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We provide two proofs, one of which is combinatorial via partitions... A bijective proof of 1.1 and 1.3 using binary trees and colored ternary trees has been given by Sun [3] himself.. U

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A q-analogue of some binomial coefficient

identities of Y Sun Victor J W Guo1 and Dan-Mei Yang2

Department of Mathematics, East China Normal University

Shanghai 200062, People’s Republic of China

1jwguo@math.ecnu.edu.cn, 2plain dan2004@126.com

Submitted: Dec 1, 2010; Accepted: Mar 24, 2011; Published: Mar 31, 2011

Mathematics Subject Classifications: 05A10, 05A17

Abstract

We give a q-analogue of some binomial coefficient identities of Y Sun [Electron

J Combin 17 (2010), #N20] as follows:

⌊n/2⌋

X

k=0

m + k k



q 2

 m + 1

n− 2k



q

q(

n

−2k

2 ) =m + n

n



q

,

⌊n/4⌋

X

k=0

m + k k



q 4

 m + 1

n− 4k



q

q(

n

−4k

2 ) =⌊n/2⌋X

k=0

(−1)km + k

k



q 2

m + n − 2k

n− 2k



q

,

where n

k



q stands for the q-binomial coefficient We provide two proofs, one of which is combinatorial via partitions

1 Introduction

Using the Lagrange inversion formula, Mansour and Sun [2] obtained the following two binomial coefficient identities:

⌊n/2⌋

X

k=0

1 2k + 1

3k k

n + k 3k



n + 1

2n n



⌊(n−1)/2⌋

X

k=0

1 2k + 1

3k + 1

k + 1

 n + k 3k + 1



n + 1

2n n

 (n > 1) (1.2)

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In the same way, Sun [3] derived the following binomial coefficient identities

⌊n/2⌋

X

k=0

1

3k + a

3k + a k

n + a + k − 1

n − 2k



2n + a

2n + a n



⌊n/4⌋

X

k=0

1 4k + 1

5k k

n + k 5k



=

⌊n/2⌋

X

k=0

(−1)k

n + 1

n + k k

2n − 2k

n



⌊n/4⌋

X

k=0

n + a + 1

4k + a + 1

5k + a k

n + a + k 5k + a



=

⌊n/2⌋

X

k=0

(−1)kn + a + k

k

2n + a − 2k

n + a

 (1.5)

It is not hard to see that both (1.1) and (1.2) are special cases of (1.3), and (1.4) is the a = 0 case of (1.5) A bijective proof of (1.1) and (1.3) using binary trees and colored ternary trees has been given by Sun [3] himself Using the same model, Yan [4] presented

an involutive proof of (1.4) and (1.5), answering a question of Sun

Multiplying both sides of (1.3) by n + a and letting m = n + a − 1, we may write it as

⌊n/2⌋

X

k=0

m + k k

 m + 1

n − 2k



=m + n n



while letting m = n + a, we may write (1.5) as

⌊n/4⌋

X

k=0

m + k k

 m + 1

n − 4k



=

⌊n/2⌋

X

k=0

(−1)km + k

k

m + n − 2k

m



The purpose of this paper is to give a q-analogue of (1.6) and (1.7) as follows:

⌊n/2⌋

X

k=0

m + k

k



q 2

 m + 1

n − 2k



q

q(

n

−2k

2 ) = m + n

n



q

⌊n/4⌋

X

k=0

m + k

k



q 4

 m + 1

n − 4k



q

q(n−4k

2 ) =

⌊n/2⌋

X

k=0

(−1)km + k

k



q 2

m + n − 2k

n − 2k



q

where the q-binomial coefficientx

k



q is defined by

x k



q

=

k

Y

i=1

1 − qx−i+1

1 − qi , if k > 0,

We shall give two proofs of (1.8) and (1.9) One is combinatorial and the other algebraic

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2 Bijective proof of (1.8)

Recall that a partition λ is defined as a finite sequence of nonnegative integers (λ1, λ2, , λr) in decreasing order λ1 > λ2 >· · · > λr A nonzero λi is called a part of λ The number of parts of λ, denoted by ℓ(λ), is called the length of λ Write |λ| = Pm

i=1λi, called the weight of λ The sets of all partitions and partitions into distinct parts are denoted by P and D respectively For two partitions λ and µ, let λ ∪ µ be the partition obtained by putting all parts of λ and µ together in decreasing order

It is well known that (see, for example, [1, Theorem 3.1])

X

λ 1 6m+1 ℓ(λ)=n

q|λ|= qnm + n

n



q

,

X

λ∈D

λ 1 6m+1 ℓ(λ)=n

q|λ|=m + 1

n



q

q(n+12 )

Therefore,

X

µ∈D

λ 1 ,µ 1 6m+1 2ℓ(λ)+ℓ(µ)=n

q2|λ|+|µ| = qn

⌊n/2⌋

X

k=0

m + k k



q 2

 m + 1

n − 2k



q

q(n−2k

2 ),

where k = ℓ(λ) Let

A = {λ ∈ P : λ1 6m + 1 and ℓ(λ) = n},

B = {(λ, µ) ∈ P × D : λ1, µ1 6m + 1 and 2ℓ(λ) + ℓ(µ) = n}

We shall construct a weight-preserving bijection φ from A to B For any λ ∈ A , we associate it with a pair (λ, µ) as follows: If λi appears r times in λ, then we let λi appear

