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This article was downloaded by: [Nipissing University]On: 08 October 2014, At: 13:51Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registeredoffice: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK
Journal of Difference Equations andApplicationsPublication details, including instructions for authors andsubscription information:http://www.tandfonline.com/loi/gdea20
Finite difference calculus foralternating permutationsDominique Foataa & Guo-Niu Hanb
a Institut Lothaire, 1, rue Murner, F-67000, Strasbourg, Franceb IRMA UMR 7501, Université de Strasbourg et CNRS, 7, rue René-Descartes, F-67084, Strasbourg, FrancePublished online: 11 Jun 2013.
To cite this article: Dominique Foata & Guo-Niu Han (2013) Finite difference calculus foralternating permutations, Journal of Difference Equations and Applications, 19:12, 1952-1966, DOI:10.1080/10236198.2013.794794
To link to this article: http://dx.doi.org/10.1080/10236198.2013.794794
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Finite difference calculus for alternating permutations
Dominique Foataa1 and Guo-Niu Hanb*
aInstitut Lothaire, 1, rue Murner, F-67000 Strasbourg, France; bIRMA UMR 7501, Universite deStrasbourg et CNRS, 7, rue Rene-Descartes, F-67084 Strasbourg, France
(Received 8 January 2013; final version received 3 April 2013)
The finite difference equation system introduced by Christiane Poupard in the study oftangent trees is reinterpreted in the alternating permutation environment. It makes itpossible to make a joint study of both tangent and secant trees and calculate thegenerating polynomial for alternating permutations by a new statistic, referred to asbeing the greater neighbour of the maximum.
Keywords: difference equation system; Poupard triangle; tangent numbers; secantnumbers; alternating permutations; strictly ordered binary trees
1. Introduction
Let f ¼ ðf nðkÞÞ ðn $ 1; 1 # k # 2n 2 1Þ be a family of rational numbers, displayed in a
triangular array of the form
f ¼
f 1ð1Þ
f 2ð1Þ f 2ð2Þ f 2ð3Þ
f 3ð1Þ f 3ð2Þ f 3ð3Þ f 3ð4Þ f 3ð5Þ
f 4ð1Þ f 4ð2Þ f 4ð3Þ f 4ð4Þ f 4ð5Þ f 4ð6Þ f 4ð7Þ
· · · · · · · · · · · · · · · · · · · · · · · · · · ·
ð1:1Þ
and consider the finite difference equation system
D2f nðkÞ þ 4f n21ðkÞ ¼ 0 ðn $ 2; 1 # k # 2n 2 3Þ; ð1:2Þ
where D stands for the classical finite difference operator (see, e.g. [18])
Df nðkÞ :¼ f nðk þ 1Þ2 f nðkÞ; ð1:3Þ
so that
D2f nðkÞ ¼ f nðk þ 2Þ2 2f nðk þ 1Þ þ f nðkÞ: ð1:4Þ
If at each step n $ 2 the two entries f nð1Þ and f nð2Þ are given explicit values, the whole
system (1.2) has a unique solution, as the equation D2f nð1Þ þ 4f n21ð1Þ ¼ 0 yields the value
of f nð3Þ, then D2f nð2Þ þ 4f n21ð2Þ ¼ 0 the value of f nð4Þ, etc.
q 2013 Taylor & Francis
*Corresponding author. Email: [email protected]
Journal of Difference Equations and Applications, 2013
Vol. 19, No. 12, 1952–1966, http://dx.doi.org/10.1080/10236198.2013.794794
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The same conclusion holds if the two bordered diagonals
ðf 1ð1Þ; f 2ð1Þ; f 3ð1Þ; f 4ð1Þ; . . . ; f nð1Þ; . . . Þ; f 1ð1Þ; f 2ð3Þ; f 3ð5Þ; f 4ð7Þ; . . . ; f nð2n 2 1Þ; . . . Þ
are taken as initial values. To see this we first note that the equation f 2ð1Þ2 2f 2ð2Þ þ
f 2ð3Þ þ 4f 1ð1Þ ¼ 0 determines f 2ð2Þ uniquely. Assuming that the triangle ðf n0 ðmÞÞ ð1 #
m # 2n0 2 1; n0 # nÞ has been determined, the system D2f nþ1ðmÞ þ 4f nðmÞ ¼ 0 ð1 #
m # 2n 2 1Þ consists of ð2n 2 1Þ linear equations with ð2n 2 1Þ unknowns, namely,
f nþ1ð2Þ; f nþ1ð3Þ; . . . ; f nþ1ð2nÞ, the underlying matrix being trigonal of the form
Fnþ1 :¼
22 1 0 0 · · · 0 0 0
1 22 1 0 · · · 0 0 0
0 1 22 1 · · · 0 0 0
..
. ... ..
. ... . .
. ... ..
. ...
0 0 0 0 · · · 1 22 1
0 0 0 0 · · · 0 1 22
0BBBBBBBBBBB@
1CCCCCCCCCCCA:
As det Fnþ1 ¼ 22n ðn $ 1Þ, the system has a unique solution.
The purpose of this paper is to solve (1.2) in four cases, when the sets of initial values
called [tan1], [tan2], [sec1], [sec2] are the following:
[tan1] f 1ð1Þ ¼ 1; f nð1Þ ¼ 0 and f nð2Þ ¼ 2P
kf n21ðkÞ for n $ 2;
[tan2] f 1ð1Þ ¼ 1; f nð1Þ ¼ f nð2n 2 1Þ ¼ 0 for n $ 2;
[sec1] f 1ð1Þ ¼ 1; f nð1Þ ¼P
kf n21ðkÞ and f nð2Þ ¼ 3P
kf n21ðkÞ for n $ 2;
[sec2] f 1ð1Þ ¼ 1; f nð1Þ ¼ f nð2n 2 1Þ ¼P
kf n21ðkÞ for n $ 2;
It will be proved (see Theorem 1.5) that both initial values [tan1] and [tan2](respectively, [sec1] and [sec2]) in fact lead to the same solution of the system and,
furthermore, that the solutions found for the f nðkÞ’s are non-negative integral values. To
avoid any confusion the solutions of (1.2) will be denoted by ðgnðkÞÞ (respectively, ðhnðkÞÞ)
when using [tan1] (respectively, [sec1]). The first numerical values of those
solutions are displayed in Figure 1.1.
