# Secant tree calculusAcademic research paper on "Mathematics"

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## Academic research paper on topic "Secant tree calculus"

﻿Cent. Eur. J. Math. • 12(12) •2014 • DOI: 10.2478/s11533-014-0429-7

1852-1870

VERS ITA

Central European Journal of Mathematics

Secant tree calculus

Research Article

Dominique Foata1*, Guo-Niu Han2^

1 Institut Lothaire, 1, rue Murner, F-67000 Strasbourg, France

2 I.R.M.A. UMR 7501, Université de Strasbourg et CNRS, 7, rue René-Descartes, F-67084 Strasbourg, France

Received 19 April 2013; accepted 9 October 2013

Abstract: A true Tree Calculus is being developed to make a joint study of the two statistics "eoc" (end of minimal chain) and "pom" (parent of maximum leaf) on the set of secant trees. Their joint distribution restricted to the set {eoc - pom < 1} is shown to satisfy two partial difference equation systems, to be symmetric and to be expressed in the form of an explicit three-variable generating function.

MSC: 05A15, 05A30, 11B68

Keywords: Tree Calculus • Partial difference equations • Binary increasing labeled trees • Secant and tangent trees • End of minimal chain • Parent of maximum leaf • Bivariate distributions • Secant numbers • Entringer distribution • Alternating permutations © Versita Sp. z o.o.

1. Introduction

Recently, a revival of studies on arithmetical and combinatorial properties of both tangent and secant numbers has taken place, using Désiré André's old model of alternating permutations [1, 2], or the equivalent structure of binary increasing labeled trees, called tangent and secant trees. See, e.g., the papers [8, 10-12, 14, 15, 21, 22, 24], or the memoir [23] and our studies on the doubloon model [5-7]. Recall that the secant numbers, denoted by E2n, appear in the Taylor expansions of sec u:

1 U2n u2 u4 u6 u8 .10

secu = - = £ E2n = 1 + -1 + -5+ -61 + -1385+ — 50521 + ... (1)

cos u L—(2n)! 2! 4! 6! 8! 10!

n>0 v '

(see, e.g., [17, p. 177-178], [3, p. 258-259]).

* E-mail: foata@unistra.fr f E-mail: guoniu.han@unistra.fr

Springer

The study of the secant numbers E2n (n > 0) developed In the present paper makes use of the combinatorial model of binary increasing labeled trees (T2n) (n > 0) with an even number of nodes, also called secant trees, as #T2n = E2n. Their definitions are recalled in Section 2; just note that in such a tree each node has no child (it is then called a leaf), or two children, except the rightmost one that has one left child only. See, e.g., the whole set T4 displayed in Fig. 1.

The refinement of each secant number E2n, we are concerned with, takes the form of a double sum E2n = ^mk f2n(m,k) of positive integers, where each summand f2n(m,k) is defined by means of two statistics "eoc" ("end of minimal chain") and "pom" ("parent of maximum leaf"), whose domain is T2n, in the following manner:

f2n(m,k) := #{t G T2n : eoc(t) = m and pom(t) = k}. (2)

The end-of-minimal-chain eoc(t) of a tree t is defined to be the label of the leaf that is reached, when, starting with the tree root, one gets along a chain of nodes, always going from parent to the smaller child, if any. The parent-of-maximum-leaf pom(t) of t from T2n is simply defined to be the label of the father of leaf 2n. Again, check Fig. 1 where the values of "eoc" and "pom" are given for all trees from T4.

Those two statistics have been introduced by Poupard [19] on tangent trees (n odd). She proved that "eoc" and "1+pom" were equidistributed on each set T2n+i of tangent trees, also that their common univariable distribution satisfied a difference equation system;she further calculated their generating function. In [9] it was proved that the equidistribution actually holds on every set Tn (n > 1) of tangent and secant trees by constructing an explict bijection ^ of Tn onto itself with the property that: 1 +pom(t) = eoc t\$>(t) for all t.

The purpose of this paper is then to calculate the joint distribution of "eoc" and "pom" on each T2n and in particular derive an explicit three-variable exponential generating function (Theorem 1.5). To achieve this we prove that the coefficients f2n(m,k) are solutions of two partial difference equation systems. The proof is based on an original true Tree Calculus, which consists of partitioning each set of secant trees into smaller subsets and developing a natural algebra on those subsets (Sections 3, 4 and 5).

