Total positivity of some polynomial matrices that enumerate labeled trees and forests II. Rooted labeled trees and partial functional digraphs

We study three combinatorial models for the lower-triangular matrix with entries t n,k = (cid:0) nk (cid:1) n n − k : two involving rooted trees on the vertex set [ n + 1], and one involving partial functional digraphs on the vertex set [ n ]. We show that this matrix is totally positive and that the sequence of its row-generating polynomials is coefficientwise Hankel-totally positive. We then generalize to polynomials t n,k ( y, z ) that count improper and proper edges, and further to polynomials t n,k ( y, ϕ ) in infinitely many indeterminates that give a weight y to each improper edge and a weight m ! ϕ m for each vertex with m proper children. We show that if the weight sequence ϕ is Toeplitz-totally positive, then the two foregoing total-positivity results continue to hold. Our proofs use production matrices and exponential Riordan arrays.


Introduction and statement of results
It is well known [59,79] that the number of rooted trees on the vertex set [n+1] def = {1, . . . , n + 1} is t n = (n + 1) n ; and it is also known (though perhaps less well so) [12,13,75] that the number of rooted trees on the vertex set [n + 1] in which exactly k children of the root are lower-numbered than the root is t n,k = n k n n−k . (1.1) The first few t n,k and t n are n There is a second combinatorial interpretation of the numbers t n,k , also in terms of rooted trees: namely, t n,k is the number of rooted trees on the vertex set [n + 1] in which some specified vertex i has k children. 1 1 This fact ought to be well known, but to our surprise we have been unable to find any published reference. Let us therefore give two proofs: First proof. Let T • n denote the set of rooted trees on the vertex set [n], and let deg T (i) denote the number of children of the vertex i in the rooted tree T . Rooted trees T ∈ T • n+1 are associated bijectively to Prüfer sequences (s 1 , . . . , s n ) ∈ [n+1] n , in which each index i ∈ [n+1] appears deg T (i) times [79, pp. 25-26]. There are n k n n−k sequences in which the index i appears exactly k times. Equivalently, by [ Replacing n → n + 1 and then setting x i = x and x j = 1 for j = i, we have Extracting the coefficient of x k yields n k n n−k . Second proof. There are f n,k = n k k n n−k−1 k-component forests of rooted trees on n labeled vertices (see the references cited in [76, footnote 1]). By adding a new vertex 0 and connecting it to the roots of all the trees, we see that f n,k is also the number of unrooted trees on n + 1 labeled vertices in which some specified vertex (here vertex 0) has degree k. Now choose a root: if this root is 0, then vertex 0 has k children; otherwise vertex 0 has k − 1 children. It follows that the number of rooted trees on n + 1 labeled vertices in which some specified vertex has k children is f n,k + nf n,k+1 = n k n n−k . The second proof was found independently by Ira Gessel (private communication).
And finally, there is a third combinatorial interpretation of the numbers t n,k [20] that is even simpler than the preceding two. Recall first that a functional digraph is a directed graph in which every vertex has out-degree 1; the terminology comes from the fact that such digraphs are in obvious bijection with functions f from the vertex set to itself [namely, − → ij is an edge if and only if f (i) = j]. Let us now define a partial functional digraph to be a directed graph in which every vertex has out-degree 0 or 1; and let us write PFD n,k for the set of partial functional digraphs on the vertex set [n] in which exactly k vertices have out-degree 0. (So PFD n,0 is the set of functional digraphs.) A digraph in PFD n,k has n − k edges. It is easy to see that |PFD n,k | = t n,k : there are n k choices for the out-degree-0 vertices, and n n−k choices for the edges emanating from the remaining vertices.
We will use all three combinatorial models at various points in this paper. The unit-lower-triangular matrix (t n,k ) n,k≥0 has the exponential generating function ∞ n=0 n k=0 t n,k t n n! x k = e xT (t) 1 − T (t) (1.2) where is the tree function [19]. 2 An equivalent statement is that the unit-lower-triangular matrix (t n,k ) n,k≥0 is the exponential Riordan array [3,22,24,70] R[F, G] with F (t) = ∞ n=0 n n t n /n! = 1/[1 − T (t)] and G(t) = T (t); we will discuss this connection in Section 4.1.
The principal purpose of this paper is to prove the total positivity of some matrices related to (and generalizing) t n and t n,k . Recall first that a finite or infinite matrix of real numbers is called totally positive (TP) if all its minors are nonnegative, and strictly totally positive (STP) if all its minors are strictly positive. 3 Background information on totally positive matrices can be found in [28,34,46,64]; they have applications to many areas of pure and applied mathematics. 4 Our first result is the following: (a) The unit-lower-triangular matrix T = (t n,k ) n,k≥0 is totally positive.
It is known [35,64] that a Hankel matrix of real numbers is totally positive if and only if the underlying sequence is a Stieltjes moment sequence, i.e. the moments of a positive measure on [0, ∞). And it is also known that (n n ) n≥0 is a Stieltjes moment sequence. 5 So Theorem 1.1(b) is equivalent to this known result. But our proof here is combinatorial and linear-algebraic, not analytic.
However, this is only the beginning of the story, because our main interest [73,74,77] is not with sequences and matrices of real numbers, but rather with sequences and matrices of polynomials (with integer or real coefficients) in one or more indeterminates x: in applications they will typically be generating polynomials that enumerate some combinatorial objects with respect to one or more statistics. We equip the polynomial ring R[x] with the coefficientwise partial order: that is, we say that P is nonnegative (and write P 0) in case P is a polynomial with nonnegative coefficients. We then say that a matrix with entries in R[x] is coefficientwise totally positive if all its minors are polynomials with nonnegative coefficients; and we say that a sequence a = (a n ) n≥0 with entries in R[x] is coefficientwise Hankeltotally positive if its associated infinite Hankel matrix H ∞ (a) = (a n+n ) n,n ≥0 is coefficientwise totally positive.
Returning now to the matrix T = (t n,k ) n,k≥0 , let us define its row-generating polynomials in the usual way: shows that n n /n! is a Stieltjes moment sequence. Moreover, n! = ∞ 0 x n e −x dx is a Stieltjes moment sequence. Since the entrywise product of two Stieltjes moment sequences is easily seen to be a Stieltjes moment sequence, it follows that n n is a Stieltjes moment sequence. But we do not know any simple formula (i.e. one involving only a single integral over a real variable) for its Stieltjes integral representation.
But this is not the end of the story, because we want to generalize these polynomials further by adding further variables. Given a rooted tree T and two vertices i, j of T , we say that j is a descendant of i if the unique path from the root of T to j passes through i. (Note in particular that every vertex is a descendant of itself.) Now suppose that the vertex set of T is totally ordered (for us it will be [n + 1]), and let e = ij be an edge of T , ordered so that j is a descendant of i. We say that the edge e = ij is improper if there exists a descendant of j (possibly j itself) that is lowernumbered than i; otherwise we say that e = ij is proper . We denote by imprope(T ) [resp. prope(T )] the number of improper (resp. proper) edges in the tree T .
We now introduce these statistics into our second combinatorial model. Let T i;k n denote the set of rooted trees on the vertex set [n] in which the vertex i has k children. For the identity |T i;k n+1 | = t n,k , we can use any i ∈ [n + 1]; but for the following we specifically want to take i = 1. With this choice we observe that the k edges from the vertex 1 to its children are automatically proper. We therefore define t n,k (y, z) = T ∈T 1;k n+1 y imprope(T ) z prope(T )−k . (1.7) Clearly t n,k (y, z) is a homogeneous polynomial of degree n−k with nonnegative integer coefficients; it is a polynomial refinement of t n,k in the sense that t n,k (1, 1) = t n,k .
In Section 4.2 we will show that the unit-lower-triangular matrix T(y, z) = t n,k (y, z) n,k≥0 is an exponential Riordan array R[F, G], and we will compute F (t) and G(t).
We now generalize (1.4) by defining the row-generating polynomials T n (x, y, z) = n k=0 t n,k (y, z) x k (1.8) or in other words where deg T (1) is the number of children of the vertex 1 in the rooted tree T . Note that T n (x, y, z) is a homogeneous polynomial of degree n in x, y, z, with nonnegative integer coefficients; it reduces to T n (x) when y = z = 1. Our third result is then: (a) The unit-lower-triangular polynomial matrix T(y, z) = t n,k (y, z) n,k≥0 is coefficientwise totally positive (jointly in y, z).
(b) The polynomial sequence T = T n (x, y, z) n≥0 is coefficientwise Hankel-totally positive (jointly in x, y, z).
Theorem 1.4 strengthens Theorems 1.1(a) and 1.2, and reduces to them when y = z = 1. The proof of Theorem 1.4(b) will be based on studying the binomial rowgenerating matrix T(y, z)B x , using the representation of T(y, z) as an exponential Riordan array.
Finally, let us consider our third combinatorial model, which is based on partial functional digraphs. Recall that a functional digraph (resp. partial functional digraph) is a directed graph in which every vertex has out-degree 1 (resp. 0 or 1). Each weakly connected component of a functional digraph consists of a directed cycle (possibly of length 1, i.e. a loop) together with a collection of (possibly trivial) directed trees rooted at the vertices of the cycle (with edges pointing towards the root). The weakly connected components of a partial functional digraph are trees rooted at the outdegree-0 vertices (with edges pointing towards the root) together with components of the same form as in a functional digraph. We say that a vertex of a partial functional digraph is recurrent (or cyclic) if it lies on one of the cycles; otherwise we call it transient (or acyclic). If j and k are vertices of a digraph, we say that k is a predecessor of j if there exists a directed path from k to j (in particular, every vertex is a predecessor of itself). 6 Note that "predecessor" in a digraph generalizes the notion of "descendant" in a rooted tree, if we make the convention that all edges in the tree are oriented towards the root. Indeed, if j is a transient vertex in a partial functional digraph, then the predecessors of j are precisely the descendants of j in the rooted tree (rooted at either a recurrent vertex or an out-degree-0 vertex) to which j belongs. On the other hand, if j is a recurrent vertex, then the predecessors of j are all the vertices in the weakly connected component containing j.
Now consider a partial functional digraph on a totally ordered vertex set (which for us will be [n]). We say that an edge − → ji (pointing from j to i) is improper if there exists a predecessor of j (possibly j itself) that is ≤ i; otherwise we say that the edge − → ji is proper . When j is a transient vertex, this coincides with the notion of improper/proper edge in a rooted tree. When j is a recurrent vertex, the edge − → ji is always improper, because one of the predecessors of j is i. (This includes the case i = j: a loop is always an improper edge.) We denote by imprope(G) [resp. prope(G)] the number of improper (resp. proper) edges in the partial functional digraph G. We then define the generating polynomial t n,k (y, z) = G∈PFD n,k y imprope(G) z prope(G) . (1.10) Since G ∈ PFD n,k has n − k edges, t n,k (y, z) is a homogeneous polynomial of degree n − k with nonnegative integer coefficients. By bijection between our second and third combinatorial models, we will prove: The row-generating polynomials (1.8)/(1.9) thus have the alternate combinatorial interpretation where deg 0(G) is the number of out-degree-0 vertices in G.
We also have an interpretation of the polynomials t n,k (y, z) in our first combinatorial model (rooted trees in which the root has k lower-numbered children); but since this interpretation is rather complicated, we defer it to Appendix A.
But this is still not the end of the story, because we can add even more variables into our second combinatorial model -in fact, an infinite set. Given a rooted tree T on a totally ordered vertex set and vertices i, j ∈ T such that j is a child of i, we say that j is a proper child of i if the edge e = ij is proper (that is, j and all its descendants are higher-numbered than i). Now let φ = (φ m ) m≥0 be indeterminates, and let t n,k (y, φ) be the generating polynomial for rooted trees T ∈ T 1;k n+1 with a weight y for each improper edge and a weight φ m def = m! φ m for each vertex i = 1 that has m proper children: where pdeg T (i) denotes the number of proper children of the vertex i in the rooted tree T . We will see later why it is convenient to introduce the factors m! in this definition. Observe also that the variables z are now redundant and therefore omitted, because they would simply scale φ m → z m φ m . And note finally that, in conformity with (1.7), we have chosen to suppress the weight φ k that would otherwise be associated to the vertex 1. We call the polynomials t n,k (y, φ) the generic rooted-tree polynomials, and the lower-triangular matrix T(y, φ) = t n,k (y, φ) n,k≥0 the generic rooted-tree matrix . Here φ = (φ m ) m≥0 are in the first instance indeterminates, so that t n,k (y, φ) belongs to the polynomial ring Z[y, φ]; but we can then, if we wish, substitute specific values for φ in any commutative ring R, leading to values t n,k (y, φ) ∈ R[y]. (Similar substitutions can of course also be made for y.) When doing this we will use the same notation t n,k (y, φ), as the desired interpretation for φ should be clear from the context. The polynomial t n,k (y, φ) is homogeneous of degree n in φ; it is also quasihomogeneous of degree n − k in y and φ when φ m is assigned weight m and y is assigned weight 1. By specializing t n,k (y, φ) to φ m = z m /m! and hence φ m = z m , we recover t n,k (y, z).
We remark that the matrix T(y, φ), unlike T(y, z), is not unit-lower-triangular: rather, it has diagonal entries t n,n (y, φ) = φ n 0 , corresponding to the tree in which 1 is the root and has all the vertices 2, . . . , n + 1 as children. More generally, the polynomial t n,k (y, φ) is divisible by φ k 0 , since the vertex 1 always has at least k leaf descendants. So we could define a unit-lower-triangular matrix T (y, φ) = t n,k (y, φ) n,k≥0 by t n,k (y, φ) = t n,k (y, φ)/φ k 0 . (Alternatively, we could simply choose to normalize to φ 0 = 1.) In Section 4.3 we will show that T(y, φ) is an exponential Riordan array R[F, G], and we will compute F (t) and G(t).
We can also define the corresponding polynomials t n,k (y, φ) in the partial-functionaldigraph model, as follows: If G is a partial functional digraph on a totally ordered vertex set, and i is a vertex of G, we define the proper in-degree of i, pindeg G (i), to be the number of proper edges − → ji in G. We then define Then, generalizing Proposition 1.5, we will prove: Now define the row-generating polynomials or in other words The main result of this paper is then the following: Let R be a partially ordered commutative ring, and let φ = (φ m ) m≥0 be a sequence in R that is Toeplitz-totally positive of order r. Then: (a) The lower-triangular polynomial matrix T(y, φ) = t n,k (y, φ) n,k≥0 is coefficientwise totally positive of order r (in y).
(b) The polynomial sequence T = T n (x, y, φ) n≥0 is coefficientwise Hankel-totally positive of order r (jointly in x, y).
(The concept of Toeplitz-total positivity in a partially ordered commutative ring will be explained in detail in Section 2.1. Total positivity of order r means that the minors of size ≤ r are nonnegative.) Specializing Theorem 1.8 to r = ∞, R = Q and φ m = z m /m! (which is indeed Toeplitz-totally positive: see (2.1) below), we recover Theorem 1.4. The method of proof of Theorem 1.8 will, in fact, be the same as that of Theorem 1.4, suitably generalized.
We now give an overview of the contents of this paper. The main tool in our proofs will be the theory of production matrices [23,24] as applied to total positivity [77], combined with the theory of exponential Riordan arrays [3,22,24,70]. Therefore, in Section 2 we review some facts about total positivity, production matrices and exponential Riordan arrays that will play a central role in our arguments. This development culminates in Corollary 2.28; it is the fundamental theoretical result that underlies all our proofs. In Section 3 we give bijective proofs of Propositions 1.3, 1.5, 1.6 and 1.7. In Section 4 we show that the matrices T, T(y, z) and T(y, φ) are exponential Riordan arrays R[F, G], and we compute their generating functions F and G. In Section 5 we combine the results of Sections 2 and 4 to complete the proofs of Theorems 1.1, 1.2, 1.4 and 1.8.
This paper is a sequel to our paper [76] on the total positivity of matrices that enumerate forests of rooted labeled trees. The methods here are basically the same as in this previous paper, but generalized nontrivially to handle exponential Riordan arrays R[F, G] with F = 1. Zhu [87,89] has employed closely related methods. See also Gilmore [39] for some total-positivity results for q-generalizations of tree and forest matrices, using very different methods.

