Formal Calculation of Q-Binomial

1. Introduction

The q-binomial coefficient is a fundamental object in combinatorics, appearing in partition theory, representation theory, and the study of quantum groups. Its classical identities—such as the q-binomial theorem, the q-Vandermonde convolution, and the summation formulas of Gauss, Cauchy, and Jacobi—are cornerstones of q-analysis. However, evaluating nested sums that involve products of q-binomial coefficients and q-integers ${\left[n\right]}_q$ often requires ad hoc techniques, and a unified computational framework has been lacking. Recent work has further explored q-binomial coefficients in various generalized settings, including q-binomial identities with nested sums and their applications in q-series [1,2]. Comprehensive surveys of classical q-identities can be found in [3].

While various methods exist for special cases, a unified framework for arbitrary nested q-binomial sums has remained elusive. We develop such a framework via a recursive operator $SU{M}_q$ that captures sums with arbitrary nesting depth and coefficients, together with a shift operator ${\nabla }_a^p$ that handles the nesting structure uniformly. Our main result (Theorem 1) establishes that every such sum admits three equivalent q-binomial expansions, related by explicit transformations (Theorem 6) that not only provide computational flexibility but also reveal deep algebraic structures, including inversion formulas, orthogonality relations, and a closed-form expression for weighted infinite series that unifies many classical q-identities.

The power of our framework is demonstrated through a wide range of applications. On one hand, many classical theorems—including Rothe’s q-binomial theorem, MacMahon’s q-binomial theorem, Gauss’s q-binomial theorem, Jacobi’s q-binomial theorem, and Cauchy’s identity—follow directly from our three-form equivalence, with proofs that are remarkably short and uniform. Beyond unifying the proofs of numerous classical q-binomial theorems, the framework presented here serves as a systematic identity generator: arbitrary specializations of the free parameters yield a vast family of new summation formulas, of which only a representative sample is included. This dual nature—both a unifying principle and a discovery tool—distinguishes our approach from previous ad hoc treatments.

The paper is organized as follows. Section 2 (Calculation Formula) recalls the necessary definitions, proves a key summation lemma, introduces the recursive operator $SU{M}_q$, and presents the main calculation theorem. Section 3 (Properties) establishes the relations among the three forms. Section 4 (Inferences) derives simple consequences of the framework that nevertheless reveal important results, including inversion formulas and orthogonality relations. Section 5 (Applications) then presents a wide range of applications, including numerous new results and finite summation identities. Section 6 (Unified Conversion Formulas) establishes a unified framework and demonstrates its utility.

2. Calculation Formula

The q-binomial coefficient is defined for non-negative integers $M,N$ by

$${{[}_M^N]}_q={\displaystyle \frac{(q^N-1)(q^{N-1}-1)\dots (q^{N-M+1}-1)}{(q^M-1)(q^{M-1}-1)\dots (q^1-1)}},q\ne 0,1.$$

For convenience, we write $G_M^N={{[}_M^N]}_q$ and ${\left[N\right]}_q=G_1^N$. The q-shifted factorial is given by

$${(a;q)}_n=\left\{\begin{array}{cc}1,\hfill & n=0,\hfill \\ (1-a)(1-aq)\dots (1-a{q}^{n-1}),\hfill & n>0.\hfill \end{array}\right.$$

The Kronecker delta ${\delta }_{ij}$ is defined as usual: ${\delta }_{ij}=1$ if $i=j$ and 0 otherwise.

The following elementary identities are fundamental:

$$\begin{array}{cc}\hfill & G_0^N=1,G_M^N=G_{N-M}^N,G_M^N=0\mathrm{whenever}M>N\mathrm{or}M<0,\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill & G_M^N=q^M{G}_M^{N-1}+G_{M-1}^{N-1}=G_M^{N-1}+q^{N-M}{G}_{M-1}^{N-1},\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=0}^{N-1}}{q}^n{G}_M^{n+K}=q^{M-K}{G}_{M+1}^{N+K},M\ge K\ge 0,\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill & G_K^M=\sum _{w\in \Omega (0^{M-K},1^K)}{q}^{inv\left(w\right)},\hfill \end{array}$$

(4)

where $inv\left(w\right)$ denotes the inversion number of the binary word w (see [4]).

Lemma 1.

For integers $M\ge K\ge 0$ and $N>0$, the following three equivalent forms hold:

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{\left[n\right]}_q{G}_M^{n+K}=q^{2(M-K)+1}{G}_1^{M+1}{G}_{M+2}^{N+K}+q^{M-K}{G}_1^{M-K}{G}_{M+1}^{N+K}$$

$$=q^{M-2K-1}{G}_1^{M+1}{G}_{M+2}^{N+K+1}+q^{M-K}\left(G_1^{M-K}-q^{-K-1}{G}_1^{M+1}\right)G_{M+1}^{N+K}$$

$$=\left(q^{2(M-K)+1}{G}_1^{M+1}-q^{2M-K+2}{G}_1^{M-K}\right)G_{M+2}^{N+K}+q^{M-K}{G}_1^{M-K}{G}_{M+2}^{N+K+1}.$$

Proof. 

We begin with the identity ${\left[n\right]}_q={\displaystyle \frac{q^n-1}{q-1}}$. Decompose it as

$${\left[n\right]}_q={\displaystyle \frac{q^n-q^{n-M+K}}{q-1}}+{\displaystyle \frac{q^{n-M+K}-1}{q-1}}=q^{n-M+K}{\displaystyle \frac{q^{M-K}-1}{q-1}}+{\displaystyle \frac{q^{n-M+K}-1}{q-1}}.$$

Using the identity $(q^{n-M+K}-1)G_M^{n+K}=(q^{M+1}-1)G_{M+1}^{n+K}$, then the sum becomes

$$\begin{array}{cc}\hfill \sum _n{q}^n{\left[n\right]}_q{G}_M^{n+K}& =\sum _n{q}^n{q}^{n-M+K}{\displaystyle \frac{q^{M-K}-1}{q-1}}{G}_M^{n+K}+\sum _n{q}^n{\displaystyle \frac{q^{n-M+K}-1}{q-1}}{G}_M^{n+K}\hfill \\ \hfill & =\sum _n{q}^n(q^{n-M+K}-1)G_1^{M-K}{G}_M^{n+K}+\sum _n{q}^n{G}_1^{M-K}{G}_M^{n+K}+\sum _n{q}^n{G}_1^{M+1}{G}_{M+1}^{n+K}\hfill \\ \hfill & =G_1^{M-K}(q^{M+1}-1)\sum _n{q}^n{G}_{M+1}^{n+K}+G_1^{M-K}\sum _n{q}^n{G}_M^{n+K}+G_1^{M+1}\sum _n{q}^n{G}_{M+1}^{n+K}\hfill \\ \hfill & =G_1^{M+1}(q^{M-K}-1)q^{M+1-K}{G}_{M+2}^{N+K}+G_1^{M-K}{q}^{M-K}{G}_{M+1}^{N+K}+G_1^{M+1}{q}^{M+1-K}{G}_{M+2}^{N+K}\hfill \\ \hfill & =q^{2(M-K)+1}{G}_1^{M+1}{G}_{M+2}^{N+K}+q^{M-K}{G}_1^{M-K}{G}_{M+1}^{N+K}.\hfill \end{array}$$

This proves the first equation.

Next, from the recurrence relation $G_M^N=q^M{G}_M^{N-1}+G_{M-1}^{N-1}$, we have

$$G_{M+2}^{N+K}=q^{-M-2}\left(G_{M+2}^{N+K+1}-G_{M+1}^{N+K}\right).G_{M+1}^{N+K}=G_{M+2}^{N+K+1}-q^{M+2}{G}_{M+2}^{N+K}.$$

Substituting this into the first equation yields the second and third equations. □

Definition 1.

We now introduce a recursive operator and a family of nested sums that will serve as our main objects.

For $p\in \mathbb{Z}$, define the operator ${\nabla }_q^p$ on functions $f\left(n\right)$ by

$${\nabla }_q^{0}f\left(n\right)=f\left(n\right),{\displaystyle \sum _{n=0}^{N-1}}{q}^n{\nabla }_q^{1}f(n+1)=f\left(N\right),{\displaystyle \sum _{n=0}^{N-1}}{q}^{n}f(n+1)={\nabla }_q^{-1}f\left(N\right),{\nabla }_q^1={\nabla }_q.$$

Remark 1.

Equivalently, the operator ${\nabla }_q^1$ can be written explicitly as

$${\nabla }_q^{1}f\left(N\right)=q^{-(N-1)}\left(f\left(N\right)-f(N-1)\right),$$

which is a normalized backward difference.

Let $K_i,D_i\in \mathbb{C}$ and $T_i\in \mathbb{N}$. Define the nested sum $SU{M}_q\left(N\right)=SU{M}_q(N,PS,PT)$ recursively as follows:

$$SU{M}_q(N,[K_1:D_1],[T_1=1])={\displaystyle \sum _{n=0}^{N-1}}{q}^n(K_1+{\left[n\right]}_q{D}_1),$$

and for $M\ge 2$,

$$SU{M}_q(N,[K_1:D_1,\dots ,K_M:D_M],[T_1,\dots ,T_M])$$

$$={\displaystyle \sum _{n=0}^{N-1}}{q}^n(K_M+{\left[n\right]}_q{D}_M){\nabla }^{p}SU{M}_q(n+1,[K_1:D_1,\dots ,K_{M-1}:D_{M-1}],[T_1,\dots ,T_{M-1}]),$$

where $T_i=T_{i-1}+2-p$(with $p\in \mathbb{Z}$) and $T_i<T_{i+1}$. Thus, p is determined by the common difference of the $T_i$’s.

To simplify notation, we adopt the following abbreviations:

$$[K_1:D,K_2:D,\dots ,K_M:D]=[K_1,K_2,\dots ,K_M]:D,$$

and if $D=1$, we simply write $[K_1,K_2,\dots ,K_M]$. In this paper, unless otherwise stated, we assume

$$PS=[K_1:D_1,K_2:D_2,\dots ,K_M:D_M],PT=[T_1,T_2,\dots ,T_M].$$

The following special cases are particularly useful:

$$\begin{array}{cc}\hfill SU{M}_q(N,PS,[1,2,\dots ,M])& ={\displaystyle \sum _{n=0}^{N-1}}{q}^n{\displaystyle \prod _{i=1}^M}(K_i+{\left[n\right]}_q{D}_i),\hfill \\ \hfill SU{M}_q(N,PS,[1,3,\dots ,2M-1])& ={\displaystyle \sum _{n_M=0}^{N-1}}{q}^{n_M}(K_M+{\left[n_M\right]}_q{D}_M)\dots {\displaystyle \sum _{n_2=0}^{n_3}}{q}^{n_2}(K_2+{\left[n_2\right]}_q{D}_2){\displaystyle \sum _{n_1=0}^{n_2}}{q}^{n_1}(K_1+{\left[n_1\right]}_q{D}_1),\hfill \\ \hfill SU{M}_q(N,PS,[1,2,4])& ={\displaystyle \sum _{n_3=0}^{N-1}}{q}^{n_3}(K_3+{\left[n_3\right]}_q{D}_3){\displaystyle \sum _{n=0}^{n_3}}{q}^n(K_1+{\left[n\right]}_q{D}_1)(K_2+{\left[n\right]}_q{D}_2),\hfill \\ \hfill SU{M}_q(N,PS,[1,3,4])& ={\displaystyle \sum _{n_3=0}^{N-1}}{q}^{n_3}(K_3+{\left[n_3\right]}_q{D}_3)(K_2+{\left[n_3\right]}_q{D}_2){\displaystyle \sum _{n=0}^{n_3}}{q}^n(K_1+{\left[n\right]}_q{D}_1),\hfill \\ \hfill SU{M}_q(N,PS,[1,4])& ={\displaystyle \sum _{n_3=0}^{N-1}}{q}^{n_3}(K_2+{\left[n_3\right]}_q{D}_2){\displaystyle \sum _{n_2=0}^{n_3}}{q}^{n_2}{\displaystyle \sum _{n_1=0}^{n_2}}{q}^{n_1}(K_1+{\left[n_1\right]}_q{D}_1).\hfill \end{array}$$

Now, let $\mathbb{K}=\left\{K_i\right\}$ and $\mathbb{T}=\left\{T_i\right\}$. For each i, we choose $X_i\in \{T_i,K_i\}$. For all combinations in $\{X_1,\dots ,X_M\}$, define

$$X\left(T\right)=card\{i:X_i=T_i\},X\left(K\right)=M-X\left(T\right).$$

For each i, write $X_T,X_K$ for the counts of T-choices and K-choices among the first i variables, and $X_{T-1},X_{K-1}$ for those among the first $i-1$ variables. Then

$$X_{T-1}+X_{K-1}=i-1,X_T+X_K=i,X_T=\left\{\begin{array}{cc}{X}_{T-1}+1,\hfill & X_i=T_i,\hfill \\ X_{T-1},\hfill & X_i=K_i.\hfill \end{array}\right.$$

Using this notation, we can now state the main calculation theorem.

Theorem 1.

Let $M=\left|\right\{T_i\left\}\right|$, $H=T_M-M$, and set $T_0=0,X_{T-1}=0$ for $j=1$. Then

$$SU{M}_q(N,PS,PT)=For{m}_1=For{m}_2=For{m}_3.$$

Where

$$For{m}_1={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)G_{H+1+g}^{N+H},For{m}_2={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right)G_{H+1+g}^{N+H+g},For{m}_3={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)G_{T_M+1}^{N+T_M-g},$$

and ${\displaystyle H_r^q\left(g\right)=H_r^q(g,PS,PT)=H_r^q(g,M)=\sum _{X\left(T\right)=g}{\displaystyle \prod _{i=1}^M}{B}_i}$, with

$$\begin{array}{c}{H}_1^q\left(g\right):B_i=\left\{\begin{array}{cc}{q}^{1+(T_i-T_{i-1})X_{T-1}}{G}_1^{T_i-X_{K-1}}{D}_i,\hfill & X_i=T_i,\hfill \\ q^{(T_i-T_{i-1}-1)X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i),\hfill & X_i=K_i,\hfill \end{array}\right.\hfill \\ H_2^q\left(g\right):B_i=\left\{\begin{array}{cc}{q}^{-(T_i-X_{K-1})}{G}_1^{T_i-X_{K-1}}{D}_i,\hfill & X_i=T_i,\hfill \\ K_i-q^{-(T_i-X_{K-1})}{G}_1^{T_i-X_{K-1}}{D}_i,\hfill & X_i=K_i,\hfill \end{array}\right.\hfill \\ H_3^q\left(g\right):B_i=\left\{\begin{array}{cc}{q}^{1+(T_i-T_{i-1}-1)X_{T-1}}\left((q^{X_{T-1}}{G}_1^{T_i}-q^{T_i}{G}_1^{X_{T-1}})D_i-K_i{q}^{T_i}\right),\hfill & X_i=T_i,\hfill \\ q^{(T_i-T_{i-1}-1)X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i),\hfill & X_i=K_i.\hfill \end{array}\right.\hfill \end{array}$$

Proof. 

