Recurrence relation with two indices and plane compositions

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Recurrence relation with two indices and plane compositionsArnold Knopfmachera; Toufik Mansourb; Augustine Munagia

a The John Knopfmacher Centre for Applicable Analysis and Number Theory, School of Mathematics,University of the Witwatersrand, Johannesburg, South Africa b Department of Mathematics, Universityof Haifa, Haifa, Israel

First published on: 11 December 2009

To cite this Article Knopfmacher, Arnold , Mansour, Toufik and Munagi, Augustine(2011) 'Recurrence relation with twoindices and plane compositions', Journal of Difference Equations and Applications, 17: 1, 115 — 127, First published on:11 December 2009 (iFirst)To link to this Article: DOI: 10.1080/10236190902919301URL: http://dx.doi.org/10.1080/10236190902919301

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Recurrence relation with two indices and plane compositions

Arnold Knopfmacherb1, Toufik Mansoura* and Augustine Munagib2

aDepartment of Mathematics, University of Haifa, 31905 Haifa, Israel; bThe John KnopfmacherCentre for Applicable Analysis and Number Theory, School of Mathematics, University of the

Witwatersrand, Johannesburg, South Africa

(Received 27 August 2008; final version received 19 March 2009)

The aim of this paper is to study analytical and combinatorial methods to solve a specialtype of recurrence relation with two indices. It is shown that the recurrence relationenumerates a natural combinatorial object called a plane composition. In addition, furtherinteresting recurrence relations arise in the study of statistics for these plane compositions.

Keywords: difference equations; plane compositions; asymptotics; generating functions

2000 Mathematics Subject Classification: Primary: 10A35, 65Q05; Secondary:05A05, 05A15, 05A16, 05A17

1. Introduction

The aim of this paper is to study analytical and combinatorial methods to solve a special

type of recurrence relation with two indices. It is well known that there is no general

procedure for solving recurrence relations, which is why it is an art, see for example

[6,8,9,12,13]. In this paper, we study the following recurrence relations:

an;m ¼Xnj¼m

j2 1

m2 1

!ðan2j;0 þ · · ·þ an2j;mÞ ð1:1Þ

with the initial conditions a0;0 ¼ 1 and an;0 ¼ 0 for n $ 1, as well as several

generalizations and variations of this recurrence. Our study focuses on presenting two

different solutions, one, an analytical solution and another, a combinatorial solution.

In fact, we show that this recurrence relation and its variations enumerate the objects in

several combinatorial families. To this end, we need the following definitions.

A partition l ¼ l1· · ·lk of a positive integer n is a finite non-increasing sequence of

positive integers l1; . . . ; lk such thatPk

i¼1li ¼ n. The li are called the parts of the

partition. For example, the partitions of four are 4, 31, 22, 211, and 1111. We denote

the number of partitions of n by pðnÞ. It is well known that the generating function for the

number of partitions of n is given by the infinite productQ

j$1ð1=ð12 x iÞÞ, see [1].

A composition s ¼ s1s2 . . .sm of n [ N is an ordered collection of one or more

positive integers whose sum is n (i.e. s is a partition of n where parts are ordered). The

number of summands, namely m, is called the number of parts of the composition. For

example, the compositions of four are 4, 31, 13, 22, 211, 121, 112, and 1111. It is well

known that the number of compositions of n is given by 2n21.

ISSN 1023-6198 print/ISSN 1563-5120 online

q 2010 Taylor & Francis

DOI: 10.1080/10236190902919301

http://www.informaworld.com

*Corresponding author. Email: [email protected]

Journal of Difference Equations and Applications

Vol. 17, No. 1, January 2011, 115–127

Downloaded By: [University of Witwatersrand] At: 09:39 11 January 2011

The paper is organized as follows. In Section 2, we solve the recurrence relation (1.1)

analytically, by using generating function techniques and present a combinatorial

interpretation of the numbers. An asymptotic estimate for the solution is also derived, and

some related recurrences are discussed. In Section 3, we study statistics related to our

combinatorial objects, which leads to refinements of the generating functions obtained in

the previous section.

2. Recurrence relation (1.1)

This section is organized in the followingway.As a first step, we solve the recurrence relation

(1.1) analytically, by using generating function techniques. Then, in the next subsection, we

present a combinatorial interpretation of this recurrence. In the last subsection, we generalize

our method to study several types of recurrence relations of a similar nature.