Note that the ranges of "eoc" and "pom" on T2n are the intervals [2, 2n] and [1, 2n — 1], respectively. The first values of the matrices M2n := (f2n(m,k)) (2 < m < 2n;1 < k < 2n — 1) are listed in Table 1. For each matrix have been evaluated the row sums f2n(m, •) := ^k f2n(m, k) = #{t G T2n : eoc(t) = m} (resp. column sums f2n(•, k) := ^m f2n(m, k) = #{t G T2n :pom(t) = k}) on the rightmost column (resp. the bottom row). In the South-East-corner is written the total sum

f2n(; •) = ^ fm(m,k) = E2n. (3)

As "1+pom" and "eoc" are equidistributed, we have:

f2n(-,k — 1) = f2n(k, •) (2 < k < 2n). (4)

Table 1. The matrices M2n (1 < n < 5)

k = 1 f2(m,.)

m = 2 1 1

f2(.,k) 1 E2 = 1

k = 1 2 3 f4(m,.)

m = 2 3 4 . . 1 1 2 . . 1 . 1 3 1

f4(.,k) 1 3 1 E4 = 5

k = 1 2 3 4 5 fa(m,.)

m = 2 1 3 1 5

3 1 2 9 3 15

4 3 7 10 1 21

5 1 4 8 2 15

6 2 2 1 5

fa(.,k) 5 15 21 15 5 Ea = 61

k = 1 2 3 4 5 6 7 fs(m,.)

m = 2 5 15 21 15 5 61

3 5 10 45 63 45 15 183

4 15 35 50 101 63 21 285

5 21 54 86 106 45 15 327

6 15 46 82 87 50 5 285

7 5 22 46 60 40 10 183

8 16 16 14 10 5 61

f8(.,k) 61 183 285 327 285 183 61 E8 = 1385

k = 1 2 3 4 5 6 7 8 9 f10(m,.)

m = 2 . 61 183 285 327 285 183 61 1385

3 61 122 . 549 855 981 855 549 183 4155

4 183 427 610 . 1405 1575 1341 855 285 6681

5 285 720 1132 1466 . 1989 1575 981 327 8475

6 327 884 1460 1863 2050 . 1405 855 285 9129

7 285 836 1448 1838 1870 1466 . 549 183 8475

8 183 606 1110 1466 1490 1155 610 61 6681

9 61 288 588 854 950 804 488 122 4155

10 272 272 256 224 178 122 61 1385

f 10 (., k) 1385 4155 6681 8475 9129 8475 6681 4155 1385 E10 = 50521

On each entry f2n(m, k) may be defined two partial differences with respect to m and k as follows:

A f2n(m, k) := f2n(m + 1, k) - hn(m,k);(5)

A f2n(m,k) := f2n(m,k + 1) - f2n(m,k). (6)

By convention f2n(m,k) := 0 if (m, k) ^ [2, 2n] x [1, 2n — 1]. Our main results are the following.

Theorem 1.1.

The finite difference equation systems hold:

A2f2n(m, k) + 4 f2n—2(m,k — 2) = 0 (2 < m < k — 3 < k < 2n — 1);(R1)

A2f2n(m,k) + 4 f2n—2(m, k) = 0(2 < m < k — 1 <k < 2n — 3). (R2)

The first two top rows and rightmost two columns of the upper triangles {f2n(m,k) : 2 < m < k < 2n — 1} (n > 2) can be evaluated in function of the row or column sums f2n—2(•, k), f2n—2(m, •), as is now stated.

Theorem 1.2.

We have:

f2n(2,k) = f2n—2(', k — 2) = f2n—2(k — 1, •) (3 < k < 2n — 1); [First top row] (7)

f2n(3,k) = 3 f2n(2,k) (4 < k < 2n — 1); [Second top row] f2n(m, 2n — 1) = f2n—2(m, •) = f2n—2(', m — 1) (2 < m < 2n — 2); [Rightmost column] f2n(m, 2n — 2) = 3 f2n(m, 2n — 1) (2 < m < 2n — 3). [Next to rightmost column]

Proposition 1.3.

We further have:

fi(: 1) = 1;f2n(., 1) = f2n-2(., •) = E2n-2(n > 2);(8) f2n(., 2) = 3f2n-2(-, •) = 3E2n-2 (n > 2); (9)

A2 f2n(m, •) + 4 f2n-2(m,.) = 0 (2 < m < 2n - 2);(R3)

A2 f2n(.,k) + 4 f'2n--2(.,k) = 0 (1 < k < 2n - 3). (R4)

Theorem 1.4.

The previous two theorems and Proposition 1.3 provide an explicit algorithm for calculating the entries of the upper triangles of the matrices M2n = (f2n(m, k)) (n > 1).

The entries of the lower triangles {f2n(m,k) :1 < k < m < 2n} in Tables 1.1 have been calculated directly by means of formula (2). Contrary to the upper triangles we do not have any explicit numerical algorithm to get them;only the entries situated on the three sides of those lower triangles can be directly evaluated, as shown in Section 6. See, in particular, Table 2. For the upper triangles, we can derive an explicit generating function, as stated in the next theorem.

Theorem 1.5.

The triple exponential generating function for the upper triangles of the matrices (f2n(m, k)) is given by

1— f , ,, x2n-k-1 yk-m-1 zm-2 _ cos(2y) + 2 cos(2(x - z)) - cos(2(z + x))

^kin^ fln(m,k)l2n-k-^y. Jk-m-^y. (m-W = 2 cos3(x + y + z) . (10)

The right-hand side of (10) is symmetric in x,z. Hence, the change x ^ z in the left-hand side of (10) shows that

f2n(2n + 1 — k, 2n +1 — m) = f-2n(m, k). (11)

The upper triangles already mentioned are then symmetric with respect to their counter-diagonals. This result can also be extended as follows.