Preliminaries
Here we review some definitions and results from [62,77] that will be needed in the sequel. We also include a brief review of ordinary and exponential Riordan arrays [3,22,24,69,70,78] and Lagrange inversion [38].
The treatment of exponential Riordan arrays in Section 2.6 contains one novelty: namely, the rewriting of the production matrix in terms of new series Φ and Ψ (see (2.23) ff. and Proposition 2.23). This is the key step that leads to Corollary 2.28.

Partially ordered commutative rings and total positivity
In this paper all rings will be assumed to have an identity element 1 and to be nontrivial (1 = 0).
A partially ordered commutative ring is a pair (R, P) where R is a commutative ring and P is a subset of R satisfying (a) 0, 1 ∈ P.
(b) If a, b ∈ P, then a + b ∈ P and ab ∈ P.
We call P the nonnegative elements of R, and we define a partial order on R (compatible with the ring structure) by writing a ≤ b as a synonym for b − a ∈ P. Please note that, unlike the practice in real algebraic geometry [11,50,57,65], we do not assume here that squares are nonnegative; indeed, this property fails completely for our prototypical example, the ring of polynomials with the coefficientwise order, since Now let (R, P) be a partially ordered commutative ring and let x = {x i } i∈I be a collection of indeterminates. A (finite or infinite) matrix with entries in a partially ordered commutative ring is called totally positive (TP) if all its minors are nonnegative; it is called totally positive of order r (TP r ) if all its minors of size ≤ r are nonnegative. It follows immediately from the Cauchy-Binet formula that the product of two TP (resp. TP r ) matrices is TP (resp. TP r ). 7 This fact is so fundamental to the theory of total positivity that we shall henceforth use it without comment.
We say that a sequence a = (a n ) n≥0 with entries in a partially ordered commutative ring is Hankel-totally positive (resp. Hankel-totally positive of order r) if its associated infinite Hankel matrix H ∞ (a) = (a i+j ) i,j≥0 is TP (resp. TP r ). We say that a is Toeplitz-totally positive (resp. Toeplitz-totally positive of order r) if its associated infinite Toeplitz matrix T ∞ (a) = (a i−j ) i,j≥0 (where a n def = 0 for n < 0) is TP (resp. TP r ). 8 When R = R, Hankel-and Toeplitz-total positivity have simple analytic characterizations. A sequence (a n ) n≥0 of real numbers is Hankel-totally positive if and only if it is a Stieltjes moment sequence [35,Théorème 9] [64, section 4.6]. And a sequence (a n ) n≥0 of real numbers is Toeplitz-totally positive if and only if its ordinary generating function can be written as ∞ n=0 a n t n = Ce γt t m with m ∈ N, C, γ, α i , β i ≥ 0, α i < ∞ and β i < ∞: this is the celebrated Aissen-Schoenberg-Whitney-Edrei theorem [46,Theorem 5.3,p. 412]. However, in a general partially ordered commutative ring R, the concepts of Hankel-and Toeplitz-total positivity are more subtle.
We will need a few easy facts about the total positivity of special matrices: Lemma 2.1 (Bidiagonal matrices). Let A be a matrix with entries in a partially ordered commutative ring, with the property that all its nonzero entries belong to two consecutive diagonals. Then A is totally positive if and only if all its entries are nonnegative.
Proof. The nonnegativity of the entries (i.e. TP 1 ) is obviously a necessary condition for TP. Conversely, for a matrix of this type it is easy to see that every nonzero minor is simply a product of some entries.
Lemma 2.2 (Toeplitz matrix of powers). Let R be a partially ordered commutative ring, let x ∈ R, and consider the infinite Toeplitz matrix Then every minor of T x is either zero or else a power of x.
In particular, if x is an indeterminate, then T x is totally positive in the ring Z[x] equipped with the coefficientwise order.
Proof. Consider a submatrix A = (T x ) IJ with rows I = {i 1 < . . . < i k } and columns J = {j 1 < . . . < j k }. We will prove by induction on k that det A is either zero or a power of x. It is trivial if k = 0 or 1. If A 12 = A 22 = 0, then A 1s = A 2s = 0 for all s ≥ 2 by definition of T x , and det A = 0. If A 12 and A 22 are both nonzero, then the first column of A is x j 2 −j 1 times the second column, and again det A = 0. Finally, if A 12 = 0 and A 22 = 0 (by definition of T x this is the only other possibility), then A 1s = 0 for all s ≥ 2; we then replace the first column of A by the first column minus x j 2 −j 1 times the second column, so that the new first column has x i 1 −j 1 in its first entry (or zero if i 1 < j 1 ) and zeroes elsewhere. Then det A equals x i 1 −j 1 (or zero if i 1 < j 1 ) times the determinant of its last k − 1 rows and columns, so the claim follows from the inductive hypothesis.
See also Example 2.9 below for a second proof of the total positivity of T x , using production matrices. Proof. It is well known that the binomial matrix B is totally positive, and this can be proven by a variety of methods: e.g. using production matrices [46, pp. [28, p. 63], or by an application of the Lindström-Gessel-Viennot lemma [33, p. 24].
Then B x,y = DBD where D = diag (x n ) n≥0 and D = diag (x −k y k ) k≥0 . By Cauchy-Binet, B x,y is totally positive in the ring Z[x, x −1 , y] equipped with the coefficientwise order. But because B is lower-triangular, the elements of B x,y actually lie in the subring Z[x, y].
See also Example 2.10 below for an ab initio proof of Lemma 2.3 using production matrices.
Finally, let us show that the sufficiency half of the Aissen-Schoenberg-Whitney-Edrei theorem holds (with a slight modification to avoid infinite products) in a general partially ordered commutative ring. We give two versions, depending on whether or not it is assumed that the ring R contains the rationals: Lemma 2.4 (Sufficient condition for Toeplitz-total positivity). Let R be a partially ordered commutative ring, let N be a nonnegative integer, and let α 1 , . . . , α N , β 1 , . . . , β N and C be nonnegative elements in R. Define the sequence a = (a n ) n≥0 in R by ∞ n=0 a n t n = C Then the Toeplitz matrix T ∞ (a) is totally positive.
Of course, it is no loss of generality to have the same number N of alphas and betas, since some of the α i or β i could be zero.
Define the sequence a = (a n ) n≥0 in R by ∞ n=0 a n t n = C e γt Then the Toeplitz matrix T ∞ (a) is totally positive.
Proof of Lemma 2.4. We make a series of elementary observations: 1) The sequence a = (1, α, 0, 0, 0, . . .), corresponding to the generating function A(t) = 1 + αt, is Toeplitz-totally positive if and only if α ≥ 0. The "only if" is trivial, and the "if" follows from Lemma 2.1 because the Toeplitz matrix T ∞ (a) is bidiagonal.
3) If a and b are sequences with ordinary generating functions A(t) and B(t), then the convolution c = a * b, defined by c n = n k=0 a k b n−k , has ordinary generating function C(t) = A(t) B(t); moreover, the Toeplitz matrix T ∞ (c) is simply the matrix product T ∞ (a) T ∞ (b). It thus follows from the Cauchy-Binet formula that if a and b are Toeplitz-totally positive, then so is c.
4) A Toeplitz-totally positive sequence can be multiplied by a nonnegative constant C, and it is still Toeplitz-totally positive.
Combining these observations proves the lemma.
Proof of Lemma 2.5. We add to the proof of Lemma 2.4 the following additional observation: 5) The sequence a = (γ n /n!) n≥0 , corresponding to the generating function A(t) = e γt , is Toeplitz-totally positive if and only if γ ≥ 0. The "only if" is again trivial, and the "if" follows from Lemma 2.3 because γ n−k /(n − k)! = n k γ n−k × k!/n! and hence T ∞ (a) = D −1 B γ,1 D where D = diag( (n!) n≥0 ).

