q -Enumeration of alternating permutations of odd length

This article was downloaded by: [University of Waterloo]On: 31 October 2014, At: 08:49Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registered office: MortimerHouse, 37-41 Mortimer Street, London W1T 3JH, UK

Journal of Discrete Mathematical Sciences andCryptographyPublication details, including instructions for authors and subscription information:http://www.tandfonline.com/loi/tdmc20

q-Enumeration of alternating permutations of oddlengthMatodzi S. Mulaudzi a & O. Akinyemi ba Department of Mathematics and Applied Mathematics , University of Venda , P. BagX5050, Thohoyandou , 0950 , South Africab Department of Statistics , University of Venda , P. Bag X5050, Thohoyandou , 0950 ,South AfricaPublished online: 03 Jun 2013.

To cite this article: Matodzi S. Mulaudzi & O. Akinyemi (2010) q-Enumeration of alternating permutations of odd length,Journal of Discrete Mathematical Sciences and Cryptography, 13:1, 45-67, DOI: 10.1080/09720529.2010.10698276

To link to this article: http://dx.doi.org/10.1080/09720529.2010.10698276

PLEASE SCROLL DOWN FOR ARTICLE

Taylor & Francis makes every effort to ensure the accuracy of all the information (the “Content”) containedin the publications on our platform. However, Taylor & Francis, our agents, and our licensors make norepresentations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose ofthe Content. Any opinions and views expressed in this publication are the opinions and views of the authors,and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be reliedupon and should be independently verified with primary sources of information. Taylor and Francis shallnot be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and otherliabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to orarising out of the use of the Content.

This article may be used for research, teaching, and private study purposes. Any substantial or systematicreproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in anyform to anyone is expressly forbidden. Terms & Conditions of access and use can be found at http://www.tandfonline.com/page/terms-and-conditions

http://www.tandfonline.com/loi/tdmc20

http://www.tandfonline.com/action/showCitFormats?doi=10.1080/09720529.2010.10698276

http://dx.doi.org/10.1080/09720529.2010.10698276

http://www.tandfonline.com/page/terms-and-conditions

http://www.tandfonline.com/page/terms-and-conditions

q -Enumeration of alternating permutations of odd length

Matodzi S. Mulaudzi 1 ∗2

O. Akinyemi 2 †

1 Department of Mathematics and Applied Mathematics4

2 Department of StatisticsUniversity of Venda6

P. Bag X5050, Thohoyandou, 0950South Africa8

Abstract

A permutation π = π1π2 . . . πn is called alternating (up-down ) permutation if10

π < π > π < π > π . . . . For odd n , their number is given by n![zn] tan z , and for even n byn![zn] sec z . It was shown in [5] that the number of alternating permutations of odd length12

with i inversions is given by[qi xn

nq !

]secq z and the number of alternating permutations

of even length with i inversions is given by[qi xn

nq !

]tanq z . In [7], several q -analogues of14

both secant and tangent numbers are given. In [4], Cristea and Prodinger consider the q -enumeration of up-down words by number of rises. In this project we give complete proofs16

of the q -analogues of the tangent numbers using the method explained in [7].

Keywords and phrases : Permutations, tangent numbers, secant numbers, up-down words.18

Introduction

A permutation σ1σ2σ3 . . .σn on {1, 2, 3, . . . , n} is said to be alternat-20

ing (or up-down) if σ1 < σ2 > σ3 < . . . or σ1 > σ2 < σ3 > . . . . The emptystring ε is an alternating permutation of length zero.22

The determination of the number of alternating permutations of thefirst n integers {1, 2, . . . , n} is known as Andre’s problem. The numbers24

An of alternating permutations on the integers from 1 to n for n =1, 2, . . . are 1, 2, 3, 4, 10, 32, 122, 544, . . . (A001250 in [11]). For example,26

∗E-mail: [email protected]†E-mail: [email protected]

——————————–Journal of Discrete Mathematical Sciences & CryptographyVol. 13 (2010), No. 1, pp. 45–67c© Taru Publications

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4

46 M. S. MULAUDZI AND O. AKINYEMI

the permutations on integers for small n are summarized in the followingtable.2

n An alternating permutations

1 1 {1}1 1 {1, 2}, {2, 1}3 4 {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}4 10 {1, 3, 2, 4}, {1, 4, 2, 3}, {2, 1, 4, 3}, {2, 3, 1, 4}, {2, 4, 1, 3},

{3, 1, 4, 2}, {3, 2, 4, 1}, {3, 4, 1, 2}, {4, 1, 3, 2}, {4, 2, 3, 1}

The number of alternating permutation An on {1, . . . n} function (see [3])4

∞∑

n=0An

xn

n!= sec(x) + tan(x).

Consequently the number A2n with even indices are called Secant num-6

bers and those with odd indices A2n+1 are called Tangent numbers. Thatis, there are8

[xn

n!

]{∑

n≥0(−1)n x2n

(2n)!

}−1

=[

xn

n!

]sec z

alternating permutations on {1, 2, . . . n} of even length, and10

[xn

n!

