asymptotics of the expected number of edges of a random acyclic digraph with indegree and outdegree at most one A recent discussion, which may be found here, examined the problem of counting the number of acyclic digraphs on $n$ labelled nodes and having $k$ edges and indegree and outdegree at most one. It was established that the bivariate mixed generating function of this class $\mathcal{G}$ of graphs on $n$ nodes and with $k$ edges is
$$ G(z, u) = \exp \left(\frac{z}{1-uz} \right).$$
This immediately implies that the expected number of edges of a random graph from $\mathcal{G}$ is
$$\epsilon_n = [z^n]\left. \frac{d}{du} G(z,u) \right|_{u=1}.$$
Evaluating this quantity we obtain
$$ \left. \exp \left(\frac{z}{1-uz} \right) z (-1)\frac{1}{(1-uz)^2} (-z)\right|_{u=1}
= \left. \exp \left(\frac{z}{1-uz} \right) \frac{z^2}{(1-uz)^2} \right|_{u=1}
= \exp \left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^2$$
Continuing with this calculation we find
$$ \epsilon_n = [z^n] \sum_{m\ge 0} \frac{1}{m!}  \left(\frac{z}{1-z}\right)^{m+2} =
\sum_{m=0}^{n-2} \frac{1}{m!} [z^n]  \left(\frac{z}{1-z}\right)^{m+2} =
\sum_{m=0}^{n-2} \frac{1}{m!} [z^{n-m-2}] \left(\frac{1}{1-z}\right)^{m+2} =
\sum_{m=0}^{n-2} \frac{1}{m!} \binom{n-m-2+m+1}{m+1} =
\sum_{m=0}^{n-2} \frac{1}{m!} \binom{n-1}{m+1}.$$
This closed form is actually quite nice, but it does not answer the question that is the most obvious one for this problem, namely Is there an asymptotic expansion of $\epsilon_n$ and if yes, what is the first term?
 A: Thanks Joriki for this important pointer correcting my mistake. In fact the number of graphs in $\mathcal{G}$ is given by $$ n! [z^n] G(z, 1) .$$
But we have
$$[z^n] \exp \left( \frac{z}{1-z} \right)
= [z^n] \sum_{m\ge 0} \frac{1}{m!} \left( \frac{z}{1-z} \right)^m$$
which is
$$ \sum_{m=1}^n  \frac{1}{m!} [z^n] \left( \frac{z}{1-z} \right)^m
= \sum_{m=1}^n  \frac{1}{m!} [z^{n-m}] \left( \frac{1}{1-z} \right)^m
= \sum_{m=1}^n  \frac{1}{m!} \binom{n-m+m-1}{m-1} =
\sum_{m=1}^n  \frac{1}{m!} \binom{n-1}{m-1}.$$
It follows that the value $\epsilon_n$ is given by
$$\epsilon_n = \frac{n! \sum_{m=0}^{n-2}  \frac{1}{m!} \binom{n-1}{m+1}}{n! \sum_{m=1}^n  \frac{1}{m!} \binom{n-1}{m-1}} =
\frac{\sum_{m=0}^{n-2}  \frac{1}{m!} \binom{n-1}{m+1}}{\sum_{m=1}^n  \frac{1}{m!} \binom{n-1}{m-1}} =
\frac{\sum_{m=2}^n  \frac{1}{(m-2)!} \binom{n-1}{m-1}}{\sum_{m=1}^n  \frac{1}{m!} \binom{n-1}{m-1}}.$$
It would appear that this implies $$\epsilon_n \sim n,$$ answering my original question. This even includes the coefficient on $n$ and it means that small components having $o(n)$ edges do not contribute in the limit. I will wait a few hours to give you a chance to prove this last asymptotic result, which does not look all that difficult.
A: We can expand this calculation to the variance in addition to the expected value. To do so we compute $$E[X(X-1)] = \frac{ [z^n] \left. \left(\frac{d}{du}\right)^2 G(z, u) \right|_{u=1}}{[z^n] G(z, 1)}$$ where $X$ is the RV representing the number of edges.
Now $$\left. \left(\frac{d}{du}\right)^2 G(z, u) \right|_{u=1} =
\left.\left( \frac{d}{du}\right)\exp\left(\frac{z}{1-uz}\right) \left(\frac{z}{1-uz}\right)^2 \right|_{u=1} =
\left.\exp\left(\frac{z}{1-uz}\right) \left(\frac{z}{1-uz}\right)^4 +
2 \exp\left(\frac{z}{1-uz}\right) \left(\frac{z}{1-uz}\right)^3 \right|_{u=1} =
\exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^4 +
2 \exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^3$$
Now expand the two terms in turn.
Left term,
$$[z^n] \exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^4  =
[z^n] \sum_{m\ge 0} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+4} =
[z^n] \sum_{m=0}^{n-4} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+4} $$ or $$
\sum_{m=0}^{n-4} \frac{1}{m!} [z^{n-m-4}] \left( \frac{1}{1-z}\right)^{m+4} =
\sum_{m=0}^{n-4} \frac{1}{m!} \binom{n-m-4+m+3}{m+3} = 
\sum_{m=0}^{n-4} \frac{1}{m!} \binom{n-1}{m+3}.$$
Right term,
$$2 [z^n] \exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^3  =
2 [z^n] \sum_{m\ge 0} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+3} =
2 [z^n] \sum_{m=0}^{n-3} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+3} $$ or $$
2 \sum_{m=0}^{n-3} \frac{1}{m!} [z^{n-m-3}] \left( \frac{1}{1-z}\right)^{m+3} =
2 \sum_{m=0}^{n-3} \frac{1}{m!} \binom{n-m-3+m+2}{m+2} = 
2 \sum_{m=0}^{n-3} \frac{1}{m!} \binom{n-1}{m+2}.$$
It follows that $$E[X(X-1)] = 
\frac{\sum_{m=0}^{n-4} \frac{1}{m!} \binom{n-1}{m+3} + 2 \sum_{m=0}^{n-3} \frac{1}{m!} \binom{n-1}{m+2}}{\sum_{m=1}^n \frac{1}{m!} \binom{n-1}{m-1}}.$$
The question now becomes, what is the asymptotic expansion of this quotient of sums? That would put the variance within reach, given that $E[X^2] = E[X(X-1)] + E[X].$
A: The following admittedly quite naive program makes it possible to calculate $\epsilon_n$ for $n$ as large as $300000.$ The value that we obtain for the constant term of the expansion is $0.2496.$ This would suggest the conjecture that
$$\epsilon_n \sim n - \sqrt{n} + \frac{1}{4}.$$
The code follows.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

