Combinatorics of non-ambiguous trees

# Combinatorics of non-ambiguous trees

Jean-Christophe Aval, Adrien Boussicault, Mathilde Bouvel, Matteo Silimbani 111LaBRI - CNRS, Université de Bordeaux, 351 Cours de la Libération, 33405 Talence, France. All authors are supported by ANR – PSYCO project (ANR-11-JS02-001).

###### Abstract

This article investigates combinatorial properties of non-ambiguous trees. These objects we define may be seen either as binary trees drawn on a grid with some constraints, or as a subset of the tree-like tableaux previously defined by Aval, Boussicault and Nadeau. The enumeration of non-ambiguous trees satisfying some additional constraints allows us to give elegant combinatorial proofs of identities due to Carlitz, and to Ehrenborg and Steingrímsson. We also provide a hook formula to count the number of non-ambiguous trees with a given underlying tree. Finally, we use non-ambiguous trees to describe a very natural bijection between parallelogram polyominoes and binary trees.

## 1 Introduction

It is well known that the Catalan numbers enumerate many combinatorial objects, such as binary trees and parallelogram polyominoes. Several bijective proofs in the literature show that parallelogram polyominoes are enumerated by Catalan numbers, the two most classical being Delest-Viennot’s bijection with Dyck paths [DV84] and Viennot’s bijection with bicolored Motzkin paths [DV84].

In this paper we demonstrate a bijection – which we believe is more natural – between binary trees and parallelogram polyominoes. In some sense, we show that parallelogram polyominoes may be seen as two-dimensional drawings of binary trees. This point of view gives rise to a new family of objects – we call them non-ambiguous trees – which are particular compact embeddings of binary trees in a grid.

The tree structure of these objects leads to a hook formula for the number of non-ambiguous trees with a given underlying tree. Unlike the classical hook formula for trees due to Knuth (see [Knu98], , Exercise 20), this one is defined on the edges of the tree.

Non-ambiguous trees are in bijection with permutations such that all their (strict) excedances stand at the beginning of the permutation word. Ehrenborg and Steingrímsson in [ES00] give a closed formula (involving Stirling numbers of the second kind) for the number of such permutations. We show that this formula can be easily proved using non-ambiguous trees and a variation of the insertion algorithm for tree-like tableaux introduced in [ABN11]. Indeed, non-ambiguous trees can also be seen as a subclass of tree-like tableaux, objects defined in [ABN11] and that are in bijection with permutation tableaux [SW07] or alternative tableaux [Nad11, Vie07].

A particular subclass of non-ambiguous trees leads to unexpected combinatorial interpretations. We study complete non-ambiguous trees, defined as non-ambiguous trees such that their underlying binary tree is complete, and show that their enumerating sequence is related to the formal power series of the logarithm of the Bessel function of order . This gives rise to new combinatorial interpretations of some identities due to Carlitz [Car63], and to the proof of a related identity involving Catalan numbers, which had been conjectured by P. Hanna.

The paper is organized as follows: in Section 2 we define non-ambiguous trees. Then, in Section 3 we give the enumeration of non-ambiguous trees satisfying certain constraints: those contained in a given rectangular box, and those with a fixed underlying tree. Section 4 introduces the family of complete non-ambiguous trees, studies the relations between this family and the Bessel function, and proves combinatorial identities involving Catalan numbers and the sequence enumerating complete non-ambiguous trees. In Section 5 we describe our new bijection between binary trees and parallelogram polyominoes. To conclude, we present in Section 6 some perspectives related to our study.

## 2 Definitions and notations

In this paper, trees are embedded in a bidimensional grid . The grid is not oriented as usual: the -axis has south-west orientation, and the -axis has south-east orientation, as shown on Figure 2.

Every -oriented (resp. -oriented) line will be called column (resp. row). Each column (resp. row) on this grid is numbered with an integer corresponding to its (resp. ) coordinate. A vertex located on the intersection of two lines has the coordinate representation: .