Production matrices
The method of production matrices [23,24] has become in recent years an important tool in enumerative combinatorics. In the special case of a tridiagonal production matrix, this construction goes back to Stieltjes' [81,82] work on continued fractions: the production matrix of a classical S-fraction or J-fraction is tridiagonal. In the present paper, by contrast, we shall need production matrices that are lower-Hessenberg (i.e. vanish above the first superdiagonal) but are not in general tridiagonal. We therefore begin by reviewing briefly the basic theory of production matrices. The important connection of production matrices with total positivity will be treated in the next subsection.
Let P = (p ij ) i,j≥0 be an infinite matrix with entries in a commutative ring R. In order that powers of P be well-defined, we shall assume that P is either row-finite (i.e. has only finitely many nonzero entries in each row) or column-finite.
Let us now define an infinite matrix A = (a nk ) n,k≥0 by a nk = (P n ) 0k (2.5) (in particular, a 0k = δ 0k ). Writing out the matrix multiplications explicitly, we have so that a nk is the total weight for all n-step walks in N from i 0 = 0 to i n = k, in which the weight of a walk is the product of the weights of its steps, and a step from i to j gets a weight p ij . Yet another equivalent formulation is to define the entries a nk by the recurrence a nk = ∞ i=0 a n−1,i p ik for n ≥ 1 (2.7) with the initial condition a 0k = δ 0k . We call P the production matrix and A the output matrix , and we write A = O(P ). Note that if P is row-finite, then so is O(P ); if P is lower-Hessenberg, then O(P ) is lower-triangular; if P is lower-Hessenberg with invertible superdiagonal entries, then O(P ) is lower-triangular with invertible diagonal entries; and if P is unit-lower-Hessenberg (i.e. lower-Hessenberg with entries 1 on the superdiagonal), then O(P ) is unit-lower-triangular. In all the applications in this paper, P will be lower-Hessenberg.
The matrix P can also be interpreted as the adjacency matrix for a weighted directed graph on the vertex set N (where the edge ij is omitted whenever p ij = 0). Then P is row-finite (resp. column-finite) if and only if every vertex has finite outdegree (resp. finite in-degree).
This iteration process can be given a compact matrix formulation. Let us define the augmented production matrix P def = 1 0 0 0 · · · P . (2.8) Then the recurrence (2.7) together with the initial condition a 0k = δ 0k can be written as This identity can be iterated to give the factorization A = · · · I 3 0 0 P where I k is the k × k identity matrix; and conversely, (2.10) implies (2.9). Now let ∆ = (δ i+1,j ) i,j≥0 be the matrix with 1 on the superdiagonal and 0 elsewhere. Then for any matrix M with rows indexed by N, the product ∆M is simply M with its zeroth row removed and all other rows shifted upwards. (Some authors use the notation M def = ∆M .) The recurrence (2.7) can then be written as (2.11) It follows that if A is a row-finite matrix that has a row-finite inverse A −1 and has first row a 0k = δ 0k , then P = A −1 ∆A is the unique matrix such that A = O(P ). This holds, in particular, if A is lower-triangular with invertible diagonal entries and a 00 = 1; then A −1 is lower-triangular and P = A −1 ∆A is lower-Hessenberg. And if A is unit-lower-triangular, then P = A −1 ∆A is unit-lower-Hessenberg. We shall repeatedly use the following easy fact: Lemma 2.6 (Production matrix of a product). Let P = (p ij ) i,j≥0 be a row-finite matrix (with entries in a commutative ring R), with output matrix A = O(P ); and let B = (b ij ) i,j≥0 be a lower-triangular matrix with invertible (in R) diagonal entries.
That is, up to a factor b 00 , the matrix AB has production matrix B −1 P B.
Proof. Since P is row-finite, so is A = O(P ); then the matrix products AB and B −1 P B arising in the lemma are well-defined. Now But B is lower-triangular with invertible diagonal entries, so B is invertible and B −1 is lower-triangular, with (B −1 ) 0j = b −1 00 δ j0 . It follows that AB = b 00 O(B −1 P B).

Production matrices and total positivity
Let P = (p ij ) i,j≥0 be a matrix with entries in a partially ordered commutative ring R. We will use P as a production matrix; let A = O(P ) be the corresponding output matrix. As before, we assume that P is either row-finite or column-finite.
When P is totally positive, it turns out [77] that the output matrix O(P ) has two total-positivity properties: firstly, it is totally positive; and secondly, its zeroth column is Hankel-totally positive. Since [77] is not yet publicly available, we shall present briefly here (with proof) the main results that will be needed in the sequel.
The fundamental fact that drives the whole theory is the following: Proposition 2.7 (Minors of the output matrix). Every k × k minor of the output matrix A = O(P ) can be written as a sum of products of minors of size ≤ k of the production matrix P .
In this proposition the matrix elements p = {p ij } i,j≥0 should be interpreted in the first instance as indeterminates: for instance, we can fix a row-finite or column-finite set S ⊆ N × N and define the matrix P S = (p S ij ) i,j∈N with entries Then the entries (and hence also the minors) of both P and A belong to the polynomial ring Z[p], and the assertion of Proposition 2.7 makes sense. Of course, we can subsequently specialize the indeterminates p to values in any commutative ring R.
Proof of Proposition 2.7. For any infinite matrix X = (x ij ) i,j≥0 , let us write X N = (x ij ) 0≤i≤N −1, j≥0 for the submatrix consisting of the first N rows (and all the columns) of X. Every k × k minor of A is of course a k × k minor of A N for some N , so it suffices to prove that the claim about minors holds for all the A N . But this is easy: the fundamental identity (2.9) implies So the result follows by induction on N , using the Cauchy-Binet formula.
If we now specialize the indeterminates p to values in some partially ordered commutative ring R, we can immediately conclude: Theorem 2.8 (Total positivity of the output matrix). Let P be an infinite matrix that is either row-finite or column-finite, with entries in a partially ordered commutative ring R. If P is totally positive of order r, then so is A = O(P ).  [36]. However, all of these results concerned only special cases: [1,17,51,85] treated the case in which the production matrix P is tridiagonal; [86] treated a (special) case in which P is upper bidiagonal; [16] treated the case in which P is the production matrix of a Riordan array; [18,36] treated (implicitly) the case in which P is upper-triangular and Toeplitz. But the argument is in fact completely general, as we have just seen; there is no need to assume any special form for the matrix P .

Remarks
3. A slightly different version of this proof was presented in [62,63]. The simplified reformulation given here, using the augmented production matrix, is due to Mu and Wang [60]. Example 2.9 (Toeplitz matrix of powers). Let P = xe 00 + y∆, where x and y are indeterminates (here e ij denotes the matrix with an entry 1 in position ij and 0 elsewhere). By Lemma 2.1, P is TP in the ring Z[x, y] equipped with the coefficientwise order. An easy computation shows that O(xe 00 + y∆) nk = x n−k y k I[k ≤ n]. When y = 1, this is the Toeplitz matrix of powers (2.2). So Theorem 2.8 implies that T x is TP in the ring Z[x] equipped with the coefficientwise order. This gives a second proof of the total positivity stated in Lemma 2.2.
Example 2.10 (Binomial matrix). Let P be the upper-bidiagonal Toeplitz matrix xI + y∆, where x and y are indeterminates. By Lemma 2.1, P is TP in the ring Z[x, y] equipped with the coefficientwise order. An easy computation shows that O(xI + y∆) = B x,y , the weighted binomial matrix with entries (B x,y ) nk = x n−k y k n k . So Theorem 2.8 implies that B x,y is TP in the ring Z[x, y] equipped with the coefficientwise order. This gives an ab initio proof of Lemma 2.3.
Then the Hankel matrix of O 0 (P ) has matrix elements (Note that the sum over k has only finitely many nonzero terms: if P is row-finite, then there are finitely many nonzero (P n ) 0k , while if P is column-finite, there are finitely many nonzero (P n ) k0 .) We have therefore proven: Lemma 2.11 (Identity for Hankel matrix of the zeroth column). Let P be a row-finite or column-finite matrix with entries in a commutative ring R. Then Combining Proposition 2.7 with Lemma 2.11 and the Cauchy-Binet formula, we obtain: Corollary 2.12 (Hankel minors of the zeroth column). Every k × k minor of the infinite Hankel matrix H ∞ (O 0 (P )) = ((P n+n ) 00 ) n,n ≥0 can be written as a sum of products of the minors of size ≤ k of the production matrix P .
And specializing the indeterminates p to nonnegative elements in a partially ordered commutative ring, in such a way that P is row-finite or column-finite, we deduce: Theorem 2.13 (Hankel-total positivity of the zeroth column). Let P = (p ij ) i,j≥0 be an infinite row-finite or column-finite matrix with entries in a partially ordered commutative ring R, and define the infinite Hankel matrix H ∞ (O 0 (P )) = ((P n+n ) 00 ) n,n ≥0 . If P is totally positive of order r, then so is H ∞ (O 0 (P )).
One might hope that Theorem 2.13 could be strengthened to show not only Hankel-TP of the zeroth column of the output matrix A = O(P ), but in fact Hankel-TP of the row-generating polynomials A n (x) for all x ≥ 0 (at least when R = R)or even more strongly, coefficientwise Hankel-TP of the row-generating polynomials. Alas, this hope is vain, for these properties do not hold in general : Example 2.14 (Failure of Hankel-TP of the row-generating polynomials). Let P = e 00 + ∆ be the upper-bidiagonal matrix with 1 on the superdiagonal and 1, 0, 0, 0, . . . on the diagonal; by Lemma 2.1 it is TP. Then A = O(P ) is the lower-triangular matrix will all entries 1 (see Example 2.9), so that Nevertheless, in one important special case -namely, exponential Riordan arrays R[1, G] -the total positivity of the production matrix does imply the coefficientwise Hankel-TP of the row-generating polynomials of the output matrix: this was shown [76,Theorem 2.20]. That result will be generalized here, in Corollary 2.28, to provide a more general sufficient (but not necessary) condition for the coefficientwise Hankel-TP of the row-generating polynomials of the output matrix.

Binomial row-generating matrices
Let A = (a nk ) n,k≥0 be a row-finite matrix with entries in a commutative ring R. (In most applications, including all those in the present paper, the matrix A will be lower-triangular.) We define its row-generating polynomials in the usual way: where the sum is actually finite because A is row-finite. More generally, let us define its binomial partial row-generating polynomials The polynomials A n,k (x) are the matrix elements of the binomial row-generating matrix AB x : where B x = B x,1 is the weighted binomial matrix defined in (1.6). The zeroth column of the matrix AB x consists of the row-generating polynomials A n (x) = A n,0 (x). In this paper the matrix A will be either the matrix T = (t n,k ) n,k≥0 or one of its polynomial generalizations.
We can now explain the method that we will use to prove Theorems 1.2 and 1.4: Proposition 2.15. Let P be a row-finite matrix with entries in a partially ordered commutative ring R, and let A = O(P ).
(a) If P is totally positive of order r, then so is A.
equipped with the coefficientwise order, then the sequence (A n (x)) n≥0 of row-generating polynomials is Hankel-totally positive of order r in the ring R[x] equipped with the coefficientwise order.
Indeed, (a) is just a restatement of Theorem 2.8; and (b) is an immediate consequence of Lemma 2.6 and Theorem 2.13 together with the fact that the zeroth column of the matrix AB x consists of the row-generating polynomials A n (x).

Riordan arrays
Let R be a commutative ring, and let f (t) = ∞ n=0 f n t n and g(t) = ∞ n=1 g n t n be formal power series with coefficients in R; note that g has zero constant term (for clarity we set g 0 = 0). Then the (ordinary) Riordan array associated to the pair (f, g) is the infinite lower-triangular matrix R(f, g) = (R(f, g) nk ) n,k≥0 defined by That is, the kth column of R(f, g) has ordinary generating function Warning. We have interchanged the letters f and g compared to the notation of Shapiro et al. [69,70] and Barry [3]. This notation seems to us more natural, but the reader should be warned.
We shall use an easy but important result that is sometimes called the fundamental theorem of Riordan arrays (FTRA): Lemma 2.16 (Fundamental theorem of Riordan arrays). Let b = (b n ) n≥0 be a sequence with ordinary generating function B(t) = ∞ n=0 b n t n . Considering b as a column vector and letting R(f, g) act on it by matrix multiplication, we obtain a sequence R(f, g)b whose ordinary generating function is f (t) B(g(t)).

Proof. We compute
We can now determine the production matrix of a Riordan array R(f, g). Let a = (a n ) n≥0 and z = (z n ) n≥0 be sequences in a commutative ring R, with ordinary generating functions A(t) = ∞ n=0 a n t n and Z(t) = ∞ n=0 z n t n . We then define the AZ matrix associated to the sequences a and z by We also write AZ(A, Z) as a synonym for AZ(a, z). It is convenient to define also , (2.27) which is well-defined if a 0 is invertible in R. We then have [24,42]  Theorem 2.17 (Production matrices of Riordan arrays). Let L be a lower-triangular matrix (with entries in a commutative ring R) with invertible diagonal entries and L 00 = 1, and let P = L −1 ∆L be its production matrix. Then L is a Riordan array if and only if P is an AZ-matrix. More precisely, L = R(f, g) if and only if P = AZ(a, z), where the generating functions f (t), g(t) and A(t), Z(t) are connected by or equivalently , Proof [42, p. 18]. Suppose that L = R(f, g). The hypotheses on L imply that f 0 = 1 and that g 1 is invertible in R; so g(t) has a compositional inverseḡ(t). Now let (p k (t)) k≥0 be the column generating functions of P = L −1 ∆L. Applying the FTRA to each column of P , we see that R(f, g)P is a matrix whose column generating functions are f (t) p k (g(t)) k≥0 . On the other hand, ∆ R(f, g) is the matrix R(f, g) with its zeroth row removed and all other rows shifted upwards, so it has column generating functions [f (t) − 1]/t for column 0 and f (t)g(t) k /t for columns k ≥ 1.
Comparing these two results, we see that ∆ R(f, g) = R(f, g) P if and only if and The latter equation can be rewritten as , (2.32) which means that the columns k ≥ 1 of the production matrix P are identical with those of AZ(a, z), when a is given by (2.29). And (2.30) then states that column 0 of the production matrix P is identical with that of AZ(a, z), when z is given by (2.29). ) where a and z are given by (2.29). Conversely, suppose that P = AZ(a, z). Let g(t) be the unique formal power series in R[[t]] with g(0) = 0 that satisfies the functional equation g(t) = t A(g(t)), and then let f (t) = 1/[1 − tZ(g(t))]. Then running the foregoing computation backwards shows that ∆ R(f, g) = R(f, g) P . Since by hypothesis L 00 = 1, it follows that L = O(P ) = R(f, g).