]{∑

n≥0(−1)n x2n+1

(2n + 1)!

}{∑

n≥0(−1)n x2n

(2n)!

}−1

=[

xn

n!

]tan x

alternating permutations on {1, 2, . . . n} on odd length (see [5]). In [5]12

there are[

qi xn

nq!

]{∑

n≥0(−1)n x2n

(2n)q!

}−1

(1)14

alternating permutations on {1, 2, . . . , n} of even length with i inversions,and16

[qi xn

nq!

]{∑

n≥0(−1)n x2n+1

(2n + 1)q!

}{∑

n≥0(−1)n x2n

(2n)q!

}−1

(2)

alternating permutations on {1, 2, 3, . . . n} of odd length with i inver-18

sions. These are also referred to as the q -analogues of the secant numbersand tangent numbers, respectively [5]. The generating functions (1) and20

(2) are denoted by secq x and tanq x , respectively.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4

q -ENUMERATION OF ALTERNATING PERMUTATIONS 47

Prodinger [7] gives a complete list of q -analogues of both secant andtangent numbers. This paper gives complete proofs of all the q -analogues2

of the tangent numbers that are listed in [7] namely,

f≥≤(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!qn(n+1)

/∑

n≥0

(−1)nz2n

[2n]q!qn(n−1), (3)4

f≥<(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!

/∑

n≥0

(−1)nz2n

[2n]q!, (4)

f ><(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!qn2

/∑

n≥0

(−1)nz2n

[2n]q!, qn2

, (5)6

f≤≥(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!qn2

/∑

n≥0

(−1)nz2n

[2n]q!qn(n−1), (6)

f≤>(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!

/∑

n≥0

(−1)nz2n

[2n]q!, (7)8

and

f <>(z) = ∑n≥0

= ∑n≥0

(−1)nz2n+1

[2n + 1]q!qn(n+1)

/∑

n≥0

(−1)nz2n

[2n]q!qn2

. (8)10

Definitions and notations

Definition 1. Let σ = σ1σ2σ3 . . .σn be a permutation on {1, 2, 3, . . . , n} .12

An “up” is an occurrence of σi < σi+1 . A “down” is an occurrence of atσi > σi+1 . An “up-down” (peak) is an occurrence of pi−1 < pi > pi+1 .14

A “down-up” (valley) is an occurrence of pi−1 > pi < pi+1 .

Let ω1ω2ω3 . . . ωn be an arbitrary word of length n over an infinite16

alphabet {1, 2, 3, . . .} . Since in a word an “up” is an occurrence of eitherωi < ωi+1 or ωi ≤ ωi+1 and a “down” is an occurrence of either18

ωi ≥ ωi+1 or ωi ≥ ωi+1 , then this gives four possibilities for “up-down” words or “down-up” words. However reading from right to left,20

the instance “≤>≤> . . . ” coincides with the instance “<≥<≥ . . .”, andsimilarly for “≥<≥< . . .” and “>≤>≤ . . .”, which gives us 6 q -tangent22

numbers.

Theorem 2 (Binomial Theorem).24

(1 + x)n =n

∑j=0

(nj

)x j.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


The coefficient (nj) is called the binomial coefficient and is defined by

(rk

)=

rk

k!, integer k ≥ 0

0, integer k < 0real r,2

where

rk = r(r− 1)(r− 2) . . . (r− k + 1),4

real r , integer k ≥ 0 is the k th falling factorial power of r . We cangeneralize the binomial theorem by considering analogous series in which6

n! is replaced by

(1 + q)(1 + q + q2)(1 + q + q2 + q3) . . . (1 + q + q + . . . + qn−1),8

which equals n! when q = 1. The finite geometric series is

1 + q + q2 + . . . + qn−1 =1− qn

1− q.10

We set:

[n]q = 1 + q + q2 + . . . + qn−1 =1− qn

1− q,12

and

[n]q! = [1]q[2]q[3]q . . . [n]q14

=(1− q)(1− q2)(1− q3) . . . (1− qn)

(1− q)n .

We also denote by16

(x; q)n = (1− x)(1− xq)(1− xq2)(1− xqn−1),

so that18

[n]q =(q; q)n

(1− q)n .

Without confusion, we shall write (x)n instead of (x; q)n (see [1]).20

Methodology

The classical book [5] offers a general frame-work to deal with words22

and patterns (“pattern algebra”). However, we decide to use a different ap-proach that works particularly well in the present context. The technique24

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4

q-ENUMERATION OF ALTERNATING PERMUTATIONS 49

that we employ has been used successfully by Prodinger [7, 9], Prodingerand Tshifhumulo [8], and more recently by Cristea and Prodinger [4].2

We consider alternating random words (instead of permutations) overthe infinite alphabet {1, 2, 3 . . .} , with the assumption that each letter j4

occurs with (geometric) probability pq j−1 , independently, for 0 < q < 1and p = 1 − q . In the limit case when q tends to 1 all letters to occur6

with the same probability, and thus approach the model for randompermutations. This technique is called “adding-a-new-slice”. In doing8

so, recursion is obtained and functional equations are derived thereof.Iterating the functional equations, we obtain probability generating functions10

(PGF) of interest, with parameter q , as in [4, 7, 8, 9].