We prove the theorem by induction on M. Using the elementary summation formula (3):

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{G}_M^{n+K}=q^{M-K}{G}_{M+1}^{N+K}.$$

Indeed, for $M=1$, we have $PS=[K_1:D_1]$ and $PT=\left[T_1\right]$. Then

$$SU{M}_q\left(N\right)={\displaystyle \sum _{n=0}^{N-1}}{q}^n(K_1+{\left[n\right]}_q{D}_1),$$

and using ${\left[n\right]}_q=G_1^n$ together with (3), we obtain

$$SU{M}_q\left(N\right)=q^1{D}_1{G}_2^N+K_1{G}_1^N=q^{-1}{D}_1{G}_2^{N+1}+(K_1-q^{-1}{D}_1)G_1^N=(q^1{D}_1-K_1{q}^2)G_2^N+K_1{G}_2^{N+1},$$

which matches the three forms.

Assume now that the theorem holds for M variables. Consider the $(M+1)$-variable case with

$$P{S}_1=[PS,K_{M+1}:D_{M+1}],P{T}_1=[PT,T_{M+1}],$$

where $T_{M+1}=T_M+2-p$. Set

$$X=T_M-M+1-p=T_{M+1}-(M+1).$$

By definition,

$$SU{M}_q(N,P{S}_1,P{T}_1)={\displaystyle \sum _{n=0}^{N-1}}{q}^n(K_{M+1}+{\left[n\right]}_q{D}_{M+1}){\nabla }_q^{p}SU{M}_q(n+1,PS,PT).$$

By the induction hypothesis, $SU{M}_q(n+1,PS,PT)$ has the forms with $H_i^q\left(g\right)$; applying (3) to ${\nabla }_q^p$ gives

$${\nabla }_q^{p}SU{M}_q(n+1,PS,PT)={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)q^{-pg}{G}_{X+g}^{n+X}={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right)G_{X+g}^{n+X+g}={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)q^{-pg}{G}_{X+M}^{n+X+M-g}.$$

Now substitute the first form into the definition:

$$SU{M}_q(N,P{S}_1,P{T}_1)={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)q^{-pg}{\displaystyle \sum _{n=0}^{N-1}}{q}^n(K_{M+1}+{\left[n\right]}_q{D}_{M+1})G_{X+g}^{n+X}.$$

Using (3) and Lemma 1, we have

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{G}_{X+g}^{n+X}=q^g{G}_{X+g+1}^{N+X}.{\displaystyle \sum _{n=0}^{N-1}}{q}^n{\left[n\right]}_q{G}_{X+g}^{n+X}=q^{2g+1}{G}_1^{X+g+1}{G}_{X+g+2}^{N+X}+q^g{G}_1^g{G}_{X+g+1}^{N+X}.$$

Therefore,

$$SU{M}_q(N,P{S}_1,P{T}_1)={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)q^{(1-p)g}(K_{M+1}+G_1^g{D}_{M+1})G_{X+g+1}^{N+X}+{\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)q^{(2-p)g+1}{G}_1^{X+g+1}{D}_{M+1}{G}_{X+g+2}^{N+X}.$$

Since $X+g+1=T_{M+1}-(M+1)+g+1$ and $N+X=N+T_{M+1}-(M+1)$, the first sum contributes to

$$G_{T_{M+1}-(M+1)+g+1}^{N+T_{M+1}-(M+1)}.$$

After shifting the index g to $g+1$ in the second sum and re-indexing, we obtain

$$H_1^q(g,M+1)=H_1^q\left(g\right)q^{(T_{M+1}-T_M-1)g}(K_{M+1}+G_1^g{D}_{M+1})+H_1^q(g-1)q^{1+(T_{M+1}-T_M)(g-1)}{G}_1^{T_{M+1}-(M+1-g)}{D}_{M+1},$$

with the convention $H_1^q(-1)=H_1^q(M+1)=0$. This matches the definition of $B_{1,j}$ for $For{m}_1$ in the theorem. Hence

$$SU{M}_q(N,P{S}_1,P{T}_1)={\displaystyle \sum _{g=0}^{M+1}}{H}_1^q(g,M+1)G_{T_{M+1}-(M+1)+1+g}^{N+T_{M+1}-(M+1)}.$$

The proofs for $For{m}_2$ and $For{m}_3$ are analogous. □

Example 1.

For $PS=[A,B,C]$, $PT=[1,3,5]$, direct expansion yields explicit coefficient:

$$\begin{array}{cc}\hfill H_1^q\left(0\right)& =ABC,\hfill \\ \hfill H_1^q\left(1\right)& =AB{q}^1{G}_1^3+A{q}^1{G}_1^2(C+1)+q^1{G}_1^1{q}^1(B+1)q^1(C+1),\hfill \\ \hfill H_1^q\left(2\right)& =q^1{G}_1^1{q}^3{G}_1^3{q}^2(C+G_1^2)+q^1{G}_1^1{q}^1(B+1)q^3{G}_1^4+A{q}^1{G}_1^2{q}^3{G}_1^4,\hfill \\ \hfill H_1^q\left(3\right)& =q^1{G}_1^1{q}^3{G}_1^3{q}^5{G}_1^5.\hfill \end{array}$$

Here, $A{q}^1{G}_1^2(C+1)$ in $H_1^q\left(1\right)$ corresponds to $K_1{T}_2{K}_3$, $B_1=A$, $B_2=q^{1+X_{T-1}}{G}_1^{3-X_{K-1}}=q^1{G}_1^2$, $B_3=(C+1)$.

$$\begin{array}{cc}\hfill H_2^q\left(0\right)& =(A-q^{-1}{G}_1^1)(B-q^{-2}{G}_1^2)(C-q^{-3}{G}_1^3),\hfill \\ \hfill H_2^q\left(1\right)& =(A-q^{-1}{G}_1^1)(B-q^{-2}{G}_1^2)q^{-3}{G}_1^3+(A-q^{-1}{G}_1^1)q^{-2}{G}_1^2(C-q^{-4}{G}_1^4)+q^{-1}{G}_1^1(B-q^{-3}{G}_1^3)(C-q^{-4}{G}_1^4),\hfill \\ \hfill H_2^q\left(2\right)& =q^{-1}{G}_1^1{q}^{-3}{G}_1^3(C-q^{-5}{G}_1^5)+q^{-1}{G}_1^1(B-q^{-3}{G}_1^3)q^{-4}{G}_1^4+(A-q^{-1}{G}_1^1)q^{-2}{G}_1^2{q}^{-4}{G}_1^4,\hfill \\ \hfill H_2^q\left(3\right)& =q^{-1}{G}_1^1{q}^{-3}{G}_1^3{q}^{-5}{G}_1^5.\hfill \end{array}$$

$$\begin{array}{cc}\hfill H_3^q\left(0\right)& =ABC,\hfill \\ \hfill H_3^q\left(1\right)& =AB{q}^1(G_1^5-q^{5}C)+A{q}^2(G_1^3-q^{3}B)(C+1)+q^3(1-q^{1}A)(B+1)(C+1),\hfill \\ \hfill H_3^q\left(2\right)& =q^5(1-q^{1}A)(q^1{G}_1^3-q^3-q^{3}B)(C+G_1^2)+q^4(1-q^{1}A)(B+1)(q^1{G}_1^5-q^5-q^{5}C)+A{q}^3(G_1^3-q^{3}B)(q^1{G}_1^5-q^5-q^{5}C),\hfill \\ \hfill H_3^q\left(3\right)& =q^6(1-q^{1}A)(q^1{G}_1^3-q^3-q^{3}B)(q^2{G}_1^5-q^5{G}_1^2-q^{5}C).\hfill \end{array}$$

Specifically, we have

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{\displaystyle \prod _{i=1}^M}(K_i+D_i{q}^n)=SU{M}_q(N,[K_i+D_i:D_i(q-1)],[1,2,\dots ,M]).$$

Proceeding analogously to the $H_1^q\left(g\right)$ case, we obtain by induction:

Theorem 2.

For constants $K_i,D_i$,

$${\displaystyle \sum _{n=0}^{N-1}}{\displaystyle \prod _{i=1}^M}(K_i+D_i{q}^n)={\displaystyle \sum _{g=1}^M}\left(\sum _{X\left(T\right)=g}{\displaystyle \prod _{j=1}^M}{B}_j\right)G_g^N+N{\displaystyle \prod _{i=1}^M}{K}_i,$$

where

$${ϵ}_j=\left\{\begin{array}{cc}0,\hfill & X_{T-1}=0,\hfill \\ 1,\hfill & X_{T-1}>0.\hfill \end{array}\right.,B_j=\left\{\begin{array}{cc}{K}_j+{ϵ}_j{q}^{X_{T-1}-1}{D}_j,\hfill & X_j=K_j,\hfill \\ D_j\left(1+{ϵ}_j(q^{2X_{T-1}}-q^{X_{T-1}}-1)\right),\hfill & X_j=T_j.\hfill \end{array}\right.$$

Now, taking the limit $q\to 1$ . We define

$$\underset{q\to 1}{lim}{\nabla }_q^p={\nabla }^p,\underset{q\to 1}{lim}{H}_r^q\left(g\right)=H_r\left(g\right),\underset{q\to 1}{lim}SU{M}_q\left(N\right)=SUM\left(N\right).$$

Since ${lim}_{q\to 1}{\left[n\right]}_q=n$ and ${lim}_{q\to 1}{G}_M^N=\left({\displaystyle \genfrac{}{}{0pt}{}{N}{M}}\right)$, the nested summation formula for $K_i+n{D}_i$ follows immediately.

Lemma 2

(Closed form for finite q-binomial sums). Let $0\le A\le M$, $N>M-A$, $a\ne 0,1,q^{-1},q^{-2},\dots$. Then

$${\displaystyle \sum _{n=0}^{N-1}}{a}^n{G}_M^{n+A}=-a^N{\displaystyle \sum _{g=0}^M}{\displaystyle \frac{q^{(N+A-M)g}{G}_{M-g}^{N+A-1-g}}{{(a;q)}_{g+1}}}+{\displaystyle \frac{a^{M-A}}{{(a;q)}_{M+1}}}.$$

Proof. 

We divide the proof into two parts.

• Step 1: The case $A=M$.

It suffices to prove

$${\displaystyle \sum _{n=0}^{N-1}}{a}^n{G}_M^{n+M}=-a^N{\displaystyle \sum _{g=0}^M}{\displaystyle \frac{q^{Ng}{G}_{M-g}^{N+M-1-g}}{{(a;q)}_{g+1}}}+{\displaystyle \frac{1}{{(a;q)}_{M+1}}}.$$

We proceed by induction on M and, for each fixed M, by induction on N.

Base case $N=1$: The left-hand side is $G_M^M=1$. For the right-hand side, we need

$$-a{\displaystyle \sum _{g=0}^M}{\displaystyle \frac{q^g}{{(a;q)}_{g+1}}}+{\displaystyle \frac{1}{{(a;q)}_{M+1}}}=1,$$

which follows from the identity

$${\displaystyle \sum _{g=0}^M}{\displaystyle \frac{q^g}{{(a;q)}_{g+1}}}={\displaystyle \frac{1-{(a;q)}_{M+1}}{a{(a;q)}_{M+1}}},$$

obtained by induction on M using ${\displaystyle \frac{1}{{(a;q)}_{g+1}}}={\displaystyle \frac{1}{{(a;q)}_g}}+{\displaystyle \frac{a{q}^g}{{(a;q)}_{g+1}}}$.

Inductive step on N: Assume (2.0.0.1) holds for $N-1$. Using $G_M^{n+M}=q^M{G}_M^{n-1+M}+G_{M-1}^{n-1+M}$ for $n\ge 1$, we get

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=0}^{N-1}}{a}^n{G}_M^{n+M}& =1+a{q}^M{\displaystyle \sum _{n=0}^{N-2}}{a}^n{G}_M^{n+M}+a{\displaystyle \sum _{n=0}^{N-2}}{a}^n{G}_{M-1}^{n+M-1}.\hfill \end{array}$$

Using induction on N for the first sum and on $M-1$ for the second, then simplifying via the q-binomial recurrence, gives (2.0.0.1) for N, completing the case $A=M$.

• Step 2: Reduction to general $A<M$.

Shifting $m=n-(M-A)$, we have

$${\displaystyle \sum _{n=0}^{N-1}}{a}^n{G}_M^{n+A}=a^{M-A}{\displaystyle \sum _{m=0}^{N-M+A-1}}{a}^m{G}_M^{m+M}.$$

Applying (2.0.0.1) (with N replaced by $N-M+A$) and simplifying the exponent of q yields the desired identity. □

Theorem 3.

Let $X=T_M-M-p\ge -1$, $Y\in \{0,1\}$, and $f\left(g\right)={(a{q}^{2+X+g};q)}_{M-g}$. Then:

(1) 

The three coefficient forms share the common value

$$A_a^q(PS,PT,p):={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)a^g{q}^{-pg}f\left(g\right)={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right)f\left(g\right)={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)a^g{q}^{-pg}.$$

(2) 

For finite N,

$${\displaystyle \sum _{n=0}^{N-1}}{a}^n{\nabla }_q^p{\mathrm{SUM}}_q(n+Y)=-a^N{\displaystyle \sum _{k=0}^M}{\displaystyle \frac{q^{(N+Y-1)k}{\nabla }_q^{p+k}{\mathrm{SUM}}_q(N+Y-1)}{{(a;q)}_{k+1}}}+{\displaystyle \frac{a^{1-Y}{A}_a^q(PS,PT,p)}{{(a;q)}_{T_M+2-p}}}.$$

(3) 

If $\left|a\right|,\left|q\right|<1$, then

$${\displaystyle \sum _{n=0}^{\infty }}{a}^n{\nabla }_q^p{\mathrm{SUM}}_q(n+Y)={\displaystyle \frac{a^{1-Y}{A}_a^q(PS,PT,p)}{{(a;q)}_{T_M+2-p}}},{\mathrm{SUM}}_q(\infty ):=\underset{N\to \infty }{lim}{\mathrm{SUM}}_q\left(N\right)={\displaystyle \frac{A_q^q(PS,PT,1)}{{(q;q)}_{T_M+1}}}.$$

Proof. 

We first record the three expansions from Theorem 1 with $X=T_M-M-p$,

$${\nabla }_q^p{\mathrm{SUM}}_q(n+Y)={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)q^{-pg}{G}_{1+X+g}^{n+Y+X}={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right)G_{1+X+g}^{n+Y+X+g}={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)q^{-pg}{G}_{1+X+M-p}^{n+Y+X+M-g}.$$

• Proof of (2).

Apply Lemma 2 to $For{m}_1$ of ${\mathrm{SUM}}_q(n+Y)$. For fixed g, we have

$${\displaystyle \sum _{n=0}^{N-1}}{a}^n{G}_{1+X+g}^{n+Y+X}=-a^N{\displaystyle \sum _{k=0}^{1+X+g}}{\displaystyle \frac{q^{(N+Y-1)k}{G}_{1+X+g-k}^{N+Y+X-1-k}}{{(a;q)}_{k+1}}}+{\displaystyle \frac{a^{1+g-Y}}{{(a;q)}_{2+X+g}}}.$$

Multiplying by $H_1^q\left(g\right)q^{-pg}$ and summing over g, the first term on the right becomes

$$-a^N{\displaystyle \sum _{k=0}^M}{\displaystyle \frac{q^{(N+Y-1)k}}{{(a;q)}_{k+1}}}{\nabla }_q^{p+k}{\mathrm{SUM}}_q(N+Y-1)$$

The second term equals

$$a^{1-Y}{\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)a^g{q}^{-pg}{(a{q}^{2+X+g};q)}_{M-g}.$$

Define this quantity as $a^{1-Y}{A}_a^q(PS,PT,p)$; this is a temporary definition. Thus (2) holds.

• Proof of (1).