Exponential Riordan arrays
Let R be a commutative ring containing the rationals, and let F (t) = ∞ n=0 f n t n /n! and G(t) = ∞ n=1 g n t n /n! be formal power series with coefficients in R; we set g 0 = 0. Then the exponential Riordan array [3,22,24,70]  Please note that the exponential Riordan array R[F, G] is nothing other than a diagonal similarity transform of the ordinary Riordan array R(F, G) associated to the same power series F and G: that is, be a sequence with exponential generating function B(t) = ∞ n=0 b n t n /n!. Considering b as a column vector and letting R[F, G] act on it by matrix multiplication, we obtain a sequence R[F, G]b whose exponential generating function is F (t) B(G(t)).
Let us now consider the product of two exponential Riordan arrays R[F 1 , G 1 ] and R[F 2 , G 2 ]. Applying the FTERA to the kth column of R[F 2 , G 2 ], whose exponential generating function is F 2 (t)G 2 (t) k /k!, we readily obtain: Lemma 2.19 (Product of two exponential Riordan arrays). We have We can now determine the production matrix of an exponential Riordan array R[F, G]. Let a = (a n ) n≥0 and z = (z n ) n≥0 be sequences in a commutative ring R, with ordinary generating functions A(s) = ∞ n=0 a n s n and Z(s) = ∞ n=0 z n s n . We then define the exponential AZ matrix associated to the sequences a and z by EAZ(a, z) nk = n! k! (z n−k + k a n−k+1 ) , (2.40) or equivalently (if R contains the rationals) where D = diag (n!) n≥0 . We also write EAZ(A, Z) as a synonym for EAZ(a, z).
Remark. We have the exponential generating functions ∞ n,k=0 (2.43) Theorem 2.22 (Production matrices of exponential Riordan arrays). Let L be a lower-triangular matrix (with entries in a commutative ring R containing the rationals) with invertible diagonal entries and L 00 = 1, and let P = L −1 ∆L be its production matrix. Then L is an exponential Riordan array if and only if P is an exponential AZ matrix.
or equivalently Proof (mostly contained in [3, pp. 217-218]). Suppose that L = R[F, G]. The hypotheses on L imply that f 0 = 1 and that g 1 is invertible in R; so G(t) has a compositional inverseḠ(s). Now let P = (p nk ) n,k≥0 be a matrix; its column exponential generating functions are, by definition, P k (t) = ∞ n=0 p nk t n /n!. Applying the FTERA to each column of P , we see that R[F, G]P is a matrix whose column exponential generating functions are F (t) P k (G(t)) k≥0 . On the other hand, ∆ R[F, G] is the matrix R[F, G] with its zeroth row removed and all other rows shifted upwards, so it has column exponential generating functions  47) or in other words where a = (a n ) n≥0 and z = (z n ) n≥0 are given by (2.45). Conversely, suppose that P = EAZ(A, Z). Define F (t) and G(t) as the unique solutions (in the formal-power-series ring R[[t]]) of the differential equations (2.44) with initial conditions F (0) = 1 and G(0) = 0. Then running the foregoing computation backwards shows that ∆ We refer to A(s) = ∞ n=0 a n s n and Z(s) = ∞ n=0 z n s n as the A-series and Zseries associated to the exponential Riordan array R[F, G].
Remark. The identity A(s) = G (Ḡ(s)) can equivalently be written as A(s) = 1/(Ḡ) (s). This is useful in comparing our work with that of Zhu [87,89], who uses the latter formulation.
Let us now show how to rewrite the production matrix (2.41) in a new way, which will be useful in what follows. Define so that F (t) = Ψ(G(t)) and Ψ(0) = F (0) = 1. Then a simple computation using (2.44)/(2.45) shows that And conversely, given any pair (A, Z) of formal power series (over a commutative ring R containing the rationals) such that A(0) is invertible in R, there is a unique pair (Φ, Ψ) satisfying (2.52) together with the normalization Ψ(0) = 1, namely [Here the integral of a formal power series is defined by It is the unique formal power series with zero constant term whose derivative is the given series.] We refer to Φ(s) and Ψ(s) as the Φ-series and Ψ-series associated to the exponential Riordan array R[F, G].
Rewriting the production matrix (2.41) in terms of the pair (Φ, Ψ) provides a beautiful -and as we shall see, very useful -factorization. For reasons that shall become clear shortly (see Lemma 2.27 below), it is convenient to study the more general quantity EAZ(A, Z + ξA): ψ n s n be formal power series with coefficients in R, and let A(s) and Z(s) be defined by (2.52). Now let ξ be any element of R (or an indeterminate). Then To prove Proposition 2.23, we need a lemma. Given a sequence ψ = (ψ n ) n≥0 in R with ordinary generating function Ψ(s) = ∞ n=0 ψ n s n , we define ψ = (ψ n ) n≥0 by ψ n = (n + 1)ψ n+1 , so that Ψ (s) = ∞ n=0 ψ n s n . We then have: Lemma 2.24. Let ψ and ψ be as above, and let D = diag (n!) n≥0 . Then Proof. All three matrices in (2.56) are lower-Hessenberg, and their (n, k) matrix elements are (for 0 ≤ k ≤ n + 1) Remarks. 1. The identity (2.56) can also be written as (2.58) The definitions (2.52) imply where the next-to-last step used Lemma 2.24.
As an immediate consequence of Proposition 2.23, we have: Remark. The hypothesis that the ring R contains the rationals can be removed, by using Lemma 2.30 (see Section 2.7) together with the reasoning used in the proof of Theorem 1.8 (see Section 5.3).
It is worth observing that the converse to Corollary 2.25 is false: Example 2.26. Let A(s) = 1 + s and Z(s) = (λ + µ) + µs. Then P = EAZ(A, Z) is the tridiagonal matrix with p n,n+1 = 1, p n,n = λ + µ + n and p n,n−1 = nµ, which can be written in the form P = LU + λI, where L is the lower-bidiagonal matrix with 1 on the diagonal and 1, 2, 3, . . . on the subdiagonal, U is the upper-bidiagonal matrix with 1 on the superdiagonal and µ on the diagonal, and I is the identity matrix; so by the tridiagonal comparison theorem [77]  which are not Toeplitz-TP coefficientwise in λ and µ. Indeed, even for real λ and µ, the sequence φ (resp. ψ) is Toeplitz-TP only for λ ∈ {0, 1} and µ ≤ 0 (resp. λ ∈ {0, 1} and µ ≥ 0). So all nonnegative λ, µ other than λ ∈ {0, 1} and µ = 0 yield counterexamples to the converse to Corollary 2.25, and even to its restriction to the case ξ = 0. In this example F (t) = e λt+µ(e t −1) and G(t) = e t − 1, so the exponential Riordan So the condition of Corollary 2.25 is sufficient but not necessary for its conclusion.
Finally, a central role will be played in this paper by a simple but remarkable identity for B −1 ξ EAZ(a, z) B ξ , where B ξ is the ξ-binomial matrix defined in (1.6) and EAZ(a, z) is the exponential AZ matrix defined in (2.40)/(2.41).
Let a = (a n ) n≥0 , z = (z n ) n≥0 and ξ be indeterminates. Then The special case z = 0 of this lemma was proven in [62, Lemma 3.6]; a simpler proof was given in [76,Lemma 2.16]. Here we give the easy generalization to include z. We will give two proofs: a first proof by direct computation from the definition (2.40)/(2.41), and a second proof using exponential Riordan arrays. where D = diag (n!) n≥0 . Since EAZ(a, z) = EAZ(a, 0) + EAZ(0, z), it suffices to consider separately the two contributions.
The key observation is that B ξ = D T ∞ (ξ n /n!) n≥0 D −1 . Now two Toeplitz matrices always commute: On the other hand, the classic recurrence for binomial coefficients implies ∆B ξ = B ξ (ξI + ∆) (2.65) (cf. Example 2.10). Therefore   Corollary 2.28 will be the main theoretical tool in this paper. Finally, it is worth singling out a subclass of Riordan arrays that will occur in the cases to be studied in the present paper:   The case c = 0 (that is, Ψ = 1 and hence F = 1) corresponds to the associated subgroup (or Lagrange subgroup) of exponential Riordan arrays; it arose in our earlier work [62,76] on generic Lah and rooted-forest polynomials. Using criterion (a), we can already see that the matrix T defined in (1.1) will correspond to c = 1, while the matrices T(y, z) and T(y, φ) defined in (1.7)/(1.12) will correspond, according to Propositions 1.3 and 1.6, to c = y. Of course, in order to apply Lemma 2.29 we will first need to prove that these matrices are indeed exponential Riordan arrays: that will be done in Section 4. But we can see now that, once we do this, the Ψ-series will be Ψ(s) = 1/(1 − cs).

A lemma on diagonal scaling
Given a lower-triangular matrix A = (a nk ) n,k≥0 with entries in a commutative ring R, let us define the matrix A = (a nk ) n,k≥0 by a nk = n! k! a nk ; (2.72) this is well-defined since a nk = 0 only when n ≥ k, in which case n!/k! is an integer.
If R contains the rationals, we can of course write A = DAD −1 where D = diag (n!) n≥0 . And if R is a partially ordered commutative ring that contains the rationals and A is TP r , then we deduce immediately from A = DAD −1 that also A is TP r . The following simple lemma [62,Lemma 3.7] shows that this conclusion holds even when R does not contain the rationals: Lemma 2.30. Let A = (a ij ) i,j≥0 be a lower-triangular matrix with entries in a partially ordered commutative ring R, and let d = (d i ) i≥1 . Define the lower-triangular

Lagrange inversion
We will use Lagrange inversion in the following form [38]: If φ(u) is a formal power series with coefficients in a commutative ring R containing the rationals, then there exists a unique formal power series f (t) with zero constant term satisfying and it is given by and more generally, if H(u) is any formal power series, then In particular, taking H(u) = u k with integer k ≥ 0, we have T a improper · · · · · · T r Figure 1: Bijection between T and T .

Bijective proofs
In this section we give bijective proofs of Propositions 1.3, 1.5, 1.6 and 1.7. This section can be skipped on a first reading, as it is not needed for proving the main theorems of the paper.

Proof of Propositions 1.3 and 1.6
Here we will prove Proposition 1.3, which asserts that the polynomials t n,k (y, z) defined in (1.7) satisfy t n,0 (y, z) = y t n,1 (y, z) for all n ≥ 1; and more generally Proposition 1.6, which asserts that the polynomials t n,k (y, φ) defined in (1.12) satisfy t n,0 (y, φ) = y t n,1 (y, φ) for all n ≥ 1.
We will prove these results by constructing, for each n ≥ 1, a bijection from the set T This construction is illustrated in Figure 1. Since the weight in (1.12) is y for each improper edge and φ m = m! φ m for each vertex i = 1 with m proper children, this proves t n,0 (y, φ) = y t n,1 (y, φ). Specializing to φ m = z m /m! then yields t n,0 (y, z) = y t n,1 (y, z).
Proof of Proposition 1.6. Fix n ≥ 1, and let T be a rooted tree on the vertex set [n + 1] in which r is the root and the vertex 1 has precisely one child a. Let T a be the subtree rooted at a, and let T r the subtree obtained from T by removing T a and the edge 1a. The vertex 1 is a leaf in T r . Now we create a new tree T , rooted at a, as follows: we start with T a and then graft T r by making r a child of a. In the tree T , the vertex 1 is a leaf. The map T → T map is a bijection, since this construction can be reversed. (The vertex r can be identified in T as the child of a that has 1 as a descendant.) Clearly, all the proper (resp. improper) edges in T are still proper (resp. improper) in T , except that: (i) The edge 1a in T is proper, which is deleted in T ; and (ii) The edge ar in T is new and improper, since the vertex 1 is a descendant of r.
In particular, the number of vertices with m proper children is the same in T and T , provided that in T one ignores the vertex 1.