Probability generating functions12

In this section we derive the q -analogues of the exponential gener-ating functions for the tangent numbers. The following instances will be14

considered, ≥≤ , ≥< , >< , ≤≥ , ≤> and <> .

The instance ≥≤16

We enumerate words with pattern ≥≤ . . . ≥≤ , which leads toq -tangent functions. Adding a new slice means adding a pair (k, i) with18

1 ≤ k ≤ i , j ≥ k , replacing ui by 1 and providing a factor u j .

For the pattern ≥≤ , this is done via20

i

∑k=1

pqk−1∞∑j=k

pq j−1u j =p2u

(1− qu)(1− qu2)

− p2u(1− qu)(1− q2u)

(q2u)i .22

So, there are two substitutions, u → 1 and u → q2u .

Hence24

T≥≤2n+1(u) =(pz)2u

(1− qu)(1− qu2)T≥≤2n−1(1)

− (pz)2u(1− qu)(1− q2u)

T≥≤2n−1(q2u). (9)26

The starting value is just

∑j≥1

pq j−1u j =pu

1− qu,28

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


which shows that

T1(u) =(pz)u1− qu

.2

Let us define f (z, u) as follows:

f (z, u) = ∑n≥0

T≥≤2n+1(u)z2n+1.4

Multiplying both sides of (9) by z2n+1 and summing over n ≥ 1, weobtain6

∑n≥1

T≥≤2n+1(u)z2n+1 = ∑n≥1

(pz2)u(qu)2

T≥≤2n−1(1)

− ∑n≥1

(pz)2u(qu)2

T≥≤2n−1(z, q2u).8

This reduces to

f (z, u)− T≥≤1 (u)z =(pz2)u

(1− qu)(1− qu2)f (z, 1)10

− (pz)2u(1− qu)(1− q2u)

f (z, q2u),

so that12

f (z, u) =(pz)u

(1− qu)+

(pz2)u(1− qu)(1− qu2)

f (z, 1)

− (pz)2u(1− qu)(1− q2u)

f (z, q2u). (10)14

Iterating (10) we obtain

f (z, u) =(pz)u(qu)1

+(pz)2u(qu)2

f (z, 1)− (pz)3q2u2

(qu)316

− (pz)4q2u2

(qu)4f (z, 1) + . . . +

(−1)n−1(pz)2n−1qn(n−1)un

(qu)2n−1

+(−1)n−1(pz)2nqn(n−1)un

(qu)2nf (z, 1) + . . .18

which may be written in the form

f (z, u) = ∑n≥0

(−1)n(pz)2n+1qn(n+1)un+1

(qu)2n+120

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


+ f (z, 1) ∑n≥1

(−1)n−1(pz)2nqn(n−1)un

(qu)2n.

Setting u = 1, we obtain2

f (z, 1) = ∑n≥0

(−1)n(pz)2n+1qn(n+1)

(q)2n+1

+ f (z, 1) ∑n≥1

(−1)n−1(pz)2nqn(n−1)

(q)2n,4

which reduces to

F(z, 1)[

1 + ∑n≤1

(−1)n(pz)2nqn(n−1)

(q)2n

]6

= ∑n≥0

(−1)n(pz)2n+1qn(n+1)

(q)2n+1.

This shows that8

f (z, 1) ∑n≥0

(−1)n(pz)2nqn(n−1)

(q)2n= ∑

n≥0

(−1)n(pz)2n+1qn(n+1)

(q)2n+1.

Hence10

f≤≥(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!qn(n+1)

/∑

n≥0

(−1)nz2n

[2n]q!qn(n−1).

The instance ≥<12

We enumerate words with pattern ≥< . . . ≥< , which leads toq -tangent functions. Adding a new slice means adding a pair (k, i) with14

1 < k ≤ i , j > k , replacing ui by 1 and providing a factor u j .For the pattern ≥< , we have16

i

∑k=1

pqk−1∞∑

j=k+1pq j−1u j

= p2qu2

(1−qu)(1−qu2) −p2qu

(1−qu)(1−q2u) (q2u)i ,18

which shows that there are two substitutions, u → 1 and u → q2uinvolved.20

Hence

T≥<2n+1(u) =

(pz)2qu2

(1− qu)(1− qu2)T≥<

2n−1(1)22

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4

52 M. S. MULAUDZI AND O. AKINYEMI2

− (pz)2qu(1− qu)(1− q2u)

T≥<2n−1(z, q2u). (11)

The starting value is just4

∑j≥1

pq j−1u j =pu

1− qu,

which show that6

T1(u) =(pz)u

(1− qu).

Let us denned f (z, u) as follow:8

f (z, u) = ∑n≥0

T≥<2n+1(u)z2n+1.