Now apply the same argument to $For{m}_2$ of ${\mathrm{SUM}}_q(n+Y)$. For each fixed g, Lemma 2 gives

$${\displaystyle \sum _{n=0}^{N-1}}{a}^n{G}_{1+X+g}^{n+Y+X+g}=-a^N{\displaystyle \sum _{k=0}^{1+X+g}}{\displaystyle \frac{q^{(N+Y-1)k}{G}_{1+X+g-k}^{N+Y+X+g-1-k}}{{(a;q)}_{k+1}}}+{\displaystyle \frac{a^{1-Y}}{{(a;q)}_{2+X+g}}}.$$

After multiplying by $H_2^q\left(g\right)$ and summing, the first term again becomes

$$-a^N{\displaystyle \sum _{k=0}^M}{\displaystyle \frac{q^{(N+Y-1)k}}{{(a;q)}_{k+1}}}{\nabla }_q^{p+k}{\mathrm{SUM}}_q(N+Y-1),$$

because the same expression ${\nabla }_q^{p+k}{\mathrm{SUM}}_q$ is obtained (with p shifted). The second term becomes

$$a^{1-Y}{\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right){(a{q}^{2+X+g};q)}_{M-g}.$$

Since two identities are both equal to ${\sum }_{n=0}^{N-1}{a}^n{\nabla }_q^p{\mathrm{SUM}}_q(n+Y)$, thus the second terms are equal:

$${\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)a^g{q}^{-pg}{(a{q}^{2+X+g};q)}_{M-g}={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right){(a{q}^{2+X+g};q)}_{M-g}.$$

Applying Lemma 2 to $For{m}_3$ similarly completes the proof of (1).

• Proof of (3).

Taking $N\to \infty$ in (2) (with $\left|a\right|<1$) gives

$${\displaystyle \sum _{n=0}^{\infty }}{a}^n{\nabla }_q^p{\mathrm{SUM}}_q(n+Y)={\displaystyle \frac{a^{1-Y}{A}_a^q(PS,PT,p)}{{(a;q)}_{T_M+2-p}}}.$$

Setting $a=q$ and using the definition of ${\mathrm{SUM}}_q$ (where the standard case is $Y=1,p=1$) yields

$${SUM}_q(\infty )={\displaystyle \sum _{n=0}^{\infty }}{q}^n{\nabla }_q^1{\mathrm{SUM}}_q(n+1)={\displaystyle \frac{A_q^q(PS,PT,1)}{{(q;q)}_{T_M+1}}}.$$

□

3. Properties

In this section, we collect several useful properties of $SU{M}_q$ and the coefficients $H_i^q\left(g\right)$.

Definition 2.

Define the q-numbers

$${\left[n\right]}_{q-}=q^{-n}{G}_1^n,{\left[n\right]}_{q+}=q^n{G}_1^n,$$

and their factorials:

$${[n!]}_{q-}={\left[n\right]}_{q-}{[n-1]}_{q-}\dots {\left[1\right]}_{q-},{[n!]}_{q+}={\left[n\right]}_{q+}{[n-1]}_{q+}\dots {\left[1\right]}_{q+},{[0!]}_{q-}={[0!]}_{q+}=1.$$

Theorem 4

(Basic Properties). The following properties hold.

(1) 

For $PT=[1,2,\dots ,M]$,

$${\nabla }_q{\mathrm{SUM}}_q(n+1)={\displaystyle \prod _{i=1}^M}(K_i+{\left[n\right]}_q{D}_i).$$

(2) 

The pairs $(K_i:D_i)$ may be permuted freely within each consecutive block of the $T_i$’s ($T_{i+1}=T_i+1$).

(3) 

For any integer p, with $X=T_M-M-p$,

$${\nabla }_q^p{\mathrm{SUM}}_q\left(N\right)={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)q^{-pg}{G}_{X+1+g}^{N+X}={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right)G_{X+1+g}^{N+X+g}={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)q^{-pg}{G}_{X+M+1}^{N+X+M-g}.$$

(4) 

For any positive integers $L_1,\dots ,L_Q$,

$${\mathrm{SUM}}_q(N,[{\left[L_1\right]}_{q-},\dots ,{\left[L_Q\right]}_{q-},P{S}_1],[L_1,\dots ,L_Q,P{T}_1])=\prod {\left[L_i\right]}_{q-}{\mathrm{SUM}}_q(N,P{S}_1,P{T}_1).$$

Thus $T_1$ can be greater than 1.

(5) 

In particular,

$${\mathrm{SUM}}_q(N,[{\left[T_1\right]}_{q-},\dots ,{\left[T_M\right]}_{q-}],[T_1,\dots ,T_M])=\prod {\left[T_i\right]}_{q-}{G}_{T_M+1}^{N+T_M}.$$

(6) 

For $H_r^q\left(g\right)$ as defined in Theorem 1,

$$\sum _{X\left(T\right)=g,X_i\in \mathbb{K}}\prod q^{X_T}=G_{M-g}^M=G_g^M.$$

Proof. 

(1) Follows directly from the definition of ${\nabla }_q$.

(2)

Follows directly from the definition of ${\mathrm{SUM}}_q$ (factors within the same nesting level commute).

(3)

This was established during the proof of Theorem 1; included here only for reference.

(4)

Observe the $For{m}_2$: when Q factors ${\left[L_1\right]}_{q-},\dots ,{\left[L_Q\right]}_{q-}$ are prepended to $PS$, we have $H_2^q\left(g\right)=0$ for $g<Q$, and for $g\ge Q$ they equal ${\prod }_{i=1}^Q{\left[L_i\right]}_{q-}{H}_2^q(g-Q,P{S}_1,P{T}_1)$. Hence the whole sum is multiplied by ${\prod }_{i=1}^Q{\left[L_i\right]}_{q-}$.

(5)

This is the special case of (4).

(6)

This is the combinatorial identity

$$G_K^M=\sum _{w\in \Omega (0^{M-K},1^K)}{q}^{inv\left(w\right)},$$

stated at the beginning of Section 1. With$K=M-g$, this sum is precisely the generating function ${\sum }_w{q}^{inv\left(w\right)}$ over binary words with $M-g$ zeros and g ones.

□

Using property (5) with $PT=[1,3\dots 2M-1]$, we obtain a nice closed form:

$${\displaystyle \sum _{n_M=0}^{N-1}}\dots {\displaystyle \sum _{n_1=0}^{n_2}}{q}^{\sum n_i}{[1+n_1]}_q{[3+n_2]}_q\dots {[2M-1+n_M]}_q={\left[1\right]}_q{\left[3\right]}_q\dots {[2M-1]}_q{G}_{2M}^{N+2M-1}.$$

Theorem 5

(Basic q-binomial identities). The following identities hold:

$$\begin{array}{cc}\hfill \sum _{0\le {\lambda }_1\le \dots \le {\lambda }_M\le N}{q}^{\sum {\lambda }_i}& =G_M^{N+M},\hfill \end{array}$$

(A1)

$$\begin{array}{cc}\hfill \sum _{A\le {\lambda }_1<\dots <{\lambda }_M\le B}{q}^{\sum {\lambda }_i}& =q^{\left({\displaystyle \genfrac{}{}{0pt}{}{M}{2}}\right)+AM}{G}_M^{B-A+1},(A\le B),\hfill \end{array}$$

(A2)

$$\begin{array}{cc}\hfill {\displaystyle \prod _{i=1}^M}(a+q^{A+i}z)& ={\displaystyle \sum _{g=0}^M}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g}{2}}\right)+(A+1)g}{G}_g^M{z}^g{a}^{M-g}.\hfill \end{array}$$

(A3)

Proof. 

(1) Take $PS=[1,\dots ,1]:0$ and $PT=[1,3,\dots ,2M-1]$. By definition,

$${\mathrm{SUM}}_q(N+1,PS,PT)=\sum _{0\le {\lambda }_1\le \dots \le {\lambda }_M\le N}{q}^{\sum {\lambda }_i}.$$

Only $H_1^q\left(0\right)=1$ is nonzero (all other $H_1^q\left(g\right)=0$ since $D_i=0$). Hence the sum equals $G_M^{N+M}$.

(2) follows from (A1) by shifting ${\lambda }_i\mapsto {\lambda }_i-(A+i-1)$.

(3) follows by expanding the product:

$${\displaystyle \prod _{i=1}^M}(a+q^{A+i}z)={\displaystyle \sum _{g=0}^M}{z}^g{a}^{M-g}\sum _{1\le i_1<\dots <i_g\le M}{q}^{{\sum }_{j=1}^g(A+i_j)}.$$

The inner sum is evaluated by (2) as $q^{Ag+\left({\displaystyle \genfrac{}{}{0pt}{}{g}{2}}\right)+g}{G}_g^M$. □

Remark 2.

Part (1) is Cayley’s theorem:

$${\displaystyle \sum _{n=0}^{MN}}p(n;M,N)q^n=G_M^{N+M},$$

where $p(n;M,N)$ denotes the number of partitions of n into at most M parts, each at most N (see [5]). Part (3) is Rothe’s q-binomial theorem($A=-1$): ${\prod }_{i=1}^M(a+q^{i-1}z)={\sum }_{g=0}^M{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g}{2}}\right)}{G}_g^M{z}^g{a}^{M-g}$.

We show that MacMahon’s q-binomial theorem follows directly from (2).

$${\displaystyle \prod _{i=1}^A}(a+q^{(2i-1)D}z)={\displaystyle \sum _{g=0}^A}{q}^{g^{2}D}{\left[\begin{array}{c}A\\ g\end{array}\right]}_{q^{2D}}{z}^g{a}^{A-g},{\displaystyle \prod _{i=1}^B}(a^{-1}+q^{(2i-1)D}{z}^{-1})={\displaystyle \sum _{h=0}^B}{q}^{h^{2}D}{\left[\begin{array}{c}B\\ h\end{array}\right]}_{q^{2D}}{z}^{-h}{a}^{-(B-h)}.$$

Multiplying these two expressions and extracting the coefficient of $z^k$ yields

$$f\left(k\right)={\displaystyle \sum _{i=0}^{A+B}}{\left[\begin{array}{c}A\\ i\end{array}\right]}_{q^{2D}}{\left[\begin{array}{c}B\\ i-k\end{array}\right]}_{q^{2D}}{q}^{i^{2}D+{(i-k)}^{2}D}{a}^{A-i}{a}^{-(B-i+k)}.$$

After simplification and application of the q-Vandermonde identity, this reduces to

$$f\left(k\right)=q^{k^{2}D}{a}^{A-B-k}{\left[\begin{array}{c}A+B\\ A-k\end{array}\right]}_{q^{2D}}.$$

Thus we obtain the two-parameter identity

$${\displaystyle \prod _{i=1}^A}(a+q^{(2i-1)D}z){\displaystyle \prod _{i=1}^B}(a^{-1}+q^{(2i-1)D}{z}^{-1})={\displaystyle \sum _{k=-B}^A}{q}^{k^{2}D}{a}^{A-B-k}{\left[\begin{array}{c}A+B\\ A-k\end{array}\right]}_{q^{2D}}{z}^k.$$

Setting $a=D=1$ gives MacMahon’s q-binomial theorem in its standard form:

$${(-qz;q^2)}_A{(-q{z}^{-1};q^2)}_B={\displaystyle \sum _{k=-B}^A}{q}^{k^2}{\left[\begin{array}{c}A+B\\ A-k\end{array}\right]}_{q^2}{z}^k.$$

Two variations of the q-binomial theorem.

The following identities, which will be used later in this paper, can be proved by induction on M using the standard recurrence for $G_k^M$:

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=0}^M}{(-1)}^{k+g}{G}_k^M{G}_g^k{q}^{{\displaystyle \frac{k(k-1)}{2}}-gk+Ak}{z}^k& =q^{-{\displaystyle \frac{g(g+1)}{2}}+Ag}{G}_g^M{(z{q}^A;q)}_{M-g}{z}^g;\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=0}^M}{(-1)}^k{G}_k^M{G}_{M-g}^{M-k}{q}^{{\displaystyle \frac{k(k-1)}{2}}+Ak}{z}^k& =G_g^M{(z{q}^A;q)}_g.\hfill \end{array}$$

(6)

In light of the above preparations, we now present one of the main theorems of this paper.

Theorem 6

(Mutual transformations). For $PT=[1,2,\dots ,M]$,

$$\begin{array}{cc}\hfill H_1^q\left(g\right)& =q^{g(g+1)}\sum _k{H}_2^q\left(k\right)G_g^k,\hfill \end{array}$$

(A1)

$$\begin{array}{cc}\hfill H_1^q\left(g\right)& =\sum _k{H}_3^q\left(k\right)G_{M-g}^{M-k}{q}^{(g+1)(g-k)},\hfill \end{array}$$

(A2)

$$\begin{array}{cc}\hfill H_2^q\left(g\right)& =\sum _k{(-1)}^{k+g}{G}_g^k{q}^{{\displaystyle \frac{g(g+1)-k(k+3)}{2}}-kg}{H}_1^q\left(k\right),\hfill \end{array}$$

(A3)

$$\begin{array}{cc}\hfill H_3^q\left(g\right)& =\sum _k{(-1)}^{k+g}{G}_{M-g}^{M-k}{q}^{{\displaystyle \frac{g(g+3)-k(k+3)}{2}}}{H}_1^q\left(k\right),\hfill \end{array}$$

(A4)

$$\begin{array}{cc}\hfill H_3^q\left(g\right)& ={(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}}\sum _k{H}_2^q\left(k\right)G_g^{M-k}{q}^{gk},\hfill \end{array}$$

(A5)

$$\begin{array}{cc}\hfill H_2^q\left(g\right)& ={(-1)}^{M-g}{q}^{{\displaystyle \frac{(M-g)(M-g-1)}{2}}}\sum _k{H}_3^q\left(k\right)G_{M-g}^k{q}^{-(M+1)k}.\hfill \end{array}$$

(A6)

Proof. 

We prove (1) in detail; the proofs of (2)–(4) are analogous.

Since the identities are homogeneous in each pair $(K_i,D_i),D_i\ne 0$, we may first set $D_i=1$ for all i by replacing $K_i$ with $K_i/D_i$; the general case then follows by scaling back. Thus we prove (1) under the normalization $D_i=1$.

For $M=1$, direct computation verifies the identity. Assume (1) holds for M. Append a new pair $(K_{M+1},1)$ with $T_{M+1}=M+2-p$. Let $H_1^q\left(g\right)$ and $H_2^q\left(g\right)$ denote the coefficients for M variables. For $M+1$, by the definition of $H_1^q$,

$$H_1^q(g,M+1)=H_1^q(g-1)q^{1+X_T}{G}_1^{1+X_T}+H_1^q\left(g\right)(K_{M+1}+G_1^{X_T}),$$

Substituting the induction hypothesis for $H_1^q(g-1)$ and $H_1^q\left(g\right)$, we get

$$\begin{array}{cc}\hfill H_1^q(g,M+1)& =q^g{G}_1^g·q^{(g-1)g}\sum _k{H}_2^q\left(k\right)G_{g-1}^k+(K_{M+1}+G_1^g)q^{g(g+1)}\sum _k{H}_2^q\left(k\right)G_g^k\hfill \\ & =q^{g(g+1)}\sum _k{H}_2^q\left(k\right)\left(q^{-g}{G}_1^g{G}_{g-1}^k+(K_{M+1}+G_1^g)G_g^k\right).\hfill \end{array}$$

On the other hand, for $M+1$,

$$H_2^q(k,M+1)=H_2^q(k-1)q^{-k}{G}_1^k+H_2^q\left(k\right)(K_{M+1}-q^{-(k+1)}{G}_1^{k+1}).$$

Hence

$$q^{g(g+1)}\sum _k{H}_2^q(k,M+1)G_g^k=q^{g(g+1)}\sum _k{H}_2^q\left(k\right)\left(q^{-(k+1)}{G}_1^{k+1}{G}_g^{k+1}+(K_{M+1}-q^{-(k+1)}{G}_1^{k+1})G_g^k\right).$$

The difference between the two expressions vanishes: after substituting $G_{g-1}^k=G_g^{k+1}-q^g{G}_g^k$, the $K_{M+1}$-terms cancel and the remaining terms reduce to zero by direct simplification of the q-binomial coefficients. Therefore $H_1^q(g,M+1)=q^{g(g+1)}{\sum }_k{H}_2^q(k,M+1)G_g^k$, proving (1). The remaining identities (2)–(4) are obtained by the same induction.