Proof of Propositions 1.5 and 1.7
Now we will prove Proposition 1.5, which asserts the equality of the polynomials t n,k (y, z) defined in (1.7) using rooted trees and the polynomials t n,k (y, z) defined in (1.10) using partial functional digraphs. We will then show that the same argument proves the more general Proposition 1.7, which asserts the equivalence of the polynomials t n,k (y, φ) defined in (1.12) and the polynomials t n,k (y, φ) defined in (1.13).
We recall that T • n denotes the set of rooted trees on the vertex set [n], while T 1;k n denotes the subset in which the vertex 1 has k children. Similarly, PFD n denotes the set of partial functional digraphs on the vertex set [n], while PFD n,k denotes the subset in which there are exactly k vertices of out-degree 0.
To prove Proposition 1.5, we will construct, for each fixed n, a bijection φ : T • n+1 → PFD n with the following properties: We observe that (c) is an immediate consequence of (a) and (b), since trees in T • n+1 have n edges, while digraphs in PFD n,k have n − k edges.
Proof of Proposition 1.5. (The reader may wish to follow, along with this proof, the example shown in Figure 2.) Let T be a rooted tree on the vertex set [n + 1] in which the vertex 1 has k children. Note that the k edges from vertex 1 to its children are all proper. Now let P = v 1 · · · v +1 ( ≥ 0) be the unique path in T from the root v 1 = r to the vertex v +1 = 1; we call it the "backbone". (Here = 0 corresponds to the case in which vertex 1 is the root.) Removing from T the edges of the path P , we obtain a collection of (possibly trivial) trees T 1 , . . . , T +1 rooted at the vertices v 1 , . . . , v +1 . (c 1 ,c 2 ) Partial functional digraphs G and G in the third model, where the two vertices 10 and 12 (resp. 9 and 11) have out-degree 0. Improper edges arising from the cycles of the permutation σ are shown in red; the other improper edges are shown in black; the proper edges are shown in blue. Now regard P as a permutation σ (written in word form) of its elements written in increasing order. 10 In particular, σ(1) = r and σ(v max ) = 1 where v max = max(v 1 , . . . , v +1 ). Let D P be the digraph whose vertex set is {v 1 , . . . , v +1 }, with edges − → ij whenever j = σ(i). Then D P consists of disjoint directed cycles (possibly of length 1); it is the representation in cycle form of the permutation σ. Now let D be the digraph obtained from D P by attaching the trees T 1 , . . . , T +1 to D P (identifying vertices with the same label) and directing all edges of those trees towards the root. Then D is a functional digraph on the vertex set [n + 1]. Furthermore, the map T → D is a bijection, since all the above steps can be reversed. Now let G be the digraph obtained from D by deleting the vertex 1 and the k tree edges incident on vertex 1, and contracting the edges Then G is a digraph on the vertex set {2, . . . , n + 1} in which every vertex has out-degree 1 except for the k children of vertex 1 in T , which have outdegree 0. Relabeling all vertices i → i − 1, we obtain a partial functional digraph G = φ(T ) ∈ PFD n,k .
The step from D to G can also be reversed: given a partial functional digraph G = PFD n,k , we relabel the vertices i → i+1 and then insert the vertex 1 immediately after the largest cyclic vertex of G (if any; otherwise 1 becomes a loop in D ); all the vertices of out-degree 0 in G are made to point to the vertex 1 in D .
It follows that the map φ : T → G is a bijection from T • n+1 to PFD n that maps T 1;k n+1 onto PFD n,k .
Clearly, in the rooted tree T , all the edges in the path P = v 1 · · · v +1 are improper, since each vertex in P has v +1 = 1 as its descendant. These edges correspond, after relabeling, to + 1 cyclic edges in the functional digraph D . These latter edges in turn correspond, after removal of vertex 1 and contraction of its edges, to cyclic edges in the partial functional digraph G (and hence also G). Because they are cyclic edges, they are necessarily improper. All the other improper/proper edges in T coincide with improper/proper edges − → ij in the partial functional digraph G (and hence G) where i is a transient vertex.
Remark. The first part of this proof (namely, the map T → D ) is the wellknown bijection from doubly-rooted trees to functional digraphs on the same vertex set [49, pp. 224-225] [79, p. 26]. In our application we need the second step to remove the vertex 1 and thereby obtain a map from rooted trees on the vertex set [n + 1] to partial functional digraphs on the vertex set [n].
Proof of Proposition 1.7. In the preceding proof, each vertex i = 1 in the rooted tree T corresponds to a vertex i − 1 in the partial functional digraph G = φ(T ). And for each proper child j of i in T , the proper edge ij in T corresponds to a proper edge − −−−−−− → j − 1 i − 1 in G; and those are the only proper edges in G. Therefore, if the vertex i = 1 in T has m proper children, then the vertex i − 1 in G has m proper incoming edges. This proves that t n,k (y, φ) = t n,k (y, φ).

The matrices T, T(y, z) and T(y, φ) as exponential Riordan arrays
In this section we show that the matrices T, T(y, z) and T(y, φ) are exponential Riordan arrays R[F, G], and we compute their generating functions F and G as well as their A-, Z-, Φ-and Ψ-series. Then the unit-lower-triangular matrix T = (t n,k ) n,k≥0 is the exponential Riordan array R[F, G] with F (t) = ∞ n=0 n n t n /n! and G(t) = ∞ n=1 n n−1 t n /n!.

The matrix T
Before proving Proposition 4.1, let us use it to compute the A-, Z-, Φ-and Ψseries: Proof. We observe that G(t) is the tree function T (t) [19], which satisfies the functional equation T (t) = te T (t) . Furthermore, we have F (t) = 1/[1 − T (t)]: this wellknown fact can be proven using the Lagrange inversion formula [see (4.4) below specialized to x = 0] or by various other methods. 11 We now apply Theorem 2.22 to determine the functions A(s) and Z(s). Implicit differentiation of the functional equation yields T (t) = e T (t) /[1−T (t)], which implies that A(s) = e s /(1−s). On the other hand, it follows immediately from the relation between F and G that Ψ(s) = 1/(1−s). This implies that Φ(s) = e s and Z(s) = e s /(1 − s) 2 .
We will give five proofs of Proposition 4.1: a direct algebraic proof using Lagrange inversion and an Abel identity; an inductive algebraic proof, using a different Abel identity; a third algebraic proof using the A-and Z-sequences of an ordinary Riordan array; a combinatorial proof using exponential generating functions based on the 11 Algebraic proof.
where the first equality used the power series defining F (t) and T (t), the second equality interpretation of t n,k as counting partial functional digraphs; and a bijective combinatorial proof based on the interpretation of t n,k as counting rooted labeled trees according to the number of children of the root that are lower-numbered than the root. In Section 4.2 we will give yet another combinatorial proof (also using exponential generating functions), this time based on the interpretation of t n,k as counting rooted labeled trees according to the number of children of a specified vertex i; but this proof will be given in the more general context of the polynomials t n,k (y, z).  Second Proof of Proposition 4.1. It is immediate that the zeroth column of T has exponential generating function F (t) = ∞ n=0 n n t n /n!. We now show by induction on k that the kth column has egf F (t) G(t) k /k! where G(t) = ∞ n=0 n n−1 t n /n!: that is, we need to show that the kth column has egf equal to G(t)/k times the egf of the (k − 1)st column, or in other words for k ≥ 1. We start from the Abel identity [67, p. 18, eq. (13a)] Now substitute x = 1 and y → y −n−1, divide both sides by n!, and relabel i = j −1: the result is Next substitute n → n − k and then set y = n: Multiplying this by n! yields which is (4.5).
Third Proof of Proposition 4.1. Showing that (t n,k ) n,k≥0 equals the exponential Riordan array R[F, G] is equivalent to showing that ((k!/n!) t n,k ) n,k≥0 equals the ordinary Riordan array R(F, G). We write r n,k def = (k!/n!) t n,k = n n−k /(n − k)! and R = (r n,k ) n,k≥0 . By the binomial theorem we have And the ordinary generating function of the zeroth column of R is obviously F (t) = ∞ n=0 n n t n /n!.
Remark. The identity (4.8) in the second proof can be written for k ≥ 1 as 13) which shows that the ordinary generating function of the kth column of R equals G(t) = ∞ j=1 j j−1 t j /j! times the ordinary generating function of the (k −1)st column. Combining this with the fact that the ordinary generating function of the zeroth column of R is F (t) = ∞ n=0 n n t n /n! gives an alternate proof that R = R(F, G).
Fourth Proof of Proposition 4.1. We begin from the fact that t n,k = n k n n−k counts partial functional digraphs on n labeled vertices that have k vertices of outdegree 0. Such a partial functional digraph is the disjoint union of k rooted trees (rooted at the vertices of out-degree 0) together with a functional digraph on the remaining vertices. Standard enumerative arguments then imply that the exponential generating function for the numbers t n,k is where F (t) = ∞ n=0 n n t n /n! is the exponential generating function for functional digraphs and T (t) = ∞ n=1 n n−1 t n /n! is the exponential generating function for rooted trees.
Fifth Proof of Proposition 4.1. We begin from the fact [12,13,75] that t n,k equals the number of rooted trees on the vertex set [n + 1] in which exactly k children of the root are lower-numbered than the root. We will prove (4.5) in the form for k ≥ 1. We interpret k t n,k as the number of triplets (T, r, v ) in which (T, r) is a rooted tree on the vertex set [n + 1] in which exactly k children of the root are lowernumbered than the root, and v is one of those lower-numbered children (we call it the "marked vertex"). See Figure 3. We interpret j j−1 as the number of rooted trees on j labeled vertices. So the summand on the right-hand side of (4.15) enumerates quintuplets (A, T 1 , r 1 , T 2 , r 2 ) where A is a subset of [n] of cardinality j, (T 1 , r 1 ) is a rooted tree on the vertex set A, and (T 2 , r 2 ) is a rooted tree on the vertex set [n+1]\A in which exactly k − 1 children of the root are lower-numbered than the root. See Figure 4.
Bijection RHS =⇒ LHS. Given the quintuplet (A, T 1 , r 1 , T 2 , r 2 ), we construct a triplet (T, r, v ) as follows. We distinguish two cases: • Case I: r 1 < r 2 . We let r 2 be the new root and add an edge making r 1 a child of r 2 ; this gives (T, r). We then mark the vertex v = r 1 . Please note that in this case the vertex n + 1 is not a descendant of v (see Figure 5a).
• Case II: r 1 > r 2 . We let r 1 be the new root and add an edge making r 2 a child of r 1 ; we then interchange the lower-numbered children of r 1 (together with all their descendants) with the lower-numbered children of r 2 (and their descendants). This gives (T, r). We observe (see Figure 5b) that r 2 is the largest-numbered among all the lower-numbered children of r in T . We observe also that the vertex n+1 must be a descendant of some lower-numbered child of r 1 in T ; we set the marked vertex v to be this lower-numbered child. Note that v must either belong to the set S 2 (consisting of the lower-numbered children of r 2 in T 2 , which became lower-numbered children of r 1 in T ) or else be the vertex r 2 .
In both cases, in the rooted tree (T, r), exactly k children of the root are lowernumbered than the root.
Tree T on vertex set [n + 1] Figure 3: A triplet (T, r, v ), where v 1 , . . . , v k are the children of the root r that are lower-numbered than r, and H depicts the children of r that are higher-numbered than r.
Tree T 2 on vertex set [n + 1] \ A Figure 4: A quintuplet (A, T 1 , r 1 , T 2 , r 2 ). Here L i (resp. H i ) depicts the children of the root r i that are lower-numbered (resp. higher-numbered) than r i . In this figure and the following ones, the shaded parts are the possible locations of the vertex n + 1.   Figure 6: Bijection LHS ⇒ RHS: Case I. Figure 7: Bijection LHS ⇒ RHS: Case II.
We now describe the inverse bijection: Bijection LHS =⇒ RHS. Given the triplet (T, r, v ), we reconstruct the quintuplet (A, T 1 , r 1 , T 2 , r 2 ) as follows. We distinguish two cases: • Case I: n + 1 is not a descendant of v . We delete the edge between r (the root of T ) and v . Then (T 1 , r 1 ) is the tree whose root is v , and A is its vertex set; (T 2 , r 2 ) is the tree whose root is r, and its vertex set is [n + 1] \ A. Since n + 1 is not a descendant of v , it n + 1 belongs to T 2 , so that A ⊆ [n]. And we have r 1 = v < r = r 2 . See Figure 6.
• Case II: n + 1 is a descendant of v . The root r of T has k (≥ 1) lowernumbered children; let v • be the largest-numbered of these. We delete the edge between r and v • ; then we interchange the lower-numbered children of r (together with all their descendants) with the lower-numbered children of v • (and their descendants). Then (T 1 , r 1 ) is the tree whose root is r, and A is its vertex set; (T 2 , r 2 ) is the tree whose root is v • , and its vertex set is [n + 1] \ A. Please observe that the marked vertex v was a lower-numbered child of r in T ; therefore, it is either equal to v • = r 2 or else becomes a lower-numbered child of v • = r 2 in T 2 . Since the vertex n + 1 was a descendant of v , it must belong to T 2 ; therefore A ⊆ [n]. And we have r 1 = r > v • = r 2 . See Figure 7.
In both cases, in the rooted tree (T 2 , r 2 ), exactly k − 1 children of the root are lowernumbered than the root.