2.1 Generating functions technique

Lemma 2.1. For all s ¼ 0; 1; 2; . . . ;m,

Xsi¼0

ð21Þis

i

!an2i;m ¼

Xnj¼m

j2 12 s

m2 12 s

!ðan2j;0 þ · · ·þ an2j;mÞ:

Proof. We carry out the proof by induction on s. From s ¼ 0 the lemma holds by (1.1).

Assume that the lemma holds for s2 1. Then by the well-known identitya

b

!2

a2 1

b

!¼

a2 1

b2 1

!and the induction hypothesis, we find that

Xsi¼0

ð21Þis

i

!an2i;m

¼Xsi¼0

ð21Þis2 1

i

!þ

s2 1

i2 1

! !an2i;m

¼Xs21

i¼0

ð21Þis2 1

i

!an2i;m þ

Xsi¼1

ð21Þis2 1

i2 1

!an2i;m

¼Xs21

i¼0

ð21Þis2 1

i

!an2i;m 2

Xs21

i¼0

ð21Þis2 1

i

!an212i;m

¼Xnj¼m

j2 s

m2 s

!ðan2j;0 þ · · ·þ an2j;mÞ2

Xn21

j¼m

j2 s

m2 s

!ðan212j;0 þ · · ·þ an212j;mÞ

¼Xnj¼m

j2 s

m2 s

!ðan2j;0 þ · · ·þ an2j;mÞ2

Xnj¼mþ1

j2 12 s

m2 s

!ðan2j;0 þ · · ·þ an2j;mÞ

¼Xnj¼m

j2 s

m2 s

!2

j2 12 s

m2 s

! !ðan2j;0 þ · · ·þ an2j;mÞ

¼Xnj¼m

j2 12 s

m2 12 s

!ðan2j;0 þ · · ·þ an2j;mÞ;

which completes the proof. A

A. Knopfmacher et al.116


Lemma 2.1. For s ¼ m gives

Xmi¼0

ð21Þim2 1

i

!an2i;m ¼ an2m;0 þ · · ·þ an2m;m: ð2:1Þ

Define AmðxÞ ¼P

n$man;mxn. Rewriting (2.1) in terms of generating functions AmðxÞ, we

obtain

ð12 xÞmAmðxÞ ¼Xmi¼0

ð21Þi x im

i

!AmðxÞ ¼ xmðA0ðxÞ þ · · ·þ AmðxÞÞ;

which is equivalent to

AmðxÞ ¼xm

ð12 xÞm 2 xmðA0ðxÞ þ · · ·þ Am21ðxÞÞ:

Hence,

ðð12 xÞm 2 xmÞÞAmðxÞ2 xðð12 xÞm21 2 xm21ÞAm21ðxÞ ¼ xmAm21ðxÞ

which implies that

AmðxÞ ¼xð12 xÞm21

ð12 xÞm 2 xmAm21ðxÞ:

Using the initial condition A0ðxÞ ¼ 1 and induction on m we get the following result.

Proposition 2.2. The generating function AmðxÞ satisfies the recurrence relation

AmðxÞ ¼xð12 xÞm21

ð12 xÞm 2 xmAm21ðxÞ

with the initial condition A0ðxÞ ¼ 1. Moreover,

AmðxÞ ¼xmð12 xÞ

m

2

!

Qmj¼1ð12 xÞ j 2 x j

¼ðx=ð12 xÞÞmQm

j¼1ð12 ðx=ð12 xÞÞ jÞ:

Hence, summing over all m $ 0 we obtain the following result.

Theorem 2.3. The generating functionP

m$0AmðxÞ is given by

Yj$1

1

12 ðx j=ð12 xÞ jÞ:

Journal of Difference Equations and Applications 117


Proof. Proposition 2.2 gives that

Xm$0

AmðxÞ ¼Xm$0

ðx=ð12 xÞÞmQmj¼1ð12 ðx=ð12 xÞÞ jÞ

:

Using Euler’s partition identity ([1], Corollary 2.2)

Xm$0

t mQmj¼1ð12 t jÞ

¼1Q

j$1ð12 t jÞ;

we obtain our claim. A

We observe that the generating function of Theorem 2.3 is related to the partition

generating function

Xn$0

pðnÞt n ¼Yj$1

1

12 t j:

Indeed, Theorem 2.3 gives that

Xn$m$0

an;mxn ¼

Xi$0

pðiÞx i

ð12 xÞi¼ 1þ

Xi;j$0

pðiþ 1Þiþ j

j

!x iþjþ1;

which leads to the following result.

