October 4, 203 Secant Tree Calculus Doinique Foata and Guo-Niu Han Abstract. A true Tree Calculus is being developed to ae a joint study of the two statistics eoc (end of inial chain) and po (parent of axiu leaf) on the set of secant trees. Their joint distribution restricted to the set {eoc po } is shown to satisfy two partial difference equation systes, to be syetric and to be expressed in the for of an explicit three-variable generating function.. Introduction Recently, a revival of studies on arithetical and cobinatorial properties of both tangent and secant nubers has taen place, using Désiré André s old odel of alternating perutations [An979, An88], or the equivalent structure of binary increasing labeled trees, called tangent and secant trees. See, e.g., the papers [GHZ0, Pr08, Pr00, Fu00, HRZ0, Jos0, SZ0, KR2, FH2], or the eoir [St0] and our studies on the doubloon odel [FH0, FH0a, FH]. Recall that the secant nubers, denoted by E, appear in the Taylor expansions of sec u: (.) sec u = cos u = n 0 u ()! E = + u2 2! + u4 4! 5 + u6 u8 u0 6 + 385 + 6! 8! 0! 5052 + (see, e.g., [Ni23, p. 77-78], [Co74, p. 258-259]). The study of the secant nubers E (n 0) developed in the present paper aes use of the cobinatorial odel of binary increasing labeled trees (T ) (n 0) with an even nuber of nodes, also called secant trees, as #T = E. 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 rightost one that has one left child only. See, e.g., the whole set T 4 displayed in Fig... The refineent of each secant nuber E, we are concerned with, taes the for of a double su E =, f (, ) of positive integers, where each suand f (, ) is defined by eans of two statistics eoc ( end of inial chain ) and po ( parent of axiu leaf ), whose doain is T, in the following anner: (.2) f (, ) := #{t T :eoc(t) = and po(t) = }. Key words and phrases. Tree Calculus, partial difference equations, binary increasing labeled trees, secant and tangent trees, end of inial chain, parent of axiu leaf, bivariate distributions, secant nubers, Entringer distribution, alternating perutations. Matheatics Subject Classifications. 05A5, 05A30, B68.
DOMINIQUE FOATA AND GUO-NIU HAN The end-of-inial-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 fro parent to the saller child, if any. The parent-of-axiu-leaf po(t) of t fro T is siply defined to be the label of the father of leaf. Again, chec Fig.. where the values of eoc and po are given for all trees fro T 4. 3 4 4 4 3 2 4 3 3 4 2 2 2 2 3 4 3 2 eoc = po = 3 3 4 2 4 2 4 2 3 3 2 Fig.. 3 2 4 2 3 2 4 3 3 2 Those two statistics have been introduced by Poupard [Po89] on tangent trees (n odd). She proved that eoc and +po were equidistributed on each set T + of tangent trees, also that their coon univariable distribution satisfied a difference equation syste; she further calculated their generating function. In [FH3] it was proved that the equidistribution actually holds on every set T n (n ) of tangent and secant trees by constructing an explict bijection φ of T n onto itself with the property that: +po(t) = eoc φ(t) for all t. The purpose of this paper is then to calculate the joint distribution of eoc and po on each T and in particular derive an explicit threevariable exponential generating function (Theore.5). To achieve this we prove that the coefficients f (, ) are solutions of two partial difference equation systes. The proof is based on an original true Tree Calculus, which consists of partitioning each set of secant trees into saller subsets and developing a natural algebra on those subsets (Sections 3, 4 and 5). Note that the ranges of eoc and po on T are the intervals [2, ] and [, ], respectively. The first values of the atrices M := (f (, )) (2 ; ) are listed in Table.2. For each atrix have been evaluated the row sus f (, ) := f (, ) = #{t T : eoc(t) = } (resp. colun sus f (, ) := f (, ) = #{t T :po(t) = }) on the rightost colun (resp. the botto row). In the South-East-corner is written the total su (.3) f (, ) =, f (, ) = E. As +po and eoc are equidistributed, we have: (.4) f (, ) = f (, ) (2 ). 2
