 07 '18 at 18:00
  • I took the liberty to take your answer as nice example to answer this question. If you don't like it, I will remove it of course. Best regards, – Markus Scheuer May 09 '18 at 21:17
  • I find that these examples are quite diverse and the above is rather different from the rest. This seems a matter of opinion. Note that the Iverson bracket $[[j\le n|j\ge 0]]$ is not the same as $[[0\le j\le n]]$ because the power series also includes a term for $j\lt 0.$ So it may only be used if it is known that $j\ge 0$ (conditional Iverson bracket). – Marko Riedel May 09 '18 at 21:41
  • Many thanks for the hint, Marko. I've corrected the usage of the Iverson brackets. – Markus Scheuer May 10 '18 at 05:43
  • could you give some information how did you choose the function? – thetha May 10 '18 at 08:04
  • How can one see that that's Function?Do one need to know a lot of them? – thetha May 10 '18 at 11:42
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Let see, $n+1\brace m+1$ means that you take $n+1$ elements and you partition them in $n+1$ disjoint and non empty blocks. Denote the partitions in the following way $${[n+1]\brace m+1}=\{\pi \vdash [n+1]:|\pi|=m+1\},$$ the set of partitions. Given a partition $\pi ,$ let $j$ the maximum integer such that there is not block $B\in \pi$ such that $\{n+1,k\}\subseteq B,$ hence $\pi = (\pi _{[j]},f:[n]\setminus [j+1]\longrightarrow [m+1]),$ where $\pi _{[j]}$ is the partition $\pi$ but restricted to the set $[j]$ =because we know that $j+1$ has to be in the same block as $n+1$ and the function sends an element to one of the blocks(notice that blocks can be ordered by lexicografic order.) and hence $${[n+1]\brace m+1} = \bigcup _{j = m}^{n-1}{[j]\brace m}\times [m+1]^{n-(j+1)}$$

Phicar
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  • I am not sure about the notation: $ \pi \vdash [n+1]$ what doest this means? Is : $ {[n+1]\brace m+1}= {n+1\brace m+1} $ Is there extra meaning in the [ n+1]? – thetha May 06 '18 at 18:04
  • $\pi \vdash [n+1]$ means that $\pi = {B_1,\cdots ,B_k}\subseteq 2^{[n+1]}$ such that $\cup B_i = [n+1]$ and $B_i\cap B_j = \emptyset$ for $i\neq j$ – Phicar May 06 '18 at 22:12