A Unified Algorithmic Framework for Nested q-Binomial and Binomial Sums

1. Introduction

Consider the sum of products of M distinct numbers chosen from $\{1,2,\dots ,N\}$. If two consecutive chosen numbers differ by more than 1, we call it a gap. Grouping products by the position and number (named g) of gaps yields $\left({\displaystyle \genfrac{}{}{0pt}{}{M-1}{g}}\right)$ groups for each g, with $2^{M-1}$ groups in total.

We find that the sum over each group has a simple closed form,

$$\sum _{\mathrm{one}\mathrm{group}}=(\mathrm{The}\mathrm{smallest}\mathrm{product}\mathrm{in}\mathrm{the}\mathrm{group})\left({\displaystyle \genfrac{}{}{0pt}{}{N+M+g}{M+1+g}}\right).$$

When the $g=0$, the result reduces to the classical identity

$${\displaystyle \sum _{n=0}^{N-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n+M}{M}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{N+M}{M+1}}\right)={\displaystyle \frac{1}{M!}}(1·2\dots M+2·3\dots (M+1)+\dots ).$$

and that for groups with the same number g of gaps the coefficients can be combined. This yields the final formula for the M-distinct sum, stated in Section 7. Observing that each group is in fact a nested sum, we are led to extend this structure to binomial coefficients and then to q-binomial. In doing so, we find that every such sum admits three equivalent expansions, which together form the unified framework developed in this paper.

The core contributions of the present work consist of five facets: computational machinery for nested summations, the three-form equivalence property, mutual transformation relations among the coefficient families, the capability to match arbitrary target sequences, and an entirely new methodological viewpoint.

Many identities emerge as by-products of this machinery. A subset of these are well-known classical q-binomial theorems, which we extend to finite-form settings; this task is substantially more difficult and conceptually distinct from the original infinite-series treatments. We employ these classical results primarily as consistency checks: recovering them serves to verify the correctness of our formalism, rather than constituting primary objectives of this paper. Beyond such benchmark checks, the same framework systematically yields numerous apparently new identities.

Although the binomial framework was established earlier and q-binomial analogues are more complex, we develop the theory for q-binomial objects first in this manuscript. This presentation strategy avoids many duplicate definitions and proofs.

To place this observation in context, recall that the q-binomial coefficient is fundamental in combinatorics, featuring in partition theory, representation theory and quantum-group studies. Classical identities including the q-binomial theorem, q-Vandermonde convolution, and summations of Gauss, Cauchy and Jacobi form cornerstones of q-analysis. Nevertheless, nested sums built from q-binomial-coefficient and q-integer products are mostly treated by ad-hoc tricks, with no general computational framework for arbitrary such sums yet available [1,2,3]. We construct this missing framework using the recursive operator ${\mathrm{SUM}}_q$ (for arbitrary-depth nested sums) together with the shift operator ${\nabla }_q^p$. Our main results (Theorems 2.1, 3.3) prove that every admissible nested sum possesses three equivalent q-binomial expansions connected by explicit transformations, yielding inversion formulas, orthogonality relations, and closed-form evaluations for weighted infinite series which subsume 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 framework, however, extends well beyond the q-domain. In the classical limit $q\to 1$, the three-form equivalence reduces to explicit summation formulas for the Stirling numbers of both kinds, as well as for the Eulerian numbers and their associated variants, providing a direct combinatorial interpretation in terms of set partitions and permutations. Moreover, since the framework yields exact summation formulas, it automatically enables the determination of congruences by simply reading off the modular behavior of the resulting binomial expansions. This yields a broad family of prime congruences that arise uniformly from the same mechanical procedure, revealing a natural yet unexpected bridge between nested summation and modular arithmetic.

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. Section 7 (The limiting case $q=1$) examines the classical degeneration and its connections to Stirling numbers, Eulerian numbers, and modular congruences.

2. Calculation Formula

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

$${\left[\begin{array}{c}M\\ N\end{array}\right]}_q={\displaystyle \frac{(q^N-1)(q^{N-1}-1)\dots (q^{N-M+1}-1)}{(q^M-1)(q^{M-1}-1)\dots (q^1-1)}},q\ne 0,1.$$

For convenience, we write $G_M^N={\left[\begin{array}{c}M\\ N\end{array}\right]}_q$ 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.

$x^{\underline{n}}$ denotes the falling factorial and $x^{\overline{n}}$ denotes the rising factorial:

$$x^{\underline{n}}=x(x-1)\dots (x-n+1),x^{\overline{n}}=x(x+1)\dots (x+n-1),$$

with the empty-product convention $x^{\underline{0}}=x^{\overline{0}}=1$.

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 2.1. 

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 2.1. 

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 2.1. 

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 $M\ge 2$,

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

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

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 2.1. 

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 2.1, 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 2.1. 

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 2.2. 

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 2.2 

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

Theorem 2.3. 

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 2.1 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 2.2 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 2.2 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 2.2 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 3.1. 

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 3.1 (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 2.1,

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

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 2.1; 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 3.2 (Basic q-binomial identities) 

The following identities hold:

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

(A3)

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

(A3)

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

(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 3.1. 

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] [ p. 116]). 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 $G_k^M$:

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

(8)

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

(9)

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

Theorem 3.3 (Mutual transformations) 

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

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

(B6)

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

(B6)

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

(B6)

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

(B6)

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

(B6)

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

(B6)

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 2.3, 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,

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

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).

The inductive proof above assumes $D_i\ne 0$ via a scaling substitution. When some $D_i=0$, every selection with $X_i=T_i$ vanishes in the definitions of $H_r^q\left(g\right)$, so only branches with $X_i=K_i$ contribute. One may verify by direct substitution that all transformation identities (B1)–(B6) continue to hold in this degenerate setting. □

We also define the symmetric functions

$$F_g^{\mathbb{K}}=\sum _{1\le {\lambda }_1<\dots <{\lambda }_g\le M}{\displaystyle \prod _{i=1}^g}{K}_{{\lambda }_i},E_g^{\mathbb{K}}=\sum _{1\le {\lambda }_1\le \dots \le {\lambda }_g\le M}{\displaystyle \prod _{i=1}^g}{K}_{{\lambda }_i},F_g^N=F_g^{\{1,2,\dots ,N\}},E_g^N=E_g^{\{1,2,\dots ,N\}}$$

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 3.4. 