Theorem 1.6.

Let Up(2n) be the set of all (m,k) from [2, 2n]x[1, 2n—1] such that either m — 1 < k, (m,k) = (3,1), or (m,k) = (2n, 2n — 2). Then (11) holds for every (m, k) G Up(2n).

Those theorems and proposition are proved in the next sections, once the main ingredients on Tree Calculus have been developed.

2. Tree Calculus

We use the traditional vocabulary on trees, such as node, leaf, child, root, ... When a node is not a leaf, it is called an internal node.

Definition. For each positive integer n a secant tree of size 2n, that is, an element of T2n, is defined by the following axioms:

(1) it is a labeled tree with 2n nodes, labeled 1, 2,..., 2n; the node labeled 1 being the root;

(2) each node has no child (then called a leaf), or one child, or two children;

(3) the label of each node is smaller than the label of its children, if any;

(4) the tree is planar and each child of each node is, either on the left (it is then called the left child), or on the right (the right child);moreover, the tree can be embedded on the Euclidean plane as follows: the root has coordinates (0, 0), the left child (if any) (—1,1), the right child (if any) (1,1), the grandchildren (if any) (—3/2, 2), (—1/2, 2), (1/2, 2), (3/2, 2), the greatgrandchildren (if any) (—7/4, 3), (—5/4, 3), ... , (7/4, 3), etc. With this convention all the nodes have different abscissas. The node having the maximum abscissa is then defined in a unique manner. Call it the rightmost node.

(5) every node is, either a leaf, or a node with two children, except the rightmost node, which has one left child, but no right child. This rightmost node will then be referred to as being the one-child node.

The orthogonal projections of the 2n nodes of a secant tree t from T2n onto a horizontal axis yields a permutation a = a(1)a(2) • • • a(2n) of 12 • • • 2n having the property that ^ > x2, x2 < x3, x3 > x4, ... , in an alternating way, usually named alternating (see [16, 23, 25]). The corresponding alternating permutation has been indicated for each t e in Fig. 1

Let t e T2n (n > 1). If a node labeled a has two children labeled b and c, define min a := min{b, c}; if it has one child b, let min a := b. The minimal chain of t is defined to be the sequence a^ —> a2 —> a3 ^ • • • aJ—1 ^ aj, with the following properties: (i) a^ = 1 is the label of the root;(ii) for i = 1, 2,..., j — 2 the (i + 1)-st term ai+1 is the label of an internal node and a^ = min af; (iii) aj is the node of a leaf. Define the "end of the minimal chain" of t to be eoc(t) := aj. If the leaf with the maximum label 2n is incident to a node labeled k, define its "parent of the maximum leaf" to be pom(t) := k. See Fig. 1.

We adopt the following notations and conventions: for each triple (n,m,k) let T2n m k (resp. T2n m,., resp. T2ni.ik) denote the subset of T2n of all trees t such that eoc(t) = m and pom(t) = k (resp. eoc(t) = m, resp. pom(t) = k). Also, symbols representing families of trees will also designate their cardinalities. With this convention Tn,m,k := #Tn m k. The matrix of the upper triangles T2n m k (2 < m < k < 2n — 1) will be denoted by Upper(T2n).

Subtrees (possibly empty) are indicated by the symbols "O," "V", or 'Q." The notation "1" (resp. "O," resp. "V") is used to indicate that the subtree "Q" (resp. "O," resp. "V") contains the one-child node or is empty. Letters occurring below or next to subtrees are labels of their roots. The end of the minimal chain in each tree is represented by a bullet

In the sequel certain families of secant trees will be represented by symbols, called trunks. For example, the symbol

is the trunk that designates the family of all trees t from the underlying set T2n having a node b, parent of both a subtree "Q" and the leaf m, which is also the end of the minimal chain. Notice that unlike the secant trees which are ordered, the trunk is unordered, so that

When the subtree "Q" contains the one-child node or is empty, we let

A(1) :=

be the family of all trees t from the underlying set T2n having a node b, parent of the right child "1X1" and the left child leaf m. Let A(*) be the set of all trees from A such that "Q" is not empty and does not contain the one-child node. We have the following decomposition

A = A(*) + A(1). (13)

When a further condition (C) is imposed on a trunk A we shall use the notation [A, (C)]. For example, the symbol

is the trunk that has the same characteristic of the trunk A as above with the further property that the node labeled a belongs, neither to the subtree of root b, nor to the path going from root 1 to b.

In our Tree Calculus we shall mostly compare the cardinalities of certain pairs of trunks, as shown in the following two examples.

Example 1.

