Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

PURE MATHEMATICS | RESEARCH ARTICLE

Taylor expansions for the generating function of Catalan-like numbersLily Li Liu1* and Xiaoli Li1

Abstract: In 2002, Eu, Liu and Yeh introduced new Taylor expansions of the gen-erating function of Catalan and Motzkin numbers. And they presented that this Taylor style expansion can be applied to more generating functions satisfying some relations (Advances in Applied Mathematics, 29 (2002) 345–357). In this paper, we focus on this Taylor expansion of the generating function of Catalan-like numbers, which are common generalizations of many classic counting coefficients, such as the Catalan numbers, the Motzkin numbers and the Schröder numbers. We present the recurrence relations of the coefficients and bivariate generating functions of the remainders of the new Taylor expansion of the generating function of Catalan-like numbers.

Subjects: Advanced Mathematics; Analysis - Mathematics; Mathematics & Statistics; Pure Mathematics; Science

Keywords: Catalan-like numbers; Catalan numbers; Schröder numbers; Motzkin numbers; Taylor expansion

AMS subject classifications: 05A15; 05A19; 05A20; 11B83; 15A45

*Corresponding author: Lily Li Liu, School of Mathematical Science, Qufu Normal University, Qufu 273165, P.R. China E-mail: liulily@mail.qfnu.edu.cn

Reviewing editor:Hari M. Srivastava, University of Victoria, Canada

Additional information is available at the end of the article

ABOUT THE AUTHORSLily Li Liu is working as an associate professor in School of Mathematical Science at Qufu Normal University in China. She is pursuing her PhD degree in Mathematics from Dalian University of Technology in 2009. She has published several research papers in national and international journals. Her area of interest is inequalities, the unimodality property and the properties of combinatorial sequences.

Xiaoli Li is a graduate student of the first author in School of Mathematical Science at Qufu Normal University in China. Her research interests are the properties of combinatorial sequences.

PUBLIC INTEREST STATEMENTEu, Liu and Yeh introduced new Taylor expansions, whose remainders involve the function itself, of the generating function of Catalan and Motzkin numbers. In this paper, we focus on this Taylor expansion of the generating function of Catalan-like numbers, which are the 0th column of special cases of Riordan arrays. Riordan arrays play an important unifying role in enumerative combinatorics. There have been quite a few papers concerned with combinatorics of Riordan arrays. Our concern is one class of special interesting Riordan arrays—the recursive matrix introduced by Aigner. The 0th column of such recursive matrix includes many classical combinatorial sequences, such as the Catalan numbers, the Motzkin numbers, the large and little Schröder numbers. We present the recurrence relations of the coefficients and bivariate generating functions of the remainders of the new Taylor expansion of the generating function of Catalan-like numbers.

Received: 22 February 2016Accepted: 07 June 2016First Published: 10 June 2016

© 2016 The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license.

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Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

1. IntroductionThe Taylor expansion of a real function f(x), which is infinitely differential, at the real number 0 is

where Rn(x) is the nth remainder, which is the error incurred in approximating the function f(x) by its (n − 1)th Taylor polynomial

∑n−1

i=0

f (i)(0)

i!xi.

Traditionally, the remainders of Taylor expansions play a central role in the theory of functions, numberical approximations, asymptotic expansions, etc. They are used mainly for quantitative or numerical purposes (Eu, Liu, & Yeh, 2002). Unlike the usual Taylor expansions, the remainders in Taylor expansions considered in this paper involve the generating function itself. Such expansions are quite different from the usual binomial expansions or continued fraction expansions but are not exceptions for combinatorial structures. Eu et al. (2002) introduced these new Taylor expansions for the Catalan numbers Ci and the Motzkin numbers Mi. They showed that

and

where the fn, gn and hn are recursively defined polynomials. This new Taylor expansions can be gen-eralized to the Catalan-like numbers (Eu et al., 2002).

