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Research ArticleRemarks on Homogeneous Al-Salam and Carlitz Polynomials

Jian-Ping Fang

School of Mathematical Science, Huaiyin Normal University, Huaian, Jiangsu 223300, China

Correspondence should be addressed to Jian-Ping Fang; [email protected]

Received 2 April 2014; Accepted 4 July 2014; Published 17 July 2014

Academic Editor: Fawang Liu

Copyright © 2014 Jian-Ping Fang. This is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Several multilinear generating functions of the homogeneous Al-Salam and Carlitz polynomials are derived from 𝑞-operator. Inaddition, two interesting relationships of product of this kind of polynomials are obtained.

1. Introduction

The Al-Salam and Carlitz polynomials

𝜙(𝑎)

𝑛 (𝑥) =

𝑛

∑

𝑘=0

[𝑛

𝑘] (𝑎; 𝑞)

𝑘𝑥𝑘 (1)

have been studied by many researchers for a long time. Thehistory of these polynomials may go back to Al-Salam andCarlitz in 1965. Since then, these polynomials have beenstudied by many mathematicians [1–14].

Recently, Cao [7] used Carlitz’s 𝑞-operators to study thefollowing homogeneous Al-Salam and Carlitz polynomials:

𝜙(𝑎)

𝑛 (𝑥, 𝑦) =

𝑛

∑

𝑘=0

[𝑛

𝑘] (𝑎; 𝑞)

𝑘𝑥𝑘𝑦𝑛−𝑘, 𝑦 ̸= 0 (2)

and he gave some linear generating functions of them. Inthis paper, we will research these polynomials by someconstruction of 𝑞-operator. With this method, some newmultilinear generating functions can be easily derived. Firstly,we give the following three results which originated from theresults about 𝜙(𝑎)𝑛 (𝑥) which appeared in [1, 2, 5–7, 9, 10].

Theorem 1. If 𝑚, 𝑛 ∈ 𝑁, max{|𝑥𝑋𝑧|, |𝑥𝑌𝑧|, |𝑦𝑋𝑧|, |𝑦𝑌𝑧|} <1, then

∞

∑

𝑛=0

𝜙(𝑎)

𝑚+𝑛 (𝑥, 𝑦) 𝜙(𝑏)

𝑛 (𝑋, 𝑌)𝑧𝑛

(𝑞; 𝑞)𝑛

=𝑦𝑚(𝑞; 𝑞)𝑚(𝑏𝑦𝑋𝑧, 𝑎𝑥𝑋𝑧; 𝑞)

∞

(𝑦𝑌𝑧, 𝑦𝑋𝑧, 𝑥𝑋𝑧; 𝑞)∞

×

∞

∑

𝑘=0

(𝑥/𝑦)𝑘(𝑎, 𝑦𝑌𝑧, 𝑦𝑋𝑧; 𝑞)

𝑘

(𝑞, 𝑏𝑦𝑋𝑧, 𝑎𝑥𝑋𝑧; 𝑞)𝑘(𝑞; 𝑞)𝑚−𝑘

× 3Φ2(𝑎𝑞𝑘, 𝑦𝑋𝑧𝑞

𝑘,

𝑏𝑋

𝑌

𝑏𝑦𝑋𝑧𝑞𝑘, 𝑎𝑥𝑋𝑧𝑞

𝑘; 𝑞, 𝑥𝑌𝑧) .

(3)

Theorem 2. If 𝑚, 𝑛 ∈ 𝑁, max{|𝑥𝑡|, |𝑦𝑡|, |𝑋𝑧|, |𝑌𝑧|, |𝑦𝑌𝛾|,|𝑦𝑋𝛾|, |𝑥𝑌𝛾|} < 1, then

∞

∑

𝑚,𝑛,𝑘=0

𝜙(𝑎)

𝑚+𝑘(𝑥, 𝑦) 𝜙

(𝑏)

𝑛+𝑘(𝑋, 𝑌)

𝑡𝑚𝑧𝑛𝛾𝑘

(𝑞; 𝑞)𝑚(𝑞; 𝑞)𝑛(𝑞; 𝑞)𝑘

=(𝑏𝑋𝑧, 𝑎𝑥𝑌𝛾; 𝑞)

∞

(𝑌𝑧,𝑋𝑧, 𝑦𝑡, 𝑦𝑌𝛾, 𝑥𝑌𝛾; 𝑞)∞

×

∞

∑

𝑚=0

(𝑦𝑋𝛾)𝑚(𝑏, 𝑌𝑧; 𝑞)

𝑚

(𝑏𝑋𝑧; 𝑞)𝑚

Hindawi Publishing CorporationJournal of MathematicsVolume 2014, Article ID 523013, 12 pageshttp://dx.doi.org/10.1155/2014/523013

2 Journal of Mathematics

×

∞

∑

𝑘=0

(𝑥/𝑦)𝑘(𝑎, 𝑦𝑡, 𝑦𝑌𝛾; 𝑞)

𝑘

(𝑞, 𝑎𝑥𝑌𝛾; 𝑞)𝑘(𝑞; 𝑞)𝑚−𝑘

× 2Φ1 (𝑎𝑞𝑘, 𝑦𝑌𝛾𝑞

𝑘

𝑎𝑥𝑌𝛾𝑞𝑘 ; 𝑞, 𝑥𝑡) .

(4)

Theorem 3 (cf. [7, Equation (1.9)]). If 𝑘,𝑚, 𝑛 ∈ 𝑁,max{|𝑥𝑌𝑧|, |𝑥𝑋𝑧|, |𝑋𝑦𝑧|, |𝑌𝑦𝑧|} < 1, then

∞

∑

𝑘=0

𝜙(𝑎)

𝑛+𝑘(𝑥, 𝑦) 𝜙

(𝑏)

𝑚+𝑘(𝑋, 𝑌)

𝑧𝑘

(𝑞; 𝑞)𝑘

=𝑦𝑛𝑌𝑚(𝑞; 𝑞)𝑛(𝑞; 𝑞)𝑚(𝑦𝑏𝑋𝑧, 𝑥𝑎𝑋𝑧; 𝑞)

∞


×

∞

∑

𝑘=0

(𝑋/𝑌)𝑘(𝑏, 𝑦𝑌𝑧; 𝑞)

𝑘

(𝑞, 𝑦𝑏𝑋𝑧; 𝑞)𝑘(𝑞; 𝑞)𝑚−𝑘

×

∞

∑

𝑗=0

(𝑥/𝑦)𝑗(𝑎, 𝑦𝑌𝑧𝑞

𝑘, 𝑦𝑋𝑧; 𝑞)

𝑗

(𝑞, 𝑦𝑏𝑋𝑧𝑞𝑘, 𝑥𝑎𝑋𝑧; 𝑞)𝑗(𝑞; 𝑞)𝑛−𝑗

× 3Φ2(𝑎𝑞𝑗, 𝑦𝑋𝑧𝑞

𝑗,

𝑏𝑋

𝑌

𝑏𝑦𝑋𝑧𝑞𝑘+𝑗, 𝑎𝑥𝑋𝑧𝑞

𝑗; 𝑞, 𝑥𝑌𝑧𝑞

𝑘) .

(5)

Now further using our method, we can deduce moreresults of multilinear generating functions.

Theorem 4. If 𝐺 ∈ 𝑍+, 𝑎1 = 𝑞−𝐺, |𝑎1𝑥1/𝑦1| < 1, then

∞

∑

𝑚1,𝑚2,𝑚3=0

𝜙(𝑎1)

𝑚1+𝑚2+𝑚3

(𝑥1, 𝑦1) 𝜙(𝑎2)

𝑚1+𝑚3

(𝑥2, 𝑦2)

× 𝜙(𝑎3)

𝑚1+𝑚2

(𝑥3, 𝑦3)𝑡𝑚1

1 𝑡𝑚2

2 𝑡𝑚3

3

(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

=(𝑎1𝑥1/𝑦1, 𝑎3𝑥3𝑡2𝑦1, 𝑎2𝑥2𝑡1𝑦1𝑦3; 𝑞)∞

(𝑥1/𝑦1, 𝑡2𝑦1𝑦3, 𝑡3𝑦1𝑦2, 𝑡1𝑦1𝑦2𝑦3, 𝑥3𝑡2𝑦1, 𝑥2𝑡1𝑦1𝑦3; 𝑞)∞

×

∞

∑

𝑗1=0

(𝑎3; 𝑞)𝑗1

(𝑥3𝑡1𝑦1𝑦2)𝑗1

(𝑞; 𝑞)𝑗1

(𝑞; 𝑞)𝑗1

×

∞

∑

𝑗2=0

(𝑥2/𝑦2)𝑗2

(𝑎2; 𝑞)𝑗2

(𝑞; 𝑞)𝑗2

(𝑞; 𝑞)𝑗1−𝑗2

∞

∑

𝑗3=0

(𝑥2𝑡3𝑦1)𝑗3

(𝑎2𝑞𝑗2 ; 𝑞)𝑗3

(𝑞; 𝑞)𝑗3

×

∞

∑

𝑗4=0

(𝑎1, 𝑥3𝑡2𝑦1, 𝑥2𝑡1𝑦1𝑦3; 𝑞)𝑗4

(−1)𝑗4𝑞−(𝑗4−1)𝑗4/2

(𝑞, 𝑥1/ (𝑦1𝑞𝑗4) ; 𝑞)𝑗4

×(𝑥1𝑞𝑗1+𝑗3

𝑦1

)

𝑗4 (𝑡3𝑦1𝑦2; 𝑞)𝑗

2+𝑗4

(𝑡2𝑦1𝑦3; 𝑞)𝑗1+𝑗4

(𝑎3𝑥3𝑡2𝑦1; 𝑞)𝑗1+𝑗4

×

(𝑡1𝑦1𝑦2𝑦3; 𝑞)𝑗2+𝑗3+𝑗4

(𝑎2𝑥2𝑡1𝑦1𝑦3; 𝑞)𝑗2+𝑗3+𝑗4

,

(6)

provided that max{|𝑡𝑖𝑥𝑗𝑦𝑙|, |𝑡1𝑥𝑖𝑦𝑗𝑦𝑙|, |𝑡1𝑦1𝑦2𝑦3|} < 1, where1 ≤ 𝑖 ̸= 𝑗 ̸= 𝑙 ≤ 3.