To the right of each triangle the row sums have been calculated, which are equal, as
stated in the next theorem, to the tangent numbers (respectively, the secant numbers).
Those classical numbers, denoted by T2nþ1 and E2n, appear in the Taylor expansions of
g1(1)=1 T1=1g2(1)=0 g2(2)=2 g2(3)=0 T3=2
g3(1)=0 g3(2)=4 g3(3)=8 g3(4)=4 g3(5)=0 T5=16T7=272g4(1)=0 g4(2)=32 g4(3)=64 g4(4)=80 g4(5)=64 g4(6)=32 g4(7)=0
h1(1)=1 E2=1E4=5E6=61
h2(1)=1 h2(2)=3 h2(3)=1h3(1)=5 h3(2)=15 h3(3)=21 h3(4)=15 h3(5)=5
h4(1)=61 h4(2)=183 h4(3)=285 h4(4)=327 h4(5)=285 h4(6)=183 h4(7)=61 E8=1385
Figure 1. The two triangles of gnðkÞ’s and hnðkÞ’s.
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tan u and sec u:
tan u ¼Xn$0
u2nþ1
ð2n þ 1Þ!T2nþ1
¼u
1!1 þ
u3
3!2 þ
u5
5!16 þ
u7
7!272 þ
u9
9!7936 þ · · ·
ð1:5Þ
sec u ¼1
cos u
¼Xn$0
u2n
ð2nÞ!E2n
¼ 1 þu2
2!1 þ
u4
4!5 þ
u6
6!61 þ
u8
8!1385 þ
u10
10!50521 þ · · ·
ð1:6Þ
(see, e.g. [5], pp. 258–259]; [22], p. 177–178).
Theorem 1.1. Let ðgnðkÞÞ (respectively, ðhnðkÞÞ be the unique solution of the finite
difference equation system (1.2) when using the initial values [tan1] (respectively, [sec1]).
Then, the row sums of the solutions are equal toXk
gnðkÞ ¼ T2n21 ðn $ 1Þ; ð1:7Þ
Xk
hnðkÞ ¼ E2n ðn $ 1Þ: ð1:8Þ
As further mentioned, Theorem 1.1 will appear as a consequence of Theorem 1.4. It
will also be shown that the generating functions for the coefficients gnðkÞ and hnðkÞ can be
evaluated in the following forms.
Theorem 1.2. Let
Zðx; yÞ :¼ 1 þXn$1
X1#k#2nþ1
f nþ1ðkÞx2nþ12k
ð2n þ 1 2 kÞ!
yk21
ðk 2 1Þ!
and Z tanðx; yÞ (respectively, Z secðx; yÞ) when f nðkÞ :¼ gnðkÞ (respectively, f nðkÞ :¼ hnðkÞ).
Then,
Z tanðx; yÞ ¼ secðx þ yÞcosðx 2 yÞ; ð1:9Þ
Z secðx; yÞ ¼ sec2ðx þ yÞcosðx 2 yÞ: ð1:10Þ
As Z tanðy; xÞ ¼ Z tanðx; yÞ and Z secðy; xÞ ¼ Z secðx; yÞ, this implies the following Corollary.
Corollary 1.3. The entries gnðkÞ and hnðkÞ have the symmetry property:
gnðkÞ ¼ gnð2n 2 kÞ; hnðkÞ ¼ hnð2n 2 kÞ ð1 # k # 2n 2 1Þ: ð1:11Þ
In view of (1.7) and (1.8), two finite sets A2n21 and A2n, of cardinalities T2n21 and E2n,
are to be found, together with a statistic, call it ‘grn’, defined on those sets with the
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property that
Xs[A2n21
xgrns ¼X
k
gnðkÞxk; ð1:12Þ
Xs[A2n
xgrns ¼X
k
hnðkÞxk: ð1:13Þ
We shall use Desire Andre’s old result [1,2], who introduced the notion of alternating
permutation, as being a permutation s ¼ sð1Þsð2Þ· · ·sðnÞ of 12· · ·n with the property that
sð1Þ . sð2Þ, sð2Þ , sð3Þ, sð3Þ . sð4Þ, etc. in an alternating way. For each n $ 1 let An
denote the set of all alternating permutations of 12· · ·n. He proved that #A2n21 ¼ T2n21,
#A2n ¼ E2n. The desired statistic ‘grn’ is then the following.
Definition. Let s ¼ sð1Þsð2Þ· · ·sðnÞ be an alternating permutation from An, so that
sðiÞ ¼ n for a certain i ð1 # i # nÞ. By convention, let sð0Þ ¼ sðn þ 1Þ :¼ 0. Define the
greater neighbor of n in s to be
grnðsÞ :¼ maxfsði 2 1Þ;sði þ 1Þ}: ð1:14Þ
Also, let
An;k :¼ fs [ An : grnðsÞ ¼ k} ð0 # k # n 2 1Þ: ð1:15Þ
Theorem 1.4. Under the same assumptions as in Theorem 1.1 we have
gnðkÞ ¼ #A2n21;k21 ðn $ 1; 1 # k # 2n 2 1Þ; ð1:16Þ
hnðkÞ ¼ #A2n;k ðn $ 1; 1 # k # 2n 2 1Þ: ð1:17Þ
Example. There are T3 ¼ 2 alternating permutations of length 3, namely, 213 and 312, and
grnð213Þ ¼ grnð312Þ ¼ 1, so that g2ð1Þ ¼ #A3;0 ¼ 0, g2ð2Þ ¼ #A3;1 ¼ 2, g2ð3Þ ¼
#A3;2 ¼ 0; there are E4 ¼ 5 alternating permutations of length 4, namely, 4132, 4231,
3142, 3241, 2143 and grnð4132Þ ¼ 1, grnð4231Þ ¼ grnð3142Þ ¼ grnð3241Þ ¼ 2,
grnð2143Þ ¼ 3, so that h2ð1Þ ¼ 1, h2ð2Þ ¼ 3, h2ð3Þ ¼ 1; in accordance with the numerical
values in Figure 1.1.