The Catalan-like numbers considered in this paper are the 0th columns of special cases of Riordan arrays. Riordan arrays play an important unifying role in enumerative combinatorics (Shapiro, Getu, Woan, & Woodson, 1991). A (proper) Riordan array, denoted by (g(x), f(x)), is an infinite lower trian-gular matrix whose generating function of the kth column is xkf k(x)g(x) for k = 0, 1, 2,… , where g(0) = 1 and f (0) ≠ 0. A Riordan array R = [dn,k]n,k≥0 can also be characterized by two sequences (an)n≥0 and (zn)n≥0 such that

for n, k ≥ 0 (see Cheon, Kim, & Shapiro, 2012; He & Sprugnoli, 2009 for instance). Call (an)n≥0 and (zn)n≥0 the A- and Z-sequences of R, respectively. The 0th column of such a Riordan array includes many classical combinatorial sequences, such as the Catalan numbers, the Motzkin numbers, the large and the little Schröder numbers. There have been quite a few papers concerned with combina-torics of Riordan arrays (see Cheon et al., 2012; Ehrenfeucht, Harju, ten Pas, & Rozenberg, 1998; He & Sprugnoli, 2009; Shapiro et al., 1991 for instance). Our concern in the present paper is one class of special interesting Riordan arrays—the recursive matrix introduced by Aigner (1999, 2001). Let p, s, t be three nonnegative numbers. Denote by R(p; s, t) = [dn,k]n,k≥0, the Riordan array with Z = (p, t, 0, 0,…) and A = (1, s, t, 0, 0,…). More precisely,

(1)f (x) =

n−1∑

i=0

f (i)(0)

i!xi + Rn(x),

C =∑

i≥0

Cixi=

n−1∑

i=0

Cixi+ xnfn(C),

M =∑

i≥0

Mixi=

n−1∑

i=0

Mixi+ xngn(M) + x

n+1hn(M),

(2)d0,0

= 1, dn+1,0 =∑

j≥0

zjdn,j , dn+1,k+1 =∑

j≥0

ajdn,k+j

(3)

d0,0

= 1, d0,k = 0 (k > 0);

dn,0 = pdn−1,0 + tdn−1,1 (n ≥ 1);

dn,k = dn−1,k−1 + sdn−1,k + tdn−1,k+1 (n, k ≥ 1).

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Following Aigner (1999, 2001), the matrix R(p; s, t) is called the recursive matrix, and the numbers C(p; s, t) = dn,0 are called the Catalan-like numbers corresponding to (�, �), where � = (p, s, s,⋯), � = (t, t, t,⋯).

The Catalan-like numbers unify many famous counting coefficients. For example, the numbers dn,0 are:

(1) the Catalan numbers Cn= C(1; 2, 1) corresponding to � = (1, 2, 2,⋯), � = (1, 1, 1,⋯);

(2) the Motzkin numbers Mn= C(1; 1, 1) corresponding to � = (1, 1, 1,⋯), � = (1, 1, 1,⋯);

(3) the large Schröder numbers Rn= C(2; 3, 2) corresponding to � = (2, 3, 3,⋯), � = (2, 2, 2,⋯);

(4) the little Schröder numbers Sn= C(1; 3, 2) corresponding to � = (1, 3, 3,⋯), � = (2, 2, 2,⋯);

(5) the restricted hexagonal numbers rn= C(3; 3, 1) corresponding to

� = (3, 3, 3,⋯), � = (1, 1, 1,⋯).

In this paper, we discuss the remainders of new Taylor expansions for the generating function of Catalan-like numbers. More precisely, we obtain the recurrence relations of the coefficients and bi-variate generating functions of the remainders.

2. The Catalan-like numbers’ Taylor expansionLet dn,0 = cn be the nth Catalan-like numbers. Denote by C(x) =

∑

n≥0 cnxn the generating function

of Catalan-like numbers. Then by the theory of Riordan array, we have

Let a = (s − p)x + (p2 − ps + rt)x2, b = −1 − (s − 2p)x, c = 1. So C(x) satisfies the following recur-rence relation

Also let a1= s − p,a

2= p2 − ps + rt, b

1= −(s − 2p).

Then we have

After arrangement successively, we have

(4)

C(x) =2

1 + (s − 2p)x +

√

1 − 2sx + (s2 − 4rt)x2

=1 + (s − 2p)x −

√

1 − 2sx + (s2 − 4rt)x2

2(s − p)x + 2(p2 − ps + rt)x2

[

(s − p)x + (p2 − ps + rt)x2]

C2(x) −[

1 + (s − 2p)x]

C(x) + 1 = 0.

C = 1 + b1xC +

(

a1x + a

2x2)

C2

= 1 + x(

b1C + a

1C2)

+ x2(

a2C2)

;

C2= C + x

(

b1C2+ a

1C3)

+ x2(

a2C3)

= 1 + x(

b1C + b

1C2+ a

1C2+ a

1C3)

+ x2(

a2C2+ a

2C3)

;

C3= C + x

(

b1C2+ b

1C3+ a

1C3+ a

1C4)

+ x2(

a2C3+ a

2C4)

= 1 + x(

b1C + b

1C2+ b

1C3+ a

1C2+ a

1C3+ a

1C4)

+ x2(

a2C2+ a

2C3+ a

2C4)

.