Theorem 2.4. For all n $ m $ 0

an :¼Xnm¼1

an;m ¼Xni¼1

n2 1

i2 1

!pðiÞ:

The asymptotic growth of an will be determined in Theorem 2.7.

2.2 Combinatorial interpretation

In this section, we use combinatorial methods to solve our recurrence relation (1.1).

In order to do that, we need the following notations and definitions.

A partition l ¼ l1· · ·lk can be represented as a Ferrers diagram. We will use the term

diagram to refer to any finite set of the cells of the two-dimensional square grid. We define

the Ferrers diagram corresponding to a partition l, also called Ferrers shape, as a left-

justified array of unit squares with lk2jþ1 squares in row j. For example, the Ferrers shape

of the partition 54,211 of 13 is



A plane composition of n is a filling of a Ferrers diagram such that the sum of

integers in the cells is n, where a filling of a diagram is an assignment of positive

integers to the cells of the diagram. If rows and columns of the Ferrers diagram are

filled with non-increasing sequences we have plane partitions. While plane partitions are

well-studied objects in combinatorics (for example [2,3,11]), plane compositions do not

seem to have been studied until now. For example, the plane compositions of three are

given by

3 , 2 1 , 1 2 , 2

1

, 1

2

, 1 1 1 , 1 1

1

, 1

1

1

.

Thus, there are eight plane compositions of three (of which all but the third and fourth

are plane partitions).

Let cn;m be the number of plane compositions of n where its diagram has first row with

exactly m cells.

Lemma 2.5. For all n $ m $ 0, cn;m satisfies recurrence (1.1), that is,

cn;m ¼Xnj¼m

j2 1

m2 1

!ðcn2j;0 þ · · ·þ cn2j;mÞ

with the initial conditions c0;0 ¼ 1 and cn;0 ¼ 0 for n $ 1.

Proof. Let l be a Ferrers diagram with at most n cells and with a first row containing

exactly m cells. We want to fill l so that the sum of its entries is exactly n. Begin by filling

the first row with positive integers with sum j. Since there arej2 1

m2 1

!compositions of j

with m parts, there are preciselyj2 1

m2 1

!possible fillings of the first row. Thus the

remaining rows can be filled with the parts of a composition of n2 j, where the number of

cells in the second row is i; 0 # i # m. Therefore, we havePn

j¼m

j2 1

m2 1

!ðcn2j;0 þ · · ·þ

cn2j;mÞ possibilities to fill l by positive integers with sum n and first row having exactly m

cells. A

Lemma 2.5 shows that the number cn;m of plane compositions of n with the first row of

length exactly m satisfies the recurrence relation (1.1). Thus Theorem 2.4 implies the

following corollary, for which we give a direct combinatorial proof below.



Corollary 2.6. The number of all plane compositions of n is given by

cn :¼Xnm¼1

cn;m ¼Xni¼1

n2 1

i2 1

!pðiÞ:

Proof. As already noted a nonempty plane composition s of n with j parts corresponds to a

Ferrers diagram l of j cells and a unique composition s 0 of n with j parts. One way to

obtain the plane composition uniquely is to fill l from bottom to the top and from left to the

right with the parts of s 0. For example,

1 2 1

1 1 3

2

,

λ =σ′ = 1211132

Note that l is not empty, and so represents the Ferrers diagram of the partition of some

j $ 1. Thus the number of plane partitions of n with fixed shape l is given byn2 1

j2 1

!pðjÞ,

wheren2 1

j2 1

!counts the number of compositions of n with j parts. Summing over all

possible diagrams we obtain that the number of plane compositions of n is given byPnj¼1

n2 1

j2 1

!pðjÞ. A

2.3 Asymptotic estimation of cn

We use Stirling’s formula for factorials, together with the Hardy-Ramanujan asymptotic

estimate for the number of partitions, (see e.g. [10]),

pðnÞ ,1

4ffiffiffi3

pnep

ffiffiffi23n

pð2:2Þ

to estimate the asymptotic growth of cn in Theorem 2.4 as n!1.