As #A2n21 ¼ T2n21, #A2n ¼ E2n, following Desire Andre’s result, it is now clear that
Theorem 1.1 is a consequence of Theorem 1.4. Thus, an analytical result is proved by
combinatorial methods.
In Proposition 2.1 it will be proved that #A1;0 ¼ 1, #A2n21;0 ¼ #A2n21;2n22 ¼ 0 ðn $
2Þ and #A2n21;1 ¼ 2T2n23 ðn $ 2Þ; also, #A2;1 ¼ 1, #A2n;1 ¼ #A2n;2n21 ¼ E2n22 ðn $ 2Þ
and #A2n;2 ¼ 3E2n22 ðn $ 2Þ. In view of Theorem 1.4 this implies that conditions
[tan2] and [sec2] are fulfilled as soon as conditions [tan1] and [sec1] hold, and
then the following theorem.
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Theorem 1.5. The entries ðgnðkÞÞ and ðhnðkÞÞ given by (1.16) and (1.17) are also solutions
of the finite difference equation system (1.2) when the initial values [tan2] and [sec2] are
used, respectively.
There are other combinatorial models which are also counted by tangent and secant
numbers, or in a one-to-one correspondence with alternating permutations, in particular,
the labelled, binary, increasing, topological trees, also called ‘arbres binaires croissants
complets’ by Viennot [32, chapter 3, p. 111]. The set of those trees having n labelled nodes
is denoted by Tn. The statistic ‘pom’ (parent of the maximum leaf), introduced by Poupard
[24] for her strictly ordered, binary trees can be extended to all of Tn. The usual bijection
(see [32]) g : Tn ! An, called projection, has the property: pomðtÞ ¼ grnðgðtÞÞ, as proved
in Theorem 5.1. We then have another combinatorial interpretation for the polynomialsPk gnðkÞx
k andP
k hnðkÞxk.
As such, the triangle ðgnðkÞÞ ðn $ 1; 1 # k # 2n 2 1Þ does not appear in Sloane’s
On-Line Encyclopedia of Integer Sequences [29], but the triangle ðgnðkÞ=2n21Þ does under
reference A008301 and is called Poupard’s triangle, after her pioneering work on strictly
ordered binary trees [24]. It is banal to verify that 2n21 divides gnðkÞ when dealing with the
combinatorial model Tn for n odd.
In contrast to Christiane Poupard [24], who showed that the distribution of the strictly
ordered binary trees satisfied the finite difference equation system D2f 2nþ1ðkÞþ
2f 2n21ðkÞ ¼ 0, we have used the multiplicative factor ‘4’ in equation (1.2) to make a
unified study of the tangent and secant cases and deal with objects in one-to-one
correspondence with alternating permutations. Mutatis mutandis, identities (1.16), as well
as (1.11) concerning the tangent numbers, are due to her. She obtains the generating
function for her trees in the form: secððx þ yÞ=ffiffiffi2
pÞ cosððx 2 yÞ=
ffiffiffi2
pÞ instead of (1.9).
However, the alternating permutation development in Sections 2 and 3, identity (1.10) and
the combinatorial properties of the entries hnðkÞ are new.
The triangle of hnðkÞ’s appears in Sloane’s [29] as sequence A125053. It was deposited
there by Paul D. Hanna. The entries have been calculated by using a procedure equivalent
to (1.2) and the initial condition [tan2]. No combinatorial interpretation is given and no
generating function calculated.
Theorem 1.4 will be proved in Section 3, once evaluations of some cardinalities such
as #An;k will be made, as done in Section 2. The proof of Theorem 1.2 is given in Section 4,
together with further identities on gnðkÞ and hnðkÞ’s.
2. Some special values
The evaluations of #A2n21;k21 and #A2n;k made in the next proposition for some values of n
and k will facilitate the derivation of the proof of Theorem 1.2. They also have their own
combinatorial interests.
Proposition 2.1. The following relations hold:
#A1;0 ¼ 1; #A2n21;0 ¼ #A2n21;2n22 ¼ 0 ðn $ 2Þ; ð2:1Þ
#A2n21;1 ¼ #A2n21;2n23 ¼ 2T2n23 ðn $ 2Þ; ð2:2Þ
#A2n21;2 ¼ #A2n21;2n24 ¼ 4T2n23 ðn $ 3Þ; ð2:3Þ
#A2n21;3 ¼ #A2n21;2n25 ¼ 6T2n23 2 8T2n25 ðn $ 3Þ; ð2:4Þ
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Xk$0
#A2n21;k ¼ T2n21 ðn $ 1Þ: ð2:5Þ
#A2;1 ¼ 1; ð2:6Þ
#A2n;1 ¼ #A2n;2n21 ¼ E2n22 ðn $ 2Þ; ð2:7Þ
#A2n;2 ¼ #A2n;2n22 ¼ 3E2n22 ðn $ 2Þ; ð2:8Þ
#A2n;3 ¼ #A2n;2n23 ¼ 5E2n22 2 4E2n24 ðn $ 2Þ; ð2:9ÞXk$1
#A2n;k ¼ E2n ðn $ 1Þ: ð2:10Þ
Proof. (2.1) Set A1 is the singleton 1 and grnð1Þ ¼ 0 by definition, so that #A1;0 ¼ 1. For
n $ 2 all alternating permutations from A2n21 have a ‘grn’ at least equal to 1. Hence,
#A2n21;0 ¼ 0. Finally, each alternating permutation of length ð2n 2 1Þ ðn $ 2Þ contains
neither the factor ð2n 2 2Þð2n 2 1Þ nor ð2n 2 1Þð2n 2 2Þ. Hence, #A2n21;2n22 ¼ 0.
