Compute the following sum $ \sum_{i=0}^{n} \binom{n}{i}(i+1)^{i-1}(n - i + 1) ^ {n - i - 1}$?

Question

I have the sum $$ \sum_{i=0}^{n} \binom{n}{i}\cdot (i+1)^{i-1}\cdot(n - i + 1) ^ {n - i - 1},$$ but I don't know how to compute it. It's not for a homework, it's for a graph theory problem that I try to solve.

Answer 1

What follows makes no claim to originality and is based on material from OEIS A000272. This computation is very similar to the one at this MSE link.

We can prove this using the labelled tree function that is known from combinatorics.

The idea that we use a convolution is sound but we actually have to do the algebra.

This will provide a closed form of the exponential generating function of the two terms that are involved.

We seek to show that $$\sum_{k=0}^n {n\choose k} (k+1)^{k-1} (n-k+1)^{n-k-1} = 2 \times (n+2)^{n-1}.$$

The combinatorial class of labelled rooted trees has the specification $$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{T} = \mathcal{Z} \times \textsc{SET}(\mathcal{T})$$ which gives the functional equation $$T(z) = z \exp T(z).$$

We have $n! [z^n] T(z) = n^{n-1}$, for a proof consult the link from the introduction. This implies that $$\frac{1}{z} T(z) = \frac{1}{z} \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!} = \sum_{n\ge 1} n^{n-1} \frac{z^{n-1}}{n!} = \sum_{n\ge 1} n^{n-2} \frac{z^{n-1}}{(n-1)!}.$$

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 $$A(z) = B(z) = \sum_{q\ge 0} \frac{(q+1)^{q-1}}{q!} z^q.$$

The equality that we seek to prove is the convolution of the two exponential generating functions $A(z)$ and $B(z)$ and to verify it we must show that $$n! [z^n] A(z) B(z) = 2 \times (n+2)^{n-1}$$ But we have $$A(z) = B(z) = \frac{1}{z} T(z).$$

It follows that $$n! [z^n] A(z) B(z) = n! \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{T(z)^2}{z^2} dz.$$ Using the same substitution as before (consult link) this becomes $$n! \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(n+3))}{w^{n+3}} \times w^2\times (\exp(-w) - w\exp(-w)) dw \\ = n! \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(n+2))}{w^{n+1}} (1-w) dw \\ = n! \left(\frac{(n+2)^n}{n!} - \frac{(n+2)^{n-1}}{(n-1)!}\right) \\ = (n+2)^n - n \times (n+2)^{n-1} = (n+2)^{n-1} (n+2-n) = 2\times (n+2)^{n-1}.$$

The labelled tree function recently appeared at this MSE link.

Answer 2

Look up Abel's binomial theorem, for example, here:

http://en.wikipedia.org/wiki/Abel's_binomial_theorem

I find this funny because, immediately preceding this, I answered a question by giving a reference to Abel's summation formula.

This means, of course, that somebody will ask about solving the quintic.