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 paper 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 2.1 ($For{m}_1$) we have

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

The elementary symmetric functions $F_g^{\mathbb{K}}$ are algebraically independent, so the parameter $K_i$ provide M degrees of freedom. By Theorem 3.4, the values $H_1^q\left(g\right)$ for $0\le g<M$ are independent and can be assigned arbitrarily by choosing the parameters $K_i$ . Therefore, given any sequence ${\left\{a_g\right\}}_{g=0}^M,a_M\ne 0$, 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 3.3 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.

For $a_M^{*}\ne 0$, the construction above with $D_i\equiv 1$ and global scaling works. For the case $a_M^{*}=0$, we may select some $D_i=0$ to achieve the target sequence. The key point is that any target sequence can be matched to $H_r^q\left(g\right)$, and we may then employ the recursive relations in Theorem 2.1 together with Theorem 3.3.

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

Theorem 3.5. 

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 3.3. 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 4.1 (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 4.1 and Theorem 3.3 to get:

Theorem 4.2 (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 3.3 to get:

Theorem 4.3 (Generalized q-Vandermonde identities) 

For $0\le k\le M$ and $p\ge -1$:

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

(C6)

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

(C6)

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

(C6)

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

(C6)

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

(C6)

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

(C6)

Binomial coefficient relations (limit $q\to 1$).

Assume all binomial arguments are non-negative integers. 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 4.3 together with Theorem 3.3, we obtain:

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

Theorem 2.3(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 `multiplying by $x^M$, this 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 identity (A), ${(-1)}^g{a}_g$ is paired with $\left({\displaystyle \genfrac{}{}{0pt}{}{X+g}{Y+g}}\right)$ (corresponding to $For{m}_2$), while ${(-1)}^g{b}_g$ is paired with $\left({\displaystyle \genfrac{}{}{0pt}{}{X}{Y+g}}\right)$ (corresponding to $For{m}_1$). Accordingly

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

Analogous substitutions hold for the remaining relations.

5. Applications

Illustration of the computational technique.

As an application of Theorem 3.2 (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 3.1(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 3.2 (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 3.2 (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 3.1 (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 3.2 (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 3.1 (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

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

(22)

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.

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

The first equation extends the Gauss q-binomial theorem [5] [p. 61]: ${\sum }_{k=0}^M{(-1)}^k{G}_k^M=\left\{\begin{array}{cc}0,\hfill & M\mathrm{odd},\hfill \\ {(q;q^2)}_{{\displaystyle \frac{M}{2}}},\hfill & M\mathrm{even}.\hfill \end{array}\right.$.

The fifth equation extends another Gauss identity [5] [p. 65]: ${\sum }_{k=0}^M{\left[\begin{array}{c}k\\ M\end{array}\right]}_{q^2}{q}^k={(-q;q)}_M.$

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

Proposition 5.1. 

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 3.1(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 3.4 and (2) ($T_1\ne 1$ does not affect the result).

For (4), by Theorem 3.1 (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 5.1. 

We define a new number,

$${〈\begin{array}{c}M\\ M\end{array}〉}^q=0,{〈\begin{array}{c}g\\ M\end{array}〉}^q=\sum _{{\lambda }_1+\dots +{\lambda }_{g+1}=M-g-1}{\displaystyle \prod _{i=1}^{g+1}}{\left[i\right]}_q^{{\lambda }_i}{[1+{\lambda }_1]}_q{[1+{\lambda }_1+{\lambda }_2]}_q\dots {[1+{\lambda }_1+{\lambda }_2+\dots +{\lambda }_g]}_q,$$

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

$${〈\begin{array}{c}g\\ M\end{array}〉}^q={[M-g]}_q{〈\begin{array}{c}g-1\\ M-1\end{array}〉}^q+{[g+1]}_q{〈\begin{array}{c}g\\ M-1\end{array}〉}^q,{〈\begin{array}{c}g\\ M\end{array}〉}^q={〈\begin{array}{c}M-g-1\\ M\end{array}〉}^q.$$

Proposition 5.2 (Three expansions of ${\left[N\right]}_q^M$ )

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

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 3.1(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 ${\lambda }_1:={\lambda }_1-1$,

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

Thus

$$H_3^q\left(g\right)=q^{-(M-g)+{\displaystyle \frac{g(g+1)}{2}}}{〈\begin{array}{c}g\\ M\end{array}〉}^q.$$

Substituting it into $SU{M}_q$ gives

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

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

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

Substituting $g\mapsto M-1-g$ and ${〈\begin{array}{c}M-g-1\\ M\end{array}〉}^q={〈\begin{array}{c}g\\ M\end{array}〉}^q$, 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 $q\to 1$,

$$\underset{q\to 1}{lim}{〈\begin{array}{c}g\\ M\end{array}〉}^q=〈\begin{array}{c}g\\ M\end{array}〉=\sum _{{\lambda }_1+\dots +{\lambda }_{g+1}=M-g-1}{\displaystyle \prod _{i=1}^{g+1}}{i}^{{\lambda }_i}(1+{\lambda }_1)(1+{\lambda }_1+{\lambda }_2)\dots (1+{\lambda }_1+{\lambda }_2+\dots +{\lambda }_g)$$

$$N^M={\displaystyle \sum _{g=1}^M}g!S_2(M,g)\left({\displaystyle \genfrac{}{}{0pt}{}{N}{g}}\right)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g!S_2(M,g)\left({\displaystyle \genfrac{}{}{0pt}{}{N+g-1}{g}}\right)={\displaystyle \sum _{g=0}^{M-1}}〈\begin{array}{c}g\\ M\end{array}〉\left({\displaystyle \genfrac{}{}{0pt}{}{N+g}{M}}\right).$$

Here $S_2(M,g)$ and $〈\begin{array}{c}g\\ M\end{array}〉$ denoting Stirling and Eulerian numbers (the expression of the latter is the main result of [9]). Theorem 2.3(1) gives three expressions for the Eulerian polynomial (here q denotes the weight parameter a from Theorem 2.3).

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{n}^M=q^N{\displaystyle \sum _{k=0}^M}{\displaystyle \frac{{(-1)}^k}{{(q-1)}^{k+1}}}{\nabla }^{1+k}{\mathrm{SUM}}_q(N-1)+{\displaystyle \frac{A_q\left(M\right)q}{{(1-q)}^{M+1}}}.$$

Here $A_q\left(M\right)$ is the classical Eulerian polynomial, with $PS=[1,1,\dots ,1],PT=[1,2,\dots ,M]$,and $A_q\left(M\right)=A_q^1(PS,PT,0)$. ${\mathrm{Form}}_3$ yields the classical representation.

