TITLE: How many "good" graphs of size $n$ are there?
QUESTION [10 upvotes]: Let's a call a directed simple graph $G$ on $n$ labelled vertices good if every vertex has outdegree 1 and, when considered as if it were undirected, it is connected. How many good graphs of size $n$ are there?
Here's my work so far. Let's call this number $T(n)$. Clearly, $T(2) = 1$: there's only the loop on two vertices.

We also have that $T(3) = 8$. We can count them using the following argument: let's call a possible shape a directed simple graph on $n$ unlabelled vertices which is good. For $n = 3$ we have the following shapes:

There are $3!$ labelled good graphs of the first shape: fix the outside vertex in 3 possible ways, then fix the loop in two possible ways. There are also $\frac{3!}{3}$ labelled good graphs of the second shape: it's simply the number of cycles on 3 elements. So in total we have: $$T(3) = 3! + \frac{3!}{3} = 8\text{.}$$
We also know that $T(4) = 78$. Let's list all possible shapes:

From top left to bottom right, it's easy to check that we have $4!$ labelled good graphs of the first shape, $2\cdot {4 \choose 2}$ of the second, $2\cdot {4 \choose 2}$ of the third, $4!$ of the fourth and $\frac{4!}{4}$ of the last. In total: $$T(4) = 4! + \left(2\cdot {4 \choose 2} + 2\cdot {4 \choose 2}\right) + 4! + \frac{4!}{4} = 3\cdot 4! + \frac{4!}{4}\text{.}$$
I think that $T(5) = 884$, but I won't draw all possible shapes or count their labelings for brevity.
I computed $T(5)$ again, and now I get 944. This also invalidates the following conjecture.
CONJECTURE [DISPROVEN]: I'm conjecturing that there's a simple-ish formula for $T(n)$. It's something like $$T(n) = (2^{n-2} - 1) \cdot n! + \frac{n!}{n} + S(n)$$ where $S(n)$ is some function I currently don't understand such that $S(2) = S(3) = S(4) = 0$, while $S(5) = 5\cdot 4$.

REPLY [9 votes]: I would like to contribute some ideas even though I don't have as much
time as I'd like at the moment. If I understand this problem correctly
then the class of graphs  under consideration call it $\mathcal{Q}$ is
in  a set-of  relationship with  the  class of  endofunctions call  it
$\mathcal{E}$ with the latter being sets of the former.
The following MSE link A has closely related material, as does this MSE link B where aspects of "Random Mapping Statistics" by Flajolet and Odlyzko are discussed.

This gives the species equation
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{E} = \textsc{SET}(\mathcal{Q})$$
which is in terms of generating functions
$$E(z) = \exp Q(z)
\quad\text{or}\quad
Q(z) = \log E(z).$$
Recall the popular labeled rooted tree function which represents the
species
$$\mathcal{T} = \mathcal{Z} \times \textsc{SET}(\mathcal{T})$$
and has the functional equation
$$T(z) = z \exp T(z).$$
We also have that $$T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}$$
(Cayley's formula)
and since there are $n^n$ endofunctions we obtain
$$E(z) = 1 + z\frac{d}{dz} T(z) = 1 + z T'(z).$$
But from the functional equation we get
$$T'(z) = \exp T(z) + z \exp T(z) T'(z) = \frac{T(z)}{z} + T(z) T'(z)$$
so that
$$T'(z) = \frac{1}{z} \frac{T(z)}{1-T(z)}.$$
This finally yields
$$E(z) = 1 + \frac{T(z)}{1-T(z)}$$
and hence $$Q(z) = \log\left(1+\frac{T(z)}{1-T(z)}\right).$$
This gives the sequence
$$1, 3, 17, 142, 1569, 21576, 355081, 6805296,
\\ 148869153, 3660215680,\ldots$$
which points us to OEIS A001865
where we  learn that indeed  we have the right  exponential generating
function. Note that the formula for $E(z)$ also proves that
$\mathcal{E} = \textsc{SEQ}(\mathcal{T}).$