In Theorem 3, take $PT=[1,2,\dots ,M]$ and $p=0$. Define

$$f\left(g\right)=(1-a{q}^{2+g})(1-a{q}^{3+g})\dots (1-a{q}^{M+1})={(a{q}^{2+g};q)}_{M-g}.$$

By the standard Rothe identity,

$${(a{q}^{2+g};q)}_{M-g}={\displaystyle \sum _{k=0}^{M-g}}{G}_k^{M-g}{(-1)}^k{q}^{{\displaystyle \frac{k(k-1)}{2}}}{\left(a{q}^{2+g}\right)}^k={\displaystyle \sum _{k=0}^{M-g}}{G}_k^{M-g}{(-1)}^k{q}^{{\displaystyle \frac{k(k-1)}{2}}+(2+g)k}{a}^k.$$

and Theorem 3(1) ($p=0$), Thus

$${\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right)f\left(g\right)={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right){\displaystyle \sum _{k=0}^{M-g}}{G}_k^{M-g}{(-1)}^k{q}^{{\displaystyle \frac{k(k-1)}{2}}+(2+g)k}{a}^k={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)a^g.$$

Comparing coefficients of $a^g$ on both sides (interchanging the order of summation in the double sum) yields

$$H_3^q\left(g\right)={\displaystyle \sum _{k=0}^M}{H}_2^q\left(k\right){(-1)}^g{G}_g^{M-k}{q}^{{\displaystyle \frac{g(g-1)}{2}}+(2+k)g},$$

which is (5).

From (5), replace g by r and set $k\mapsto M-k$ to obtain

$$H_3^q\left(r\right)={(-1)}^r{q}^{{\displaystyle \frac{r(r-1)}{2}}+(M+2)r}{\displaystyle \sum _{k=0}^M}{H}_2^q(M-k)G_r^k{q}^{-rk}.$$

Multiply both sides by $G_g^r{q}^{-(M+1)r}$ and sum over r:

$${\displaystyle \sum _{r=0}^M}{H}_3^q\left(r\right)G_g^r{q}^{-(M+1)r}={\displaystyle \sum _{k=0}^M}{H}_2^q(M-k){\displaystyle \sum _{r=0}^M}{(-1)}^r{G}_g^r{G}_r^k{q}^{{\displaystyle \frac{r(r-1)}{2}}-(1-k)r}.$$

By (8) (with $z=1$, $A=g+1-k$, and $M=k$),

$${\displaystyle \sum _{r=0}^M}{(-1)}^r{G}_g^r{G}_r^k{q}^{{\displaystyle \frac{r(r-1)}{2}}-gr+(g+1-k)r}={(-1)}^g{q}^{-{\displaystyle \frac{g(g+1)}{2}}+(g+1-k)g}{G}_g^k{(q^{g+1-k},q)}_{k-g}={(-1)}^g{q}^{-{\displaystyle \frac{g(g-1)}{2}}}{\delta }_{g,k}.$$

(For $k>g$, ${(q^{g+1-k},q)}_{k-g}=0$; the only non-zero term occurs when $k=g$, yielding ${(-1)}^g{q}^{-{\displaystyle \frac{g(g-1)}{2}}}$.)

Substituting this into the previous equation yields

$$H_2^q(M-g)={(-1)}^g{q}^{{\displaystyle \frac{g(g-1)}{2}}}{\displaystyle \sum _{r=0}^M}{H}_3^q\left(r\right)G_g^r{q}^{-(M+1)r}.$$

Applying the substitution $g\mapsto M-g$ and relabelling r as k gives (6). □

We also define the symmetric functions

$$F_g^{\mathbb{K}}=\sum _{1\le {\lambda }_1<\dots <{\lambda }_g\le M}{\displaystyle \prod _{i=1}^g}{K}_{{\lambda }_i},E_g^{\mathbb{K}}=\sum _{1\le {\lambda }_1\le \dots \le {\lambda }_g\le M}{\displaystyle \prod _{i=1}^g}{K}_{{\lambda }_i},$$

and their q-analogues

$$E_g^{N,q}=\sum _{1\le {\lambda }_1\le \dots \le {\lambda }_g\le N}{\displaystyle \prod _{i=1}^g}{\left[{\lambda }_i\right]}_q=S_2^q(N+g,N),E_g^{N,q-}=\sum _{1\le {\lambda }_1\le \dots \le {\lambda }_g\le N}{\displaystyle \prod _{i=1}^g}{\left[{\lambda }_i\right]}_{q-}=S_2^{q-}(N+g,N).$$

Theorem 7.

Assume $PT=[1,2,\dots ,M]$ and $D_i=1$. Then, for $H_1^q\left(g\right)$,

$$\sum _{X\left(T\right)=g,X_i=K_i}\prod B_i=F_{M-g}^{\mathbb{K}}{E}_0^{g,q}+F_{M-g-1}^{\mathbb{K}}{E}_1^{g,q}+...+F_0^{\mathbb{K}}{E}_{M-g}^{g,q}.$$

Proof. 

Recall the general expression for $B_i$ in Form 1:

$$B_i=\left\{\begin{array}{cc}{q}^{(T_i-T_{i-1}-1)X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i),\hfill & X_i=K_i,\hfill \\ q^{1+(T_i-T_{i-1})X_{T-1}}{G}_1^{T_i-X_{K-1}}{D}_i,\hfill & X_i=T_i.\hfill \end{array}\right.$$

Since $T_i+1=T_{i+1}$ and $D_i=1$, $T_i-X_{K-1}=i-((i-1)-X_{T-1})=1+X_{T-1}$, the factor $B_i$ simplifies to

$$B_i=\left\{\begin{array}{cc}{K}_i+G_1^{X_T},\hfill & X_i=K_i,\hfill \\ q^{X_T}{G}_1^{X_T},\hfill & X_i=T_i.\hfill \end{array}\right.$$

For a fixed choice of g indices with $X_i=T_i$, let their positions be ${\tau }_1<\dots <{\tau }_g$. Then $X_T=j$ for the j-th such index, so

$$\prod _{X_i=T_i}{B}_i={\displaystyle \prod _{j=1}^g}{q}^j{G}_1^j={[g!]}_{q+}.$$

Thus the T-terms contribute a common factor ${[g!]}_{q+}$ independent of the specific positions. Therefore,

$$H_1^q\left(g\right)={[g!]}_{q+}\sum _{X_i=K_i}\prod B_i.$$

Now let the K-term indices be ${\lambda }_1<\dots <{\lambda }_{M-g}$. For these indices, $X_T$ equals the number of T-terms preceding ${\lambda }_j$. Expanding the product over the K-terms,

$$\prod _{X_i=K_i}(K_i+G_1^{X_T})={\displaystyle \sum _{r=0}^{M-g}}\left(\sum _{{\mu }_1<\dots <{\mu }_r}{K}_{{\mu }_1}\dots K_{{\mu }_r}\right)\left(\sum _{\begin{array}{c}1\le {\nu }_1\le \dots \le {\nu }_{M-g-r}\\ {\nu }_j\le X_T\left({\mu }_j\right)\end{array}}{G}_1^{{\nu }_1}\dots G_1^{{\nu }_{M-g-r}}\right).$$

The second sum is independent of the specific positions of the T-terms and equals $E_{M-g-r}^{g,q}$. Hence

$$\sum _{X_i=K_i}\prod B_i={\displaystyle \sum _{r=0}^{M-g}}{F}_r^{\mathbb{K}}{E}_{M-g-r}^{g,q}.$$

□

In this article we frequently use the following triple representation: for $1+p\ge 0$,

$${\displaystyle \sum _{g=0}^M}{a}_g{G}_{1+p+g}^X={\displaystyle \sum _{g=0}^M}{b}_g{G}_{1+p+g}^{X+g}={\displaystyle \sum _{g=0}^M}{c}_g{G}_{1+p+M}^{X+M-g},$$

where we set $a_g^{*}:=a_g{q}^{-pg}$ and $c_g^{*}:=c_g{q}^{-pg}$.

Now consider the special case $D_i=1$ and $PT=[1,2,\dots ,M]$. From Theorem 1 ($For{m}_1$) we have

$${\nabla }_q^{-p}SU{M}_q(X-p)={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)q^{pg}{G}_{1+p+g}^X,H_1^q\left(M\right)={[M!]}_{q+}.$$

By Theorem 7, the values $H_1^q\left(g\right)$ for $0\le g<M$ are independent and can be assigned arbitrarily by choosing the parameters $K_i$ (since they are expressed as a convolution of elementary and complete homogeneous symmetric functions, whose Jacobian is non-singular). Therefore, given any sequence ${\left\{a_g\right\}}_{g=0}^M$, we can choose $K_i$ and a constant c such that

$$a_g^{*}=c{H}_1^q\left(g\right)(0\le g<M),$$

and then the remaining coefficient $a_M^{*}$ determines c via $c=a_M^{*}/{[M!]}_{q+}$ (because $H_1^q\left(M\right)={[M!]}_{q+}$). Consequently,

$${\displaystyle \sum _{g=0}^M}{a}_g{G}_{1+p+g}^X={\displaystyle \frac{a_M^{*}}{{[M!]}_{q+}}}{\nabla }_q^{-p}SU{M}_q(X-p).$$

Although this reduction is derived from $For{m}_1$, it extends naturally to all three forms. We obtain

$$a_g^{*}=c{H}_1^q\left(g\right),b_g=c{H}_2^q\left(g\right),c_g^{*}=c{H}_3^q\left(g\right),PT=[1,2,\dots ,M]$$

for the same constant c. Since c is irrelevant in our applications, we identify $a_g^{*}=H_1^q\left(g\right),b_g=H_2^q\left(g\right),c_g^{*}=H_3^q\left(g\right)$. Thus, for any sequences $a_g^{*},b_g,c_g^{*}$ obtained in this way, the relations in Theorem 6 are universally valid.

For any admissible $PT$, we may choose a new set of parameters $P{S}_1=\left\{K_i^{\prime }\right\}$,

$$SU{M}_q(N,PS,PT)=C·{\nabla }_q^{M-T_M}SU{M}_q(N,P{S}_1,[1,2,\dots ,M]),$$

where C is a constant.

As a natural inference, we obtain the following necessary and sufficient condition for reducing the number of terms.

Theorem 8.

For a reduction by R terms ($0<R\le M$), the sum

$${\displaystyle \sum _{g=0}^M}{a}_g{G}_{1+p+g}^X={\displaystyle \sum _{g=0}^{M-R}}{x}_g{G}_{1+p+R+g}^{X+R}$$

is possible iff,

$${\displaystyle \sum _{k=0}^M}{(-1)}^k{G}_g^k{q}^{{\displaystyle \frac{-k(k+3)}{2}}-kg}{a}_k^{*}=0(0\le g<R),\mathrm{or}{\displaystyle \sum _{k=0}^M}{(-1)}^k{G}_{M-g}^{M-k}{q}^{{\displaystyle \frac{-k(k+3)}{2}}}{a}_k^{*}=0(0\le M-g<R).$$

Here the conditions are equivalent to $b_g=0$ or $c_{M-g}^{*}=0$. Analogous conditions hold for the other two forms.

4. Inferences

The following theorems are direct algebraic consequences of Theorem 6. They are self-contained structural results of the three-form framework, revealing the algebraic duality among the coefficient spaces and of independent interest.

Simplifying the mutual expressions yields the inversion formulas.

Theorem 9

(Inversion formula). The following inversion relations hold; equivalently, in dual form:

1. 

$a_g={\sum }_{k=0}^M{b}_k{G}_g^k$. Then

$$b_g={\displaystyle \sum _{k=0}^M}{(-1)}^{k+g}{G}_g^k{q}^{{\displaystyle \frac{k(k-1)+g(g+1)}{2}}-gk}{a}_k.$$

2. 

$a_g={\sum }_{k=0}^M{c}_k{G}_{M-g}^{M-k}{q}^{-gk}$. Then

$$c_g={\displaystyle \sum _{k=0}^M}{(-1)}^{k+g}{G}_{M-g}^{M-k}{q}^{{\displaystyle \frac{k(k-1)+g(g+1)}{2}}}{a}_k.$$

3. 

$c_g={\sum }_{k=0}^M{b}_k{G}_g^{M-k}{q}^{gk}$. Then

$$b_g={\displaystyle \sum _{k=0}^M}{(-1)}^{M-g-k}{c}_k{G}_{M-g}^k{q}^{{\displaystyle \frac{M^2+k^2+g^2-M+k+g}{2}}-Mg-Mk}.$$

The inversions are not equivalent to these standard formulas, (see [6]).

$$a_n={\displaystyle \sum _{k=0}^n}{G}_k^n{b}_k,b_n={\displaystyle \sum _{k=0}^n}{(-1)}^{n-k}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{G}_k^n{a}_k.$$

Set $a_g$, $b_g$, or $c_g$ to ${\delta }_{rg}$ and applying the inversion formulas in Theorem 9 and Theorem 6 to get:

Theorem 10

(Orthogonality relations). For $0\le r,g\le M$, the following identities hold:

$$\begin{array}{cc}\hfill \sum _k{(-1)}^{k+r}{G}_g^k{G}_k^r{q}^{{\displaystyle \frac{r(r-1)+k(k+1)}{2}}-rk}& ={\delta }_{rg}.\hfill \\ \hfill \sum _k{(-1)}^{k+r}{G}_{M-g}^{M-k}{G}_{M-k}^{M-r}{q}^{{\displaystyle \frac{r(r-1)+k(k+1)}{2}}-gk}& ={\delta }_{rg}.\hfill \\ \hfill \sum _k{(-1)}^{M-g-k}{G}_{M-g}^k{G}_k^{M-r}{q}^{{\displaystyle \frac{M^2+g^2+k^2-M+g+k}{2}}-Mg-Mk+rk}& ={\delta }_{rg}.\hfill \end{array}$$

Using the three-form equivalence and setting $a_g$, $b_g$, or $c_g$ to ${\delta }_{gk}$ and Theorem 6 to get:

Theorem 11

(Generalized q-Vandermonde identities). For $0\le k\le M$ and $p\ge -1$:

$$\begin{array}{cc}\hfill \sum _g{q}^{g(g+1+p)}{G}_g^k{G}_{1+p+g}^N& =G_{1+p+k}^{N+k},\hfill \end{array}$$

(A1)

$$\begin{array}{cc}\hfill \sum _g{q}^{g(g+1+p)-k(g+1+p)}{G}_{M-g}^{M-k}{G}_{1+p+g}^N& =G_{1+p+M}^{N+M-k},\hfill \end{array}$$

(A2)

$$\begin{array}{cc}\hfill \sum _g{(-1)}^{k+g}{q}^{{\displaystyle \frac{g(g+1)-k(k+3)}{2}}-gk-pk}{G}_g^k{G}_{1+p+g}^{N+g}& =G_{1+p+k}^N,\hfill \end{array}$$

(A3)

$$\begin{array}{cc}\hfill \sum _g{(-1)}^{M-g}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{M-g}{2}}\right)-(M+1+p)k}{G}_{M-g}^k{G}_{1+p+g}^{N+g}& =G_{1+p+M}^{N+M-k},\hfill \end{array}$$

(A4)

$$\begin{array}{cc}\hfill \sum _g{(-1)}^{k+g}{q}^{{\displaystyle \frac{g(g+3)-k(k+3)}{2}}-pk+pg}{G}_{M-g}^{M-k}{G}_{1+p+M}^{N+M-g}& =G_{1+p+k}^N,\hfill \end{array}$$

(A5)

$$\begin{array}{cc}\hfill \sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}+gk+pg}{G}_g^{M-k}{G}_{1+p+M}^{N+M-g}& =G_{1+p+k}^{N+k}.\hfill \end{array}$$

(A6)

Binomial coefficient relations.