Theorem 2.7. As n!1, the number of plane compositions

cn ,2n22e ð1=24Þpð8

ffiffi3

p ffiffin

pþpÞffiffiffi

3p

n:

Proof. It is clear that the terms in the sum, namelyn2 1

i2 1

!pðiÞ are strictly increasing at

least for i , n=2. To find the location of the maximal terms of the sum, we consider where

the ratio of consecutive terms is equal to one. Now using (2.2),

n2 1

i2 1

!pðiÞ

n2 1

i

!pðiþ 1Þ

, 12pffiffiffiffi6i

p

� �1þ i

n2 i:



Setting this right hand expression equal to one and solving for i in tems of n, we find that

asymptotically i ¼ k where k , n=2þ ðffiffiffin

pp=4

ffiffiffi3

pÞ. To estimate the size of the maximal

term, we use Stirling’s formula for the binomial coefficient and (2.2) for pðkÞ. We find

pðkÞ ,1

2ffiffiffi3

pnep

ffiffin3

pþp 2

12 ð2:3Þ

and

n2 1

k2 1

!,

2n2ð1=2Þe2ðp 2=24Þffiffiffin

p ffiffiffiffip

p : ð2:4Þ

Now, we consider the behavior of the terms in the neighbourhood of the maximal term

with index k. For j jj , n1=2þ e with small e . 0, ðpðk þ jÞ=pðkÞÞ , expð jp=ffiffiffiffiffi3n

pÞ and

n2 1

k þ j

!

n2 1

k2 1

! , exp 2jpffiffiffiffiffi3n

p 22j2

n

� �:

The contributions to the sum for values of j outside this range are negligible. Hence, we

have

cn :¼Xni¼1

n2 1

i2 1

!pðiÞ ,

n2 1

k2 1

!pðkÞ

Xj jj,n 1=2þ1

e22j 2

n

The sum is asymptotic to the Gaussian integralð121

e22x 2

n dx ¼

ffiffiffiffiffiffipn

p ffiffiffi2

p :

Using this together with the asymptotic estimates (2.3) and (2.4), we obtain the result. A

In Table 1, we compare the asymptotics estimate of Theorem 2.7 with the exact value

of an, as well as the ratio of these numbers, for values of n between 200 and 1000.

2.4 Variations of recurrence (1.1)

We now study the sequence bn;m that satisfies the recurrence relation

bn;m ¼Xnj¼m

j2 1

m2 1

!ðbn2j;0 þ · · ·þ bn2j;m21Þ; ð2:5Þ

Table 1. Asymptotics of the number of plane compositions.

n Theorem 2.7 an The ratio

200 2.41563 £ 1068 2.25114 £ 1068 1.07307400 7.98684 £ 10132 7.59399 £ 10132 1.05173600 2.97179 £ 10196 2.85127 £ 10196 1.04227800 3.45983 £ 10259 3.3376 £ 10259 1.036621000 1.89653 £ 10322 1.83636 £ 10322 1.03276



where n $ m $ 0. To solve this, it is convenient to introduce strict plane compositions.

These are plane compositions in which the lengths of the rows in the Ferrers diagram are

strictly increasing from top to bottom. In this subsection we show that the number of strict

>plane compositions of n is given byPn

j¼1

n2 1

j2 1

!qðjÞ, where qðjÞ is the number of

partitions of j into distinct parts with generating functionXj$0

qð jÞx j ¼Yj$1

ð1þ x jÞ:

Let bn;m be the number of strict plane compositions of n where its diagram has first row

with exactly m cells. In the same way as in the proof of Lemma 2.5 one shows that the

sequence bn;m satisfies the recurrence relation (2.5). Then similar techniques as used in the

proof of Theorem 2.4 lead to the following result.

Theorem 2.8. The generating function for the number of strict plane compositions of n is

given by

Yj$1

1þx j

ð12 xÞj

� �:

Moreover, bn :¼Pn

m¼1bn;m ¼Pn

j¼1

n2 1

j2 1

!qð jÞ.

Another example is the sequence tn;m that satisfies the recurrence relation

tn;m ¼Xnj¼m

m

j2 m

!ðtn2j;0 þ · · ·þ tn2j;mÞ: ð2:6Þ

This is related to the enumeration of plane compositions of n in which the entries belong to

the set {1,2}. Let tn;m be the number of such plane compositions such that the first row of

each diagram hasm cells. Then, in a similar manner to the previous reasoning, we find that

tn;m satisfies the recurrence relation (2.6). Hence we obtain the following result.

Theorem 2.9. The generating function for the number tn of plane compositions of n with

entries in the set {1,2} is given by

Yj$1

1

12 x j 2 x2j

� �:

Moreover, tn ¼Pn

j¼1

j

n2 j

!pð jÞ.