Now to extract coefficients from this for a closed formula we expand
the logarithm to get for the count $q_n$ the formula
$$n! [z^n] \sum_{k\ge 1} \frac{(-1)^{k+1}}{k}
\left(\frac{T(z)}{1-T(z)}\right)^k.$$
Observe  that we  can  restrict  this to  $k\le  n$  because the  tree
function term starts at $z,$ getting
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\left(\frac{T(z)}{1-T(z)}\right)^k.$$
We still need the terms of the fraction in the tree function,
which can be done by Lagrange inversion. We have
$$[z^n] \left(\frac{T(z)}{1-T(z)}\right)^k
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\left(\frac{T(z)}{1-T(z)}\right)^k \; dz.$$
Put $T(z) = w$ so that $z=w/\exp(w)=w\exp(-w)$ and
$dz=(\exp(−w)−w\exp(−w))\; dw$ to get
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n+1))}{w^{n+1}}
\left(\frac{w}{1-w}\right)^k 
(\exp(−w)−w\exp(−w))\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(wn)}{w^{n+1-k}}
\frac{1}{(1-w)^{k-1}} \; dw .$$
Extracting the residue at zero yields for $k\ge 2$
$$\sum_{q=0}^{n-k} \frac{n^{n-k-q}}{(n-k-q)!}
{q+k-2\choose k-2}$$
and for $k=1,$ $$\frac{n^{n-1}}{(n-1)!}.$$
Collecting these into one formula finally yields
$$n^n +  
n! \sum_{k=2}^n \frac{(-1)^{k+1}}{k}
\sum_{q=0}^{n-k} \frac{n^{n-k-q}}{(n-k-q)!}
{q+k-2\choose k-2}.$$
This computation is closely related to the material at this
MSE link.
Addendum. I just noticed in one of the other posts that fixed points are not admitted in those endofunctions. The above material admits fixed points as in the discussion and the diagram at the Wikipedia entry.
Addendum Thu Dec 18 19:57:09 CET 2014.
The case when there are no fixed points goes as follows.
There are now $(n-1)^n$ endofunctions with no fixed points, 
which gives $$E(z) = 1 + \sum_{n\ge 1} (n-1)^n \frac{z^n}{n!}.$$
Now observe that when we apply Lagrange inversion to
$$\frac{\exp(-T(z))}{1-T(z)}$$ we obtain
$$[z^n] \frac{\exp(-T(z))}{1-T(z)}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{\exp(-T(z))}{1-T(z)}\; dz$$
which using the same substitution as before becomes
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n+1))}{w^{n+1}}
\frac{\exp(-w)}{1-w}
(\exp(−w)−w\exp(−w))\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n-1))}{w^{n+1}}\; dw
= \frac{(n-1)^n}{n!}.$$
But $\exp(-T(z)) = \frac{z}{T(z)}$ and we finally get for
$E(z)$ the closed form
$$\frac{z}{T(z)(1-T(z))}.$$
This gives for $Q(z)$ that
$$Q(z) = \log\left(\frac{z}{T(z)(1-T(z))}\right)$$
(note that $E(z)$  has a constant term in its  expansion at zero which
is one) which produces the sequence 
$$0, 1, 8, 78, 944, 13800, 237432, 4708144, 105822432, 2660215680, 
\\ 73983185000, 2255828154624,\ldots $$
which  points us  to OEIS A000435, confirming 
the  result from  the accepted answer.  Note that  the OEIS
says  that this  sequence was  the first  in the  database, so  we are
content to be referencing it in this computation.
To get a closed form re-write $Q(z)$ as follows:
$$Q(z) = \log\left(1 + \frac{z}{T(z)(1-T(z))} -1\right)$$
to get the formula
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\left(\frac{z}{T(z)(1-T(z))} -1\right)^k$$
This is
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\sum_{q=0}^k {k\choose q} (-1)^{k-q}
\left(\frac{z}{T(z)(1-T(z))}\right)^q.$$
We can drop the term for $q=0$ when $n\ge 1,$ getting
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\sum_{q=1}^k {k\choose q} (-1)^{k-q}
\left(\frac{z}{T(z)(1-T(z))}\right)^q.$$
Use Lagrange inversion to extract  coefficients from the tree function
term.
$$[z^n] \left(\frac{z}{T(z)(1-T(z))}\right)^q
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\left(\frac{z}{T(z)(1-T(z))}\right)^q
\; dz$$
which using the same substitution as before becomes
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n+1-q))}{w^{n+1-q}}
\left(\frac{1}{w(1-w)}\right)^q
(\exp(−w)−w\exp(−w))\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n-q))}{w^{n+1}}
\frac{1}{(1-w)^{q-1}} \; dw$$
Extracting coefficients we get
$$\sum_{p=0}^n \frac{(n-q)^{n-p}}{(n-p)!} {p+q-2\choose q-2}$$
when $q\ge 2$ and for $q=1$
$$\frac{(n-1)^n}{n!}.$$
Substituting this into the sum formula yields
$$n! \times \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\times k \times (-1)^{k-1} \frac{(n-1)^n}{n!}
\\ + n! \times \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\sum_{q=2}^k {k\choose q} (-1)^{k-q}
\sum_{p=0}^n \frac{(n-q)^{n-p}}{(n-p)!} {p+q-2\choose q-2}.$$
This simplifies to
$$n\times (n-1)^n
+ n! \times \sum_{k=2}^n \frac{(-1)^{k+1}}{k}
\sum_{q=2}^k {k\choose q} (-1)^{k-q}
\sum_{p=0}^n \frac{(n-q)^{n-p}}{(n-p)!} {p+q-2\choose q-2}.$$
This formula  can be  used to  compute the count  of these  graphs for
large  $n$  where  the  tree  function  formula  would  no  longer  be
practicable.