Theorem 5. If 𝐺 ∈ 𝑍+, 𝑎1 = 𝑞−𝐺, then

∞

∑

𝑚1,𝑚2,𝑚3=0

𝜙(𝑎1)

𝑚1+𝑚2+𝑚3

(𝑥1, 𝑦1) 𝜙(𝑎2)

𝑚1+𝑚2

(𝑥2, 𝑦2)

× 𝜙(𝑎3)

𝑚3

(𝑥3, 𝑦3)𝑡𝑚1

1 𝑡𝑚2

2 𝑡𝑚3

3

(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

=(𝑎1𝑥1/𝑦1, 𝑎3𝑥3𝑡3𝑦1, 𝑎2𝑥2𝑡2𝑦1; 𝑞)∞

(𝑥1/𝑦1, 𝑡3𝑦1𝑦3, 𝑡3𝑦1𝑥3, 𝑡1𝑦1𝑦2, 𝑡2𝑦1𝑦2, 𝑡2𝑥2𝑦1; 𝑞)∞

×

∞

∑

𝑗1,𝑗2=0

(𝑎2; 𝑞)𝑗1

(𝑡1𝑥2𝑦1)𝑗1

(−1)𝑗2𝑞−(𝑗2−1)𝑗2/2

(𝑞; 𝑞)𝑗1

×

(𝑎1, 𝑡3𝑦3𝑦1, 𝑡3𝑥3𝑦1, 𝑡2𝑥2𝑦1, 𝑡1𝑦2𝑦1; 𝑞)𝑗2

(𝑞, 𝑞𝑥1/ (𝑦1𝑞𝑗2) , 𝑎3𝑡2𝑥3𝑦1; 𝑞)𝑗

2

×

(𝑡2𝑦2𝑦1; 𝑞)𝑗1+𝑗2

(𝑎2𝑡2𝑥2𝑦1; 𝑞)𝑗1+𝑗2

(𝑞𝑗1𝑥1

𝑦1

)

𝑗2

,

(7)

provided that max{|𝑦1𝑦2𝑡1|, |𝑦1𝑦2𝑡2|, |𝑦1𝑦3𝑡3|, |𝑥3𝑦1𝑡3|,|𝑥2𝑦1𝑡1|, |𝑥1𝑦1𝑡2|, |𝑎1𝑥1/𝑦1|} < 1.

Theorem 6. If𝑚𝑖 ∈ 𝑁,max{|𝑥1|, |𝑦1𝑡𝑖|, |𝑥1𝑡𝑖|} < 1, 𝑖 = 1, 2, 3,then

∞

∑

𝑚1,𝑚2,𝑚3=0

𝜙(𝑎)

𝑚1+𝑚2+𝑚3

(𝑥1, 𝑦1)

×𝑡𝑚1

1 𝑡𝑚2

2 (−𝑡3)𝑚3

𝑞𝑚3(𝑚3−1)/2

(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

=(𝑦1𝑡3, 𝑎𝑥1𝑡2; 𝑞)∞

(𝑦1𝑡1, 𝑦1𝑡2, 𝑥1𝑡2; 𝑞)∞

× 3Φ2(𝑎, 𝑦1𝑡2,

𝑡3

𝑡1𝑦1𝑡3, 𝑎𝑥1𝑡2

; 𝑞, 𝑥1𝑡1) .

(8)

Journal of Mathematics 3

Theorem 7. If 𝑚𝑖 ∈ 𝑁, 𝑛, 𝐺 ∈ 𝑍+, 𝑎 = 𝑞−𝐺, max{|𝑦𝑡𝑖|,|𝑎𝑥/𝑦|} < 1, 𝑖 = 1, 2, . . . , 𝑛, then

∞

∑

𝑚1,...,𝑚𝑛=0

𝜙(𝑎)

𝑚1+⋅⋅⋅+𝑚

𝑛

(𝑥, 𝑦)𝑡𝑚1

1 ⋅ ⋅ ⋅ 𝑡𝑚𝑛

𝑛

(𝑞; 𝑞)𝑚1

⋅ ⋅ ⋅ (𝑞; 𝑞)𝑚𝑛

=(𝑎𝑥/𝑦; 𝑞)

∞

(𝑥/𝑦, 𝑦𝑡1, 𝑦𝑡2, . . . , 𝑦𝑡𝑛; 𝑞)∞

×

∞

∑

𝑗=0

(𝑎, 𝑦𝑡1, 𝑦𝑡2, . . . , 𝑦𝑡𝑛; 𝑞)𝑗(−1)𝑗𝑞−(𝑗−1)𝑗/2

(𝑞, 𝑥/ (𝑦𝑞𝑗) ; 𝑞)𝑗

(𝑥

𝑦)

𝑗

.

(9)

Theorem 8. If 𝑚𝑖 ∈ 𝑁, 𝑡𝑖 ̸= 0, max{|𝑥|, |𝑦𝑡𝑖|, |𝑥𝑡𝑖|} < 1, 𝑖 =1, 2, 3, 4, then

∞

∑

𝑚1,𝑚2,𝑚3,𝑚4=0

𝜙(𝑎)

𝑚1+𝑚2+𝑚3+𝑚4

(𝑥, 𝑦)

×(−𝑡1)𝑚1

𝑡𝑚2

2 𝑡𝑚3

3 (−𝑡4)𝑚4

𝑞𝑚1(𝑚1−1)/2

𝑞𝑚4(𝑚4−1)/2

(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

(𝑞; 𝑞)𝑚4

=(𝑦𝑡1, 𝑦𝑡4; 𝑞)∞

(𝑦𝑡2, 𝑦𝑡3; 𝑞)∞

∞

∑

𝑗=0

(𝑎, 𝑡1/𝑡2, 𝑦𝑡3; 𝑞)𝑗(𝑥𝑡2)𝑗

(𝑞, 𝑦𝑡1, 𝑦𝑡4; 𝑞)𝑗

× 2Φ1(𝑎𝑞𝑗,

𝑡4

𝑡3

𝑦𝑡4𝑞𝑗; 𝑞, 𝑥𝑡3) .

(10)

Theorem 9. If 𝑚𝑖 ∈ 𝑁, max{|𝑦𝑖𝑦𝑡𝑖|, |𝑥𝑖𝑦𝑡𝑖|, |𝑎0/𝑥0|} < 1, 𝑖 =0, 1, 2, 3, then

∞

∑

𝑚,𝑚1,𝑚2,𝑚3=0

𝜙(𝑎)

𝑚1+𝑚2+𝑚3+𝑠 (𝑥, 𝑦) 𝜙

(𝑎)

𝑚+𝑟

× (𝑥0, 𝑦0) 𝜙(𝑎1)

𝑚1

(𝑥1, 𝑦1) 𝜙(𝑎2)

𝑚2

(𝑥2, 𝑦2)

× 𝜙(𝑎3)

𝑚3

(𝑥3, 𝑦3)

×𝑡𝑚𝑡1𝑚1𝑡𝑚2

2 𝑡𝑚3

3

(𝑞; 𝑞)𝑚(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

= (𝑦𝑠𝑦𝑟

0(𝑞; 𝑞)𝑟(𝑎𝑥

𝑦, 𝑎1𝑦𝑥1𝑡1, 𝑎2𝑦𝑥2𝑡2, 𝑎3𝑦𝑥3𝑡3; 𝑞)

∞

)

× ((𝑥

𝑦, 𝑦𝑦1𝑡1, 𝑦𝑥1𝑡1, 𝑦𝑦2𝑡2, 𝑦𝑥2𝑡2,

𝑦𝑦3𝑡3, 𝑦𝑥3𝑡3, 𝑦𝑦0𝑡; 𝑞)

∞

)

−1

×

∞

∑

𝑗1,𝑗2=0

(𝑥0/𝑦0)𝑗1

(𝑦𝑥0𝑡)𝑗2

(𝑎0; 𝑞)𝑗1+𝑗2

(𝑞; 𝑞)𝑗1

(𝑞; 𝑞)𝑗2

(𝑞; 𝑞)𝑟−𝑗1

×

∞

∑

𝑗3=0

((𝑎, 𝑦𝑦1𝑡1, 𝑦𝑥1𝑡1, 𝑦𝑦2𝑡2, 𝑦𝑥2𝑡2, 𝑦𝑦3𝑡3, 𝑦𝑥3𝑡3; 𝑞)𝑗3

× (𝑦𝑦0𝑡; 𝑞)𝑗1+𝑗3

)

× ((𝑞𝑦

𝑥, 𝑎1𝑦𝑥1𝑡1, 𝑎2𝑦𝑥2𝑡2, 𝑎3𝑦𝑥3𝑡3; 𝑞)

𝑗3

)

−1

× 𝑞𝑗3(𝑗2+𝑠+1)

,

(11)

where 𝑡0 = 𝑡,𝑚0 = 𝑚, and 𝐹 ∈ 𝑍+, 𝑎0 = 𝑞−𝐹.