Multiplying both sides of (11) by z2n+1 and summing over n ≥ 1 we10

obtain

∑n≥1

T≥<2n+1(u)z2n+1 = ∑

n≥1

p2qu2

(qu)2T≥<

2n−1(1)− ∑n≥1

p2qu2

(qu)2T≥<

2n−1(q2u).12

This reduce to

f (z, u)− T1(u)z =(pz)2qu2

(1− qu)(1− q2u)f (z, 1)14

− (pz)2qu2

(1− qu)(1− q2u)f (z, q2u),

so that16

f (z, u) =(pz)u

(1− qu)+

(pz)2qu2

(1− qu)(1− q2u)f (z, 1)

− (pz)2qu2

(1− qu)(1− q2u)f (z, q2u). (12)18


f (z, u) =(pz)u(qu)

+(pz)2u(qu)2

f (z, 1)− (pz)3q3u3

(qu)3− (pz)4q6u3

(qu)4f (z, 1) + . . .20

+(−1)n−1(pz)2n−1q2n2−3n+1u2n+1

(qu)2n−1

+(−1)n−1(pz)2nqn(n−1)u2n

(qu)2nf (z, 1) . . .22

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4



f (z, u) = ∑n≥0

(−1)n(pz)2n+1qn(2n+1)u2n+1

(qu)2n+12

+ f (z, 1) ∑n≥1

(−1)n−1(pz)2nqn(n−1)u2n

(qu)2n.


f (z, 1) = ∑n≥0

(−1)n(pz)2n+1qn(2n+1)

(q)2n+1

+ f (z, 1) ∑n≥1

(−1)n−1(pz)2nqn(n−1)

(q)2n,6

which reduce to,

f (z, 1)[

1 + ∑n≥0

(−1)n−1(pz)2nqn(n−1)

(q)2n

]8

= ∑n≥0

(−1)n(pz)2n+1qn(2n+1)

(q)2n+1.

Therefore10

f (z, 1) ∑n≥0

(−1)n−1(pz)2nqn(n−1)

(q)2n= ∑

n≥0

(−1)n(pz)2n+1qn(2n+1)

(q)2n+1.

Hence12

f≥<(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!qn(2n+1)

/∑

n≥0

(−1)nz2n

[2n]q!qn(2n−1).

The instance ><14

We enumerate words with pattern >< . . . >< , which leads toq -tangent functions. Adding a new slice means adding a pair (k, i) with16

1 < k < i , j > k , replacing ui by 1 and providing a factor u j .For the pattern >< , we have18

i−1

∑k=1

pqk−1∞∑

j=k+1pq j−1u j =

p2qu2

(1− qu)(1− qu2)

− p2qu(1− qu)(1− q2u)

(q2u)i ,20

so, there are two substitution, u → 1 and u → q2u involved.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


Hence

T><2n+1(u) =

(pz)2qu2

(1− qu)(1− qu2)T><

2n−1(1)2

− (pz)2qu(1− qu)(1− q2u)

T><2n−1(z, q2u). (13)

The starting value is just4

∑j≥1

pq j−1u j =pu

1− qu,

which shows that6

T1(u) =(pz)u

(1− qu).

Let us define f (z, u) as follow:8

f (z, u) = ∑n≥0

T><2n+1(u)z2n+1.

Multiplying both sides of (13) z2n+1 by summing over n ≥ 1 we obtain10

∑n≥1

T><2n+1(u)z2n+1 = ∑

n≥1

p2qu2

(qu)2T><

2n−1(1)− ∑n≥1

p2uq(qu)2

T><2n−1(q2u).

This reduces to12

f (z, u)− T><1 (u)z =

(pz)2qu2

(1− qu)(1− q2u)

− (pz)2u2

q(1− qu)(1− q2u)f (q2u)14

so that

f (z, u) =(pz)u

(1− qu)=

(pz)2qu2

(1− qu)(1− q2u)16

− (pz)2u2

q(1− qu)(1− q2u)f (q2u) . (14)

Iterating (14) we obtain18

f (z, u) =(pz)u(qu)

+(pz)2qu2

(qu)2f (z, 1)

− (pz)3qu2

(qu)3− (pz)4q4u3

(qu)4f (z, 1) + . . .20

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


+(−1)n−1(pz)2n−1qn2−2n+1un

(qu)2n−1

+(−1)n−1(pz)2nqn2

un

(qu)2n f (z, 1) . . .4

which can be written in the form

f (z, u) = ∑n≥0

(−1)n−1(pz)2n+1qn2un+1

(qu)2n+16

+ f (z, 1) ∑n≥1

(−1)n−1(pz)2nqn2un

(qu)2n.


f (z, 1) = ∑n≥0

(−1)n−1(pz)2n+1qn2

(q)2n+1+ f (z, 1) ∑

n≥1

(−1)n−1(pz)2nqn2

(q)2n,

which reduces to10

f (z, 1)[

1 + ∑n≥0

(−1)n(pz)2nqn2

(q)2n

]= ∑

n≥0

(−1)n−1(pz)2n+1qn2

(q)2n+1.

Hence12

f ><(z) = ∑n≥0

(−1)n−1z2n+1

[2n + 1]q!qn2

/∑

n≥0

(−1)nz2n

[2n]q!qn2

.