The two trunks

, m] and [

have the same cardinalities, by using the bijection [ m+2 mm+1 ^++1).

m+1 \ /m+2

, m +2]

Example 2.

To compare the cardinalities of the following two trunks

we decompose them according to the location of the one-child node:

k+2 X/ V k+2 X/ k+21/ V k+2s k+1 k+1 k+1 k+1 C = k -L- k -L- Yk _L. Yk

:= C (*) + C (V) + C (1) + C (©);

k+2 k+1

D = 1 k

:= D(*) + D(V) + D(1).

In the above seven trunks of the two decompositions the symbols "EH," "V," "O" without slash are not empty and do not contain the one-child node. Furthermore, note that D = D(*) + D(V) + D(O also holds.

Pivoting the two subtrees on each of the nodes k, (k + 1), (k + 2) in C(*) (resp. in D(*)) yields the same trunk, as in (12). We then say that subtree pivoting is permitted on those nodes. Furthermore, the three subtrees "O," "EH" and "V" in C(*) play a symmetric role, while in D(*) only "O" and "V," on the one hand, and "EH" and "V," on the other hand, have a symmetric role. Hence,

C(*) = 2 D(*).

Now, remember that in each secant tree the one-child node lies at the rightmost position. This implies that subtree pivoting on each of the ancestors of the one-child is not permitted. Thus, subtree pivoting in C(V) and D(V) is only permitted on nodes k + 2 and k + 1. On the other hand, the two subtrees "O" and '"EH" play a symmetric role in C(V), but not in D(V). Hence,

C(V) = 2 D(V).

In C(1\[) and D(1\|) the two subtrees "O" and "V" play a symmetric role. Moreover, subtree pivoting on nodes k + 2 is permitted in C(1), but not in D(\). Hence,

C (\) = 2 D(\),

so that

C — 2 D = C(\).

3. Proof that (R1) holds

The decomposition

m \ /m+1 n

, m +1]

means that in each tree from T2nmtk the node (m + 1) is, or is not, the sibling of the leaf m. In the next decomposition the node m is, or is not, the parent of the leaf (m +1):

-2n,m+1,k —

As already explained in Example 1, Section 2, we can write:

AT2n,m,k = T'2n,m+1,k — T'2n,m,k =

m \ /m+1

so that

^ T2n,m,k = (T2n,m+2,k — T2n,m+1,k) — (T2n,m+1,k — T2n,m,k)

m+1 \ /m+2 m+-

m \ /m+1

Depending on the mutual positions of nodes m, (m + 1) and (m + 2) the further decompositions prevail, as again k still remains attached to (2n):

, m] := A + A2;

m+1 m+2 m+1 \/m+2

B = Im + [ I , m] := B1 + B2;

m + 2] := C + C2 + C3;

,m +2] := D1+ D2.

In the above decompositions the subsets Bt and C are identical. Furthermore, A2 = C3 and B2 = D2. Accordingly, A2 T2n+1,m,k = A — 2 — C2 + D1. A further decomposition of those four terms, depending upon the occurrence of the

one-child node, is to be worked out:

B1 C2 D1

= Ai(*)+ Ai(1)+ Ai(O); = Bi(*) + Bi (©);

= C2(*) + C2(1) + C2(O); = Di(*) + Di(1) + Di(O).

Now, Bi(*) can be decomposed into

m+l\/m+2 m+1\/ m+2 m+1\/m+2

Bi(*)= lm = lm + lm = Bi,i(*) + Bi,2(*),

where "Q" and "O" are supposed to be non-empty in B12(*). By the Tree Calculus techniques developed in Section 2, we have Ai(*) = 2 B12(*). On the other hand, B1(O) can also be written

Bi(O) =

so that Ai(1) = 2 Bi(O) and Ai (CO) = Bi(O). Furthermore, C2M = Di(*), C2I = 2 Di(1) and C2O = Di Altogether, Ai — 2Bi — C2 + Di = (Ai(*) + Ai(1) + Ai(O)) — 2(Bi,i(*) + Bi,2(*) + Bi(©)) — (C2M + C2I + C2(O)) (Di (*) + Di (1) + Di (©)) = Ai (O) — 2 Biii (*) — Di (1).

As m is supposed to be at least equal to 2, we can write

Ai(O) =

, Di(1) =

which shows that those two families are equal. Thus,

A2 T2n,m,k = —2 B^(*) = —2

This expression is also equal to — 4T2n—2tmtk—2, because in each tree t from B1 the nodes (m + 1) and (m + 2) are both leaves. Remove them, as well as the two edges going out of m, and subtract 2 from all the remaining nodes greater than (m + 2). The tree thereby derived belongs to T2n—2mk—2- d

4. Tree Calculus for proving that (R 2) holds

With n > 3 and 2 < m < k — 1 < k < 2n — 3 we have:

T2n,m,k

,m] + [ I ,m,k + 1],

meaning that each tree from T2nsmsk has one of the two forms: either k + 1 is incident to k, or not, and the leaf m is the end of the minimal chain.