(2.2) When n $ 2, each alternating permutation from A2n21;1 starts with ð2n2 1Þ1, or
ends with 1ð2n2 1Þ. After removal of those two letters, there remains an alternating
permutation on f2;3; . . . ;2n2 2}. Hence, #A2n21;1 ¼ 2T2n23. Next, each permutation from
A2n21;2n23 must contain, either the three-letter factor ð2n2 1Þð2n2 3Þð2n2 2Þ, or
ð2n2 2Þð2n2 3Þð2n2 1Þ. The removal of the factor ð2n2 1Þð2n2 3Þ (respectively,
ð2n2 3Þð2n2 1Þ) yields an alternating permutation of the set f1;2; . . . ; ð2n2 4Þ; ð2n2 2Þ},
of cardinality ð2n2 3Þ. This proves the relation #A2n21;2n23 ¼ 2T2n23.
(2.3) Start with an alternating permutation on f1; 3; 4; . . . ; 2n 2 2}, then having ð2n 2
3Þ elements. There are four possibilities to generate a permutation from A2n21;2: (1) insert
ð2n 2 1Þ2 to the left; (2) insert 2ð2n 2 1Þ to the right; (3) insert 2ð2n 2 1Þ just before 1; (4)
insert ð2n 2 1Þ2 just after 1. For the second identity in (2.3) proceed in the same way: in
each alternating permutation on f1; 2; . . . ; 2n 2 1}nf2n 2 4; 2n 2 1} the two letters
ð2n 2 3Þ, ð2n 2 2Þ are necessarily local maxima. There are four possibilities to obtain a
permutation from A2n21;2n24: insert ð2n 2 1Þð2n 2 4Þ just before, either ð2n 2 3Þ or
ð2n 2 2Þ; also insert ð2n 2 4Þð2n 2 1Þ just after, either ð2n 2 3Þ, or ð2n 2 2Þ.
(2.4) Each permutation from A2n21;3 containing factor 2 1 (respectively, 1 2) starts with
2 1 (respectively, ends with 1 2). Dropping factor 2 1 (respectively, 1 2) and subtracting 2
from the remaining letters yields an alternating permutation from A2n23;1. There are then
2ð2T2n25Þ permutations from A2n21;3 containing either 2 1 or 1 2.
If a permutation from A2n21;3 contains neither one of those two factors, it has one of the
six properties: it starts with ð2n2 1Þ3, or contains one of the three-letter factor 1ð2n2 1Þ3,
3ð2n2 1Þ1, 2ð2n2 1Þ3, 3ð2n2 1Þ2, or still ends with 3ð2n2 1Þ. After removal of the
two-letter factor ð2n2 1Þ3 or 3ð2n2 1Þ there remains an alternating permutation on
f1;2;4; . . . ; ð2n2 2Þ} not starting with 2 1 and not ending with 1 2. There are then
6ðT2n23 2 2T2n25Þ such permutations. Altogether, #A2n21;3 ¼ 4T2n25 þ 6ðT2n23 2
2T2n25Þ ¼ 6T2n23 2 8T2n25.
The proof of the second identity in (2.4) follows a different pattern. If the letter
ð2n 2 4Þ is a local minimum (i.e. less than its two adjacent letters) in a permutation s
from A2n21;2n25, then s necessarily contains one of the four five-letter factors
ð2n 2 1Þð2n 2 5Þð2n 2 2Þð2n 2 4Þð2n 2 3Þ, ð2n 2 1Þð2n 2 5Þð2n 2 3Þð2n 2 4Þð2n 2 2Þ,
ð2n 2 2Þð2n 2 4Þð2n 2 3Þð2n 2 5Þð2n 2 1Þ, ð2n 2 3Þð2n 2 4Þð2n 2 2Þð2n 2 5Þð2n 2 1Þ.
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Replacing this five-letter factor by ð2n 2 5Þ yields a permutation from A2n25. Thus, there
are 4T2n25 permutations from A2n21;2n25 in which ð2n 2 4Þ is a local minimum.
In the other permutations from A2n21;2n25, all the four letters ð2n24Þ, ð2n23Þ,
ð2n22Þ, ð2n21Þ are local maxima (i.e. greater than their adjacent letters). Let A02n21;2n25
be the set of those permutations. When the two-letter factor ð2n21Þð2n25Þ or ð2n25Þ�
ð2n21Þ is deleted from such a permutation, there remains a permutation on f1;2; . . . ;ð2n2
1Þ}nfð2n25Þ;ð2n21Þ} in which the third largest letter ð2n24Þ is not a local minimum. Let
A002n23 be the set of those permutations. But the alternating permutations on the latter set in
which ð2n24Þ is a local minimum necessarily contain the three-letter factor ð2n23Þ�
ð2n24Þð2n22Þ or ð2n22Þð2n24Þð2n23Þ. There are then 2T2n25 such permutations.
Hence, #A002n23 ¼ T2n23 22T2n25. To obtain a permutation from A0
2n21;2n25 it suffices to
start from a permutation s00 from A002n23 and insert ð2n21Þð2n25Þ (respectively,
ð2n25Þð2n21Þ) just before (respectively, just after) each one of the three letters ð2n24Þ,
ð2n23Þ, ð2n22Þ (which are all local maxima). There are then 6ðT2n23 22T2n25Þ such
permutations. Altogether, #A2n21;2n25 ¼ 4T2n25 þ6ðT2n23 22T2n25Þ ¼ 6T2n23 28T2n25.
