Formal Calculation of Q-Binomial

Peng Ji

doi:10.20944/preprints202505.0208.v1

Submitted:

04 May 2025

Posted:

06 May 2025

Read the latest preprint version here

Abstract

This article offers formulas for computing various q-binomial nested sums, give three forms of results. More importantly, it reveals the three forms of q-binomial and their interrelationships. It is a powerful tool for q-analysis, which can prove and generalize many classic conclusions in a simple way. This article also utilized it to obtain a large number of new results, including formulas for q-Eulerian numbers and polynomials. By taking the limit of q to 1, it can calculate general nested sums and analyze binomial coefficients.

Keywords:

Formal Calculation

;

q-nested sum

;

q-binomial

;

q-analysis

;

q-calculus

Subject:

Computer Science and Mathematics - Discrete Mathematics and Combinatorics

MSC: 05A30

1. Calculation Formula

q-binomial:

{[\begin{matrix} M \\ N \end{matrix}]}_{q} = \frac{(q^{N} - 1) (q^{N - 1} - 1) . . . (q^{N - M + 1} - 1)}{(q^{M} - 1) (q^{M - 1} - 1) . . . (q^{1} - 1)}, q \neq 0, 1

, abbreviated as

G_{M}^{N}, {[N]}_{q} = G_{1}^{N}

.

{(a; q)}_{n} = \{؜_{1, n = 0}^{(1 - a) (1 - a q) . . . (1 - a q^{n - 1}), n > 0}

.

G_{g_{1}, g_{2} . . . g_{p}}^{M} = \frac{{(q; q)}_{M}}{\prod_{i = 1}^{p} {(q; q)}_{g_{p}}}, \sum g_{i} = M

. The following relationship is established:

\begin{matrix} G_{0}^{N} = 1, G_{M}^{N} = G_{N - M}^{N}, G_{M}^{N} = 0, M > N, M < 0 . \end{matrix}

(1)

\begin{matrix} 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} . \end{matrix}

(2)

\begin{matrix} \sum_{n = 0}^{N - 1} q^{n} G_{M}^{n + K} = q^{M - K} G_{M + 1}^{N + K} . K \leq M . \end{matrix}

(3)

\begin{matrix} G_{K}^{M} = \sum_{w \in Ω (0^{M - K}, 1^{K})} q^{i n v (w)}, inv (.) denotes the inversion statistic . [1] \end{matrix}

(4)

Lemma 1.1.

\begin{matrix} \sum_{n = 0}^{N - 1} q^{n} {[n]}_{q} G_{M}^{n + K}, M > 0, M \geq 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} (1) \\ = q^{M - 2 K - 1} G_{1}^{M + 1} G_{M + 2}^{N + K + 1} + q^{M - K} (G_{1}^{M - K} - q^{- K - 1} G_{1}^{M + 1}) G_{M + 1}^{N + K} (2) \\ = (q^{2 (M - K) + 1} G_{1}^{M + 1} - q^{2 M - K + 2} G_{1}^{M - K}) G_{M + 2}^{N + K} + q^{M - K} G_{1}^{M - K} G_{M + 2}^{N + K + 1} (3) . \end{matrix}

Proof.

\begin{matrix} \sum q^{n} {[n]}_{q} G_{M}^{n + K} = \sum q^{n} \frac{q^{n} - q^{n - M + K}}{q - 1} G_{M}^{n + K} + \sum q^{n} \frac{q^{n - M + K} - 1}{q - 1} G_{M}^{n + K} \\ = \sum q^{n} (q^{n - M + K} - 1) \frac{q^{M - K} - 1}{q - 1} G_{M}^{n + K} + \sum q^{n} \frac{q^{M - K} - 1}{q - 1} G_{M}^{n + K} + \sum q^{n} \frac{q^{n - M + K} - 1}{q - 1} G_{M}^{n + K} \\ = \sum q^{n} (q^{M + 1} - 1) G_{1}^{M - K} G_{M + 1}^{n + K} + \sum q^{n} G_{1}^{M - K} G_{M}^{n + K} + \sum q^{n} G_{1}^{M + 1} G_{M + 1}^{n + K} \\ = (q^{M - K} - 1) G_{1}^{M + 1} G_{M + 2}^{N + K} q^{M + 1 - K} + G_{1}^{M - K} G_{M + 1}^{N + K} q^{M - K} + G_{1}^{M + 1} G_{M + 2}^{N + K} q^{M + 1 - 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} (1) \\ = q^{2 (M - K) + 1} G_{1}^{M + 1} q^{- M - 2} (G_{M + 2}^{N + K + 1} - G_{M + 1}^{N + K}) + q^{M - K} G_{1}^{M - K} G_{M + 1}^{N + K} \\ = q^{M - 2 K - 1} G_{1}^{M + 1} G_{M + 2}^{N + K + 1} + q^{M - K} (G_{1}^{M - K} - q^{- K - 1} G_{1}^{M + 1}) G_{M + 1}^{N + K} (2) \\ = q^{2 (M - K) + 1} G_{1}^{M + 1} G_{M + 2}^{N + K} + q^{M - K} G_{1}^{M - K} (G_{M + 2}^{N + K + 1} - q^{M + 2} G_{M + 2}^{N + K}) \\ = (q^{2 (M - K) + 1} G_{1}^{M + 1} - q^{2 M - K + 2} G_{1}^{M - K}) G_{M + 2}^{N + K} + q^{M - K} G_{1}^{M - K} G_{M + 2}^{N + K + 1} (3) . \end{matrix}

□

Definition 1.1.

Recursively define

\nabla_{q}^{p}

,

p \in Z

;

S U M_{q} (N) = S U M_{q} (N, P S, P T)

,

K_{i}, D_{i} \in C

,

T_{i} \in N

.

\begin{matrix} \nabla_{q}^{0} f (n) = f (n), \sum_{n = 0}^{N - 1} q^{n} \nabla_{q}^{1} f (n + 1) = f (N), \sum_{n = 0}^{N - 1} q^{n} f (n + 1) = \nabla_{q}^{- 1} f (N), \nabla_{q}^{1} = \nabla_{q} . \\ S U M_{q} (N, [K_{1} : D_{1}], [T_{1} = 1]) = \sum_{n = 0}^{N - 1} q^{n} (K_{1} + {[n]}_{q} D_{1}) . \\ S U M_{q} (N, [K_{1} : D_{1}, K_{2} : D_{2}], [T_{1}, T_{2} = T_{1} + 2 - p]) = \sum_{n = 0}^{N - 1} q^{n} (K_{2} + {[n]}_{q} D_{2}) \nabla^{p} S U M_{q} (n + 1, [K_{1} : D_{1}], [T_{1}]) . \\ S U M_{q} (N, P S, [1, 2 . . . M]) = \sum_{n = 0}^{N - 1} \prod_{i = 1}^{M} q^{n} (K_{i} + {[n]}_{q} D_{i}) . \\ S U M_{q} (N, P S, [1, 3 . . . 2 M - 1]) = \sum_{n_{M} = 0}^{N - 1} q^{n_{M}} (K_{M} + {[n_{M}]}_{q} D_{M}) . . . \sum_{n_{2} = 0}^{n_{3}} q^{n_{2}} (K_{2} + {[n_{2}]}_{q} D_{2}) \sum_{n_{1} = 0}^{n_{2}} q^{n_{1}} (K_{1} + {[n_{1}]}_{q} D_{1}) . \\ S U M_{q} (N, P S, [1, 2, 4]) = \sum_{n_{3} = 0}^{N - 1} q^{n_{3}} (K_{3} + {[n_{3}]}_{q} D_{3}) \sum_{n = 0}^{n_{3}} q^{n} (K_{1} + {[n]}_{q} D_{1}) (K_{2} + {[n]}_{q} D_{2}) . \\ S U M_{q} (N, P S, [1, 3, 4]) = \sum_{n_{3} = 0}^{N - 1} q^{n_{3}} (K_{3} + {[n_{3}]}_{q} D_{3}) (K_{2} + {[n_{3}]}_{q} D_{2}) \sum_{n = 0}^{n_{3}} q^{n} (K_{1} + {[n]}_{q} D_{1}) . \\ S U M_{q} (N, P S, [1, 4]) = \sum_{n_{3} = 0}^{N - 1} q^{n_{3}} (K_{2} + {[n_{3}]}_{q} D_{2}) \sum_{n_{2} = 0}^{n_{3}} q^{n_{2}} \sum_{n_{1} = 0}^{n_{2}} q^{n_{1}} (K_{1} + {[n_{1}]}_{q} D_{1}) . \\ Abbreviations : [K_{1} : D, K_{2} : D . . . K_{M} : D] = [K_{1}, K_{2} . . . K_{M}] : D, [K_{1}, K_{2} . . . K_{M}] : 1 = [K_{1}, K_{2} . . . K_{M}] . \\ In this paper, the default P S = [K_{1} : D_{1}, K_{2} : D_{2} . . . K_{M} : D_{M}], P T = [T_{1}, T_{2} . . . T_{M}], T_{i} < T_{i + 1} . \\ Use K, T to represent the set {K_{i}}, {T_{i}} . (K_{1} + T_{1}) (K_{2} + T_{2}) . . . (K_{M} + T_{M}) = \sum \prod_{i = 1}^{M} X_{i}, X_{i} = T_{i} o r K_{i} \end{matrix}

Definition 1.2.

X (T)

=Number of

{X_{1}, X_{2} . . . X_{M}} \in T

.

X_{T - 1}

=Number of

{X_{1}, X_{2} . . . X_{i - 1}} \in T

,

X_{K - 1}

=Number of

{X_{1}, X_{2} . . . X_{i - 1}} \in K

.

X_{T}

=Number of

{X_{1}, X_{2} . . . X_{i}} \in T

,

X_{K}

=Number of

{X_{1}, X_{2} . . . X_{i}} \in K

.

Obviously,

X_{T - 1} + X_{K - 1} = i - 1

,

X_{T} + X_{K} = i

. Use the auxiliary form and each

X_{i}

cannot be exchanged:

Theorem 1.1.

H = T_{M} - M, S U M_{q} (N, P S, P T) =

\begin{matrix} F o r m_{1} \to \sum_{g = 0}^{M} H_{1}^{q} (g) G_{H + 1 + g}^{N + H}, B_{i} = \{؜_{q^{(T_{i} - T_{i - 1} - 1) X_{T - 1}} (K_{i} + G_{1}^{X_{T - 1}} D_{i}), X_{i} = K_{i}}^{q^{1 + (T_{i} - T_{i - 1}) X_{T - 1}} G_{1}^{T_{i} - X_{K - 1}} D_{i}, X_{i} = T_{i}}, \\ F o r m_{2} \to \sum_{g = 0}^{M} H_{2}^{q} (g) G_{H + 1 + g}^{N + H + g}, B_{i} = \{؜_{K_{i} - q^{- (T_{i} - X_{K - 1})} G_{1}^{T_{i} - X_{K - 1}} D_{i}, X_{i} = K_{i}}^{q^{- (T_{i} - X_{K - 1})} G_{1}^{T_{i} - X_{K - 1}} D_{i}, X_{i} = T_{i}} \\ F o r m_{3} \to \sum_{g = 0}^{M} H_{3}^{q} (g) G_{T_{M} + 1}^{N + T_{M} - g}, B_{i} = \{؜_{q^{(T_{i} - T_{i - 1} - 1) X_{T - 1}} (K_{i} + G_{1}^{X_{T - 1}} D_{i}), X_{i} = K_{i}}^{q^{1 + (T_{i} - T_{i - 1} - 1) X_{T - 1}} {(q^{X_{T - 1}} G_{1}^{T_{i}} - q^{T_{i}} G_{1}^{X_{T - 1}}) D_{i} - K_{i} q^{T_{i}}}, X_{i} = T_{i}} . \\ H_{i}^{q} (g) = H_{i}^{q} (g, P S, P T) = H_{i}^{q} (g, M), is defined as \sum_{X (T) = g} \prod_{i = 1}^{M} B_{i} . \end{matrix}

Proof.

\begin{matrix} S U M_{q} (1, [K_{1} : D_{1}], [1]) = \sum_{n = 0}^{N - 1} q^{n} (K_{1} + {[n]}_{q} D_{1}) = \sum_{n = 0}^{N - 1} q^{n} G_{1}^{n} D_{1} + \sum_{n = 0}^{N - 1} q^{n} K_{1} \\ = 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} . \\ It ’ s holds when M = 1, suppose that holds when M . \\ P S 1 = [P S, K_{M + 1} : D_{M + 1}], P T 1 = [P T, T_{M + 1} = T_{M} + 2 - p], X = 1 + T_{M} - M - p . \\ If f (n + 1) = \sum A_{g} G_{H + 1 + g}^{n + H + 1 + g}, then \nabla_{q}^{p} f (n + 1) = \sum A_{g} G_{H + 1 + g - p}^{n + H + 1 + g - p} . \\ S U M_{q} (N, P S 1, P T 1) \\ = \sum_{n = 0}^{N - 1} q^{n} (K_{M + 1} + {[n]}_{q} D_{M + 1}) \nabla_{q}^{p} S U M_{q} (n + 1) = \sum_{n = 0}^{N - 1} q^{n} (K_{M + 1} + {[n]}_{q} D_{M + 1}) \sum_{g = 0}^{M} H_{2}^{q} (g) G_{X + g}^{n + X + g} \\ = \sum_{g = 0}^{M} (K_{M + 1} - q^{- X - g - 1} G_{1}^{X + g + 1} D_{M + 1}) H_{2}^{q} (g) G_{X + 1 + g}^{N + X + g} + \sum_{g = 0}^{M} q^{- X - g - 1} G_{1}^{X + g + 1} D_{M + 1} H_{2}^{q} (g) G_{X + 2 + g}^{N + X + g + 1} \\ = \sum_{g = 0}^{M} (K_{M + 1} - q^{- (T_{M + 1} - (M - g))} G_{1}^{T_{M + 1} - (M - g)} D_{M + 1}) H_{2}^{q} (g) G_{T_{M + 1} - (M + 1) + g + 1}^{N + T_{M + 1} - (M + 1) + g} \\ + \sum_{g = 0}^{M} q^{- (T_{M + 1} - (M - g))} G_{1}^{T_{M + 1} - (M - g)} D_{M + 1} H_{2}^{q} (g) G_{T_{M + 1} - (M + 1) + g + 2}^{N + T_{M + 1} - (M + 1) + g + 1} \\ = \sum_{g = 0}^{M + 1} H_{2}^{q} (g, P S 1, P T 1) G_{T_{M + 1} - (M + 1) + 1 + g}^{N + T_{M + 1} - (M + 1) + g} . Proof of F o r m_{2} completion . \\ If f (n + 1) = \sum A_{g} G_{H + 1 + g}^{n + H + 1}, then \nabla_{q}^{p} f (n + 1) = \sum A_{g} G_{H + 1 + g - p}^{n + H + 1 - p} q^{- p g} . \\ S U M_{q} (N, P S 1, P T 1) \\ = \sum_{n = 0}^{N - 1} q^{n} (K_{M + 1} + {[n]}_{q} D_{M + 1}) \sum_{g = 0}^{M} H_{1}^{q} (g) G_{X + g}^{n + X} q^{- p g} \\ = \sum_{g = 0}^{M} (K_{M + 1} q^{g} + q^{g} G_{1}^{g} D_{M + 1}) H_{1}^{q} (g) G_{X + 1 + g}^{N + X} q^{- p g} + \sum_{g = 0}^{M} q^{2 g + 1} G_{1}^{X + g + 1} D_{M + 1} H_{1}^{q} (g) G_{X + 2 + g}^{N + X} q^{- p g} \\ = \sum_{g = 0}^{M} (K_{M + 1} + G_{1}^{g} D_{M + 1}) H_{1}^{q} (g) G_{T_{M + 1} - (M + 1) + 1 + g}^{N + T_{M + 1} - (M + 1)} q^{(1 - p) g} + \sum_{g = 0}^{M} q^{(2 - p) g + 1} G_{1}^{T_{M + 1} - (M - g)} D_{M + 1} H_{1}^{q} (g) G_{X + 2 + g}^{N + X} \\ = \sum_{g = 0}^{M + 1} H_{1}^{q} (g, P S 1, P T 1) G_{T_{M + 1} - (M + 1) + 1 + g}^{N + T_{M + 1} - (M + 1)} . Proof of F o r m_{1} completion . \\ If f (n + 1) = \sum A_{g} G_{T_{M} + 1}^{n + 1 + T_{M} - g}, then \nabla_{q}^{p} f (n + 1) = \sum A_{g} G_{T_{M} + 1 - p}^{n + 1 + T_{M} - g - p} q^{- p g} . \\ S U M_{q} (N, P S 1, P T 1) \\ = \sum_{n = 0}^{N - 1} q^{n} (K_{M + 1} + {[n]}_{q} D_{M + 1}) \sum_{g = 0}^{M} H_{3}^{q} (g) G_{T_{M} + 1 - p}^{n + 1 + T_{M} - p - g} q^{- p g} \\ = \sum_{g = 0}^{M} K_{M + 1} q^{g} H_{3}^{q} (g) G_{T_{M} + 2 - p}^{N + 1 + T_{M} - p - g} q^{- p g} + \sum_{g = 0}^{M} q^{g} G_{1}^{g} D_{M + 1} H_{3}^{q} (g) G_{T_{M} + 3 - p}^{N + T_{M} - p + 2 - g} q^{- p g} \\ + \sum_{g = 0}^{M} (q^{2 g + 1} G_{1}^{T_{M} + 2 - p} - q^{T_{M} + 3 - p + g} G_{1}^{g}) D_{M + 1} H_{3}^{q} (g) G_{T_{M} + 3 - p}^{N + T_{M} - p + 1 - g} q^{- p g} \\ = \sum_{g = 0}^{M} (K_{M + 1} q^{g} + q^{g} G_{1}^{g} D_{M + 1}) H_{3}^{q} (g) G_{T_{M} + 3 - p}^{N + T_{M} - p + 2 - g} q^{- p g} \\ + \sum_{g = 0}^{M} ((q^{2 g + 1} G_{1}^{T_{M} + 2 - p} - q^{T_{M} + 3 - p + g} G_{1}^{g}) D_{M + 1} - q^{T_{M} + 3 - p} K_{M + 1} q^{g}) H_{3}^{q} (g) G_{T_{M} + 3 - p}^{N + T_{M} - p + 1 - g} q^{- p g} \\ = \sum_{g = 0}^{M} (K_{M + 1} + G_{1}^{g} D_{M + 1}) H_{3}^{q} (g) G_{T_{M + 1} + 1}^{N + T_{M + 1} - g} q^{(1 - p) g} \\ + \sum_{g = 0}^{M} ((q^{g} G_{1}^{T_{M + 1}} - q^{T_{M + 1}} G_{1}^{g}) D_{M + 1} - q^{T_{M + 1}} K_{M + 1}) H_{3}^{q} (g) G_{T_{M + 1} + 1}^{N + T_{M + 1} - (g - 1)} q^{1 + (1 - p) g} \\ = \sum_{g = 0}^{M + 1} H_{3}^{q} (g, P S 1, P T 1) G_{T_{M + 1} + 1}^{N + T_{M + 1} - g} . Proof of F o r m_{3} completion . \end{matrix}

□

\sum_{n = 0}^{N - 1} q^{n} \prod_{i = 1}^{M} (K_{i} + D_{i} q^{n}) = S U M (N, [K_{1} + D_{1} : D_{1} (q - 1) . . . K_{M} + D_{M} : D_{M} (q - 1)], [1, 2 . . . M])

= \sum_{g = 0}^{M} H_{1}^{q} (g) G_{1 + g}^{N} .