Using the same dichotomy,

T2n!m!k+1 =

,m,k].

As the second terms of the above two equations are in one-to-one correspondence by the transposition (k,k + 1) we have:

T2n!m!k+1 — T2n,m,k =

, m] — [

,m]:= B — 4.

In the same manner,

T2n!m!k+2 — T2n,m,k+1 =

, m] — [

,m]:= D — C.

A T2n,m,k = (T2n,m,k+2 — T'2n,m,k+1 ) — (T'2n,m,k+1 — T'2n,m,k) = D — C — B + A.

The further decompositions of the components of the previous sum depend on the mutual positions of the nodes k, (k + 1),

(k + 2);

, m] = [

, m,k ] + [

,m] := D1 + D2;

, m] = [

, m, k ] +

, m] := C| + C2;

k+2 k+i1

, m | + | | k , m] := Bi + B2 + B3 + B4 + B5;

, m] := Ai + A2,

where "O" and "Q" cannot be both empty in B3 and B5.

Obviously, = and C, = A-,, so that the sum D — C—B +A may be written (B2 — B4) — (C2 —A2) + (D2 — B3 — B5) — 2 B2, showing that the components fall in three categories: (1) B2 and B4 having one subtree to determine;(2) C2 and A2 having two such subtrees; (3) D2, B3 and B5 having three of them.

(1) With the same notation as in (13) write: B2 = B2M + B2(1) and B4 = B4(*) + B4(1). First, B2(*) = B4M, but B2(1) = 2 B4(1). Hence,

B2 — B4 = B4(

(2) Next, C2 = C2(*) + C2(1) + C2(O) and A2 = A2M+ 4,(D + 4,(®|. As before, C2M = A2M and also C2O)) = 4,(. However, C2(1) = 2A2(1). Altogether,

C2 — A2 = A2(1) =

(3) The calculation of the third sum requires the following decompositions:

= D2M + D2(V) + D2(1) + D2(©); = Bs(*) + B3(V) + Bs(1); = B5(*) + B5CV) + B5(1);

By the Tree Calculus techniques developed in Section 2, especially Example 2, we have

D2(*) = 2 BS(*) = 2 B5 (*); D2(V) = 2 Bs(V) = 2 Bs(V); D2(1) = 2 Bs(1) = 2 BS(1).

Hence,

D2 - B3 - B5 = D2(O = [

^2(S) = [ I '™] =

where the two subtrees "O" and "v" are non-empty. The last Identity can be rewritten: A2(1\) = B4(\) + D2(O). Hence, A2 T2nmM = D - C - B + A = (B2 - B4) - (C2 - A2) + (D - B3 - B5) - 2 B2 = B4I - A2I + D2(O - 2 B2 =

-2 B2 = -2[ I , m], The last term is also equal to -4%2n-2,m,k, because in each tree t from B2 the nodes (k + 2) and (2n) are both leaves. Remove them, as well as the two edges going out of (k + 1), change (k + 1) into (2n - 2) and subtract 2 from all the remaining nodes greater than (k + 2). The tree thereby derived belongs to %2n-2,m,k. □

5. Proofs of Theorem 1.2, Proposition 1.3 and Theorem 1.4

Proof of Theorem 1.2. First top row. The trees t from T2n such that eoc(t) = 2 contain the edge 2 ^ 1. Remove it and change each remaining node label j by j - 2. We get a tree t' from T2n-2 such that pom(t') = k - 2. The second equality is a consequence of (4).

Second top row. As k > 4, the subtrees "Q" and "O" below are non-empty, so that

f2n(3,k ) =

2\ / \ /2 V + \ 11

A(\) + A(O);

2 ^ ^^ 2. ^y

fa,(2, k) = "\\/ 3 := B = "\\/ 3 = B(\),

since B(*) is empty. Therefore, A(\) = 2 B(\), A(( = B(\) and then f2jn(3, k) = 3 f-2n(2, k).

Rightmost column. The trees t from T2n such that eoc(t) = m and pom(t) = 2n - 1 contain the rightmost path 2n ^ (2n - 1) —> . Remove it. What is left is a tree t' from T2n-2 such that eoc(t') = m. The second equality is a consequence of (4).

Next to rightmost column. If t e T2n and pom(t) = 2n - 1, then t contains the rightmost path 2n ^ (2n - 1) as already noted. On the other hand, the node with label (2n - 2) is necessarily a leaf. The transposition (2n - 2,2n - 1) transforms t into a tree t| such that pom^) = 2n - 2. Also, removing the path 2n ^ (2n - 1) and rooting either the subtree 2n-1\y2n, or the subtree 2n\y2n, onto the leaf labeled (2n - 2) gives rise to two trees t2, t3 such that pom(t2) = pom(t3) = 2n - 2. □

Proof of Proposition 1.3. For (8) and (9) we only have to reproduce the proofs made in Theorem 1.2 for the first and second top rows, the parent of the maximum node playing no role.