No comment for (2.5) and (2.6).
(2.7) Simply note that the only alternating permutations from A2n;1 and A2n;2n21 are,
respectively, of the form: ð2nÞ1sð3Þ· · ·sð2nÞ and sð1Þsð2Þ· · ·ð2nÞð2n 2 1Þ.
(2.8) Same proof as for (2.3): start with an alternating permutation on
f1; 3; 4; . . . ; ð2n 2 1Þ}. There are exactly three possibilities to generate a permutation
from A2n;2: insert ð2nÞ2 to the left, or just after letter 1, or still insert 2ð2nÞ just before letter
1. For the second identity start with a permutation on f1; 2; . . . ; 2n}nf2n 2 2; 2n} and
insert ð2nÞð2n 2 2Þ either to the right or just before ð2n 2 1Þ, or still insert ð2n 2 2Þð2nÞ
just after ð2n 2 1Þ.
(2.9) Each permutation from A2n;3 containing factor 21 is necessarily of the form
s ¼ 21sð3Þ· · ·sð2nÞ, so that the alternating permutation s0 :¼ ðsð3Þ2 2Þ· · ·ðsð2nÞ2 2Þ
belongs to A2n22;1. There are then E2n24 such permutations. If a permutation from A2n;3
does not contain 21, it has one of the five properties: it starts with ð2nÞ 3, or contains one of
the three-letter factor 1 ð2nÞ 3, 3ð2nÞ1, 2ð2nÞ3, 3ð2nÞ2. After removal of the two-letter
factor ð2nÞ3 or 3ð2nÞ there remains an alternating permutation on f1; 2; 4; . . . ; ð2n 2 1Þ}
not starting with 21. There are then 5ðE2n22 2 E2n24Þ such permutations. Altogether,
#A2n;3 ¼ E2n24 þ 5ðE2n22 2 E2n24Þ ¼ 5E2n22 2 4E2n24.
The proof for the second identity in (2.9) is quite similar. Each permutation from
A2n;2n23 containing the factor ð2n21Þð2n22Þ necessarily ends with the four-letter factor
ð2nÞð2n23Þð2n21Þð2n22Þ. There are then E2n24 such permutations. The other
permutations from A2n;2n23 contain one of the four three-letter factors
ð2nÞð2n23Þð2n22Þ, ð2n22Þð2n23Þð2nÞ, ð2nÞð2n23Þð2n21Þ and ð2n21Þð2n2
3Þð2nÞ, or ends with ð2nÞð2n23Þ. After removal of the two-letter factor ð2nÞð2n23Þ or
ð2n23Þð2nÞ there remains an alternating permutation on f1;2; . . . ;ð2n24Þ;ð2n22Þ; ð2n2
1Þ}, not ending with the two-letter factor ð2n21Þð2n22Þ. There are E2n22 2E2n24 such
permutations. Altogether, #A2n;2n23 ¼E2n24 þ5ðE2n22 2E2n24Þ.
No comment for (2.10). A
3. Proof of Theorem 1.4
Let anðkÞ :¼ #A2n21;k21 and bnðkÞ :¼ #A2n;k. From Proposition 2.1 it follows that the
initial conditions [tan1] and [tan2] hold when f nðkÞ ¼ anðkÞ, and [sec1] and also
[sec2] when f nðkÞ ¼ bnðkÞ. It remains to prove that in each case (1.2) holds.
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By means of identities (2.2)–(2.4) and (2.7)–(2.9) we easily verify that (1.2) holds for
both anðkÞ and bnðkÞ when n ¼ 2; 3 and 1 # k # 2n 2 3. It also holds for anðkÞ when n $ 4
and k ¼ 1; 2; 2n 2 4; 2n 2 3, and for bnðkÞ when n $ 4 and k ¼ 1; 2n 2 3.
What is left to prove is D2anðkÞ þ 4an21ðkÞ ¼ 0, that is, D2A2n21;k21 þ 4A2n23;k21 ¼ 0
for n $ 4 and 3 # k # 2n 2 5 [by identifying each finite set with its cardinality] and also
D2bnðkÞ þ 4bn21ðkÞ ¼ 0, that is, D2A2n;k þ 4A2n22;k ¼ 0 for n $ 4 and 2 # k # 2n 2 4;
altogether,
D2An;k þ 4An22;k ¼ 0
for n $ 7 and 2 # k # n 2 4 ðn evenÞ 2 # k # n 2 5 ðn oddÞ:ð2:11Þ
Let v ¼ y1; . . . ; ym be a non-empty word with distinct letters from the set
f0; 1; 2; . . . ; n} and ~v ¼ ym; . . . ; y1 be its mirror image. If m ¼ 1 and y1 ¼ 0, let ½v� ¼
½0� be the empty set. If m $ 2 and y1 ¼ 0 (respectively, ym ¼ 0), let ½v� be the set of all
alternating permutations from An, if any, whose left factors are equal to y2½ . . . �ym, or
whose right factors are equal to ym½ . . . �y2. When y1 $ 1, let ½v� be the set of all alternating
permutations from An, if any, containing either factor v or factor ~v. Finally, let ½~v� :¼ ½v�.