B_{i} = \{؜_{K_{i} + D_{i} + G_{1}^{X_{T - 1}} D_{i} (q - 1) = K_{i} + D_{i} + G_{1}^{X_{T}} D_{i} (q - 1) = K_{i} + q^{X_{T}} D_{i}, X_{i} = K_{i}}^{q^{1 + X_{T - 1}} G_{1}^{i - X_{K - 1}} D_{i} (q - 1) = q^{X_{T}} G_{1}^{1 + X_{T - 1}} D_{i} (q - 1) = q^{X_{T}} G_{1}^{X_{T}} D_{i} (q - 1) = q^{X_{T}} (q^{X_{T}} - 1) D_{i}, X_{i} = T_{i}}

.

Following a similar form, we are able to prove inductively:

Theorem 1.2.

\sum_{n = 0}^{N - 1} \prod_{i = 1}^{M} (K_{i} + D_{i} q^{n}) = \sum_{g = 1}^{M} f (g) G_{g}^{N} + N \prod K_{i}, f (g) = \sum \prod B_{i}

,

B_{i} = \{؜_{K_{i}, X_{T - 1} = 0; K_{i} + q^{X_{T - 1} - 1} D_{i}, X_{T - 1} > 0, X_{i} = K_{i}}^{D_{i}, X_{T - 1} = 0; q^{X_{T - 1}} (q^{X_{T - 1}} - 1) D_{i}, X_{T - 1} > 0, X_{i} = T_{i}}

.

Definition 1.3.

{lim}_{q \to 1} H^{q} (g) = H (g), S U M_{q} (N) = S U M (N)

.

{lim}_{q \to 1} {[n]}_{q} = n, G_{M}^{N} = (؜_{M}^{N}) . which yields the nested summation formula for K_{i} + n D_{i}

.

2. Property

Definition 2.1.

{[n]}_{q -} = q^{- n} G_{1}^{n}, {[n!]}_{q -} = {[n]}_{q -} . . . {[2]}_{q -} {[1]}_{q -}, {[0!]}_{q -} = 1, {[n]}_{q +} = q^{n} G_{1}^{n},

similarly defining

{[n!]}_{q +}

.

Theorem 2.1.

(1).

\nabla_{q} S U M_{q} (n + 1, P S, [1, 2 . . . M]) = \prod_{i = 1}^{M} (K_{i} + {[n]}_{q} D_{i})

.

(2). At

S U M_{q} (N, [. . . P S . . .], [. . . T, T + 1 . . . T + M - 1 . . .])

,

K_{i} : D_{i}

can exchange orders.

(3).

\nabla_{q}^{p} S U M_{q} (N) = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{X + 1 + g}^{N + X} q^{- g p} = \sum_{g = 0}^{M} H_{2}^{q} (g) G_{X + 1 + g}^{N + X + g} = \sum_{g = 0}^{M} H_{3}^{q} (g) G_{X + M + 1}^{N + X + M - g} q^{- g p}

,

X = T_{M} - M - p

.

(4).

S U M_{q} (N, [{[L_{1}]}_{q -}, {[L_{2}]}_{q -} . . . {[L_{Q}]}_{q -}, P S], [L_{1}, L_{2} . . . L_{Q}, P T]) = \prod_{i = 1}^{Q} {[L_{i}]}_{q -} S U M_{q} (N)

. So

T_{1}

can great than 1,

T_{i} \in N

.

(5).

S U M_{q} (N, [{[T_{1}]}_{q -}, {[T_{2}]}_{q -} . . . {[T_{M}]}_{q -}], [T_{1}, T_{2} . . . T_{M}]) = \prod_{i = 1}^{M} {[T_{i}]}_{q -} G_{T_{M} + 1}^{N + T_{M}}

.

(6) At

H^{q} (g)

,

\sum_{X_{i} \in K} \prod q^{X_{T}} = G_{M - g}^{M} = G_{g}^{M}

.

Proof. (1) and (2) is derived from the definition of

S U M_{q}

.

(3) is derived from

\sum_{n = 0}^{N - 1} q^{n} G_{M}^{n + K} = q^{M - K} G_{M + 1}^{N + K}

, which has already used in the proof of [1.1].

P S 1 = [{[L_{1}]}_{q -}, {[L_{2}]}_{q -} . . . {[L_{Q}]}_{q -}, P S], P T 1 = [L_{1}, L_{2} . . . L_{Q}, P T])

,

H_{2}^{q} (g < Q, P S 1, P T 1) = 0

,

H_{2}^{q} (g \geq Q, P S 1, P T 1) = \prod_{i = 1}^{Q} {[L_{i}]}_{q -} H_{2}^{q} (g - Q)

, which proves (4) and (5).

(6) is actually

G_{K}^{M} = \sum_{w \in Ω (0^{M - K}, 1^{K})} q^{i n v (w)}

. □

\begin{matrix} {[x]}_{q -} + {[n]}_{q} = q^{- x} \frac{q^{x} - 1}{q - 1} + \frac{q^{n} - 1}{q - 1} = \frac{q^{n + x} - 1}{q^{x} (q - 1)} = q^{- x} {[n + x]}_{q} . From (5) : \\ P T = [1, 2 . . . M], P S = [{[T_{i}]}_{q -}] \to \sum_{n = 0}^{N - 1} q^{n} {[n + 1]}_{q} {[n + 2]}_{q} . . . {[n + M]}_{q} = {[1]}_{q} {[2]}_{q} . . . {[M]}_{q} G_{M + 1}^{N + M} . \\ P T = [1, 3 . . . 2 M - 1], P S = [{[T_{i}]}_{q -}] \to \sum_{n_{M} = 0}^{N - 1} . . . \sum_{n_{1} = 0}^{n_{2}} q^{\sum_{i = 1}^{M} n_{i}} \prod_{i = 1}^{M} {[n_{i} + 2 i - 1]}_{q} = {[1]}_{q} {[3]}_{q} . . . {[2 M - 1]}_{q} G_{2 M}^{N + 2 M - 1} . \\ P S = [0, - {[1]}_{q}, - {[2]}_{q} . . . - {[M - 1]}_{q}] \to H_{1}^{q} (g < M) = H_{3}^{q} (g < M) = 0, \\ H_{1}^{q} (M) = H_{3}^{q} (M) = q^{M + (T_{2} - T_{1}) + 2 (T_{3} - T_{2}) + . . . + (M - 1) (T_{M} - T_{M - 1})} \prod G_{1}^{T_{i}} = q^{M + (M - 1) T_{M} - \sum_{i = 1}^{M - 1} T_{i}} \prod G_{1}^{T_{i}} . \\ P T = [1, 2 . . . M] \to \sum_{n = 0}^{N - 1} q^{n} {[n]}_{q} {[n - 1]}_{q} . . . {[n - M + 1]}_{q} = q^{M} {[1]}_{q} {[2]}_{q} . . . {[M]}_{q} G_{M + 1}^{N} . \\ P T = [1, 3 . . . 2 M - 1] \to q^{0 + 1 + 2 + . . . + (M - 1)} \sum_{n_{M} = 0}^{N - 1} . . . \sum_{n_{1} = 0}^{n_{2}} q^{\sum_{i = 1}^{M} n_{i}} \prod_{i = 1}^{M} {[n_{i} + 1 - i]}_{q} \\ = q^{M + (M - 1) (2 M - 1) - {(M - 1)}^{2}} {[1]}_{q} {[3]}_{q} . . . {[2 M - 1]}_{q} G_{2 M}^{N + M - 1} \to \\ \sum_{n_{M} = 0}^{N - 1} . . . \sum_{n_{1} = 0}^{n_{2}} q^{\sum_{i = 1}^{M} n_{i}} \prod_{i = 1}^{M} {[n_{i} + 1 - i]}_{q} = q^{\frac{M (M + 1)}{2}} {[1]}_{q} {[3]}_{q} . . . {[2 M - 1]}_{q} G_{2 M}^{N + M - 1} . \end{matrix}

Theorem 2.2.

P T = [1, 2 . . . M]

,

(1).

H_{1}^{q} (g) = q^{g (g + 1)} \sum_{k = g}^{M} H_{2}^{q} (k) G_{g}^{k}

.

(2).

H_{1}^{q} (g) = \sum_{k = 0}^{g} H_{3}^{q} (k) G_{M - g}^{M - k} q^{(g + 1) (g - k)}

.

(3).

H_{2}^{q} (g) = \sum_{k = g}^{M} {(- 1)}^{k + g} G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} H_{1}^{q} (k) = \sum_{k = g}^{M} {(- 1)}^{k + g} G_{g}^{k} q^{\frac{g (g + 1) - k (k + 3)}{2} - k g} H_{1}^{q} (k)

.

(4).

H_{3}^{q} (g) = \sum_{k = 0}^{g} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} H_{1}^{q} (k) = \sum_{k = 0}^{g} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{\frac{g (g + 3) - k (k + 3)}{2}} H_{1}^{q} (k)

.

Proof.

\begin{matrix} Direct verification when M = 1, assuming M holds . P S 1 = [P S, K_{M + 1} : D_{M + 1}], P T 1 = [P T, T_{M + 1}] . \\ H_{1}^{q} (g, M + 1) = H_{1}^{q} (g - 1) q^{1 + X_{T}} G_{1}^{1 + X_{T}} D_{M + 1} + H_{1}^{q} (g) (K_{M + 1} + G_{1}^{X_{T}} D_{M + 1}) \\ = q^{1 + X_{T}} G_{1}^{1 + X_{T}} D_{M + 1} q^{g (g - 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) G_{g - 1}^{k} + (K_{M + 1} + G_{1}^{X_{T}} D_{M + 1}) q^{g (g + 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) G_{g}^{k} \\ = q^{g} G_{1}^{g} D_{M + 1} q^{g (g - 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) (G_{g}^{k + 1} - q^{g} G_{g}^{k}) + (K_{M + 1} + G_{1}^{g} D_{M + 1}) q^{g (g + 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) G_{g}^{k} \\ = q^{- g} G_{1}^{g} D_{M + 1} q^{g (g + 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) G_{g}^{k + 1} + K_{M + 1} q^{g (g + 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) G_{g}^{k} . (1 *) \\ H_{2}^{q} (g, M + 1) = H_{2}^{q} (g - 1) q^{- g} G_{1}^{g} D_{M + 1} + H_{2}^{q} (g) (K_{M + 1} - q^{- (g + 1)} G_{1}^{g + 1} D_{M + 1}) . \\ q^{g (g + 1)} \sum_{k = 0}^{M + 1} H_{2}^{q} (k, M + 1) G_{g}^{k} = \\ = q^{g (g + 1)} \sum_{k = 0}^{M + 1} (H_{2}^{q} (k - 1) q^{- k} G_{1}^{k} D_{M + 1} + H_{2}^{q} (k) (K_{M + 1} - q^{- (k + 1)} G_{1}^{k + 1} D_{M + 1})) G_{g}^{k} \\ = q^{g (g + 1)} \sum_{k = 1}^{M + 1} (H_{2}^{q} (k - 1) q^{- k} G_{1}^{k} D_{M + 1} G_{g}^{k} + q^{g (g + 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) (K_{M + 1} - q^{- (k + 1)} G_{1}^{k + 1} D_{M + 1})) G_{g}^{k} \\ = q^{g (g + 1)} \sum_{k = 0}^{M} (H_{2}^{q} (k) q^{- (1 + k)} G_{1}^{1 + k} D_{M + 1} G_{g}^{k + 1} + q^{g (g + 1)} \sum_{k = 0}^{M} H_{2}^{q} (k) (K_{M + 1} - q^{- (k + 1)} G_{1}^{k + 1} D_{M + 1})) G_{g}^{k} . (1 * *) \\ ((1 *) - (1 * *)) / (q^{g (g + 1)} D_{M + 1}) = \\ \sum_{k = 0}^{M} H_{2}^{q} (k) (q^{- g} G_{1}^{g} G_{g}^{k + 1} - q^{- (1 + k)} G_{1}^{1 + k} G_{g}^{k + 1} + q^{- (k + 1)} G_{1}^{k + 1} G_{g}^{k}) = \sum_{k = 0}^{M} H_{2}^{q} (k) (. . .) . \\ (. . .) = q^{- g} G_{1}^{g} G_{g}^{k + 1} - q^{k + 1 - g} q^{- (1 + k)} G_{1}^{1 + k} G_{g - 1}^{k} = 0 . Proof of (1) is complete . \\ H_{1}^{q} (g, M + 1) = \\ q^{g} G_{1}^{g} D_{M + 1} \sum_{k = 0}^{M} H_{3}^{q} (k) G_{M - g + 1}^{M - k} q^{g (g - 1 - k)} + (K_{M + 1} + G_{1}^{g} D_{M + 1}) \sum_{k = 0}^{M} H_{3}^{q} (k) G_{M - g}^{M - k} q^{(g + 1) (g - k)} . (2 *) \\ \sum_{k = 0}^{M + 1} H_{3}^{q} (k, M + 1) G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k)} \\ = \sum_{k = 0}^{M} q ((q^{k} G_{1}^{M + 1} - q^{M + 1} G_{1}^{k}) D_{M + 1} - K_{M + 1} q^{M + 1}) H_{3}^{q} (k) G_{M - g + 1}^{M - k} q^{(g + 1) (g - k - 1)} \\ + \sum_{k = 0}^{M} (K_{M + 1} + G_{1}^{k} D_{M + 1}) H_{3}^{q} (k) G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k)} . (2 * *) \\ Items containing K_{M + 1} : \\ K_{M + 1} G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k)} - K_{M + 1} q^{M + 2} G_{M + 1 - g}^{M - k} q^{(g + 1) (g - k - 1)} = K_{M + 1} G_{M - g}^{M - k} q^{(g + 1) (g - k)} . \\ Items does not contain K_{M + 1} in (2 *) : \\ = q^{g} G_{1}^{g} D_{M + 1} G_{M - g + 1}^{M - k} q^{g (g - 1 - k)} + G_{1}^{g} D_{M + 1} G_{M - g}^{M - k} q^{(g + 1) (g - k)} . \\ Divide by D_{M + 1} q^{g} q^{g (g - 1 - k)} {(q - 1)}^{- 1} \\ = (q^{g} - 1) G_{M + 1 - g}^{M - k} + (q^{g} - 1) G_{M - g}^{M - k} q^{g - k} = (q^{g} - 1) G_{M + 1 - g}^{M + 1 - k} . \\ Items does not contain K_{M + 1} in (2 * *) : \\ = \frac{q^{M + 2} - q^{k + 1}}{q - 1} D_{M + 1} G_{M - g + 1}^{M - k} q^{(g + 1) (g - k - 1)} + G_{1}^{k} D_{M + 1} G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k)} . \\ Divide by D_{M + 1} q^{g} q^{g (g - 1 - k)} {(q - 1)}^{- 1} \\ = (q^{M + 1 - k} - 1) G_{M + 1 - g}^{M - k} + (q^{g} - q^{g - k}) G_{M + 1 - g}^{M + 1 - k} = (q^{g} - 1) G_{M + 1 - g}^{M + 1 - k} . \\ (2 *) = (2 * *), proof of (2) is complete . \\ H_{2}^{q} (g, M + 1) = H_{2}^{q} (g - 1) q^{- g} G_{1}^{g} D_{M + 1} + H_{2}^{q} (g) (K_{M + 1} - q^{- (g + 1)} G_{1}^{g + 1} D_{M + 1}) . \\ = q^{- g} G_{1}^{g} D_{M + 1} \sum_{k = 0}^{M} {(- 1)}^{k + g - 1} H_{1}^{q} (k) G_{g - 1}^{k} q^{- k (k + 1) + (؜_{2}^{k - g + 1})} \\ + (K_{M + 1} - q^{- (1 + g)} G_{1}^{g + 1} D_{M + 1}) \sum_{k = 0}^{M} {(- 1)}^{k + g} H_{1}^{q} (k) G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} . (3 *) \\ \sum_{k = 0}^{M + 1} {(- 1)}^{k + g} H_{1}^{q} (k, M + 1) G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} \\ = \sum_{k = 0}^{M + 1} {(- 1)}^{k + g} (q^{k} G_{1}^{k} D_{M + 1} H_{1}^{q} (k - 1) + (K_{M + 1} + G_{1}^{k} D_{M + 1}) H_{1}^{q} (k)) G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} \\ = \sum_{k = 0}^{M} {(- 1)}^{k + g - 1} q^{1 + k} G_{1}^{1 + k} D_{M + 1} H_{1}^{q} (k) G_{g}^{1 + k} q^{- (k + 1) (k + 2) + (؜_{2}^{k - g + 1})} \\ + \sum_{k = 0}^{M} {(- 1)}^{k + g} (K_{M + 1} + G_{1}^{k} D_{M + 1}) H_{1}^{q} (k) G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} . (3 * *) \end{matrix}