To obtain (R4) simply write (R2) for m =2 and rewrite it using the first identity of Theorem 1.2 dealing with the "first top row." Next, (R3) is deduced from (R4) by using identity (4). □

Proof of Theorem 1.4. By induction, the row and column sums f2n(m, •) and f2n(',k) of the matrices M2n can be calculated by means of relations (8), (9) and, either (R3), or (R4). The first and second top rows (resp. rightmost and next to rightmost columns) of the matrix M2n are known by Theorem 1.2. It then suffices to apply, either rule (R1), from top to bottom, or rule (R2) from right to left to obtain the remaining entries of the upper triangles of M2n. □

6. The lower triangles

Observe that for 2n > 6 the non-zero entries of the first row, first column and last column of M2n are identical and they differ from the entries in the bottom row. For instance, the sequence 5,15, 21,15, 5 in M8 appears three times, but the bottom row reads: 16,16,14,10, 5.

By Theorem 1.2 we already know that f2n(2, k) = f2n—2(k — 1, •) = f2n(k — 1, 2n — 1) (3 < k < 2n — 1). We also have: f2n(m, 1) = f2n—2(m — 1, •) (3 < m < 2n — 1), by using this argument: each tree t from T2n satisfying pom(t) = 1 has its leaf node 2n incident to root 1. Just remove the edge 2n ^ 1. Change each remaining label j to j — 1. We get a secant tree t' belonging to T2n—2. The mapping t ^ t' is bijective;moreover, eoc(t') = eoc(t) — 1. As a summary,

f2n(2,k) = f2n(k — 1,2n — 1) = f2n(k, 1) (3 < k < 2n — 1). (14)

Introduce a further statistic "ent" (short hand for "Entringer") on each tree t from Tn (even for tangent trees) as follows: ent(t) is the label of the rightmost node of t. The distribution of "ent" is well-known, mostly associated with the model of the alternating permutations and traditionally called Entringer distribution (see, e.g., [4,11,18,19]). We also know how to calculate the generating function of that distribution and build up the Entringer triangle (Entn(j)) (1 < j < n — 1;n > 2), as done in Table 2.

For instance, the leftmost 61 on the row n =7 is the sum of all the entries in the previous row (including an entry 0 on the right!); the second 61 is the sum of the leftmost five entries;56, the sum of the leftmost four entries;46, the sum of the leftmost three entries;32, the sum of the leftmost two entries and 16 is equal to the leftmost entry. Let Entn(j) = #{t e Tn : ent(t) = j}.

Table 2. The Entringer distribution

j = 1 2 3 4 5 6

n =2 1

4 2 2 1

5 5 5 4 2

6 16 16 14 10 5

7 61 61 56 46 32 16

Proposition 6.1.

For 2 < k < 2n — 2 we have:

f2n(2n,k) = Ent2n-2(k - 1). (15)

Proof. Note that in a tree t e T2n such that eoc(t) = 2n and pom(t) = k the leaf labeled 2n is the unique son of the node labeled k, which is also the rightmost node. Let (1 = a-j) — a2 — a3 — ■ ■ ■ — (a- = k) — (aj = 2n) be the minimal chain of t. Form a new tree t' by means of the following changes:

(i) delete the path —> aj-1 — 2n;

(ii) for i = 1, 2,..., j - 2 replace each node label at of the minimal chain by ai+i - 1;

(iii) replace each other node label b by b - 1.

The label of the rightmost node of t' is then equal to the pom(t) -1 = k -1. The mapping t i—> t' is obvious bijective. □

Finally, the upper diagonal {f2n(k +1, k) : (1 < k < 2n - 1)} of the lower triangle in each matrix M2n can also be fully evaluated. First, in an obvious manner,

(also equal to f2n—4(-, •) = E2n—4) can be proved as follows. To each tree t G T2n such that eoc(t) = 3, pom(t) = 1 there correspond two trees, whose "eoc" and "pom" are equal to 2 and 3, respectively, as Illustrated by the diagram:

Likewise, each tree t G T2n such that eoc(t) = 2n and pom(t) = 2n — 2 necessarily has its rightmost four nodes equal to (2n — 1),b, 2n, (2n — 2) in that order, with b being a node less than (2n — 2). In particular, the latter node is its rightmost (one-child) node. To such a tree there correspond two trees, whose "eoc" and "pom" are equal to (2n — 1), (2n — 2), respectively, as illustrated by the next diagram. In particular, the node b becomes the rightmost node of those two such trees.

f2n (2,1) = f2n(2n, 2n — 1) = 0.

Second, the identities

f2n(3,2) = 2 f2n(3,1) = f2n(2n — 1,2n — 2) = 2 fa,(2n, 2n — 2)

Proposition 6.2.

We have the crossing equalities:

f2n(k — 1,k) + f2n(k + 1,k)= f2n(k, k — 1) + f2n(k, k + 1),

for 3 < k < 2n — 2.