TREE SECANT CALCULUS = 2 3 f 4 (,.) = f 2 (,.) = 2.. M 2 = = 2 M 4 = 3 2. 3 f 2 (., ) E 2 = 4.. f 4 (., ) 3 E 4 = 5 M 6 = = 2 3 4 5 f 6 (,.) = 2.. 3 5 3 2. 9 3 5 4 3 7 0. 2 5 4 8 2. 5 6. 2 2. 5 f 6 (., ) 5 5 2 5 5 E 6 = 6 M 8 = = 2 3 4 5 6 7 f 8 (,.) = 2.. 5 5 2 5 5 6 3 5 0. 45 63 45 5 83 4 5 35 50. 0 63 2 285 5 2 54 86 06. 45 5 327 6 5 46 82 87 50. 5 285 7 5 22 46 60 40 0. 83 8. 6 6 4 0 5. 6 f 8 (., ) 6 83 285 327 285 83 6 E 8 = 385 M 0 = = 2 3 4 5 6 7 8 9 f 0 (,.) = 2.. 6 83 285 327 285 83 6 385 3 6 22. 549 855 98 855 549 83 455 4 83 427 60. 405 575 34 855 285 668 5 285 720 32 466. 989 575 98 327 8475 6 327 884 460 863 2050. 405 855 285 929 7 285 836 448 838 870 466. 549 83 8475 8 83 606 0 466 490 55 60. 6 668 9 6 288 588 854 950 804 488 22. 455 0. 272 272 256 224 78 22 6. 385 f 0 (., ) 385 455 668 8475 929 8475 668 455 385 E 0 = 5052 Table.2: the atrices M ( n 5) 3
DOMINIQUE FOATA AND GUO-NIU HAN On each entry f (, ) ay be defined two partial differences with respect to and as follows: (.5) f (, ) := f ( +, ) f (, ); (.6) f (, ) := f (, + ) f (, ). By convention f (, ) := 0 if (, ) [2, ] [, ]. Our ain results are the following. Theore.. The finite difference equation systes hold: (R ) 2 f (, ) + 4 f 2 (, 2) = 0 (2 3 < ); (R 2) 2 f (, ) + 4 f 2 (, ) = 0 (2 < 3). The first two top rows and rightost two coluns of the upper triangles {f (, ) : 2 < } (n 2) can be evaluated in function of the row or colun sus f 2 (, ), f 2 (, ), as is now stated. Theore.2. We have: (.7) f (2, ) = f 2 (, 2) = f 2 (, ) (3 ); [First top row] f (3, ) = 3 f (2, ) (4 ); [Second top row] f (, ) = f 2 (, ) = f 2 (, ) (2 2); f (, 2) = 3 f (, ) (2 3). Proposition.3. We further have: (.8) (.9) (R 3) [Rightost colun] [Next to rightost colun] f 2 (, ) = ; f (, ) = f 2 (, ) = E 2 (n 2); f (, 2) = 3f 2 (, ) = 3E 2 (n 2); 2 f (, ) + 4 f 2 (, ) = 0 (2 2); (R 4) 2 f (, ) + 4 f 2 (, ) = 0 ( 3). Theore.4. The previous two theores and Proposition.3 provide an explicit algorith for calculating the entries of the upper triangles of the atrices M = (f (, )) (n ). The entries of the lower triangles {f (, ) : < } in Tables. have been calculated directly by eans of forula (.2). Contrary to the upper triangles we do not have any explicit nuerical algorith to get the; 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 6.. For the upper triangles, we can derive an explicit generating function, as stated in the next theore. 4
TREE SECANT CALCULUS Theore.5. The triple exponential generating function for the upper triangles of the atrices (f (, )) is given by x y z 2 (.0) f (, ) ( )! ( )! ( 2)! 2 < = cos(2y) + 2 cos(2(x z)) cos(2(z + x)) 2 cos 3. (x + y + z) The right-hand side of (.0) is syetric in x, z. Hence, the change x z in the left-hand side of (.0) shows that (.) f ( +, + ) = f (, ). The upper triangles already entioned are then syetric with respect to their counter-diagonals. This result can also be extended as follows. Theore.6. Let Up() be the set of all (, ) fro [2, ] [, ] such that either, (, ) = (3, ), or (, ) = (, 2). Then (.) holds for every (, ) Up(). Those theores and proposition are proved in the next sections, once the ain 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, that is, an eleent of T, is defined by the following axios: () it is a labeled tree with nodes, labeled, 2,..., ; the node labeled 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 saller 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); oreover, the tree can be ebedded on the Euclidean plane as follows: the root has coordinates (0, 0), the left child (if any) (, ), the right child (if any) (, ), the grandchildren (if any) ( 3/2, 2), ( /2, 2), (/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 axiu abscissa is then defined in a unique anner. Call it the rightost node. (5) every node is, either a leaf, or a node with two children, except the rightost node, which has one left child, but no right child. This rightost node will then be referred to as being the one-child node. 5