A non-ambiguous tree may be seen as a binary tree embedded in the grid in such a way that its edges may be recovered from the embedding of its vertices in the grid (Figure 2). In other words, the vertices determine the tree without ambiguity, whence the name of these objects.

Formally, a non-ambiguous tree of size is a set of points such that:

1. ; we call this point the root of ;

2. given a non-root point , there exists one point such that and , or one point such that , , but not both (which means that the pattern is avoided);

3. there is no empty line between two given points: if there exists a point such that (resp. ), then for every (resp. ) there exists such that (resp. ).

Figure 3 shows some examples and counterexamples of non-ambiguous trees.

It is straightforward that a non-ambiguous tree has a tree structure: except for the root, every point has a unique parent, which is the nearest point preceding in the same row (resp. column). In this case, we will say that is the right child (resp. left child) of . In this paper, we orient every edge of a tree from the root to the leaves. We shall denote by the underlying binary tree associated to .

Figure 4 shows all the non-ambiguous trees of size , grouping inside a rectangle those having the same underlying binary tree.

###### Remark 1.

A tree-like tableau [ABN11] of size is a set of points placed in the boxes of a Ferrers diagram such that conditions 1, 2, 3 defining non-ambiguous trees are satisfied. Figure 5 shows an example of a tree-like tableau of size . It should be clear that non-ambiguous trees are in bijection with tree-like tableaux with rectangular shape.

## 3 Enumeration of non-ambiguous trees

Eventhough non-ambiguous trees are new combinatorial objects, their enumeration has been previously studied. Indeed, non-ambiguous trees of size are in bijection with permutations of size with all their strict excedances at the beginning, whose enumeration has been studied by Ehrenborg and Steingrímsson in [ES00]. The size preserving bijection (that we do not detail here) between these two families of combinatorial objects is a consequence of Lemma in [SW07] and of results proved in [ABN13]. The sequence counting the number of non-ambiguous tree of size is referenced in [Slo] as , but no simple formula is known. In this section, we provide enumerative formulas for non-ambiguous trees with additional constraints.

### 3.1 Non-ambiguous trees inside a fixed rectangle

In this section, we make extensive use of Remark 1 and we view non-ambiguous trees as tree-like tableaux of rectangular shape. For brevity, we write TLT for tree-like tableau in the sequel. Of particular importance for our purpose is the insertion procedure for TLTs, which gives to these objects a recursive structure. Since it is not the central purpose of the present paper, we shall not recall here all the definitions and properties of TLTs, and we refer the reader to [ABN11]. Figure 5 shows an example of a TLT of size 7. In full generality, the size of TLT is given by the numbers of dotted cells.

Moreover, in this section, we use a different drawing convention for non-ambiguous trees, inherited from the drawing convention for TLTs: we will draw non-ambiguous trees with the -axis vertical, and the -axis horizontal.

Given a non-ambiguous tree, its -size (resp. -size) may be defined as the maximum of the -coordinate (resp. -coordinate) of its points. The aim of this subsection is to give a formula for the number of non-ambiguous trees with -size equal to and -size equal to . Because the size of a TLT is given by its semi-perimeter, remark that the size of such a non-ambiguous tree is given by .

We denote by the unsigned Stirling numbers of the first kind, i.e. the number of permutations of size with exactly disjoint cycles.

We shall prove that for every integers , one has:

 n∑k=1c(n,k)A(k,ℓ)=nℓ−1n!. (1)

Inverting Equation (1), we obtain that it is equivalent to Equation (2) below:

 A(k,ℓ)=k∑i=1(−1)k−iS(k,i)i!iℓ−1, (2)

where denotes the Stirling numbers of the second kind, i.e. the number of partitions of a set of elements into non-empty parts. It is known [SW07, ABN13] that is equal to the number of permutations of size with exactly strict excedances in position . Consequently, Corollary 6.6 in [ES00] provides a proof of Equation (1). More precisely, in [ES00], Ehrenborg and Steingrímsson prove (2) by an inclusion-exclusion argument, and then deduce (1) by inversion.