\begin{matrix} Items containing K_{M + 1} in (3 *) and (3 * *) = {(- 1)}^{k + g} K_{M + 1} G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} . \\ Items does not contain K_{M + 1} in (3 *) : \\ = q^{- g} G_{1}^{g} D_{M + 1} {(- 1)}^{k + g - 1} G_{g - 1}^{k} q^{- k (k + 1) + (؜_{2}^{k - g + 1})} - q^{- (1 + g)} G_{1}^{1 + g} D_{M + 1} {(- 1)}^{k + g} G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} . \\ Divide by D_{M + 1} q^{- (1 + g)} {(- 1)}^{k + g - 1} q^{- k (k + 1) + (؜_{2}^{k - g})} {(q - 1)}^{- 1} \\ = (q^{g} - 1) G_{g - 1}^{k} q^{k - g + 1} + (q^{1 + g} - 1) G_{g}^{k} = q^{k + 1} G_{g - 1}^{k} - q^{k - g + 1} G_{g - 1}^{k} + q^{1 + g} G_{g}^{k} - G_{g}^{k} . (A *) \\ Items does not contain K_{M + 1} in (3 * *) : \\ = {(- 1)}^{k + g - 1} q^{1 + k} G_{1}^{1 + k} D_{M + 1} G_{g}^{1 + k} q^{- (k + 1) (k + 2) + (؜_{2}^{k - g + 1})} + {(- 1)}^{k + g} G_{1}^{k} D_{M + 1} G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} . \\ Divide by D_{M + 1} q^{- (1 + g)} {(- 1)}^{k + g - 1} q^{- k (k + 1) + (؜_{2}^{k - g})} {(q - 1)}^{- 1} \\ = (q^{1 + k} - 1) (q^{g} G_{g}^{k} + G_{g - 1}^{k}) - q^{1 + g} (q^{k} - 1) G_{g}^{k} = q^{1 + k} G_{g - 1}^{k} - q^{g} G_{g}^{k} - G_{g - 1}^{k} + q^{1 + g} G_{g}^{k} . (A * *) \\ (A *) - (A * *) = - q^{k - g + 1} G_{g - 1}^{k} - G_{g}^{k} - (- q^{g} G_{g}^{k} - G_{g - 1}^{k}) = - G_{g}^{k + 1} + G_{g}^{k + 1} = 0 . \\ (3 *) = (3 * *), proof of (3) is complete . \\ H_{3}^{q} (g, M + 1) = ((q^{g} G_{1}^{M + 1} - q^{M + 2} G_{1}^{g - 1}) D_{M + 1} - K_{M + 1} q^{M + 2}) H_{3}^{q} (g - 1) + (K_{M + 1} + G_{1}^{g} D_{M + 1}) H_{3}^{q} (g) \\ = ((q^{g} G_{1}^{M + 1} - q^{M + 2} G_{1}^{g - 1}) D_{M + 1} - K_{M + 1} q^{M + 2}) \sum_{k = 0}^{M} {(- 1)}^{k + g - 1} G_{M - g + 1}^{M - k} q^{g (g - k - 1) - (؜_{2}^{g - k - 1})} H_{1}^{q} (k) \\ + (K_{M + 1} + G_{1}^{g} D_{M + 1}) \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} H_{1}^{q} (k) . (4 *) \\ \sum_{k = 0}^{M + 1} {(- 1)}^{k + g} H_{1}^{q} (k, M + 1) G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} \\ = \sum_{k = 0}^{M + 1} {(- 1)}^{k + g} (q^{k} G_{1}^{k} D_{M + 1} H_{1}^{q} (k - 1) + (K_{M + 1} + G_{1}^{k} D_{M + 1}) H_{1}^{q} (k)) G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} \\ = \sum_{k = 0}^{M} {(- 1)}^{k + g - 1} q^{1 + k} G_{1}^{1 + k} D_{M + 1} H_{1}^{q} (k) G_{M + 1 - g}^{M - k} q^{(g + 1) (g - k - 1) - (؜_{2}^{g - k - 1})} \\ + \sum_{k = 0}^{M} {(- 1)}^{k + g} (K_{M + 1} + G_{1}^{k} D_{M + 1}) H_{1}^{q} (k) G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} . (4 * *) \\ Items containing K_{M + 1} in (4 *) : \\ = - K_{M + 1} q^{M + 2} {(- 1)}^{k + g - 1} G_{M - g + 1}^{M - k} q^{g (g - k - 1) - (؜_{2}^{g - k - 1})} + {(- 1)}^{k + g} K_{M + 1} G_{M - g}^{M - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} . \\ Divide by K_{M + 1} {(- 1)}^{k + g} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} = q^{M + 1 - g} G_{M - g + 1}^{M - k} + G_{M - g}^{M - k} = G_{M - g + 1}^{M - k + 1} \\ = Items containing K_{M + 1} in (4 * *) divided by K_{M + 1} {(- 1)}^{k + g} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} . \\ Items does not contain K_{M + 1} in (4 *) : \\ = (q^{g} G_{1}^{M + 1} - q^{M + 2} G_{1}^{g - 1}) D_{M + 1} {(- 1)}^{k + g - 1} G_{M - g + 1}^{M - k} q^{g (g - k - 1) - (؜_{2}^{g - k - 1})} + G_{1}^{g} D_{M + 1} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} . \\ Divide by D_{M + 1} {(- 1)}^{k + g - 1} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} {(q - 1)}^{- 1} \\ = [(q^{g} G_{1}^{M + 1} - q^{M + 2} G_{1}^{g - 1}) G_{M - g + 1}^{M - k} q^{- g - 1} - G_{1}^{g} G_{M - g}^{M - k}] {(q - 1)}^{- 1} = (q^{M - g - 1} - q^{- 1}) G_{M - g + 1}^{M - k} - (q^{g} - 1) G_{M - g}^{M - k} . (B *) \\ Items does not contain K_{M + 1} in (4 * *) : \\ = {(- 1)}^{k + g - 1} q^{1 + k} G_{1}^{1 + k} D_{M + 1} G_{M + 1 - g}^{M - k} q^{(g + 1) (g - k - 1) - (؜_{2}^{g - k - 1})} + {(- 1)}^{k + g} G_{1}^{k} D_{M + 1} G_{M + 1 - g}^{M + 1 - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} . \\ Divide by D_{M + 1} {(- 1)}^{k + g - 1} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} {(q - 1)}^{- 1} \\ = (q^{1 + k} G_{1}^{1 + k} G_{M + 1 - g}^{M - k} q^{- k - 2} - G_{1}^{k} G_{M + 1 - g}^{M + 1 - k}) {(q - 1)}^{- 1} = (q^{k} - q^{- 1}) G_{M + 1 - g}^{M - k} - (q^{k} - 1) G_{M + 1 - g}^{M + 1 - k} . (B * *) \\ (B *) - (B * *) = (q^{M - g - 1} - q^{k}) G_{M - g + 1}^{M - k} - (q^{g} - 1) G_{M - g}^{M - k} + (q^{k} - 1) (G_{M - g + 1}^{M - k} + q^{g - k} G_{M - g}^{M - k}) \\ = (q^{M - g - 1} - 1) G_{M - g + 1}^{M - k} - (q^{g - k} - 1) G_{M - g}^{M - k} = 0 . \\ (4 *) = (4 * *), proof of (4) is complete . \end{matrix}

□

In the calculation of

H^{q} (g)

,

\prod X_{i} = (\prod X_{i}, X_{i} \in T) (\prod X_{i}, X_{i} \in K)

.

Definition 2.2.

H^{q} (g, \sum T) = H^{q} (g, \sum T, P S, P T) = H^{q} (g, \sum T, M) = \sum \prod_{X_{i} \in T} B_{i}, H^{q} (g, \sum K) = \sum \prod_{X_{i} \in K} B_{i}

Definition 2.3.

F_{0}^{K} = 1, E_{0}^{K} = 1

,

\begin{matrix} F_{g}^{K} = \sum_{1 \leq λ_{1} < . . . < λ_{g} \leq M} \prod_{i = 1}^{g} K_{λ_{i}}, F_{g}^{N} = F_{g}^{{1, 2 . . . N}} . E_{g}^{K} = \sum_{1 \leq λ_{1} \leq . . . \leq λ_{g} \leq M} \prod_{i = 1}^{g} K_{λ_{i}}, E_{g}^{N} = E_{g}^{{1, 2 . . . N}} . \\ E_{g}^{N, q} = \sum_{1 \leq λ_{1} \leq . . . \leq λ_{g} \leq N} \prod_{i = 1}^{g} {[λ_{i}]}_{q} = S_{2}^{q} (N + g, N) . E_{g}^{N, q -} = \sum_{1 \leq λ_{1} \leq . . . \leq λ_{g} \leq N} \prod_{i = 1}^{g} {[λ_{i}]}_{q -} = S_{2}^{q -} (N + g, N) . \end{matrix}

Theorem 2.3.

PT = [T,T+1...T+M-1] and

D_{i} = 1

,

H_{1}^{q} (g, \sum K) = F_{M - g}^{K} E_{0}^{g, q} + F_{M - g - 1}^{K} E_{1}^{g, q} + . . . + F_{0}^{K} E_{M - g}^{g, q}

.

Proof.

\begin{matrix} P S 1 = [P S, K_{M + 1}], P T 1 = [P T, T_{M + 1}] . Using induction to prove . \\ H_{1}^{q} (g, \sum K, M + 1) = H_{1}^{q} (g - 1, \sum K) + H_{1}^{q} (g, \sum K) (K_{M + 1} + G_{1}^{g}) . \\ F_{A}^{{K_{1}, K_{2} . . . K_{M}, K_{M + 1}}} i n H_{1}^{q} (g, \sum K, M + 1) has three sources . \\ = F_{A}^{K} E_{M - (g - 1) - A}^{g - 1, q} + F_{A - 1}^{K} E_{M - g - (A - 1)}^{g, q} K_{M + 1} + F_{A}^{K} E_{M - g - A}^{g, q} G_{1}^{g} \\ = F_{A}^{K} (E_{M + 1 - g - A}^{g - 1, q} + E_{M - g - A}^{g, q} G_{1}^{g}) + F_{A - 1}^{K} E_{M + 1 - g - A}^{g, q} K_{M + 1} \\ = F_{A}^{K} E_{M + 1 - g - A}^{g, q} + F_{A - 1}^{K} E_{M + 1 - g - A}^{g, q} K_{M + 1} = F_{A}^{{K_{1}, K_{2} . . . K_{M}, K_{M + 1}}} E_{M + 1 - g - A}^{g, q} . \end{matrix}

□

So

P T = [1, 2 . . . M]

, we can always choose

K

,

\nabla_{q}^{- p} S U M_{q} (X - p, P S, P T) = \sum_{g = 0}^{M} H_{1}^{q} (g) q^{p g} G_{1 + p + g}^{X}

,

H_{1}^{q} (M) = {[M!]}_{q +}

,

H_{1}^{q} (g < M) p^{p g}

can take any value. So

\sum_{g = 0}^{M} a_{g} G_{1 + p + g}^{X}, 1 + p \geq 0,

can be converted to

\frac{a_{M}}{{[M!]}_{q +}} q^{- p M} \nabla_{q}^{- p} S U M_{q} (X - p, P S, P T)

. For an arbitrary PT1,

S U M_{q} (N, P S 1, P T 1)

can be converted into constant

\times \nabla_{q}^{- (T_{M} - M)} S U M_{q} (N, P S, P T)

.

In this article, if

\sum_{g = 0}^{M} a_{g} G_{1 + p + g}^{X} = \sum_{g = 0}^{M} b_{g} G_{1 + p + g}^{X + g} = \sum_{g = 0}^{M} c_{g} G_{1 + p + M}^{X + M - g}, 1 + p \geq 0

, then c is a constant,

a_{g} = c \times H_{1}^{q} (g, P S, P T) q^{p g}

,

b_{g} = c \times H_{2}^{q} (g, P S, P T)

,

c_{g} = c \times H_{3}^{q} (g, P S, P T) q^{p g}

. From [2.2], [2.1(3)]:

Theorem 2.4.

X, K \in N, M > K, 1 + p \geq 0, a_{g}^{*} = a_{g} q^{- p g}, c_{g}^{*} = c_{g} q^{- p g}

.

If

\sum_{g = 0}^{M} a_{g} G_{1 + p + g}^{X} = \sum_{g = 0}^{M} b_{g} G_{1 + p + g}^{X + g} = \sum_{g = 0}^{M} c_{g} G_{1 + p + M}^{X + M - g}

, then

\begin{matrix} a_{g}^{*} = q^{g (g + 1)} \sum_{k = g}^{M} b_{k} G_{g}^{k} = \sum_{k = 0}^{g} G_{M - g}^{M - k} q^{(g + 1) (g - k)} c_{k}^{*} . \\ b_{g} = {(- 1)}^{g} G_{g}^{k} q^{\frac{g (g + 1)}{2}} \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{\frac{- k (k + 3)}{2} - k g} a_{k}^{*} . c_{g}^{*} = {(- 1)}^{g} G_{g}^{k} q^{\frac{g (g + 3)}{2}} \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{\frac{- k (k + 3)}{2}} a_{k}^{*} . \end{matrix}

\sum_{g = 0}^{M} a_{g} G_{1 + p + g}^{X}

can be coverted to

\sum_{g = 0}^{M - K} (. . .) G_{1 + p + K + g}^{X + K}

is equivalent to

\begin{matrix} \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{- k (k + 1) + (؜_{2}^{k - g})} a_{k}^{*} = 0, 0 \leq g < K . \\ \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} a_{k}^{*} = 0, 0 \leq M - g < K . \end{matrix}

The latter part refers to the necessary and sufficient conditions for merging, which correspond to

b_{g} = c_{M - g} = 0

.

3. Application

Proposition 3.1.

(1).

\sum_{0 \leq λ_{1} \leq λ_{2} \leq . . . \leq λ_{M} \leq N} q^{\sum_{i = 1}^{M} λ_{i}} = G_{M}^{N + M} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{M g + \frac{g (g + 1)}{2}} G_{g}^{M} G_{2 M}^{N + 2 M - g}

.

(2).

\sum_{A \leq λ_{1} < λ_{2} < . . . < λ_{M} \leq B} q^{\sum_{i = 1}^{M} λ_{i}} = q^{(؜_{2}^{M}) + A M} G_{M}^{B - A + 1}, A, B \in Z, \sum_{1 \leq λ_{1} < . . . < λ_{g} \leq M} q^{\sum_{i = 1}^{M} λ_{i}} = q^{(؜_{2}^{g + 1})} G_{g}^{M}

.

(3).

\prod_{i = 1}^{M} (a + q^{A + i} z) = \sum_{g = 0}^{M} q^{(؜_{2}^{g}) + (A + 1) g} G_{g}^{M} z^{g} a^{M - g}

.

(4).

G_{M - K}^{N - K} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{(M - K) g + \frac{g (g + 1)}{2}} G_{g}^{M} G_{2 M - K}^{N + M - K - g}, N \geq M \geq K \geq 0

.

(5).

\sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} G_{K}^{N - g} = {(q^{M}; q^{- 1})}_{M - K}, N \geq M \geq K \geq 0

.

(6).

\sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} G_{M + 1}^{N - g} = \sum_{i = 0}^{N - M - 1} q^{(M + 1) i}, N \geq M + 1, M > 0

.

(7).

{[؜_{M}^{N + M}]}_{q^{2}} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{g^{2}} {[؜_{g}^{M}]}_{q^{2}} G_{2 M}^{N + 2 M - g}

.

Proof.

\begin{matrix} P S = [1, 1 . . . 1] : 0, P T = [1, 3 . . . 2 M - 1], H_{1}^{q} (g > 0) = 0, H_{1}^{q} (0) = 1, S U M (N + 1) = G_{M}^{N + M} \to pre - equation of (1) . \\ \sum_{A \leq λ_{1} < . . . < λ_{M} \leq B} q^{\sum_{i = 1}^{M} λ_{i}} = q^{A + (A + 1) + . . . + (A + M - 1)} \sum_{0 \leq λ_{1} \leq . . . \leq λ_{M} \leq B - A - M + 1} q^{\sum_{i = 1}^{M} λ_{i}} \to (2) . \\ At H_{3}^{q} (g), B_{i} = \{؜_{q^{X_{T}} = q^{i - X_{K}}, X_{i} = K_{i}}^{q^{X_{T}} (- q^{2 i - 1}), X_{i} = T_{i}}, Extract q^{i} form B_{i}, after extraction, B_{i} = \{؜_{q^{- X_{K}}, X_{i} = K_{i}}^{q^{X_{T}} (- q^{i - 1}), X_{i} = T_{i}}, \\ H_{3}^{q} (g) = q^{(؜_{2}^{M + 1}) - (؜_{2}^{M - g + 1})} {(- 1)}^{g} q^{(؜_{2}^{g + 1}) - g} \sum_{1 \leq λ_{1} < . . . < λ_{g} \leq M} q^{\sum_{i = 1}^{g} λ_{i}} = {(- 1)}^{g} q^{(؜_{2}^{M + 1}) - (؜_{2}^{M - g + 1}) + (؜_{2}^{g}) + (؜_{2}^{g + 1})} G_{g}^{M} . \\ S U M (N + 1) = G_{M}^{N + M} = \sum_{g = 0}^{M} H_{3}^{q} (g) G_{2 M}^{N + 2 M - g} \to post - equation of (1) . \\ Comparing the coefficients of z^{g} on both sides of (3), combined with (2), proves that (3) . \\ A = - 1, it ’ s Rothe ’ s q - Binomial Theorem \prod_{i = 1}^{M} (a + q^{i - 1} z) = \sum_{g = 0}^{M} q^{(؜_{2}^{g})} G_{g}^{M} z^{g} a^{M - g} . \\ Using the post - equation of (1) and \nabla_{q}^{K} to obtain (4) \to K = M, \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} G_{M}^{N - g} = 1 \\ = \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} (q^{M} G_{M}^{N - g - 1} + G_{M - 1}^{N - g - 1}) = q^{M} + \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} G_{M - 1}^{N - g - 1} \\ \to \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} G_{M - 1}^{N - g - 1} = 1 - q^{M} . Continuing the same process \to (5) . \\ K = 0, \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} = {(q; q)}_{M}; K = M, \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{M} G_{M}^{N - g} = 1, it ’ s a special case of (4) . \\ Using the same recursive and inductive methods can obtain (6) . \\ S U M_{q} (N + 1, [1, 1 . . . 1] : q - 1, [1, 3 . . . 2 M - 1]) = \sum_{0 \leq λ_{1} \leq . . . \leq λ_{M} \leq N} q^{2 \sum λ_{i}} = {[؜_{M}^{N + M}]}_{q^{2}} . \\ A t H_{1}^{q} (g), B_{i} = \{؜_{q^{(T_{i} - T_{i - 1} - 1) X_{T - 1}} (K_{i} + G_{1}^{X_{T - 1}} D_{i}) = q^{2 X_{T - 1}} = q^{2 X_{T}}, X_{i} = K_{i}}^{q^{1 + (T_{i} - T_{i - 1}) X_{T - 1}} G_{1}^{T_{i} - X_{K - 1}} D_{i} = q^{1 + 2 X_{T - 1}} (q^{2 i - 1 - X_{K - 1}} - 1) = q^{2 X_{T} - 1} (q^{i + X_{T - 1}} - 1), X_{i} = T_{i}}, \\ Unable to derive a concise expression for H_{1}^{q} (g), the same applies to H_{2}^{q} (g) . \\ A t H_{3}^{q} (g), B_{i} = \{؜_{q^{X_{T - 1}} (K_{i} + G_{1}^{X_{T - 1}} D_{i}) = q^{X_{T - 1}} (1 + G_{1}^{X_{T - 1}} (q - 1)) = q^{2 X_{T - 1}} = q^{2 X_{T}}, X_{i} = K_{i}}^{q^{1 + X_{T - 1}} {(q^{X_{T - 1}} G_{1}^{T_{i}} - q^{T_{i}} G_{1}^{X_{T - 1}}) D_{i} - K_{i} q^{T_{i}}} = - q^{1 + 2 X_{T - 1}} = - q^{- 1 + 2 X_{T}}, X_{i} = T_{i}} . \\ H_{3}^{q} (g) = {(- 1)}^{g} q^{- g + 2 (\sum_{i = 1}^{g} i)} \sum_{X_{i} \in K} \prod q^{2 X_{T}} = {(- 1)}^{g} q^{g^{2}} {[؜_{g}^{M}]}_{q^{2}} . \end{matrix}