The instance ≤≥14

We enumerate words with pattern ≤≥ . . . ≤≥ , which leads toq -tangent functions. Adding a new slice means adding a pair (k, i) with16

k = 1, . . . ∞ , j = 1, . . . k and i ≤ k , k ≥ j replacing ui by 1 and providinga factor u j .18

For the pattern ≤≥ , we have

∞∑k=i

pqk−1k

∑j=1

pq j−1u j =pu

q(1− qu)qi − p2u

q(1− qu)(1− q2u)(q2u)i ,20

which shows that there are two substitutions, u → q and u → q2u .Hence22

T≤≥2n+1(u) =pu

q(1− qu)T≤≥2n−1(q)

− p2uq(1− qu)(1− q2u)

T≤≥2n−1(q2u). (15)24

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4



∑k≥i

pqk−1u j = qi−1,2

which shows that

T≤≥1 (u) =pu

1− qu.4


f (z, u) = ∑n≥0

T≤≥2n+1(u)z2n+16

Multiplying both sides of (15) by z2n+1 and summing over n ≥ 1 weobtain8

∑n≥1

T≤≥2n+1(u)z2n+1 = ∑n≥1

puq(qu)

T≤≥2n−1(q)z2n+1

− ∑n≥1

p2

q(qu)2T≤≥2n−1(q2u)z2n−1.10

This reduces to

f (z, u)− T1(u)z =puz2

q(1− qu)f (z, q)12

− (pz)2uq(1− qu)(1− q2u)

f (z, q2u),

so that14

f (z, u) =puz

(1− qu)+

puz2

q(1− qu)f (z, q)

− (pz)2

q(1− qu)(q− q2u)f (z, q2u). (16)16

Iterating (16), we obtain

f (z, u) =puz(qu)

+puz2

q(qu)f (z, q)− (pz)2

q(qu)2f (z, q2u)18

=puz(qu)

+puz2

q(qu)f (z, q)− (pz)3q2u2

q(qu)3

− p3q2uz4

q2(qu)3f (z, q) +

(pz)5q6u3

q2(qu)5+ . . .20

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


+(−1)n−1(pz)2n−1u2n−1qn(n−1)

qn−1(qu)2n−1

+(−1)n−1 p2n−1z2nunqn2−n

qn(qu)2n−1f (z, q) + . . .4


f (z, u) = ∑n≥0

(−1)n(pz)2n+1u2n+1qn(n+1)

qn(qu)2n+16

+ f (z, q) ∑n≥1

(−1)n−1 p2n−1z2nunqn(n−1)

qn(qu)2n−1.

which reduce to8

f (z, u)[

1 + ∑n≥1

(−1)n p2n−1z2nunqn(n−1)

qn(qu)2n−1

]

= ∑n≥0

(−1)n(pz)2n+1u2n+1qn(n+1)

qn(qu)2n+110

and so,

f (z, u) ∑n≥0

(−1)n p2n−1z2nunqn(n−1)

qn(qu)2n−112

= ∑n≥0

(−1)n(pz)2n+1u2n+1qn(n+1)

qn(qu)2n+1.

Plugging in u = q and solving the resulting equation f (z, q) we obtain,14

f (z, q) = ∑n≥0

(−1)n(pz)2n+1q(n+1)2

qn(q2)2n+1

/∑

n≥0

(−1)n p2nz2nqn2

qn(1− q)(q2)2n−1. (17)

Set16

M = ∑n≥0

(−1)n p2nz2nqn2

qn(q)2n,

and18

N = ∑n≥0

(−1)n(pz)2n+1qn+12

qn(q2)2n+1.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


Substituting (17) into f (z, u) we obtain

f (z, u) =[

∑n≥0

(−1)n(pz)2n+1u2n+1qn(n+1)

qn(qu)2n+12

+N × ∑n≥1

(−1)n−1 p2n−1z2nunqn2−n

qn(qu)2n−1

]/M,

f (z, u) =[

∑n≥0

(−1)n(pz)2n+1un+1qn(n+1)

qn(qu)2n+1× M4

+N × ∑n≥1

(−1)n−1 p2n−1z2nunqn(n−1)

qn(qu)2n−1

]/M,

The coefficient of (pz)2n+1 in the first sum on the right side of the equal6

sign is given by

u

∑k=0

(−1)kqk(k+1)uk+1

qk(qu)2k+1

(−1)n−kq(n−k)2

qn−k(q)2n−2k8

= (−1)nqn2−n ∑n≥0

q2k2−2nk+kuk+1

(qu)2k+1(q)2n−2k

= (−1)nqn2−nn

∑k=0

q2k2−2nk+kuk+1

(qu)2k+1(q)2n−2k. (18)10

The coefficient of (pz)2n+1 in the second sum on the right side of the equalsign is given by12

n

∑k=1

(−1)n−k q(n−k)2+2(n−k)+1

(1− q)qn−k(q2)2(n−k)+1× (−1)k−1 ukqk2−k

qk(qu)2k−1

= (−1)n+1qn2n

∑k=1

q2k2−2nk−3k+1uk

(q)2n−2k+2(qu)2k−114

= (−1)n+1qn2+nn−1

∑k=0

q2(k2+2k+1)−3(k+1)−2n(k+1)+1uk+1

(q)2n−2(k+1)+2(qu)2(k+1)−1

= (−1)n+1qn2+nn−1

∑k=0

q2k2+k−2nkuk+1

(q)2n−2k(qu)2k+116

= (−1)n+1qn2−nn

∑k=0

q2k2+k−2nkuk+1

(q)2n−2k(qu)2k+1+ (−1)n qn2

un+1

(qu)2n+1. (19)