Theorem 10. If𝑀𝑖 ∈ 𝑍+, 𝑎𝑖 = 𝑞−𝑀𝑖 , 𝑥𝑖 ̸= 0, and 𝑦𝑖 ̸= 0, then

∞

∑

𝑚1,𝑚2,𝑚3,𝑚4=0

𝜙(𝑎1)

𝑚2+𝑚3+𝑚4

(𝑥1, 𝑦1)

× 𝜙(𝑎2)

1+𝑚3+𝑚4

(𝑥2, 𝑦2) 𝜙(𝑎3)

𝑚1+𝑚2+𝑚4

× (𝑥3, 𝑦3) 𝜙(𝑎4)

𝑚1+𝑚2+𝑚3

(𝑥4, 𝑦4)

×(−𝑡1)𝑚1

𝑡𝑚2

2 𝑡𝑚3

3 (−𝑡4)𝑚4

𝑞𝑚1(𝑚1−1)/2

𝑞𝑚4(𝑚4−1)/2

(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

(𝑞; 𝑞)𝑚4

= (𝑎4𝑥4

𝑦4

,𝑎3𝑥3

𝑦3

,𝑎2𝑥2

𝑦2

,𝑎1𝑥1

𝑦1

, 𝑡1𝑦2𝑦3𝑦4, 𝑡4𝑦1𝑦2𝑦3; 𝑞)

∞

× ((𝑥4

𝑦4

,𝑥3

𝑦3

,𝑥2

𝑦2

,𝑥1

𝑦1

, 𝑡2𝑦1𝑦3𝑦4, 𝑡3𝑦1𝑦2𝑦4; 𝑞)

∞

)

−1

×

∞

∑

𝑘1,𝑘2,𝑘3,𝑘4=0

((𝑎4; 𝑞)𝑘4

(𝑎3; 𝑞)𝑘3

(𝑎2; 𝑞)𝑘2

× (𝑎1; 𝑞)𝑘1

𝑞𝑘1+𝑘2+𝑘3+𝑘4)

× ((𝑞,𝑞𝑦4

𝑥4

; 𝑞)

𝑘4

(𝑞,𝑞𝑦3

𝑥3

; 𝑞)

𝑘3

×(𝑞,𝑞𝑦2

𝑥2

; 𝑞)

𝑘2

(𝑞,𝑞𝑦1

𝑥1

; 𝑞)

𝑘1

)

−1

×

(𝑡1𝑦2𝑦3𝑦4; 𝑞)𝑘2+𝑘3+𝑘4

(𝑡4𝑦1𝑦2𝑦3; 𝑞)𝑘1+𝑘2+𝑘3

(𝑡2𝑦1𝑦3𝑦4; 𝑞)𝑘1+𝑘3+𝑘4

(𝑡3𝑦1𝑦2𝑦4; 𝑞)𝑘1+𝑘2+𝑘4

,

(12)

provided thatmax{|𝑎𝑖𝑥𝑖/𝑦𝑖|, |𝑡𝑖𝑦𝑗𝑦𝑘𝑦𝑙|} < 1, where 1 ≤ 𝑖 ̸= 𝑗 ̸=𝑘 ̸= 𝑙 ≤ 4.


Polynomials (2) evidently reduce to the Rogers-Szegöpolynomials (cf. [9])

𝜙(0)

𝑛 (𝑥) = 𝐻𝑛 (𝑥) =

𝑛

∑

𝑘=0

[𝑛

𝑘] 𝑥𝑘 (13)

when 𝑎 = 0 and 𝑦 = 1. And when 𝑦 = 1 they reduce to thecommon Al-Salam and Carlitz polynomials (1) (cf. [9]).

So now we take some special cases for checking.Let 𝑎 = 𝑏 = 0, 𝑥 = 𝑌 = 1 in (3); we have the following.

Corollary 11 (cf. [6, Theorem 1.1] or [9, Equation (4.1)]). If𝑚, 𝑛 ∈ 𝑁, then

∞

∑

𝑛=0

𝐻𝑚+𝑛 (𝑦)𝐻𝑛 (𝑋)𝑧𝑛

(𝑞; 𝑞)𝑛

=

𝑦𝑚(𝑦𝑋𝑧2; 𝑞)∞

(𝑧, 𝑋𝑧, 𝑦𝑧, 𝑦𝑋; 𝑞)∞

× 3Φ1 (𝑞−𝑚, 𝑦𝑧, 𝑦𝑋𝑧

𝑦𝑋𝑧2 ; 𝑞,

𝑞𝑚

𝑦) ,

(14)

provided thatmax{|𝑧|, |𝑦𝑧|, |𝑋𝑧|, |𝑦𝑋𝑧|} < 1.

Remark 12 (from [6, Equation (3.1)]). We know thatCorollary 11 is equivalent to Theorem 1.1 given in [6]. And ifwe take𝑚 = 0 and 𝑦 = 𝑌 = 1, (3) turns to [9, Equation (1.2)].

Taking 𝑎 = 𝑏 = 0, 𝑥 = 𝑌 = 1 in (4), we have the following.

Corollary 13 (cf. [6, Theorem 1.2] or [9, Equation (1.3)]). If𝑚, 𝑛 ∈ 𝑁, then

∞

∑

𝑚,𝑛,𝑘=0

𝐻𝑚+𝑘 (𝑦)𝐻𝑛+𝑘 (𝑋)𝑡𝑚𝑧𝑛𝛾𝑘

(𝑞; 𝑞)𝑚(𝑞; 𝑞)𝑛(𝑞; 𝑞)𝑘

=(𝑦𝑡𝛾, 𝑦𝑋𝑧𝛾; 𝑞)

∞

(𝑡, 𝑧, 𝑋𝑧, 𝑦𝑡, 𝛾, 𝑦𝛾, 𝑦𝑋𝛾; 𝑞)∞

× 3Φ2 (𝑧, 𝑦𝑡, 𝑦𝛾

𝑦𝑡𝛾, 𝑦𝑋𝑧𝛾; 𝑞, 𝑋𝛾) ,

(15)

provided that max{|𝑡|, |𝑧|, |𝛾|, |𝑦𝑡|, |𝑦𝛾|, |𝑋𝑧|, |𝑋𝛾|, |𝑦𝑋𝛾|} <1.

Remark 14. UsingHall’s transformation [15, Equation (3.2.7)]

3Φ2 (𝑎, 𝑏, 𝑐

𝑑, 𝑒; 𝑞,

𝑑𝑒

(𝑎𝑏𝑐))

=(𝑒/𝑎, 𝑑𝑒/𝑏𝑐; 𝑞)

∞

(𝑒, 𝑑𝑒/𝑎𝑏𝑐; 𝑞)∞

3Φ2(

𝑎,𝑑

𝑏,

𝑑

𝑐

𝑑,𝑑𝑒

(𝑏𝑐)

; 𝑞,𝑒

𝑎) ,

(16)

we find that Corollary 13 is equivalent to Theorem 1.2 givenin [6]. And if we take 𝑦 = 𝑌 = 1, then, with simplifying, (4)turns to [9, Equation (3.5)].

Taking 𝑎 = 𝑏 = 0 and 𝑥 = 𝑋 = 1 in (5), we have thefollowing.

Corollary 15 (cf. [6, Theorem 1.3] or [9, Equation (1.4)]). If𝑚, 𝑛, 𝑘 ∈ 𝑁, then

∞

∑

𝑘=0

𝐻𝑛+𝑘 (𝑦)𝐻𝑚+𝑘 (𝑌)𝑧𝑘

(𝑞; 𝑞)𝑘

=

(𝑦𝑌𝑧2; 𝑞)∞

(𝑧, 𝑦𝑧, 𝑌𝑧, 𝑦𝑌𝑧; 𝑞)∞

×

𝑚

∑

𝑘=0

𝑛

∑

𝑗=0

[𝑚

𝑘] [𝑚

𝑗]

𝑌𝑚−𝑘

𝑦𝑛−𝑗(𝑦𝑌𝑧; 𝑞)

𝑘+𝑗

(𝑦𝑌𝑧2; 𝑞)𝑘+𝑗

× (𝑌𝑧; 𝑞)𝑘(𝑦𝑧; 𝑞)

𝑗,

(17)

provided thatmax{|𝑧|, |𝑦𝑧|, |𝑌𝑧|, |𝑦𝑌𝑧|} < 1.

The rest of this paper is organized as follows. In Section 2,we will give some notations and lemmas. In Section 3, wegive the proofs of theorems. Section 4 describes the relation-ship between Ψ𝑛 (𝑎, 𝑏; 𝑥1, 𝑦1; 𝑧) (cf. [4, Equation (2.2)]) and𝜙(𝑎)𝑛 (𝑥, 𝑦). In addition, an interesting relationship between the

polynomial multiplications is given.

2. Notations and Some Lemmas

In this paper, we apply the standard notations that followfrom [15] and assume that 0 < |𝑞| < 1; the 𝑞-shifted factorialand its compact factorials are defined, respectively, by

(𝑎; 𝑞)0= 1;

(𝑎; 𝑞)𝑛=

𝑛−1

∏

𝑘=0

(1 − 𝑎𝑞𝑘) ,

(𝑎; 𝑞)∞=

∞

∏

𝑘=0

(1 − 𝑎𝑞𝑘)

(18)

and (𝑎1, . . . , 𝑎𝑚; 𝑞)𝑛 = (𝑎1; 𝑞)𝑛, . . . , (𝑎𝑚; 𝑞)𝑛, where 𝑛 =0, 1, . . . ,∞. And we use 𝑁 to denote the set of nonnegativeintegers and 𝑍+ to denote the set of positive integers.

For any complex number 𝛼, we have

(𝑎; 𝑞)𝛼=

(𝑎; 𝑞)∞

(𝑎𝑞𝛼; 𝑞)∞

. (19)

The 𝑞-binomial coefficient and the 𝑞-binomial theoremare given by

[𝑛

𝑘] =

(𝑞; 𝑞)𝑛

(𝑞; 𝑞)𝑘(𝑞; 𝑞)𝑛−𝑘

,

∞

∑

𝑚=0

(𝑎; 𝑞)𝑚

(𝑞; 𝑞)𝑚

𝑦𝑚=(𝑎𝑦; 𝑞)

∞

(𝑦; 𝑞)∞

,

(20)

respectively.


The basic hypergeometric series 𝑟Φ𝑠 is defined by

𝑟Φ𝑠 (𝑎1, . . . , 𝑎𝑟

𝑏1, . . . , 𝑏𝑠; 𝑞, 𝑥)

=

∞

∑

𝑛=0

(𝑎1, 𝑎2, . . . , 𝑎𝑟; 𝑞)𝑛

(𝑞, 𝑏1, . . . , 𝑏𝑠; 𝑞)𝑛

× [(−1)𝑛𝑞𝑛(𝑛−1)/2

]1+𝑠−𝑟

𝑥𝑛.

(21)

The 𝑞-derivative operator𝐷𝑞 and 𝑞-shifted operator 𝜂 (cf.[1, 6, 7, 9, 10, 16–23]), acting on the variable 𝑥, are defined by

𝐷𝑞 {𝑓 (𝑥)} =𝑓 (𝑥) − 𝑓 (𝑥𝑞)

𝑥,

𝜂 {𝑓 (𝑥)} = 𝑓 (𝑥𝑞) .