□

\begin{matrix} \sum_{A \leq λ_{1} < λ_{2} < . . . < λ_{M} \leq B} q^{\sum_{i = 1}^{M} λ_{i} D} = q^{(؜_{2}^{M}) D + A M D} {[؜_{M}^{B - A + 1}]}_{q^{D}} . \\ \prod_{i = 1}^{A} (1 + q^{C + D i} z) = \sum_{g = 0}^{A} q^{g C} \sum_{1 \leq λ_{1} \leq . . . \leq λ g \leq A} q^{\sum_{i = 1}^{g} λ_{i} D} z^{g} = \sum_{g = 0}^{A} q^{(؜_{2}^{g}) D + g D + g C} {[؜_{g}^{A}]}_{q^{D}} z^{g} . \\ \prod_{i = 1}^{A} (a + q^{(2 i - 1) D} z) = \sum_{g = 0}^{A} q^{g^{2} D} {[؜_{g}^{A}]}_{q^{2 D}} z^{g} a^{A - g}, \prod_{i = 1}^{B} (a^{- 1} + q^{(2 i - 1) D} z^{- 1}) = \sum_{g = 0}^{B} q^{g^{2} D} {[؜_{g}^{B}]}_{q^{2 D}} z^{- g} a^{- (B - g)} . \\ \prod_{i = 1}^{A} (a + q^{(2 i - 1) D} z) \prod_{i = 1}^{B} (a^{- 1} + q^{(2 i - 1) D} z^{- 1}) = \sum_{k = - B}^{A} z^{k} f (k) . \\ f (k) = \sum_{i = 0}^{A + B} {[؜_{i}^{A}]}_{q^{2 D}} {[؜_{i - k}^{B}]}_{q^{2 D}} q^{i^{2} D} q^{{(i - k)}^{2} D} a^{A - i} a^{- (B - (i - k))} \\ = q^{k^{2} D} a^{A - B - k} \sum_{i = 0}^{A + B} {[؜_{i}^{A}]}_{q^{2 D}} {[؜_{i - k}^{B}]}_{q^{2 D}} q^{2 D (i (i - k))} = q^{k^{2} D} a^{A - B - k} {[؜_{A - k}^{A + B}]}_{q^{2 D}} . \end{matrix}

\begin{matrix} The last step used q - Vandermorde identity . a = D = 1, it ’ s MacMahon ’ s q - binomial theorem . [2] pp 74 \\ | q |, | x | < 1, \frac{1}{{(x; q)}_{M + 1}} = (1 + x + x^{2} + . . .) (1 + x q + x^{2} q^{2} + . . .) (1 + x q^{2} + + x^{2} q^{4} + . . .) . . . (1 + x q^{M} + + x^{2} q^{2 M} + . . .) . \\ When we multiply this out, the coefficient of x^{2} will be \sum q^{a} q^{b}, 0 \leq a \leq b \leq M, and so forth . \\ (1) \to \frac{1}{{(x; q)}_{M + 1}} = \sum_{k = 0}^{\infty} G_{k}^{M + k} x^{k}, \frac{1}{{(x; q)}_{\infty}} = \sum_{k = 0}^{\infty} \frac{x^{k}}{{(q; q)}_{k}}, it ’ s Euler and Cauchy ’ s identity [2] pp 121 - 123 \\ (3) \to {(- x, q)}_{M} = \sum_{g = 0}^{M} q^{(؜_{2}^{g})} G_{g}^{M} x^{g} \to {(- x, q)}_{\infty} = \sum_{k = 0}^{\infty} \frac{q^{(؜_{2}^{k})} x^{k}}{{(q; q)}_{k}}, it ’ s Euler ’ s identity [2] pp 129 \\ C \geq 2, P S = [1, 1 . . . 1] : q - 1, P T = [1, C + 1, 2 C + 1 . . . C (M - 1) + 1], A t H_{3}^{q} (g), B_{i} = \{؜_{q^{C X_{T}}, X_{i} = K_{i}}^{- q^{C X_{T - 1} + 1}, X_{i} = T_{i}} \to \\ X = C (M - 1) - M + 2, M > 1, \sum_{0 \leq λ_{1} \leq . . . \leq λ_{X} \leq N - 1} q^{\sum_{i = 1}^{X} f_{i} λ_{i}} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{g + C \frac{g (g - 1)}{2}} {[؜_{g}^{M}]}_{q^{C}} G_{C (M - 1) + 2}^{N + C (M - 1) + 1 - g}, \\ f_{i} = 2, i \equiv 1 (mod C - 1); f_{i} = 1, i \neg \equiv 1 (mod C - 1) . \end{matrix}

Proposition 3.2.

(1).

G_{M + 1}^{N + M} = \sum_{g = 0}^{M} q^{(g + 1) g} G_{g}^{M} G_{1 + g}^{N}; G_{M + 1 + p}^{N + M + p} = \sum_{g = 0}^{M} q^{(g + 1 + p) g} G_{g}^{M} G_{p + 1 + g}^{N + p} = \sum_{g = 0}^{M + p} q^{(g + 1) g} G_{g}^{M + p} G_{1 + g}^{N}

,

M + p \geq 0

.

(2).

\sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} G_{k}^{M} q^{(؜_{2}^{k}) - g k} = 0, g < M; \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} G_{k}^{M} q^{(؜_{2}^{k})} = 0, g > 0

.

(3).

M > g + x, \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} G_{k}^{M} q^{(؜_{2}^{k}) - (g + x) k} = 0; M > g, \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} G_{k}^{M} q^{(؜_{2}^{k}) - (M - 1) k} = 0

.

(4).

g > x, \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} G_{k}^{M} q^{(؜_{2}^{k}) - x k} = 0; g > 0, \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} G_{k}^{M} q^{(؜_{2}^{k}) - (g - 1) k} = 0

.

Proof.

\begin{matrix} P S = [{[1]}_{q -}, {[2]}_{q -} . . . {[M]}_{q -}], P S 1 = [{[M]}_{q -}, {[M - 1]}_{q -} . . . {[1]}_{q -}], P T = [1, 2 . . . M] \\ S U M_{q} (N, P S, P T) = {[M!]}_{q -} G_{M + 1}^{N + M} = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{1 + g}^{N} = S U M_{q} (N, P S 1, P T) . \\ I n H_{1}^{q} (g, P S 1, P T), X_{i} \in K, B_{i} = {[M + 1 - i]}_{q -} + G_{1}^{X_{T - 1}} = q^{- (M - i + 1)} G_{1}^{M + 1 - i + X_{T - 1}} . \\ H_{1}^{q} (g, \sum K) = G_{1}^{M} G_{1}^{M - 1} . . . G_{1}^{g + 1} q^{- (M + 1) (M - g)} \sum_{1 \leq λ_{1} < . . . < λ_{M - g} \leq M} q^{λ_{1} + . . . + λ_{M - g}} . \\ = G_{1}^{M} G_{1}^{M - 1} . . . G_{1}^{g + 1} q^{(؜_{2}^{M - g + 1}) - (M + 1) (M - g)} G_{M - g}^{M} . H_{1}^{q} (g) = q^{\sum_{i = 1}^{g} i} {[g!]}_{q} H_{1}^{q} (g, \sum K) . \\ \frac{H_{1}^{q} (g)}{{[M!]}_{q -}} = q^{(؜_{2}^{M - g + 1}) - (M + 1) (M - g) + \sum_{i = 1}^{M} i + \sum_{i = 1}^{g} i} G_{M - g}^{M} = q^{g (g + 1)} G_{g}^{M}, this and [2.1 (3)] yields (1) . \\ It ’ s q - Vandermorde identity : G_{M + t}^{N + M} = G_{N - t}^{N + M} = \sum_{g = 0}^{M} q^{g (g + t)} G_{g}^{M} G_{g + t}^{N} . (2) is obtained from [2.4] . \\ It is obvious from q - Binomial theorem that \sum_{k = 0}^{M} {(- 1)}^{k} q^{(؜_{2}^{k})} G_{k}^{M} = 0 . \\ \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - g k} G_{k}^{M} = \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - g k} G_{k}^{M - 1} + q^{M} \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - g k} G_{k - 1}^{M - 1} q^{- k} . \\ M > 1 + g, \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - g k} G_{k}^{M - 1} = 0 \to \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - (g + 1) k} G_{k - 1}^{M - 1} = 0 . \\ \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - g k} G_{k}^{M} q^{- k} = \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - g k} G_{k}^{M - 1} + \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k}) - (g + 1) k} G_{k - 1}^{M - 1} = 0 . \\ Similarly, M > g + x, \sum_{k = g}^{M} {(- 1)}^{k} G_{g}^{k} q^{(؜_{2}^{k})} G_{k}^{M} q^{- (g + x) k} = 0 \to (3) . \\ \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} G_{k}^{M} q^{(؜_{2}^{k})} = \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{(؜_{2}^{k})} G_{k}^{M - 1} + q^{M} \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{(؜_{2}^{k}) - k} G_{k - 1}^{M - 1} = 0 . \\ {(- 1)}^{k} G_{M - g}^{M - k} q^{(؜_{2}^{k}) - k} G_{k - 1}^{M - 1} = q^{- 1} {(- 1)}^{k} G_{M - 1 - (g - 1)}^{M - 1 - (k - 1)} q^{(؜_{2}^{k - 1})} G_{k - 1}^{M - 1} \to g > 0, \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{(؜_{2}^{k})} G_{k}^{M - 1} = 0 . \\ \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{(؜_{2}^{k}) - k} G_{k}^{M} = \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{(؜_{2}^{k})} G_{k}^{M - 1} + q^{- 1} \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - 1 - (g - 1)}^{M - 1 - (k - 1)} q^{(؜_{2}^{k - 1})} G_{k - 1}^{M - 1} = 0 . \\ Similarly, g > x, \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} q^{(؜_{2}^{k}) - x k} G_{k}^{M} = 0 . \end{matrix}

□

Proposition 3.3.

If

\sum_{g = 0}^{M} a_{g} G_{g}^{n} = \prod_{i = 1}^{M} (K_{i} + {[n]}_{q} D_{i})

then

\sum_{g = 0}^{M} a_{g} G_{g}^{T + g} G_{T + g}^{n + T} = \prod_{i = 1}^{M} (K_{i} + {[n]}_{q} D_{i}) G_{T}^{n + T}

.

Proof.

\begin{matrix} P T 1 = [1, 2 . . . M], P T 2 = [T + 1, T + 2 . . . T + M], \\ \nabla_{q} S U M (n + 1, P S, P T 1) = \prod_{i = 1}^{M} (K_{i} + {[n]}_{q} D_{i}), \nabla_{q} S U M (n + 1, P S, P T 2) = \prod_{i = 1}^{M} (K_{i} + {[n]}_{q} D_{i}) G_{T}^{n + T}, \\ At H_{1}^{q} (g), B_{i} of P T 1 = \{؜_{(K_{i} + G_{1}^{X_{T}} D_{i}), X_{i} = K_{i}}^{q^{X_{T}} G_{1}^{X_{T}} D_{i}, X_{i} = T_{i}}, B_{i} of P T 2 = \{؜_{(K_{i} + G_{1}^{X_{T}} D_{i}), X_{i} = K_{i}}^{q^{X_{T}} G_{1}^{T + X_{T}} D_{i}, X_{i} = T_{i}} . \\ H_{1}^{q} (g, P S, P T 2) = H_{1}^{q} (g, P S, P T 1) \frac{G_{1}^{T} . . . G_{1}^{T + g}}{G_{1}^{1} . . . G_{1}^{g}} = H_{1}^{q} (g, P S, P T 1) G_{g}^{T + g}, a_{g} = constant \times H_{1}^{q} (g, P S, P T 1) q^{- g} \end{matrix}

□

For

D_{i} = (q - 1) K_{i}, P S = [T, T + D . . . T + (M - 1) D], H_{3}^{q} (g)

and T are not related, similar equations exist.

Proposition 3.4.

x \geq 0,

(1).

\sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k}^{M} G_{g}^{k} q^{\frac{k (k - 1)}{2} - g k + x k} z^{k} = q^{\frac{g (g - 1)}{2} - g^{2} + x g} G_{g}^{M} {(z q^{x}; q)}_{M - g} z^{g}

.

(2).

\sum_{k = 0}^{M} {(- 1)}^{k} G_{k}^{M} G_{M - g}^{M - k} q^{\frac{k (k - 1)}{2} + x k} z^{k} = G_{g}^{M} {(z q^{x}; q)}_{g}

.

Proof.

\begin{matrix} For (1) : g = 0, it ’ s [3.1 (3)] . When g = M, it clearly holds . Prove by induction . \\ \sum_{k = 0}^{M} {(- 1)}^{k + g} (G_{k}^{M - 1} + q^{M - k} G_{k - 1}^{M - 1}) G_{g}^{k} q^{\frac{k (k - 1)}{2} - g k + x k} z^{k} \\ = q^{\frac{g (g - 1)}{2} - g^{2} + x g} G_{g}^{M - 1} {(z q^{x}; q)}_{M - 1 - g} z^{g} + q^{M} \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k - 1}^{M - 1} G_{g}^{k} q^{\frac{k (k - 1)}{2} - g k + x k - k} z^{k} \\ = q^{\frac{g (g - 1)}{2} - g^{2} + x g} G_{g}^{M - 1} {(z q^{x}; q)}_{M - 1 - g} z^{g} - q^{M} \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k}^{M - 1} (G_{g}^{k} + q^{k + 1 - g} G_{g - 1}^{k}) q^{\frac{k (k - 1)}{2} - g (k + 1) + x (k + 1) - 1} z^{k + 1} \\ = q^{\frac{g (g - 1)}{2} - g^{2} + x g} (G_{g}^{M - 1} {(z q^{x}; q)}_{M - 1 - g} z^{g} - z^{g + 1} q^{M + x - g - 1} G_{g}^{M - 1} {(z q^{x}; q)}_{M - 1 - g}) \\ - q^{M} z \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k}^{M - 1} q^{k + 1 - g} G_{g - 1}^{k} q^{\frac{k (k - 1)}{2} - g (k + 1) + x (k + 1) - 1} z^{k} \\ = q^{\frac{g (g - 1)}{2} - g^{2} + x g} G_{g}^{M - 1} {(z q^{x}; q)}_{M - g} z^{g} + q^{M + x - 2 g} z \sum_{k = 0}^{M} {(- 1)}^{k + g - 1} G_{k}^{M - 1} G_{g - 1}^{k} q^{\frac{k (k - 1)}{2} - (g - 1) k + x k} z^{k} \\ = {(z q^{x}; q)}_{M - g} z^{g} (q^{\frac{g (g - 1)}{2} - g^{2} + x g} G_{g}^{M - 1} + q^{\frac{(g - 1) (g - 2)}{2} - {(g - 1)}^{2} + x (g - 1)} q^{M + x - 2 g} G_{g - 1}^{M - 1}) \\ = q^{\frac{g (g - 1)}{2} - g^{2} + x g} {(z q^{x}; q)}_{M - g} z^{g} (G_{g}^{M - 1} + q^{M - g} G_{g - 1}^{M - 1}) \to (1) . \\ For (2) : g = M, it ’ s [3.1 (3)] . When g = 0, it clearly holds . \\ \sum_{k = 0}^{M} {(- 1)}^{k} (q^{k} G_{k}^{M - 1} + G_{k - 1}^{M - 1}) G_{M - g}^{M - k} q^{\frac{k (k - 1)}{2} + x k} z^{k} \\ = \sum_{k = 0}^{M} {(- 1)}^{k} G_{k}^{M - 1} G_{M - g}^{M - k} q^{\frac{k (k - 1)}{2} + (x + 1) k} z^{k} + \sum_{k = 0}^{M} {(- 1)}^{k} G_{k - 1}^{M - 1} G_{M - 1 - (g - 1)}^{M - 1 - (k - 1)} q^{\frac{k (k - 1)}{2} + x k} z^{k} \\ = \sum_{k = 0}^{M} {(- 1)}^{k} G_{k}^{M - 1} G_{M - 1 - (g - 1)}^{M - 1 - k} q^{\frac{k (k - 1)}{2} + (x + 1) k} z^{k} + \sum_{k = 0}^{M} {(- 1)}^{k} G_{k}^{M - 1} q^{g - k} G_{M - 1 - g}^{M - 1 - k} q^{\frac{k (k - 1)}{2} + (x + 1) k} z^{k} \\ - z \sum_{k = 0}^{M} {(- 1)}^{k} G_{k}^{M - 1} G_{M - 1 - (g - 1)}^{M - 1 - k} q^{\frac{k (k + 1)}{2} + x (k + 1)} z^{k} \\ = G_{g - 1}^{M - 1} {(z q^{1 + x}; q)}_{g - 1} - z q^{x} G_{g - 1}^{M - 1} {(z q^{1 + x}; q)}_{g - 1} + q^{g} G_{g}^{M - 1} {(z q^{x}; q)}_{g} \to (2) . \end{matrix}

□

Proposition 3.5.

\sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k}^{M} G_{g}^{k} = \{؜_{0, M + g \equiv 1 (mod 2)}^{G_{g}^{M} {(q; q^{2})}_{\frac{M - g}{2}}, M + g \equiv 0 (mod 2)}

;

\sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k}^{M} G_{M - g}^{M - k} = \{؜_{0, g \equiv 1 (mod 2)}^{G_{g}^{M} {(q; q^{2})}_{\frac{g}{2}}, g \equiv 0 (mod 2)}

.