Using those notations we get
An;k ¼X
0#y#k21
½ynk�
¼X
0#y#k21
½ynkðk þ 1Þ� þX
0#y#k21kþ2#z#n21
½ynkz� þX
1#y#k21
½ynk0�;
An;kþ1 ¼X
0#y#k
½ynðk þ 1Þ�
¼ ½knðk þ 1Þ� þX
0#y#k21kþ2#z#n21
½ynðk þ 1Þz� þX
1#y#k21
½ynðk þ 1Þ0�:
The transposition ðk; k þ 1Þ maps the set ½ynkz� onto the set ½ynðk þ 1Þz� for
z [ fk þ 2; . . . ; n 2 1} < f0}, so that we may write
DAn;k ¼ An;kþ1 2 An;k
¼ ½knðk þ 1Þ�2X
0#y#k21
½ynkðk þ 1Þ�
¼ ½knðk þ 1Þ�2X
0#y1;y2#k21y1–y2
½y1nkðk þ 1Þy2�;
DAn;kþ1 ¼An;kþ2 2An;kþ1 ¼½ðkþ1Þnðkþ2Þ�2X
0#y#k
½ynðkþ1Þðkþ2Þ�
¼ ½ðkþ1Þnðkþ2Þ�2 ½knðkþ1Þðkþ2Þ�
2X
0#y#k21
½ynðkþ1Þðkþ2Þk�
2X
0#y1;y2#k21y1–y2
½y1nðkþ1Þðkþ2Þy2�:
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For 2 # k # n 2 4 the permutation
k k þ 1 k þ 2
k þ 1 k þ 2 k
!
maps ½y1nkðk þ 1Þy2� onto ½y1nðk þ 1Þðk þ 2Þy2� in a bijective manner. Hence,
D2An;k ¼ DAn;kþ1 2 DAn;k
¼ ½ðk þ 1Þnðk þ 2Þ�2 ½knðk þ 1Þðk þ 2Þ�
2X
0#y#k21
½ynðk þ 1Þðk þ 2Þk�2 ½knðk þ 1Þ�:
But
½knðk þ 1Þ� ¼ ½ðk þ 2Þknðk þ 1Þ� þ ½knðk þ 1Þðk þ 2Þ�
þX
z1;z2[fkþ3; ... ;n21}<f0}z1–z2
½z1knðk þ 1Þz2�:
Again, the permutation
k k þ 1 k þ 2
k þ 1 k þ 2 k
!
maps the last sum onto the set ½ðk þ 1Þnðk þ 2Þ�. Altogether, as ½ðk þ 2Þknðk þ 1Þ� ¼
½knðk þ 1Þðk þ 2Þ�, we have
D2An;k ¼ 23½knðk þ 1Þðk þ 2Þ�2X
0#y#k21
½ynðk þ 1Þðk þ 2Þk�:
When removing the factor ðk þ 1Þðk þ 2Þ and replacing each integer z $ k þ 2
by ðz 2 2Þ, in each alternating permutation, both sets ½knðk þ 1Þðk þ 2Þ� andP0#y#k21½ynðk þ 1Þðk þ 2Þk� are transformed into An22;k; so that D2An;k ¼ 24An22;k.
4. The bivariate generating functions
Let f ¼ ðf nðkÞÞ ðn $ 1; 1 # k # 2n 2 1Þ be the family of rational numbers, as displayed
in (1.1), that satisfies the finite-difference equation system (1.2) under the initial conditions
[tan2] of [sec2]. We know that the system has then a unique solution. With the
triangle f associate the infinite matrix
G ¼ ðgijÞði$0;j$0Þ :¼
f 1ð1Þ 0 f 2ð3Þ 0 f 3ð5Þ 0 f 4ð7Þ · · ·
0 f 2ð2Þ 0 f 3ð4Þ 0 f 4ð6Þ · · ·
f 2ð1Þ 0 f 3ð3Þ 0 f 4ð5Þ · · ·
0 f 3ð2Þ 0 f 4ð4Þ · · ·
f 3ð1Þ 0 f 4ð3Þ · · ·
0 f 4ð2Þ · · ·
f 4ð1Þ · · ·
0BBBBBBBBBBBBB@
1CCCCCCCCCCCCCA: ð4:1Þ
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In other words, define gij :¼ 0 when i þ j is odd, and gij :¼ f nðkÞ with k :¼ j þ 1, 2n ¼
2 þ i þ j when i þ j is even. For i þ j even the mapping ði; jÞ 7! ðn; kÞ is one-to-one, the
reverse mapping being for n $ 1, 1 # k # 2n 2 1 given by i ¼ 2n 2 1 2 k, j ¼ k 2 1.
In terms of the entries gij relation (1.2) may be written in the form
gi;j ¼ 2gi21;j21 þ1
2ðgi21;jþ1 þ giþ1;j21Þ ði $ 1; j $ 1Þ; ð4:2Þ
gij ¼ 0; if i þ j odd: ð4:3Þ
Furthermore, the full matrix G ¼ ðgi;jÞ ði $ 0; j $ 0Þ is completely determined as soon as
its first row ðg0;jÞ ðj $ 0Þ and first column ðgi;0Þ ði $ 0Þ are known. Let f 7! G denote the
above correspondence between those triangles and matrices.
Let Zðx; yÞ :¼P
i$0;j$0gi;jðxi=i!Þðy j=j!Þ. It is easily verified that Zðx; yÞ satisfies the
partial differential equation
›2Zðx; yÞ
›x›y¼ 2Zðx; yÞ þ
1
2
›2Zðx; yÞ
›x2þ
1
2
›2Zðx; yÞ
›y2; ð4:4Þ
if and only if the coefficients gi;j satisfy relation (4.2). Hence, Zðx; yÞ is fully determined by
(4.4) and by the generating functions Zðx; 0Þ ¼P
i$0gi;0x i=i! and Zð0; yÞ ¼P
j$0g0;jyj=j!
for the first column and first row of matrix G.