Taking $q\to 1$. If

$$\sum _g{a}_g\left({\displaystyle \genfrac{}{}{0pt}{}{X}{Y+g}}\right)=\sum _g{b}_g\left({\displaystyle \genfrac{}{}{0pt}{}{X+g}{Y+g}}\right)=\sum _g{c}_g\left({\displaystyle \genfrac{}{}{0pt}{}{X+M-g}{Y+M}}\right),$$

Applying Theorem 11 together with Theorem 6, we obtain:

$$\begin{array}{cc}\hfill & \sum _g{(-1)}^g{a}_g\left({\displaystyle \genfrac{}{}{0pt}{}{X+g}{Y+g}}\right)=\sum _g{(-1)}^g{b}_g\left({\displaystyle \genfrac{}{}{0pt}{}{X}{Y+g}}\right)={(-1)}^M\sum _g{c}_{M-g}\left({\displaystyle \genfrac{}{}{0pt}{}{X+M-g}{Y+M}}\right).\hfill \\ \hfill & \sum _g{(-1)}^g{a}_g\left({\displaystyle \genfrac{}{}{0pt}{}{X+M-g}{Y+M}}\right)=\sum _g{b}_{M-g}\left({\displaystyle \genfrac{}{}{0pt}{}{X+g}{Y+g}}\right)=\sum _g{(-1)}^g{c}_g\left({\displaystyle \genfrac{}{}{0pt}{}{X}{Y+g}}\right).\hfill \\ \hfill & \sum _g{b}_g{(}_{Y+M}^{X+M-g})=\sum {(-1)}^{M-g}{a}_{M-g}{(}_{Y+g}^{X+g})=\sum {(-1)}^{M-g}{c}_{M-g}{(}_{Y+g}^X).\hfill \\ \hfill & \sum _g{c}_g{(}_{Y+g}^{X+g})=\sum {(-1)}^g{a}_{M-g}{(}_{Y+M}^{X+M-g})=\sum {(-1)}^g{b}_{M-g}{(}_{Y+g}^X).\hfill \end{array}$$

(*)

Theorem 3(1) gives, for $a\ne 0,1$,

$${\displaystyle \sum _{g=0}^M}{a}_g{a}^g{(1-a)}^{M-g}={\displaystyle \sum _{g=0}^M}{b}_g{(1-a)}^{M-g}={\displaystyle \sum _{g=0}^M}{c}_g{a}^g.$$

Under the substitution $a\mapsto z/x$ and multiply by $x^M$, it becomes

$$\sum _g{a}_g{(x-z)}^{M-g}{z}^g=\sum _g{b}_g{(x-z)}^{M-g}{x}^g=\sum _g{c}_g{x}^{M-g}{z}^g.$$

In (*), ${(-1)}^g{a}_g$ is paired with $\left({\displaystyle \genfrac{}{}{0pt}{}{X+g}{Y+g}}\right)$ (role of $For{m}_2$), while ${(-1)}^g{b}_g$ is paired with $\left({\displaystyle \genfrac{}{}{0pt}{}{X}{Y+g}}\right)$ (role of $For{m}_1$). Thus

$$\sum _g{(-1)}^g{a}_g{(x-z)}^{M-g}{x}^g=\sum _g{(-1)}^g{b}_g{(x-z)}^{M-g}{z}^g={(-1)}^M\sum _g{c}_{M-g}{x}^{M-g}{z}^g.$$

Other formulas also have such substitutions.

5. Applications

• Illustration of the computational technique.

As an application of Theorem 5 (1), consider the sum

$${\mathrm{SUM}}_q(N+1,[1,\dots ,1]:q-1,[1,3,\dots ,2M-1])=\sum _{0\le {\lambda }_1\le \dots \le {\lambda }_M\le N}{q}^{2\sum {\lambda }_i}={\left[\begin{array}{c}N+M\\ M\end{array}\right]}_{q^2},$$

Form 1 (detailed computation).

Recall the general expression for $B_i$ in Form 1:

$$B_i=\left\{\begin{array}{cc}{q}^{(T_i-T_{i-1}-1)X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i),\hfill & X_i=K_i,\hfill \\ q^{1+(T_i-T_{i-1})X_{T-1}}{G}_1^{T_i-X_{K-1}}{D}_i,\hfill & X_i=T_i.\hfill \end{array}\right.$$

Substituting $K_i=1,D_i=q-1,T_i=2i-1$:

• Case $X_i=K_i$:

$$B_i=q^{X_{T-1}}·\left(1+G_1^{X_{T-1}}(q-1)\right)=q^{X_{T-1}}·q^{X_{T-1}}=q^{2X_{T-1}}.$$

Since $X_i=K_i$ implies $X_T=X_{T-1}$, we have $B_i=q^{2X_T}$.

• Case $X_i=T_i$:

$$B_i=q^{1+2X_{T-1}}{G}_1^{T_i-X_{K-1}}(q-1)=q^{1+2X_{T-1}}(q^{T_i-X_{K-1}}-1).$$

Now $T_i-X_{K-1}=(2i-1)-((i-1)-X_{T-1})=i+X_{T-1}$, $X_i=T_i$ implies $X_T=X_{T-1}+1$. Hence

$$B_i=q^{1+2X_{T-1}}(q^{i+X_{T-1}}-1)=q^{2X_T-1}(q^{i+X_{T-1}}-1),$$

Therefore, Form 1 gives

$$B_i=\left\{\begin{array}{cc}{q}^{2X_T},\hfill & X_i=K_i,\hfill \\ q^{2X_T-1}(q^{i+X_{T-1}}-1),\hfill & X_i=T_i.\hfill \end{array}\right.$$

The second branch contains i and $X_{T-1}$ coupled, so summing over all choices of T-positions is nontrivial and a closed form for $H_1^q\left(g\right)$ is not easily obtained. The same complication occurs in $For{m}_2$.

• Form 3 (detailed computation).

The general expression for $B_i$ in Form 3 is

$$B_i=\left\{\begin{array}{cc}{q}^{2X_T},\hfill & X_i=K_i,\hfill \\ -q^{2X_T-1},\hfill & X_i=T_i.\hfill \end{array}\right.$$

For a fixed choice of g indices assigned to $\mathbb{T}$, say $i_1<\dots <i_g$, we have $X_T=j$ at position $i_j$. The product over all i is

$${\displaystyle \prod _{i=1}^M}{B}_i={(-1)}^g{q}^{{\sum }_{X_i=T_i}(2X_T-1)}{q}^{{\sum }_{X_i=K_i}2X_T}.$$

Now

$$\sum _{X_i=T_i}(2X_T-1)=2\sum _{X_i=T_i}{X}_T-g=2(1+2+\dots +g)-g=g^2.$$

Thus, by Theorem 4(6),

$$H_3^q\left(g\right)=\sum _{X\left(T\right)=g}{\displaystyle \prod _{i=1}^M}{B}_i={(-1)}^g{q}^{g^2}\sum _{X\left(T\right)=g,X_i\in \mathbb{K}}\prod q^{2X_T}={(-1)}^g{q}^{g^2}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^2}.$$

Substituting this into Form 3, where $T_M+1=2M$, gives

$${\left[\begin{array}{c}N+M\\ M\end{array}\right]}_{q^2}={\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{g^2}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^2}{G}_{2M}^{N+2M-g}.$$

Compared to Theorem 5 (1).

$${\mathrm{SUM}}_q(N+1,[1,\dots ,1]:0,[1,3,\dots ,2M-1])=\sum _{0\le {\lambda }_1\le \dots \le {\lambda }_M\le N}{q}^{\sum {\lambda }_i}=G_M^{N+M}.$$

The Form 3 $B_i$ factors are

$$B_i=\left\{\begin{array}{cc}{q}^{X_{T-1}}=q^{X_T}=q^{i-X_K},\hfill & X_i=K_i,\hfill \\ q^{X_T}(-q^{2i-1})=-q^{2i+X_T-1},\hfill & X_i=T_i,\hfill \end{array}\right.=q^i\left\{\begin{array}{cc}{q}^{-X_K},\hfill & X_i=K_i,\hfill \\ -q^{i+X_T-1},\hfill & X_i=T_i.\hfill \end{array}\right..$$

The product of the extracted factors $q^i$ over all $i=1,\dots ,M$ gives $q^{\left({\displaystyle \genfrac{}{}{0pt}{}{M+1}{2}}\right)}$. The remaining factors contribute as follows:

• ${\prod }_{X_i=K_i}{q}^{-X_K}$ gives $q^{-\left({\displaystyle \genfrac{}{}{0pt}{}{M-g+1}{2}}\right)}$;
• ${\prod }_{X_i=T_i}(-q^{X_T-1})$ gives ${(-1)}^g{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)-g}$;
• ${\prod }_{X_i=T_i}{q}^i$, summed over all choices of g indices, gives ${\sum }_{1\le {\lambda }_1<\dots <{\lambda }_g\le M}{q}^{\sum {\lambda }_i}.$

Hence(Theorem 5 (2)),

$$H_3^q\left(g\right)=q^{\left({\displaystyle \genfrac{}{}{0pt}{}{M+1}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{M-g+1}{2}}\right)}{(-1)}^g{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)-g}\sum _{1\le {\lambda }_1<\dots <{\lambda }_g\le M}{q}^{\sum {\lambda }_i}={(-1)}^g{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{M+1}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{M-g+1}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)-g}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g}{2}}\right)+g}{G}_g^M.$$

Using $\left({\displaystyle \genfrac{}{}{0pt}{}{M+1}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{M-g+1}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{g}{2}}\right)=Mg+{\displaystyle \frac{g(g+1)}{2}}$, we obtain

$$G_M^{N+M}={\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{Mg+{\displaystyle \frac{g(g+1)}{2}}}{\left[\begin{array}{c}M\\ g\end{array}\right]}_q{G}_{2M}^{N+2M-g}.$$

The classical q-Vandermonde identity.

$$PS=\left[{\left[M\right]}_{q-},{[M-1]}_{q-},\dots ,{\left[1\right]}_{q-}\right],PT=[1,2,\dots ,M].$$

By Theorem 4 (2) (4),

$$SU{M}_q(N,PS,PT)=SU{M}_q(N,[{\left[1\right]}_{q-},{\left[2\right]}_{q-},\dots ,{\left[M\right]}_{q-}],PT)={[M!]}_{q-}{G}_{M+1}^{N+M}={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)G_{1+g}^N.$$

We now compute $H_1^q\left(g\right)$. For the i-th factor, $K_i={[M+1-i]}_{q-}$. When $X_i=K_i$,

$$B_i={[M+1-i]}_{q-}+G_1^{X_{T-1}}=q^{-(M+1-i)}{G}_1^{M+1-i+X_{T-1}}.$$

By Theorem 5 (2),

$$\sum _{X_i\in K}\prod B_i=q^{-(M+1)(M-g)}\left({\displaystyle \prod _{i=g+1}^M}{G}_1^i\right)\sum _{1\le {\lambda }_1<\dots <{\lambda }_{M-g}\le M}{q}^{\sum {\lambda }_i}=q^{-(M+1)(M-g)}\left({\displaystyle \prod _{i=g+1}^M}{G}_1^i\right)q^{\left({\displaystyle \genfrac{}{}{0pt}{}{M-g+1}{2}}\right)}{G}_{M-g}^M,$$

and multiplying by the contribution from the gT-terms, which is ${[g!]}_{q+}$, we obtain

$$H_1^q\left(g\right)={[g!]}_{q+}·q^{-(M+1)(M-g)+\left({\displaystyle \genfrac{}{}{0pt}{}{M-g+1}{2}}\right)}\left({\displaystyle \prod _{i=g+1}^M}{G}_1^i\right)G_{M-g}^M.$$

Since ${[M!]}_{q-}=q^{-\left({\displaystyle \genfrac{}{}{0pt}{}{M+1}{2}}\right)}{\prod }_{i=1}^M{G}_1^i$ and $G_{M-g}^M=G_g^M$, a direct simplification of the exponents gives

$${\displaystyle \frac{H_1^q\left(g\right)}{{[M!]}_{q-}}}=q^{g(g+1)}{G}_g^M.$$

Therefore,

$$G_{M+1}^{N+M}={\displaystyle \sum _{g=0}^M}{q}^{g(g+1)}{G}_g^M{G}_{1+g}^N.$$

Now apply ${\nabla }_q^{-p}$ to both sides. By Theorem 4 (3), ${\nabla }_q^{-p}{G}_{M+1}^{N+M}=G_{M+1+p}^{N+M+p}$ and ${\nabla }_q^{-p}\left(q^{g(g+1)}{G}_g^M{G}_{1+g}^N\right)=q^{(1+p+g)g}{G}_g^M{G}_{1+p+g}^{N+p}$. and replacing M with $M+p$ in the above identity, we obtain

$$\begin{array}{c}\hfill G_{M+1+p}^{N+M+p}={\displaystyle \sum _{g=0}^M}{q}^{(1+p+g)g}{G}_g^M{G}_{1+p+g}^{N+p}={\displaystyle \sum _{g=0}^{M+p}}{q}^{g(g+1)}{G}_g^{M+p}{G}_{1+g}^N.\end{array}$$

(7)

Extension of Classical Theorems.

In view of the recurring occurrence of products such as $G_k^M{G}_g^k$ in the preceding sections, we obtain the following generalizations.