Using the definition of a difference,

$${\nabla }^{1+k}SUM(N-1)={\nabla }^k{(N-1)}^M={\displaystyle \sum _{j=0}^k}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){(N-1-j)}^M,$$

$$={\displaystyle \sum _{j=0}^k}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\displaystyle \sum _{g=0}^M}\left({\displaystyle \genfrac{}{}{0pt}{}{M}{g}}\right)N^g{(j+1)}^{M-g}{(-1)}^{M-g}={\displaystyle \sum _{g=0}^M}\left({\displaystyle \genfrac{}{}{0pt}{}{M}{g}}\right)N^g{(-1)}^{M-g-k}{\displaystyle \sum _{j=0}^k}{(-1)}^{j+k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){(j+1)}^{M-g}.$$

Recall the identity for Stirling numbers of the second kind

$$S_2(M+1,k+1)={\displaystyle \frac{1}{k!}}{\displaystyle \sum _{j=0}^k}{(-1)}^{j+k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){(j+1)}^M,$$

we obtain

$${\nabla }^{1+k}SUM\left(n\right)={\displaystyle \sum _{g=0}^M}{(-1)}^{M-g-k}\left({\displaystyle \genfrac{}{}{0pt}{}{M}{g}}\right)N^g{S}_2(M-g+1,k+1)k!.$$

Therefore

$$q^N{\displaystyle \sum _{k=0}^M}{\displaystyle \frac{{(-1)}^k}{{(q-1)}^{k+1}}}{\nabla }^{1+k}SUM(N-1)$$

$$={\displaystyle \frac{q^N}{{(q-1)}^{M+1}}}{\displaystyle \sum _{g=0}^M}{(-1)}^{M-g}{N}^g{(q-1)}^g\left({\displaystyle \genfrac{}{}{0pt}{}{M}{g}}\right){\displaystyle \sum _{k=0}^M}{(q-1)}^{M-k-g}{S}_2(M-g+1,k+1)k!.$$

At $N=0$, we have ${\sum }_{n=0}^{N-1}{q}^n{n}^M=0$, so

$${\displaystyle \sum _{k=0}^M}{(-1)}^M{(q-1)}^{M-k}{S}_2(M+1,k+1)k!=-A_q\left(M\right)q{(-1)}^{M+1}.$$

This gives us the fourth expression for $A_q\left(M\right)$.

$$q{A}_q\left(M\right)={\displaystyle \sum _{k=0}^M}{(q-1)}^{M-k}{S}_2(M+1,k+1)k!,$$

and more generally,

$$q{A}_q(M-g)={\displaystyle \sum _{k=0}^{M-g}}{(q-1)}^{M-g-k}{S}_2(M-g+1,k+1)k!.$$

Therefore

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{n}^M={\displaystyle \frac{q^{N+1}}{{(q-1)}^{M+1}}}{\displaystyle \sum _{g=0}^M}{(-1)}^{M-g}{N}^g{(q-1)}^g\left({\displaystyle \genfrac{}{}{0pt}{}{M}{g}}\right)A_q(M-g)+{\displaystyle \frac{A_q\left(M\right)q}{{(1-q)}^{M+1}}}.$$

From Theorem 2.3, we see that the Eulerian polynomial is merely a special case.

Proposition 5.3. 

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 5.1(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 3.1(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 5.1(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 3.5.

(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 3.3(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 5.4 (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 5.1(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 $M=1$,

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

Propositions 5.4 and 5.3 illustrate the new proof method developed in this paper.

6. Unified Conversion Formulas

The following theorem is a direct consequence of Theorem 3.3 and Theorem 3.2(3); it also generalizes Theorem 2.3(1).

Theorem 6.1. 

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 3.3 and (8)–(), 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 3.1(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 6.1(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 6.1(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 6.1(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 6.1(3).

From Theorem 6.1(3) and Theorem 3.2(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 6.1(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

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

() 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 6.1(1)(2) yields:

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

(4)

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

(4)

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

(4)

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

(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 2.1 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 6.1(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 2.2)

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

we obtain from Theorem 6.1(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 6.1(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.

7. The Limiting Case $\mathbf{q}=\mathbf{1}$

Taking the limit $q\to 1$ in Theorem 2.1, the three expansions reduce to the following binomial forms. Let $H=T_M-M$.

$$SUM(N,PS,PT)={\displaystyle \sum _{g=0}^M}{H}_1\left(g\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+H}{H+1+g}}\right)={\displaystyle \sum _{g=0}^M}{H}_2\left(g\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+H+g}{H+1+g}}\right)={\displaystyle \sum _{g=0}^M}{H}_3\left(g\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+T_M-g}{T_M+1}}\right).$$

where

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

We have

$$SUM(N,[1,1,\dots ,1],[1,3,\dots ,2M-1])=SUM(N,[1,1,\dots ,1],[3,5,\dots ,2M-1])=E_M^N=S_2(N+M,N).$$

$$SUM(N,[1,2,\dots ,M],[1,3,\dots ,2M-1])=SUM(N,[2,3,\dots ,M],[3,5,\dots ,2M-1])=F_M^{N+M-1}=S_1(N+M,N).$$

Here $S_1(n,k)$ and $S_2(n,k)$ denote the unsigned Stirling numbers of the first and second kind, respectively. These formulas enable the computation of $S_1(N,N-M)$ and $S_2(N,N-M)$.

For completeness, we recall the definitions of the associated Stirling numbers used in this section.

The associated Stirling numbers of the first kind $S_{1,r}(n,k)$ are defined by

$$S_{1,r}(n,k)={\displaystyle \frac{n!}{k!}}\sum _{\begin{array}{c}{i}_1+\dots +i_k=n\\ i_j\ge r\end{array}}{\displaystyle \frac{1}{i_1\dots i_k}},$$

and satisfy the recurrence

$$S_{1,r}(n+1,k)=n{S}_{1,r}(n,k)+{\left(n\right)}_{r-1}{S}_{1,r}(n-r+1,k-1),n\ge kr.$$

The associated Stirling numbers of the second kind $S_{2,r}(n,k)$ are defined by

$$S_{2,r}(n,k)={\displaystyle \frac{n!}{k!}}\sum _{\begin{array}{c}{i}_1+\dots +i_k=n\\ i_j\ge r\end{array}}{\displaystyle \frac{1}{i_1!\dots i_k!}},$$

and satisfy

$$S_{2,r}(n+1,k)=k{S}_{2,r}(n,k)+\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r-1}}\right)S_{2,r}(n-r+1,k-1),n\ge kr.$$

The terminology and systematic treatment of associated Stirling numbers are due to Howard [10].