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


The sum of both equations (18) and (19) is

(−1)nqn2un+1

(qu)2n+1.2

Therefore

f (z, u) = ∑n≥0

(−1)n(pz)2n+1qn2un+1

(qu)2n+1

/∑

n≥0

(−1)n(pz)2nqn2

qn(qu)2n.4

Setting u = 1, we obtain

f (z) = ∑n≥0

(−1)n(pz)2n+1qn2

(q)2n+1

/∑

n≥0

(−1)n(pz)2nqn2

qn(q)2n.6

Hence

f≤≥(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!qn2

/∑

n≥0

(−1)nz2n

[2n]q!qn(n−1).8

The instance ≤>

We enumerate words with pattern ≤> . . . ≤> , which leads to10

q -tangent functions. Adding a new slice means adding a pair (k, i) withi ≤ k , k > j , replacing ui by 1 and providing a factor u j .12

For the pattern ≤> , we have

∞∑k=i

pqk−1k−1

∑j=1

pq j−1u j =pu

q(1− qu)qi − p2

q2(1− qu)(1− q2u)(q2u)i ,14

which shows that there are two substitutions, u → q and u → q2u .

Hence16

T≤>2n+1(u) =

puq(1− qu)

T≤>2n−1(q)

− p2

q(1− qu)(1− q2u)T≤≥2n−1(q2u). (20)18


∑k≥i

pqk−1u j =pu

(1− qu),20

which shows that

T≤>1 (u) =

pu(1− qu)

.22

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4



f (z, u) = ∑n≥0

T≤>2n+1(u)z2n+1.2


∑n≥1

T≤>2n+1(u)z2n+1 = ∑

n≥1

puq(qu)

T≤>2n−1(q)z2n+1

− ∑n≥1

p2

q2(qu)2T≤>

2n−1(q2u)z2n+1.6

This reduces to

f (z, u)− T1(u)z =puz2

q(1− qu)f (z, q)8

− (pz)2

q2(1− qu)(1− q2u)f (z, q2u),

so that10

f (z, u) =(pz)u

(1− qu)puz2

q(1− qu)f (z, q)

− (pz)2

q2(1− qu)(1− q2u)f (z, q2u). (21)12


f (z, u) =puz(qu)

+puz2

q(qu)f (z, q)− (pz)3u

(qu)314

− p3uz4

q(qu)3− p3uz4

q(qu)3f (z, q) +

(pz)5u(qu)5

+ . . .

+(−1)n−1(pz)2n−1u

(qu)2n−1+

(−1)n−1 p2n−1z2nuq(qu)2n−1

f (z, q) . . .16


f (z, u) = ∑n≥0

(−1)n(pz)2n+1u(qu)2n+1

+ f (z, q) ∑n≥1

(−1)n−1 p2n−1z2n

q(qu)2n−1.18

Plugging in u = q and solving the resulting equation f (z, q) we obtain

f (z, q) = ∑n≥0

(−1)n(pz)2n+1q(q2)2n+1

+ f (z, q) ∑n≥1

(−1)n−1 p2n−1z2n

(q2)2n−1,20

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


which reduce to

f (z, q)[

1 + ∑n≥0

(−1)n p2nz2n

(q2)2n

]= ∑

n≥0

(−1)n(pz)2n+1q(q2)2n+1

,2

and so,

f (z, q) = ∑n≥0

(−1)n(pz)2n+1q(q2)2n+1

/∑

n≥0

(−1)n p2nz2n

(q2)2n. (22)4

Set

α = ∑n≥0

(−1)n (pz)2n+1q(q2)2n+1

,6

and

β = ∑n≥0

(−1)n p2nz2n

(q2)2n.8


f (z, u) = ∑n≥0

(−1)n(pz)2n+1q(qu)2n+1

+α × ∑n≥1

(−1)n−1 p2n−1z2nuq(qu)2n−1

/β10

=(

∑n≥0

(−1)n(pz)2n+1q(qu)2n+1

×β

+α × ∑n≥1

(−1)n−1 p2n−1z2nuq(qu)2n−1

)/∑

n≥0

(−1)n p2nz2n

(qu)2n.12

The coefficient of (pz)2n+1 in the first sum on the right side of the equalsign is given by14

n

∑k=0

(−1)k u(qu)2k+1

× (−1)n−k 1(qu)2n−2k

= (−1)nn

∑k=0

u(qu)2k+1(q)2n−2k

. (23)16

The coefficient of (pz)2n+1 in the second sum on the right side of the equalsign is given by18