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^M}{(-1)}^{k+g}{G}_k^M{G}_g^k=\left\{\begin{array}{cc}0,\hfill & M+g\mathrm{odd},\hfill \\ G_g^M{(q;q^2)}_{{\displaystyle \frac{M-g}{2}}},\hfill & M+g\mathrm{even}.\hfill \end{array}\right..\hfill \\ \hfill & {\displaystyle \sum _{k=0}^M}{(-1)}^{k+g}{G}_k^M{G}_{M-g}^{M-k}=\left\{\begin{array}{cc}0,\hfill & g\mathrm{odd},\hfill \\ G_g^M{(q;q^2)}_{{\displaystyle \frac{g}{2}}},\hfill & g\mathrm{even}.\hfill \end{array}\right..\hfill \\ \hfill & {\displaystyle \sum _{k=0}^M}{(-1)}^{k+g}{G}_k^M{G}_g^k{q}^k=\left\{\begin{array}{cc}{q}^g{G}_g^M{(q;q^2)}_{{\displaystyle \frac{M-g+1}{2}}},\hfill & M+g\mathrm{odd},\hfill \\ q^g{G}_g^M{(q;q^2)}_{{\displaystyle \frac{M-g}{2}}},\hfill & M+g\mathrm{even}.\hfill \end{array}\right..\hfill \\ \hfill & {\displaystyle \sum _{k=0}^M}{(-1)}^{k+g}{G}_k^M{G}_g^k{q}^{-k}=\left\{\begin{array}{cc}{(-1)}^{M-g}{q}^{-M}{G}_g^M{(q;q^2)}_{{\displaystyle \frac{M-g+1}{2}}},\hfill & M+g\mathrm{odd},\hfill \\ {(-1)}^{M-g}{q}^{-M}{G}_g^M{(q;q^2)}_{{\displaystyle \frac{M-g}{2}}},\hfill & M+g\mathrm{even}.\hfill \end{array}\right..\hfill \\ \hfill & {\displaystyle \sum _{k=0}^M}{{[}_k^M]}_{q^2}{{[}_g^k]}_{q^2}{q}^k=q^g{{[}_g^M]}_{q^2}{(-q;q)}_{M-g}.\hfill \\ \hfill & {\displaystyle \sum _{k=0}^M}{{[}_k^M]}_{q^2}{{[}_g^k]}_{q^2}{q}^{-k}=q^{-M}{{[}_g^M]}_{q^2}{(-q;q)}_{M-g}.\hfill \end{array}$$

The first equation extends the Gauss q-binomial theorem [5]: ${\sum }_{k=0}^M{(-1)}^k{G}_k^M=\left\{\begin{array}{cc}0,\hfill & M\mathrm{odd},\hfill \\ {(q;q^2)}_{{\displaystyle \frac{M}{2}}},\hfill & M\mathrm{even}.\hfill \end{array}\right.$.

The fifth equation extend another Gauss identity [5]: ${\sum }_{k=0}^M{{[}_k^M]}_{q^2}{q}^k={(-q;q)}_M.$

All identities above are derived by elementary induction and are presented here for reference.

Proposition 1.

The following identities hold:

1. 

For $PS=[{\left[T_1\right]}_{q-},{\left[T_2\right]}_{q-},\dots ,{\left[T_M\right]}_{q-}]$ and $PT=[T_1,T_2,\dots ,T_M]$,

$$H_1^q\left(g\right)=\left({\displaystyle \prod _{i=1}^M}{\left[T_i\right]}_{q-}\right)q^{(g+1+p)g}{G}_g^M,p=T_M-M.$$

2. 

For $PS=[{\left[K\right]}_{q-},{[K+1]}_{q-},\dots ,{[K+M-1]}_{q-}]$ and $PT=[T,T+1,\dots ,T+M-1]$,

$$H_1^q\left(g\right)=q^{\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)}\left({\displaystyle \prod _{i=1}^g}{[T-1+i]}_q\right)\left({\displaystyle \prod _{i=1}^{M-g}}{[K+M-i]}_{q-}\right)G_g^M.$$

3. 

With the same $\mathbb{K}$ as above,

$${\displaystyle \sum _{r=0}^{M-g}}{F}_{M-g-r}^{\mathbb{K}}{E}_r^{g,q}=\left({\displaystyle \prod _{i=1}^{M-g}}{[K+M-i]}_{q-}\right)G_g^M.$$

4. 

For general $PS$ and $PT$, if we prepend a factor ${\left[T\right]}_{q-}$ with a new T, we have

$$H_1^q(g,[{\left[T\right]}_{q-},PS],[T,PT])={\left[T\right]}_{q-}\left(H_1^q\left(g\right)+q^{p+g}{H}_1^q(g-1)\right),p=T_M-M.$$

Proof. 

For (1), from Theorem 4(3), we have

$$SU{M}_q(N,PS,PT)={\displaystyle \prod _{i=1}^M}{\left[T_i\right]}_{q-}{G}_{T_M+1}^{N+T_M}={\displaystyle \prod _{i=1}^M}{\left[T_i\right]}_{q-}{G}_{M+p+1}^{N+M+p}={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)G_{1+p+g}^{N+p}.$$

Using $G_{M+p+1}^{N+M+p}={\sum }_g{q}^{(g+1+p)g}{G}_g^M{G}_{1+p+g}^{N+p}$ (22) , yields (1).

For (2), the proof is analogous to that of (22).

For (3), it follows directly from Theorem 7 and (2) ($T_1\ne 1$ does not affect the result).

For (4), by Theorem 4 (4), $SU{M}_q(N,[{\left[T\right]}_{q-},PS],[T,PT])={\left[T\right]}_{q-}SU{M}_q(N,PS,PT).$ Expanding both sides in $For{m}_1$ and comparing the coefficients of $G_{1+p+g}^{N+p}$. □

Definition 3.

We define a new number,

$${{\langle }_M^M\rangle }^q=0,{{\langle }_g^M\rangle }^q=\sum _{{\lambda }_1+\dots +{\lambda }_{g+1}=M-g-1}{\displaystyle \prod _{i=1}^{g+1}}{\left[i\right]}_q^{{\lambda }_i}{[1+{\lambda }_1]}_q{[1+{\lambda }_1+{\lambda }_2]}_q\dots {[1+{\lambda }_1+{\lambda }_2+\dots +{\lambda }_g]}_q,$$

with all ${\lambda }_i\ge 0$. They satisfy

$${{\langle }_g^M\rangle }^q={[M-g]}_q{{\langle }_{g-1}^{M-1}\rangle }^q+{[g+1]}_q{{\langle }_g^{M-1}\rangle }^q,{{\langle }_g^M\rangle }^q={{\langle }_{M-g-1}^M\rangle }^q.$$

Proposition 2

(Three expansions of ${\left[N\right]}_q^M$ ). 

$${\left[N\right]}_q^M={\displaystyle \sum _{g=1}^M}{[g!]}_{q+}{S}_2^q(M,g)q^{-g}{G}_g^N={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{[g!]}_{q-}{S}_2^{q-}(M,g)q^g{G}_g^{N+g-1}={\displaystyle \sum _{g=0}^{M-1}}{{\langle }_g^M\rangle }^q{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{M-g}{2}}\right)}{G}_M^{N+g}.$$

Proof. 

Take

$$PS=\left[{\left[1\right]}_{q-},{\left[1\right]}_{q-},\dots ,{\left[1\right]}_{q-}\right],PT=[2,3,\dots ,M],$$

with $K_i={\left[1\right]}_{q-}=q^{-1}$, $D_i=1$, and note that here $card\left\{T_i\right\}=M-1,H=M-(M-1)=1$.

First, by Theorem 4(1)(4),

$${\nabla }_q^{1}SU{M}_q(N,[{\left[1\right]}_{q-},{\left[1\right]}_{q-},\dots ,{\left[1\right]}_{q-}],[1,2,3,\dots ,M])={\displaystyle \prod _{i=1}^M}(q^{-1}+{[N-1]}_q)=q^{-M}{\left[N\right]}_q^M.$$

$$=q^{-1}{\nabla }_q^{1}SU{M}_q(N,PS,PT)=q^{-1}{\displaystyle \sum _{g=0}^{M-1}}{H}_1^q\left(g\right)q^{-g}{G}_{1+g}^N=q^{-1}{\displaystyle \sum _{g=0}^{M-1}}{H}_2^q\left(g\right)G_{1+g}^{N+g}.$$

We now compute $H_1^q\left(g\right)$. The $B_i$ factors are

$$B_i=\left\{\begin{array}{cc}{q}^{-1}{G}_1^{1+X_T},\hfill & X_i=K_i,\hfill \\ q^{X_T}{G}_1^{1+X_T}=q^{-1}{q}^{1+X_T}{G}_1^{1+X_T},\hfill & X_i=T_i.\hfill \end{array}\right.$$

The product over the T-terms is independent of their positions and equals

$$\prod _{X_i=T_i}{B}_i={[(g+1)!]}_{q+}{q}^{-g}{\left[1\right]}_{q+}^{-1}={[(g+1)!]}_{q+}{q}^{-g-1}.$$

Thus

$$H_1^q\left(g\right)={[(g+1)!]}_{q+}{q}^{-g-1}\sum _{X_i=K_i}\prod B_i={[(g+1)!]}_{q+}{q}^{-g-1}{q}^{-(M-1-g)}{E}_{M-1-g}^{g+1,q},$$

$E_r^{g,q}={\sum }_{1\le {\nu }_1\le \dots \le {\nu }_r\le g}{\left[{\nu }_1\right]}_q\dots {\left[{\nu }_r\right]}_q=S_2^q(r+g,g)$. Substituting it into $SU{M}_q$ gives

$${\left[N\right]}_q^M=q^{-1}{\displaystyle \sum _{g=0}^{M-1}}{[(g+1)!]}_{q+}{S}_2^q(M,g+1)q^{-g}{G}_{1+g}^N$$

The substitution $g\mapsto g+1$ gives the first identity; the proof of the second is identical.

Now Set

$$PS=\left[{\left[1\right]}_{q-},{\left[1\right]}_{q-},\dots ,{\left[1\right]}_{q-}\right],PT=[1,2,3,\dots ,M],$$

Thus

$${\nabla }_q^{1}SU{M}_q\left(N\right)=q^{-M}{\left[N\right]}_q^M={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)q^{-g}{G}_M^{N+M-1-g}.$$

the $B_i$ factors are

$$B_i=\left\{\begin{array}{cc}{q}^{-1}{G}_1^{1+X_T},\hfill & X_i=K_i,\hfill \\ q^{X_T}{G}_1^{i-X_T}=q^{X_T}{G}_1^{X_K},\hfill & X_i=T_i.\hfill \end{array}\right.$$

${\prod }_{X_i=K_i}{q}^{-1}$ gives $q^{-(M-g)}$, ${\prod }_{X_i=T_i}{q}^{X_T}$ gives $q^{{\displaystyle \frac{g(g+1)}{2}}}$, ${\sum }_{X_i=K_i}\prod G_1^{1+X_T}$ gives $E_{M-g}^{g+1,q}$.

When $X_j=K_j,j\le {\lambda }_1$, the factor $T_{{\lambda }_1+1}$ first appears in $\prod B_i$. All leading $G_1^{1+X_T}$ are $G_1^1$, and all $1+X_T(X_i=K_i)$ after it are greater than 1. Thus

$$\prod G_1^{1+X_T}{G}_1^{X_K}={\left(G_1^1\right)}^{{\lambda }_1}{[0+{\lambda }_1]}_q\dots .$$

When $T_{{\lambda }_1+1+{\lambda }_2+1}$ appears for the second time, $G_1^{1+X_T}=G_1^2$ for $X_j\in \mathbb{K}$ with ${\lambda }_1+1<j<{\lambda }_1+1+{\lambda }_2+1$, and all later $1+X_T(X_i=K_i)$ are greater than 2. At this point, $X_K={\lambda }_1+{\lambda }_2$, so

$$\prod G_1^{1+X_T}{G}_1^{X_K}={\left(G_1^1\right)}^{{\lambda }_1}{\left(G_1^2\right)}^{{\lambda }_2}{[0+{\lambda }_1]}_q{[0+{\lambda }_1+{\lambda }_2]}_q\dots .$$

Continuing this way yields

$$\prod G_1^{1+X_T}{G}_1^{X_K}={\left(G_1^1\right)}^{{\lambda }_1}{\left(G_1^2\right)}^{{\lambda }_2}\dots {\left(G_1^{1+g}\right)}^{{\lambda }_{1+g}}{[0+{\lambda }_1]}_q{[0+{\lambda }_1+{\lambda }_2]}_q\dots {[0+{\lambda }_1+{\lambda }_2+\dots +{\lambda }_g]}_q,\sum {\lambda }_i=M-g.$$

Since the term vanishes when ${\lambda }_1=0$, we may assume ${\lambda }_1>0$ and set ${\lambda }_1:={\lambda }_1-1$,

$$\prod G_1^{1+X_T}{G}_1^{X_K}={\left(G_1^1\right)}^{{\lambda }_1}{\left(G_1^2\right)}^{{\lambda }_2}\dots {\left(G_1^{1+g}\right)}^{{\lambda }_{1+g}}{[1+{\lambda }_1]}_q{[1+{\lambda }_1+{\lambda }_2]}_q\dots {[1+{\lambda }_1+{\lambda }_2+\dots +{\lambda }_g]}_q={{\langle }_g^M\rangle }^q,\sum {\lambda }_i=M-g-1.$$

Thus

$$H_3^q\left(g\right)=q^{-(M-g)+{\displaystyle \frac{g(g+1)}{2}}}{{\langle }_g^M\rangle }^q.$$

Substituting it into $SU{M}_q$ gives

$$q^{-M}{\left[N\right]}_q^M={\displaystyle \sum _{g=0}^M}{q}^{-(M-g)+{\displaystyle \frac{g(g+1)}{2}}}{{\langle }_g^M\rangle }^q{q}^{-g}{G}_M^{N+M-1+g}.$$

$H_3^q\left(M\right)=0$, thus

$${\left[N\right]}_q^M={\displaystyle \sum _{g=0}^{M-1}}{q}^{{\displaystyle \frac{g(g+1)}{2}}}{{\langle }_g^M\rangle }^q{G}_M^{N+M-1-g}.$$

Substituting $g\mapsto M-1-g$ and ${{\langle }_{M-g-1}^M\rangle }^q={{\langle }_g^M\rangle }^q$, gives the third identity. □

The first and third expansions correspond, respectively, to the classical q-Stirling numbers of the second kind and the q-Eulerian numbers (Carlitz [6,7,8]). The second expansion, however, appears less common in the literature.

Take the limit $q\to 1$,

$$\underset{q\to 1}{lim}{{\langle }_g^M\rangle }^q={\langle }_g^M\rangle =\sum _{{\lambda }_1+\dots +{\lambda }_{g+1}=M-g-1}{\displaystyle \prod _{i=1}^{g+1}}{i}^{{\lambda }_i}(1+{\lambda }_1)(1+{\lambda }_1+{\lambda }_2)\dots (1+{\lambda }_1+{\lambda }_2+\dots +{\lambda }_g)$$

$$N^M={\displaystyle \sum _{g=1}^M}g!S_2(M,g)\left({\displaystyle \genfrac{}{}{0pt}{}{N}{g}}\right)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g!S_2(M,g)\left({\displaystyle \genfrac{}{}{0pt}{}{N+g-1}{g}}\right)={\displaystyle \sum _{g=0}^{M-1}}{\langle }_g^M\rangle \left({\displaystyle \genfrac{}{}{0pt}{}{N+g}{M}}\right).$$

Here $S_2(M,g)$ and ${\langle }_g^M$ denoting Stirling and Eulerian numbers (the expression of the latter is the main result of [9]), Theorem 3(1) gives three expressions for the Eulerian polynomial.

Proposition 3.

For nonnegative integers satisfying the stated conditions, the following identities hold:

1. 

If $0\le A<M$ and $T\ge 0$, then

$${\displaystyle \sum _{g=0}^M}{G}_g^M{G}_A^{A+T+g}{G}_{A+T+1+g}^{N+A+T}{q}^{g(g+1+T)}={\displaystyle \sum _{k=0}^A}{G}_{k+T}^{A+T}{G}_{M+T}^{M+T+k}{G}_{M+T+1+k}^{N+M+T}{q}^{k(k+1+T)}.$$

2. 