Definition 7.1. 

$p,q\in \mathbb{N}$. Define the operation ⊙ by

$$E_p^q\odot ([I_1,I_2,\dots ,I_M],C)=\sum _{{\lambda }_1+{\lambda }_2+\dots +{\lambda }_q=p,{\lambda }_i\ge 0}{1}^{{\lambda }_1}{2}^{{\lambda }_2}\dots q^{{\lambda }_q}(I_1+{\lambda }_{1}C)(I_2+{\lambda }_{1}C+{\lambda }_{2}C)\dots (I_{q-1}+{\lambda }_{1}C+\dots +{\lambda }_{q-1}C).$$

Case 1: $PS=[1,1,\dots ,1]$, $PT=[1,2,\dots ,M]$.

For $H_1$, when $X_i=K_i$, we have $B_i=1+X_{T-1}=1+X_T$. Hence

$$\sum _{X_i=K_i}\prod B_i=E_{M-g}^{g+1}=S_2(M+1,g+1),$$

so

$$H_1\left(g\right)=g!S_2(M+1,g+1).$$

For $H_2$, when $X_i=K_i$, we have $B_i=1+X_{K-1}-i=-X_{T-1}=-X_T$. Hence

$$H_2\left(g\right)=g!\sum _{X_i=K_i}\prod B_i=g!{(-1)}^{M-g}{E}_{M-g}^g={(-1)}^{M-g}g!S_2(M,g).$$

For $H_3$, we have already obtained in Proposition 5.2:

$$H_3\left(g\right)=E_{M-g}^{g+1}\odot ([0,0,\dots ,0],1)=E_{M-g-1}^{g+1}\odot ([1,1,\dots ,1],1)=〈\begin{array}{c}g\\ M\end{array}〉.$$

Case 2: $PS=[2,3,\dots ,M]$, $PT=[3,5,\dots ,2M-1]$

(Row 6 of Table 1).

Set

$$P{S}_1=[PS,M+1],P{T}_1=[PT,2M+1].$$

For $H_1$, the recursive definition gives

$$\begin{array}{cc}\hfill H_1(g,P{S}_1,P{T}_1)& =H_1(g-1)(2M+1-[(M-1)-(g-1)])+H_1\left(g\right)(M+1+g)\hfill \\ & =(M+g+1)H_1\left(g\right)+(M+g+1)H_1(g-1).\hfill \end{array}$$

With the initial condition at $M=1$ and the standard recurrence for $S_{1,2}$, we obtain

$$H_1\left(g\right)=S_{1,2}(M+g+1,g+1).$$

A combinatorial interpretation is also available. Since

$$SUM(N,[2,\dots ,M],[3,\dots ,2M-1])=F_M^{N+M-1},$$

it represents the sum of products of M distinct numbers ${\prod }_{i=1}^M{k}_i$ chosen from $\{1,2,\dots ,N+M-1\}$.

If $k_{i+1}-k_i>1$, we call it a gap. Let $P{X}_g=[x_1,x_2,\dots ,x_M]$ denote the pattern in a group with g gaps, normalized so that $x_1=1$ and $x_{i+1}-x_i=1$ or 2. For example, when $M=4$,

$$\begin{array}{c}P{X}_0=[1,2,3,4],\\ P{X}_1=[1,3,4,5],[1,2,4,5],[1,2,3,5],\\ P{X}_2=[1,3,4,6],[1,3,5,6],[1,2,4,6],\\ P{X}_3=[1,3,5,7].\end{array}$$

By Theorem 3.1(5), the sum over each group is

$$SUM(N-g,P{X}_g,P{X}_g)=\left({\displaystyle \prod _{i=1}^M}{x}_i\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+M}{M+1+g}}\right).$$

The gap patterns allow one to read off the products $P{X}_g$ entirely by visual inspection, without invoking any algebraic formula for the coefficients.

Thus $H_1\left(g\right)={\sum }_{P{X}_g}\prod x_i$, and each $1·\prod B_i$ corresponds to a unique $\prod x_i$ in some $P{X}_g$. This gives the explicit form

$$S_{1,2}(M+g+1,g+1)=\sum {\displaystyle \frac{(M+g)!}{i_1{i}_2\dots i_g}},$$

where the sum is over $2\le i_1<i_2<\dots <i_g\le M+g-1$ with $i_{j+1}-i_j\ge 2$. This is precisely the conclusion of [11].

For $H_2$, the recurrence becomes

$$\begin{array}{cc}\hfill H_2(g,P{S}_1,P{T}_1)& =H_2(g-1)(M+g+1)+H_2\left(g\right)\left(M+1+\left[\right(M-1)-g]-(2M+1)\right)\hfill \\ & =-(g+1)H_2\left(g\right)+(M+g+1)H_2(g-1).\hfill \end{array}$$

With the initial condition at $M=1$ and the recurrence for $S_{2,2}$, we obtain

$$H_2\left(g\right)={(-1)}^{M-1-g}{S}_{2,2}(M+g+1,g+1).$$

Indeed, substituting this into the recurrence gives

$$\begin{array}{cc}\hfill {(-1)}^{M-g}{S}_{2,2}(M+g+2,g+1)& =-(g+1){(-1)}^{M-1-g}{S}_{2,2}(M+g+1,g+1)\hfill \\ & +(M+g+1){(-1)}^{M-1-(g-1)}{S}_{2,2}(M+g,g),\hfill \end{array}$$

which simplifies to

$$S_{2,2}(M+g+2,g+1)=(g+1)S_{2,2}(M+g+1,g+1)+(M+g+1)S_{2,2}(M+g,g),$$

matching the standard recurrence.

For $H_3$, using the method from Proposition 5.2 , we can obtain

$$H_3\left(g\right)=E_g^{M-g}\odot ([2,3,\dots ,M],2).$$

Table 1. Summary of $H\left(g\right)$ for various choices of $PS$ and $PT$.