⌊r/2⌋ times in λ and r − 2⌊r/2⌋ times in µ For example, if λ = (7, 5, 5, 4, 4, 4, 4, 2, 2, 2, 1), then λ = (5, 4, 4, 2) and µ = (7, 2, 1) Clearly, (λ, µ) ∈ B and |λ| = 2|λ| + |µ| It is easy

to see that φ : λ 7→ (λ, µ) is a bijection This proves that

X

λ∈A

q|λ|= X

(λ,µ)∈B

q2|λ|+|µ|

Namely, the identity (1.8) holds

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3 Involutive proof of (1.9)

It is easy to see that

qn

⌊n/2⌋

X

k=0

(−1)km + k

k



q 2

m + n − 2k

n − 2k



q

=

⌊n/2⌋

X

k=0

(−1)k X

λ 1 6m+1 ℓ(λ)=k

q2|λ| X

µ 1 6m+1 ℓ(µ)=n−2k

q|µ|

λ 1 ,µ 1 6 m+1 2ℓ(λ)+ℓ(µ)=n

(−1)ℓ(λ)q2|λ|+|µ|, (3.1)

and

qn

⌊n/4⌋

X

k=0

m + k k



q 4

 m + 1

n − 4k



q

q(n−4k

2 ) = X

µ∈D

λ 1 ,µ 1 6m+1 4ℓ(λ)+ℓ(µ)=n

Let

U = {(λ, µ) ∈ P × P : λ1, µ1 6m + 1 and 2ℓ(λ) + ℓ(µ) = n},

V = {(λ, µ) ∈ U : each λi appears an even number of times and µ ∈ D}

We shall construct an involution θ on the set U \ V with the properties that θ preserves 2|λ| + |µ| and reverses the sign (−1)ℓ(λ)

For any (λ, µ) ∈ U \ V , notice that either some λi appears an odd number of times

in λ, or some µj is repeated in µ, or both are true Choose the largest such λi and µj if they exist, denoted by λi 0 and µj 0 respectively Define

θ((λ, µ)) =

( ((λ \ λi 0), µ ∪ (λi 0, λi 0)), if λi 0 >µj 0 or µ ∈ D, ((λ ∪ µj 0), µ \ (µj 0, µj 0)), if λi 0 < µj 0 or λi 0 does not exist

For example, if λ = (5, 5, 4, 4, 4, 3, 3, 3, 1, 1) and µ = (5, 3, 2, 2, 1), then

θ(λ, µ) = ((5, 5, 4, 4, 3, 3, 3, 1, 1), (5, 4, 4, 3, 2, 2, 1))

It is easy to see that θ is an involution on U \ V with the desired properties This proves that

X

(λ,µ)∈U

(−1)ℓ(λ)q2|λ|+|µ| = X

(λ,µ)∈V

(−1)ℓ(λ)q2|λ|+|µ|

µ∈D

τ 1 ,µ 1 6m+1 4ℓ(τ )+ℓ(µ)=n

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4 Generating function proof of (1.8) and (1.9)

Recall that the q-shifted factorial is defined by

(a; q)0 = 1, (a; q)n =

n−1

Y

k=0

(1 − aqk), n = 1, 2,

Then we have

1 (z2; q2)m+1(−z; q)m+1 =

1

1 (z4; q4)m+1

(−z; q)m+1 = 1

(z; q)m+1

1 (−z2; q2)m+1

By the q-binomial theorem (see, for example, [1, Theorem 3.3]), we may expand (4.1) and (4.2) respectively as follows:

X

k=0

m + k k



q 2

z2k

! m+1

X

k=0

m + 1 k



q

q(

k

2)zk

!

=

X

k=0

m + k k



q

zk, (4.3)

X

k=0

m + k k



q 4

z4k

! m+1

X

k=0

m + 1 k



q

q(

k

2)zk

!

=

X

k=0

m + k k



q

zk

! ∞

X

k=0

m + k k



q 2

(−1)kz2k

!

Comparing the coefficients of zn in both sides of (4.3) and (4.4), we obtain (1.8) and (1.9) respectively

Finally, we give the following special cases of (1.8):

⌊n/2⌋

X

k=0

n + k k



q 2

 n + 1 2k + 1



q

q(

n

−2k

2 ) =2n

n



q

⌊n/2⌋

X

k=0

n + k

k + 1



q 2

 n 2k + 1



q

q(n−2k−1

2 ) = 2n

n − 1



q

When q = 1, the identities (4.5) and (4.6) reduce to (1.1) and (1.2) respectively

Acknowledgments This work was partially supported by the Fundamental Research Funds for the Central Universities, Shanghai Rising-Star Program (#09QA1401700), Shanghai Leading Academic Discipline Project (#B407), and the National Science Foun-dation of China (#10801054)

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[1] G E Andrews, The Theory of Partitions, Cambridge University Press, Cambridge, 1998 [2] T Mansour and Y Sun, Bell polynomials and k-generalized Dyck paths, Discrete Appl Math 156 (2008), 2279–2292

[3] Y Sun, A simple bijection between binary trees and colored ternary trees, Electron J Combin 17 (2010), #N20

[4] S H F Yan, Bijective proofs of identities from colored binary trees, Electron J Combin

15 (2008), #N20

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