The involved entries are located on the four bullets drawn in the following diagramme.

k—1 k

k—1 k

Proof. Let i, j be two different Integers from the set {(k — 1), k, (k + 1)}. Say that i and j are connected In a tree t, If the tree contains the edge i—j, or if i and j are brothers and of them is the end of the minimal chain of t. Each of the four ingredients of the previous identity is now decomposed into five terms, depending on whether the nodes (k — 1), k, (k + 1) are connected or not, namely: no connectedness;only k, (k + 1) connected;(k — 1), k connected;(k — 1), (k + 1) connected;all connected. Thus,

Now, the following identities hold: A, = Cv A2 = C4, A3 = D2, A4 = C2, B1 = D1, B3 = D4, B4 = D3, so that ZiA + Bi) — Zi(Q + D) = (B5 — D5) — (C3 — A5 — B2).

As before, we may write B5 = B5(*) + B5(©), D5 = D5(*) + D5(©). As B5(*) = D5(*) and B5(©) = 2 D5^©), we get:

B5 — D5 = D

5 - - -

as k is supposed to be greater than 3. Next,

2n • 2n

k\ / k+1"

C3 — B2 = [ I ,k + 1] —

,k — 1]

C3 - B2 - a5 =

:= E - F

,k - 1] =

= F (\) =

By comparing the evaluations (19) and (20) we get: (B5 — D5) — (C3 — A5 — B2) = 0. This completes the proof of (18). □

The entries f2n(3,2) and f2n(2n — 1,2n — 2), belonging to the upper diagonal of the lower triangle matrix M2n, being evaluated by identity (17), and the entries of the upper triangle being known by Theorem 1.4, we can apply the crossing equalities (18), starting with k = 3, and obtain all the values of the entries of that upper diagonal. Altogether, besides the entries of the upper triangle, all the entries lying on the border of the lower triangle can be calculated, as illustrated in boldface in the following matrix M8.

Table 3. The matrix M8, bold-faced entries analytically evaluated

k = 1 2 3 4 5 6 7 f8(m.)

m = 2 5 15 21 15 5 61

3 5 10 45 63 45 15 183

4 15 35 50 101 63 21 285

5 21 54 86 106 45 15 327

6 15 46 82 87 50 5 285

7 5 22 46 60 40 10 183

8 16 16 14 10 5 61

fa(.,k) 61 183 285 327 285 183 61 E8 = 1385

7. Generating functions for the f2n (m, k)

The calculation of the generating functions for the f2n(m, k) is similar to the calculation made for the tangent tree case in a our previous paper [9]. Recall the definition and some basic properties of the Poupard matrix. Let G = (gisj) (i > 0, j > 0) be an infinite matrix with nonnegative integral entries. Say that G is a Poupard matrix, if for every i > 0, j > 0 the following identity holds:

gij+2 — 2 gi+ij+1 + gi+2,j + 4 gLij = 0. (21)

Remark.

Last coefficient Is 4, not 2 as In [9]. Let

G(x,y):= £ gy (xi/i!)(y'/j!);

i>0J>0

R(y) := £ gy (yj/j!) (¡ > 0);

Cj(x) := £ gy (x£/i!) (j > 0)

be the exponential generating functions for the matrix itself, its rows and columns, respectively. The next two lemmas, serving to prove Theorem 1.5, are replica of Propositions 9.1 and 9.2 in [9]. Their proofs are omitted.

Lemma 7.1.

The following four properties are equivalent.

(i) G = (g,) (i > 0, j > 0) is a Poupard matrix;

(ii) Rf(y) - 2 R+ (y) + R^(y) + 4 Rl(y) = 0 for all i > 0;

(iii) C"(x) — 2 C+ (x) + Cj+2 (x) + 4 Cj (x) for all j > 0;

(iv) G(x, y) satisfies the partial differential equation:

topy - 2 + ^+4 G(x, y) = 0. (22)

ox2 dx dy dy2

Lemma 7.2.

Let G(x, y) be the exponential generating function for a Poupard matrix G. Then,

G(x, y) = A(x + y) cos(2 y) + B(x + y) sin(2 y), (23)

where A(y) and B(y) are two arbitrary series.

The entries of the upper triangles of the matrices (M2n) (see Table 1) are now recorded as entries of infinite matrices (Q^) (p > 1) by

0, if i + j = p mod 2;

f2n(m, k), if i + j = p mod 2;

,,(p) ._ Wl,j : =

with m := p +1, k := p + j + 2, 2n := p + i + j + 3. Conversely, i := 2n — k — 1, j := k — m — 1, p := m — 1. In particular, the first one Q(1) = («,') (i,j > 0) contains the first rows of the upper triangles, displayed as counter-diagonals. Furthermore, a counter-diagonal with zero entries is placed between two successive rows.