$\mathit{P}\mathit{S}$	$\mathit{P}\mathit{T}$	${\mathit{H}}_1\left(\mathit{g}\right)$	${\mathit{H}}_2\left(\mathit{g}\right)$	${\mathit{H}}_3\left(\mathit{g}\right)$
$[1,1,\dots ,1]$	$[1,2,\dots ,M]$	$g!S_2(M+1,g+1)$	${(-1)}^{M-g}g!S_2(M,g)$	$〈\begin{array}{c}g\\ M\end{array}〉$
$[1,1,\dots ,1]$	$[2,3,\dots ,M]$	$(g+1)!S_2(M,g+1)$	${(-1)}^{M-1-g}(g+1)!S_2(M,g+1)$	$〈\begin{array}{c}g\\ M\end{array}〉$
$[1,1,\dots ,1]$	$[1,3,\dots ,2M-1]$	$E_{M-g}^{g+1}\odot (PT,1)$	${(-1)}^{M-g}{S}_{1,2}(M+g,g)$	$E_{M-g}^{g+1}\odot ([0,1,\dots ],2)$
$[1,1,\dots ,1]$	$[3,5,\dots ,2M-1]$	$S_{2,2}(M+g+1,g+1)$	${(-1)}^{M-1-g}{S}_{1,2}(M+g+1,g+1)$	$E_{M-1-g}^{g+1}\odot ([2,3,\dots ],2)$
$[1,2,\dots ,M]$	$[1,3,\dots ,2M-1]$	$\{*\}$	${(-1)}^{M-g}{E}_{M-g}^g\odot ([3,5,\dots ],1)$	$E_g^{M-g}\odot ([2,3,\dots ],2)$
$[2,3,\dots ,M]$	$[3,5,\dots ,2M-1]$	$S_{1,2}(M+g+1,g+1)$	${(-1)}^{M-1-g}{S}_{2,2}(M+g+1,g+1)$	$E_g^{M-g}\odot ([2,3,\dots ],2)$

Here $\{*\}=S_{1,2}(M+g+1,g+1)+S_{1,2}(M+g,g)$. See Proposition 5.1(4).

Table 1. Summary of $H\left(g\right)$ for various choices of $PS$ and $PT$.

$\mathit{P}\mathit{S}$	$\mathit{P}\mathit{T}$	${\mathit{H}}_1\left(\mathit{g}\right)$	${\mathit{H}}_2\left(\mathit{g}\right)$	${\mathit{H}}_3\left(\mathit{g}\right)$
$[1,1,\dots ,1]$	$[1,2,\dots ,M]$	$g!S_2(M+1,g+1)$	${(-1)}^{M-g}g!S_2(M,g)$	$〈\begin{array}{c}g\\ M\end{array}〉$
$[1,1,\dots ,1]$	$[2,3,\dots ,M]$	$(g+1)!S_2(M,g+1)$	${(-1)}^{M-1-g}(g+1)!S_2(M,g+1)$	$〈\begin{array}{c}g\\ M\end{array}〉$
$[1,1,\dots ,1]$	$[1,3,\dots ,2M-1]$	$E_{M-g}^{g+1}\odot (PT,1)$	${(-1)}^{M-g}{S}_{1,2}(M+g,g)$	$E_{M-g}^{g+1}\odot ([0,1,\dots ],2)$
$[1,1,\dots ,1]$	$[3,5,\dots ,2M-1]$	$S_{2,2}(M+g+1,g+1)$	${(-1)}^{M-1-g}{S}_{1,2}(M+g+1,g+1)$	$E_{M-1-g}^{g+1}\odot ([2,3,\dots ],2)$
$[1,2,\dots ,M]$	$[1,3,\dots ,2M-1]$	$\{*\}$	${(-1)}^{M-g}{E}_{M-g}^g\odot ([3,5,\dots ],1)$	$E_g^{M-g}\odot ([2,3,\dots ],2)$
$[2,3,\dots ,M]$	$[3,5,\dots ,2M-1]$	$S_{1,2}(M+g+1,g+1)$	${(-1)}^{M-1-g}{S}_{2,2}(M+g+1,g+1)$	$E_g^{M-g}\odot ([2,3,\dots ],2)$

Here $\{*\}=S_{1,2}(M+g+1,g+1)+S_{1,2}(M+g,g)$. See Proposition 5.1(4).

We observe that classical numbers such as Stirling and Eulerian numbers are not isolated special cases within our framework. Instead, they arise naturally as a specific realization of the general formulas we derived. This perspective enables straightforward generalizations to broader families of sequences. As an illustration of possible generalizations, we present the following generalized sequences.

$${\mathit{F}}_{\mathit{T}}(\mathit{N}+\mathit{M},\mathit{N})=F_M^{\{T,T+1,\dots ,T+N+M-2\}}=\mathrm{SUM}\left(N,[T,T+1,\dots ,T+M-1],[1,3,\dots ,2M-1]\right)$$

For $H_1\left(g\right)$ of this $SUM$, there is no direct counterpart for $S_{1,2}$. Nevertheless, using the notion of gaps, we obtain

$$H_1\left(g\right)=\sum _{{\widetilde{PX}}_g}+\sum _{{\widetilde{PX}}_{g-1}}$$

where $\widetilde{PX}$ denotes gap-modified patterns, obtained by substituting selected entries $X_i\in \mathbb{K}$ with $T+X_i-1$. For illustration, we write the product expansions for $M=3$:

$$\sum _{P{X}_1}+\sum _{P{X}_2}=(1,2,4)+(1,3,4)+(1,3,5)\to H_1\left(g\right)=(T,2,4)+(1,T+2,4)+(1,3,T+4)$$

We can also define

$${\mathit{E}}_{\mathit{T}}(\mathit{N}+\mathit{M},\mathit{N})=E_M^{\{T,T+1,\dots ,T+N-1\}}=\mathrm{SUM}\left(N,[T,T,\dots ,T],[1,3,\dots ,2M-1]\right)$$

Clearly,

$$\nabla \mathrm{SUM}\left(n,[T,T,\dots ,T],[1,2,\dots ,M]\right)={(T+n-1)}^M={\displaystyle \sum _{g=0}^M}g!E_T(M+1,g+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{g}}\right)$$