Our goal is to prove Equation (1) directly and combinatorially. For this purpose, we provide combinatorial interpretations of the numbers and that appear in Equation (1) with TLTs and non-ambiguous trees respectively. As mentioned in Remark 1, non-ambiguous trees are nothing but TLTs with a rectangular shape, so that our interpretation of and is then with unified objects. This is the key in proving Equation (1) by a combinatorial approach.

Recall that by definition is the number of non-ambiguous trees with -size equal to and -size equal to . On the other hand, we claim that counts the number of TLTs of size with exactly points in their first row. Indeed, this follows by definition of and the fact that TLTs of size with exactly points in their first row are in bijection with permutations of size with exactly disjoint cycles. This is a consequence of Theorem in [Bur07], re-formulated in terms of TLTs as in [ABN13].

###### Proposition 2.

For every integers , we have the following identity:

 n∑k=1c(n,k)A(k,ℓ)=nℓ−1n! (3)

where denotes the number of TLTs of size with exactly points in their first row.

The proof consists in two steps:

• introduce a set whose cardinality will be proved to be (Lemma 3);

• prove that the elements of are in bijection with pairs enumerated by the left-hand side of (3) (Lemma 4).

We define as the set of TLTs of size whose first rows are of equal length. Figure 6 shows an element of .

To compute the cardinality of , we will use a slight variation of the insertion procedure defined in [ABN11]. This new insertion procedure depends on , and we shall call it the -insertion procedure.

##### The ℓ-insertion procedure.

Let be an element of . The -special box of is defined as

• the right-most box of the -th row, if the -st row is strictly smaller;

• the right-most dotted box at the bottom of a column, otherwise.

Given an integer in , we associate to it an edge of the South-East border of by labeling these edges from South-West to North-East. Note that we exclude the (vertical) edges at the right of the first rows of . As in the insertion procedure of [ABN11], if is a horizontal (resp. vertical) edge of the border of , we add a row (resp. column) to below (resp. to the right of) , composed of empty boxes except for one dotted box: the one below (resp. to the right of) . Next, if the dotted box added is to the left of the -special box of , we add a ribbon (a connected set of empty boxes without any square) adjacent simultaneously to the -special box of and to the dotted box added.

Figure 7 illustrates the -insertion procedure.

###### Lemma 3.

The cardinality of is given by

 #Tn,ℓ=nℓ−1n!. (4)
###### Proof.

We prove the above by induction on . By [ABN11, Theorem 2.2], the number of unrestricted TLTs of size is equal to , thus (4) is true for .

For the inductive step, we claim that the -insertion procedure gives a bijection between and . To prove this fact, we first notice that the result of the -insertion procedure on and with is a TLT whose first rows are of equal length, thus an element of . Next, as explained in [ABN11], we observe that the new box added by -insertion in an element of is easy to recognize: it is the rightmost among dotted boxes at the bottom of a column. Thus we are able to invert the -insertion procedure. This ends the proof of the lemma. ∎

Now we will send bijectively an element of on pairs of objects enumerated by the left-hand side of (3). The bijection relies on the -cut procedure described below, and illustrated in Figure 8.

##### The ℓ-cut procedure.

Let be an element of . We first cut by putting the first rows in and the next rows in . Now we see as part of a TLT. We add to it a first row whose length equals the width of . We observe that there is exactly one way to put dots in this row to obtain a TLT, that we denote : we are forced to put dots in the boxes corresponding to non-empty columns in . Next we remove empty columns in to get a non-ambiguous tree with rows, denoted .

###### Lemma 4.

Elements of are in bijection with pairs such that

1. is a TLT of size ,

2. is a non-ambiguous tree with rows,

3. the number of dotted boxes in the first row of and the width of are equal.

###### Proof.

It should be clear the -cut procedure is invertible, which implies the lemma. ∎

Putting together Lemmas 3 and 4 gives a bijective proof of Proposition 2.