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Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

Then following Eu et al. (2002), Taylor expansions for the generating function of Catalan-like numbers are

where gn and hn are polynomials, and xngn(C) + xn+1hn(C) is the nth remainder.

So in order to get the uniqueness of gn and hn, it suffices to prove that the equation a(C) + xb(C) = 0 only has the trivial solution for polynomials a(y) and b(y). Note that

is a continuous function. For any y, there exist different �, �, so that C(�) = C(�) = y. By inserting x = �, � into a(C) + xb(C) = 0, we can get a(y) = b(y) = 0. As to the polynomial, a and b must be the constant zero.

Since

we set

where

Now, we replace C with 1 + x(b1C + a

1C2) + x2(a

2C2). So we have

Then we set

(5)

Ck = 1 + x

(

k∑

i=1

b1Ci +

k∑

i=1

a1Ci+1

)

+ x2k∑

i=1

a2Ci+1

= 1 + x

k∑

i=1

(

b1+ a

1C)

Ci + x2k∑

i=1

a2Ci+1.

(6)C =

n−1∑

k=0

ckxk+ xngn(C) + x

n+1hn(C).

C(x) =1 + (s − 2p)x −

√

1 − 2sx + (s2 − 4rt)x2

2(s − p)x + 2(p2 − ps + rt)x2

(7)C = 1 + x(b1C + a

1C2) + x2(a

2C2),

C = 1 + xg1(C) + x2h

1(C)

= 1 + x(

g1,1C + g

1,2C2

)

+ x2h1,1C2,

g1,1

= b1, g

1,2= a

1, h

1,1= a

2.

C = 1 + x(b1C + a

1C2) + x2(a

2C2)

= 1 + x(b1+ a

1C) + x2

[

b21C + (2b

1a1+ a

2)C2 + a2

1C3

]

+ x3(b1a2C2 + a

1a2C3)

= 1 + b1x + a

1x[

1 + x(b1C + a

1C2) + x2(a

2C2)

]

+ x2[

b21C + (2b

1a1+ a

2)C2 + a2

1C3

]

+ x3(b1a2C2 + a

1a2C3)

= 1 + x(a1+ b

1) + x2

[

(b21+ a

1b1)C + (2b

1a1+ a

2+ a2

1)C2 + a2

1C3

]

+ x3[

(b1a2+ a

1a2)C2 + a

1a2C3

]

.

(8)C = 1 + c

1x + x2g

2(C) + x3h

2(C)

= 1 + c1x + x2(g

2,1C + g

2,2C2 + g

2,3C3) + x3(h

2,1C2 + h

2,2C3),

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Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

where

We replace C with 1 + x(b1C + a

1C2) + x2(a

2C2). And we have

So inserting (Equation (9)) into (Equation (8)), we get

Then we set

where

After arrangement successively, we can get the nth Taylor expansion for the generating function of Catalan-like numbers as follows (Eu et al., 2002).

g2,1

= b21+ a

1b1= b

1(g1,1

+ g1,2);

g2,2

= 2b1a1+ a

2+ a2

1= b

1g1,2

+ a1(g1,1

+ g1,2) + h

1,1;

g2,3

= a21= a

1g1,2;

h2,1

= b1a2+ a

1a2= a

2(g1,1

+ g1,2);

h2,2

= a1a2= a

2g1,2.

(9)

g2(C) = g

2,1C + g

2,2C2 + g

2,3C3

= g2,1C + g

2,2C2 + g

2,3

[

C2 + x(b1C3 + a

1C4) + x2(a

2C4)

]

= g2,1C + (g

2,2+ g

2,3)C2 + x(b

1g2,3C3 + a

1g2,3C4) + x2(a

2g2,3C4)

= g2,1C + (g

2,2+ g

2,3)

[

C + x(b1C2 + a

1C3) + x2(a

2C3)

]

+ x(b1g2,3C3 + a

1g2,3C4)

+ x2(a2g2,3C4)

= (g2,1

+ g2,2

+ g2,3)C + x

{

b1(g2,2

+ g2,3)C2 +

[

a1(g2,2

+ g2,3) + b

1g2,3

]

C3 + a1g2,3C4

}

+ x2[

a2(g2,2

+ g2,3)C3 + a

2g2,3C4

]

= (g2,1

+ g2,2

+ g2,3)

[

1 + x(b1C + a

1C2) + x2(a

2C2)

]

+ x2[

a2(g2,2

+ g2,3)C3 + a

2g2,3C4

]

+ x{

b1(g2,2

+ g2,3)C2 +

[

a1(g2,2

+ g2,3) + b

1g2,3

]

C3 + a1g2,3C4

}

.