Moreover, it follows easily from the definition that $H_3(g,[T,PS],[T,PT])=T·H_3(g,PS,PT)$,

$${〈\begin{array}{c}g\\ M\end{array}〉}_T=H_3\left(g,[T,1,1,\dots ,1],[T,T+1,\dots ,T+M-1]\right)=T·H_3\left(g,[1,1,\dots ,1],[T+1,\dots ,T+M-1]\right).$$

$$=T\sum 1^{{\lambda }_1}{2}^{{\lambda }_2}\dots {(g+1)}^{{\lambda }_{g+1}}(T+{\lambda }_1)(T+{\lambda }_1+{\lambda }_2)\dots (T+{\lambda }_1+\dots +{\lambda }_g),{\lambda }_1+\dots +{\lambda }_{g+1}=M-g,{\lambda }_i\ge 0.$$

which matches

$${〈\begin{array}{c}j\\ n\end{array}〉}_T=(T-1+n-j){〈\begin{array}{c}j-1\\ n-1\end{array}〉}_T+(j+1){〈\begin{array}{c}j\\ n-1\end{array}〉}_T,{〈\begin{array}{c}0\\ 1\end{array}〉}_T=T,{〈\begin{array}{c}1\\ 1\end{array}〉}_T=0,{〈\begin{array}{c}j\\ n\end{array}〉}_T=0,j<0,j>n.$$

From Theorem 3.3, we readily obtain

$${\displaystyle \sum _{g=0}^M}{(-1)}^{g}g!E_T(M+1,g+1)={(T-1)}^M,{\displaystyle \sum _{g=0}^M}{〈\begin{array}{c}g\\ M\end{array}〉}_T=T^{\overline{M}},{\displaystyle \sum _{g=0}^M}〈\begin{array}{c}g\\ M\end{array}〉=M!.$$

Proof of a classical identity

$$\mathrm{SUM}(N,\left[K\right],\left[4\right])=\mathrm{SUM}(N,[1,K],[1,4])={\displaystyle \frac{1}{2!}}\mathrm{SUM}(N,[1,2,K],[1,2,4])={\displaystyle \frac{1}{3!}}\mathrm{SUM}(N,[1,2,3,K],[1,2,3,4])$$

$$=\mathrm{SUM}(N,[1,1:0,K],[1,3,5\left]\right),\mathrm{same}\mathrm{as}\mathrm{SUM}(N,[1,K],[1,4\left]\right),$$

$$=\mathrm{SUM}(N,[1:0,1:0,1:0,K],[1,3,5,7])={\displaystyle \sum _{n_4=0}^{N-1}}{\displaystyle \sum _{n_3=0}^{n_4}}{\displaystyle \sum _{n_2=0}^{n_3}}{\displaystyle \sum _{n_1=0}^{n_2}}(K+n_4).$$

We can verify that

$$P{S}_1=[1:0,1:0,1:0,K],P{T}_1=[1,3,5,7],H_1(g>1,P{S}_1,P{T}_1)=0,H_1(g\le 1,P{S}_1,P{T}_1)=H_1(g,\left[K\right],\left[4\right]).$$

For $1\le j\le r$ and $N\in \mathbb{N}$, the following identity holds:

$$\underset{j\mathrm{nested}\mathrm{sums}}{\underbrace{{\sum }_{k_j=1}^N\dots {\sum }_{k_2=1}^{k_3}{\sum }_{k_1=1}^{k_2}}}\nabla \mathrm{SUM}(k_j,PS,[1,2,\dots ,M])=\mathrm{SUM}(N+1,PS,[j,j+1,\dots ,M+j-1]),$$

$$\underset{r\mathrm{nested}\mathrm{sums}}{\underbrace{{\sum }_{k_r=1}^N\dots {\sum }_{k_2=1}^{k_3}{\sum }_{k_1=1}^{k_2}}}\nabla \mathrm{SUM}(k_j,PS,[1,2,\dots ,M])={\nabla }^{j-r}\mathrm{SUM}(N+1,PS,[j,j+1,\dots ,M+j-1]),$$

As an example:

$${\displaystyle \sum _{k_r=1}^N}\dots {\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}\left({\displaystyle \genfrac{}{}{0pt}{}{k_j}{M}}\right)={\displaystyle \frac{1}{M!}}{\nabla }^{j-r}\mathrm{SUM}(N+1,[0,-1,\dots ,-(M-1)],[j,j+1,\dots ,M+j-1]).$$

Since $H_1(g<M)=0$ and $H_1\left(M\right)={\displaystyle \frac{(M+j-1)!}{(j-1)!}}$, we obtain

$${\displaystyle \frac{1}{M!}}{\nabla }^{j-r}{\displaystyle \frac{(M+j-1)!}{(j-1)!}}\left({\displaystyle \genfrac{}{}{0pt}{}{N+j}{M+j}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{M+j-1}{j-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+r}{M+r}}\right).$$

This recovers the main result of [12].

We now derive classical congruences modulo a prime P. $K_i,D_i\in \mathbb{Z}$ with $(P,D_i)=1$, and all congruences are modulo P. We write $D_i^{-1}$ for the modular inverse of $D_i$, so that $D_i{D}_i^{-1}\equiv 1$.

Theorem 7.1. 

For $0\le Y\le T_M-M$:

• If $T_M<P-1$, then

  $$SUM(P-Y)\equiv 0.$$
• If $T_M=P-1$, then

  $$SUM(P-Y)\equiv \prod T_i{D}_i.$$
• If $T_M=P$, then

  $$SUM(P-Y)\equiv H_1(M-1).$$
  For $PT=[1,2,\dots ,P]$ and $P>2$,

  $$SUM\left(P\right)\equiv -(\prod D_i)\sum K_i{D}_i^{-1}.$$

Proof. 

Let $H=T_M-M$, so $0\le H<T_M\le P$ and $0\le Y\le H$. Then

$$SUM(P-Y)=H_1\left(M\right)\left({\displaystyle \genfrac{}{}{0pt}{}{P+H-Y}{T_M+1}}\right)+\dots +H_1\left(0\right)\left({\displaystyle \genfrac{}{}{0pt}{}{P+H-Y}{H+1}}\right),$$

where $2P>P+H-Y\ge P$.

If $T_M<P-1$, then every binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{P+H-Y}{H+1+g}}\right)$ is divisible by P, hence $SUM(P-Y)\equiv 0$.

If $T_M=P-1$, then only the $g=M$ term survives, $\left({\displaystyle \genfrac{}{}{0pt}{}{P+H-Y}{P}}\right)={\displaystyle \frac{(P+1)(P+2)...}{1·2...}}\equiv 1$, $SUM(P-Y)\equiv H_1\left(M\right)\equiv \prod T_i{D}_i$.

If $T_M=P$, then $H_1\left(M\right)=\prod T_i{D}_i\equiv 0$, so the leading term vanishes and $SUM(P-Y)\equiv H_1(M-1)$.