(22)

The 𝑞-Leibnitz rule for the product of two functions (cf.[17–25]) is given by

𝐷𝑛

𝑞 {𝑓 (𝑥) 𝑔 (𝑥)} =

𝑛

∑

𝑘=0

[𝑛

𝑘] 𝑞𝑘(𝑘−𝑛)

𝐷𝑘

𝑞 {𝑓 (𝑥)}𝐷𝑛−𝑘

𝑞 {𝑔 (𝑥𝑞𝑘)} .

(23)

From definition, we can easily obtain𝐷𝑞 = 𝑥−1(1 − 𝜂); by

mathematical induction, we have the following proposition.

Proposition 16 (cf. [20, Equation (6)]). If 𝑓(𝑥) is any analyt-ical function, then

𝐷𝑛

𝑞 {𝑓 (𝑥)} = 𝑥−𝑛𝑛

∑

𝑘=0

(𝑞−𝑛; 𝑞)𝑘

(𝑞; 𝑞)𝑘

(𝑞𝜂)𝑘{𝑓 (𝑥)} . (24)

In [19, 20], we had established the following 𝑞-operatorstructure:

1Φ0 (𝑏

−; 𝑞, 𝑐𝐷𝑞) =

∞

∑

𝑛=0

(𝑏; 𝑞)𝑛(𝑐𝐷𝑞)

𝑛

(𝑞; 𝑞)𝑛

. (25)

For convenient, we use 𝐹(𝑎; 𝑥𝐷𝑞)𝑦 to denote the operator

1Φ0 (𝑎− ; 𝑞, 𝑥𝐷𝑞) acting on variable 𝑦. From this definition we

can easily get the following Lemma.

Lemma 17. One has𝜙(𝑎)

𝑛 (𝑥, 𝑦) = 𝐹(𝑎; 𝑥𝐷𝑞)𝑦{𝑦𝑛} . (26)

Proof. For

𝐷(𝑘)

𝑞 {𝑦𝑛} =

(𝑞; 𝑞)𝑛

(𝑞; 𝑞)𝑛−𝑘

𝑦𝑛−𝑘, (27)

we have𝐹(𝑎; 𝑥𝐷𝑞)𝑦

{𝑦𝑛}

=

∞

∑

𝑘=0

(𝑎; 𝑞)𝑘𝑥𝑘

(𝑞; 𝑞)𝑘

(𝑞; 𝑞)𝑛


𝑦𝑛−𝑘

=

𝑛

∑

𝑘=0

[𝑛

𝑘] (𝑎; 𝑞)

𝑘𝑥𝑘𝑦𝑛−𝑘,

(28)

which completes the proof.

It is the reason why we employ this operator to study theproperties of homogeneous Al-Salam and Carlitz polynomi-als.

From Proposition 16, we may easily obtain the followingidentity.

Lemma 18. If 𝐺 ∈ 𝑍+, 𝑎 = 𝑞−𝐺, |𝑎𝑥/𝑦| < 1 and 𝑓(𝑥) is anyanalytical function, then

𝐹(𝑎; 𝑥𝐷𝑞)𝑦{𝑓 (𝑦)}


∞

(𝑥/𝑦; 𝑞)∞

∞

∑

𝑘=0

(−1)𝑘(𝑎; 𝑞)𝑘(𝑥/𝑦)

𝑘

(𝑞, 𝑥/ (𝑦𝑞𝑘) ; 𝑞)𝑘

× 𝑞−𝑘(𝑘−1)/2

𝑓 (𝑦𝑞𝑘) .

(29)

Proof. Applying Proposition 16, LHS of Lemma 18 comes to

∞

∑

𝑛=0

(𝑎; 𝑞)𝑛(𝑥/𝑦)

𝑛

(𝑞; 𝑞)𝑛

×

𝑛

∑

𝑘=0

(𝑞; 𝑞)𝑛𝑞𝑘


(−1)𝑘𝑞𝑘(𝑘−1)/2−𝑛𝑘

𝑓 (𝑦𝑞𝑘)

=

∞

∑

𝑘=0

(−1)𝑘(𝑎; 𝑞)𝑘(𝑥/𝑦)

𝑘

(𝑞; 𝑞)𝑘

𝑞−𝑘(𝑘−1)/2

𝑓 (𝑦𝑞𝑘)

×

∞

∑

𝑛=0

(𝑎𝑞𝑘; 𝑞)𝑛

(𝑞; 𝑞)𝑛

(𝑥

𝑦𝑞𝑘)

𝑛

=

∞

∑

𝑘=0

(−1)𝑘(𝑎; 𝑞)𝑘(𝑥/𝑦)

𝑘

(𝑞; 𝑞)𝑘

𝑞−𝑘(𝑘−1)/2

𝑓 (𝑦𝑞𝑘)

×(𝑎𝑥/𝑦; 𝑞)

∞

(𝑥/ (𝑦𝑞𝑘) ; 𝑞)∞


∞

(𝑥/𝑦; 𝑞)∞

∞

∑

𝑘=0

(−1)𝑘(𝑎; 𝑞)𝑘(𝑥/𝑦)

𝑘

(𝑞, 𝑥/ (𝑦𝑞𝑘) ; 𝑞)𝑘

× 𝑞−𝑘(𝑘−1)/2

𝑓 (𝑦𝑞𝑘) .

(30)

We complete the proof.

Remark 19. To calculate the inner sum of the above, underthe condition 𝑎 = 𝑞−𝐺, 𝑞-binomial theorem is usable.

Lemma 20 (cf. [20, Lemma 1]). If |𝑠𝑐| < 1, |𝑡𝑐| < 1, then

𝐹(𝑏; 𝑐𝐷𝑞)𝑎{(𝑎𝜔; 𝑞)

∞

(𝑎𝑠, 𝑎𝑡; 𝑞)∞

}

=(𝑎𝜔, 𝑏𝑐𝑡; 𝑞)

∞

(𝑎𝑠, 𝑎𝑡, 𝑐𝑡; 𝑞)∞

3Φ2(𝑏, 𝑎𝑡,

𝜔

𝑠𝑎𝜔, 𝑏𝑐𝑡

; 𝑞, 𝑠𝑐) .

(31)


Proof. LHS of Lemma 20 equates to

∞

∑

𝑛=0

(𝑏; 𝑞)𝑛𝑐𝑛

(𝑞; 𝑞)𝑛

𝑛

∑

𝑘=0

[𝑛


𝐷𝑘

𝑞 {(𝑎𝜔; 𝑞)

∞

(𝑎𝑠; 𝑞)∞

}

× 𝐷𝑛−𝑘

𝑞 {1

(𝑎𝑡𝑞𝑘; 𝑞)∞

}

=

∞

∑

𝑛=0


(𝑞; 𝑞)𝑛

×

𝑛

∑

𝑘=0

[𝑛


𝑠𝑘(𝜔/𝑠; 𝑞)

𝑘(𝑎𝜔𝑞𝑘; 𝑞)∞

(𝑎𝑠; 𝑞)∞

(𝑡𝑞𝑘)𝑛−𝑘

(𝑎𝑡𝑞𝑘; 𝑞)∞

=(𝑎𝜔; 𝑞)

∞


∞

∑

𝑘=0

(𝑎𝑡, 𝜔/𝑠; 𝑞)𝑘

(𝑞, 𝑎𝜔; 𝑞)𝑘

𝑠𝑘

×

∞

∑

𝑛=𝑘

(𝑏; 𝑞)𝑛


𝑐𝑛𝑡𝑛−𝑘

=(𝑎𝜔; 𝑞)

∞


∞

∑

𝑘=0

(𝑎𝑡, 𝜔/𝑠; 𝑞)𝑘


(𝑠𝑐)𝑘

×

∞

∑

𝑚=0

(𝑏; 𝑞)𝑚+𝑘

(𝑞; 𝑞)𝑚

(𝑡𝑐)𝑚

=(𝑎𝜔; 𝑞)

∞


∞

∑

𝑘=0

(𝑎𝑡, 𝜔/𝑠, 𝑏; 𝑞)𝑘


(𝑠𝑐)𝑘

×

∞

∑

𝑚=0

(𝑏𝑞𝑘; 𝑞)𝑚

(𝑞; 𝑞)𝑚

(𝑡𝑐)𝑚

=(𝑎𝜔; 𝑞)

∞


∞

∑

𝑘=0

(𝑎𝑡, 𝜔/𝑠, 𝑏; 𝑞)𝑘


(𝑠𝑐)𝑘(𝑏𝑐𝑡𝑞𝑘; 𝑞)∞

(𝑐𝑡; 𝑞)∞

=(𝑎𝜔, 𝑏𝑐𝑡; 𝑞)

∞


3Φ2(𝑏, 𝑎𝑡,

𝜔

𝑠𝑎𝜔, 𝑏𝑐𝑡

; 𝑞, 𝑠𝑐) .

(32)


The special cases of Lemma 20 as taking 𝑡 = 0, 𝜔 = 𝑡 = 0,and 𝜔 = 0, respectively, will be frequently used.

Lemma 21. For |𝑠𝑐| < 1,

𝐹(𝑏; 𝑐𝐷𝑞)𝑎{(𝑎𝜔; 𝑞)

∞

(𝑎𝑠; 𝑞)∞

} =(𝑎𝜔; 𝑞)

∞

(𝑎𝑠; 𝑞)∞

2Φ1(𝑏,

𝜔

𝑠𝑎𝜔

; 𝑞, 𝑠𝑐) .

(33)

Lemma 22. If |𝑠𝑐| < 1, then

𝐹(𝑏; 𝑐𝐷𝑞)𝑎{

1

(𝑎𝑠; 𝑞)∞

} =(𝑏𝑐𝑠; 𝑞)

∞

(𝑎𝑠, 𝑐𝑠; 𝑞)∞

. (34)

Lemma 23. For |𝑠𝑐| < 1, |𝑡𝑐| < 1,

𝐹(𝑏; 𝑐𝐷𝑞)𝑎{

1


}

=(𝑏𝑐𝑡; 𝑞)

∞


2Φ1 (𝑏, 𝑎𝑡

𝑏𝑐𝑡; 𝑞, 𝑠𝑐) .

(35)

Throughout this paper, we also often use the followingproperty.