C = 1 + c1x + x2g

2(C) + x3h

2(C)

= 1 + c1x + x2(g

2,1+ g

2,2+ g

2,3) + x4

[

a2(g2,1

+ g2,2

+ g2,3)C2 + a

2(g2,2

+ g2,3)C3 + a

2g2,3C4

]

+ x3{

b1(g2,1

+ g2,2

+ g2,3)C +

[

a1(g2,1

+ g2,2

+ g2,3) + b

1(g2,2

+ g2,3)]

C2}

+ x3{

[

a1(g2,2

+ g2,3) + b

1g2,3

]

C3 + a1g2,3C4 + x3(h

2,1C2 + h

2,2C3)

}

.

C = 1 + c1x + c

2x2 + x3g

3C + x4h

3C

= 1 + c1x + c

2x2 + x3

(

g3,1C + g

3,2C2 + g

3,3C3 + g

3,4C4

)

+ x4(

h3,1C2 + h

3,2C3 + h

3,3C4

)

,

g3,1

= b1(g2,1

+ g2,2

+ g2,3);

g3,2

= a1(g2,1

+ g2,2

+ g2,3) + b

1(g2,2

+ g2,3) + h

2,1;

g3,3

= a1(g2,2

+ g2,3) + b

1g2,3

+ h2,2;

g3,4

= a1g2,3;

h3,1

= a2(g2,1

+ g2,2

+ g2,3);

h3,2

= a2(g2,2

+ g2,3);

h3,3

= a2g2,3.

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where

and gn,i = hn,j = 0, for i > n + 1 and j > n. Now we present the main result of this paper based on the method proposed by Eu et al. (2002).

Theorem 1 Let

be the nth Taylor expansion for the generating fuction of Catalan-like numbers defined by Equation (3). Then we have the following.

(i) gn,1 = b1cn−1, hn,1 = a2cn−1, for n ≥ 1;

(ii) gn,2 = (b1+ a

1)cn−1 −

(

b21− a

2

)

cn−2 and hn,2 = a2(cn−1 − b1cn−2), for n ≥ 2;

(iii) gn,k =b1

a2

hn,k +a1

a2

hn,k−1 + hn−1,k−1, for n, k ≥ 2, where hn−1,0 = 0;

(iv) gn,k and hn,k satisfy the following recurrence relations.

where g1,1

= b1, g

1,2= a

1,h

1,1= a

2.

Proof It is trivial for n = 1. Let n ≥ 2. Replacing each Ck in (n − 1)th Taylor expansion of Equation (6) with 1 + x(b

1

∑k

i=1 Ci+ a

1

∑k

i=1 Ci+1

) + x2∑k

i=1 a2Ci+1, we get

Note that

Hence comparing coefficients of each term of the remainder on the above two equations, we can get

C =

n−1∑

k=0

ckxk+ xn

n+1∑

k=1

gn,kCk+ xn+1

n∑

k=1

hn,kCk+1,

gn(C) =

n+1∑

k=1

gn,kCk, hn(C) =

n∑

k=1

hn,kCk+1,

C =

n−1∑

k=0

ckxk+ xn

n+1∑

k=1

gn,kCk+ xn+1

n∑

k=1

hn,kCk+1

gn,k = gn,k+1 + b1gn−1,k + a1gn−1,k−1 + a2gn−2,k−1, for n ≥ 3 and 3 ≤ k ≤ n;

hn,k = hn,k+1 + b1hn−1,k + a1hn−1,k−1 + a2hn−2,k−1, for n ≥ 3 and 2 ≤ k ≤ n − 1;

C =

n−2∑

k=0

ckxk+ xn−1

n∑

k=1

gn−1,kCk+ xn

n−1∑

k=1

hn−1,kCk+1

=

n−1∑

k=0

ckxk+ xn

(

n−1∑

k=1

hn−1,kCk+1

+

n∑

i=1

b1

n∑

k=i

gn−1,kCi+

n∑

i=1

a1

n∑

k=i

gn−1,kCi+1

)

+ xn+1n∑

i=1

a2

n∑

k=i

gn−1,kCi+1.

C =

n−1∑

k=0

ckxk+ xn

n+1∑

k=1

gn,kCk+ xn+1

n∑

k=1

hn,kCk+1.