DOMINIQUE FOATA AND GUO-NIU HAN The orthogonal projections of the nodes of a secant tree t fro T onto a horizontal axis yields a perutation σ = σ()σ(2) σ() of 2 having the property that x > x 2, x 2 < x 3, x 3 > x 4,..., in an alternating way, usually naed alternating (see [Vi88], [KPP94], [St99]). The corresponding alternating perutation has been indicated for each t T 4 in Fig.. Let t T (n ). If a node labeled a has two children labeled b and c, define in a := in{b, c}; if it has one child b, let in a := b. The inial chain of t is defined to be the sequence a a 2 a 3 a j a j, with the following properties: (i) a = is the label of the root; (ii) for i =, 2,..., j 2 the (i+)-st ter a i+ is the label of an internal node and a i+ = in a i ; (iii) a j is the node of a leaf. Define the end of the inial chain of t to be eoc(t) := a j. If the leaf with the axiu label is incident to a node labeled, define its parent of the axiu leaf to be po(t) :=. See Fig... We adopt the following notations and conventions: for each triple (n,, ) let T,, (resp. T,,, resp. T,,) denote the subset of T of all trees t such that eoc(t) = and po(t) = (resp. eoc(t) =, resp. po(t) = ). Also, sybols representing failies of trees will also designate their cardinalities. With this convention T n,, := #T n,,. The atrix of the upper triangles T,, (2 < ) will be denoted by Upper(T ). Subtrees (possibly epty) are indicated by the sybols,, or. The notation (resp. \, resp. \ ) is used to indicate that the subtree (resp., resp. ) contains the one-child node or is epty. Letters occurring below or next to subtrees are labels of their roots. The end of the inial chain in each tree is represented by a bullet. In the sequel certain failies of secant trees will be represented by sybols, called truns. For exaple, the sybol A = b is the trun that designates the faily of all trees t fro the underlying set T having a node b, parent of both a subtree and the leaf, which is also the end of the inial chain. Notice that unlie the secant trees which are ordered, the trun is unordered, so that (2.) b b. 6
TREE SECANT CALCULUS When the subtree contains the one-child node or is epty, we let b A() := b, be the faily of all trees t fro the underlying set T having a node b, parent of the right child and the left child leaf. Let A( ) be the set of all trees fro A such that is not epty and does not contain the one-child node. We have the following decoposition (2.2) A = A( ) + A(). When a further condition (C) is iposed on a trun A we shall use the notation [A, (C)]. For exaple, the sybol B = [ b, a ] is the trun that has the sae characteristic of the trun 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 fro root to b. In our Tree Calculus we shall ostly copare the cardinalities of certain pairs of truns, as shown in the following two exaples. Exaple. The two truns + +2 [, ] and [ +, + 2] have the sae cardinalities, by using the bijection ( + +2 +2 + ). Exaple 2. To copare the cardinalities of the following two truns +2 +2 C = and D =, we decopose the according to the location of the one-child node: +2 C = + +2 \ + 7 +2 + \ +2
DOMINIQUE FOATA AND GUO-NIU HAN := C( ) + C( \ ) + C() + C(\ ); +2 D = + +2 := D( ) + D( \ ) + D(). \ + +2 In the above seven truns of the two decopositions the sybols,, without slash are not epty and do not contain the one-child node. Furtherore, note that D = D( ) + D( \ ) + D(\ ) also holds. Pivoting the two subtrees on each of the nodes, (), (+2) in C( ) (resp. in D( )) yields the sae trun, as in (2.). We then say that subtree pivoting is peritted on those nodes. Furtherore, the three subtrees, and in C( ) play a syetric role, while in D( ) only and, on the one hand, and and, on the other hand, have a syetric role. Hence, C( ) = 2 D( ). Now, reeber that in each secant tree the one-child node lies at the rightost position. This iplies that subtree pivoting on each of the ancestors of the one-child is not peritted. Thus, subtree pivoting in C( \ ) and D( \ ) is only peritted on nodes + 2 and +. On the other hand, the two subtrees and play a syetric role in C( \ ), but not in D( \ ). Hence, C( \ ) = 2 D( \ ). In C() and D() the two subtrees and play a syetric role. Moreover, subtree pivoting on nodes + 2 is peritted in C(), but not in D(). Hence, so that C() = 2 D(), C 2 D = C(\ ). The decoposition 3. Proof that (R ) holds T,, = + + [ 8, + ]