Proof.

\begin{matrix} M + g is odd, {(- 1)}^{k + g} G_{k}^{M} G_{g}^{k} = \frac{{(- 1)}^{k + g} {(q; q)}_{M}}{{(q; q)}_{g} {(q; q)}_{k - g} {(q; q)}_{M - k}} = - {(- 1)}^{g + M - k + g} G_{M - k + g}^{M} G_{g}^{M - k + g} \to S u m = 0 . \\ M + g is even, use induction to prove . Sum = \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k}^{M - 1} G_{g}^{k} + \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k - 1}^{M - 1} G_{g}^{k} q^{M - k} \\ = 0 + \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k - 1}^{M - 1} G_{g}^{k} q^{M - k} = \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k - 1}^{M - 1} G_{g}^{k - 1} q^{M - k} + \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k - 1}^{M - 1} G_{g - 1}^{k - 1} q^{M - g} \\ = \sum_{k = 0}^{M} {(- 1)}^{k + g - 1} G_{k - 1}^{M - 1} G_{g}^{k - 1} (1 - q^{M - k}) + \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{k - 1}^{M - 1} G_{g - 1}^{k - 1} q^{M - g} \\ = \sum_{k = 0}^{M} {(- 1)}^{k + (g + 1)} G_{k}^{M - 1} G_{g + 1}^{k} (1 - q^{g + 1}) + \sum_{k = 0}^{M} {(- 1)}^{(k - 1) + (g - 1)} G_{k - 1}^{M - 1} G_{g - 1}^{k - 1} q^{M - g} \\ = G_{g + 1}^{M - 1} {(q; q^{2})}_{(M - g - 2) / 2} (1 - q^{g + 1}) + q^{M - g} G_{g - 1}^{M - 1} {(q; q^{2})}_{(M - g) / 2} = (G_{g}^{M - 1} + q^{M - g} G_{g - 1}^{M - 1}) {(q; q^{2})}_{(M - g) / 2} . \\ Replace k by M - k, g by M - g to complete the proof . g = 0, It ’ s Gauss ’ s q - Binomial theorems [2] pp 61 \end{matrix}

□

Proposition 3.6.

\prod_{i = 1}^{M} (b + q^{A + i} z) = \sum_{g = 0}^{M} \prod_{i = 1}^{g} (q^{M + 1 - i} - 1) \prod_{i = 1}^{M - g} (b + q^{A + 1 - i} z) G_{g}^{M} z^{g} q^{g (g + A - M)}

.

Proof.

\begin{matrix} P S = [b + q^{A} z : (q - 1) q^{A} z, b + q^{A - 1} z : (q - 1) q^{A - 1} z . . . b + q^{A - M + 1} z : (q - 1) q^{A - M + 1} z], P T = [1, 2 . . . M] . \\ \nabla_{q}^{1} S U M_{q} (M + 1) = \prod_{i = 1}^{M} (b + q^{M + A - i + 1} z) = \prod_{i = 1}^{M} (b + q^{A + i} z) = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{g}^{M} q^{- g} . \\ A t H_{1}^{q} (g), B_{i} = \{؜_{b + q^{X_{T - 1}} q^{A - i + 1} z = b + q^{A - X_{K - 1}} z, X_{i} = K_{i}}^{q^{X_{T}} (q^{X_{T}} - 1) q^{A - i + 1} z, X_{i} = T_{i}} . \\ H_{1}^{q} (g) = q^{g (A + 1) + \sum_{i = 1}^{g} i} z^{g} \prod_{i = 1}^{g} (q^{i} - 1) [\sum_{- M \leq λ_{1} < . . . < λ_{g} \leq - 1} q^{\sum_{i = 1}^{g} λ_{i}}] \prod_{i = 1}^{M - g} (a + q^{A + 1 - i} z) \\ = q^{g (A + 1) + (؜_{2}^{g + 1})} q^{(؜_{2}^{g}) - M g} G_{g}^{M} z^{g} \prod_{i = 1}^{g} (q^{i} - 1) \prod_{i = 1}^{M - g} (b + q^{A + 1 - i} z) . \\ \prod_{i = 1}^{M} (b + q^{A + i} z) = \sum_{g = 0}^{M} \prod_{i = 1}^{g} (q^{i} - 1) \prod_{i = 1}^{M - g} (b + q^{A + 1 - i} z) G_{g}^{M} G_{g}^{M} q^{g (g + A - M + 1)} q^{- g} z^{g} . \end{matrix}

□

\begin{matrix} b = 1, z = - 1, \prod_{i = 1}^{M} (1 - q^{A + i}) = \sum_{g = 0}^{M} \prod_{i = 1}^{g} (1 - q^{M + 1 - i}) \prod_{i = 1}^{M - g} (1 - q^{A + 1 - i}) G_{g}^{M} q^{g (g + A - M)} \\ \to G_{M}^{A + M} = \sum_{g = 0}^{M} G_{M - g}^{A} G_{g}^{M} q^{g (g + A - M)}, it ’ s q - Vandermorde identity . \\ \prod_{i = 1}^{M} (b - q^{A + i} z) = \sum_{g = 0}^{M} \prod_{i = 1}^{g} (z q^{g + A - M} - q^{1 - i + g + A} z) \prod_{i = 1}^{M - g} (b - q^{A + 1 - i} z) G_{g}^{M} . \\ = \sum_{g = 0}^{M} \prod_{i = 1}^{g} (z q^{g + A - M} - q^{g - i} q^{A + 1} z) \prod_{i = 1}^{M - g} (b - q^{A + 1 - i - (g + A - M)} z q^{g + A - M}) G_{g}^{M} . \\ a = q^{A + 1} z, d = z q^{A - M} \to \prod_{i = 1}^{M} (b - q^{i - 1} a) = \sum_{g = 0}^{M} \prod_{i = 1}^{g} (d q^{g} - q^{g - i} a) \prod_{i = 1}^{M - g} (b - q^{M - g - i} d q^{g + 1}) G_{g}^{M} . \\ It can be compared to Jacobi ’ s q - binomial theorem [2] pp 71 \\ \prod_{i = 1}^{M} (b - q^{i - 1} a) = \sum_{g = 0}^{M} \prod_{i = 1}^{g} (c - q^{g - i} a) \prod_{i = 1}^{M - g} (b - q^{M - g - i} c) G_{g}^{M} . \end{matrix}

Proposition 3.7.

P S = [{[T_{i}]}_{q -}]

,

P S 1 = [{[K]}_{q -}, {[K + 1]}_{q -} . . . {[K + M - 1]}_{q -}], P T 1 = [T, T + 1 . . . T + M - 1]

.

(1).

H_{1}^{q} (g, P S, P T) = \prod_{i = 1}^{M} {[T_{i}]}_{q -} q^{(g + 1 + p) g} G_{g}^{M}, p = T_{M} - M

.

(2).

H_{1}^{q} (g, P S 1, P T 1) = \prod_{i = 1}^{g} {[T - 1 + i]}_{q} \prod_{i = 1}^{M - g} {[K + M - i]}_{q} G_{g}^{M} q^{(؜_{2}^{g + 1}) - (K + M) (M - g) + (؜_{2}^{M - g + 1})}

.

(3).

K = {P S 1}, F_{M - g}^{K} E_{0}^{g, q} + F_{M - g - 1}^{K} E_{1}^{g, q} + . . . + F_{0}^{K} E_{M - g}^{g, q} = \prod_{i = 1}^{M - g} {[K + M - i]}_{q} G_{g}^{M} q^{(؜_{2}^{M - g + 1}) - (K + M) (M - g)}

.

Proof.

\begin{matrix} \sum_{g = 0}^{M} H_{1}^{q} (g) G_{1 + p + g}^{N + p} = \prod {[T_{i}]}_{q -} G_{T_{M} + 1}^{N + T_{M}} = \prod {[T_{i}]}_{q -} G_{M + p + 1}^{N + M + p}, [3.2] yields (1), [2.4] can also reach the conclusion . \\ At (2), K_{i} can exchange order, let P S = [{[K + M - 1]}_{q -} . . . {[K + 1]}_{q -}, {[K]}_{q -}] . \\ B_{i} o f H_{1}^{q} (g) = \{؜_{q^{- (K + M - i)} G_{1}^{K + M - i + X_{T}}, X_{i} = K_{i}}^{q^{X_{T}} G_{1}^{T + X_{T - 1}}, X_{i} = T_{i}} . H_{1}^{q} (g) = q^{(؜_{2}^{g + 1})} \prod_{i = 1}^{g} {[T - 1 + i]}_{q} H_{1}^{q} (g, \sum K) . \\ H_{1}^{q} (g, \sum K) = \prod_{i = 1}^{M - g} {[K + M - i]}_{q} q^{- (K + M) (M - g)} q^{(؜_{2}^{M - g + 1})} G_{g}^{M} and [2.3] yields (2) and (3) . \\ If K = T, then p = T - 1, q^{(؜_{2}^{g + 1}) - (T + M) (M - g) + (؜_{2}^{M - g + 1})} = q^{g (g + 1 + p) - \sum_{i = 1}^{M} (T - 1 + i)}, it ’ s compatible with (1) . \end{matrix}

□

\begin{matrix} F o r H_{1} (g), (1) = \prod T_{i} (؜_{g}^{M}), (2) = \prod_{i = 1}^{g} (T - 1 + i) \prod_{i = 1}^{M - g} (K + M - i) (؜_{g}^{M}) . \\ \frac{{(1 - q)}^{A + M}}{{(q; q)}_{A} {(q; q)}_{M}} \nabla_{q} S U M (N, [{[1]}_{q -}, {[2]}_{q -} . . . {[A]}_{q -}, {[B + 1]}_{q -}, {[B + 2]}_{q -} . . . {[B + M]}_{q -}], [1, 2 . . . A, A + 1, A + 2 . . . A + M]) \\ = G_{A}^{n + A} G_{M}^{n + M + B} = \frac{{(1 - q)}^{A + M}}{{(q; q)}_{A} {(q; q)}_{M}} {[A!]}_{q -} \nabla_{q} S U M (N, [{[B + 1]}_{q -} . . . {[B + M]}_{q -}], [A + 1 . . . A + M]) . \\ \frac{{(1 - q)}^{A + M}}{{(q; q)}_{A} {(q; q)}_{M}} \nabla_{q} S U M (N, [{[X]}_{q -}, {[X - 1]}_{q -} . . . {[X - A + 1]}_{q -}, {[Y]}_{q -}, {[Y + 1]}_{q -} . . . {[Y + M - 1]}_{q -}], [1, 2 . . . A + M]) \\ 0 \leq Y \leq M, P S 1 = [{[1]}_{q -}, {[2]}_{q -} . . . {[Y]}_{q -}, {[0]}_{q -}, {[- 1]}_{q -}, {[- 2]}_{q -} . . . {[- (M - Y) + 1]}_{q -}, {[X]}_{q -}, {[X - 1]}_{q -} . . . {[X - A + 1]}_{q -}], \\ = G_{A}^{n + X} G_{M}^{n + Y} = \frac{{(1 - q)}^{A + M}}{{(q; q)}_{A} {(q; q)}_{M}} \nabla_{q} S U M (N, P S 1, [1, 2 . . . A + M]) \\ = \frac{{(1 - q)}^{A + M}}{{(q; q)}_{A} {(q; q)}_{M}} {[Y!]}_{q -} \nabla_{q} S U M (N, [{[0]}_{q -} . . . {[- (M - Y) + 1]}_{q -}, {[X - A + 1]}_{q -} . . . {[X]}_{q -}], [Y + 1 . . . A + M]) . \\ If H_{1}^{q} (g) > 0, then X_{1}, X_{2} . . . X_{M - Y} \in T, so H_{1}^{q} (g < M - Y) = 0 . It ’ s not difficult to deduce H_{1}^{q} (g \geq M - Y) . \\ They can obtain the formula for G_{A}^{n + A} G_{M}^{n + M + B} and G_{A}^{n + X} G_{M}^{n + Y}, 0 \leq Y \leq M . \\ H_{1}^{q} (g, [{[1]}_{q -}, {[2]}_{q -} . . . {[M]}_{q -}], [1, 2 . . . M]) = \prod_{i = 1}^{M} {[T_{i}]}_{q -} q^{(g + 1) g} G_{g}^{M} = q^{\frac{- M (M + 1)}{2} + g (g + 1)} \prod_{i = 1}^{M} G_{1}^{i} G_{g}^{M} . \\ B_{i} o f H_{1}^{q} (g) = \{؜_{q^{- T_{i}} G_{1}^{T_{i} + X_{T - 1}}, X_{i} = K_{i}}^{q^{- T_{i}} q^{T_{i} + X_{T}} G_{1}^{T_{i} - X_{K - 1}}, X_{i} = T_{i}} \to B_{i} = \{؜_{G_{1}^{i + X_{T - 1}}, X_{i} = K_{i}}^{q^{i} G_{1}^{i - X_{K - 1}}, X_{i} = T_{i}}, \sum_{X (T) = g} \prod_{i = 1}^{M} B_{i} = q^{\frac{g (g + 1)}{2}} \prod_{i = 1}^{M} G_{1}^{i} G_{g}^{M}, \\ \to B_{i} = \{؜_{G_{1}^{i + X_{T - 1}}, X_{i} = K_{i}}^{q^{i - X_{T}} G_{1}^{i - X_{K - 1}} = q^{X_{K}} G_{1}^{i - X_{K - 1}}, X_{i} = T_{i}}, \sum_{X (T) = g} \prod_{i = 1}^{M} B_{i} = \prod_{i = 1}^{M} G_{1}^{i} G_{g}^{M} . Promote this conclusion : \end{matrix}

Definition 3.1.

Set

T

come from p Source:

S_{1}, S_{2} . . . S_{p}

.

D i f f (S_{x}, S_{x}) = 0, D i f f (S_{x}, S_{y}) = - D i f f (S_{y}, S_{x}) = 1, x > y

.

D i f f (T_{i}, T_{j}) = D i f f (S_{x}, S_{y}), T_{i} \in S_{x}, T_{j} \in S_{y}

.

Proposition 3.8.

\sum_{g_{1} + . . . + g_{p} = M, g_{i} = | S_{i} |} \prod_{i = 1}^{M} ؜ G_{1}^{T_{i} + \sum_{j < i} D i f f (T_{j}, T_{i})} q^{\sum_{j < i, D i f f (T_{j}, T_{i}) = - 1} 1} = G_{g_{1}, g_{2} . . . g_{p}}^{M} \prod_{i = 1}^{M} G_{1}^{T_{i}}

,

T_{i} \geq i

.

Proof.

\begin{matrix} Record the sum as W_{q} (g_{1}, g_{2}, . . . g_{p}, P T) . \\ W (1, 1, [T_{1}, T_{2}]) = G_{1}^{T_{1}} G_{1}^{T_{2} + 1} + G_{1}^{T_{1}} G_{1}^{T_{2} - 1} q = G_{1}^{T_{1}} G_{1}^{T_{2}} G_{1}^{2}, it ’ s holds . Suppose W_{q} (g_{1}, g_{2}, P T) holds, \\ W_{q} (g_{1}, g_{2} + 1, [P T, T_{M + 1}]) = T_{M + 1} \in S o u r c e_{1} + T_{M + 1} \in S o u r c e_{2} \\ = W_{q} (g_{1}, g_{2}, P T) G_{1}^{T_{M + 1} + g_{1}} + W_{q} (g_{1} - 1, g_{2} + 1, P T) G_{1}^{T_{M + 1} - (g_{2} + 1)} q^{g_{2} + 1} . \\ = (\prod_{i = 1}^{M} G_{1}^{T_{i}}) G_{g_{1}, g_{2}}^{M} G_{1}^{T_{M + 1} + g_{1}} + (\prod_{i = 1}^{M} G_{1}^{T_{i}}) G_{g_{1} - 1, g_{2} + 1}^{M} G_{1}^{T_{M + 1} - (g_{2} + 1)} q^{g_{2} + 1} . \\ Just need to prove : G_{g_{1}}^{M} G_{1}^{T_{M + 1} + g_{1}} + G_{g_{1} - 1}^{M} G_{1}^{T_{M + 1} - (M - g_{1} + 1)} q^{M - g_{1} + 1} = G_{1}^{T_{M + 1}} G_{g_{1}}^{M + 1} . \\ (R i g h t s i d e) \times (q^{M - g_{1} + 1} - 1) / G_{g_{1}}^{M} = (q^{T_{M + 1} - 1} + . . . + q + 1) (q^{M + 1} - 1) = (1) . \\ (L e f t s i d e) \times (q^{M - g_{1} + 1} - 1) / G_{g_{1}}^{M} = (q^{M - g_{1} + 1} - 1) G_{1}^{T_{M + 1} + g_{1}} + (q^{g_{1}} - 1) G_{1}^{T_{M + 1} - (M - g_{1} + 1)} q^{M - g_{1} + 1} \\ = (q^{M - g_{1} + 1} - 1) (q^{T_{M + 1} + g_{1} - 1} + . . . + q + 1) + (q^{g_{1}} - 1) (q^{T_{M + 1} - 1} + . . . + q^{M - g_{1} + 2} + q^{M - g_{1} + 1}) = (2) . \\ (1) - (2) = 0 \to It ’ s holds when p = 2. W_{q} (g_{1}, g_{2} + g_{3}, P T) = G_{g_{1}, g_{2} + g_{3}}^{g_{1} + g_{2} + g_{3}} \prod_{i = 1}^{M} G_{1}^{T_{i}} . \\ Every product has g_{2} + g_{3} factors come from S o u r c e_{2}, divide them to g_{2} \times S o u r c e_{2} + g_{3} \times S o u r c e_{3}, \\ g_{1} - factors are invariant, (g_{2} + g_{3}) - factors are variant . \\ \sum \prod (variant factors) = W_{q} (g_{2}, g_{3}, [X_{1}, X_{2} . . . X_{g_{2} + g_{3}}]) = G_{g_{2}, g_{3}}^{g_{2} + g_{3}} \prod_{i = 1}^{g_{2} + g_{3}} G_{1}^{X_{i}} . \\ W_{q} (g_{1}, g_{2}, g_{3}, [T_{1}, T_{2} . . . T_{M}]) = G_{g_{1}, g_{2} + g_{3}}^{g_{1} + g_{2} + g_{3}} G_{g_{2}, g_{3}}^{g_{2} + g_{3}} \prod_{i = 1}^{M} G_{1}^{T_{i}} = G_{g_{1}, g_{2}, g_{3}}^{g_{1} + g_{2} + g_{3}} \prod_{i = 1}^{M} G_{1}^{T_{i}} . \end{matrix}

□

Proposition 3.9.

K \in N, K > 0, \sum_{n = 0}^{N - 1} q^{K n} G_{M}^{n + M}

= \sum_{g = 0}^{K - 1} q^{\frac{g (g + 1)}{2}} \prod_{i = 1}^{g} (q^{M + i} - 1) G_{g}^{K - 1} G_{1 + M + g}^{N + M}

= q^{- (M + 1) (K - 1)} \sum_{g = 0}^{K - 1} q^{- \frac{g (g - 1)}{2}} \prod_{i = 1}^{g} (q^{M + i} - 1) {[؜_{g}^{K - 1}]}_{q^{- 1}} G_{1 + M + g}^{N + M + g}

= \sum_{g = 0}^{K - 1} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{K - 1} G_{M + K}^{N + M + K - 1 - g}

.