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where hn−1,0 = 0,hn−1,n = 0, gn−1,n+1 = 0.

(i) If k = 1, then we have

(ii) If k = 2, then we have

(iii) If n, i ≥ 2, then we have

Hence, we can get

(iv) If n ≥ 3, 3 ≤ i ≤ n, then we have

(10)

gn,1 = b1

n∑

k=1

gn−1,k;

gn,i = b1

n∑

k=i

gn−1,k + a1

n∑

k=i−1

gn−1,k + hn−1,i−1 (i ≥ 2,n ≥ 2);

gn,n+1 = a1gn−1,n;

hn,i = a2

n∑

k=i

gn−1,k (i ≥ 1);

gn,1 = b1

n∑

k=1

gn−1,k = b1cn−1;

hn,1 = a2

n∑

k=1

gn−1,k = a2cn−1.

gn,2 = b1

n∑

k=2

gn−1,k + a1

n∑

k=1

gn−1,k + hn−1,1

= b1(

n∑

k=1

gn−1,k − gn−1,1) + a1

n∑

k=1

gn−1,k + hn−1,1

= (b1+ a

1)

n∑

k=1

gn−1,k − b1gn−1,1 + hn−1,1

= (b1+ a

1)cn−1 − b

2

1cn−2 + a2cn−2

= (b1+ a

1)cn−1 −

(

b21− a

2

)

cn−2;

hn,2 = a2

n∑

k=2

gn−1,k = a2(

n∑

k=1

gn−1,k − gn−1,1) = a2(cn−1 − b1cn−2).

a2gn,i = a2(b1

n∑

k=i

gn−1,k + a1

n∑

k=i−1

gn−1,k + hn−1,i−1)

= a2b1

n−1∑

k=i

gn−1,k + a2a1

n−1∑

k=i−1

gn−1,k + a2hn−1,i−1

= b1hn,i + a1hn,i−1 + a2hn−1,i−1.

gn,i =b1

a2

hn,i +a1

a2

hn,i−1 + hn−1,i−1.

gn,i − gn,i+1 = b1

n∑

k=i

gn−1,k + a1

n∑

k=i−1

gn−1,k + hn−1,i−1 − (b1

n∑

k=i+1

gn−1,k + a1

n∑

k=i

gn−1,k + hn−1,i)

= b1gn−1,i + a1gn−1,i−1 + hn−1,i−1 − hn−1,i

= b1gn−1,i + a1gn−1,i−1 + a2

n−1∑

k=i−1

gn−2,k − a2

n−1∑

k=i

gn−2,k

= b1gn−1,i + a1gn−1,i−1 + a2gn−2,i−1.

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If n ≥ 3, 2 ≤ i ≤ n − 1, then we have

✷

In the proof of Theorem (1), we can get recurrence relations (Equation (10)) of gn,k and hn,k with g1,1

= b1 and h

1,1= a

2. So we can obtain a relation between gn(C) and hn(C) according to (iii).

Corollary 2 The two functions gn and hn satisfy

for n ≥ 1 with h0(C) = 0.

Proof Since

and

✷

we have

hn,i − hn,i+1 = a2

n∑

k=i

gn−1,k − a2

n∑

k=i+1

gn−1,k

= a2gn−1,i

= a2(b1

n−1∑

k=i

gn−2,k + a1

n−1∑

k=i−1

gn−2,k + hn−2,i−1)

= a2b1

n−1∑

k=i

gn−2,k + a2a1

n−1∑

k=i−1

gn−2,k + a2hn−2,i−1

= b1hn−1,i + a1hn−1,i−1 + a2hn−2,i−1.

gn(C) =(b1+ a

1C)hn(C)

a2C

+ hn−1(C),

gn(C) =

n+1∑

k=1

gn,kCk and hn(C) =

n∑

k=1

hn,kCk+1,

a2gn,i = b1hn,i + a1hn,i−1 + a2hn−1,i−1, (n, i ≥ 2),

gn(C) =

n+1∑

k=1

gn,kCk

= gn,1C +1

a2

n+1∑

k=2

a2gn,kC

k

= gn,1C +1

a2

n+1∑

k=2

(b1hn,k + a1hn,k−1 + a2hn−1,k−1)C

k

= b1cn−1C +

b1

a2

n+1∑

k=2

hn,kCk+a1

a2

n+1∑

k=2

hn,k−1Ck+

n+1∑

k=2

hn−1,k−1Ck

= b1cn−1C +

b1

a2C

n+1∑

k=2

hn,kCk+1

+a1

a2

hn(C) +

n+1∑

k=2

hn−1,k−1Ck

= b1cn−1C +

b1

a2C(

n+1∑

k=1

hn,kCk+1

− hn,1C2) +

a1

a2

hn(C) + hn−1(C)