If $A,B,T\ge 0$ and $A+B<M$, then

$${\displaystyle \sum _{g=0}^M}{(-1)}^g{G}_g^M{G}_A^{A+T+g}{G}_B^g{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g}{2}}\right)-g(A+B)}=0.$$

3. 

If $K,T\ge 0$, then

$${\displaystyle \sum _{g=0}^M}{(-1)}^g{G}_g^M{G}_{M+K}^{M+K+T+g}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{M+1-g}{2}}\right)+(M-g)K}={(-1)}^M{G}_K^{T+M+K}.$$

Proof. 

(1) Take

$$PS=\left[{[T+1]}_{q-},{[T+2]}_{q-},\dots ,{[T+M]}_{q-}\right],PT=[T+A+1,T+A+2,\dots ,T+A+M].$$

Applying Proposition 1(2) gives

$$\begin{array}{cc}\hfill SU{M}_q\left(N\right)& ={\displaystyle \sum _{g=0}^M}{G}_{A+T+1+g}^{N+A+T}{\displaystyle \prod _{i=1}^g}{[T+A+i]}_q{\displaystyle \prod _{i=1}^{M-g}}{[T+g+i]}_{q-}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)}{G}_g^M\hfill \\ \hfill & ={\displaystyle \sum _{g=0}^M}{G}_{A+T+1+g}^{N+A+T}{\displaystyle \prod _{i=1}^g}{[T+A+i]}_q{\displaystyle \prod _{i=1}^{M-g}}{[T+g+i]}_q{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)-(T+M+1)(M-g)+\left({\displaystyle \genfrac{}{}{0pt}{}{M+1-g}{2}}\right)}{G}_g^M\hfill \\ \hfill & ={\displaystyle \frac{{(q;q)}_A{(q^{T+A+1};q)}_{M-A}}{{(1-q)}^M}}{\displaystyle \sum _{g=0}^M}{G}_{A+T+1+g}^{N+A+T}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)-(T+M+1)(M-g)+\left({\displaystyle \genfrac{}{}{0pt}{}{M+1-g}{2}}\right)}{G}_A^{A+T+g}{G}_g^M.\hfill \end{array}$$

(*)

On the other hand, using Theorem 4(2)(4), the same $SU{M}_q\left(N\right)$ can be rewritten as

$$SU{M}_q\left(N\right)={\displaystyle \prod _{i=1}^{M-A}}{[T+A+i]}_{q-}·SU{M}_q\left(N,[{[T+1]}_{q-},\dots ,{[T+A]}_{q-}],[T+M+1,\dots ,T+M+A]\right).$$

Applying Proposition 1(2) again yields

$$\begin{array}{cc}\hfill SU{M}_q\left(N\right)& =q^{-{\displaystyle \frac{(M-A)(2T+A+M+1)}{2}}}{\displaystyle \sum _{k=0}^A}{G}_{M+T+1+k}^{N+M+T}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)-(T+A+1)(A-k)+\left({\displaystyle \genfrac{}{}{0pt}{}{A+1-k}{2}}\right)}{\displaystyle \prod _{i=1}^M}{[T+k+i]}_q{G}_k^A.\hfill \end{array}$$

(**)

Comparing (*) and (**), and simplifying the q-factorial factors, gives exactly (1).

(2) follows from (1) by applying Theorem 8.

(3) Take

$$PS=\left[{[T+1]}_{q-},{[T+2]}_{q-},\dots ,{[T+M]}_{q-}\right],PT=[T+K+M+1,T+K+M+2,\dots ,T+K+2M].$$

By (1), the $H_1^q\left(g\right)$ coefficients are

$$a_g=H_1^q\left(g\right)=q^{\left({\displaystyle \genfrac{}{}{0pt}{}{g+1}{2}}\right)-(T+M+1)(M-g)+\left({\displaystyle \genfrac{}{}{0pt}{}{M+1-g}{2}}\right)}{\displaystyle \prod _{i=1}^g}{[T+K+M+i]}_q{\displaystyle \prod _{i=1}^{M-g}}{[T+M+1-i]}_q{G}_g^M.$$

Using the transformation formula in Theorem 6(3), which states

$$b_g=H_2^q\left(g\right)=\sum _k{(-1)}^{k+g}{H}_1^q\left(k\right)q^{{\displaystyle \frac{g(g+1)-k(k+3)}{2}}-kg},$$

and taking $g=0$, we obtain

$$b_0=\sum _k{(-1)}^k{a}_k^{*}{q}^{-{\displaystyle \frac{k(k+3)}{2}}}=\sum _k{(-1)}^k{H}_1^q\left(k\right)q^{-{\displaystyle \frac{k(k+3)}{2}}-(T+K+M)k}=\sum _g{(-1)}^g{H}_1^q\left(g\right)q^{-{\displaystyle \frac{g(g+3)}{2}}-(T+K+M)g}.$$

On the other hand, from the definition of $H_2^q\left(0\right)$ , a direct computation gives

$$b_0=H_2^q\left(0\right)={(-1)}^M{q}^{-M(T+K+M+1)}{\displaystyle \prod _{i=1}^M}{[K+i]}_q.$$

Therefore,

$${(-1)}^M{q}^{-M(T+K+M+1)}{\displaystyle \prod _{i=1}^M}{[K+i]}_q=\sum _g{(-1)}^g{H}_1^q\left(g\right)q^{-{\displaystyle \frac{g(g+3)}{2}}-(T+K+M)g}.$$

Substituting the expression for $H_1^q\left(g\right)$ and simplifying the product gives

$${(-1)}^M{\displaystyle \prod _{i=1}^M}{[K+i]}_q=\sum _g{(-1)}^g{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{M+1-g}{2}}\right)+(M-g)K}{\displaystyle \prod _{i=1}^g}{[T+K+M+i]}_q{\displaystyle \prod _{i=1}^{M-g}}{[T+M+1-i]}_q{G}_g^M.$$

Simplifying the product

$${\displaystyle \prod _{i=1}^g}{[T+K+M+i]}_q{\displaystyle \prod _{i=1}^{M-g}}{[T+M+1-i]}_q={\displaystyle \prod _{i=1}^{K+M}}{[T+g+i]}_q/{\displaystyle \prod _{i=1}^K}{[T+M+i]}_q$$

Thus,

$${(-1)}^M{\displaystyle \prod _{i=1}^K}{[T+M+i]}_q=\sum _g{(-1)}^g{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{M+1-g}{2}}\right)+(M-g)K}{\displaystyle \prod _{i=1}^{K+M}}{[T+g+i]}_q/{\displaystyle \prod _{i=1}^M}{[K+i]}_q{G}_g^M.$$

Multiply both sides by ${(q;q)}_K^{-1}{(1-q)}^K$, we arrive at

$${(-1)}^M{G}_K^{T+M+K}={\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{M+1-g}{2}}\right)+(M-g)K}{G}_g^M{G}_{M+K}^{T+M+K+g},$$

which is exactly (3). □

Proposition 4

(A nested summation identity with step size 2). 

$${\displaystyle \sum _{n_M=0}^{N-1}}\dots {\displaystyle \sum _{n_1=0}^{n_2}}{[K+2n_1]}_q{[K+1+2n_2]}_q\dots {[K+M-1+2n_M]}_q{q}^{-{\sum }_{j=1}^M{n}_j}=q^{-(N-1)M}{\displaystyle \prod _{i=1}^M}{[K+N-2+i]}_q{G}_M^{M+N-1}.$$

Proof. 

Take

$$PS=\left[{\left[K\right]}_{q-},{[K+1]}_{q-},\dots ,{[K+M-1]}_{q-}\right],PT=[1,2,\dots ,M].$$

By Proposition 1(2), the coefficients $H_1^q\left(g\right)$ for this $SU{M}_q\left(N\right)$ satisfy

$$\sum _{X_i=K_i}\prod B_i=q^{-(K+M)(M-g)+\left({\displaystyle \genfrac{}{}{0pt}{}{M-g+1}{2}}\right)}{\displaystyle \prod _{i=1}^{M-g}}{[K+g-1+i]}_q{G}_g^M.$$

And the definition of $B_i$ is

$$B_i=q^{-(K-1+i)}{G}_1^{K-1+i}+G^{X_T}=q^{-(K-1)-i}{G}_1^{K-1+i+X_T},X_i=K_i.$$

Thus for a choice of $M-g$ indices from $\mathbb{K}$ (say, ${\lambda }_1<\dots <{\lambda }_{M-g}$), the product of the corresponding $B_i$’s is

$${\displaystyle \prod _{t=1}^{M-g}}{q}^{-(K-1)-{\lambda }_t}{G}_1^{K-1+{\lambda }_t+X_T},$$

This leads, after summing over all such choices and relabelling, to the multinomial expansion

$$\sum _{X_i=K_i}\prod B_i=\sum _{\begin{array}{c}0\le n_1\le \dots \le n_{M-g}\le g\end{array}}{q}^{-(K-1)(M-g)-{\sum }_{t=1}^{M-g}t-{\sum }_{t=1}^{M-g}{n}_t}{\displaystyle \prod _{t=1}^{M-g}}{G}_1^{K-1+t+2n_t}.$$

This follows because the inversion statistic for the K-choices produces $q^{\sum n_t}$ after the standard reduction; details are routine.

Comparing the two expressions yields

$$\sum _{\begin{array}{c}0\le n_1\le \dots \le n_{M-g}\le g\end{array}}{\displaystyle \prod _{t=1}^{M-g}}{G}_1^{K-1+t+2n_t}{q}^{g(M-g)-{\sum }_{t=1}^{M-g}{n}_t}={\displaystyle \prod _{t=1}^{M-g}}{G}_1^{K+g-1+t}{G}_g^M.$$

Now set $g=N-1$ and replace $M-g$ by M (i.e., take $g=N-1$ and reindex the sum). We get

$$\sum _{\begin{array}{c}0\le n_1\le \dots \le n_M\le N-1\end{array}}{\displaystyle \prod _{t=1}^M}{G}_1^{K-1+t+2n_t}{q}^{(N-1)M-{\sum }_{t=1}^M{n}_t}={\displaystyle \prod _{t=1}^M}{G}_1^{K+N-2+t}{G}_{N-1}^{M+N-1}.$$

Multiplying both sides by $q^{-(N-1)M}$ and noting that $G_1^a={\left[a\right]}_q$, $G_{N-1}^{M+N-1}=G_M^{M+N-1}$, we obtain

$${\displaystyle \sum _{n_M=0}^{N-1}}\dots {\displaystyle \sum _{n_1=0}^{n_2}}{\displaystyle \prod _{i=1}^M}{[K-1+i+2n_i]}_q{q}^{-{\sum }_{i=1}^M{n}_i}=q^{-(N-1)M}{\displaystyle \prod _{i=1}^M}{[K+N-2+i]}_q{G}_M^{M+N-1}.$$

which is exactly the claimed identity. □

In particular,

$$SUM(N,[1,2,\dots ,M]:2,[1,3,\dots ,2M-1])=M!{\left({\displaystyle \genfrac{}{}{0pt}{}{N+M-1}{M}}\right)}^2,$$

and for $M=1$,

$$1+3+\dots +(2N-1)=N^2.$$

6. Unified Conversion Formulas

The following theorem is a direct consequence of Theorem 6 and Theorem 5(3); it also generalizes Theorem 3(1).

Theorem 12.

For any integer A, the following hold:

1. 

${\sum }_{g=0}^M{a}_g^{*}{q}^{-{\displaystyle \frac{g(g+1)}{2}}+Ag}{z}^g={\sum }_{g=0}^M{b}_g{(-q^{A+1}z;q)}_g={\sum }_{g=0}^M{c}_g^{*}{q}^{-{\displaystyle \frac{g(g+1)}{2}}+Ag}{z}^g{(-q^{A+1}z;q)}_{M-g}$.

2. 

${\sum }_{g=0}^M{b}_g{q}^{Ag}{z}^g={\sum }_{g=0}^M{(-1)}^g{a}_g^{*}{q}^{-{\displaystyle \frac{g(g+3)}{2}}}{(q^{A-g+1}z;q)}_g=q^{AM}{z}^M{\sum }_{g=0}^M{c}_g^{*}{q}^{-(M+1)g}{(q^{-A}{z}^{-1};q)}_g$.

3. 

${\sum }_{g=0}^M{c}_g^{*}{q}^{Ag}{z}^g={\sum }_{g=0}^M{a}_g^{*}{q}^{Ag}{z}^g{(q^{A+2+g}z;q)}_{M-g}={\sum }_{g=0}^M{b}_g{(q^{A+2+g}z;q)}_{M-g}$.

Thus, any one of the three coefficient sequences determines the other two; the parameters $A,z,a_g,b_g,c_g$ may be chosen freely to produce further relations.

• Three-form equivalence from special coefficients.

Taking the special choices

$$a_g^{*,1}=q^{g(g+1)+Ag}{G}_g^M,a_g^{*,2}=a_g^{*,1}{z}^g$$

Using Theorem 6 and (5)–(6), we obtain

$$b_g^{*,1}=q^{Ag}{G}_g^M{(q^A;q)}_{M-g},b_g^{*,2}=q^{Ag}{G}_g^M{(z{q}^A;q)}_{M-g}{z}^g,$$

$$c_g^{*,1}={(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}}{G}_g^M{(q^A;q)}_g,c_g^{*,2}={(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}}{G}_g^M{(z{q}^A;q)}_g.$$

Substituting these into the three-form equivalence gives

$$\begin{array}{cc}\hfill \sum _g{q}^{g(g+1+A+p)}{G}_g^M{z}^g{G}_{1+p+g}^N& =\sum _g{q}^{Ag}{(z{q}^A;q)}_{M-g}{G}_g^M{G}_{1+p+g}^{N+g}\hfill \\ & =\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}+pg}{G}_g^M{(z{q}^A;q)}_g{G}_{1+p+M}^{N+M-g}.\hfill \end{array}$$

Generating function identities and consequences.

Applying the first identity in Theorem 4(3), we obtain

$$\sum _g{a}_g^{*,1}{q}^{-{\displaystyle \frac{g(g+1)}{2}}}{z}^g=\sum _g{a}_g^{*,2}{q}^{-{\displaystyle \frac{g(g+1)}{2}}}={(-z{q}^{1+A};q)}_M.$$

This yields the equivalent expressions (Theorem 12(1))

$$\begin{array}{cc}\hfill {(-z{q}^{1+A};q)}_M& =\sum _g{q}^{Ag}{G}_g^M{(q^A;q)}_{M-g}{(-zq;q)}_g\hfill \\ & =\sum _g{q}^{Ag}{z}^g{G}_g^M{(z{q}^A;q)}_{M-g}{(-q;q)}_g\hfill \\ & =\sum _g{(-1)}^g{q}^g{z}^g{G}_g^M{(q^A;q)}_g{(-zq;q)}_{M-g}\hfill \\ & =\sum _g{(-1)}^g{q}^g{G}_g^M{(z{q}^A;q)}_g{(-q;q)}_{M-g}.\hfill \end{array}$$

Setting $A=1$ and replacing $zq$ by z in the starred equation gives

$${\displaystyle \frac{{(-zq;q)}_{\infty }}{{(q;q)}_{\infty }}}=\sum _g{\displaystyle \frac{q^g{(-z;q)}_g}{{(q;q)}_g}}.$$

When $A=0$, we have

$$a_g^{*,2}=q^{g(g+1)}{G}_g^M{z}^g,b_g^{*,2}=G_g^M{(z;q)}_{M-g}{z}^g.$$

Computing $a_g^{*,2}$ from $b_g^{*,2}$ yields

$$\sum _g{(z;q)}_{M-g}{z}^g{G}_g^M{G}_k^g=z^k{G}_k^M,$$

which is equivalent to ($z=q$)

$$\sum _g{\displaystyle \frac{q^g}{{(q;q)}_{g-k}}}={\displaystyle \frac{q^k}{{(q;q)}_{M-k}}},$$

and hence ($k=0$), gives a known formula [5]

$$\sum _g{\displaystyle \frac{q^g}{{(q;q)}_g}}={\displaystyle \frac{1}{{(q;q)}_M}}.$$

Jacobi’s q-binomial theorem.

