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Show that $$ \sum_{r=1}^{n-1}\binom{n-2}{r-1}r^{r-1}(n-r)^{n-r-2}= n^{n-2} $$

I don't know whether such identity already exists, or has been posted here before. I discovered this identity while solving a problem but nor able to prove. And if the problem is correct then this must have to be true. Its coming out to be true for $n$ upto $10$. So please prove or disprove.( I am quite sure it is true)

Pedro
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Prahlad
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    Are you sure this isn't just the binomial theorem in disguise? – Alexander Sep 05 '13 at 17:05
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    @Alexander If it was $x^{r-1}(n-x)^{n-r-2}$, then probably – Dennis Meng Sep 05 '13 at 17:06
  • Just a remark... When $r=n-1$, the exponent in $(n-r)^{n-r-2}$ equals $-1$. – Mike F Sep 05 '13 at 17:24
  • @Mike counter-remark: $n - r$ would be $1$ – Dennis Meng Sep 05 '13 at 17:29
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    @dennis Meng I think in fact he just wanted to say we have $$\sum_{r=1}^{n-2} \binom{n-2}{r-1} r^{r-1} \cdot (n-r)^{n-r-2} + \binom{n-1}{n-2} \cdot (n-1)^{(n-2)}$$ – Dominic Michaelis Sep 05 '13 at 17:32
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    The RHS is the number of trees on $n$ nodes, and likewise the last item in each term of the sum is the number of trees on $n-r$ nodes. There may be a decompositional approach based on cutting trees along some specific edge(s) and counting by the largest piece left over... – Steven Stadnicki Sep 05 '13 at 17:35
  • @prahlad well it may be helpful if we know your original problem, so maybe we got somewhere to start – Dominic Michaelis Sep 05 '13 at 17:38
  • I tested it up to $n=1000$ and it is only wrong for $n=1$ – Dominic Michaelis Sep 05 '13 at 17:40
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    @DominicMichaelis My guess is he's trying to prove this – Dennis Meng Sep 05 '13 at 17:41
  • This looks annoyingly easy to prove by induction or even by applying the binomial theorem on $\left(\cfrac{n}2+\cfrac{n}2\right)^{n-2}$, but darn it gets messy. – Daniel R Sep 06 '13 at 06:58
  • @Daniel R: The binomial theorem would give something like $\sum_r {n-2 \choose r} \left(\frac{n}{2}\right)^{r} \left(\frac{n}{2}\right)^{n-2-r}$, I don't see how you could make the bases dependent on $r$. A straightforward inductive proof might also run into problems, since there's little connection between $n^{n-2}$ and $(n+1)^{n-1}$. – sdcvvc Sep 06 '13 at 12:39
  • @sdcvvc: Yep, that's what I figured out too, but it took me 20 minutes and two sheets of paper. :) – Daniel R Sep 06 '13 at 12:40
  • See also http://math.stackexchange.com/questions/347124 – sdcvvc Sep 06 '13 at 13:00
  • I could have sworn a straightforward proof by induction would do this, but I count 8 sheets of paper still crumpled on my floor from when the question was asked trying to connect the inductive hypothesis to the inductive step to no avail. I stopped due to lack of time in dejection but will try again eventually. I did make a quickie program in Mathematica to test this for values, and could find no contradiction. This is actually a good proof to write up. I am actually a bit shocked that I could not make the connection. – J. W. Perry Sep 07 '13 at 04:57
  • That's called Abel[-Hurwitz] identity, I believe. – Grigory M Jun 27 '15 at 22:00

2 Answers2

6

Count how many are there triples $(T,e,x)$ where $T$ is a tree on $\{1,2,\dots,n\}$, $e$ is an edge of $T$ and $x$ is a vertex of $T$.

Way 1: $n^{n-2} n (n-1)$ by Cayley's formula.

Way 2:

The edge $e$ splits the tree into two parts, left $L$ and right $R$. We will say $R$ is the one that contains $x$.

First, select two endpoints of $e$: $a \in L$ and $b \in R$, this can be done in $n(n-1)$ ways.

Let $r$ be the cardinality of $R$. Select which of $r$ numbers in $\{1,2,\dots,n\}$ belong to $R$. We already committed to $a \in L$ and $b \in R$; this means $n-2 \choose r-1$.

There are $r$ ways to choose $x$, $r^{r-2}$ ways to choose $R$ and $(n-r)^{n-r-2}$ ways to choose $L$.

Therefore

$n^{n-2} n(n-1) = n(n-1) \sum_{r=1}^{n-1} {n-2 \choose r-1} r \cdot r^{r-2} (n-r)^{n-r-2}$ QED

sdcvvc
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1

Suppose we are trying to show that $$\sum_{r=1}^{n-1} {n-2\choose r-1} r^{r-1} (n-r)^{n-r-2} = n^{n-2}$$ where $n\ge 2.$

Looking to the combinatorial proof for inspiration we present the labelled tree function $T(z)$ with functional equation $$\mathcal{T} = \mathcal{Z} \times \mathfrak{P}(\mathcal{T}) \quad\text{or}\quad T(z) = z \exp T(z)$$

which counts the number of labelled rooted trees on $n$ nodes so that $$n! [z^n] T(z) = n^{n-1}.$$

We now re-write the equation to better match the tree function. Start with $$\sum_{r=0}^{n-2} {n-2\choose r} (r+1)^r (n-r-1)^{n-r-3} = n^{n-2}$$ and continue with $$\sum_{r=0}^{n} {n\choose r} (r+1)^r (n+1-r)^{n-r-1} = (n+2)^n.$$

Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$ i.e. the product of the two generating functions is the generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$

In the present case we have $$A(z) = T'(z)$$ by inspection. Note also that $$T(z) = \sum_{n\ge 1} n\times n^{n-2} \frac{z^n}{n!} = \sum_{n\ge 1} n^{n-2} \frac{z^n}{(n-1)!} = z \sum_{n\ge 1} n^{n-2} \frac{z^{n-1}}{(n-1)!}.$$

It follows that $$B(z) = \sum_{n\ge 0} (n+1)^{n-1} \frac{z^{n}}{n!} = \frac{1}{z} T(z).$$

We thus obtain that $$A(z) B(z) = \frac{1}{z} T(z) T'(z).$$

Observe that when we differentiate the functional equation we obtain $$T'(z) = \exp T(z) + z \exp T(z) \times T'(z) = \frac{1}{z} T(z) + T(z) \times T'(z)$$ and hence $$T'(z) = \frac{1}{z} \frac{T(z)}{1-T(z)}.$$

This yields $$A(z) B(z) = \frac{1}{z^2} \frac{T^2(z)}{1-T(z)}.$$

Now to extract coefficients from this we have $$n! [z^n] A(z) B(z) = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{z^2} \frac{T^2(z)}{1-T(z)} \; dz \\ = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+3}} \frac{T^2(z)}{1-T(z)} \; dz.$$

Using a type of Lagrange inversion and putting $w=T(z)$ we get from the functional equation $w = z \exp w$ or $z = w \exp(-w)$ and hence $dz = (\exp(-w) - w\exp(-w)) \; dw$ or $dz = \exp(-w) ( 1 - w ) \; dw,$ yielding for the integral

$$\frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(n+3))}{w^{n+3}} \frac{w^2}{1-w} \exp(-w)(1-w) \; dw \\ = \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(n+2))}{w^{n+1}} \; dw.$$

Extracting the residue we find $$n! \times \frac{(n+2)^{n}}{n!} = (n+2)^n$$

as claimed, QED.

Marko Riedel
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