n

∑k=1

(−1)k−1 uq(1− q)(qu)2k−1

(−1)n−k q1(1− q)(qu)2n−2k+1

= (−1)n+1n

∑k=1

u(q)2n−2k+2(qu)2k−1

20

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


= (−1)n+1n−1

∑k=0

u(q)2n−2k(qu)2k+1

= (−1)n+1n+1

∑k=0

u(1)2n−2k(qu)2k+1

+(−1)nu(qu)2n+1

. (24)4

The sum of both equation (23) and (24) is

(−1)nu(qu)2n+1

.6

Therefore

f (z, u) = ∑n≥0

(−1)n(pz)2n+1u(qu)2n+1

/∑≥0

(−1)n(pz)2n

(q)2n.8


f (z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!

/∑

n≥0

(−1)nq2n

[2n]q!.10

Hence

f≤>(z) = ∑n≥0

(−1)nz2n+1

[2n + 1]q!

/∑

n≥0

(−1)nz2n

[2n]q!.12

The instance <>

We enumerate words with pattern <> . . . <> , which leads to14

q -tangent functions. Adding a new slice means adding a pair (k, i) withi < k , k > j , replacing ui by 1 and providing a factor u j .16

For the pattern <> , we have

∞∑

k=i+1pqk−1

k−1

∑j=1

pq j−1u j =pu

(1− qu)qi − p2u

q(1− qu)(1− q2u)(q2u)i ,18

which shows that there are two substitutions, u → q and u → q2u .Hence20

T<>2n+1(u) =

pu(1− qu)

T<>2n−1(q)

− p2uq(1− qu)(1− q2u)

T<>2n−1(q2u). (25)22


∑k≥i

pqk−1u j =pu

(1− qu),24

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


which shows that

T<>1 (u) =

pu1− qu

.2


f (z, u) = ∑n≥0

T<>2n+1(u)z2n+1.4


∑n≥1

T<>2n+1(u)z2n+1 = ∑

n≥1

pu(qu)

T<>2n−1(q)z2n+1

− ∑n≥1

pz(qu)2

T<>2n−1(q2u)z2n+1.8

This reduce to

f (z, u)− T(u)z =puz2

(1− qu)f (z, q)− (pz)2u

(1− qu)(1− q2u)f (z, q2u),10

so that

f (z, u) =(pz)u

(1− qu)+

puz2

(1− qu)f (z, q)12

− (pz)2

(1− qu)(1− q2u)f (z, q2u). (26)

Iterating (26) we obtain14

f (z, u) =(pz)u(qu)

+puz2

(qu)f (z, q)− (pz)2q2u2

(qu)3

− p3q2u2z4

(qu)3f (z, q) +

(pz)5q6u3

(qu)5+ . . .16

+(−1)n−1(pz)2n−1qn(n−1)un

(qu)2n−1

+(−1)n−1(pz)2nqn(n−1)un

(1− q)(qu)2n−1f (z, q) . . .18


f (z, u) = ∑n≥0

(−1)n(pz)2n+1qn(n+1)un+1

(qu)2n−120

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


+ f (z, q) ∑n≥1


(1− q)(qu)2n−1.

Plugging in u = q and solving the resulting equation f (z, q) we obtain4

f (z, q) = ∑n≥0

(−1)n(pz)2n+1qn(n+1)qn+1

(q2)2n−1

+ f (z, q) ∑n≥1

(−1)n−1(pz)2nqn(n−1)qn

(1− q)(q2)2n−1,6

which reduce

f (z, q)[

1 + ∑n≥1

(−1)n(pz)2nqn2

(q)2n

]8

= ∑n≥0

(−1)n(pz)2n+1qn(n+1)qn+1

(q2)2n+1.

which reduce10

f (z, q)[

1 + ∑n≥1

(−1)n(pz)2nqn2

(q)2n

]

= ∑n≥0

(−1)n(pz)2n+1qn(n+1)qn+1

(q2)2n+1.12

and so,

f (z, q) = ∑n≥0

(−1)n(pz)2n+1qn(n+1)qn+1

(q2)2n+1

/∑

n≥0

(−1)n(pz)2nqn2

(q)2n. (27)14

Set

Q = ∑n≥0

(−1)n(pz)2n+1qn(n+1)qn+1

(q2)2n+1,16

and

R = ∑n≥0

(−1)n(pz)2nqn2

(q)2n.18


f (z, u) = ∑n≥0

(−1)n(pz)2n+1qn(n+1)un+1

(qu)2n+120

+[

Q× ∑n≥1


(1− q)(qu)2n−1

]/R,

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


and then

f (z, u) =(

∑n≥0

(−1)n(pz)2n+1qn(n+1)un+1

(qu)2n+1×R2

+ Q× ∑n≥1


(1− q)(qu)2n−1

)/R.