TREE SECANT CALCULUS eans that in each tree fro T,, the node ( + ) is, or is not, the sibling of the leaf. In the next decoposition the node is, or is not, the parent of the leaf ( + ): + T,+, = + + [, ]. As already explained in Exaple, Section 2, we can write: + T,, = T,+, T,, = so that +, 2 T,, = (T,+2, T,+, ) (T,+, T,, ) +2 + +2 + + = + := A B C + D. + Depending on the utual positions of nodes, ( + ) and ( + 2) the further decopositions prevail, as again still reains attached to (): A = +2 + +2 + + [, ] := A + A 2 ; + B = +2 + +2 + [, ] := B + B 2 ; + + + +2 +2 C = + + [, + 2] := C + C 2 + C 3 ; +2 + + D = + [, + 2] := D + D 2. In the above decopositions the subsets B and C are identical. Furtherore, A 2 = C 3 and B 2 = D 2. Accordingly, 2 T +,, = 9
DOMINIQUE FOATA AND GUO-NIU HAN A 2 B C 2 +D. A further decoposition of those four ters, depending upon the occurrence of the one-child node, is to be wored out: A := A ( ) + A () + A (\ ); B := B ( ) + B (\ ); Now, B ( ) can be decoposed into C 2 := C 2 ( ) + C 2 () + C 2 (\ ); D := D ( ) + D () + D (\ ). + B ( ) = +2 + = +2 + = B, ( ) + B,2 ( ), + +2 where and are supposed to be non-epty in B,2 ( ). By the Tree Calculus techniques developed in Section 2, we have A ( ) = 2 B,2 ( ). On the other hand, B (\ ) can also be written \ + +2 B (\ ) =, so that A () = 2 B (\ ) and A (\ ) = B (\ ). Furtherore, C 2 ( ) = D ( ), C 2 () = 2 D () and C 2 (\ ) = D (\ ). Altogether, A 2 B C 2 + D = ( A ( ) + A () + A (\ ) ) 2 ( B, ( ) + B,2 ( ) + B (\ ) ) ( C 2 ( ) + C 2 () + C 2 (\ ) ) + ( D ( ) + D () + D (\ ) ) = A (\ ) 2 B, ( ) D (). As is supposed to be at least equal to 2, we can write +2\ + +2 A (\ ) =, D () = +, which shows that those two failies are equal. Thus, + 2 T,, = 2 B, ( ) = 2. +2 This expression is also equal to 4 T 2,, 2, because in each tree t fro B the nodes ( + ) and ( + 2) are both leaves. Reove the, as well as the two edges going out of, and subtract 2 fro all the reaining nodes greater than ( + 2). The tree thereby derived belongs to T 2,, 2. 0
TREE SECANT CALCULUS 4. Tree Calculus for proving that (R 2) holds With n 3 and 2 < 3 we have: T,, = [, ] + [,, + ], eaning that each tree fro T,, has one of the two fors: either + is incident to, or not, and the leaf is the end of the inial chain. Using the sae dichotoy, T,, = [, ] + [,, ]. As the second ters of the above two equations are in one-to-one correspondence by the transposition (, + ) we have: T,, T,, = [, ] [, ] := B A. In the sae anner, +2 +2 T,,+2 T,, = [, ] [, ] := D C. Thus, 2 T,, = ( T,,+2 T,, ) ( T,, T,, ) = D C B + A. The further decopositions of the coponents of the previous su depend on the utual positions of the nodes, ( + ), ( + 2); D = [ +2 +2 C = [, ] = [ +2 +2,, ] + [, ] := D + D 2 ; +2 +2, ] = [,, ] + [ := C + C 2 ;, ]
DOMINIQUE FOATA AND GUO-NIU HAN +2 B = [, ] = [,, + 2] + [, ] +2 +2 +2 + [, ] + [, ] + [, ] := B + B 2 + B 3 + B 4 + B 5 ; +2 A = [, ] = [,, + 2 ] + [ := A + A 2,, ] where and cannot be both epty in B 3 and B 5. Obviously, D = B and C = A, so that the su D C B + A ay be written (B 2 B 4 ) (C 2 A 2 ) + (D 2 B 3 B 5 ) 2 B 2, showing that the coponents fall in three categories: () B 2 and B 4 having one subtree to deterine; (2) C 2 and A 2 having two such subtrees; (3) D 2, B 3 and B 5 having three of the. () With the sae notation as in (2.2) write: B 2 = B 2 ( ) + B 2 () and B 4 = B 4 ( ) + B 4 (). First, B 2 ( ) = B 4 ( ), but B 2 () = 2 B 4 (). Hence, +2 B 2 B 4 = B 4 () = [, ]. (2) Next, C 2 = C 2 ( ) + C 2 () + C 2 (\ ) and A 2 = A 2 ( ) + A 2 () + A 2 (\ ). As before, C 2 ( ) = A 2 ( ) and also C 2 (\ ) = A 2 (\ ). However, C 2 () = 2 A 2 (). Altogether, +2 C 2 A 2 = A 2 () = [, ]. (3) The calculation of the third su requires the following decopositions: D 2 := D 2 ( ) + D 2 ( \ ) + D 2 () + D 2 (\ ); B 3 := B 3 ( ) + B 3 ( \ ) + B 3 (); B 5 := B 5 ( ) + B 5 ( \ ) + B 5 (); 2