long double term(int n, int m)
{
  long double v = (long double)1;
  while(m>1){
    v *= (long double)((n-1)-(m-2));
    v /= (long double)(m-1);
    v /= (long double)(m);

    m--;
  }

  return v;
}


int main(int argc, char **argv)
{
  int n;

  if(argc!=2 || sscanf(argv[1], "%d", &n)!=1 || n<1){
    fprintf(stderr, "expecting a positive integer\n");
    exit(-1);
  }

  long double p = 0, q = 1;

  int m;
  for(m=2; m<=n; m++){
    if(m%(n/10)==0) printf("%d%%\n", m*100/n);

    long double t = term(n, m);
    p += t*(long double)(m)*(long double)(m-1);
    q += t;
  }

  long double eps = p/q;
  printf("%.18Le %.18Le %.18Le\n", p, q,
         eps-((long double)n-sqrtl((long double)n)));

  exit(0);
}

A: The following program can be used to calculate $E[X(X-1)]$ from the closed form as a quotient of sums. It would appear to support the conjecture that
$$ E[X(X-1)] \sim n^2 - 2 n^{3/2} + \frac{1}{2} n.$$

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

long double term(int n, int m)
{
  long double v = (long double)1;
  while(m>1){
    v *= (long double)((n-1)-(m-2));
    v /= (long double)(m-1);
    v /= (long double)(m);

    m--;
  }

  return v;
}


int main(int argc, char **argv)
{
  int n;

  if(argc!=2 || sscanf(argv[1], "%d", &n)!=1 || n<1){
    fprintf(stderr, "expecting a positive integer\n");
    exit(-1);
  }

  long double p = 0, q = 0, r = 0;

  int m;
  for(m=1; m<=n; m++){
    if(m%(n/10)==0) printf("%d%%\n", m*100/n);

    long double t = term(n, m);
    p += t*
      (long double)(m)*
      (long double)(m-1)*
      (long double)(m-2)*
      (long double)(m-3);
    q += 2*t*
      (long double)(m)*
      (long double)(m-1)*
      (long double)(m-2);

    r += t;
  }

  long double eps = (p+q)/r;
  printf("%.18Le %.18Le %.18Le\n", p, q,
         (eps-(((long double)n)*((long double)n)
               -(long double)2*
               (long double)n*sqrtl((long double)n)))/
         (long double)n);

  exit(0);
}