For $PT=[1,2,\dots ,P]$, Theorem 3.4 gives

$$H_1(M-1)\equiv (P-1)!(\prod D_i)(F_1^{\left\{K_i{D}_i^{-1}\right\}}+E_1^{P-1})\equiv -(\prod D_i)F_1^{\left\{K_i{D}_i^{-1}\right\}}\equiv -(\prod D_i)\sum K_i{D}_i^{-1},$$

where we used $E_1^{P-1}={\displaystyle \frac{P(P-1)}{2}}\equiv 0$, $F_1^{\left\{K_i{D}_i^{-1}\right\}}=\sum K_i{D}_i^{-1}$, and Wilson’s theorem $(P-1)!\equiv -1(modP)$. □

Nested summations are allowed here, but generally only $PT=[1,2,\dots ,M]$ is used.

Theorem 7.1 gives a uniform congruence criterion. The key is to sum up to P and then discard all terms containing P as a factor; the number of surviving terms is determined by the number of factors. For instance, for $P>4$, we have

$$1·1·2+2·2·3+\dots +(P-2)·(P-2)·(P-1)\equiv 1·2·2+2·3·3+\dots +(P-2)·(P-1)·(P-1)\equiv 0(modP).$$

Thus the congruence is not proved case by case, but follows from the same uniform mechanism.

Congruences for Stirling numbers

Set $1\le M<P-1$, $2M-1\ge P$, and $P\le N+M\le 2P-2$.

$$\mathrm{SUM}(N,[2,3,\dots ,M],[3,5,\dots ,2M-1])={\displaystyle \sum _{g=0}^{M-1}}{S}_{1,2}(M+g+1,g+1)\left({\displaystyle \genfrac{}{}{0pt}{}{N+M}{M+g+1}}\right).$$

Recall the explicit representation

$$S_{1,2}(M+g+1,g+1)={\displaystyle \frac{(M+g+1)!}{(g+1)!}}\sum _{\begin{array}{c}{i}_1+\dots +i_{g+1}=M+g+1\\ i_j\ge 2\end{array}}{\displaystyle \frac{1}{i_1\dots i_{g+1}}}.$$

In this representation, whenever $M+g+1\ge P$, the largest possible part occurs when $i_1=\dots =i_g=2$, giving $i_{g+1}=M+1-g<P$. Hence all $i_j<P$ in all distributions, so no denominator is divisible by P. Since the numerator $(M+g+1)!$ contains a factor of P, it follows that

$$S_{1,2}(M+g+1,g+1)\equiv 0(modP)\mathrm{whenever}M+g+1\ge P.$$

Consequently, in the sum over g, the only possible nonzero contributions modulo P come from terms with $M+g+1<P$. For those terms, the binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{N+M}{M+g+1}}\right)$ is divisible by P, since $N+M<2P$ and $M+g+1<P$; Hence the entire sum is congruent to $0(modP)$. Thus

$$\mathrm{SUM}\left(N\right)=S_1(N+M,N)=F_M^{N+M-1}\equiv 0(modP).$$

Equivalently, for $P>3$, $1\le M\le P-2$, $0\le Z\le M$, and $2\le K\le P-1$,

$$S_1(P+M-Z,P-Z)\equiv 0,S_1(P,K)\equiv 0,F_g^{P-1}\equiv F_g^P\equiv 0,1\le g\le P-2.$$

Similar analysis applies elsewhere.

Proof of Wilson’s theorem.

By Theorem 3.1(1),

$$\nabla SUM(2,[1,2,\dots ,P-1],[1,2,\dots ,P-1])=H_1\left(1\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{1}}\right)+H_1\left(0\right)\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right).$$

Using Theorem 3.4,

$$\begin{array}{cc}\hfill \nabla SUM\left(2\right)& =1!\left(F_{P-2}^{P-1}{E}_0^1+\dots +F_0^{P-1}{E}_{P-2}^1\right)+(P-1)!\hfill \\ & =\left(F_{P-2}^{P-1}+\dots +F_0^{P-1}\right)+(P-1)!.\hfill \end{array}$$

Since $F_0^{P-1}=1$ and $F_g^{P-1}\equiv 0(modP)$ for $1\le g\le P-2$, we obtain

$$\nabla SUM\left(2\right)\equiv 1+(P-1)!(modP).$$

On the other hand, by the definition of ∇,

$$\nabla SUM\left(2\right)={\displaystyle \frac{P!}{1!}}\equiv 0(modP).$$

Therefore $(P-1)!\equiv -1(modP)$, which is Wilson’s theorem.

Proof of Wolstenholme’s theorem.

For $P>3$, consider

$$\begin{array}{cc}\hfill SUM(2,[2,3,\dots ,P-2],[3,5,\dots ,2P-5\left]\right)& =H_1\left(1\right)\left({\displaystyle \genfrac{}{}{0pt}{}{P}{P}}\right)+H_1\left(0\right)\left({\displaystyle \genfrac{}{}{0pt}{}{P}{P-1}}\right)\hfill \\ & =S_{1,2}(P,2)+P(P-2)!.\hfill \end{array}$$

To evaluate this modulo $P^2$, use the explicit formula for $S_{1,2}$,

$${\displaystyle \frac{S_{1,2}(P,2)}{P}}=(P-1)!\left({\displaystyle \frac{1}{2(P-2)}}+{\displaystyle \frac{1}{3(P-3)}}+\dots +{\displaystyle \frac{1}{\frac{P-1}{2}·\frac{P+1}{2}}}\right).$$

$$\equiv -1\left(-2^{-2}-3^{-2}-\dots -{\left({\displaystyle \frac{P-1}{2}}\right)}^{-2}\right)\equiv (2^2+3^2+\dots +{\left({\displaystyle \frac{P-1}{2}}\right)}^2)(modP).$$

$$(1^2+2^2+3^2+\dots +{(P-1)}^2)\equiv 2(1^2+2^2+3^2+\dots +{\left({\displaystyle \frac{P-1}{2}}\right)}^2)\equiv 0(modP).$$

Thus

$$(1^2+2^2+3^2+\dots +{\left({\displaystyle \frac{P-1}{2}}\right)}^2)\equiv 0(modP).{\displaystyle \frac{S_{1,2}(P,2)}{P}}\equiv -1(modP).$$

Hence $S_{1,2}(P,2)\equiv -P(mod{P}^2)$. Since $SUM\left(2\right)=S_{1,2}(P,2)+P(P-2)!$, and $(P-2)!\equiv 1(modP)$ (by Wilson’s theorem), we conclude

$$SUM\left(2\right)\equiv 0(mod{P}^2).$$

This is Wolstenholme’s theorem [13], equivalently ${\sum }_{i=1}^{P-1}{\displaystyle \frac{(P-1)!}{i}}\equiv 0(mod{P}^2).$