TREE SECANT CALCULUS By the Tree Calculus techniques developed in Section 2, especially Exaple 2, we have D 2 ( ) = 2 B 3 ( ) = 2 B 5 ( ); D 2 ( \ ) = 2 B 3 ( \ ) = 2 B 3 ( \ ); D 2 () = 2 B 3 () = 2 B 3 (). Hence, \ +2 D 2 B 3 B 5 = D 2 (\ ) = [, ]. Now +2 +2 +2 A 2 () = [, ] = [, ] + [, ], where the two subtrees and are non-epty. The last identity can be rewritten: A 2 () = B 4 () + D 2 (\ ). Hence, 2 T,, = D C B + A = (B 2 B 4 ) (C 2 A 2 ) + (D 2 B 3 B 5 ) 2 B 2 = B 4 () A 2 () + D 2 (\ ) 2 B 2 = 2 B 2 = 2 [, ]. The last ter is also equal to 4 T 2,,, because in each tree t fro B 2 the nodes ( + 2) and () are both leaves. Reove the, as well as the two edges going out of ( +), change ( +) into ( 2) and subtract 2 fro all the reaining nodes greater than ( + 2). The tree thereby derived belongs to T 2,,. 5. Proofs of Theore.2, Proposition.3 and Theore.4 Proof of Theore.2. First top row. The trees t fro T such that eoc(t) = 2 contain the edge 2. Reove it and change each reaining node label j by j 2. We get a tree t fro T 2 such that po(t ) = 2. The second equality is a consequence of (.4). Second top row. As 4, the subtrees and below are nonepty, so that \ 3 2 f (3, ) = 3 3 2 = + 2 +2 := A = A( ) + A(\ ); 3
DOMINIQUE FOATA AND GUO-NIU HAN f (2, ) = 3 2 := B = 2 3 = B(), since B( ) is epty. Therefore, A( ) = 2 B( ), A(\ ) = B( ) and then f 2,n (3, ) = 3 f (2, ). Rightost colun. The trees t fro T such that eoc(t) = and po(t) = contain the rightost path ( ). Reove it. What is left is a tree t fro T 2 such that eoc(t ) =. The second equality is a consequence of (.4). Next to rightost colun. If t T and po(t) =, then t contains the rightost path ( ), as already noted. On the other hand, the node with label ( 2) is necessarily a leaf. The transposition ( 2, ) transfors t into a tree t such that po(t ) = 2. Also, reoving the path ( ) and rooting either the subtree, or the subtree, onto the leaf labeled ( 2) gives rise to two trees t 2, t 3 such that po(t 2 ) = po(t 3 ) = 2. Proof of Proposition.3. For (.8) and (.9) we only have to reproduce the proofs ade in Theore.2 for the first and second top rows, the parent of the axiu node playing no role. To obtain (R 4) siply write (R 2) for = 2 and rewrite it using the first identity of Theore.2 dealing with the first top row. Next, (R 3) is deduced fro (R 4) by using identity (.4). Proof of Theore.4. By induction, the row and colun sus f (, ) and f (, ) of the atrices M can be calculated by eans of relations (.8), (.9) and, either (R 3), or (R 4). The first and second top rows (resp. rightost and next to rightost coluns) of the atrix M are nown by Theore.2. It then suffices to apply, either rule (R ), fro top to botto, or rule (R 2) fro right to left to obtain the reaining entries of the upper triangles of M. 6. The lower triangles Observe that for 6 the non-zero entries of the first row, first colun and last colun of M are identical and they differ fro the entries in the botto row. For instance, the sequence 5, 5, 2, 5, 5 in M 8 appears three ties, but the botto row reads: 6, 6, 4, 0, 5. 4
TREE SECANT CALCULUS By Theore.2 we already now that f (2, ) = f 2 (, ) = f (, ) (3 ). We also have: f (, ) = f 2 (, ) (3 ), by using this arguent: each tree t fro T satisfying po(t) = has its leaf node incident to root. Just reove the edge. Change each reaining label j to j. We get a secant tree t belonging to T 2. The apping t t is bijective; oreover, eoc(t ) = eoc(t). As a suary, (6.) f (2, ) = f (, ) = f (, ) (3 ). Introduce a further statistic ent (short hand for Entringer ) on each tree t fro T n (even for tangent trees) as follows: ent(t) is the label of the rightost node of t. The distribution of ent is well-nown, ostly associated with the odel of the alternating perutations and traditionally called Entringer distribution (see, e.g., [En66], [Po82], [Po87], [GHZ0]). We also now how to calculate the generating function of that distribution and build up the Entringer triangle (Ent n (j)) ( j n ; n 2), as done in Table 6.. For instance, the leftost 6 on the row n = 7 is the su of all the entries in the previous row (including an entry 0 on the right!); the second 6 is the su of the leftost five entries; 56, the su of the leftost four entries; 46, the su of the leftost three entries; 32, the su of the leftost two entries and 6 is equal to the leftost entry. Let Ent n (j) = #{t T n : ent(t) = j}. j = 2 3 4 5 6 n = 2 3 4 2 2 5 5 5 4 2 6 6 6 4 0 5 7 6 6 56 46 32 6 Table 6.. The Entringer distribution Proposition 6.. For 2 2 we have: (6.2) f (, ) = Ent 2 ( ). Proof. Note that in a tree t T such that eoc(t) = and po(t) = the leaf labeled is the unique son of the node labeled, which is also the rightost node. Let ( = a ) a 2 a 3 (a j = ) (a j = ) be the inial chain of t. For a new tree t by eans of the following changes: 5