Lemma 24. If |𝑠𝑐| < 1, |𝑡𝑐| < 1, then

𝐹(𝑏; 𝑐𝐷𝑞)𝑎{𝑎𝑀(𝑎𝜔; 𝑞)

∞


}

=𝑎𝑀(𝑞; 𝑞)𝑀(𝑎𝜔, 𝑏𝑐𝑡; 𝑞)

∞


×

∞

∑

𝑘=0

(𝑐/𝑎)𝑘(𝑏, 𝑎𝑠, 𝑎𝑡; 𝑞)

𝑘

(𝑞, 𝑎𝜔, 𝑏𝑐𝑡; 𝑞)𝑘(𝑞; 𝑞)𝑀−𝑘

× 3Φ2(𝑏𝑞𝑘, 𝑎𝑡𝑞

𝑘,

𝜔

𝑠

𝑎𝜔𝑞𝑘, 𝑏𝑐𝑡𝑞

𝑘; 𝑞, 𝑠𝑐) .

(36)

Proof. We find that LHS of Lemma 24 equates to

∞

∑

𝑛=0


(𝑞; 𝑞)𝑛

𝑛

∑

𝑘=0

[𝑛


𝐷𝑘

𝑞 {𝑎𝑀}

× 𝐷𝑛−𝑘

𝑞

{

{

{

(𝑎𝜔𝑞𝑘; 𝑞)∞

(𝑎𝑠𝑞𝑘, 𝑎𝑡𝑞𝑘; 𝑞)∞

}

}

}

=

∞

∑

𝑘=0

(𝑞; 𝑞)𝑀𝑎𝑀−𝑘

(𝑞; 𝑞)𝑘(𝑞; 𝑞)𝑀−𝑘

×

∞

∑

𝑛=𝑘

(𝑏; 𝑞)𝑛𝑐𝑛𝑞𝑘(𝑘−𝑛)


𝐷𝑛−𝑘

𝑞

{

{

{



}

}

}

= 𝑎𝑀(𝑞; 𝑞)𝑀

∞

∑

𝑘=0

(𝑐/𝑎)𝑘(𝑏; 𝑞)𝑘

(𝑞; 𝑞)𝑘(𝑞; 𝑞)𝑀−𝑘

𝐹(𝑏𝑞𝑘; (𝑐

𝑞𝑘)𝐷𝑞)

×{

{

{



}

}

}


=𝑎𝑀(𝑞; 𝑞)𝑀(𝑎𝜔, 𝑏𝑐𝑡; 𝑞)

∞


×

∞

∑

𝑘=0

(𝑐/𝑎)𝑘(𝑏, 𝑎𝑠, 𝑎𝑡; 𝑞)

𝑘

(𝑞, 𝑎𝜔, 𝑏𝑐𝑡; 𝑞)𝑘(𝑞; 𝑞)𝑀−𝑘

× 3Φ2(𝑏𝑞𝑘, 𝑎𝑡𝑞

𝑘,

𝜔

𝑠

𝑎𝜔𝑞𝑘, 𝑏𝑐𝑡𝑞

𝑘; 𝑞, 𝑠𝑐) .

(37)

This completes the proof.

3. Proof of Theorems

Proof of Theorem 1. LHS of (3) is equal to

∞

∑

𝑛=0

𝐹(𝑎; 𝑥𝐷𝑞)𝑦{𝑦(𝑚+𝑛)

} 𝐹(𝑏;𝑋𝐷𝑞)𝑌{𝑌(𝑛)}

𝑧𝑛

(𝑞; 𝑞)𝑛

= 𝐹(𝑎; 𝑥𝐷𝑞)𝑦𝐹(𝑏;𝑋𝐷𝑞)𝑌

{𝑦𝑚

(𝑌𝑦𝑧; 𝑞)∞

}

= 𝐹(𝑎; 𝑥𝐷𝑞)𝑦{𝑦𝑚(𝑦𝑏𝑋𝑧; 𝑞)

∞

(𝑦𝑌𝑧, 𝑦𝑋𝑧; 𝑞)∞

}

=𝑦𝑚(𝑞; 𝑞)𝑚(𝑏𝑦𝑋𝑧, 𝑎𝑥𝑋𝑧; 𝑞)

∞


×

∞

∑

𝑘=0

(𝑥/𝑦)𝑘(𝑎, 𝑦𝑌𝑧, 𝑦𝑋𝑧; 𝑞)

𝑘

(𝑞, 𝑏𝑦𝑋𝑧, 𝑎𝑥𝑋𝑧; 𝑞)𝑘(𝑞; 𝑞)𝑚−𝑘

× 3Φ2(𝑎𝑞𝑘, 𝑦𝑋𝑧𝑞

𝑘,

𝑏𝑋

𝑌

𝑏𝑦𝑋𝑧𝑞𝑘, 𝑎𝑥𝑋𝑧𝑞

𝑘; 𝑞, 𝑥𝑌𝑧) .

(38)



𝐹(𝑎; 𝑥𝐷𝑞)𝑦{

1

(𝑦𝑡 : 𝑞)∞

𝐹(𝑏;𝑋𝐷𝑞)𝑌{

1

(𝑌𝑦𝛾, 𝑌𝑧; 𝑞)∞

}}

=(𝑏𝑋𝑧; 𝑞∞)

(𝑌𝑧,𝑋𝑧; 𝑞)∞

∞

∑

𝑚=0

(𝑏, 𝑌𝑧; 𝑞)𝑚(𝑋𝛾)𝑚

(𝑞; 𝑏𝑋𝑧)𝑚

× 𝐹(𝑎; 𝑥𝐷𝑞)𝑦{

𝑦𝑚

(𝑦𝑡, 𝑦𝑌𝛾 : 𝑞)∞

} ;

(39)

then using Lemma 24 and simplifying, we can get the desiredresult.


𝐹(𝑎; 𝑥𝐷𝑞)𝑦{𝑦𝑛𝐹(𝑏;𝑋𝐷𝑞)𝑌

{𝑌𝑚

(𝑌𝑦𝑧; 𝑞)∞

}}

= 𝑌𝑚(𝑞; 𝑞)𝑚

∞

∑

𝑘=0

(𝑋/𝑌)𝑘(𝑏; 𝑞)𝑘

(𝑞; 𝑞)𝑘(𝑞; 𝑞)𝑚−𝑘

𝐹(𝑎; 𝑥𝐷𝑞)𝑦

×{

{

{

𝑦𝑛(𝑦𝑏𝑋𝑧𝑞

𝑘; 𝑞)∞

(𝑦𝑌𝑧𝑞𝑘, 𝑦𝑋𝑧; 𝑞)∞

}

}

}

=𝑦𝑛𝑌𝑚(𝑞; 𝑞)𝑛(𝑞; 𝑞)𝑚(𝑦𝑏𝑋𝑧, 𝑥𝑎𝑋𝑧; 𝑞)

∞


×

∞

∑

𝑘=0

(𝑋/𝑌)𝑘(𝑏, 𝑦𝑌𝑧; 𝑞)

𝑘

(𝑞, 𝑦𝑏𝑋𝑧; 𝑞)𝑘(𝑞; 𝑞)𝑚−𝑘

×

∞

∑

𝑗=0

(𝑥/𝑦)𝑗(𝑎, 𝑦𝑌𝑧𝑞

𝑘, 𝑦𝑋𝑧; 𝑞)

𝑗

(𝑞, 𝑦𝑏𝑋𝑧𝑞𝑘, 𝑥𝑎𝑋𝑧; 𝑞)𝑗(𝑞; 𝑞)𝑛−𝑗

× 3Φ2(𝑎𝑞𝑗, 𝑦𝑋𝑧𝑞

𝑗,

𝑏𝑋

𝑌

𝑏𝑦𝑋𝑧𝑞𝑘+𝑗, 𝑎𝑥𝑋𝑧𝑞

𝑗; 𝑞, 𝑥𝑌𝑧𝑞

𝑘) .

(40)


Proof of Theorem 4. LHS of (6) is equal to∞

∑

𝑚1,𝑚2,𝑚3=0

𝐹(𝑎1; 𝑥1𝐷𝑞)𝑦1

{𝑦𝑚1+𝑚2+𝑚3

1 } 𝐹(𝑎2; 𝑥2𝐷𝑞)𝑦2

× {𝑦𝑚1+𝑚3

2 } 𝐹(𝑎3; 𝑥3𝐷𝑞)𝑦3

{𝑦𝑚1+𝑚2

3 }

×𝑡𝑚1

1 𝑡𝑚2

2 𝑡𝑚3

3

(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

= 𝐹(𝑎1; 𝑥1𝐷𝑞)𝑦1



× {1

(𝑡1𝑦1𝑦2𝑦3, 𝑡2𝑦1𝑦3, 𝑡3𝑦1𝑦2; 𝑞)∞

}



× {1

(𝑡3𝑦1𝑦2; 𝑞)∞


× {1

(𝑡1𝑦1𝑦2𝑦3, 𝑡2𝑦1𝑦3; 𝑞)∞

}}



×{

{

{

(𝑎3𝑥3𝑡2𝑦1; 𝑞)∞

(𝑡2𝑦1𝑦3, 𝑥3𝑡2𝑦1; 𝑞)∞

∞

∑

𝑗1=0

(𝑎3, 𝑡2𝑦1𝑦3; 𝑞)𝑗1

(𝑞, 𝑎3𝑥3𝑡2𝑦1; 𝑞)𝑗1

×(𝑥3𝑡1𝑦1𝑦2)

𝑗1

(𝑡3𝑦1𝑦2, 𝑡1𝑦1𝑦2𝑦3; 𝑞)∞

}

}

}



×{

{

{

(𝑎3𝑥3𝑡2𝑦1; 𝑞)∞

(𝑡2𝑦1𝑦3, 𝑥3𝑡2𝑦1; 𝑞)∞

∞

∑

𝑗1=0

(𝑎3, 𝑡2𝑦1𝑦3; 𝑞)𝑗1

(𝑞, 𝑎3𝑥3𝑡2𝑦1; 𝑞)𝑗1

× 𝐹(𝑎2; 𝑥2𝐷𝑞)𝑦2

{(𝑥3𝑡1𝑦1𝑦2)

𝑗1

(𝑡3𝑦1𝑦2, 𝑡1𝑦1𝑦2𝑦3; 𝑞)∞

}}

}

}

=

∞

∑

𝑗1=0

(𝑎3; 𝑞)𝑗1

(𝑥3𝑡1𝑦2)𝑗1

(𝑞; 𝑞)𝑗1

(𝑞; 𝑞)𝑗1

×

∞

∑

𝑗2=0

(𝑥2/𝑦2)𝑗2

(𝑎2; 𝑞)𝑗2

(𝑞; 𝑞)𝑗2

(𝑞; 𝑞)𝑗1−𝑗2

×

∞

∑

𝑗3=0

(𝑥2𝑡3)𝑗3

(𝑎2𝑞𝑗2 ; 𝑞)𝑗3

(𝑞; 𝑞)𝑗3


× {(𝑦𝑗1+𝑗3

1 (𝑎3𝑥3𝑡2𝑦1𝑞𝑗1 , 𝑎2𝑥2𝑡1𝑦1𝑦3𝑞

𝑗2+𝑗3 ; 𝑞)∞)

× ((𝑡2𝑦1𝑦3𝑞𝑗1 , 𝑡3𝑦1𝑦2𝑞

𝑗2 , 𝑡1𝑦1𝑦2𝑦3𝑞

𝑗2+𝑗3 ,

𝑥3𝑡2𝑦1, 𝑥2𝑡1𝑦1𝑦3; 𝑞)∞)−1} .