= b1cn−1C +

b1

a2C(

n∑

k=1

hn,kCk+1

− hn,1C2) +

a1

a2

hn(C) + hn−1(C)

= b1cn−1C +

b1

a2Chn(C) −

b1

a2Ca2cn−1C

2+a1

a2

hn(C) + hn−1(C)

=b1

a2Chn(C) +

a1

a2

hn(C) + hn−1(C)

=(b1+ a

1C)hn(C)

a2C

+ hn−1(C).

Page 9 of 12

Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

Corollary 3    The generating functions

have closed forms:

Proof Since

and

we have

where

G(x, y) =∑

n≥k≥1

gn,kxn−1yk−1 +

∑

n≥1

gn,n+1xn−1yn and H(x, y) =

∑

n≥k≥1

hn,kxn−1yk−1

G =

�

b1+ a

1y

a2

+ xy

�

H and H =a2C − y

∑∞

n=1 hn,nxn−1yn−1(1 − a

1xy)

1 − y + b1xy + a

1xy2 + a

2x2y2

.

a2gn,i = b1hn,i + a1hn,i−1 + a2hn−1,i−1, (n, i ≥ 2),

gn,1 = b1cn−1, hn,1 = a2cn−1, gn,n+1 = a1gn−1,n, hn,n = a2gn−1,n,

(11)

G(x, y) =∑

n≥k≥1

gn,kxn−1yk−1 +

∑

n≥1

gn,n+1xn−1yn

=

∞∑

n=1

n∑

k=1

gn,kxn−1yk−1 +

∞∑

n=1

gn,n+1xn−1yn

=

∞∑

n=1

n∑

k=2

gn,kxn−1yk−1 +

∞∑

n=1

gn,1xn−1

+

∞∑

n=1

gn,n+1xn−1yn

=1

a2

∞∑

n=1

n∑

k=2

a2gn,kx

n−1yk−1 +

∞∑

n=1

gn,1xn−1

+

∞∑

n=1

gn,n+1xn−1yn

=1

a2

∞∑

n=1

n∑

k=2

(b1hn,k + a1hn,k−1 + a2hn−1,k−1)x

n−1yk−1 +

∞∑

n=1

gn,1xn−1

+

∞∑

n=1

gn,n+1xn−1yn

=b1

a2

∞∑

n=1

n∑

k=2

hn,kxn−1yk−1 +

a1

a2

∞∑

n=1

n∑

k=2

hn,k−1xn−1yk−1 +

∞∑

n=1

b1cn−1x

n−1

+

∞∑

n=1

n∑

k=2

hn−1,k−1xn−1yk−1 +

∞∑

n=1

a1gn−1,nx

n−1yn,

(12)

∞∑

n=1

n∑

k=2

hn,kxn−1yk−1 =

∞∑

n=1

n∑

k=1

hn,kxn−1yk−1 −

∞∑

n=1

hn,1xn−1

= H −

∞∑

n=1

hn,1xn−1

= H −

∞∑

n=1

a2cn−1x

n−1

= H − a2C,

∞∑

n=1

n∑

k=2

hn,k−1xn−1yk−1 =

∞∑

n=1

n−1∑

v=1

hn,vxn−1yv

= y

∞∑

n=1

n−1∑

v=1

hn,vxn−1yv−1

= y(H −

∞∑

n=1

hn,nxn−1yn−1) = y(H −

∞∑

n=1

a2gn−1,nx

n−1yn−1), (13)

Page 10 of 12

Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

and

Now inserting (Equation (12))–(Equation (14)) into (Equation (11)), we can get

Also since

and

we have

where

(14)∞∑

n=1

n∑

k=2

hn−1,k−1xn−1yk−1 = xy

∞∑

n=1

n∑

k=2

hn−1,k−1xn−2yk−2 = xyH.