Proof.

\begin{matrix} S U M_{q} (N, [{[1]}_{q -}, {[2]}_{q -} . . . {[M - 1]}_{q -}, 1 : q - 1, 1 : q - 1 . . . 1 : q - 1], [1, 2 . . . M - 1, M + 1, M + 2 . . . M + K - 1]) \\ = {[(M - 1)!]}_{q -} \sum_{g = 0}^{M} q^{n} (1 + q^{n} - 1) . . . (1 + q^{n} - 1) G_{M}^{n + M} = {[(M - 1)!]}_{q -} \sum_{g = 0}^{M} q^{K n} G_{M}^{n + M} . \\ = {[(M - 1)!]}_{q -} S U M_{q} (N, [1, 1 . . . 1] : q - 1, [M + 1, M + 2 . . . M + K - 1]) [(4)] \\ \sum_{g = 0}^{N - 1} q^{K n} G_{M}^{n + M} = S U M_{q} (N, [1, 1 . . . 1] : q - 1, [M + 1, M + 2 . . . M + K - 1]) \\ = \sum_{g = 0}^{K - 1} H_{1}^{q} (g) G_{1 + M + g}^{N + M} = \sum_{g = 0}^{K - 1} H_{2}^{q} (g) G_{1 + M + g}^{N + M + g} = \sum_{g = 0}^{K - 1} H_{3}^{q} (g) G_{M + K}^{N + M + K - 1 - g} . \\ B_{i} = \{؜_{1 + G_{1}^{X_{T - 1}} (q - 1) = q^{X_{T - 1}} = q^{X_{T}}, X_{i} = K_{i}}^{q^{1 + X_{T - 1}} G_{1}^{M + i - X_{K - 1}} (q - 1) = q^{X_{T}} (q^{M + X_{T}} - 1), X_{i} = T_{i}}, H_{1}^{q} (g) = q^{\frac{g (g + 1)}{2}} \prod_{i = 1}^{g} (q^{M + i} - 1) G_{g}^{K - 1} . \\ B_{i} = \{؜_{q^{- (M + 1 + X_{T - 1})} = q^{- (M + 1)} q^{- X_{T}}, X_{i} = K_{i}}^{q^{- (M + X_{T})} (q^{M + X_{T}} - 1), X_{i} = T_{i}}, H_{2}^{q} (g) = q^{- \frac{g (g + 1)}{2} - M g - (M + 1) (K - 1 - g)} \prod_{i = 1}^{g} (q^{M + i} - 1) {[؜_{g}^{K - 1}]}_{q^{- 1}} . \\ B_{i} = \{؜_{q^{X_{T - 1}} = q^{X_{T}}, X_{i} = K_{i}}^{- q^{1 + X_{T - 1}} = - q^{X_{T}}, X_{i} = T_{i}}, H_{3}^{q} (g) = {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g}^{K - 1} . \end{matrix}

□

Definition 3.2.

{〈؜_{M}^{M}〉}^{q} = 0, {〈؜_{g}^{M}〉}^{q} = \sum_{λ_{1} + . . . + λ_{g + 1} = M - g} \prod_{i = 1}^{g + 1} {[i]}_{q}^{λ_{i}} {[1 + λ_{1}]}_{q} {[1 + λ_{1} + λ_{2}]}_{q} . . . {[1 + λ_{1} + . . . + λ_{g}]}_{q}, λ_{i} \geq 0

.

Easy to obtain:

{〈؜_{g}^{M}〉}^{q} = {[M - g]}_{q} {〈؜_{g - 1}^{M - 1}〉}^{q} + {[g + 1]}_{q} {〈؜_{g}^{M - 1}〉}^{q}

, so

{〈؜_{g}^{M}〉}^{q} = {〈؜_{M - g - 1}^{M}〉}^{q}

.

Proposition 3.10.

{[N]}_{q}^{M} = \sum_{g = 1}^{M} {[g!]}_{q} S_{2}^{q} (M, g) q^{\frac{g (g - 3)}{2}} G_{g}^{N} = \sum_{g = 1}^{M} {(- 1)}^{M - g} {[g!]}_{q} S_{2}^{q -} (M, g) q^{\frac{- g (g - 3)}{2}} G_{g}^{N + g - 1} = \sum_{g = 0}^{M - 1} {〈؜_{g}^{M}〉}^{q} q^{(؜_{2}^{M - g})} G_{M}^{N + g}

.

Proof.

\begin{matrix} \nabla_{q}^{1} S U M (N, [{[1]}_{q -}, {[1]}_{q -} . . . {[1]}_{q -}], [1, 2 . . . M]) = \prod (\frac{1}{q} + \frac{q^{N - 1} - 1}{q - 1}) = q^{- M} {[N]}_{q}^{M} \\ = q^{- 1} \nabla_{q}^{1} S U M (N, {[1]}_{q -}, {[1]}_{q -} . . . {[1]}_{q -}], [2, 3 . . . M]) = q^{- 1} \sum_{g = 0}^{M - 1} H_{1}^{q} (g) G_{1 + g}^{N} q^{- g} . = q^{- 1} \sum_{g = 0}^{M - 1} H_{2}^{q} (g) G_{1 + g}^{N + g} . \\ I n H_{1}^{q} (g), B_{i} = \{؜_{q^{- 1} + G_{1}^{X_{T - 1}} = q^{- 1} G_{1}^{1 + X_{T - 1}} = q^{- 1} G_{1}^{1 + X_{T}}, X_{i} = K_{i}}^{q^{1 + X_{T - 1}} G_{1}^{(i + 1) - X_{K - 1}} = q^{X_{T}} G_{1}^{1 + X_{T}} = q^{- 1} q^{1 + X_{T}} G_{1}^{1 + X_{T}}, X_{i} = T_{i}} . \\ H_{1}^{q} (g) = q^{- g} {[(g + 1)!]}_{q} q^{\frac{g (g + 1)}{2}} / q^{1} H_{1}^{q} (g, \sum K) = {[(g + 1)!]}_{q} q^{- (g + 1)} q^{\frac{g (g + 1)}{2}} q^{- (M - 1 - g)} E_{M - 1 - g}^{g + 1, q} . \\ q^{- M} {[N]}_{q}^{M} = q^{- 1} \sum_{g = 0}^{M - 1} q^{- M} {[(g + 1)!]}_{q} q^{\frac{g (g + 1)}{2}} S_{2}^{q} (M, g + 1) G_{1 + g}^{N} q^{- g} \to F o r m_{1} . \\ I n H_{2}^{q} (g), B_{i} = \{؜_{q^{- 1} - q^{- (i + 1 - X_{K - 1})} G_{1}^{i + 1 - X_{K - 1}} = q^{- 1} - q^{- (2 + X_{T})} G_{1}^{2 + X_{T}} = - q^{- 1} q^{- 1 - X_{T}} G_{1}^{1 + X_{T}}, X_{i} = K_{i}}^{q^{- (1 + X_{T})} G_{1}^{1 + X_{T}}, X_{i} = T_{i}} . \\ H_{2}^{q} (g) = {[(g + 1)!]}_{q -} / {[1!]}_{q -} H_{2}^{q} (g, \sum K) = q {[(g + 1)!]}_{q -} {(- 1)}^{M - 1 - g} q^{- (M - 1 - g)} E_{M - 1 - g}^{g + 1, q -} . \\ q^{- M} {[N]}_{q}^{M} = q^{- 1} \sum_{g = 0}^{M - 1} q {[(g + 1)!]}_{q -} {(- 1)}^{M - 1 - g} q^{- (M - 1 - g)} E_{M - 1 - g}^{g + 1, q -} G_{1 + g}^{N + g} \to F o r m_{2} . \\ \nabla_{q}^{1} S U M (N, [{[1]}_{q -}, {[1]}_{q -} . . . {[1]}_{q -}], [1, 2 . . . M]) = \sum_{g = 0}^{M} H_{3}^{q} (g) G_{M}^{N + M - 1 - g} q^{- g} . \\ B_{i} = \{؜_{q^{- 1} + G_{1}^{X_{T - 1}} = q^{- 1} G_{1}^{1 + X_{T - 1}} = q^{- 1} G_{1}^{1 + X_{T}}, X_{i} = K_{i}}^{q (q^{X_{T - 1}} G_{1}^{i} - q^{i} G_{1}^{X_{T - 1}} - q^{i} q^{- 1}) = q \frac{q^{i - 1} - q^{X_{T - 1}}}{q - 1} = q^{X_{T}} G_{1}^{i - X_{T}}, X_{i} = T_{i}}, H_{3}^{q} (g) = q^{- (M - g)} q^{\frac{g (g + 1)}{2}} {〈؜_{g}^{M}〉}^{q}, H_{3}^{q} (M) = 0 . \\ q^{- M} {[N]}_{q}^{M} = \sum_{g = 0}^{M - 1} {〈؜_{g}^{M}〉}^{q} q^{\frac{g (g + 1)}{2}} q^{- (M - g)} q^{- g} G_{M}^{N + M - 1 - g} . \\ {[N]}_{q}^{M} = \sum_{g = 0}^{M - 1} {〈؜_{g}^{M}〉}^{q} q^{\frac{g (g + 1)}{2}} G_{M}^{N + M - 1 - g} = \sum_{g = 0}^{M - 1} {〈؜_{M - 1 - g}^{M}〉}^{q} q^{\frac{(M - 1 - g) (M - g)}{2}} G_{M}^{N + g} . \end{matrix}

□

\begin{matrix} This can be compared to N^{M} = \sum_{g = 1}^{M} g! S_{2} (M, g) (؜_{g}^{N}) = \sum_{g = 1}^{M} {(- 1)}^{M - g} g! S_{2} (M, g) (؜_{g}^{N + g - 1}) = \sum_{g = 0}^{M - 1} 〈؜_{g}^{M}〉 (؜_{M}^{N + g}) . \\ S_{2} (M, g) is Stirling number of the \sec ond kind . S_{2} (M, g) = E_{M - g}^{g}, 〈؜_{g}^{M}〉 is Eulerian number, \\ 〈؜_{g}^{M}〉 = \sum_{λ_{1} + λ_{2} + . . + λ_{g + 1} = M - g} \prod_{i = 1}^{g + 1} i^{λ_{i}} (1 + λ_{1}) (1 + λ_{1} + λ_{2}) . . . (1 + λ_{1} + . . + λ_{g}) [3] . \end{matrix}

Proposition 3.11.

\begin{matrix} q^{M n} = \sum_{g = 0}^{M} G_{g}^{M} \prod_{i = 0}^{g - 1} (q^{n} - q^{i}) = q^{- M} \sum_{g = 0}^{M} q^{g} {[؜_{g}^{M}]}_{q^{- 1}} \prod_{i = 1}^{g} (q^{n} - q^{- i}) = \sum_{g = 0}^{M} G_{g}^{M} {(- 1)}^{g} q^{(؜_{2}^{g})} G_{M}^{n + M - g} . \end{matrix}

Proof.

\begin{matrix} q^{M n} = \nabla_{q}^{1} S U M_{q} (n + 1, [1, 1 . . . 1] : q - 1, [1, 2 . . . M]) \\ = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{g}^{n} q^{- g} = \sum_{g = 0}^{M} H_{2}^{q} (g) G_{g}^{n + g} = \sum_{g = 0}^{M} H_{3}^{q} (g) G_{M}^{n + M - g} q^{- g} . \\ B_{i} = \{؜_{1 + G_{1}^{X_{T - 1}} (q - 1) = q^{X_{T - 1}} = q^{X_{T}}, X_{i} = K_{i}}^{q^{1 + X_{T - 1}} G_{1}^{i - X_{K - 1}} (q - 1) = q^{X_{T}} (q^{X_{T}} - 1), X_{i} = T_{i}}, H_{1}^{q} (g) = {(q - 1)}^{g} {[g!]}_{q +} H_{1}^{q} (g, \sum K) = {(q - 1)}^{g} {[g!]}_{q +} G_{g}^{M} . \\ B_{i} = \{؜_{q^{- (1 + X_{T - 1})} = q^{- 1} q^{- X_{T}}, X_{i} = K_{i}}^{q^{- (1 + X_{T - 1)}} (q^{1 + X_{T - 1}} - 1) = q^{- X_{T}} (q^{X_{T}} - 1), X_{i} = T_{i}}, H_{2}^{q} (g) = {(q - 1)}^{g} {[g!]}_{q -} q^{- (M - g)} {[؜_{g}^{M}]}_{q^{- 1}} . \\ B_{i} = \{؜_{q^{X_{T - 1}} = q^{X_{T}}, X_{i} = K_{i}}^{- q^{1 + X_{T - 1}} = - q^{X_{T}}, X_{i} = T_{i}}, H_{3}^{q} (g) = {(- 1)}^{g} q^{(؜_{2}^{g + 1})} H_{3}^{q} (g, \sum K) = {(- 1)}^{g} q^{(؜_{2}^{g + 1})} G_{g}^{M} . \end{matrix}

□

Proposition 3.12.

(1).

{(K + q D)}^{M} = (q - 1) D \sum_{g = 0}^{M - 1} {(K + D)}^{g} {(K + q D)}^{M - 1 - g} + {(K + D)}^{M}

.

(2).

{(K + D)}^{M} = D \sum_{g = 0}^{M - 1} K^{g} {(K + D)}^{M - 1 - g} + K^{M}

.

(3).

{(K + q D)}^{M} = q (q - 1) D^{2} \sum_{a + b + c = M - 2, a, b, c \geq 0} K^{a} {(K + D)}^{b} {(K + q D)}^{c} + q D \sum_{g = 0}^{M - 1} K^{g} {(K + D)}^{M - 1 - g} + K^{M}

.

Proof.

\begin{matrix} P S = [K + D, K + D . . . K + D] : (q - 1) D, P T = [1, 2 . . . M] . \\ B_{i} = \{؜_{K + D q^{X_{T}}, X_{i} = K_{i}}^{q^{X_{T}} (q^{X_{T}} - 1) D, X_{i} = T_{i}}, H_{1}^{q} (0) = {(K + D)}^{M}, H_{1}^{q} (1) = q^{1} (q^{1} - 1) D \sum_{a + b = M - 1, a, b \geq 0} {(K + D)}^{a} {(K + q D)}^{b} . \\ S U M_{q} (2) - S U M_{q} (1) = H_{1}^{q} (1) + H_{1}^{q} (0) G_{1}^{2} - H_{1}^{q} (0) = H_{1}^{q} (1) + q H_{1}^{q} (0) = q {(K + q D)}^{M} \to (1) . \\ From [1.2] : {(K + D)}^{M} = f (1) G_{1}^{1} + K^{M} \to (2) . \\ From [1.2] : {(K + q D)}^{M} = \sum_{n = 0}^{1} \prod (K + q^{n} D) - \sum_{n = 0}^{0} \prod (K + q^{n} D) = \\ f (2) G_{2}^{2} + f (1) G_{1}^{2} + 2 K^{M} - (f (1) G_{1}^{1} + K^{M}) = f (2) + q f (1) + K^{M} \to (3) . \end{matrix}

□

Proposition 3.13.

(1).

0 \leq A < M, 0 \leq T

,

\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)} = \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).

0 \leq A, B, T, 0 \leq A + B < M

,

\sum_{g = 0}^{M} G_{g}^{M} G_{A}^{A + T + g} G_{B}^{g} q^{(؜_{2}^{g}) - g (A + B)} {(- 1)}^{g} = 0

.

(3).

0 \leq K, T

,

\sum_{g = 0}^{M} {(- 1)}^{g} G_{g}^{M} G_{M + K}^{M + K + T + g} q^{(؜_{2}^{M + 1 - g}) + (M - g) K} = {(- 1)}^{M} G_{K}^{T + M + K}

.

Proof.

\begin{matrix} P S = [{[T + 1]}_{q -}, {[T + 2]}_{q -} . . . {[T + M]}_{q -}], P T = [T + A + 1, T + A + 2 . . . T + A + M]), \\ S U M_{q} (N) = \sum_{g = 0}^{M} G_{A + T + 1 + g}^{N + A + T} {[T + A + 1]}_{q} . . . {[T + A + g]}_{q} \times {[T + g + 1]}_{q} . . . {[T + M]}_{q} q^{\frac{g (1 + g)}{2} - (M - g) (T + M + 1) + (؜_{2}^{M + 1 - g})} G_{g}^{M} [3.7] \\ = \frac{{(q; q)}_{A} {(q^{T + A + 1}; q)}_{M - A}}{{(1 - q)}^{M}} \sum_{g = 0}^{M} G_{A + T + 1 + g}^{N + A + T} q^{\frac{g (1 + g)}{2} - (M - g) (T + M + 1) + (؜_{2}^{M + 1 - g})} G_{A}^{T + A + g} G_{g}^{M} (*) \\ = \prod_{i = 1}^{M - A} {[T + A + i]}_{q -} S U M_{q} (N, [{[T + 1]}_{q -} . . . {[T + A]}_{q -}], [T + M + 1 . . . T + M + A]) . [2.1 (4)] \\ = q^{\frac{- (M - A) (T + A + 1 + T + M)}{2}} \prod_{i = 1}^{M - A} {[T + A + i]}_{q} \sum_{k = 0}^{A} G_{M + T + 1 + k}^{N + M + T} q^{\frac{k (1 + k)}{2} - (A - k) (T + A + 1) + (؜_{2}^{A + 1 - k})} \\ {[T + M + 1]}_{q} . . . {[T + M + k]}_{q} \times {[T + 1 + k]}_{q} . . . {[T + A]}_{q} G_{k}^{A} \\ = q^{\frac{- (M - A) (T + A + 1 + T + M)}{2}} \sum_{k = 0}^{A} G_{M + T + 1 + k}^{N + M + T} q^{\frac{k (1 + k)}{2} - (A - k) (T + A + 1) + (؜_{2}^{A + 1 - k})} \prod_{i = 1}^{M} {[T + k + i]}_{q} G_{k}^{A} . (* *) \\ Compare (*) and (* *) : \\ \sum_{g = 0}^{M} G_{A + T + 1 + g}^{N + A + T} q^{\frac{g (1 + g) - (M - g) (g + M + 1 + 2 T)}{2}} G_{A}^{A + T + g} G_{g}^{M} \\ = q^{\frac{- (M - A) (T + A + 1 + T + M)}{2}} \sum_{k = 0}^{A} G_{M + T + 1 + k}^{N + M + T} q^{\frac{k (1 + k) - (A - k) (k + A + 1 + 2 T)}{2}} \frac{{(q^{T + k + 1}; q)}_{M}}{{(q^{T + A + 1}; q)}_{M - A} {(q; q)}_{k} {(q; q)}_{A - k}} \\ = q^{\frac{- (M - A) (T + A + 1 + T + M)}{2}} \sum_{k = 0}^{A} G_{M + T + 1 + k}^{N + M + T} q^{\frac{k (1 + k) - (A - k) (k + A + 1 + 2 T)}{2}} G_{M + T}^{M + T + k} G_{k + T}^{A + T} \\ Proof of (1) completed, the two sides correspond to merging and unfolding . \\ A + B < M, From [2.4] and (1) : \\ \sum_{g = 0}^{M} G_{g}^{M} G_{A}^{A + T + g} G_{B}^{g} q^{g (1 + g + T)} q^{- g (g + 1) + (؜_{2}^{g - B}) - (A + T) g} {(- 1)}^{g} \\ = \sum_{g = 0}^{M} G_{g}^{M} G_{A}^{A + T + g} G_{B}^{g} q^{- g A + (؜_{2}^{g - B})} {(- 1)}^{g} = 0 \to \sum_{g = 0}^{M} G_{g}^{M} G_{A}^{A + T + g} G_{B}^{g} q^{(؜_{2}^{g}) - g (A + B)} {(- 1)}^{g} = 0 . \\ This is (2), when A = 0, it ’ s [3.2] . Similar expressions about G_{M - B}^{M - g} can also be obtained . \\ S U M_{q} (N, [{[T + 1]}_{q -}, {[T + 2]}_{q -} . . . {[T + M]}_{q -}], [T + K + M + 1, T + K + M + 2 . . . T + K + 2 M]) . \\ H_{1}^{q} (g) = q^{\frac{g (1 + g)}{2} - (M - g) (T + M + 1) + (؜_{2}^{M - g + 1})} \prod_{i = 1}^{g} {[T + K + M + i]}_{q} \prod_{i = 1}^{M - g} {[T + M + 1 - i]}_{q} G_{g}^{M} . \end{matrix}