Also, if dn;m denotes the number of strict plane compositions of n with entries in {1,2},

where the diagrams have first rows with exactly m cells, then dn;m satisfies the recurrence

relation

dn;m ¼Xnj¼m

m

j2 m

!ðdn2j;0 þ · · ·þ dn2j;m21Þ; ð2:7Þ

which gives the following result.



Theorem 2.10. The generating function for the number dn of strict plane compositions of n

into parts in the set {1,2} is given by

Yj$1

ð1þ x j þ x2jÞ:

Moreover, dn ¼Pn

j¼1

j

n2 j

!qð jÞ.

By considering plane compositions of n with other restrictions on the set of parts used,

further examples of related recurrences can be found. For example, Chinn and Heubach

[4,5] and Knopfmacher and Mays [7] have considered various types of ordinary

compositions with restrictions on the sizes of parts.

3. Statistics on the set of plane compositions

Let s be any plane composition of n associated with shape l. We denote the number of the

values in the main diagonal of s by diag(s), which equals the number of cells in the main

diagonal of the partition l. Also, we denote the sum of the numbers in the main diagonal of

s by trace(s). We call trace(s) the trace of the plane composition. For example, if s is the

plane composition

2 2 1 2 1 1 1

1 3 4 1

2 1

1

then diagðsÞ ¼ 2 and traceðsÞ ¼ 2 þ 3 ¼ 5. Note that diagðsÞ is also the length of the

maximal square that fits inside the shape l, and is known in the case of partitions as the

Durfee square.

Let Fðx; y; p; qÞ be the generating function for the number of plane compositions of n

according to the length of the diagonal and the trace, that is

Fðx; p; qÞ ¼Xn$0

Xs

xnpdiagðsÞq traceðsÞ;

where the internal sum runs over all plane compositions of n. In this section, we first find

an explicit formula for the generating function Fðx; p; qÞ. Then we refine our result by

finding the generating function Gl;mðx; p; qÞ for the number of plane compositions of n

contained in a box of height l and width m exactly, according to the length of the diagonal

and the trace. That is

Gl;mðx; p; qÞ ¼Xn$0

Xs

xnpdiagðsÞq traceðsÞ;

where the internal sum runs over all plane compositions of n contained in a Ferrers

diagram whose first row has length m and first column has length l. These two results

include several applications.



3.1 Explicit formula for the generating function Fðx; p; qÞ

In order to find an explicit formula for the generating function Fðx; p; qÞ, we need the

following notation. Denote the generating function for all plane compositions of n

according to the statistic trace(s) with diagðsÞ ¼ k by Fkðx; qÞ.It is obvious that each diagram l can be decomposed into three pieces: the maximal

square Sl contained in the diagram, the sub diagram Al above it, and the sub diagram Rl on

the right side of it, as illustrated in Figure 1. Now let us consider the set of plane

compositions with diagðsÞ ¼ k.

Note that if diagðsÞ ¼ k for some plane composition then the length of the first row of

the As and the length of the first column of Rs is at most k. Thus,

. The contribution of the diagram of the maximal square to the generating function

Fkðx; qÞ is given by

xq

12 xq

� �kx

12 x

� �kðk21Þ

;

where ðxq=12 xqÞk is the generating function for the composition that appears in

the main diagonal and the term ðx=12 xÞkðk21Þ is the generating function for all

others numbers not in the main diagonal of the square.

. The contribution of the subdiagram As in the generating function Fkðx; qÞ is givenby the generating function for the number of all plane compositions of n with first

row of length k. By Proposition 2.2, we get the contribution as

Xkm¼0


¼1Qk


. Using the transpose symmetric operation, we get, by Proposition 2.2, that the

contribution of the subdiagram Rs in the generating function Fkðx; qÞ is also

Xkm¼0


¼1Qk


Hence, we have that the generating function for the number of all plane compositions s of

n with diagðsÞ ¼ k according to traceðsÞ is given by

Fkðx; qÞ ¼ðxq=ð12 xqÞÞkðx=ð12 xÞÞkðk21ÞQk

j¼1ð12 ðx=ð12 xÞÞ jÞ2:

Summing over all possible values of k, we get an explicit formula for Fðx; p; qÞ.

Figure 1. Decomposition of a diagram by its maximal box.