G(x, y) =b1

a2

(H − a2C) +

a1

a2

y(H −

∞∑

n=1

a2gn−1,nx

n−1yn−1) + xyH +

∞∑

n=1

b1cn−1x

n−1

+

∞∑

n=1

a1gn−1,nx

n−1yn

=b1

a2

H − b1C +

a1

a2

yH −

∞∑

n=1

a1gn−1,nx

n−1yn + xyH + b1C +

∞∑

n=1

a1gn−1,nx

n−1yn

= (b1+ a

1y

a2

+ xy)H.

hn,k = hn,k+1 + b1hn−1,k + a1hn−1,k−1 + a2hn−2,k−1, n ≥ 3 and2 ≤ k ≤ n − 1,

hn,1 = a2cn−1, hn,2 = a2(cn−1 − b1cn−2), n ≥ 2,

(15)

H(x, y) =∑

n≥k≥1

hn,kxn−1yk−1 =

∞∑

n=1

n∑

k=1

hn,kxn−1yk−1

=

∞∑

n=3

n∑

k=1

hn,kxn−1yk−1 +

2∑

n=1

n∑

k=1

hn,kxn−1yk−1

=

∞∑

n=3

n−1∑

k=2

hn,kxn−1yk−1 +

∞∑

n=3

(hn,1xn−1

+ hn,nxn−1yn−1) + h

1,1+ h

2,1x + h

2,2xy

=

∞∑

n=3

n−1∑

k=2

hn,kxn−1yk−1 +

∞∑

n=1

hn,1xn−1

+

∞∑

n=2

hn,nxn−1yn−1

=

∞∑

n=3

n−1∑

k=2

(hn,k+1 + b1hn−1,k + a1hn−1,k−1 + a2hn−2,k−1)xn−1yk−1

+ a2C +

∞∑

n=2

hn,nxn−1yn−1,

(16)

∞∑

n=3

n−1∑

k=2

hn−1,k−1xn−1yk−1 =

∞∑

u=2

u−1∑

v=1

hu,vxuyv = xy

∞∑

n=2

n−1∑

k=1

hn,kxn−1yk−1

= xy

(

∞∑

n=1

n∑

k=1

hn,kxn−1yk−1 −

1∑

n=1

n∑

k=1

hn,kxn−1yk−1 −

∞∑

n=2

hn,nxn−1yn−1

)

= xy(H −

∞∑

n=1

hn,nxn−1yn−1),

Page 11 of 12

Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

and

Now inserting (Equation (16))–(Equation (19)) into (Equation (15)), we have

Multiplying by y on both sides of (Equation 20) and then after arrangement, we have

So we get

✷

3. RemarksIn the paper, we study the remainders of new Taylor expansions for the generating function of Catalan-like numbers. Since the Catalan-like numbers, such as the Motzkin numbers (Aigner, 1998; Donaghey & Shapiro, 1977; Sulanke, 2001), the Catalan numbers (Aigner, 2001; He, 2013; Mahmoud & Qi, 2016), the large and little Schröder numbers (Ehrenfeucht et al., 1998; Qi, Shi, & Guo, 2016a, 2016b; Stanley, 1997) naturally appear in the combinatorial objects, their Taylor expansions can be interpreted in distinct combinatorial ways.

(16)

∞∑

n=3

n−1∑

k=2

hn−2,k−1xn−1yk−1 = x2y

∞∑

n=1

n∑

k=1

hn,kxn−1yk−1 = x2yH,

(18)

∞∑

n=3

n−1∑

k=2

hn,k+1xn−1yk−1 =

∞∑

n=3

n∑

u=3

hn,uxn−1yu−2 =

∞∑

n=3

n∑

k=3

hn,kxn−1yk−2

=

∞∑

n=1

n∑

k=1

hn,kxn−1yk−2 −

2∑

n=1

n∑

k=1

hn,kxn−1yk−2 −

∞∑

n=3

2∑

k=1

hn,kxn−1yk−2

=1

y(H − h

1,1− h

2,1x − h

2,2xy −

∞∑

n=3

hn,1xn−1

−

∞∑

n=3

hn,2xn−1y)

=1

y

(

H −

∞∑

n=1

hn,1xn−1

−

∞∑

n=2

hn,2xn−1y

)

=1

y

[

H −

∞∑

n=1

a2cn−1x

n−1−

∞∑

n=2

a2

(

cn−1 − b1cn−2)

xn−1y

]

=1

y(H − a

2C − a

2yC + a

2y + a

2b1xyC),

(19)

∞∑

n=3

n−1∑

k=2

hn−1,kxn−1yk−1 =

∞∑

u=2

u∑

k=2

hu,kxuyk−1 = x

∞∑

n=2

n∑

k=2

hn,kxn−1yk−1

= x

(

∞∑

n=1

n∑

k=1

hn,kxn−1yk−1 −

1∑

n=1

n∑

k=1

hn,kxn−1yk−1 −

∞∑

n=2

1∑

k=1

hn,kxn−1yk−1

)

= x

(

H − h1,1

−

∞∑

n=2

hn,1xn−1

)

= x(H −

∞∑

n=1

hn,1xn−1

) = x(H −

∞∑

n=1

a2cn−1x

n−1)

= x(H − a2C).