### 3.2 Non-ambiguous trees with a fixed underlying tree: a new hook formula

Let be a binary tree. We define as the number of non-ambiguous trees such that their underlying binary tree is . The aim of this section is to get a formula for : this will be done by Proposition 8, which shows that may be expressed by a new and elegant hook formula on the edges of . To do this, we encode any non-ambiguous tree by a triple where is a binary tree, and (resp. ) is a word called the left (resp. right) code of . To distinguish the vertices of , we label them by integers from to the size of , as shown on Figure 9.

The first entry in is the underlying binary tree associated to . Observe that we keep the labels on vertices when we extract the underlying binary tree. Now we denote by (resp. ) the set of the end points of the left (resp. right) edges of , which gives and on the example in Figure 9. The definition of non-ambiguous trees ensures that the set is the interval . Thus for , we may set as the unique label such that , and we proceed symmetrically for . On the example of Figure 9, we have: and . Our starting point is the following lemma.

###### Lemma 5.

The application which sends to the triple is injective.

###### Proof.

Consider a non-ambiguous tree with vertices and . From the definition of non-ambiguous tree, we know that the and -coordinates of the root are . Moreover, for every left edge of we have

 Y(s)=Y(t) and αL(X(t))=t,

namely, is the position where appears in . Similarly, for every right edge , we have

 X(s)=X(t) and αR(Y(t))=t,

namely, is the position where appears in . It is easy to check that we have independent equations in variables (the and -coordinates of the vertices), so we get a unique non-ambiguous tree. ∎

Lemma 5 allows us to encode a non-ambiguous tree by a triple , where is a binary tree, and (resp. is a word in which every label (resp. ) appears exactly once. Of course, is not surjective on such triples. If we take , it should be clear that is forced to be . Consequently, our next task is to characterize the pairs of codes which are compatible with a given binary tree , i.e. such that is in the image of . In order to describe this characterization, we need to define partial orders on the sets and . The pairs of compatible codes will be seen to correspond to pairs of linear extensions of the posets and . The posets are defined as follows: given (resp. ), we say that if and only if there exists a path in the oriented tree starting from and ending at . Figure 10 (with minima at the top) illustrates this notion.

The next lemma is the crucial step to prove Proposition 8.

###### Lemma 6.

Given a binary tree , the pairs of codes compatible with are exactly the pairs where is a linear extension of and is a linear extension of .

Moreover, such pairs are in bijection with non-ambiguous trees with underlying tree .

Figure 11 gives these compatible codes, together with the corresponding non-ambiguous trees, in the case of the tree of Figure 10.

###### Proof.

Given a tree , consider the map defined on the set of non-ambiguous trees with underlying tree by , where . We prove that the image of is , where we denote by the set of linear extensions of a poset . The second statement in Lemma 6 will then follow, since we deduce from the injectivity of (Lemma 5) that is also injective.

First, we prove that . Without loss of generality, we will prove that . We need to prove that, if , then precedes in , which we shall write . If , there exists a path in starting from and ending at . When we go through the path, the -coordinates of the vertices remain unchanged along right edges, while they increase along left edges. Since , we have , which is equivalent to .

Now the hard part is to prove that . Let . From the triple , we may build a set of points in the grid, denoted , as follows: we place a point (the root) at position and one point for every vertex in at coordinates

 {X(v)=iwithαL(i)=vandY(v)=Y(% parent(v)) if v∈VL;X(v)=X(parent(v))andY(v)=jwithαR(j)=v if v∈VR.

We now prove that is a non-ambiguous tree, whose underlying binary tree is of course . This will follow from the three following statements, that we prove below.

• for every left (resp. right) edge of , we have (resp. ) in ;

• avoids the pattern ;

• two different vertices in correspond to points at different positions in .

Proof of Without loss of generality, we show this property for the set of left edges. First of all, we define the oldest left (resp. right) ancestor of a vertex to be the vertex such that the path going from to contains only right (resp. left) edges and this path is the longest with this property.