(41)

Setting

𝑓 (𝑦1) = (𝑦𝑗1+𝑗3

1 (𝑎3𝑥3𝑡2𝑦1𝑞𝑗1 , 𝑎2𝑥2𝑡1𝑦1𝑦3𝑞

𝑗2+𝑗3 ; 𝑞)∞)

× ((𝑡2𝑦1𝑦3𝑞𝑗1 , 𝑡3𝑦1𝑦2𝑞

𝑗2 , 𝑡1𝑦1𝑦2𝑦3𝑞

𝑗2+𝑗3 ,

𝑥3𝑡2𝑦1, 𝑥2𝑡1𝑦1𝑦3; 𝑞)∞)−1,

(42)

then by applying Lemma 18 and simplifying, we get


{𝑓 (𝑦1)}

= (𝑦𝑗1+𝑗3

1 (𝑎1𝑥1

𝑦1

, 𝑎3𝑥3𝑡2𝑦1, 𝑎2𝑥2𝑡1𝑦1𝑦3; 𝑞)

∞

)

× ((𝑥1

𝑦1

, 𝑡2𝑦1𝑦3, 𝑡3𝑦1𝑦2, 𝑡1𝑦1𝑦2𝑦3,

𝑥3𝑡2𝑦1, 𝑥2𝑡1𝑦1𝑦3; 𝑞)

∞

)

−1

×

∞

∑

𝑗4=0

(𝑎1, 𝑥3𝑡2𝑦1, 𝑥2𝑡1𝑦1𝑦3; 𝑞)𝑗4

(−1)𝑗4𝑞−(𝑗4−1)𝑗4/2

(𝑞, 𝑥1/ (𝑦1𝑞𝑗4) ; 𝑞)𝑗4

× (𝑥1𝑞𝑗1+𝑗3

𝑦1

)

𝑗4 (𝑡3𝑦1𝑦2; 𝑞)𝑗

2+𝑗4

(𝑡2𝑦1𝑦3; 𝑞)𝑗1+𝑗4

(𝑎3𝑥3𝑡2𝑦1; 𝑞)𝑗1+𝑗4

×

(𝑡1𝑦1𝑦2𝑦3; 𝑞)𝑗2+𝑗3+𝑗4

(𝑎2𝑥2𝑡1𝑦1𝑦3; 𝑞)𝑗2+𝑗3+𝑗4

;

(43)

substituting it into (41), we complete the theorem.





× {1

(𝑦1𝑦2𝑡1, 𝑦1𝑦2𝑡2, 𝑦1𝑦3𝑡3; 𝑞)∞

}



× {(𝑎3𝑥3𝑦1𝑡3; 𝑞)∞

(𝑦1𝑦3𝑡3, 𝑦1𝑥3𝑡3; 𝑞)∞

1

(𝑦1𝑦2𝑡1, 𝑦1𝑦2𝑡2; 𝑞)∞

}

=

∞

∑

𝑗1=0

(𝑎2; 𝑞)𝑗1

(𝑡1𝑥2)𝑗1

(𝑞; 𝑞)𝑗1


× { (𝑦𝑗1

1 (𝑎3𝑦1𝑥3𝑡3, 𝑎2𝑦1𝑥2𝑡2𝑞𝑗1 ; 𝑞)∞)

× ( (𝑦1𝑦3𝑡3, 𝑦1𝑥3𝑡3, 𝑦1𝑦2𝑡1,

𝑦1𝑦2𝑡2𝑞𝑗1 , 𝑦1𝑥2𝑡2; 𝑞)∞

)−1} .

(44)

Applying Lemma 18 and simplifying, we get the theorem.This completes the proof.

Proof of Theorems 6 and 7. Using Lemma 17, thenTheorem 6can be obtained by Lemma 20 directly.

Using Lemma 17 and then applying Lemma 18, we can getTheorem 7.


𝐹(𝑎; 𝑥𝐷𝑞)𝑦{(𝑦𝑡1, 𝑦𝑡4; 𝑞)∞

(𝑦𝑡2, 𝑦𝑡3; 𝑞)∞

}

=

∞

∑

𝑛=0

(𝑎; 𝑞)𝑛𝑥𝑛

(𝑞; 𝑞)𝑛

𝐷𝑛

𝑞 {(𝑦𝑡1, 𝑦𝑡4; 𝑞)∞

(𝑦𝑡2, 𝑦𝑡3; 𝑞)∞

}


=

∞

∑

𝑛=0

(𝑎; 𝑞)𝑛𝑥𝑛

(𝑞; 𝑞)𝑛

×

𝑛

∑

𝑘=0

[𝑛


𝐷𝑘

𝑞 {(𝑦𝑡1; 𝑞)∞

(𝑦𝑡2; 𝑞)∞

}

× 𝐷𝑛−𝑘

𝑞

{

{

{

(𝑦𝑡4𝑞𝑘; 𝑞)∞

(𝑦𝑡3𝑞𝑘; 𝑞)∞

}

}

}

=(𝑦𝑡1, 𝑦𝑡4; 𝑞)∞

(𝑦𝑡2, 𝑦𝑡3; 𝑞)∞

∞

∑

𝑛=0

(𝑎; 𝑎)𝑛(𝑥𝑡3)𝑛

(𝑞; 𝑞)𝑛

×

𝑛

∑

𝑘=0

(𝑞; 𝑞)𝑛(𝑡2/𝑡3)

𝑘


(𝑡1/𝑡2, 𝑦𝑡3; 𝑞)𝑘(𝑡4/𝑡3; 𝑞)𝑛−𝑘

(𝑦𝑡𝑞; 𝑞)𝑘

=(𝑦𝑡1, 𝑦𝑡4; 𝑞)∞

(𝑦𝑡2, 𝑦𝑡3; 𝑞)∞

∞

∑

𝑗=0

(𝑎, 𝑡1/𝑡2, 𝑦𝑡3; 𝑞)𝑗(𝑥𝑡2)𝑗

(𝑞, 𝑦𝑡1, 𝑦𝑡4; 𝑞)𝑗

× 2Φ1(𝑎𝑞𝑗,

𝑡4

𝑡3

𝑦𝑡4𝑞𝑗; 𝑞, 𝑥𝑡3) ;

(45)

we complete the proof.

Proof of Theorem 9. LHS of (11) equates to

𝐹(𝑎; 𝑥𝐷𝑞)𝑦𝐹(𝑎0; 𝑥0𝐷𝑞)𝑦

0




× {𝑦𝑠𝑦𝑟0

(𝑦𝑦0𝑡, 𝑦𝑦1𝑡1, 𝑦𝑦2𝑡2, 𝑦𝑦3𝑡3; 𝑞)∞

}

= 𝐹(𝑎; 𝑥𝐷𝑞)𝑦𝐹(𝑎0; 𝑥0𝐷𝑞)𝑦

0



{𝑦𝑠𝑦𝑟0

(𝑦𝑦0𝑡, 𝑦𝑦1𝑡1, 𝑦𝑦2𝑡2; 𝑞)∞

×(𝑎3𝑦𝑥3𝑡3; 𝑞)∞

((𝑦𝑦3𝑡3, 𝑦𝑥3𝑡3; 𝑞)∞)}


0


× {𝑦𝑠𝑦𝑟0

(𝑦𝑦0𝑡, 𝑦𝑦1𝑡1; 𝑞)∞

(𝑎2𝑦𝑥2𝑡2; 𝑞)∞

((𝑦𝑦2𝑡2, 𝑦𝑥2𝑡2; 𝑞)∞)

×(𝑎3𝑦𝑥3𝑡3; 𝑞)∞

((𝑦𝑦3𝑡3, 𝑦𝑥3𝑡3; 𝑞)∞)}


0

{𝑦𝑠𝑦𝑟0

(𝑦𝑦0𝑡; 𝑞)∞

}

× {(𝑎1𝑦𝑥1𝑡1; 𝑞)∞

((𝑦𝑦1𝑡1, 𝑦𝑥1𝑡1; 𝑞)∞)

(𝑎2𝑦𝑥2𝑡2; 𝑞)∞

((𝑦𝑦2𝑡2, 𝑦𝑥2𝑡2; 𝑞)∞)

×(𝑎3𝑦𝑥3𝑡3; 𝑞)∞

((𝑦𝑦3𝑡3, 𝑦𝑥3𝑡3; 𝑞)∞)} .

(46)

If we set

𝑓 (𝑦) = 𝐹(𝑎0; 𝑥0𝐷𝑞)𝑦0

{𝑦𝑠𝑦𝑟0

(𝑦𝑦0𝑡; 𝑞)∞

}

× {(𝑎1𝑦𝑥1𝑡1; 𝑞)∞

((𝑦𝑦1𝑡1, 𝑦𝑥1𝑡1; 𝑞)∞)

(𝑎2𝑦𝑥2𝑡2; 𝑞)∞

((𝑦𝑦2𝑡2, 𝑦𝑥2𝑡2; 𝑞)∞)

×(𝑎3𝑦𝑥3𝑡3; 𝑞)∞

((𝑦𝑦3𝑡3, 𝑦𝑥3𝑡3; 𝑞)∞)} ,

(47)

using Lemma 24, we get 𝑓(𝑦); then employing Lemma 18, wecan get the desired result.