The matrix T(y, z)
We now prove that the matrix T(y, z) = (t n,k (y, z)) n,k≥0 is an exponential Riordan array R[F, G], and we compute F and G. Most of this computation was done a quarter-century ago by Dumont and Ramamonjisoa [26]: their arguments handled the case k = 0, and we extend those arguments slightly to handle the case of general k. Our presentation follows the notation of [76].
Let T • n denote the set of rooted trees on the vertex set [n]; let T in which each improper (resp. proper) edge gets a weight y (resp. z) except that in A n,k the k proper edges connecting the vertex 1 to its children are unweighted. And then define the exponential generating functions A n,k (y, z) t n n! (4.21) We will then prove the following key result, which is a slight extension of [26,Proposition 7] to handle the case k = 0: The series R, S and A k satisfy the following identities: (a) S(t; y, z) = exp z R(t; y, z) and hence (d) d dt R(t; y, z) = exp z R(t; y, z) 1 − yR(t; y, z) Solving the differential equation of Proposition 4.3(d) with the initial condition R(0; y, z) = 0, we obtain: Corollary 4.4. The series R(t; y, z) satisfies the functional equation (4.22) and hence has the solution where T (t) is the tree function (1.3).
Comparing Proposition 4.3(b) with the definition (2.33) of exponential Riordan arrays, we conclude: and R(t; y, z) is given by (4.23). n+1 , and suppose that the root vertex 1 has k (≥ 0) children. All k edges emanating from the root vertex are proper and thus get a weight z each. Deleting these edges and the vertex 1, one obtains an unordered partition of {2, . . . , n + 1} into blocks B 1 , . . . , B k and a rooted tree T j on each block B j . Standard enumerative arguments then yield the relation (a) for the exponential generating functions.
(b) Consider a tree T ∈ T 1;k n+1 with root r, and let r 1 , . . . , r l+1 (l ≥ 0) be the path in T from the root r 1 = r to the vertex r l+1 = 1. 12 All l edges of this path are improper, and all k edges from the vertex 1 to its children are proper (and unweighted). Deleting these edges and the vertex 1, one obtains a partition of {2, . . . , n + 1} into an ordered collection of blocks B 1 , . . . , B l and an unordered collection of blocks B 1 , . . . , B k , together with a rooted tree on each block. Standard enumerative arguments then yield the relation (b) for the exponential generating functions.
(c) In a tree T ∈ T • n , focus on the vertex 1 (which might be the root, a leaf, both or neither). Let T be the subtree rooted at 1, and let T be the tree obtained from T by deleting all the vertices of T except the vertex 1 (it thus has the vertex 1 as a leaf). The vertex set [n] is then partitioned as {1} ∪ V ∪ V , where {1} ∪ V is the vertex set of T and {1} ∪ V is the vertex set of T ; and T is obtained by joining T and T at the common vertex 1. Standard enumerative arguments then yield the relation (c) for the exponential generating functions.
Remarks. 1. Dumont and Ramamonjisoa also gave [26, sections 2-5] a second (and very interesting) proof of the k = 0 case of Proposition 4.3, based on a contextfree grammar [14] and its associated differential operator.
2. We leave it as an open problem to find a direct combinatorial proof of the functional equation (4.22), without using the differential equation of Proposition 4.3(d).
4. The polynomials R n and A n,0 also arise [44] as derivative polynomials for the tree function: in the notation of [44] we have R n (y, 1) = G n (y − 1) and A n,0 (y, 1) = y F n (y − 1) for n ≥ 1. The formula (4.23) is then equivalent to [44, Theorem 4.2, equation for G n ].

The matrix T(y, φ)
We now show how Proposition 4.3 can be generalized to incorporate the additional indeterminates φ = (φ m ) m≥0 . We define T • n , T [i] n and T i;k n as before, and then define the obvious generalizations of (4.16)-(4.18): where pdeg T (i) denotes the number of proper children of the vertex i in the rooted tree T , and φ m = m! φ m . (Note that in R n and S n we give weights to all the vertices, while in A n,k we do not give any weight to the vertex 1. 13 ) We then define the exponential generating functions Let us also define the generating function We then have: Proposition 4.7. The series R, S and A k defined in (4.30)-(4.32) satisfy the following identities: (a) S(t; y, φ) = Φ R(t; y, φ) and hence Proof. The proof is identical to that of Proposition 4.3, with the following modifications: (a) Consider a tree T ∈ T [1] n+1 in which the root vertex 1 has k children. Since all k edges emanating from the root vertex are proper, we get here a factor φ k /k! in place of the z k /k! that was seen in Proposition 4.3. Therefore, the function e zs in Proposition 4.3 is replaced here by the generating function Φ(s).
(b) No change is needed.
(c) No change is needed. (The tree T has vertex 1 as a leaf, but in A n,0 the vertex 1 is anyway unweighted.) Comparing Proposition 4.7(b) with the definition (2.33) of exponential Riordan arrays, we conclude: , G(t) = R(t; y, φ) (4.34) and R(t; y, φ) is the solution of the differential equation of Proposition 4.7(d) with initial condition R(0; y, φ) = 0.
We see that Ψ(s) is the same here as in (4.26); only Φ is different. In this section we will prove Theorems 1.1, 1.2, 1.4 and 1.8. The proofs are now very easy: we combine the general theory of total positivity in exponential Riordan arrays developed in Section 2 (culminating in Corollary 2.28) with the specific computations of Φ-and Ψ-series carried out in Section 4.
It suffices of course to prove Theorem 1.8, since Theorems 1.1, 1.2 and 1.4 are contained in it as special cases: take φ m = z m /m! to get Theorem 1.4; then take y = z = 1 to get Theorems 1.1 and 1.2. However, we find it instructive to work our way up, starting with Theorems 1.1 and 1.2 and then gradually adding extra parameters.

The matrix T
Proof of Theorems 1.1 and 1.2. In order to employ the theory of exponential Riordan arrays, we work here in the ring Q, even though the matrix elements actually lie in Z.
Since this proof employed the production-matrix method (hidden inside Corollary 2.28), it is worth making explicit what the production matrix is: Proposition 5.1 (Production matrix for T). The production matrix P = T −1 ∆T is the unit-lower-Hessenberg matrix where B 1 is the binomial matrix [i.e. (1.6) at x = 1], T 1 is the lower-triangular matrix of all ones [i.e. (2.2) at x = 1], and D = diag (n!) n≥0 . More generally, we have Proof. Since φ m = 1/m! and ψ m = 1, Proposition 2.23 implies and Lemma 2.27 implies (5.2).
Remarks. 1. The zeroth and first columns of the matrix P are identical: that is, p n,0 = p n,1 . This can be seen from Lemma 2.29 with c = 1, by noting either that t n,0 = t n,1 for n ≥ 1 or that Ψ(s) = 1/(1−s). Alternatively, it can be seen directly from (5.1): the zeroth and first columns of the matrix ∆ DT 1 D −1 are identical (namely, they are both equal to 1/(n+1)!); so the zeroth and first columns of M ∆ DT 1 D −1 are identical, for any row-finite matrix M . (Indeed, this would be the case if D = diag( (n!) n≥0 ) were replaced by any diagonal matrix diag(d 0 , d 1 , d 2 , . . .) satisfying d 0 = d 1 .) We will also see that p n,0 = p n,1 in the explicit formula (5.16).
The equality p n,0 = p n,1 implies, by Lemma 2.29(b) ⇐⇒ (b ), the factorization P = P ∆ T (e 00 + ∆) (5.4) where e 00 denotes the matrix with an entry 1 in position (0, 0) and all other entries zero, and P ∆ T is the lower-triangular matrix obtained from P by deleting its zeroth column.
2. Closely related to the production matrix P = B 1 ∆ DT 1 D −1 are It was shown in [76, Section 4.1] that P is the production matrix for the forest matrix F = (f n,k ) n,k≥0 where f n,k = n k k n n−k−1 counts k-component forests of rooted trees on n labeled vertices; and that P = ∆ P ∆ T is the production matrix for F = ∆F∆ T = (f n+1,k+1 ) n,k≥0 . All three production matrices correspond to the same Aseries A(s) = e s /(1 − s), but with different splittings into Φ and Ψ.
We have more to say about this production matrix P , but in order to avoid disrupting the flow of the argument we defer it to Section 5.4.

The matrix T(y, z)
Proof of Theorem 1.4. In order to employ the theory of exponential Riordan arrays, we work here in the ring Q[y, z], even though the matrix elements actually lie in Z[y, z].
Analogously to Proposition 5.1, we have: Production matrix for T(y, z)). The production matrix P (y, z) = T(y, z) −1 ∆T(y, z) is the unit-lower-Hessenberg matrix where B z is the weighted binomial matrix (1.6), T y is the Toeplitz matrix of powers (2.2), and D = diag (n!) n≥0 . More generally, Remarks. 1. The zeroth and first columns of the matrix P (y, z) satisfy p n,0 = yp n,1 . This can be seen from Lemma 2.29 with c = y, by noting either that t n,0 (y, z) = yt n,1 (y, z) for n ≥ 1 (Proposition 1.3) or that Ψ(s) = 1/(1 − ys). Alternatively, it can be seen directly from (5.1): the zeroth column of the matrix ∆ DT y D −1 is y times the first column (they are, respectively, y n+1 /(n + 1)! and y n /(n + 1)!); so the zeroth column of M ∆ DT y D −1 is y times the first column, for any row-finite matrix M .
The equality p n,0 = yp n,1 implies, by Lemma 2.29(b) ⇐⇒ (b ), the factorization P (y, z) = P (y, z) ∆ T (y e 00 + ∆) . (5.8) 2. Closely related to the production matrix P (y, z) = B z ∆ DT y D −1 are It was shown in [76,Section 4.3] that P (y, z) is the production matrix for F(y, z) = f n,k (y, z) n,k≥0 where f n,k (y, z) counts k-component forests of rooted trees on the vertex set [n] with a weight y (resp. z) for each improper (resp. proper) edge. Likewise, P (y, z) = ∆ P (y, z)∆ T is the production matrix for F (y, z) = ∆F(y, z)∆ T = f n+1,k+1 (y, z) n,k≥0 . All three production matrices correspond to the same A-series A(s) = e zs /(1 − ys), but with different splittings into Φ and Ψ.

The matrix T(y, φ)
The proof is similar to that in the preceding subsections, but a bit of care is needed to handle the case in which the ring R does not contain the rationals. We now use the definition (2.72) to rewrite this as Having done this, the equality T(y, φ)B x = O(B −1 x P B x ) is now a valid identity in the ring Z[y, φ]. We can therefore now substitute elements φ in any commutative ring R for the indeterminates φ, and the identity still holds.
By hypothesis the sequence φ is Toeplitz-totally positive in the ring R. By Lemma 2.4, the sequence ψ is Toeplitz-totally positive in the ring Z[y] equipped with the coefficientwise order. By Lemma 2.30, the matrices T ∞ (φ) and T ∞ (ψ) are also totally positive. Therefore B −1 x P B x is totally positive in the ring R[x, y] equipped with the coefficientwise order. Proposition 2.15 then yields Theorem 1.8. T(y, φ)). The production matrix P (y, φ) = T(y, φ) −1 ∆T(y, φ) is the unit-lower-Hessenberg matrix

Proposition 5.3 (Production matrix for
where T y is the Toeplitz matrix of powers (2.2), and is defined in (2.72). More generally, Remark. 1. The zeroth and first columns of the matrix P (y, φ) satisfy p n,0 = yp n,1 , for exactly the same reasons as were observed for P (y, z). This implies the factorization P (y, φ) = P (y, φ) ∆ T (y e 00 + ∆) . (5.14) 2. Closely related to the production matrix P (y, φ) = T ∞ (φ) ∆ T y are It was shown in [76,Section 4.4] that P (y, φ) is the production matrix for F(y, φ) = f n,k (y, φ) n,k≥0 where f n,k (y, φ) counts k-component forests of rooted trees on the vertex set [n] with a weight y for each improper edge and a weight φ m def = m! φ m for each vertex with m proper children. Likewise, P (y, φ) = ∆ P (y, φ)∆ T is the production matrix for F (y, φ) = ∆F(y, φ)∆ T = f n+1,k+1 (y, φ) n,k≥0 . All three production matrices correspond to the same A-series A(s) = Φ(s)/(1 − ys), but with different splittings into Φ and Ψ.