For example, in Figure 12 the vertices named are the oldest left ancestors of those named .

Let be the oldest left ancestor of in . Our construction of the set implies that . Moreover, since is a left edge, we have . Keeping in mind these two facts, we consider the following two cases:

• if is the root of the tree, then . Since , we have ;

• otherwise, is the ending point of a left edge in , and . Since in there exists a path form to , then . Moreover, we know that , so and , implying that .

Proof of We proceed by way of contradiction. Suppose that there are three points such that form the pattern . We remark that these three points are different from the root, because for each of them either the or the -coordinate is nonzero. Without loss of generality, we suppose that is the end of a left edge. Let be the oldest left ancestor of . By construction, , and hence . Since , is the end of a left edge. The points and therefore both belong to and have the same -coordinate, enventhough they are distinct. This is impossible, since is a linear extension of , and hence all points in must have different -coordinates.

Proof of We proceed by way of contradiction. Suppose that there are two different vertices whose positions in coincide. By construction, they cannot be the root of the tree. We have two cases:

• one of the vertices is the end of a left edge, and the other is the end of a right edge. In this case, we find an occurrence of the forbidden pattern ;

• and are both the end of a left edge (the other case is similar). In this case, we have and which is impossible.

Now we come to the final step toward proving Proposition 8.

###### Lemma 7.

The Hasse diagrams of and are forests.

###### Proof.

We prove this proposition by way of contradiction. Suppose that there is a cycle in the Hasse diagram of (the case of is analogous). We can deduce from the poset structure that there are two paths in starting from an element and ending at . This would imply that in the tree there are two different paths from to , and hence there would be a cycle in the tree. ∎

Figure 13 shows an example of the forests obtained by computing the Hasse diagrams of and .

###### Proposition 8.

The number of non-ambiguous trees with underlying tree is given by

 NA(T)=#{left edges}!#{right edges}!∏e∈VLne∏e∈VRne (5)

where, for every left edge (resp. right edge) , is the number of left edges (resp. right edges) contained in the subtree whose root is the ending point of , plus .

###### Proof.

Recall that Knuth’s hook formula [Knu98] gives the number of linear extensions of a poset whose Hasse diagram is a forest: namely, the number of these linear extensions is divided by the product of the hook lengths of all vertices in . Therefore, by Lemmas 6 and 7, the number of non-ambiguous trees with underlying tree is given by the product of the results of Knuth’s hook formula applied to the Hasse diagrams of and . Specifically, when applying Knuth’s formula to (resp. ), the hook length of any vertex in (resp. ) is the number of descendants of in (resp. ) including , which corresponds exactly to for the left (resp. right) edge whose end point is . ∎

This new hook formula is illustrated by Figure 14.

## 4 Complete non-ambiguous trees and combinatorial identities

A non-ambiguous tree is complete whenever its vertices have either or children. An example of complete non-ambiguous tree can be found in Figure 15. A complete non-ambiguous tree always has an odd number of vertices. Moreover, as in complete binary trees, a complete non-ambiguous tree with vertices has exactly internal vertices, leaves, right edges and left edges. Denote by the number of complete non-ambiguous trees with internal vertices. The sequence is known in [Slo] as . We give in this section the first combinatorial interpretation for this integer sequence, in terms of complete non-ambiguous trees. Moreover, we use this interpretation to give in Propositions 9 and 11 combinatorial proofs of two identities due to Carlitz [Car63].

### 4.1 Enumeration of complete non-ambiguous trees, and connection to Bessel function

Denote by the number of complete binary trees with internal vertices. It is well-known that is the -th Catalan number, and that, for every , we have the identity:

 Cn+1=∑i+j=nCiCj. (6)

Proposition 9 gives a variant of this identity for complete non-ambiguous trees:

###### Proposition 9.

For every , we have:

 bn+1=∑i+j=n(n+1i)(n+1j)bibj. (7)
###### Proof.