Proof of Theorem 10. LHS of (12) equates to





{(𝑡1𝑦2𝑦3𝑦4, 𝑡4𝑦1𝑦2𝑦3; 𝑞)∞

(𝑡2𝑦1𝑦3𝑦4, 𝑡3𝑦1𝑦2𝑦4; 𝑞)∞

} ;

(48)

then using Lemma 18 four times we can get the result.

4. Some Other Cases and a Homogeneous𝑞-Mehler’s Formula

In this section we obtain Mehler’s formula involving a 3Φ2series and then give an interesting relationship between thepolynomial multiplications.

Theorem 25 (cf. [1, Equation (1.17)] or [7, Lemma 12]). Onehas∞

∑

𝑛=0

𝜙(𝑎)

𝑛 (𝑥, 𝑦) 𝜙(𝑏)

𝑛 (𝑋, 𝑌)𝑡𝑛

(𝑞; 𝑞)𝑛

=(𝑎𝑥𝑌𝑡, 𝑏𝑦𝑋𝑡; 𝑞)

∞

(𝑥𝑌𝑡, 𝑦𝑋𝑡, 𝑦𝑌𝑡; 𝑞)∞

3Φ2 (𝑎, 𝑏, 𝑦𝑌𝑡

𝑎𝑥𝑌𝑡, 𝑏𝑦𝑋𝑡; 𝑞, 𝑥𝑋𝑡) .

(49)

Proof. LHS of Theorem 25 is equal to∞

∑

𝑛=0

𝜙(𝑏)


(𝑞; 𝑞)𝑛

𝐹(𝑎; 𝑥𝐷𝑞)𝑦{𝑦𝑛}

= 𝐹(𝑎; 𝑥𝐷𝑞)𝑦{

∞

∑

𝑛=0

𝑛

∑

𝑘=0

(𝑏; 𝑞)𝑘𝑋𝑘𝑌𝑛−𝑘(𝑡𝑦)𝑛


}

= 𝐹(𝑎; 𝑥𝐷𝑞)𝑦{(𝑏𝑡𝑋𝑦; 𝑞)

∞

(𝑡𝑋𝑦; 𝑞)∞

1

(𝑡𝑌𝑦; 𝑞)∞

}

=(𝑎𝑥𝑌𝑡, 𝑏𝑦𝑋𝑡; 𝑞)

∞

(𝑥𝑌𝑡, 𝑦𝑋𝑡, 𝑦𝑌𝑡; 𝑞)∞

× 3Φ2 (𝑎, 𝑏, 𝑦𝑌𝑡

𝑎𝑥𝑌𝑡, 𝑏𝑦𝑋𝑡; 𝑞, 𝑥𝑋𝑡) .

(50)



Remark 26. Theorem 25 can be reduced to the common𝑞-Mehler’s formula (cf. [1, Equation (1.6)]) by setting 𝑎 = 𝑏 =0 and 𝑦 = 𝑌 = 1. Employing this formula, we can show thatthe polynomials Ψ𝑛 (𝑎, 𝑏; 𝑥1, 𝑦1; 𝑧) defined by Andrews [4]can be expressed by 𝜙(𝑎)𝑛 (𝑥, 𝑦). Using Sear’s nonterminating3Φ2 transformation formula [15, page 241, Equation III.10]

3Φ2 (𝐴, 𝐵, 𝐶

𝐷, 𝐸; 𝑞,

𝐷𝐸

𝐴𝐵𝐶)

=(𝐵,𝐷𝐸/𝐴𝐵,𝐷𝐸/𝐵𝐶; 𝑞)

∞

(𝐷, 𝐸,𝐷𝐸/𝐴𝐵𝐶; 𝑞)∞

× 3Φ2(

𝐷

𝐵,𝐸

𝐵,

𝐷𝐸

𝐴𝐵𝐶𝐷𝐸

𝐴𝐵,𝐷𝐸

𝐵𝐶

; 𝑞, 𝐵) ,

(51)

then the right-hand side of (49) reduces to

(𝑎, 𝑥𝑦𝑋𝑌𝑡2, 𝑏𝑥𝑋𝑡; 𝑞)

∞

(𝑥𝑌𝑡, 𝑦𝑋𝑡, 𝑦𝑌𝑡, 𝑥𝑋𝑡; 𝑞)∞

× 3Φ2(

𝑏𝑦𝑋𝑡

𝑎, 𝑥𝑌𝑡, 𝑥𝑋𝑡

𝑥𝑦𝑋𝑌𝑡2, 𝑏𝑥𝑋𝑡

; 𝑞, 𝑎) .

(52)

Then setting 𝑥𝑋 = 𝑧, 𝑦𝑋 = 𝑦1, and 𝑥𝑌 = 𝑥1, we have (cf. [4,Theorem 4.1])

∞

∑

𝑛=0

𝜙(𝑎)

𝑛 (𝑥, 𝑦) 𝜙(𝑏)


(𝑞; 𝑞)𝑛

=

(𝑎, 𝑏𝑧𝑡, 𝑥1𝑦1𝑡2; 𝑞)∞

(𝑡, 𝑦1𝑡, 𝑧𝑡, 𝑥1𝑡; 𝑞)∞

× 3Φ2(

𝑏𝑡𝑦1

𝑎, 𝑧𝑡, 𝑥1𝑡

𝑏𝑧𝑡, 𝑥1𝑦1𝑡2; 𝑞, 𝑥𝑋𝑡) .

(53)

Comparison of the above identity with Theorem 4.1 thatappeared in [4] gives the following.

Corollary 27 (cf. [4, Corollary 3.1]). If 𝑥𝑋 = 𝑧, 𝑦𝑋 = 𝑦1, and𝑥𝑌 = 𝑥1, then

Ψ𝑛 (𝑎, 𝑏; 𝑥1, 𝑦1; 𝑧) = 𝜙(𝑎)

𝑛 (𝑥, 𝑦) 𝜙(𝑏)

𝑛 (𝑋, 𝑌) . (54)

Combined with Theorem 6, if we replace (𝑥, 𝑦, 𝑋, 𝑌, 𝑎, 𝑏, 𝑡) by(𝑥1, 𝑦1, 1, 𝑡2/𝑡1, 𝑎1, 𝑡3/𝑡1, 𝑡1), respectively, and then equate thecoefficient of 𝑡𝑁1 , we have the following.

Corollary 28. One has

𝜙(𝑎)

𝑁(𝑥1, 𝑦1) 𝜙

(𝑏)

𝑁 (1, 𝑌)

(𝑞; 𝑞)𝑁

= ∑

𝑚1+𝑚2+𝑚3=𝑁

𝜙(𝑎)

𝑚1+𝑚2+𝑚3

(𝑥1, 𝑦1)

×𝑌𝑚2𝑏𝑚3(−1)𝑚3𝑞(𝑚3−1)𝑚3/2

(𝑞; 𝑞)𝑚1

(𝑞; 𝑞)𝑚2

(𝑞; 𝑞)𝑚3

.

(55)

Theorem 29 (cf. [1, Equation (1.17)]). If𝑚, 𝑛 ∈ 𝑁, then∞

∑

𝑚,𝑛=0

𝜙(𝑎1)

𝑚+𝑛 (𝑥1, 𝑦1) 𝜙(𝑎2)

𝑚 (𝑥2, 𝑦2) 𝜙(𝑎3)

𝑛

× (𝑥3, 𝑦3)𝑡𝑚𝑧𝑛

(𝑞; 𝑞)𝑚(𝑞; 𝑞)𝑛

=(𝑎1𝑥1𝑥3𝑧, 𝑎2𝑥2𝑦1𝑡, 𝑎3𝑥3𝑦1𝑧; 𝑞)∞

(𝑥1𝑥3𝑧, 𝑥2𝑦1𝑡, 𝑥3𝑦1𝑧, 𝑦1𝑦2𝑡, 𝑦1𝑦3𝑧; 𝑞)∞

×

∞

∑

𝑗1,𝑗2,𝑗3=0

(𝑥1𝑦3𝑧)𝑗1

(𝑥1𝑥2𝑡)𝑗2

(𝑥1𝑦2𝑡)𝑗3

(𝑞; 𝑞)𝑗1

(𝑞; 𝑞)𝑗2

(𝑞; 𝑞)𝑗3

×

(𝑎2𝑥2/𝑦2, 𝑥2𝑦1𝑡; 𝑞)𝑗3

(𝑎3𝑥3𝑧/ (𝑥2𝑡) ; 𝑞)𝑗2

(𝑎2𝑥2𝑦1𝑡; 𝑞)𝑗3

×

(𝑦1𝑦3𝑧; 𝑞)𝑗2+𝑗3

(𝑎1, 𝑥3𝑦1𝑧; 𝑞)𝑗1+𝑗2+𝑗3

(𝑎3𝑥3𝑦1𝑧; 𝑞)𝑗2+𝑗3

(𝑎1𝑥1𝑥3𝑧; 𝑞)𝑗1+𝑗2+𝑗3

,

(56)

provided that max{|𝑥1𝑦3𝑧|, |𝑥3𝑦1𝑧|, |𝑥1𝑦2𝑡|, |𝑥2𝑦1𝑡|, |𝑦1𝑦2𝑡|,|𝑦1𝑦3𝑧|, |𝑎1𝑥1𝑥3𝑧|} < 1.

Proof. Applying Lemma 17 and then using 𝑞-binomial theo-rem, we get that the LHS of (56) comes to

𝐹(𝑎1, 𝑥1𝐷𝑞)𝑦1



× {1

(𝑦1𝑦2𝑡, 𝑦1𝑦3𝑧; 𝑞)∞

} .