Solving coefficients via matrices

Define an $(M+1)\times (M+1)$ matrix,

$$A_{1,2,3}(P,Q,M)=\left(\begin{array}{ccc}\left({\displaystyle \genfrac{}{}{0pt}{}{P}{Q}}\right)& \dots & \left({\displaystyle \genfrac{}{}{0pt}{}{P}{Q-M}}\right)\\ ⋮& \ddots & ⋮\\ \left({\displaystyle \genfrac{}{}{0pt}{}{P+M}{Q+M}}\right)& \dots & \left({\displaystyle \genfrac{}{}{0pt}{}{P+M}{Q}}\right)\end{array}\right),\left(\begin{array}{ccc}\left({\displaystyle \genfrac{}{}{0pt}{}{P}{Q}}\right)& \dots & \left({\displaystyle \genfrac{}{}{0pt}{}{P+M}{Q}}\right)\\ ⋮& \ddots & ⋮\\ \left({\displaystyle \genfrac{}{}{0pt}{}{P+M}{Q+M}}\right)& \dots & \left({\displaystyle \genfrac{}{}{0pt}{}{P+2M}{Q+M}}\right)\end{array}\right),\left(\begin{array}{ccc}\left({\displaystyle \genfrac{}{}{0pt}{}{P+M}{Q}}\right)& \dots & \left({\displaystyle \genfrac{}{}{0pt}{}{P}{Q-M}}\right)\\ ⋮& \ddots & ⋮\\ \left({\displaystyle \genfrac{}{}{0pt}{}{P+2M}{Q+M}}\right)& \dots & \left({\displaystyle \genfrac{}{}{0pt}{}{P+M}{Q}}\right)\end{array}\right).$$

Here $P=N+T_M-M,Q=N-1$. List $\mathrm{SUM}\left(N\right),\dots ,\mathrm{SUM}(N+M)$ from top to bottom, with g increasing from left to right. They correspond to ${\mathrm{Form}}_{1,2,3}$. It is easy to prove by Gaussian elimination: Up to sign factors arising from column permutations,

$$det{A}_1(P,Q,M)=det{A}_2(P,Q,M)=det{A}_3(P,Q,M),detA(P,Q,M)=detA(P,P-Q,M).$$

$$detA(P,0,M)=1,detA(P,1,M)=\left({\displaystyle \genfrac{}{}{0pt}{}{P+M}{1+M}}\right),detA(P,Q>1,M)={\displaystyle \prod _{g=0}^{Q-1}}{\displaystyle \frac{\left({\displaystyle \genfrac{}{}{0pt}{}{P+M-g}{1+M}}\right)}{\left({\displaystyle \genfrac{}{}{0pt}{}{1+M-g}{1+M}}\right)}}.$$

If $\mathrm{SUM}\left(N\right)$ or $\nabla \mathrm{SUM}\left(N\right)$ is easy to obtain, treat $H_r\left(g\right)$ as unknowns and apply Cramer’s rule:

$$\begin{array}{cc}\hfill H_1\left(g\right)& ={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({\displaystyle \genfrac{}{}{0pt}{}{T_M-M+1+g}{T_M-M+k}}\right)\mathrm{SUM}\left(k\right)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({\displaystyle \genfrac{}{}{0pt}{}{T_M-M+g}{T_M-M+k-1}}\right)\nabla \mathrm{SUM}\left(k\right).\hfill \end{array}$$

(M3)

$$\begin{array}{cc}\hfill z\left(k\right)& ={\displaystyle \sum _{i=1}^k}{(-1)}^{i+k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)\nabla \mathrm{SUM}\left(k\right),H_2\left(g\right)={\displaystyle \sum _{k=g+1}^{M+1}}{(-1)}^{g+k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k-1}{g}}\right)z\left(k\right).\hfill \end{array}$$

(M3)

$$\begin{array}{cc}\hfill H_3\left(g\right)& ={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({\displaystyle \genfrac{}{}{0pt}{}{2+T_M}{g+1-k}}\right)\mathrm{SUM}\left(k\right)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({\displaystyle \genfrac{}{}{0pt}{}{1+T_M}{g+1-k}}\right)\nabla \mathrm{SUM}\left(k\right).\hfill \end{array}$$

(M3)

$\nabla \mathrm{SUM}(N,[1,1,\dots ,1],[2,3,\dots ,M])=N^M$, from which the following classical formulas may be recovered.

(M1) $\to S_2(M,g)={\displaystyle \frac{1}{g!}}{\sum }_{k=0}^g{(-1)}^{g+k}\left({\displaystyle \genfrac{}{}{0pt}{}{g}{k}}\right)k^M={\displaystyle \frac{1}{g!}}{\sum }_{k=0}^g{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{g}{k}}\right){(g-k)}^M$.

(M3) $\to 〈{\displaystyle \genfrac{}{}{0pt}{}{M}{g}}〉={\sum }_{k=1}^{g+1}{(-1)}^{1+g+k}\left({\displaystyle \genfrac{}{}{0pt}{}{M+1}{g+1-k}}\right)k^M={\sum }_{k=0}^g{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{M+1}{k}}\right){(g+1-k)}^M$.

For $T\in \mathbb{N}$, we obtain the fully matching generalized formulas, $E_1(M+1,g+1)=S_2(M+1,g+1),{〈\begin{array}{c}g\\ M\end{array}〉}_1=〈\begin{array}{c}g\\ M\end{array}〉$.

$$E_T(M+1,g+1)={\displaystyle \frac{1}{g!}}{\displaystyle \sum _{k=0}^g}{(-1)}^{g+k}\left({\displaystyle \genfrac{}{}{0pt}{}{g}{k}}\right){(T+k)}^M,{〈\begin{array}{c}g\\ M\end{array}〉}_T={\displaystyle \frac{1}{(T-1)!}}{\displaystyle \sum _{k=0}^g}{(-1)}^{g+k}\left({\displaystyle \genfrac{}{}{0pt}{}{T+M}{g-k}}\right){(T+k)}^{\underline{T}}{(k+1)}^{M-1}.$$

Where ${〈\begin{array}{c}g\\ M\end{array}〉}_T$ obtains from ${\displaystyle \frac{1}{(T-1)!}}·H_3(g,[1,2,\dots ,T,1,1,\dots ,1],[1,2,\dots ,T+M-1])$.