DOMINIQUE FOATA AND GUO-NIU HAN (i) delete the path a j ; (ii) for i =, 2,..., j 2 replace each node label a i of the inial chain by a i+ ; (iii) replace each other node label b by b. The label of the rightost node of t is then equal to the po(t) =. The apping t t is obvious bijective. Finally, the upper diagonal {f ( +, ) : ( )} of the lower triangle in each atrix M can also be fully evaluated. First, in an obvious anner, (6.3) f (2, ) = f (, ) = 0. Second, the identities (6.4) f (3, 2) = 2 f (3, ) = f (, 2) = 2 f (, 2) (also equal to f 4 (, ) = E 4 ) can be proved as follows. To each tree t T such that eoc(t) = 3, po(t) = there correspond two trees, whose eoc and po are equal to 2 and 3, respectively, as illustrated by the diagra: f (3, ) = 3 2 3 2 = f (3, 2). Liewise, each tree t T such that eoc(t) = and po(t) = 2 necessarily has its rightost four nodes equal to ( ), b,, ( 2) in that order, with b being a node less than ( 2). In particular, the latter node is its rightost (one-child) node. To such a tree there correspond two trees, whose eoc and po are equal to ( ), ( 2), respectively, as illustrated by the next diagra. In particular, the node b becoes the rightost node of those two such trees. b 2 2 b. Proposition 6.2. We have the crossing equalities: (6.5) f (, ) + f ( +, )=f (, ) + f (, + ), for 3 2. 6
TREE SECANT CALCULUS The involved entries are located on the four bullets drawn in the following diagrae. + + Proof. Let i, j be two different integers fro the set {( ),, ()}. 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 the is the end of the inial chain of t. Each of the four ingredients of the previous identity is now decoposed into five ters, depending on whether the nodes ( ),, ( + ) are connected or not, naely: no connectedness; only, ( + ) connected; ( ), connected; ( ), ( + ) connected; all connected. Thus, f (, ) = [,, + ] + [ + [, + ] + [,, := A + A 2 + A 3 + A 4 + A 5 ; f ( +, ) = [,, ] + [ + [, f (, ) = [ ] + [ := B + B 2 + B 3 + B 4 + B 5 ;, +, ] + [ + [, + ] + [,, ] + ], ] ] + ] := C + C 2 + C 3 + C 4 ; f (, + ) = [,, ] + [ 7, ], ]
+ [ DOMINIQUE FOATA AND GUO-NIU HAN, ] + [, := D + D 2 + D 3 + D 4 + D 5. ] + Now, the following identities hold: A = C, A 2 = C 4, A 3 = D 2, A 4 = C 2, B = D, B 3 = D 4, B 4 = D 3, so that i (A i + B i ) i (C i + D i ) = (B 5 D 5 ) (C 3 A 5 B 2 ). As before, we ay write B 5 = B 5 ( ) + B 5 (\ ), D 5 = D 5 ( ) + D 5 (\ ). As B 5 ( ) = D 5 ( ) and B 5 (\ ) = 2 D 5 (\ ), we get: \ (6.6) B 5 D 5 = D 5 (\ ) = = \, as is supposed to be greater than 3. Next, C 3 B 2 = [, + ] [, ] = [, + ] [, ] = C 3 B 2 A 5 = := E F = (E( ) F ( )) + (E(\ ) F (\ )) ; (6.7) \ = F (\ ) = \ =. By coparing the evaluations (6.6) and (6.7) we get: (B 5 D 5 ) (C 3 A 5 B 2 ) = 0. This copletes the proof of (6.5). 8
TREE SECANT CALCULUS The entries f (3, 2) and f (, 2), belonging to the upper diagonal of the lower triangle atrix M, being evaluated by identity (6.4), and the entries of the upper triangle being nown by Theore.4, we can apply the crossing equalities (6.5), starting with = 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 atrix M 8. M 8 = = 2 3 4 5 6 7 f 8 (,.) = 2.. 5 5 2 5 5 6 3 5 0. 45 63 45 5 83 4 5 35 50. 0 63 2 285 5 2 54 86 06. 45 5 327 6 5 46 82 87 50. 5 285 7 5 22 46 60 40 0. 83 8. 6 6 4 0 5. 6 f 8 (., ) 6 83 285 327 285 83 6 E 8 = 385 Table 6.: the atrix M 8, bold-faced entries analytically evaluated. 7. Generating functions for the f (, ) The calculation of the generating functions for the f (, ) is siilar to the calculation ade for the tangent tree case in a our previous paper [FH3]. Recall the definition and soe basic properties of the Poupard atrix. Let G = (g i,j ) (i 0, j 0) be an infinite atrix with nonnegative integral entries. Say that G is a Poupard atrix, if for every i 0, j 0 the following identity holds: (7.) g i,j+2 2 g i+,j+ + g i+2,j + 4 g i,j = 0. Rear. Last coefficient is 4, not 2 as in [FH3]. Let G(x, y) := g i,j (x i /i!) (y j /j!); R i (y) := g i,j (y j /j!) i 0, j 0 (i 0); C j (x) := i 0 g i,j (x i /i!) (j 0) be the exponential generating functions for the atrix itself, its rows and coluns, respectively. The next two leas, serving to prove Theore.5, are replica of Propositions 9. and 9.2 in [FH3]. Their proofs are oitted. 9 j 0
