Number of Directed Acyclic Graphs with indegree at most one How many DAGs are there where every vertex has indegree at most one? If not known, is there a lower bound? 
 A: Let $G$ be such a DAG. $G$ must have at least one source, i.e., at least one vertex of indegree $0$. Let $R$ be the set of sources, and for each $r\in R$ let $G_r$ be the set of vertices reachable from $r$. Each $G_r$ is an arborescence. Add one new vertex $v_G$ with edges from $v_G$ to each $r\in R$ to get a new DAG $G'$; $G'$ is itself an arborescence. Conversely, deleting the root from an arborescence with at least two vertices leaves a DAG in which every vertex has indegree at most $1$.
The sequence $\langle a_n:n\in\Bbb N\rangle$ in OEIS A000081 gives the number of unlabelled rooted trees with $n$ nodes; this is the same as the number of arborescences on $n$ nodes, so it’s the number of DAGs on $n-1$ nodes, all with indegree at most $1$. No closed form is given. There is an asymptotic result,
$$a_n\sim cd^nn^{-3/2}\;,$$
where $c=0.439924\ldots$ is Otter’s asymptotic constant $\beta$, and $d=2.955765\ldots$ is Otter’s rooted tree constant.
There is also a recurrence,
$$a_{n+1}=\frac1n\sum_{k=1}^n\left(\sum_{d\mid k}da_d\right)a_{n-k+1}\;.$$
A: I would like to add to  the excellent answer by @BrianMScott by giving
a proof of the recurrence relation.   What we have here are forests of
rooted  trees.  Working  with  rooted  trees  we  obtain  the  species
$\mathcal{T}$ (multiset of trees attached to the root):
$$\mathcal{T} = \mathcal{Z} \mathfrak{M}(\mathcal{T}).$$
This translates to the generating function
$$T(z) = 
z \exp\left(\sum_{l\le 1} \frac{T(z^l)}{l}\right).$$
Differentiate to obtain
$$T'(z) = 
\exp\left(\sum_{l\ge 1} \frac{T(z^l)}{l}\right)
\\ + z \exp\left(\sum_{l\ge 1} \frac{T(z^l)}{l}\right)
\sum_{l\ge 1} T'(z^l) z^{l-1}
\\ = \frac{1}{z} T(z)
+ T(z) \sum_{l\ge 1} T'(z^l) z^{l-1}.$$
Extracting coefficients on $[z^n]$ yields
$$(n+1) T_{n+1} 
= T_{n+1} + 
\sum_{q=0}^{n-1} T_{n-q}
[z^q] \sum_{l\ge 1} T'(z^l) z^{l-1}
\\ = T_{n+1} + 
\sum_{q=0}^{n-1} T_{n-q}
\sum_{l\ge 1} [z^{q-(l-1)}] T'(z^l).$$
For the coefficient extractor to return a non-zero coefficient we must
have  $q = l-1+pl$  with $p\ge  0$ and  $l-1+lp\le n-1$  which implies
$p\le n/l-1.$ The result is
$$(n+1) T_{n+1} 
= T_{n+1} + 
\sum_{l\ge 1} \sum_{p=0}^{\lfloor n/l\rfloor-1}
T_{n+1-l(p+1)}
[z^{pl}] T'(z^l)
\\ = T_{n+1} + 
\sum_{l=1}^{n} \sum_{p=0}^{\lfloor n/l\rfloor-1}
(p+1) T_{p+1} T_{n+1-l(p+1)}.$$
This finally yields
$$\bbox[5px,border:2px solid #00A000]{
T_{n+1} =  
\frac{1}{n} 
\sum_{l=1}^{n} \sum_{p=0}^{\lfloor n/l\rfloor-1}
(p+1) T_{p+1} T_{n+1-l(p+1)}.}$$
Working to simplify this we first obtain
$$\frac{1}{n} 
\sum_{l=1}^{n} \sum_{p=1}^{\lfloor n/l\rfloor}
p T_{p} T_{n+1-lp}.$$
Now we are enumerating pairs $lp$ here where $1\le lp\le n$ so putting
$lp=k$ we may write this as
$$\frac{1}{n} 
\sum_{k=1}^{n} \sum_{p|k}
p T_{p} T_{n+1-k}$$
which is
$$\bbox[5px,border:2px solid #00A000]{
T_{n+1} =  
\frac{1}{n} 
\sum_{k=1}^{n} T_{n+1-k}
\sum_{p|k} p T_{p}.}$$
This is the form given in OEIS A000081. 
Answering  the question given  by the  OP we  return to  forests of
rooted trees and obtain the species $\mathcal{F}$ with equation
$$\mathcal{F}   =  \mathfrak{M}(\mathcal{T})$$   which  in   terms  of
generating functions yields
$$F(z) = T(z)/z.$$
This simply says that a forest is  in a bijection with a tree where we
attach the  trees to a new root  node.  Therefore a tree  on $n$ nodes
yields a forest  on $n-1$ nodes and we may use  $T_{n+1}$ to count the
DAGs that were being queried.