The proof of this proposition is similar to the classical proof of (6): the left (resp. right) subtree (resp. ) of a complete non-ambiguous tree with internal vertices is a complete non-ambiguous tree with (resp. ) internal vertices, where .

Figure 15 shows an example of left and right subtree of a complete non-ambiguous tree.

Hence, in order to construct an arbitrary complete non-ambiguous tree with internal vertices, we need to choose:

• the number of internal vertices contained in ( may range between and , the number is equal to );

• the complete non-ambiguous tree structure of (resp. ) – we have (resp. ) choices;

• the way of interlacing the right (resp. left) edges of and .

We denote by (resp. ) the end points of the right edges in (resp. ) such that if , then (resp. ), and by and the roots of and . Now, if we want to interlace the right edges in with those in , we need to decide at what positions we want to insert the vertices with respect to , saving the relative order among and . A vertex can be placed either to the left of , or between and (), or to the right of .

Hence, we must choose the positions of (multiple choices of the same position are allowed) among possible ones. This shows that there are ways of interlacing the right edges of the subtrees and , where denotes the number of way of choosing objects within , with possible repetitions.

Analogous arguments apply to left edges. In this case, we have different interlacements. This ends the proof. ∎

###### Corollary 10.

The sequence satisfies the following identity

 ∑k≥0bkx2(k+1)((k+1)!2k+1)2=−ln(J0(x)). (8)
###### Proof.

It is well known (see, e.g., [AS64]) that the Bessel function satisfies the differential equation

 d2ydx2+1xdydx+y=0, (9)

The first coefficients in its series expansion are and .

Consider now the function . Equation (7) ensures that satisfies Equation (9), i.e. the same second order differential equation as .

Setting , we have . Moreover, in only the even powers of have non-zero coefficients. Hence, since , we have for every . In particular, . These arguments imply that . ∎

### 4.2 Combinatorial identities

Corollary 10 shows that non-ambiguous trees provide a combinatorial interpretation – and to our knowledge, the first one – of sequence A002190 [Slo].

In [Car63], the author shows analytically that identities (7) and (10) below are equivalent. We give a combinatorial proof of this fact.

###### Proposition 11.

For every , we have:

 n−1∑k=0(−1)k(nk+1)(n−1k)bk=1. (10)
###### Proof.

We fix an integer and we take . We define a gridded tree of size to be a set of points placed in a grid, such that Condition 2 defining non-ambiguous trees is satisfied (which means we consider a non-ambiguous tree of size embedded in a grid) and such that the underlying tree is complete and that its root belongs to the first column. This implies that there are empty columns and empty rows, and that the first column is not empty. Figure 16 shows an example of a gridded tree of size .

It is easy to verify that there are gridded trees of size . We call trivial gridded tree the tree of size consisting of a single vertex in . Now, for every integer , we define an involution on the set of non trivial gridded trees. This involution associates a gridded tree of size with a gridded tree either of size or .

To define this involution, consider a gridded tree of size and add a virtual root at position ; the previous root becomes the left child of the virtual root. Now consider the path starting from the virtual root, going down through the tree, turning at each internal vertex, and ending at a leaf. This path is unique. There are two cases:

1. the path does not cross an empty row, nor an empty column: we erase the leaf and its parent from the tree, getting a new gridded tree of size . We can always erase the leaf and its parent, except if the parent is the virtual root. This happens only if the tree is the trivial gridded tree. As we restricted to non trivial gridded trees, this case never happens.

2. the path crosses an empty row or an empty column: we choose the first empty row or column met while visiting the path. Without loss of generality, we suppose that it is a column, say . Then, we add a new vertex at the position where crosses the path, and we add in the same column a new leaf (whose parent is ) in the topmost empty row. While visiting the path, we did not meet an empty row. Since there are as many empty rows as empty columns, there is always an empty row below . This operation gives rise to a new gridded tree of size .