\begin{matrix} Direct calculations yield H_{2}^{q} (0) = {(- 1)}^{M} q^{- M (T + K + M + 1)} \prod_{i = 1}^{M} {[K + i]}_{q} . \\ H_{2}^{q} (0) = \sum_{g = 0}^{M} {(- 1)}^{g} H_{1}^{q} (g) q^{- g (g + 1) + \frac{g (g - 1)}{2} - (T + K + M) g} . [(3)] \\ {(- 1)}^{M} \prod_{i = 1}^{M} {[K + i]}_{q} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{(؜_{2}^{M - g + 1}) + (M - g) K} \prod_{i = 1}^{g} {[T + K + M + i]}_{q} \prod_{i = 1}^{M - g} {[T + M + 1 - i]}_{q} G_{g}^{M} . \\ {(- 1)}^{M} \prod_{i = 1}^{M} {[K + i]}_{q} \prod_{i = 1}^{K} {[T + M + i]}_{q} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{(؜_{2}^{M - g + 1}) + (M - g) K} \prod_{i = 1}^{K + M} {[T + g + i]}_{q} G_{g}^{M} . \\ {(- 1)}^{M} \prod_{i = 1}^{K} {[T + M + i]}_{q} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{(؜_{2}^{M - g + 1}) + (M - g) K} \prod_{i = 1}^{K + M} {[T + g + i]}_{q} / \prod_{i = 1}^{M} {[K + i]}_{q} \times G_{g}^{M} . \\ {(- 1)}^{M} G_{K}^{T + M + K} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{(؜_{2}^{M - g + 1}) + (M - g) K} \times G_{M + K}^{T + K + M + g} G_{g}^{M} \to (3) . It ’ s the case of A + B \geq M of (2) . \end{matrix}

□

q-Vandermorde: K=T+1+M, A=0, X=N+T,

\sum_{g = 0}^{M} G_{g}^{M} G_{K - (M - g)}^{N + T} q^{g (g + 1 + T)} = \sum_{g = 0}^{M}

G_{g}^{M} G_{K - g}^{X} q^{(M - g) (K - g)} = G_{K}^{X + M}

.

Proposition 3.14.

\sum_{n_{M} = 0}^{N - 1} . . . \sum_{n_{1} = 0}^{n_{2}} \prod_{i = 1}^{M} {[K - 1 + i + 2 n_{i}]}_{q} q^{- \sum_{j = 1}^{M} n_{j}} = q^{- (N - 1) M} \prod_{i = 1}^{M} {[K + N - 2 + i]}_{q} G_{M}^{M + N - 1}

.

Proof.

\begin{matrix} P S = [{[K]}_{q -}, {[K + 1]}_{q -} . . . {[K + M - 1]}_{q -}], P S 1 = [{[K + M - 1]}_{q -}, {[K + M - 2]}_{q -} . . . {[K]}_{q -}], P T = [1, 2 . . . M] . \\ B_{i} = K_{i} o f H_{1}^{q} (g, P S, P T) = q^{- (K - 1 + i)} G_{1}^{K - 1 + i} + G^{X_{T}} = q^{- (K - 1 + i)} G_{1}^{K - 1 + i + X_{T}} . Expand by definition : \\ H_{1}^{q} (g, \sum K, P S, P T) = \sum_{n_{M - g} = 0}^{g} . . . \sum_{n_{1} = 0}^{n_{2}} \prod_{i = 1}^{M - g} {[K - 1 + i + 2 n_{i}]}_{q} q^{- (K - 1) (M - g) - \sum_{j = 1}^{M - g} j - \sum_{j = 1}^{M - g} n_{j}} \\ = H_{1}^{q} (g, \sum K, P S 1, P T) = \prod_{i = 1}^{M - g} {[K + g - 1 + i]}_{q} q^{- (M - g) (K + M)} q^{(؜_{2}^{M + 1 - g})} G_{g}^{M} . [3.7] \\ \sum_{n_{M - g} = 0}^{g} . . . \sum_{n_{1} = 0}^{n_{2}} \prod_{i = 1}^{M - g} {[K - 1 + i + 2 n_{i}]}_{q} q^{g (M - g) - \sum_{j = 1}^{M - g} n_{j}} = \prod_{i = 1}^{M - g} {[K + g - 1 + i]}_{q} G_{g}^{M} . \\ \sum_{n_{M} = 0}^{g} . . . \sum_{n_{1} = 0}^{n_{2}} \prod_{i = 1}^{M} {[K - 1 + i + 2 n_{i}]}_{q} q^{g M - \sum_{j = 1}^{M} n_{j}} = \prod_{i = 1}^{M} {[K + g - 1 + i]}_{q} G_{g}^{M + g} . \\ \sum_{n_{M} = 0}^{N - 1} . . . \sum_{n_{1} = 0}^{n_{2}} \prod_{i = 1}^{M} {[K - 1 + i + 2 n_{i}]}_{q} q^{(N - 1) M - \sum_{j = 1}^{M} n_{j}} = \prod_{i = 1}^{M} {[K + N - 2 + i]}_{q} G_{N - 1}^{M + N - 1} . \end{matrix}

□

K = 1, M = 1 \to \sum_{n = 0}^{N - 1} {[1 + 2 n]}_{q} q^{- n} = q^{- (N - 1)} {[N]}_{q}^{2}

.

4. The Extension of q-Euler Polynomials and the Relationship Between Three Forms

In this section,

a \neq 1, q^{- 1}, q^{- 2}, . . ., q^{- M}

.

Lemma 4.1.

\sum_{n = 0}^{N - 1} a^{n} G_{M}^{n + A} = - a^{N} \sum_{g = 0}^{M} \frac{q^{(N + A - M) g} G_{M - g}^{N + A - 1 - g}}{{(a; q)}_{g + 1}} + \frac{a^{M - A}}{{(a; q)}_{M + 1}}, 0 \leq A \leq M, N > M - A .

Proof.

\begin{matrix} A = M, M = 0, \sum_{n = 0}^{N - 1} a^{n} = \frac{- a^{N}}{1 - a} + \frac{1}{1 - a}, holds . \\ A = M, M = 1, \sum_{n = 0}^{N - 1} a^{n} G_{1}^{n + 1} = \sum_{n = 0}^{N - 1} a^{n} (1 + q + . . . + q^{n}) \\ = (1 + q + . . . + q^{n}) \sum_{n = 0}^{N - 1} a^{n} - q \sum_{n = 0}^{0} a^{n} - q^{2} \sum_{n = 0}^{1} a^{n} . . . - q^{N - 1} \sum_{n = 0}^{N - 2} a^{n} \\ = G_{1}^{N} \frac{1 - a^{N}}{1 - a} - \frac{q (1 - a)}{1 - a} - \frac{q^{2} (1 - a^{2})}{1 - a} . . . \frac{q^{N - 1} (1 - a^{N - 1})}{1 - a} \\ = G_{1}^{N} \frac{1 - a^{N}}{1 - a} + \frac{1 + a q + . . . + a^{N - 1} q^{N - 1}}{1 - a} - \frac{1 + q + . . . + q^{N - 1}}{1 - a} = \frac{G_{1}^{N} - G_{1}^{N} a^{N}}{1 - a} + \frac{1 - a^{N} q^{N}}{(1 - a) (1 - a q)} - \frac{G_{1}^{N}}{1 - a} \\ = \frac{- a^{N} q^{N}}{(1 - a) (1 - a q)} + \frac{- a^{N} G_{1}^{N}}{1 - a} + \frac{1}{(1 - a) (1 - a q)}, holds . \\ A = M, N = 1, - a^{1} \sum_{g = 0}^{M} \frac{q^{g}}{{(a; q)}_{g + 1}} + \frac{1}{{(a; q)}_{M + 1}} = \frac{1 - a q^{M}}{{(a; q)}_{M + 1}} - a^{1} \sum_{g = 0}^{M - 1} \frac{q^{g}}{{(a; q)}_{g + 1}} \\ = \frac{(1 - a q^{M}) (1 - a q^{M - 1})}{{(a; q)}_{M + 1}} - a^{1} \sum_{g = 0}^{M - 2} \frac{q^{g}}{{(a; q)}_{g + 1}} = . . . = 1, holds . Suppose it ’ s holds when N . \\ \sum_{n = 0}^{N} a^{n} G_{M}^{n + M} = \sum_{n = 0}^{N} a^{n} (q^{M} G_{M}^{n - 1 + M} + G_{M - 1}^{n - 1 + M}) = a q^{M} \sum_{n = 0}^{N - 1} a^{n} G_{M}^{n + M} + \sum_{n = 0}^{N} a^{n} G_{M - 1}^{n + M - 1} \\ = - q^{M} a^{N + 1} \sum_{g = 0}^{M} \frac{q^{N g} G_{M - g}^{N + M - 1 - g}}{{(a; q)}_{g + 1}} + \frac{a q^{M}}{{(a; q)}_{M + 1}} - a^{N + 1} \sum_{g = 0}^{M - 1} \frac{q^{(N + 1) g} G_{M - 1 - g}^{N + M - 1 - g}}{{(a; q)}_{g + 1}} + \frac{1}{{(a; q)}_{M}} \\ = - a^{N + 1} \sum_{g = 0}^{M} \frac{q^{N g + M} G_{M - g}^{N + M - 1 - g} + q^{(N + 1) g} G_{M - 1 - g}^{N + M - 1 - g}}{{(a; q)}_{g + 1}} + \frac{1}{{(a; q)}_{M + 1}} \\ = - a^{N + 1} \sum_{g = 0}^{M} \frac{q^{(N + 1) g} q^{M - g} G_{M - g}^{N + M - 1 - g} + q^{(N + 1) g} G_{M - 1 - g}^{N + M - 1 - g}}{{(a; q)}_{g + 1}} + \frac{1}{{(a; q)}_{M + 1}} \\ = - a^{N + 1} \sum_{g = 0}^{M} \frac{q^{(N + 1) g} G_{M - g}^{(N + 1) + M - 1 - g}}{{(a; q)}_{g + 1}} + \frac{1}{{(a; q)}_{M + 1}} . Proof of A = M completed . \\ \sum_{n = 0}^{N - 1} a^{n} G_{M}^{n + A} = a^{M - A} \sum_{n = 0}^{N - M + A - 1} a^{n} G_{M}^{n + M}, complete the remaining proofs . \end{matrix}

□

\begin{matrix} Taking the limit q \to 1, \sum_{n = 0}^{N - 1} a^{n} (؜_{M}^{n + A}) = - a^{N} \sum_{g = 0}^{M} \frac{(؜_{M - g}^{N + A - 1 - g})}{{(1 - a)}^{g + 1}} + \frac{a^{M - A}}{{(1 - a)}^{M + 1}}, 0 \leq A \leq M . \\ \sum_{n = 0}^{N - 1} q^{n} G_{M}^{n + A} = q^{M - A} G_{M + 1}^{N + A} \to {(q^{N + A - M}; q)}_{M + 1} = - \sum_{g = 0}^{M} q^{(N + A - M) (g + 1)} G_{M - g}^{N + A - 1 - g} {(q^{g + 2}; q)}_{M - g} + 1 . \end{matrix}

Theorem 4.1.

X = T_{M} - M - y \geq - 1, 0 \leq Y \leq 1

,

f (g) = {(a q^{2 + X + g}; q)}_{M - g} = {(a q^{2 + T_{M} - M - y + g}; q)}_{M - g}

,

(1).

\sum_{g = 0}^{M} H_{1}^{q} (g) a^{g} q^{- y g} f (g) = \sum_{g = 0}^{M} H_{2}^{q} (g) f (g) = \sum_{g = 0}^{M} H_{3}^{q} (g) a^{g} q^{- y g}

,define as

A_{a}^{q} (P S, P T, y)

.

(2).

\sum_{n = 0}^{N - 1} a^{n} \nabla_{q}^{y} S U M_{q} (n + Y) = - a^{N} \sum_{k = 0}^{M} \frac{q^{(N + Y - 1) k} \nabla_{q}^{y + k} S U M_{q} (N + Y - 1)}{{(a; q)}_{k + 1}} + \frac{a^{1 - Y} A_{a}^{q} (P S, P T, y)}{{(a; q)}_{T_{M} + 2 - y}}

.

(3).

| a |, | q | < 1, \sum_{n = 0}^{\infty} a^{n} \nabla_{q}^{y} S U M_{q} (n + Y) = \frac{a^{1 - Y} A_{a}^{q} (P S, P T, y)}{{(a; q)}_{T_{M} + 2 - y}}

.

(4).

S U M_{q} (\infty) = \frac{A_{q}^{q} (P S, P T, 1)}{{(q; q)}_{T_{M} + 1}}, \nabla_{q}^{y} S U M_{q} (\infty) = \frac{A_{q}^{q} (P S, P T, 1 + y)}{{(q; q)}_{T_{M} + 1 - y}}

.

Proof.

\begin{matrix} \sum_{n = 0}^{N - 1} a^{n} \nabla_{q}^{y} S U M_{q} (n + Y) = \sum_{g = 0}^{M} H_{1}^{q} (g) \sum_{n = 0}^{N - 1} a^{n} G_{1 + X + g}^{n + Y + X} q^{- y g} \\ = \sum_{g = 0}^{M} H_{1}^{q} (g) (- a^{N} \sum_{k = 0}^{1 + X + g} \frac{q^{(N + Y - 1 - g) k} q^{- y g} G_{1 + X + g - k}^{N + Y - 1 + X - k}}{{(a; q)}_{k + 1}} + \frac{a^{1 + g - Y} q^{- y g}}{{(a; q)}_{2 + X + g}}) . \\ = - a^{N} \sum_{k = 0}^{M} \frac{q^{(N + Y - 1) k} \sum_{g = 0}^{M} H_{1}^{q} (g) q^{- (y + k) g} G_{1 + X + g - k}^{N + Y - 1 + X - k}}{{(a; q)}_{k + 1}} + \frac{\sum_{g = 0}^{M} H_{1}^{q} (g) a^{1 + g - Y} q^{- y g} f (g)}{{(a; q)}_{T_{M} + 2 - y}} \\ = - a^{N} \sum_{k = 0}^{M} \frac{q^{(N + Y - 1) k} \nabla_{q}^{y + k} S U M_{q} (N + Y - 1)}{{(a; q)}_{k + 1}} + \frac{\sum_{g = 0}^{M} H_{1}^{q} (g) a^{1 + g - Y} q^{- y g} f (g)}{{(a; q)}_{T_{M} + 2 - y}} . \\ \sum_{n = 0}^{N - 1} a^{n} \nabla_{q}^{y} S U M_{q} (n + Y) = \sum_{g = 0}^{M} H_{2}^{q} (g) \sum_{n = 0}^{N - 1} a^{n} G_{1 + X + g}^{n + Y + X + g} \\ = \sum_{g = 0}^{M} H_{2}^{q} (g) (- a^{N} \sum_{k = 0}^{1 + X + g} \frac{q^{(N + Y - 1) k} G_{1 + X + g - k}^{N + Y - 1 + X + g - k}}{{(a; q)}_{k + 1}} + \frac{a^{1 - Y}}{{(a; q)}_{2 + X + g}}) . \\ = - a^{N} \sum_{k = 0}^{M} \frac{q^{(N + Y - 1) k} \nabla_{q}^{y + k} S U M_{q} (N + Y - 1)}{{(a; q)}_{k + 1}} + \frac{a^{1 - Y} \sum_{g = 0}^{M} H_{2}^{q} (g) f (g)}{{(a; q)}_{T_{M} + 2 - y}} . \\ \sum_{n = 0}^{N - 1} a^{n} \nabla_{q}^{y} S U M_{q} (n + Y) = \sum_{g = 0}^{M} H_{3}^{q} (g) \sum_{n = 0}^{N - 1} a^{n} G_{T_{M} + 1 - p}^{n + Y + T_{M} - p - g} q^{- y g} \\ = \sum_{g = 0}^{M} H_{3}^{q} (g) (- a^{N} \sum_{k = 0}^{T_{M} + 1 - p} \frac{q^{(N + Y - 1) k} q^{- (y + k) g} G_{T_{M} + 1 - p - k}^{N + Y - 1 + T_{M} - p - g - k}}{{(a; q)}_{T_{M} + 2 - p}} + \frac{a^{1 + g - Y} q^{- y g}}{{(a; q)}_{T_{M} + 2 - y}}) . \\ = - a^{N} \sum_{k = 0}^{M} \frac{q^{(N + Y - 1) k} \nabla_{q}^{y + k} S U M_{q} (N + Y - 1)}{{(a; q)}_{T_{M} + 2 - y}} + \frac{\sum_{g = 0}^{M} H_{3}^{q} (g) a^{1 + g - Y} q^{- y g}}{{(a; q)}_{T_{M} + 2 - y}} . \\ The three summations are identical, which leads to (1) and (2) . (3) is obvious . \\ S U M_{q} (N) = \sum_{n = 0}^{N - 1} q^{n} \nabla_{q}^{1} S U M_{q} (n + 1) \to (4) . {lim}_{N \to \infty} G_{M}^{N} = \frac{1}{{(q; q)}_{M}} can draw the same conclusion . \end{matrix}

□

\sum_{n = 0}^{\infty} a^{n} {[n]}_{q}^{M} = \frac{E_{M} (a, q)}{{(a; q)}_{M + 1}}

,

E_{M} (a, q)

is q-Eularian polynomials.[2] pp 332. From [3.10], we can get three expressions for

E_{M} (a, q)

. In particular, we can get expressions for Eularian polynomials:

\sum_{n \geq 1} t^{n} n^{M} = \frac{t A_{M} (t)}{{(1 - t)}^{M + 1}}

.