(20)

H(x, y) =1

y(H − a

2C − a

2yC + a

2y + a

2b1xyC) + b

1x(H − a

2C)

+ a1xy(H −

∞∑

n=1

hn,nxn−1yn−1) + a

2x2yH + a

2C +

∞∑

n=2

hn,nxn−1yn−1.

Hy = H − a2C + b

1xyH + a

1xy2H + a

2x2y2H + y

∞∑

n=1

hn,nxn−1yn−1(1 − a

1xy).

H =a2C − y

∑∞

n=1 hn,nxn−1yn−1(1 − a

1xy)

1 − y + b1xy + a

1xy2 + a

2x2y2

.

(17)

Page 12 of 12

Liu & Li, Cogent Mathematics (2016), 3: 1200305http://dx.doi.org/10.1080/23311835.2016.1200305

© 2016 The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license.You are free to: Share — copy and redistribute the material in any medium or format Adapt — remix, transform, and build upon the material for any purpose, even commercially.The licensor cannot revoke these freedoms as long as you follow the license terms.

Under the following terms:Attribution — You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. No additional restrictions You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits.

AcknowledgementsThe authors thank the anonymous referees for their constructive comments and helpful suggestions which have greatly improved the original manuscript.

FundingThis work was supported in part by the Domestic Visiting Scholar Program of the Young Teacher of Shandong Province and the Program for Scientific Research Innovation Team in Applied Probability and Statistics of Qufu Normal University [grant number 0230518].

Author detailsLily Li Liu1

E-mail: liulily@mail.qfnu.edu.cnXiaoli Li1

E-mail: 971253130@qq.com1 School of Mathematical Science, Qufu Normal University,

Qufu 273165, P.R. China.

Citation informationCite this article as: Taylor expansions for the generating function of Catalan-like numbers, Lily Li Liu & Xiaoli Li, Cogent Mathematics (2016), 3: 1200305.

ReferencesAigner, M. (1998). Motzkin numbers. European Journal of

Combinatorics, 19, 663–675.Aigner, M. (1999). Catalan-like numbers and determinants.

Journal of Combinatorial Theory, Series A, 87, 33–51.Aigner, M. (2001). Catalan and other numbers—A recurrent

theme. In H. Crapo & D. Senato (Eds.), Algebraic combinatorics and computer science (pp. 347–390). Berlin: Springer.

Cheon, G.-S., Kim, H., & Shapiro, L. W. (2012). Combinatorics of Riordan arrays with identical A and Z sequences. Discrete Mathematics, 312, 2040–2049.

Donaghey, R., & Shapiro, L. W. (1977). Motzkin numbers. Journal of Combinatorial Theory, Series, 23, 291–301.

Ehrenfeucht, A., Harju, T., ten Pas, P., & Rozenberg, G. (1998). Permutations, parenthesis words, and Schröder numbers. Discrete Mathematics, 190, 259–264.

Eu, S.-P., Liu, S.-C., & Yeh, Y.-N. (2002). Taylor expansions for Catalan and Motzkin numbers. Advances in Applied Mathematics, 29, 345–357.

He, T.-X. (2013). Parametric Catalan numbers and Catalan triangles. Linear Algebra and its Applications, 438, 1467–1484.

He, T.-X., & Sprugnoli, R. (2009). Sequence characterization of Riordan arrays. Discrete Mathematics, 309, 3962–3974.

Mahmoud, M., & Qi, F. (2016). Three identities of the Catalan–Qi numbers. Mathematics, 4, Article 35, 7 p. doi:10.3390/math4020035

Qi, F., Shi, X.-T. & Guo, B.-N. (2016a). Integral representations of the large and little Schröder numbers. doi:10.13140/RG.2.1.1988.3288

Qi, F., Shi, X.-T., & Guo, B.-N. (2016b). Two explicit formulas of the Schröder numbers. Integers, 16, Paper No. A23, 15 p.

Shapiro, L. W., Getu, S., Woan, W.-J., & Woodson, L. C. (1991). The Riordan group. Discrete Applied Mathematics, 34, 229–239.

Stanley, R. P. (1997). Hipparchus, Plutarch, Schröder and Hough. The American Mathematical Monthly, 104, 344–350.

Sulanke, R. A. (2001). Bijective recurrences for Motzkin paths. Advances in Applied Mathematics, 27, 627–640.