Remark that adding (resp. removing) a leaf and its parent in (resp. from) a gridded tree following the previous algorithm does not remove (resp. add) any empty row or column that crosses the path from the virtual root to . For this reason, this operation is an involution. Figure 17 shows how the involution acts on two examples. ∎

We now state and prove an identity satisfied by Catalan numbers, whose proof is similar to Proposition 11, and which will be used to get Corollary 13.

###### Proposition 12.

For every , we have:

 n∑k=0(−1)n+k(n+kn−k)Ck=0. (11)
###### Proof.

We define a -labeled binary tree as a complete binary tree with internal vertices together with a labeling of its (internal or external) vertices by non-negative integers, such that the sum of these labels is equal to . Since is the number of ways to choose elements with repetition in a set of cardinality , we may interpret combinatorially as the number of -labeled binary trees. Indeed, a -labeled binary tree may be described by a complete binary tree with internal vertices, together with a choice of vertices (with possible repetition) among its (internal or external) vertices, any vertex receiving label when it has been chosen times.

Now the proof is based on an involution defined on -labeled binary trees, for , which modifies the parameter by . This involution is defined in a way similar to the one used in the proof of Proposition 11. Let us consider a -labeled binary tree. We add to it a virtual root and consider the path starting from the virtual root, going down through the tree, turning at each internal vertex, and ending at a leaf. Two cases occur:

1. If all the vertices on this path are labeled by zero, then we erase the leaf at the end of this path, together with its parent, and we increase by one the label of the sibling of . This produces a -labeled binary tree.

2. Otherwise we consider the first time this path encounters a vertex that has a non-zero label. We decrease by one the label of and we add a new vertex on the edge between and its parent, having two children: and a new leaf (the new sibling of ). The labels associated to the new vertices and are zero, so that the tree obtained is a -labeled binary tree.

Figure 18 illustrates this operation on two examples (labels zero are omitted on this figure). It is clear that we define in this way an involution that modifies the parameter by , which implies (11). ∎

We are now able to obtain a new identity involving Catalan numbers and the sequence , which has been conjectured by P. Hanna (see [Slo], sequence A002190).

###### Corollary 13.

For every , we have:

 n∑k=0(−1)kbkCk(n+kn−k)2=0. (12)
###### Proof.

We fix an integer and define the matrix as follows:

 D(n0)i,j={0if i=0(−1)j(i+ji−j)2Cjotherwise.

We also set to be the column vector whose entries are . With this notations, Identity (12) holds for every if and only if where is the column vector whose entries are all . Analogously, we define the matrix and set to be the column vector whose entries are all . Then, Identity (10) for all from to yields that . Because the matrix is lower triangular with entries on the diagonal, it is invertible, and we can write . Hence, setting , to conclude the proof of Corollary 13, it is enough to prove that:

 H(n0)⋅1––(n0)=0––(n0). (13)

Notice first that for any and . This follows immediately from the definition of and the fact that for all . The main part of the proof is now to show that for , we have . Indeed, Proposition 12 then ensures that

 for all i≥1,i∑j=0(−1)i+j(i+ji−j)Cj=0,hence, for all 1≤i≤n0,i∑j=0H(n0)i,j=0.

Notice that for , . Consequently, the above equality is equivalent to

 for all 1≤i≤n0,n0∑j=0H(n0)i,j=0.

Combined with for all , this yields as desired.

So let us focus on proving that for , .

Since is characterized by , we have to prove that, for every such that and ,

 n0∑k=0(−1)i+k(i+ki−k)(k+1j+1)(kj)Ck=(i+ji−j)2Cj. (14)

If , the summand vanishes. Moreover, it is easy to check that, if , both sides in Identity (14) vanish, since all terms are . Hence, to conclude the proof, it is sufficient to prove that the following identity holds for any and :

 i∑k=0(−1)i+k(i+ki−k)(k+1j+1)(kj)Ck=(i+ji−j)2Cj (15)

For this purpose, for any , and , we define

 Fj(i,k)=(−1)i+k(i+ki−k)(k+1j+1)(kj)Ck(i+ji−j)2Cj