The sequence for $n=30$ to $n=34$ reads

38086159100543376291945674612050231296000000,
3117962569860399657478478640723143576082043800,
263711778692997479722657378560127779200642842624,
23019602620026625886784119896351926037410391377792,
2071846675499818842878197235287956993753027358752768

Additional observations.
The OEIS entry OEIS A000435
says that this sequence is the  normalized total height of all labeled
rooted trees  on $n$ nodes  (sum of height of  all nodes in  all trees
scaled by $n$). The question arises on how to prove this.

The  height  parameter  is  represented by  the  bivariate  generating
function  $T(z,  u)$  where  $T(z,  1)  =  T(z)$  (the  ordinary  tree
function) and we have the functional equation
$$T(z, u) = z
+ z\frac{T(uz,u)^1}{1!}
+ z\frac{T(uz,u)^2}{2!}
+ z\frac{T(uz,u)^3}{3!}
+ z\frac{T(uz,u)^4}{4!}
+ \cdots$$
or
$$T(z, u) = z \exp T(uz, u).$$
The exponential generating  function for the sum of the  height of all
nodes of all rooted labeled trees by the number of nodes is given by
$$G(z) = \left.\frac{\partial}{\partial u} T(z, u)\right|_{u=1}.$$
We thus differentiate the functional equation, getting
$$G(z) = \left. z \exp T(uz, u)
\left(z \frac{\partial}{\partial z} T(z, u) +
\frac{\partial}{\partial u} T(z, u) \right)\right|_{u=1}$$ 
which becomes
$$G(z) = T(z) (z T'(z) + G(z))$$
so that
$$G(z) (1-T(z)) = z T(z) T'(z)$$
or
$$G(z) = \frac{z T(z)}{1-T(z)} T'(z).$$
We  may substitute  the  expression  for the  derivative  of the  tree
function that we obtained earlier into this to get
$$G(z) = \frac{z T(z)}{1-T(z)} 
\frac{1}{z} \frac{T(z)}{1-T(z)} 
= \left(\frac{T(z)}{1-T(z)}\right)^2.$$
We  have computed  the exponential  generating function  of the  total
height of  all labelled trees on  $n$ nodes, which gives the sequence
$$0, 2, 24, 312, 4720, 82800, 1662024, 37665152, 952401888, 
\\ 26602156800, 813815035000 $$
which is OEIS A001864.
The normalized height is this sequence  divided by $n.$ To verify that
this is  indeed the same as  the count of endofunctions  with no fixed
point we must show that
$$z \frac{d}{dz} Q(z) = G(z)$$
i.e.  that  the endofunctions  times  $n$  give  the total  height  or
alternatively, the total height divided by $n$ give the endofunctions.

But the left is
$$ z \frac{T(z)(1-T(z))}{z} \\ \times
\left(\frac{1}{T(z)(1-T(z))} 
- z \frac{1-2T(z)}
{T(z)^2 (1-T(z))^2} T'(z) \right)$$
which gives
$$1- z \frac{1 - 2T(z)}{T(z)(1-T(z))} T'(z)
= 1- \frac{1 - 2T(z)}{T(z)(1-T(z))} \frac{T(z)}{1-T(z)}
= 1- \frac{1 - 2T(z)}{(1-T(z))^2}
\\ = \frac{1-2T(z)+T(z)^2 -(1- 2T(z))}{(1-T(z))^2}
= \left(\frac{T(z)}{1-T(z)}\right)^2,$$
thus concluding the proof.