DOMINIQUE FOATA AND GUO-NIU HAN Lea 7.. The following four properties are equivalent. (i) G = (g i,j ) (i 0, j 0) is a Poupard atrix; (ii) R i (y) 2 R i+ (y) + R i+2(y) + 4 R i (y) = 0 for all i 0; (iii) C j (x) 2 C j+ (x) + C j+2(x) + 4 C j (x) for all j 0; (iv) G(x, y) satisfies the partial differential equation: 2 G(x, y) (7.2) x 2 2 2 G(x, y) + 2 G(x, y) x y y 2 + 4 G(x, y) = 0. Lea 7.2. Let G(x, y) be the exponential generating function for a Poupard atrix G. Then, (7.3) G(x, y) = A(x + y) cos(2 y) + B(x + y) sin(2 y), where A(y) and B(y) are two arbitrary series. The entries of the upper triangles of the atrices (M ) (see Table.2) are now recorded as entries of infinite atrices (Ω (p) ) (p ) by (7.4) ω (p) i,j := { 0, if i + j p od 2; f (, ), if i + j p od 2; with := p +, := p + j + 2, := p + i + j + 3. Conversely, i :=, j :=, p :=. In particular, the first one Ω () = (ω () i,j ) (i, j 0) contains the first rows of the upper triangles, displayed as counter-diagonals. Furtherore, a counter-diagonal with zero entries is placed between two successive rows. Ω () = 0 2 3 4 5 6 7 0 f 4 (2, 3) 0 f 6 (2, 5) 0 f 8 (2, 7) 0 f 0 (2, 9) 0 0 f 6 (2, 4) 0 f 8 (2, 6) 0 f 0 (2, 8) 0 2 f 6 (2, 3) 0 f 8 (2, 5) 0 f 0 (2, 7) 0 3 0 f 8 (2, 4) 0 f 0 (2, 6) 0 4 f 8 (2, 3) 0 f 0 (2, 5) 0 5 0 f 0 (2, 4) 0 6 f 0 (2, 3) 0 7 0 = 0 2 3 4 5 6 7 0 0 0 5 0 6 0 0 3 0 5 0 83 0 2 0 2 0 285 0 3 0 5 0 327 0 4 5 0 285 0 5 0 83 0 6 6 0 7 0. 20
TREE SECANT CALCULUS Proposition 7.3. Every atrix Ω (p) (p ) is a Poupard atrix. Proof. Using Definition (7.4) we have ω (p) i,j+2 2 ω(p) i+,j+ + ω(p) i+2,j + 4 ω(p) i,j = f +2 (, + 2) 2 f +2 (, + ) + f +2 (, ) + 4 f (, ) = f +2 (, ) + 4 f (, ) = 0, by rule (R 2). The row labeled i of Ω (p) will be denoted by Ω (p) i, and the exponential generating function for that row by Ω (p) i, (y) = j 0 ω(p) i,j yj /j!. Also, Ω (p) (x, y) := i 0 Ω(p) i, (y)xi /i! will be the double exponential generating function for the atrix Ω (p). Proposition 7.4. For all p we have: Ω (p+) 0, (y) = Ω () p,(y) and Ω (p), (y) = 3 d dy Ω(p) 0, (y) = 3 d dy Ω() p, (y). Proof. For the first identity it suffices to prove ω (p+) 0,j = ω () p,j, that is f (, ) = f (2, + ). This is true by the syetry property of the Poupard triangle, as proved in Corollary.3 in [FH2]. For the second identity it suffices to prove ω (p),j = 3ω(p) 0,j+, that is This is true by Theore.2. f (, 2) = 3f (, ). As x and y play a syetric role in (7.3), the solution in (7.3) ay also be written G(x, y) = A(x + y) cos(2 x) + B(x + y) sin(2 x), so that the generating function of each atrix Ω (p) is of the for Ω (p) (x, y) = A(x + y) cos(2 x) + B(x + y) sin(2 x). Hence, Ω (p) (x, y) = Ω (p) (y) = A(y). Also, {x=0} 0, ( / x)ω (p) (x, y) = (( / x)a(x + y)) cos(2 x) + A(x + y)( 2) sin(2 x) + (( / x)b(x + y)) sin(2 x) + B(x + y) 2 cos(2 x) 2
and x Ω(p) (x, y) DOMINIQUE FOATA AND GUO-NIU HAN {x = 0} = x A(x + y) {x = 0} +2 B(x + y) = d A(y) + 2 B(y) dy = Ω (p), (y). {x = 0} By Proposition 7.4 we have A(y) = Ω () (y) and B(y) = (Ω(p) (y) p,, d ( dy A(y))/2 = 3 d dy Ω() (y) d ) p, dy Ω() (y) /2 = d p, dy Ω() (y). Hence, p, (7.5) Ω (p) (x, y) = cos(2 x) Ω () (x + y) + sin(2 x) p, y Ω() (x + y). p, First, ae the evaluation of Ω () (x, y). The row labeled 0 of the atrix Ω () reads:, 0,, 0, 5, 0, 6,..., which is the sequence of the coefficients of the Taylor expansion of sec y. Thus, Ω () (y) = sec y. Taing p = in 0, (7.5) we get Ω () (x, y) = sec(x + y) cos(2 x) + sec(x + y) tan(x + y) sin(2 x) (7.6) cos(x y) = cos 2 (x + y). For further use let us also calculate the partial derivative of Ω () (x, y) with respect of y: y Ω() (x, y) = (7.7) (7.8) = Now, define cos 3 (x + y) 2 cos 3 (x + y) ( ) sin(x y) cos(x + y) + 2 cos(x y) sin(x + y) ( sin(2y) + 3 sin(2x) ). zp Ω(x, y, z) := Ω (p) (x, y) (p )! p and ae use of (7.5) (7.8): Ω(x, y, z) = cos(2x) p Ω () p, + sin(2x) y (x + y) zp (p )! ( p 22 Ω () p, ) (x + y) zp (p )!