Applying Theorem 12(1) to $a_g^{*,2}=q^{g(g+1)}{G}_g^M{z}^g,b_g^{*,2}=G_g^M{(z;q)}_{M-g}{z}^g$, we get

$$\sum _g{a}_g^{*,2}{q}^{-{\displaystyle \frac{g(g+1)}{2}}-g}{z}^{-g}{x}^g{(-1)}^g=\sum _g{b}_g^{*,2}{(x/z;q)}_g.$$

Thus

$$(1-x)(1-xq)\dots (1-x{q}^{M-1})=\sum _g{G}_g^M(1-z)(1-zq)\dots (1-z{q}^{M-g-1})(z-x)(z-xq)\dots (z-x{q}^{g-1}).$$

Replacing x by $a/b$ and z by $c/b$, then multiplying through by $b^M$, recovers Jacobi’s q-binomial theorem (see [5], p. 71):

$$(b-a)(b-aq)\dots (b-a{q}^{M-1})=\sum _g{G}_g^M(b-c)(b-cq)\dots (b-c{q}^{M-g-1})(c-a)(c-aq)\dots (c-a{q}^{g-1}).$$

Finite form of Jacobi’s Durfee square identity.

Take $c_g^{*,1}={(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}}{G}_g^M{(q^0;q)}_g={\delta }_{g0}$, Theorem 12(3) gives

$$\sum _g{a}_g^{*,1}{q}^{(r-1)g}{a}^g{(a{q}^{1+r+g};q)}_{M-g}=1,$$

or equivalently($a_g^{*,1}=q^{g(g+1)}{G}_g^M$),

$${\displaystyle \frac{{\sum }_g{G}_g^M{q}^{g^2+gr}{a}^g{(a{q}^{1+r+g};q)}_{M-g}}{{(aq;q)}_M}}={\displaystyle \frac{1}{{(aq;q)}_M}}.$$

Letting $M\to \infty$, this becomes the finite form of Jacobi’s Durfee square identity (see [5], pp. 158–159):

$$\sum _g{\displaystyle \frac{q^{g^2+gr}{a}^g}{{(q;q)}_g{(aq;q)}_{g+r}}}={\displaystyle \frac{1}{{(aq;q)}_{\infty }}}.$$

Special case $b_g=q^g$.

Taking $b_g=q^g$ and using the identity

$${\displaystyle \sum _{k=0}^M}{G}_g^k{q}^{k-g}=G_{g+1}^{M+1},$$

we obtain

$$a_g^{*}=q^{g(g+2)}{G}_{g+1}^{M+1}.$$

Similarly, applying (see [5], p. 22)

$$\sum _k{G}_g^{M-k}{q}^{(g+1)k}=G_{g+1}^{M+1}$$

yields

$$c_g^{*}={(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}}{G}_{g+1}^{M+1}.$$

Substituting these into the three-form equivalence gives

$$\sum _g{q}^{g(g+2+p)}{G}_{g+1}^{M+1}{G}_{1+p+g}^N=\sum _g{q}^g{G}_{1+p+g}^{N+g}=\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g+3)}{2}}+pg}{G}_{g+1}^{M+1}{G}_{1+p+M}^{N+M-g}.$$

Consequences of the conversion formulas.

Computing $b_g$ from $a_g^{*}$ and $c_g^{*}$ gives

$$\sum _k{(-1)}^{k+g}{G}_g^k{q}^{{\displaystyle \frac{g(g+1)-k(k+1)}{2}}-gk}{G}_{k+1}^{M+1}=q^{\left({\displaystyle \genfrac{}{}{0pt}{}{M-g}{2}}\right)}\sum _k{(-1)}^{M+k+g}{G}_{M-g}^k{q}^{{\displaystyle \frac{k(k+3)}{2}}-(M+1)k}{G}_{k+1}^{M+1}=q^g.$$

Conversely, computing $a_g^{*}$ and $c_g^{*}$ from $b_g$ yields

$$G_{g+1}^{M+1}=\sum _k{(-1)}^k{G}_{M-g}^{M-k}{G}_{k+1}^{M+1}{q}^{{\displaystyle \frac{k(k+1)}{2}}}=\sum _k{(-1)}^k{G}_{M-g}^{M-k}{G}_{k+1}^{M+1}{q}^{{\displaystyle \frac{k(k+1)}{2}}-g(k+1)}.$$

Application of Theorem 12(3).

From Theorem 12(3) and Theorem 5(3), we have

$$\sum _g{q}^g{(q^{1+g};q)}_{M-g}=\sum _g{q}^{g(g+1)}{G}_{g+1}^{M+1}{(q^{1+g};q)}_{M-g}=\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g+1)}{2}}}{G}_{g+1}^{M+1}=1-{(1;q)}_{M+1}=1.$$

Also,

$$\sum _g{c}_g^{*}{q}^{-g}{z}^g=z^{-1}-z^{-1}{(z;q)}_{M+1}=\sum _g{q}^g{(z{q}^{1+g};q)}_{M-g},$$

which is equivalent to

$${(z;q)}_{M+1}+{\displaystyle \sum _{g=0}^M}z{q}^g{(z{q}^{1+g};q)}_{M-g}=1\to 1+{\displaystyle \sum _{g=0}^M}{\displaystyle \frac{z{q}^g}{{(z;q)}_{g+1}}}={\displaystyle \frac{1}{{(z;q)}_{M+1}}},$$

a generalization of the identity ${\sum }_g{\displaystyle \frac{q^g}{{(q;q)}_g}}={\displaystyle \frac{1}{{(q;q)}_M}}$ (see [5], p. 113).

• Euler’s identity.

Finally, by Theorem 12(2),

$$\sum _g{b}_g{q}^{-g}=M+1=q^{-M}\sum _g{c}_g^{*}{q}^{-(M+1)g}{(q;q)}_g=\sum _g{(-1)}^g{G}_{g+1}^{M+1}{q}^{{\displaystyle \frac{g(g+1)}{2}}-M(g+1)}{(q;q)}_g.$$

Replacing q by $q^{-1}$ recovers Euler’s identity (see [5], p. 83): ${\sum }_{g=1}^M{G}_g^M{(q;q)}_{g-1}=M$.

• Parallel derivation from $c_g^{*}$.

Starting from

$$c_g^{*,1}={(-1)}^g{q}^{{\displaystyle \frac{g(g+1)}{2}}+Bg}{G}_g^M,c_g^{*,2}=c_g^{*,1}{z}^g,$$

(6) gives

$$a_g^{*,1}=q^{g(g+1)}{G}_g^M{(q^{B-g};q)}_g,a_g^{*,2}=q^{g(g+1)}{G}_g^M{(q^{B-g}z;q)}_g.$$

Substituting these into Theorem 12(1)(2) yields:

$$\begin{array}{cc}\hfill \sum _g{(-1)}^g{q}^{Ag+Bg}{G}_g^M{z}^g{(-q^{A+1}z;q)}_{M-g}& =\sum _g{q}^{{\displaystyle \frac{g(g+1)}{2}}+Ag}{G}_g^M{(q^{B-g};q)}_g{z}^g,\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill \sum _g{(-1)}^g{q}^{Ag+Bg}{G}_g^M{z}^g{(-q^{A+1};q)}_{M-g}& =\sum _g{q}^{{\displaystyle \frac{g(g+1)}{2}}+Ag}{G}_g^M{(q^{B-g}z;q)}_g,\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill q^{AM}{z}^M\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g+1)}{2}}+Bg-(M+1)g}{G}_g^M{(q^{-A}{z}^{-1};q)}_g& =\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g-1)}{2}}}{G}_g^M{(q^{B-g};q)}_g{(q^{A+1-g}z;q)}_g,\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill q^{AM}\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g+1)}{2}}+Bg-(M+1)g}{G}_g^M{z}^g{(q^{-A};q)}_g& =\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g-1)}{2}}}{G}_g^M{(q^{B-g}z;q)}_g{(q^{A+1-g};q)}_g.\hfill \end{array}$$

(4)

$B=1,A=1$, (1) yields

$$\sum _g{(-1)}^g{G}_g^M{z}^g{(-z;q)}_{M-g}=1.z=-q\to \sum _g{\displaystyle \frac{q^g}{{(q;q)}_g}}={\displaystyle \frac{1}{{(q;q)}_M}}.$$

$B=1,A=0$, (3) yields

$$z^M\sum _g{(-1)}^g{q}^{{\displaystyle \frac{g(g+1)}{2}}-Mg}{G}_g^M{(z^{-1};q)}_g=1.$$

Set $z=q^{-1}$, replace g by $M-g$ yields

$$q^{-M}\sum _g{(-1)}^{M-g}{q}^{{\displaystyle \frac{(M-g)(M-g+1)}{2}}-M(M-g)}{G}_g^M{(q;q)}_{M-g}=q^{-{\displaystyle \frac{M(M+1)}{2}}}\sum _g{(-1)}^{M-g}{q}^{{\displaystyle \frac{g(g-1)}{2}}}{G}_g^M{(q;q)}_{M-g}=1.$$

Thus

$${\displaystyle \sum _{g=0}^M}{\displaystyle \frac{{(-1)}^g{q}^{\frac{g(g-1)}{2}}}{{(q;q)}_g}}={\displaystyle \frac{{(-1)}^M{q}^{\frac{M(M+1)}{2}}}{{(q;q)}_M}}.$$

Generalization to arbitrary $T_i$ and Cauchy’s identity.

In the preceding discussion we assumed $T_i\in \mathbb{N}$. However, apart from the combinatorial interpretation of $SU{M}_q\left(N\right)$, $T_i$ may be arbitrary numbers. To illustrate this flexibility, set

$$a=b{q}^{-T},PS=[a,a,\dots ,a]:a(q-1),PT=[T,T+1,\dots ,T+M-1].$$

Here a and b are independent by virtue of the free parameter T.

A direct computation using Theorem 1 gives the $H_1^q\left(g\right)$ and $H_3^q\left(g\right)$ coefficients:

$$H_1^q\left(g\right)=G_g^M{q}^{{\displaystyle \frac{g(g+1)}{2}}}(b-a)(bq-a)\dots (b{q}^{g-1}-a)a^{M-g},$$

and

$$H_3^q\left(g\right)=G_g^M{(-1)}^g{q}^{{\displaystyle \frac{g(g+1)}{2}}}{a}^M.$$

With $a_g^{*}=H_1^q\left(g\right)q^{-(T-1)g}$ and $c_g^{*}=H_3^q\left(g\right)q^{-(T-1)g}$, Thus

$$\begin{array}{cc}\hfill \sum _g{c}_g^{*}{\left(bx\right)}^g{q}^{(T-1)g-(T+1)g}& =\sum _g{H}_3^q\left(g\right){\left(bx\right)}^g{q}^{-(T+1)g}\hfill \\ & =a^M\sum _g{G}_g^M{(-1)}^g{q}^{{\displaystyle \frac{g(g-1)}{2}}}{\left(b{q}^{-T}x\right)}^g\hfill \\ & =a^M{(ax;q)}_M.\hfill \end{array}$$

On the other hand (Theorem 12(3)), expressing the same sum in terms of $a_g^{*}$ gives ($b^g{q}^{-Tg}=a^g$)

$$\begin{array}{cc}\hfill \sum _g{a}_g^{*}{\left(bx\right)}^g{q}^{(T-1)g-(T+1)g}{(bx{q}^g;q)}_{M-g}& =\sum _g{H}_1^q\left(g\right){\left(bx\right)}^g{q}^{-(T+1)g}{(bx{q}^g;q)}_{M-g}\hfill \\ & =a^M\sum _g{G}_g^M{q}^{{\displaystyle \frac{g(g-1)}{2}}}(b-a)(bq-a)\dots (b{q}^{g-1}-a)x^g{(bx{q}^g;q)}_{M-g}.\hfill \end{array}$$

Equating the two expressions and dividing by ${(bx;q)}_M$, we obtain

$${\displaystyle \frac{{(ax;q)}_M}{{(bx;q)}_M}}={\displaystyle \frac{{\sum }_g{G}_g^M{q}^{\frac{g(g-1)}{2}}(b-a)(bq-a)\dots (b{q}^{g-1}-a)x^g{(bx{q}^g;q)}_{M-g}}{{(bx;q)}_M}}.$$

Letting $M\to \infty$ and using the standard limiting argument recovers Cauchy’s identity (see [5], p. 260):

$${\displaystyle \frac{{(ax;q)}_{\infty }}{{(bx;q)}_{\infty }}}={\displaystyle \sum _{g=0}^{\infty }}{\displaystyle \frac{q^{\frac{g(g-1)}{2}}{x}^g(b-a)(bq-a)\dots (b{q}^{g-1}-a)}{{(q;q)}_g{(bx;q)}_g}}.$$

The case $M=\infty$.

Letting $M\to \infty ,\left|x\right|<1$ and taking $b_g=x^g$ then (Lemma 2)

$$a_g^{*}=q^{g(g+1)}{\displaystyle \sum _{k=0}^{\infty }}{x}^k{G}_g^k={\displaystyle \frac{q^{g(g+1)}{x}^g}{{(x;q)}_{g+1}}}.$$

we obtain from Theorem 12(1):

$${\displaystyle \sum _{g=0}^{\infty }}{\displaystyle \frac{q^{\frac{g(g+1)}{2}-Ag}{x}^g{z}^g}{{(x;q)}_{g+1}}}={\displaystyle \sum _{g=0}^{\infty }}{x}^g{(-q^{-A+1}z;q)}_g.$$

For $A\ge 0,z=-q^{-1}$, the right-hand side collapses to a finite sum:

$${\displaystyle \sum _{g=0}^{\infty }}{(-1)}^g{\displaystyle \frac{q^{\frac{g(g-1)}{2}-Ag}{x}^g}{{(x;q)}_{g+1}}}={\displaystyle \sum _{g=0}^A}{x}^g{(q^{-A};q)}_g.$$

Theorem 12(2):

$${\sum }_{g=0}^{\infty }{x}^g{z}^g{q}^{Ag}={\displaystyle \frac{1}{1-xz{q}^A}}={\sum }_{g=0}^{\infty }{\displaystyle \frac{{(-1)}^g{q}^{\frac{g(g-1)}{2}}{x}^g}{{(x;q)}_{g+1}}}{(q^{A+1-g}z;q)}_g$$

For $A\ge 0,z=1$, the right-hand side collapses to a finite sum:

$${\displaystyle \frac{1}{1-x{q}^A}}={\sum }_{g=0}^A{\displaystyle \frac{{(-1)}^g{q}^{\frac{g(g-1)}{2}}{x}^g}{{(x;q)}_{g+1}}}{(q^A;q^{-1})}_g.$$

In summary, the framework above not only provides short proofs of classical q-binomial theorems (and their finite extensions), but also yields a substantial collection of apparently new identities. Moreover, by freely specializing the free parameters, one can systematically generate a vast further family of such identities, of which only a few representative examples are given here.