(57)

Twice applying Lemma 22, the above identity is equal to


{(𝑎2𝑥2𝑦1𝑡, 𝑎3𝑥3𝑦1𝑧; 𝑞)∞

(𝑦1𝑦2𝑡, 𝑥2𝑦1𝑡, 𝑦1𝑦3𝑧, 𝑥3𝑦1𝑧; 𝑞)∞

}

=

∞

∑

𝑛=0

(𝑎1; 𝑞)𝑛𝑥𝑛1

(𝑞; 𝑞)𝑛

𝐷𝑛

𝑞

× {(𝑎2𝑥2𝑦1𝑡, 𝑎3𝑥3𝑦1𝑧; 𝑞)∞

(𝑦1𝑦2𝑡, 𝑥2𝑦1𝑡, 𝑦1𝑦3𝑧, 𝑥3𝑦1𝑧; 𝑞)∞

} .

(58)

For (58), by using 𝑞-Leibnitz rule three times and using𝑞-binomial theorem, we can achieve the proof after simplifi-cation.


Taking 𝑎1 = 𝑎2 = 𝑎3 = 0, 𝑦1 = 𝑦2 = 𝑦3 = 1, we have thefollowing.

Corollary 30 (cf. [9, Equation (4.4)]). One has∞

∑

𝑚,𝑛=0

𝐻𝑚+𝑛 (𝑥1)𝐻𝑚 (𝑥2)𝐻𝑛 (𝑥3)𝑡𝑚𝑧𝑛

(𝑞; 𝑞)𝑚(𝑞; 𝑞)𝑛

=

(𝑥1𝑥3𝑧2; 𝑞)∞

(𝑥1𝑧, 𝑥1𝑥3𝑧, 𝑥2𝑡, 𝑥3𝑧, 𝑡, 𝑧; 𝑞)∞

×

∞

∑

𝑗1,𝑗2=0

(𝑥2𝑡; 𝑞)𝑗2

(𝑧, 𝑥3𝑧; 𝑞)𝑗1+𝑗2

(𝑥1𝑡)𝑗1+𝑗2

𝑥2𝑗1

(𝑞; 𝑞)𝑗1

(𝑞; 𝑞)𝑗2

(𝑥1𝑥3𝑧2; 𝑞)𝑗1+𝑗2

.

(59)

Remark 31. Corollary 30 is equivalent to [9, Equation (4.4)]after applying 𝑞-Heine transformation formula to simplify.Note that the process of proof is only using 𝑞-Leibnitz rule,so we can remove the restrictive condition 𝑎 = 𝑞−𝐹, 𝑏 = 𝑞−𝐺in [7, Theorem 3]. Employing the same method, we recoverTheorem 7 as follows.

Theorem 32. If 𝑚𝑖 ∈ 𝑁, max{|𝑦𝑡𝑖|, |𝑥𝑡𝑖|} < 1, 𝑖 = 1, 2, . . . , 𝑛,then

∞

∑

𝑚1,...,𝑚𝑛=0

𝜙(𝑎)

𝑚1+⋅⋅⋅+𝑚

𝑛

(𝑥, 𝑦)𝑡𝑚1

1 ⋅ ⋅ ⋅ 𝑡𝑚𝑛

𝑛

(𝑞; 𝑞)𝑚1

⋅ ⋅ ⋅ (𝑞; 𝑞)𝑚𝑛

=(𝑎𝑥𝑡1; 𝑞)∞

(𝑥𝑡1, 𝑦𝑡1, 𝑦𝑡2, . . . , 𝑦𝑡𝑛; 𝑞)∞

×

∞

∑

𝑗1,𝑗2,...,𝑗𝑛−1=0

(𝑎, 𝑦𝑡1; 𝑞)𝑗1+𝑗2+⋅⋅⋅+𝑗

𝑛−1

𝑥𝑗1+𝑗2+⋅⋅⋅+𝑗

𝑛−1

(𝑎𝑥𝑡1; 𝑞)𝑗1+𝑗2+⋅⋅⋅+𝑗

𝑛−1

(𝑞; 𝑞)𝑗1

(𝑞; 𝑞)𝑗2

× ((𝑦𝑡2; 𝑞)𝑗2+⋅⋅⋅+𝑗

𝑛−1

, . . . , (𝑦𝑡𝑛−2; 𝑞)𝑗𝑛−2+𝑗𝑛−1

×(𝑦𝑡𝑛−1; 𝑞)𝑗𝑛−1

𝑡𝑗1

2 𝑡𝑗2

3 , . . . , 𝑡𝑗𝑛−1

𝑛 )

× ((𝑞; 𝑞)𝑗3

, . . . , (𝑞; 𝑞)𝑗𝑛−1

)−1.

(60)

Proof . Using Lemma 17, LHS of (60) is equal to

𝐹(𝑎; 𝑥𝐷𝑞)𝑦{

1

(𝑦𝑡𝑛, 𝑦𝑡𝑛−1, . . . , 𝑦𝑡1; 𝑞)∞

}

=

∞

∑

𝑘=0


(𝑞; 𝑞)𝑘

𝐷𝑘

𝑞 {1

(𝑦𝑡𝑛, 𝑦𝑡𝑛−1, . . . , 𝑦𝑡1; 𝑞)∞

}

=

∞

∑

𝑘=0


(𝑞; 𝑞)𝑘

×

𝑘

∑

𝑗1=0

[𝑘

𝑗1] 𝑞𝑗1(𝑗1−𝑘)𝐷𝑗1

𝑞 {1

(𝑦𝑡𝑛, 𝑦𝑡𝑛−1, . . . , 𝑦𝑡2; 𝑞)∞

}

× 𝐷𝑘−𝑗1

𝑞

{

{

{

1

(𝑦𝑡𝑗1

1 ; 𝑞)∞

}

}

}

=

∞

∑

𝑘=0


(𝑞; 𝑞)𝑘

×

𝑘

∑

𝑗1=0

[𝑘

𝑗1]𝐷𝑗1

𝑞 {1

(𝑦𝑡𝑛, 𝑦𝑡𝑛−1, . . . , 𝑦𝑡2; 𝑞)∞

}

×

𝑡𝑘−𝑗1

1 (𝑦𝑡1; 𝑞)𝑗1

(𝑦𝑡1; 𝑞)∞

.

(61)

Exchanging the order of summation and using 𝑞-bino-mial, theorem (61) is equal to

(𝑎𝑥𝑡1; 𝑞)∞

(𝑥𝑡1, 𝑦𝑡1; 𝑞)∞

∞

∑

𝑗1=0

(𝑎, 𝑦𝑡1; 𝑞)𝑗1

𝑥𝑗1

(𝑞, 𝑎𝑥𝑡1; 𝑞)𝑗1

× 𝐷𝑗1

𝑞 {1

(𝑦𝑡𝑛, 𝑦𝑡𝑛−1, . . . , 𝑦𝑡2; 𝑞)∞

}

=(𝑎𝑥𝑡1; 𝑞)∞

(𝑥𝑡1, 𝑦𝑡1; 𝑞)∞

∞

∑

𝑗1=0

(𝑎, 𝑦𝑡1; 𝑞)𝑗1

𝑥𝑗1

(𝑞, 𝑎𝑥𝑡1; 𝑞)𝑗1

× 𝐷𝑗1

𝑞 {1

(𝑦𝑡𝑛, 𝑦𝑡𝑛−1, . . . , 𝑦𝑡3; 𝑞)∞

×1

(𝑦𝑡2; 𝑞)∞

} .

(62)

Applying 𝑞-Leibnitz and exchanging the order, (62) is equalto

(𝑎𝑥𝑡1; 𝑞)∞

(𝑥𝑡1, 𝑦𝑡1, 𝑦𝑡2; 𝑞)∞

×

∞

∑

𝑗2=0

(𝑦𝑡2; 𝑞)𝑗2

(𝑞; 𝑞)𝑗2

𝐷𝑗2

𝑞 {1

(𝑦𝑡𝑛, 𝑦𝑡𝑛−1, . . . , 𝑦𝑡3; 𝑞)∞

}

×

∞

∑

𝑗1=0

(𝑎, 𝑦𝑡1; 𝑞)𝑗1+𝑗2

𝑥𝑗1+𝑗2𝑡𝑗2

2

(𝑞; 𝑞)𝑗2

(𝑎𝑥𝑡1; 𝑞)𝑗1+𝑗2

.

(63)

By induction, similar proof can be performed to get thedesired result.

Taking 𝑎 = 0, 𝑦 = 1 in (60), we have the following.

Corollary 33. If 𝑚𝑖 ∈ 𝑁, max{|𝑡𝑖|, |𝑥𝑡𝑖|} < 1, 𝑖 = 1, 2, . . . , 𝑛,then

∞

∑

𝑚1,...,𝑚𝑛=0

𝐻𝑚1+⋅⋅⋅+𝑚

𝑛(𝑥)

𝑡𝑚1

1 ⋅ ⋅ ⋅ 𝑡𝑚𝑛

𝑛

(𝑞; 𝑞)𝑚1

⋅ ⋅ ⋅ (𝑞; 𝑞)𝑚𝑛

=1

(𝑥𝑡1, 𝑡1, 𝑡2, . . . , 𝑡𝑛; 𝑞)∞


×

∞

∑

𝑗1,𝑗2,...,𝑗𝑛−1=0

(𝑡1; 𝑞)𝑗1+𝑗2+⋅⋅⋅+𝑗

𝑛−1

𝑥𝑗1+𝑗2+⋅⋅⋅+𝑗

𝑛−1

(𝑞; 𝑞)𝑗1

(𝑞; 𝑞)𝑗2

× ( (𝑡2; 𝑞)𝑗2+⋅⋅⋅+𝑗

𝑛−1

⋅ ⋅ ⋅ (𝑡𝑛−2; 𝑞)𝑗𝑛−2+𝑗𝑛−1

× (𝑡𝑛−1; 𝑞)𝑗𝑛−1

𝑡𝑗1

2 𝑡𝑗2

3 ⋅ ⋅ ⋅ 𝑡𝑗𝑛−1

𝑛 )

× ((𝑞; 𝑞)𝑗3

⋅ ⋅ ⋅ (𝑞; 𝑞)𝑗𝑛−1

)−1.

(64)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper.

Acknowledgments

The author is supported by Jiangsu Overseas Researchand Training Program for University Prominent Youngand Middle-Aged Teachers and Presidents and Univer-sity National Natural Science Foundation of Jiangsu (no.14KJB110002). The author is also supported by the NationalNatural Science Foundation of China (nos. 11371163 and11001098).

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