Theorem 3.1. The generating function for the number of all plane compositions s of n

according diag(s) and trace(s) is given by

Fðx; p; qÞ ¼Xk$0

pkðxq=ð12 xqÞÞkðx=ð12 xÞÞkðk21ÞQkj¼1ð12 ðx j=ð12 xÞjÞÞ2

:

Note the special case (cf. Theorem 2.3):

Fðx; 1; 1Þ ¼Xk$0

ðx=ð12 xÞÞk2Qk

j¼1ð12 ðx j=ð12 xÞ jÞÞ2¼Yj$1

1

12 ðx j=ðð12 xÞ jÞÞ:

This corresponds to another partition identity of Euler (see [1], Corollary 2.6). A plane

composition is said to be symmetric if it has a symmetric diagram and the numbers filled in

the diagram are symmetric to the main diagonal. For example, there is only one symmetric

plane composition of 2 which is 2. Modifying the proof of Theorem 3.1 (note that in this

case Al and Rl are the same plane composition up to transpose, and the maximal square of

the plane partition is symmetric to its main diagonal), we get the following result.

Theorem 3.2. The generating function for the number of all symmetric plane compositions

to s of n according diag(s) and trace(s) is given by

Xk$0

pkðxq=ð12 xqÞÞkðx2=ð12 x2ÞÞ

k

2

!

Qkj¼1ð12 x2j=ðð12 x2ÞjÞÞ

¼Yn$0

1þpxq

12 xq

x2

12 x 2

� �n� �:

The product form above is by means of Euler’s partition identity, [1], Corollary 2.2.

3.2 The generating function Gl;mðx; p; qÞ

Let s be any plane composition, and denote by s0 the plane composition that is obtained

from s be deleting the first row and the first column. Rewriting the generating function

Gl;mðx; p; qÞ for the number of plane compositions s in terms of the generating function for

the plane compositions s0, we obtain the recurrence relation

Gl;mðx; p; qÞ ¼xpq

12 xq

x

12 x

� �mþl22

1þXl21

i¼1

Xm21

j¼1

Gi;jðx; p; qÞ

!; ð3:1Þ

where l;m $ 1. Now we are interested in solving this recurrence. At first, we note that

(3.1) gives

Gl;mðx; p; qÞ2x

12 xGl;m21ðx; p; qÞ

� �2

x

12 xGl21;mðx; p; qÞ2

x

12 xGl21;m21ðx; p; qÞ

� �

¼x

12 x

� �mþ l2 2

xpq

12 xqGl21;m21ðx; p; qÞ;

for all l;m $ 2.



Define GlðuÞ ¼ Glðx; p; q; uÞ ¼P

m$0Gl;mðx; p; qÞum. Multiplying the above recur-

rence relation by u m, and summing over all m $ 2 and applying the initial conditions

Gl;0ðx; p; qÞ ¼ 0 and Gl;1ðx; p; qÞ ¼xpq

12 xq

x

12 x

� �l21

;

we obtain that

12xu

12 x

� �GlðuÞ2

x

12 x12

xu

12 x

� �Gl21ðuÞ ¼

xpqu

12 xq

x

12 x

� �l21

Gl21

xu

12 x

� �;

for all l $ 2. The initial conditions can be obtained directly from the definitions, which are

G0ðuÞ ¼ 1 and G1ðuÞ ¼xpqu

12 xq

1

12 xu=ð12 xÞ:

Now define Gðv; uÞ ¼ Gðx; p; q; v; uÞ ¼P

l$0GlðuÞvl. Multiplying the last recurrence

relation by v l and summing over all possible values of l $ 2, and using the initial

conditions G0ðuÞ and G1ðuÞ, we get that

Gðv; uÞ ¼ 1þxpq

12 xq

vu

ð12 xu=ð12 xÞÞð12 xv=ð12 xÞÞG

xv

12 x;

xu

12 x

� �:

Iterating the above functional equation an infinite number of times, we get the following

result.

Theorem 3.3. The generating function Gðx; p; q; v; uÞ is given by

Xj$0

ðxpq=ð12 xqÞÞjðx=ð12 xÞÞ jð j21ÞuvQji¼1ð12 x i=ðð12 xÞiÞuÞð12 x i=ðð12 xÞiÞvÞ

:

The above theorem for u ¼ v ¼ 1 reduces to Theorem 3.1.

Acknowledgements

The authors would like to thank the referee for careful reading of the previous version of the presentpaper and for suggesting the references [12,13].

Notes

1. Email: [email protected]. Email: [email protected]

References

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Recurrence relation with two indices and plane compositions

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