More on the production matrix for T
We now wish to say a bit more about the production matrix P for the tree matrix T. We begin by giving an explicit formula: These matrix elements satisfy in particular p n,0 = p n,1 = nS n + 1 for all n ≥ 0.
The formula (5.16) has a very easy proof, based on the theory of exponential Riordan arrays together with our formulae for A(s) and Z(s); we begin by giving this proof. On the other hand, it is also of some interest to see that this production matrix can be found by "elementary" algebraic methods, without relying on the machinery of exponential Riordan arrays or on any combinatorial interpretation; this will be our second proof.
and a little algebra leads to (5.16a,b). It is then easy to see that p n,0 = p n,1 = nS n +1. It follows that P = T −1 ∆T has matrix elements Setting N = n − k + 1 and j = k − 1 + gives which after a bit of playing with the binomial coefficients gives We now use the Abel identity [67, p. 22, eq. (27)] with x = k and y = 1 − N − k = −n: this gives which is (5.16b).
Remarks. 1. The first few rows of this production matrix are This matrix P (or its lower-triangular variant P ∆ T in which the zeroth column is deleted) is not currently in [61]. However, the zeroth and first columns are [61, A001339], and the second column p n,2 = nS n /2 is [61, A036919]. 2. As mentioned earlier, it is not an accident that p n,0 = p n,1 : by Lemma 2.29 this reflects the fact that Ψ(s) = 1/(1 − s), or equivalently that t n,0 = t n,1 . For the same reason, the production matrices P (y, z) and P (y, φ) satisfy p n,0 = yp n,1 .
3. The ordered subset numbers satisfy the recurrence S m = mS m−1 + 1.
Let us now state some further properties of the matrix elements p n,k : Proposition 5.5. Define the matrix P = (p n,k ) n,k≥0 by (5.16)/(5.17). Then: (a) The p n,k are nonnegative integers that satisfy the backward recurrence p n,k = (k + 1)p n,k+1 + n k − 1 (5.28) with initial condition p n,n+1 = 1.
(b) The p n,k are also given by are polynomials in n with integer coefficients. In particular, Q 0 (n) = Q 1 (n) = −1 and Q 2 (n) = 0, so that p n,0 = p n,1 = nS n + 1 and p n,2 = nS n /2.
Proof. (a) It is immediate from (5.16a)/(5.17) that the p n,k are nonnegative integers.
And it is easy to verify, using the recurrence S m = mS m−1 + 1, that the quantities (5.16) indeed satisfy the recurrence (5.28).
(b) Introducing the Ansatz (5.29), a simple computation shows that the recurrence (5.28) for p n,k is equivalent to the recurrence for Q k (n). Furthermore, simple computations show that p n,0 = p n,1 = nS n + 1, so that Q 0 (n) = Q 1 (n) = −1. It is then easy to see that the unique solution of the recurrence (5.31) with initial condition Q 0 (n) = −1 is (5.30).
Remarks. 1. The first few polynomials Q k (n) are This triangular array is apparently not in [61]. In any case it follows immediately from (5.30b) that for k ≥ 3 the leading term in Q k (n) is (k−1)n k−2 . And it also follows from (5.30b) that for k ≥ 4 the next-to-leading term in 2. Before we found either of the two proofs of Proposition 5.4, we initially guessed the formulae (5.16) for p n,k , as follows: Comparison of successive columns of (5.27) suggested the backwards recurrence (5.28) for each row of (5.27), with initial condition p n,n+1 = 1. On the other hand, by looking at the diagonals (n − k = constant) successively for n − k = −1, 0, 1, 2, . . . , a little experimentation led to the formula (5.16b).
3. The factorization (5.4) implies that the unit-lower-Hessenberg matrix P is totally positive if and only if the unit-lower-triangular matrix P def = P ∆ T , obtained from P by deleting its zeroth column, is totally positive. Now, the production matrix of P -namely, the unit-lower-Hessenberg matrix Q = P −1 ∆ P -appears to have a very simple form: q n,n = 2 for n ≥ 1 (5.34c) q n,n+1 = 1 (5.34d) q n,k = 0 for k > n + 1 (5.34e) (We have not proven this formula for Q, but it is probably not too difficult.) Alas, this matrix Q is not even TP 2 (for instance, q 10 q 21 − q 11 q 20 = 2 · 3 − 2 · 6 = −6), so we cannot use this method to prove the total positivity of P . Nor does it help to subtract of the identity matrix from Q (which would correspond to factoring out a binomial matrix from P on the left): there does not exist any c ∈ R for which the leading 3 × 3 principal submatrix of Q − cI is TP 2 .
has k lower-numbered children). However, in order to make this interpretation most natural, we modify the model slightly, by now considering rooted trees in which the root has k higher -numbered children (this is of course equivalent by reversing all the labels). We denote by T • n+1,k the set of rooted trees on the vertex set [n + 1] in which exactly k children of the root are higher-numbered than the root.
We will therefore be defining a bijection between two models on the vertex set [n + 1]: Model 1. Rooted trees in which the root has k higher-numbered children.
Model 2. Rooted trees in which the vertex 1 has k children.
We begin with some definitions.
Let T be a tree on a totally ordered vertex set (for us it will be [n + 1]), and let e = ij be an edge of T , where i is the parent and j is the child. We say that the edge e = ij is increasing if i < j, and decreasing if i > j. We recall that the edge e = ij is improper if there exists a descendant of j (possibly j itself) that is lower-numbered than i; otherwise it is proper . Clearly, every decreasing edge is necessarily improper; an increasing edge can be either proper or improper, depending on the behavior of the descendants of j.
We now classify edges in a tree T ∈ T • n+1,k (that is, Model 1) as either regular or irregular , as follows: Definition A.1. Let e = ij be an edge in a tree T ∈ T • n+1,k , where i is the parent and j is the child. We classify this edge as follows: (I1) If ij is decreasing, then it is irregular.
(I2) If ij is increasing and improper, and i is not the root, then ij is irregular.
(I3) If ij is increasing and i is the root, then ij is regular. (That is, the k increasing edges emanating from the root are all regular.) (I4) Suppose that all the children of vertex 1 are higher-numbered than the root. If i = 1 and there is a descendant of j that is lower-numbered than the root, then ij is irregular. (Note that in this case the root cannot be vertex 1; so this rule does not contradict rule (I3).) (I5) Suppose that vertex 1 has at least one child that is lower-numbered than the root ρ. (Note that this implies ρ = 1.) Let T 1 be the maximal increasing subtree of T rooted at vertex 1, whose vertices are where v +1 < ρ < v +2 (of course ρ / ∈ T 1 ). Then: (a) all the edges on the path from vertex 1 to v +1 are irregular; and (b) an edge ij ∈ T 1 with parent i = v s and child j = v t (s < t) is irregular in case one of the following is satisfied: (b1) + 2 ≤ s < t and there is a descendant of v t in T that is < v s ; (b2) s ≤ < + 2 ≤ t and there is a descendant of v t in T that is < v s+1 ; (b3) s = + 1 < + 2 ≤ t and there is a descendant of v t in T that is < ρ; (b4) s < t ≤ and there is a descendant v τ of v t in T 1 such that v τ +1 < ρ and there is a descendant of v τ +1 in T that is < v s+1 ; (b5) s < t ≤ and there is a descendant v τ of v t in T 1 that is > ρ, and a descendant of v τ in T that is < v s+1 .
(I6) All other edges are regular.
(We apologize for the complexity of this definition; but these are the cases that seem to be needed.) We recall that the polynomials t n,k (y, z) enumerate trees in Model 2 with a weight y (resp. z) for each improper (resp. proper) edge, except that the k proper edges emanating from vertex 1 are unweighted. We now assert -and this is the main result of this appendix -that the same polynomials t n,k (y, z) enumerate trees in Model 1 with a weight y (resp. z) for each irregular (resp. regular) edge, except that the k regular edges emanating from the root are unweighted: Proposition A.2. The polynomials t n,k (y, z) defined in (1.7) satisfy To prove Proposition A.2, we will construct, for each fixed n and k, a bijection σ from Model 2 (namely, the set T 1;k n+1 ) to Model 1 (namely, the set T • n+1,k ), with the property that the number of proper (resp. improper) edges in T equals the number of regular (resp. irregular) edges in σ(T ). Moreover, we will be able to say which edge in T corresponds to which edge in σ(T ): that is, for each T ∈ T 1;k n+1 we will construct a bijection ψ T : E(T ) → E(σ(T )) such that e ∈ E(T ) is proper (resp. improper) if and only if ψ T (e) ∈ E(σ(T )) is regular (resp. irregular). We summarize this as follows: Proposition A.3. There are bijections (σ, ψ T ) from Model 2 to Model 1 that map proper (resp. improper) edges in Model 2 to regular (resp. irregular) edges in Model 1.
The remainder of this appendix is devoted to proving Proposition A.3.
Given a tree T rooted at r in Model 2, let v 1 < v 2 < · · · < v k be the k children of the vertex 1. If vertex 1 is the root, then its k children are obviously higher-numbered than the root. In this situation we define σ(T ) = T , which also belongs to Model 1; and we define ψ T (e) = e for all e ∈ E(T ). Now suppose that the vertex 1 is not the root. First consider the case k = 0. Since we will use this special case as a tool in handling the general case, we here denote the bijection φ instead of σ.
The following construction is inspired by [13, proof of Lemma 1].
Proof of Lemma A.4. Let T be a tree in Model 2, in which r is the root and vertex 1 is a leaf. Since we have already handled the case r = 1, we assume henceforth Let T max be the maximal increasing subtree of T rooted at r, and let T 0 , . . . , T p be the trees obtained from T by deleting all the edges in T max . Let r j be the root of T j for 0 ≤ j ≤ p. In particular, we choose r 0 to be the root r. Note that each r j is a vertex in T max (otherwise it would not become a root when we delete the edges in T max ); and conversely, every vertex in T max becomes a root r j (though its tree T j might be trivial). Therefore, all of the higher-numbered children of r j belong to T max , while all of the lower-numbered children of r j belong to T j . We denote the set of those lower-numbered children by L j . Of course, L 0 = L. See Figure 8(b). Furthermore, since T max is an increasing tree, it is rooted at its smallest label (namely, r); therefore r j > r for 1 ≤ j ≤ p.
Notation: Let S be a sequence of increasing numbers. For a ∈ S and b ∈ S, define ρ −b +a as an operator acting on S such that ρ −b +a (S) := S ∪ {a}\{b} is still increasingly ordered. For example, ρ −5 +2 (1, 3, 5, 7) = (1, 2, 3, 7). We observe that the inverse of ρ −b +a is ρ −a +b . Further, if T is a tree whose vertex set is S, we write ρ −b +a (T ) to denote the tree with vertex set ρ −b +a (S) that is obtained from T by relabeling the vertices according to the map ρ −b +a . We now make the trivial observation that, since r > 1 by assumption, the vertex 1 is not in T max . Let T i be the tree containing vertex 1; since k = 0, the vertex 1 is a leaf in T i . The bijection φ is defined in three steps: Step 1. Take the tree T max , and relabel its vertices to obtain ρ −r i +1 (T max ). Since T max is an increasing tree, it is rooted at its smallest label (namely, r); therefore, the relabeled tree ρ −r i +1 (T max ) is rooted at its smallest label, which is the vertex 1.
[If i = 0, we relabeled r → 1 and left all other labels unaffected. If i = 0, we relabeled r → 1 and relabeled the second-smallest label of T max (that is, the lowest-numbered child of r) to r -among other relabelings, the details of which will be worked out below.] φ(T ) Step 2. Graft ρ −r i +1 (T max ) onto T i by identifying the two vertices 1; call the result T i .
Step 3. Graft each tree T j (j = i) onto T i by identifying the two vertices r j ; call the result φ(T ). See Figure 9(a).
In this way we obtain a tree φ(T ) rooted at r i , in which all the children of r i are lower-numbered, and in which ρ −r i +1 (T max ) is the maximal increasing subtree of φ(T ) rooted at the vertex 1. Furthermore, if i = 0, then the lowest-numbered child of the vertex 1 is r, which is smaller than the root r i of φ(T ); while if i = 0, then, as shown in Figure 9(b), the children of vertex 1 are precisely the set H (which did not undergo any relabeling in Step 1), which are all larger than the root r of φ(T ).