TREE SECANT CALCULUS = cos(2x) Ω () (z, x + y) + sin(2x) y Ω() (z, x + y) (7.9) cos(x + y z) = cos(2x) cos 2 (x + y + z) sin(2(x + y)) + 3 sin(2z) + sin(2x) 2 cos 3 (x + y + z) = (cos(2x) ( 2 cos 3 cos(2(x + y)) + cos(2z) ) (x + y + z) + sin(2x) ( sin(2(x + y)) + 3 sin(2z) )) ( ) = 2 cos 3 cos(2y) + 2 cos(2(x z)) cos(2(z + x)). (x + y + z) By definition of the ω (p) i,j s we get,,n ω (p) i,j z p Ω(x, y, z) = (p, i 0, j 0); (p )! p,i,j = x y (7.0) f (, ) ( )! ( )! x i i! y j j! z 2 ( 2)!, the latter su over the set {3 + }. We have proved Theore.5. References [An879] Désiré André. Développeent de sec x et tan x, C. R. Math. Acad. Sci. Paris, vol. 88, 879, p. 965 979. [An88] Désiré André. Sur les perutations alternées, J. Math. Pures et Appl., vol. 7, 88, p. 67 84. [Co74] Louis Cotet. Advanced Cobinatorics. D. Reidel/Dordrecht-Holland, Boston, 974. [En66] R.C. Entringer. A cobinatorial interpretation of the Euler and Bernoulli nubers, Nieuw Arch. Wis., vol. 4, 966, p. 24 246. [FH0] Doinique Foata; Guo-Niu Han. The doubloon polynoial triangle, Raanujan J., vol. 23, 200, p. 07 26 (The Andrews Festschrift). [FH0a] Doinique Foata; Guo-Niu Han. Doubloons and q-secant nubers, Münster J. of Math., vol. 3, 200, p. 29 50. [FH] Doinique Foata; Guo-Niu Han. Doubloons and new q-tangent nubers, Quarterly J. Math., vol. 62, 20, p. 47 432. [FH2] Doinique Foata; Guo-Niu Han. Finite difference calculus for alternating perutations, J. Difference Equations and Appl., 203, 6 p. http://dx.doi.org/0.080/023698.203.794794 or http://arxiv.org/abs/304.2483. 23
DOMINIQUE FOATA AND GUO-NIU HAN [FH3] Doinique Foata; Guo-Niu Han. Tree Calculus for Bivariable Difference Equations, preprint, 36 p. http://arxiv.org/abs/304.2484. [Fu00] Marus Fule. A continued fraction expansion for a q-tangent function, Sé. Lothar. Cobin., B45b (2000), 3pp. [GHZ0] Yoann Gelineau; Heesung Shin; Jiang Zeng. Bijections for Entringer failies, Europ. J. Cobin., vol. 32, 20, p. 00 5. [HRZ0] Guo-Niu Han; Arthur Randrianarivony; Jiang Zeng. Un autre q-analogue des nobres d Euler, The Andrews Festschrift. Seventeen Papers on Classical Nuber Theory and Cobinatorics, D. Foata, G.-N. Han eds., Springer-Verlag, Berlin Heidelberg, 200, pp. 39-58. Sé. Lothar. Cobin., Art. B42e, 22 pp. [Jo39] Charles Jordan. Calculus of Finite Differences. Röttig and Rowalter, Budapest, 939. [Jos0] M. Josuat-Vergès. A q-enueration of alternating perutations, Europ. J. Cobin., vol. 3, 200, p. 892 906. [KR2] Sergey Kitaev; Jeffrey Reel. Quadrant Mared Mesh Patterns in Alternating Perutations, Sé. Lothar. Cobin., B68a (202), 20pp. [KPP94] A. G. Kuznetsov; I. M. Pa; A. E. Postniov. Increasing trees and alternating perutations, Uspehi Mat. Nau, vol. 49, 994, p. 79 0. [Ni23] Niels Nielsen. Traité éléentaire des nobres de Bernoulli. Paris, Gauthier- Villars, 923. [Po82] Christiane Poupard. De nouvelles significations énuératives des nobres d Entringer, Discrete Math., vol. 38, 982, p. 265 27. [Po89] Christiane Poupard. Deux propriétés des arbres binaires ordonnés stricts, Europ. J. Cobin., vol. 0, 989, p. 369 374. [Po97] Christiane Poupard. Two other interpretations of the Entringer nubers, Europ. J. Cobin., vol. 8, 997, p. 939 943. [Pr00] Helut Prodinger. Cobinatorics of geoetrically distributed rando variables: new q-tangent and q-secant nubers, Int. J. Math. Math. Sci., vol. 24, 2000, p. 825 838. [Pr08] Helut Prodinger. A Continued Fraction Expansion for a q-tangent Function: an Eleentary Proof, Sé. Lothar. Cobin., B60b (2008), 3 pp. [St0] Richard P. Stanley. A Survey of Alternating Perutations, in Cobinatorics and graphs, 65 96, Contep. Math., 53, Aer. Math. Soc. Providence, RI, 200. [SZ0] Heesung Shin; Jiang Zeng. The q-tangent and q-secant nubers via continued fractions, Europ. J. Cobin., vol. 3, 200, p. 689 705. [Vi88] Xavier G. Viennot. Séries génératrices énuératives, chap. 3, Lecture Notes, 60 p., 988, notes de cours donnés à l École Norale Supérieure Ul (Paris), UQAM (Montréal, Québec) et Université de Wuhan (Chine) http://web.ac.co/xgviennot/xavier Viennot/cours.htl. Doinique Foata Institut Lothaire, rue Murner F-67000 Strasbourg, France foata@unistra.fr Guo-Niu Han I.R.M.A. UMR 750 Université de Strasbourg et CNRS 7, rue René-Descartes F-67084 Strasbourg, France guoniu.han@unistra.fr 24