Q(1) =

( f4(2,3) 0

fa(2,3) 0

fs(2,3) 0

fa(2,4) 0

fs(2,4) 0

fa(2,5) 0

fs(2,5) 0

fs(2, 6)

fs(2,7) 0 0 •••

0 1 2 3 4 5 6 7

0 1 0 1 0 5 0 61 0 \

1 0 3 0 15 0 183 0

2 1 0 21 0 285 0

3 0 15 0 327 0

4 5 0 285 0

5 0 183 0

6 61 0

7 0 /

Proposition 7.3.

Every matrix Q(p) (p > 1) is a Poupard matrix.

Proof. Using Definition (24) we have

wp+2 - 2 wPu+1 + ut+2J + 4 Jpj = WK k + 2) - 2 WK k + 1) + f2n+2(m, k) + 4 /^(m, k) = A f2n+2(m, k) + 4 /2«(m, k) = 0,

by rule (R2). □

The row labeled i of Q(p) will be denoted by Qpj and the exponential generating function for that row by Q(p(y) = Zj>0 o)\Pjy'/j!. Also, Qp)(x, y) := ^i>0 Qf.'(y)x7i! will be the double exponential generating function for the matrix Qp).

Proposition 7.4.

For all p > 1 we have:

Q!C1)(y) = Qj1.(y) and Qp.(y) = 3 ^Qoj.(y) = 3 ^Qj—,.(y).

Proof. For the first identity it suffices to prove (Jq+1'1 = (¡j^, that is

f2n(m, 2n — 1) = f2n(2, 2n +1 - m).

This is true by the symmetry property of the Poupard triangle, as proved in Corollary 1.3 in [8]. For the second identity it suffices to prove (Jfj = 3w0p)+1, that is

f2n(m, 2n — 2) = 3f2n(m, 2n — 1).

This is true by Theorem 1.2. □

As x and y play a symmetric role in (23), the solution in (23) may also be written

C(x, y) = A(x + y) cos(2x) + B(x + y) sln(2x), so that the generating function of each matrix Q(p) is of the form

Q(p)(x, y) = A(x + y) cos(2x) + B(x + y) sin(2x).

Hence, Q(p>(x, y) | q}= Q{0pl(y) = A(y). Also,

(d/3x)Q(p)(x,y) = ((d/ dx)A(x + y)) cos(2x) + A(x + y)(—2)sin(2x) + ((d/dx)B(x + y)) sin(2x) + B(x + y)2 cos(2x) and

ixQ(p)(x, y) I{x = 0} = dxA(x + y) I{x = 0} +2 B(x + y) I {x = 0} = TyA(y +2 B(y) = Q-(y).

By Proposition 7.4 we have A(y) = Q<1.>1,.(y) and B(y) = (Q^y) — £-A(y))/2 = (3dQ^y) — dyQ^y))/2

dy \ dy p—dy

dQ^.(y). Hence,

Q(p>(x, y) = cos(2x)Qp—1,.(x + y) + sin(2x) dQ^.(x + y). (25)

First, make the evaluation of Q(1)(x, y). The row labeled 0 of the matrix Q(1) reads: 1, 0, 1, 0, 5, 0, 61, ..., which is the sequence of the coefficients of the Taylor expansion of sec y. Thus, Qg\(y) = sec y. Taking p = 1 in (25) we get

Q(1)(x, y) = sec(x + y) cos(2x) + sec(x + y) tan(x + y) sin(2x)

h = yi. (26)

cos2 (x + y)

For further use let us also calculate the partial derivative of Q'1)(x, y) with respect of y:

-Q(1)(x, y) = -3(1+—) |sin(x — y) cos(x + y) + 2 cos(x — y) sin(x + y) j

2 cos3(x + y)

Now, define

y cos3(x + y)

1 'sin(2y)+ 3 sin(2x)). (27)

Q(x,y,z):= £ Q(p)(x,y)(pZ—^ (28)

and make use of (25)-(28):

Q(x, y,z) = cos(2x) £ Qp,^1,.(x + y) (-—^ + sin(2x) — (£ Q^x + y) (pZ—^ )

= cos(2x) Q(1)(z, x + y) + sin(2x) d Q(1)(z, x + y)

,„. . cos(x + y — z) sin(2(x + y))+3sin(2z)

= cos(2x) —^-ï-^ + sin(2x) —V—iT—-

cos2(x + y + z) 2 cos3(x + y + z)

1 |cos(2x) 'cos(2(x + y)) + cos(2z)) + sin(2x) 'sin(2(x + y)) + 3sin(2z)) j

2 cos3(x + y + z) By definition of the «y's we get

2 cos3(x + y + z)

1 |cos(2y) + 2 cos(2(x — z)) — cos(2(z + x))j. (29)

Qx y,z) = £ «in» (pz^xï^ (p >1, i >0, j >0);

y2n—k— i i ,k—m—1 zm—2

= £ f2n(m,k)(2n—k—1! (k—m—^ (m—2)!, (30)

the latter sum over the set {3 < m +1 < k < 2n — 1}. We have proved Theorem 1.5.

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