These observations allow us to obtain the inverse of φ. If the smallest-numbered child of vertex 1 is smaller than the root of φ(T ), then that child is r, and we are in the case i = 0; otherwise we are in the case i = 0, and the root of φ(T ) is r. If we delete the edges of ρ −r i +1 (T max ) from φ(T ), we recover the trees T i ; this undoes Step 3. We then undo Step 2 by separating the subtree rooted at vertex 1. And finally, we undo Step 1 by relabeling ρ −r i +1 (T max ) using the map ρ −1 +r i ; this yields T max . We can then reassemble the pieces T max and T 0 , . . . , T p to obtain T .
It is also clear how the map ψ T is defined, since each edge of T corresponds, via relabeling and grafting, to a well-defined edge of φ(T ).
We now look at how the map ψ T acts on proper and improper edges of T . We first observe that the edges in T 0 , . . . , T p do not undergo any relabeling; so an edge e in one of these subtrees is increasing (resp. decreasing) according as the edge ψ T (e) in φ(T ) is increasing (resp. decreasing). Furthermore, the descendants in these subtrees are the same as the descendants of their images in φ(T ), except that the vertex 1 in φ(T ) has acquired extra descendants, which are anyway higher-numbered and therefore do not affect properness or improperness. Therefore, an edge e in one of these subtrees is proper (resp. improper) according as the edge ψ T (e) in φ(T ) is proper (resp. improper). This means, by rules (I1), (I2) and (I6) of Definition A.1, that an edge e in one of these subtrees is proper (resp. improper) according as the edge ψ T (e) in φ(T ) is regular (resp. irregular). [Note that rule (I3) plays no role here, because k = 0. Rule (I4) does not apply because all the edges in φ(T ) emanating from vertex 1 lie in ρ −r i +1 (T max ). And rule (I5) applies only within ρ −r i +1 (T max ).] We now need to consider the edges in ρ −r i +1 (T max ). We divide the proof into two cases: Case 1: i = 0. Let r = u 1 < u 2 < · · · < u m be the vertices in T max ; then ρ −r i +1 acts as follows: ρ −r i +1 : (r, u 2 , . . . , u m ) → (1, u 2 , . . . , u m ). That is, as previously observed, ρ −r i +1 (T max ) is obtained from T max by only relabeling the root r as the vertex 1; and φ(T ) is as shown in Figure 9(b). Therefore, for all edges in T max other than those emanating from the root, properness/improperness in T corresponds to properness/improperness of their images in φ(T ); and by rules (I1), (I2) and (I6), this corresponds to regularity/irregularity of the images in φ(T ). [Rule (I4) does not apply because the parent is not vertex 1; and rule (I5) does not apply because all the children of vertex 1 are higher-numbered than the root r.] Now consider an edge e in T that emanates from the root r to a higher-numbered child h ∈ H. The bijection ψ T maps e to an edge e in φ(T ) that emanates from vertex 1 to h ∈ H. The edge e is improper in case there is a descendant of h in T that is < r; and this is equivalent to the existence of a descendant of h in φ(T ) that is < r. Since in this case r is the root of φ(T ), and all the children of vertex 1 in φ(T ) are > r, rule (I4) of Definition A.1 specifies that the edge e is irregular whenever e is improper; otherwise, by rule (I6), it is regular. [Once again, rule (I5) does not apply here.] Case 2: i = 0. Let r = u 1 < u 2 < · · · < u < u +1 = r i < u +2 < · · · < u m be the vertices in T max ; then ρ −r i +1 acts as follows: ρ −r i +1 : (r, u 2 , u 3 , . . . , u , r i , u +2 , . . . , u m ) → (1, r, u 2 , . . . , u −1 , u , u +2 , . . . , u m ) := (v 1 , v 2 , v 3 , . . . , v , v +1 , v +2 , . . . , v m ) Set u 0 := 1. Then ρ −r i +1 (T max ) is obtained from T max by relabeling each vertex u s that is ≤ r i by u s−1 , and leaving all vertices > r i unchanged; in other words, v s = u s−1 for s ≤ + 1 and v s = u s for s ≥ + 2. Therefore, each edge e = u s u t in T max ⊆ T maps onto ψ T (e) = v s v t in ρ −r i +1 (T max ) ⊆ φ(T ); and the descendants of u t in T max map via ψ T onto the descendants of v t in ρ −r i +1 (T max ). Note that in φ(T ), vertex 1 has at least one child (namely, r) that is lowernumbered than the root r i . Therefore rule (I5) applies, with ρ = r i and T 1 = ρ −r i +1 (T max ): (a) All the edges on the path from the root r to vertex r i in T max ⊆ T are improper, since vertex 1 is a descendant of r i in T . These edges map, under the relabeling ρ −r i +1 , onto the path from vertex 1 to v +1 in φ(T ). By rule (I5a) of Definition A.1, all the edges in this path are irregular.
The foregoing case needed to be treated separately, because the vertices in the path from r to r i in T max ⊆ T have descendants (in particular, the vertex 1) that do not correspond (via the relabeling) to descendants in their images in φ(T ), because the tree T i was moved from its position in T to the root in φ(T ). This problem does not arise in the remaining cases: (b1) Consider an edge e = u s u t in T max , where + 2 ≤ s < t. These vertices do not get relabeled, so ψ T (e) = e. This edge is improper in T in case there is a descendant of u t in T that is < u s . By rule (I5b1) of Definition A.1, this edge is irregular in φ(T ) in exactly the same situation.
(b2,3) Now consider an edge e = u s u t in T max , where s ≤ + 1 < + 2 ≤ t. Then vertex u s gets relabeled to u s−1 , while u t does not get relabeled; so ψ T (e) = v s v t = u s−1 u t . The edge e is improper in T in case there is a descendant of u t in T that is < u s . Now u s = v s+1 in case s ≤ , while u s = r i = ρ in case s = + 1. By rules (I5b2,3) of Definition A.1, the edge v s v t is irregular in φ(T ) exactly when e is improper in T .
Note that in cases (b1-3), the descendants of u t in T are the same as the descendants of v t in φ(T ), because the relevant trees T j were grafted in the same place (since their roots r j did not get relabeled). Things will be slightly more complicated in the remaining cases: (b4,5) Consider an edge e = u s u t in T max , where s < t ≤ . Then vertices u s and u t both get relabeled, so ψ T (e) = v s v t = u s−1 u t−1 . The edge e is improper in T in case there is a descendant of u t in T that is < u s . (Note that u s = v s+1 and u t = v t+1 because s, t ≤ .) Such a descendant cannot lie in T max , because T max is increasing, but it can lie in one of the trees T j that is attached to T max . So consider all of the descendants u τ of u t in T max . If one of these descendants is r i , then we are in the already-treated case (a); so we can assume that they are all either < r i or > r i . The images of the vertices u τ under ρ −r i +1 are the descendants v τ = ρ −r i +1 (u τ ) of v t in T 1 = ρ −r i +1 (T max ). Now consider the two cases u τ < r i and u τ > r i (recalling that r i = ρ): (b4) u τ < ρ is equivalent to u τ = v τ +1 < ρ. The edge u s u t is improper in T in case there is a descendant of u τ = v τ +1 in T that is < u s = v s+1 ; and the descendants of u τ = r j in the tree T j ⊆ T are the same as the descendants of v τ +1 = r j in the tree T j ⊆ φ(T ). By rule (I5b4) of Definition A.1, the edge v s v t is irregular in φ(T ) exactly when e is improper in T .
(b5) u τ > ρ is equivalent to u τ = v τ > ρ. The edge u s u t is improper in T in case there is a descendant of u τ = v τ in T that is < u s = v s+1 ; and the descendants of u τ = r j in the tree T j ⊆ T are the same as the descendants of v τ = r j in the tree T j ⊆ φ(T ). By rule (I5b5) of Definition A.1, the edge v s v t is irregular in φ(T ) exactly when e is improper in T . the trees T max ⊆ T and ρ −r i +1 (T max ) ⊆ φ(T ) along with some of the trees T j hanging off them (namely, those trees attached to the descendants of u t are shown). The edge e = u s u t and its image ψ T (e) = v s v t are shown in thick red. Note that tree T j is attached at vertex u j , which equals v j = ψ T (u j ) or v j+1 = ψ T (u j+1 ) according as u j > r i or u j < r i . See Figure 10 for an example illustrating the cases (b4,5).
There is, a priori , one additional case for an edge e = u s u t in T max , namely, s < t = + 1. But this corresponds to the last edge on the path from r to r i in T max , and hence was already treated on case (a).
We have now considered all the cases in which an edge e ∈ T can be improper; so by rule (I6) of Definition A.1, this completes the proof.
Remark. In case (b4,5) one might worry what happens when v τ < ρ while v τ +1 > ρ, which was not included in either (b4) or (b5). This happens if and only if τ = + 1, i.e. u τ = r i , in which case all the edges having v τ as a descendant are irregular by case (a). Now we consider the general case k ≥ 1. The following construction is inspired by [13, proof of Lemma 2].
Proof of Proposition A.3. Given a tree T rooted at r in Model 2, let v 1 < v 2 < · · · < v k be the k children of the vertex 1. For any vertex i other than the root, we define its top ancestor to be the ancestor of i (possibly i itself) that is a child of the root. We construct the bijection σ in the following three cases: Case I: v 1 < r.
Case II: v 1 > r and the top ancestor of vertex 1 is lower-numbered than the root.
Step 2. Attach the trees rooted at the vertices of H onto vertex 1 via new edges.
Step 3. Attach the trees rooted at v 1 , v 2 , . . . , v k onto the root r via new edges.
(Note that this is identical to Case I but with the roles of v 1 and r interchanged.) See Figure 12. We obtain thereby a tree σ(T ) rooted at r with k higher-numbered children v 1 , v 2 , . . . , v k , which also has the following properties: A1) all the children of vertex 1 are higher-numbered than the root; B2) the top ancestor of vertex 1 is lower-numbered than the root.
The k edges in T that emanate from vertex 1 to its children (shown in blue in Figure 12) are clearly proper. These edges are mapped to the k edges in σ(T ) that emanate from the root r to its higher-numbered children, which are regular by rule (I3) of Definition A.1.
An edge e ∈ T that emanates from the root r to a higher-numbered child h ∈ H (shown in red in Figure 12) is improper if (and only if) there is a descendant of h that is < r. Such an edge is mapped to the edge ψ T (e) ∈ σ(T ) that emanates from vertex 1 to its child h ∈ H; and since all the children of vertex 1 are higher-numbered than the root r, rule (I4) of Definition A.1 applies and says that the edge ψ T (e) is irregular if and only if e is improper.
All other edges e ∈ T (shown in black in Figure 11) have the property that ψ T (e) = e. Moreover, if e / ∈ T ({r} ∪ L ∪ D L ), then the descendants of e in T are the same as its descendants in σ(T ), so by rules (I1), (I2) and (I6) of Definition A.1, the edge e is proper/improper in T exactly when it is regular/irregular in σ(T ). Finally, all the edges e ∈ T ({r} ∪ L ∪ D L ) are improper in T (because vertex 1 is a descendant), and they are irregular in σ(T ) by rules (I1) and (I2).
We have therefore shown that the bijection ψ T maps proper/improper edges in T onto regular/irregular edges in σ(T ).
Case III: v 1 > r and the top ancestor of vertex 1 is higher-numbered than the root.
Let L (resp. H) denote the lower-(resp. higher-) numbered children of the root r, and let D L (resp. D H ) denote their descendants excluding those in L (resp. H) itself. Let u be the vertex on the path from the root r to vertex 1 such that the path from r to u is maximal increasing. Clearly, r < u. The first two steps in constructing the tree σ(T ) are as follows: Step 1. Delete the k edges from vertex 1 to its children, and denote by T 0 the subtree rooted at r in which 1 is a leaf.
Step 2. Use the bijection φ constructed in Lemma A.4 to yield a tree φ(T 0 ). Note that the vertex u plays the role of r i in Lemma A.4, so that the operator ρ −u +1 acts on the maximal increasing subtree (T 0 ) max of T 0 rooted at r. Note also that φ(T 0 ) is rooted at u, which has no higher-numbered children. See Figure 13.
Then we distinguish two subcases, according as u < v 1 or u > v 1 : Case III(a): u < v 1 . Figure 11: The trees T and σ(T ) in Case I. The blue edges in T [resp. σ(T )] are proper (resp. regular). The red edges in T [resp. σ(T )] could be either proper or improper (resp. regular or irregular), depending on the behavior of their descendants.     Figure 15: The trees T , T 1 and φ (T 1 ) in Case III(b). Note that here T is the same as in Figure 13, but the subtree T 1 is shown in more detail. Figure 16: The trees T and σ(T ) in Case III(b).