But for any given formal power series in one variable f ðxÞ ¼ 1 þPn$1f 2nðx
2n=ðð2nÞ!ÞÞ it can also be verified that the bivariate formal power series
Zðx; yÞ ¼X
i$0;j$0
gi;jx i
i!
y j
j!¼ f ðx þ yÞ secðx þ yÞ cosðx 2 yÞ ð4:5Þ
satisfies (4.4) and that the generating functions for its first column and first row are given
by f ðxÞ and f ðyÞ, respectively. This proves the following proposition.
Proposition 4.1. Let f ðxÞ ¼ 1 þP
n$1f 2nðx2n=ðð2nÞ!ÞÞ be given and G ¼ ðgijÞ ði $ 0; j $
0Þ be an infinite matrix, whose entries satisfy relations (4.2) and (4.3), on the one hand,
and such that g0;0 ¼ 1, g2nþ1;0 ¼ g0;2nþ1 ¼ 0 for n $ 0 and g2n;0 ¼ g0;2n ¼ f 2n for n $ 1,
on the other hand. Then, identity (4.5) holds.
Using the correspondence gij $ f nðkÞ mentioned above, series Zðx; yÞ can be rewritten
Zðx; yÞ ¼ 1 þXn$1
X1#k#2nþ1
f nþ1ðkÞx2nþ12k
ð2n þ 1 2 kÞ!
yk21
ðk 2 1Þ!; ð4:6Þ
which is then equal to f ðx þ yÞ secðx þ yÞ cosðx 2 yÞ under the assumptions of the previous
Proposition.
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Now, consider the two triangles described in Figure 1.1 and let Gtan ¼ ðgtanij Þ and
Gsec ¼ ðgsecij Þ be the two G-matrices attached to them:
Gtan ¼ ðgtanij Þ ¼
g1ð1Þ 0 g2ð3Þ 0 g3ð5Þ 0 g4ð7Þ · · ·
0 g2ð2Þ 0 g3ð4Þ 0 g4ð6Þ · · ·
g2ð1Þ 0 g3ð3Þ 0 g4ð5Þ · · ·
0 g3ð2Þ 0 g4ð4Þ · · ·
g3ð1Þ 0 g4ð3Þ · · ·
0 g4ð2Þ · · ·
g4ð1Þ · · ·
0BBBBBBBBBBBBB@
1CCCCCCCCCCCCCA;
Gsec ¼ ðgsecij Þ ¼
h1ð1Þ 0 h2ð3Þ 0 h3ð5Þ 0 h4ð7Þ · · ·
0 h2ð2Þ 0 h3ð4Þ 0 h4ð6Þ · · ·
h2ð1Þ 0 h3ð3Þ 0 h4ð5Þ · · ·
0 h3ð2Þ 0 h4ð4Þ · · ·
h3ð1Þ 0 h4ð3Þ · · ·
0 h4ð2Þ · · ·
h4ð1Þ · · ·
0BBBBBBBBBBBBB@
1CCCCCCCCCCCCCA:
The exponential generating function for the first row and first column of Gtan is equal to
f ðxÞ ¼ 1. On the other hand, as hnð1Þ ¼ hnð2n þ 1Þ ¼ #A2n;1 ¼ E2n22 by [sec2], (1.8)
and (1.17), the exponential generating function for the first row and first column of Gsec is
equal to h1ð1Þ þ h2ð1Þx2=2! þ h3ð1Þx
4=4! þ · · · ¼ E0 þ E2x=2! þ E4x4=4 þ · · · ¼ secðxÞ.
Theorem 1.2 is then a consequence of the previous Proposition.
When ðx; yÞ is equal to ðx; xÞ, then to ðx;2xÞ in (1.9) and (1.10), we obtain:
Z tanðx; xÞ ¼ secð2xÞ; Z tanðx;2xÞ ¼ cosð2xÞ; Z secðx; xÞ ¼ sec2ð2xÞ ¼ 1 þPn$1 4nT2nþ1x2n=ð2nÞ! and Z secðx;2xÞ ¼ cosð2xÞ. Looking for the coefficients of
x2n=ð2nÞ! on both sides in the first (respectively, last) two formulae yields four further
identities
X1#k#2nþ1
2n
k 2 1
!gnþ1ðkÞ ¼ 4nE2n ðn $ 1Þ; ð4:7Þ
X1#k#2nþ1
ð21Þk2n
k 2 1
!gnþ1ðkÞ ¼ ð21Þn4n ðn $ 1Þ; ð4:8Þ
X1#k#2nþ1
2n
k 2 1
!hnþ1ðkÞ ¼ 4nT2nþ1 ðn $ 1Þ and ð4:9Þ
X1#k#2nþ1
ð21Þk2n
k 2 1
!hnþ1ðkÞ ¼ ð21Þn4n ðn $ 1Þ: ð4:10Þ
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The last two identities are mentioned in Sloane’s Encyclopedia [29] (sequence A125053)
without proofs.
5. Alternating permutations and binary trees
In this section the traditional vocabulary on trees, such as node, leaf, child, root, etc. is
used. In particular, when a node is not a leaf, it is said to be an internal node.
Definition. An n-labelled, binary, increasing, topological tree is defined by the following
axioms:
(1) it is a labelled tree with n nodes, labelled 1; 2; . . . ; n; the node labelled 1 is called
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 a node is, either on the left (it is then called the
left child), or on the right (the right child);
(5) when n is odd, each node is either a leaf or a node with two children; when n is
even, each node is either a leaf or a node with two children, except the rightmost
node (uniquely defined) which has one left child, but no right child. It will be
referred to as being the one-son child.