The coefficient of (pz)2n+1 in the first sum on the right side of the equation4

sign is given by

n

∑k=0

(−1)k qk(k+1)uk+1

(qu)2k+1× (−1)n−k q(n−k)2

(q)2(n−k)6

= (−1)nn

∑k=0

qk2−2nk+n2+k2+kuk+1

(qu)2k+1(q)2n−2k

= (−1)nqn2n

∑k=0

q2k2−2nk+kuk+1

(qu)2k+1(q)2n−2k. (28)8

The coefficient of (pz)2n+1 in the second sum on the right side of theequation sign is given by10

∑k=1

(−1)n−k q(n−k)2+ 2(n− k) + 1

(1− q)(q2)2n−2k+1(−1)k−1 qn2−kuk

(qu)2k−1

= (−1)n+1n

∑k=1

q2k2−3k−2nk+n2+2n+1uk

(q)2n−2k+2(qu)2k−112

= (−1)n+1qn2+2nn

∑k=1

q2k2−3k−2nk+1uk

(q)2n−2k+2(qu)2k−1

= (−1)n+1qn2+2nn−1

∑k=0

q2(k2+2k+1)−3(k+1)−2n(k+1)+1uk+1

(q)2n−2(k+1)+2(qu)2k+114

= (−1)n+1qn2+2nn−1

∑k=0

q2k2+4k+2−3k−3−2nk−2n+1uk+1

(q)2n−2k(qu)2k+1

(−1)n+1qn2n

∑k=0

q2k2+k−2nkuk+1

(qu)2n+1(q)2n−2k+ (−1)nqn2 q2n2+n−2n2

un+1

(qu)2n+116

= (−1)n+1qn2n

∑k=0

q2k2k−2nkuk+1

(qu)2k+1(q)2n−2k+ (−1)n qn2+nun+1

(qu)2n+1. (29)

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


The sum of both equations (28) and (29) is

(−1)nqn2+nun+1

(qu)2n+1.2

Therefore

f (z, u) = ∑n≥0

(−1)n(pz)2n+1un+1qn2+n

(qu)2n+1

/∑

n≥0

(−1)n(pz)2nqn2

(q)2n.4


f (z) = ∑n≥0

(−1)n(pz)2n+1

(q)2n+1!qn(n+1)

/∑

n≥0

(−1)n(pz)2n

(q)2n!qn2

,6

which reduces to

f (z) = ∑n≥0

(−1)n(pz)2n+1

[2n + 1]q!qn(n+1)

/∑

n≥0

(−1)n(pz)2n

[2n]q!qn2

.8

Hence

f <>(z) = ∑n≥0

(−1)n(pz)2n+1

[2n + 1]q!qn(n+1)

/∑

n≥0

(−1)nz2n

[2n]q!qn2

.10

Conclusion

By using the adding -anew-slices techniques we obtain recursions,12

iterating them generating functions f (z) of interest comes out with q -probabilities. In the limit case as q tend to 1 all letters occur with the14

same probability and all the probability generating functions tend to thefunction16

f (z) = ∑n≥0

(−1)n z2n+1

(2n + 1)!

/∑

n≥0(−1)n z2n

(2n)!,

which is exactly the tangent function. This problem conforms with Cristea18

and Prodinger [4].

References20

[1] G. Andrews, The Theory of partitions, Volume 2 of Encyclopedia ofMathematics and its Applied, Addison-Wesley, 1976.22

[2] G. Andrews, R. Askey and R. Roy, Special Functions, Volume 71 of En-cyclopedia of Mathematics and its Applications, Cambridge University24

Press, 1999.

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4


[3] L. Carlitz and R. Scoville, Generating function for certain type ofpermutations, J. Comb. Theory (A), Vol. 18 (1975), pp. 262–275.2

[4] L. L. Cristea and H. Prodinger, q -Enumeration of up-down wordsby number of rises, preprint, available at http://math.sun.ac.za/4

∼ prodinger/pdmles/crisprod.pdf[5] I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John6

Wiley & Sons, 1983.[6] D. E. Knuth, The Art of Computer Programming, Vol. 3, 3rd edition,8

Addison Wesley, 1983.[7] H. Prodinger, Combinatorics of geometrically distributed random10

variables: the new q -tangent and q -tangent number, Internet. J.Math. & Math. Sci., Vol. 24 (12) (2000), pp. 825–838.12

[8] H. Prodinger and T. A. Tshifhumulo, On q -Oliver function, Annals ofCombinatorics, Vol. 6 (2002), pp. 181–194.14

[9] H. Prodinger, q -Enumeration of Salie permutations, preprint, avail-able at http://maths.sun.ac.za/∼ prodinger/postscriptfiles/16

salie paper.ps

[10] D. Rotem, Stack sortable permutations, Discrete Math., Vol. 46 (1981),18

pp. 185–196.[11] N. J. A. Sloane, Sequence A001250 in The On-Line Encyclopedia of Inte-20

ger Sequences, available at http://www.research.att.com/∼ njas/sequences/index.html22

Received July, 2009

Dow

nloa

ded

by [

Uni

vers

ity o

f W

ater

loo]

at 0

8:49

31

Oct

ober

201

4

Documents

q -Enumeration of alternating permutations of odd length