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How can I prove the identity

$$2(n-1)n^{n-2}=\sum_k\binom{n}{k}k^{k-1}(n-k)^{n-k-1}?$$

I know that the number of trees on $n$ vertices is $n^{n-2}$, and that a tree with $n$ vertices has $n-1$ edges, but I don't know how to continue. Could you help me please?

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The right hand side strongly suggests using the binomial theorem. Are you looking for a combinatorial proof instead? –  Srivatsan Sep 19 '11 at 0:00
    
@Sri: Note the bases are in terms of $k$ in the summation, so applying the binomial theorem won't work I guess. –  TMM Sep 19 '11 at 0:04
    
@Alex: Do you have any more context for this problem? For instance, it is not clear from the equality that it has anything to do with graphs or trees, but apparently you have reason to think it does. Is this question related to any specific kind of graphs/trees? –  TMM Sep 19 '11 at 0:09

2 Answers 2

up vote 6 down vote accepted

The OP is in the correct direction. By Cayley's formula, the number of labeled trees on $n$ vertices is $n^{n-2}$. We will count the number of labelled trees in a different way and get the equality.

Take a tree $T$ and an edge $e \in T$. Then $T \setminus \{ e \}$ contains two components, say of sizes $k$ and $n-k$. Also, each of the components is a tree by itself. With this in mind, we build the tree $T$ as follows. Partition the vertices into two parts, of sizes $k$ and $n-k$ respectively. Draw trees $T_1$ and $T_2$ on the two parts. Then pick an arbitrary vertex from each of the parts and join the pair by an edge.

The number of ways of doing this procedure is: $$ \sum_{k=1}^{n-1} \binom{n}{k} \times k^{k-2} (n-k)^{n-k-2} \times (k (n-k)) = R, $$ the right hand side of the purported equation. The binomial coefficient arises from choosing $k$ vertices. The second factor is by applying the Cayley's formula for the number of trees on $k$ and $n-k$ vertices. The $k(n-k)$ factor is the number of ways to connect these components by a single edge.

Now, this procedure overcounts the number of trees. Specifically, each tree can be cut in any of its $n-1$ edges to give a pair of components. (That's a factor of $n-1$.) Further, in selecting $k$ vertices to make two components of sizes $k$ and $n-k$, we distinguish the two components (one that is selected, and one that is not). So we should further divide the number we got by $2$.

Thus the number of labelled trees on $n$ vertices is $$ \frac{R}{2(n-1)}. $$ Equating it to $n^{n-2}$, we get the claim.

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The Lambert function has the following Maclaurin series:

$$-W(-x)=\sum_{k=1}^\infty \frac{k^{k-1}}{k!} x^k$$

(In some references, $-W(-x)$ is referred to as the "tree function", $T(x)$, as it is a generating function for rooted labeled trees.)

This series can be derived through Lagrangian inversion. In particular, the coefficient of $x^k$ in the power series of the tree function is given by the expression

$$\frac1{k!}\left.\dfrac{\mathrm d^{k-1}}{\mathrm dt^{k-1}}\left(\frac{t}{t\exp(-t)}\right)^k\right\vert_{t=0}=\frac{k^{k-1}}{k!}$$

The original equation can be re-expressed as

$$\frac{2(n-1)n^{n-2}}{n!}=\sum_{k=1}^{n-1} \frac{k^{k-1}}{k!}\frac{(n-k)^{n-k-1}}{(n-k)!}$$

The right hand side is the autoconvolution of the sequence $\dfrac{k^{k-1}}{k!}$; its ordinary generating function is thus the square of the generating function of $\dfrac{k^{k-1}}{k!}$, which is $(-W(-x))^2=W(-x)^2$.

To find an expression for the series coefficient of $W(-x)^2$, we consider first the related function $\left(\dfrac{W(-x)}{-x}\right)^2-1=\exp(-2\,W(-x))-1$. This function is the inverse of $\dfrac{\log\sqrt{1+x}}{\sqrt{1+x}}$. Applying Lagrangian inversion to $\dfrac{\log\sqrt{1+x}}{\sqrt{1+x}}$ and using the results obtained in this answer, we obtain the series

$$\left(\frac{W(-x)}{-x}\right)^2-1=\sum_{k=1}^\infty \frac{2(k+2)^{k-1}}{k!}x^k$$

which rearranges to

$$W(-x)^2=\sum_{k=0}^\infty \frac{2(k+2)^{k-1}}{k!}x^{k+2}$$

and can also be expressed as

$$W(-x)^2=\sum_{k=1}^\infty \frac{2(k-1) k^{k-2}}{k!} x^k$$

which proves the simpler expression for the autoconvolution.

The last series also appears as formula 11 of this paper by Corless, Jeffrey, and Knuth, as well as appearing in a disguised form as equation 5.60 in Concrete Mathematics.

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