Each such binary tree t may be drawn on a Euclidean plane: the root has coordinates
ð0; 0Þ, the left son of the root ð21; 1Þ and the right son ð1; 1Þ, the grandsons ð23=2; 2Þ,
ð21=2; 2Þ, ð1=2; 2Þ, ð3=2; 2Þ, respectively, the great-grandsons ð27=4; 3Þ, ð25=4; 3Þ, . . . ,
ð7=4; 3Þ, etc. With this convention all the nodes have different abscissas. Let t have n nodes
and make the orthogonal projections of those nodes on a horizontal axis. Writing the labels
of the projected n nodes yields a permutation s ¼ sð1Þsð2Þ· · ·sðnÞ of 12· · ·n. We say that
s is the projection of t and t the spreading out of s. Moreover, s is alternating. For
instance, the two trees t1 and t2 in Figure 1.2 are labelled, binary, increasing, topological
trees, with 7 and 8 nodes, respectively. Their projections s1 ¼ 6154723 and s2 ¼
61548273 are alternating.
For each n $ 1 let Tn be the set of all n-labelled, binary, increasing, topological trees.
Then, t 7! s is a bijection of Tn onto An, so that we also have: #A2nþ1 ¼ T2nþ1,
#A2n ¼ E2n. Each tree t from Tn is said to be tangent (respectively, secant), if n is odd
(respectively, even).
The two statistics ‘emc’ (‘end of minimal chain’) and ‘pom’ (‘parent of maximum
leaf’) we now define on each set Tn have been introduced by Poupard [24] in her study of
5
43
62
1
=
t1= t2=
5
4
7
62
1
3
7 8
3274516s1 = 7 3284516s2
Figure 2. Tangent, secant trees and alternating permutations.
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the strictly ordered binary trees and provide two other combinatorial interpretations for
entries gnðkÞ and hnðkÞ. Their definitions are also valid for all binary increasing trees, in
particular, for secant and tangent trees. Let n $ 2 and t be a binary increasing tree, with n
nodes labelled 1; 2; . . . ; n. Let a be the label of an internal node. If the node has two
children labelled b and c, define min a :¼ minfb; c}; if it has one child b, let min a :¼ b.
The minimal chain of t is defined to be the sequence a1 ! a2 ! a3 ! · · · ! aj21 ! aj,
with the following properties:
(i) a1 ¼ 1 is the label of the root;
(ii) for i ¼ 1; 2; . . . ; j 2 1 the i-th term ai is the label of an internal node and
aiþ1 ¼ min ai;
(iii) aj is the label of a leaf.
Define the end of the minimal chain in t to be: eocðtÞ :¼ aj. As t is increasing, there is a
unique leaf with label n. If that leaf is incident to a node labelled k, define the parent of the
maximum leaf in t to be: pomðtÞ :¼ k. By convention, eocðtÞ ¼ 1 and pomðtÞ ¼ 0 for the
unique t [ T1.
The minimal chain of tree t1 (respectively, t2) displayed in Figure 1.2 is 1 ! 2 ! 3
(respectively, 1 ! 2 ! 3 ! 7). Then eocðt1Þ ¼ 3, eocðt2Þ ¼ 7. Also, pomðt1Þ ¼ 4 and
pomðt2Þ ¼ 4.
Theorem 5.1. Let s ¼ sð1Þsð2Þ· · ·sðnÞ be the projection of the n-labelled, binary,
increasing, topological tree t. Then pomðtÞ ¼ grnðsÞ. In other words, the parent of the
maximum leaf in t is the greater neighbour of n in s.
Proof. Let sðiÞ ¼ n with 2 # i # n 2 1. The parent of the node labelled n in t is either the
node labelled sði 2 1Þ or the node labelled sði þ 1Þ. Let sðjÞ be the label of the node of the
common ancester of the previous two nodes in t. Then, sðjÞ # minfsði 2 1Þ;sði þ 1Þ},
j – i and i 2 1 # j # i þ 1. Hence, either sðjÞ ¼ sði 2 1Þ , sði þ 1Þ, or sðjÞ ¼
sði þ 1Þ , sði 2 1Þ. In the first (respectively, the second) case the parent of n is sði þ
1Þ (respectively, sði 2 1Þ) and grnðsÞ ¼ maxfsði 2 1Þ;sði þ 1Þ}. A
In [16] an explicit bijection of Tn onto itself is constructed that maps the statistic
‘eoc 2 1’ onto the statistic ‘pom.’ For each of the polynomialsP
k gnðkÞxk,P
k hnðkÞxk,
we then have three combinatorial interpretations: one on alternating permutations by the
statistic ‘grn,’ and two on labelled, binary, increasing, topological trees by ‘pom’ and
‘eoc’.
Recently, there has been a revival of studies on arithmetical and combinatorial
properties of both tangent and secant numbers. Ordering the alternating permutations
according to their leftmost elements has led to the Entringer recurrence, having interesting
properties [6,12,13,15,23,25]. The geometry of those permutations has been fully
exploited [21,31], in particular by looking at their quadrant marked mesh patterns in [20],
or for defining and studying natural q-analogues of the tangent and secant numbers
[3,4,7,10,30, p. 148–149]. Further q-analogues were also introduced, based no longer on
alternating permutations, but on the so-called doubloon model (see [8,9,11]). The classical
continued fraction expansions of secant and tangent have made possible the discovery of
other q-analogues (see [14,17,19,26–28]).
Acknowledgement
The authors thank the referee for his careful reading and his knowledgeable remarks.
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Note
1. Email: [email protected]
References
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[15] Y. Gelineau, H. Shin, and J. Zeng, Bijections for entringer families, Eur. J. Combin. 32 (2011),pp. 100–115.
[16] G.-N. Han, The Poupard Statistics on Tangent and Secant Trees, Strasbourg, preprint (2013),12 p.
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![Alternating Permutations and Symmetric Functionsrstan/pubs/pubfiles/143.pdfalternating” permutations were also considered by Ouchterlony [14] in the setting of pattern avoidance](https://img.pdfslide.net/doc/110x75/60f7ea0a7fe766688963ded5/alternating-permutations-and-symmetric-functions-rstanpubspubfiles143pdf-alternatinga.jpg)
