Counting acyclic digraphs by sources and sinks

We count labeled acyclic digraphs according to the number sources, sinks, and edges. 1. Counting acyclic digraphs by sources. Let An(t;α) = ∑ D αt, where the sum is over all acyclic digraphs D on the vertex set [n] = {1, 2, . . . , n}, e(D) is the number of edges of D, and s(D) is the number of sources of D; that is, the number of vertices of D of indegree 0. Let An(t) = An(t; 1). To find a recurrence for An(t, α), we take an acyclic digraph and add some new vertices as sources. In the digraph we obtain, the new vertices will be a subset of the set of sources. Lemma 1. n ∑ j=0 (1 + t)j(n−j) ( n j ) αAn−j(t) = An(t;α + 1). (1) Proof. We count triples (S,D,E), where S is a subset of [n], D is an acyclic digraph on [n]−S, and E is a subset of the set S × ([n] − S). We think of the elements of E as edges from S to D. To a triple (S,D,E) we assign the weight αt, where j is the size of S and e is the total number of edges in E and in D. It is clear that the sum of the weights of these triples in which |S| = j is (1 + t)j(n−j) ( n j ) αAn−j(t), and summing on j yields the left side of (1). To a triple (S,D,E) we may associate the pair (S,D′) in which D′ is the digraph on [n] whose edges are those of D together with the edges in E. Note that S is a subset of the set of sources of D′. It is easily seen that this correspondence gives a bijection from triples (S,D,E) to pairs (S,D′) in which D′ is an acyclic digraph on [n] and S is a subset of the set of sources of D′. This proves (1). In working with recurrences like (1), generating functions of the form ∞ ∑ n=0 an x (1 + t)( n 2)n! ∗partially supported by NSF grant DMS-9306297 Counting Acyclic Digraphs by Sources and Sinks 2 are useful, since the convolution n ∑ j=0 (1 + t)j(n−j) ( n j ) ajbn−j = cn is equivalent to the generating function equation [ ∞ ∑ n=0 an x (1 + t)( n 2)n! ] [ ∞ ∑ n=0 bn x (1 + t)( n 2)n! ] = ∞ ∑ n=0 cn x (1 + t)( n 2)n! . For some further applications of these generating functions, which we call graphic generating functions, see [5] and [6]. We can now express a generating function for An(t;α) in terms of the power series F (x) = ∞ ∑ n=0 x (1 + t)( n 2)n! . Theorem 1. ∞ ∑ n=0 An(t;α) x (1 + t)( n 2)n! = F ( (α− 1)x ) /F (−x)