A_{M} (a) = \sum_{g = 0}^{M} H_{1} (g) a^{g} {(1 - a)}^{M - g} = \sum_{g = 0}^{M} H_{2} (g) {(1 - a)}^{M - g} = \sum_{g = 0}^{M} H_{3} (g) a^{g}

= \sum_{g = 0}^{M} g! S_{2} (M, g) a^{g} {(1 - a)}^{M - g} = \sum_{g = 0}^{M} {(- 1)}^{M - g} g! S_{2} (M, g) {(1 - a)}^{M - g} = \sum_{g = 0}^{M} 〈؜_{g}^{M}〉 a^{g} .

More clearly, we can reformulate (1) as: if

\sum_{g = 0}^{M} a_{g} G_{1 + p + g}^{X} = \sum_{g = 0}^{M} b_{g} G_{1 + p + g}^{X + g} = \sum_{g = 0}^{M} c_{g} G_{1 + p + M}^{X + M - g}

, then

\sum_{g = 0}^{M} a_{g} q^{- y g} a^{g} {(a q^{2 + p - y + g}; q)}_{M - g} = \sum_{g = 0}^{M} b_{g} {(a q^{2 + p - y + g}; q)}_{M - g} = \sum_{g = 0}^{M} c_{g} q^{- y g} a^{g}, 2 + p - y > 0

.

At [2.4], some relationships have been obtained, and now the remaining ones can be deduced:

Theorem 4.2.

X \in N

, if

\sum_{g = 0}^{M} b_{g} G_{1 + p + g}^{X + g} = \sum_{g = 0}^{M} c_{g} G_{1 + p + M}^{X + M - g}

, then

(1).

c_{g} q^{- p g} = {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} \sum_{k = 0}^{M - g} b_{k} G_{g}^{M - k} q^{g k} = {(- 1)}^{g} q^{\frac{g (g + 3)}{2} + M g} \sum_{k = g}^{M} b_{M - k} G_{g}^{k} q^{- g k}

.

(2).

b_{M - g} = {(- 1)}^{g} q^{\frac{g (g - 1)}{2}} \sum_{k = g}^{M} c_{k} q^{- p k} G_{g}^{k} q^{- (M + 1) k}

,

b_{g} = {(- 1)}^{M - g} q^{\frac{(M - g) (M - g - 1)}{2}} \sum_{k = M - g}^{M} c_{k} q^{- p k}

G_{M - g}^{k} q^{- (M + 1) k}

.

Proof.

\begin{matrix} At [], P T = [1, 2 . . . M], f (g) = (1 - a q^{2 + g}) (1 - a q^{3 + g}) . . . (1 - a q^{M + 1}), \\ \sum_{g = 0}^{M} b_{g} f (g) = \sum_{g = 0}^{M} b_{g} \sum_{k = 0}^{M - g} G_{k}^{M - g} {(- a)}^{k} q^{\frac{g (g - 1)}{2} + (2 + g) k} = \sum_{g = 0}^{M} c_{g} a^{g} q^{- p g} . \\ Compare a^{g} on both sides and yields (1) . \\ \sum_{x = g}^{M} c_{x} = \sum_{x = g}^{M} {(- 1)}^{x} q^{\frac{x (x + 1)}{2} + (M + p + 1) x} \sum_{k = x}^{M} b_{M - k} G_{x}^{k} q^{- x k} . \\ \sum_{x = g}^{M} c_{x} G_{g}^{x} q^{- \frac{x (x + 1)}{2} - (M + p + 1) x} q^{\frac{(x - g) (x - g - 1)}{2}} = \sum_{x = g}^{M} c_{x} G_{g}^{x} q^{\frac{g (g + 1)}{2} - (M + p + 2) x - g x} . \\ \sum_{x = g}^{M} c_{x} G_{g}^{x} q^{\frac{g (g + 1)}{2} - (M + p + 2) x - g x} q^{g x + x} = \sum_{x = g}^{M} {(- 1)}^{x} G_{g}^{x} q^{\frac{(x - g) (x - g - 1)}{2} + g x + x} \sum_{k = x}^{M} b_{M - k} G_{x}^{k} q^{- x k} . \\ \sum_{x = g}^{M} c_{x} G_{g}^{x} q^{\frac{g (g + 1)}{2} - (M + p + 1) x} = \sum_{k = x}^{M} b_{M - k} \sum_{x = g}^{k} {(- 1)}^{x} G_{g}^{x} G_{x}^{k} q^{\frac{(x - g) (x - g - 1)}{2}} q^{- (k - g - 1) x} = {(- 1)}^{k} b_{M - k} q^{k} . [(3)] \end{matrix}

□

Theorem 4.3.

0 \leq g \leq M,

(1).

a_{g} = \sum_{k = g}^{M} b_{k} G_{g}^{k} \leftrightarrow b_{g} = \sum_{k = g}^{M} {(- 1)}^{k + g} G_{g}^{k} q^{(؜_{2}^{k - g})} a_{k}

.

(2).

a_{g} = \sum_{k = 0}^{g} c_{k} G_{M - g}^{M - k} q^{- (g - k) k} \leftrightarrow c_{g} = \sum_{k = 0}^{g} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{- (g - k) k - (؜_{2}^{g - k})} a_{k}

.

(3).

c_{g} = \sum_{k = g}^{M} b_{M - k} G_{g}^{k} q^{- g k} \leftrightarrow b_{M - g} q^{- \frac{g (g - 1)}{2}} = \sum_{k = g}^{M} {(- 1)}^{k + g} c_{k} G_{g}^{k} q^{\frac{k (k + 1)}{2}}

.

Proof.

\begin{matrix} By combining [2.4] and [4.2], the inversion formula can be obtained . \\ a_{g} q^{- p g} = \sum_{k = 0}^{g} c_{k} G_{M - g}^{M - k} q^{(g + 1) (g - k)} q^{- p k}, c_{g} q^{- p g} = \sum_{k = 0}^{g} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} a_{k} q^{- p k} . \\ Replace a_{g} with a_{g} q^{- p g}, c_{g} with c_{g} q^{- p g}, then \\ a_{g} = \sum_{k = 0}^{g} c_{k} G_{M - g}^{M - k} q^{(g + 1) (g - k)} = q^{(g + 1) g} \sum_{k = 0}^{g} c_{k} G_{M - g}^{M - k} q^{- (g + 1) k} = q^{(g + 1) g} \sum_{k = 0}^{g} q^{- (g - k) k} G_{M - g}^{M - k} c_{k} q^{- (k + 1) k}, \\ c_{g} = \sum_{k = 0}^{g} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{(g + 1) (g - k) - (؜_{2}^{g - k})} a_{k} = q^{(g + 1) g} \sum_{k = 0}^{g} {(- 1)}^{k + g} G_{M - g}^{M - k} q^{- (g - k) k - (؜_{2}^{g - k})} a_{k} q^{- (k + 1) k} . \\ Replace a_{g} with a_{g} q^{- (g + 1) g}, c_{g} with c_{g} q^{- (g + 1) g} \to (2) . Other proofs are similar . \end{matrix}

□

Theorem 4.4.

1 + p \geq 0

. If

\sum_{g = 0}^{M} a_{g} G_{1 + p + g}^{X} = \sum_{g = 0}^{M} b_{g} G_{1 + p + g}^{X + g} = \sum_{g = 0}^{M} c_{g} G_{1 + p + M}^{X + M - g}, a_{g}^{*} = a_{g} q^{- p g}

,

c_{g}^{*} = c_{g} p^{- p g}

, then

(1).

\sum_{g = 0}^{M} a_{g}^{*} q^{- \frac{g (g + 1)}{2} + A g} a^{M - g} z^{g} = a^{M} \sum_{g = 0}^{M} b_{g} {(- q^{A + 1} \frac{z}{a}; q)}_{g} = \sum_{g = 0}^{M} c_{g}^{*} q^{- \frac{g (g + 1)}{2} + A g} a^{M - g} z^{g} {(- q^{A + 1} \frac{z}{a}; q)}_{M - g}

.

(2).

\sum_{g = 0}^{M} b_{g} q^{A g} a^{M - g} z^{g} = a^{M} \sum_{g = 0}^{M} {(- 1)}^{g} a_{g}^{*} q^{- \frac{g (g + 3)}{2}} {(q^{A - g + 1} \frac{z}{a}; q)}_{g} = q^{A M} z^{M} \sum_{g = 0}^{M} c_{g}^{*} q^{- (M + 1) g} {(q^{- A} \frac{a}{z}; q)}_{g}

.

(3).

\sum_{g = 0}^{M} c_{g}^{*} q^{A g} a^{M - g} z^{g} = \sum_{g = 0}^{M} a_{g}^{*} q^{A g} z^{g} a^{M - g} {(q^{A + 2 + g} \frac{z}{a}; q)}_{M - g} = a^{M} \sum_{g = 0}^{M} b_{g} {(q^{A + 2 + g} \frac{z}{a}; q)}_{M - g}

.

Proof.

From [2.4] and [4.2]:

\begin{matrix} \sum_{g = 0}^{M} a_{g}^{*} q^{- g (g + 1)} q^{\frac{g (g - 1)}{2} + (A + 1) g} a^{M - g} z^{g} = \sum_{g = 0}^{M} \sum_{k = 0}^{M} b_{k} G_{g}^{k} q^{\frac{g (g - 1)}{2} + (A + 1) g} a^{M - g} z^{g} \\ = \sum_{k = 0}^{M} b_{k} a^{M - k} \sum_{g = 0}^{M} G_{g}^{k} q^{\frac{g (g - 1)}{2} + (A + 1) g} a^{k - g} z^{g} = \sum_{k = 0}^{M} b_{k} a^{M - k} \prod_{i = 1}^{k} (a + q^{A + i} z) . \\ \sum_{g = 0}^{M} a_{g}^{*} a^{M - g} z^{g} = \sum_{g = 0}^{M} \sum_{k = 0}^{M} c_{k}^{*} G_{M - g}^{M - k} q^{(g + 1) (g - k)} a^{M - g} z^{g}, q^{(g + 1) (g - k)} = q^{(؜_{2}^{g - k}) + \frac{g^{2} + 3 g - k^{2} - 3 k}{2}} . \\ \sum_{g = 0}^{M} a_{g}^{*} q^{- \frac{g (g + 3)}{2} + (A + 1) g} a^{M - g} z^{g} = \sum_{k = 0}^{M} c_{k}^{*} q^{- \frac{k (k + 3)}{2} + (A + 1) k} z^{k} \sum_{g = 0}^{M} G_{g - k}^{M - k} q^{(؜_{2}^{g - k}) + (A + 1) (g - k)} a^{(M - k) - (g - k)} z^{g - k} \\ = \sum_{k = 0}^{M} c_{k}^{*} q^{- \frac{k (k + 1)}{2} + A k} z^{k} \prod_{i = 1}^{M - k} (a + q^{A + i} z) \to (1) . \\ \sum_{g = 0}^{M} b_{g} q^{A g} a^{M - g} z^{g} = \sum_{g = 0}^{M} q^{A g} a^{M - g} z^{g} \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{g}^{k} q^{- k (k + 1) + \frac{(k - g) (k - g - 1)}{2}} a_{k}^{*} \\ = \sum_{k = 0}^{M} {(- 1)}^{k} a_{k}^{*} q^{- \frac{k (k + 3)}{2}} a^{M - k} \sum_{g = 0}^{M} G_{g}^{k} q^{(؜_{2}^{g}) + (A - k + 1) g} a^{k - g} {(- z)}^{g} = \sum_{k = 0}^{M} {(- 1)}^{k} a_{k}^{*} q^{- \frac{k (k + 3)}{2}} a^{M - k} \prod_{i = 1}^{k} (a - q^{A - k + i} z) . \\ \sum_{g = 0}^{M} b_{g} q^{A g} a^{M - g} z^{g} = \sum_{g = 0}^{M} q^{A g} a^{M - g} z^{g} {(- 1)}^{M - g} q^{\frac{(M - g) (M - g - 1)}{2}} \sum_{k = 0}^{M} G_{M - g}^{k} q^{- (M + 1) k} c_{k}^{*} \\ = \sum_{k = 0}^{M} q^{- (M + 1) k} c_{k}^{*} \sum_{g = 0}^{M} a^{M - g} z^{g} {(- 1)}^{M - g} G_{M - g}^{k} q^{\frac{(M - g) (M - g - 1)}{2} + A g}, replace g with M - g \to \\ = \sum_{k = 0}^{M} q^{- (M + 1) k} c_{k}^{*} \sum_{g = 0}^{M} z^{M - g} {(- a)}^{g} G_{g}^{k} q^{\frac{g (g - 1)}{2} + A (M - g)} = q^{A M} \sum_{k = 0}^{M} q^{- (M + 1) k} c_{k}^{*} z^{M - k} \prod_{i = 1}^{k} (z - q^{- A - 1 + i} a) \to (2) . \\ \sum_{g = 0}^{M} c_{g}^{*} q^{A g} a^{M - g} z^{g} = \sum_{g = 0}^{M} \sum_{k = 0}^{M} {(- 1)}^{k + g} a_{k}^{*} G_{g - k}^{M - k} q^{\frac{g (g + 3)}{2} - \frac{k (k + 3)}{2} + A g} a^{M - g} z^{g} \\ = \sum_{k = 0}^{M} {(- 1)}^{k} a_{k}^{*} q^{- \frac{k (k + 3)}{2}} \sum_{g = 0}^{M} {(- 1)}^{g} G_{g - k}^{M - k} q^{\frac{g (g + 3)}{2} + A g} a^{M - g} z^{g}, replace g with g + k \to \\ = \sum_{k = 0}^{M} {(- 1)}^{k} a_{k}^{*} q^{- \frac{k (k + 3)}{2}} \sum_{g = 0}^{M} {(- 1)}^{g + k} G_{g}^{M - k} q^{\frac{(g + k) (g + k + 3)}{2} + A (g + k)} z^{g + k} a^{M - k - g} \\ = \sum_{k = 0}^{M} a_{k}^{*} q^{A k} z^{k} \sum_{g = 0}^{M} G_{g}^{M - k} q^{\frac{g (g - 1)}{2} + (A + k + 2) g} {(- z)}^{g} a^{M - k - g} = \sum_{k = 0}^{M} a_{k}^{*} q^{A k} z^{k} \prod_{i = 1}^{M - k} (a - q^{A + k + 1 + i} z) . \\ \sum_{g = 0}^{M} c_{g}^{*} q^{A g} a^{M - g} z^{g} = \sum_{k = 0}^{M} b_{k} \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} G_{g}^{M - k} q^{g k + A g} a^{M - g} z^{g} \\ = \sum_{k = 0}^{M} b_{k} a^{k} \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g - 1)}{2} + (A + k + 2) g} G_{g}^{M - k} a^{M - k - g} z^{g} = \sum_{k = 0}^{M} b_{k} a^{k} \prod_{i = 1}^{M - k} (a - q^{A + k + 1 + i} z) \to (3) . \end{matrix}

□

\begin{matrix} a \neq z \neq 0, if \sum_{g = 0}^{M} a_{g} (؜_{1 + p + g}^{X}) = \sum_{g = 0}^{M} b_{g} (؜_{1 + p + g}^{X + g}) = \sum_{g = 0}^{M} c_{g} (؜_{1 + p + M}^{X + M - g}), then \\ \sum_{g = 0}^{M} a_{g} a^{M - g} z^{g} = \sum_{g = 0}^{M} b_{g} a^{M - g} {(a + z)}^{g} = \sum_{g = 0}^{M} c_{g} z^{g} {(a + z)}^{M - g} . \\ From [4.4] 4.1 [] 4.4 [] and the mutual transformation of a_{g}, b_{g}, c_{g} [2.4] [4.2], many equations can be obtained . \\ We can arbitrarily specify a set of values, calculate the other two sets, and obtain the corresponding relationships . \\ \sum_{g = 0}^{M} a_{g} G_{1 + p + g}^{X} = \sum_{g = 0}^{M} b_{g} G_{1 + p + g}^{X + g} = \sum_{g = 0}^{M} c_{g} G_{1 + p + M}^{X + M - g} . 1 + p \geq 0 . \\ a_{M} = c_{M} = 1, a_{g < M} = c_{g < M} = 0 \to b_{g} = {(- 1)}^{M + g} G_{g}^{M} q^{- M (1 + M) + \frac{(M - g) (M - g - 1)}{2}} \\ \to \sum_{g = 0}^{M} b_{g} {(- q^{A + 1} \frac{z}{a}; q)}_{g} = q^{A M - \frac{M (M + 1)}{2}} z^{M} a^{- M}; \sum_{g = 0}^{M} b_{g} {(q^{A + g + 2} \frac{z}{a}; q)}_{M - g} = q^{A M} z^{M} a^{- M} . \\ b_{M} = 1, b_{g < M} = 0 \to a_{g} = q^{g (g + p + 1)} G_{g}^{M} \\ \to \sum_{g = 0}^{M} a_{g} z^{g} {(z q^{2 + p + g}; q)}_{M - g} = 1 . \sum_{g = 0}^{M} {(- 1)}^{g} a_{g} q^{- p g} q^{- \frac{g (g + 3)}{2}} {(q^{A - g + 1} \frac{z}{a}; q)}_{g} = q^{A M} z^{M} a^{- M} . \\ a_{0}^{*} = b_{0} = 1, a_{g > 0}^{*} = b_{g > 0} = 0 \to c_{g}^{*} = {(- 1)}^{g} q^{\frac{g (g + 1)}{2} + g} G_{g}^{M} [] \to \sum_{g = 0}^{M} c_{g}^{*} q^{- (M + 1) g} {(q^{- A} \frac{a}{z}; q)}_{g} = q^{- A M} z^{- M} a^{M} . \\ \to \sum_{k = 0}^{g} {(- 1)}^{k} G_{k}^{M} G_{M - g}^{M - k} q^{\frac{k (k + 1)}{2} - k g} = 0, g > 0; \sum_{k = g}^{M} {(- 1)}^{k} G_{k}^{M} G_{g}^{k} q^{\frac{k (k + 1)}{2} - M k} = 0, g < M \to special case of [3.2 (3) (4)] . \\ b_{g} = {(- 1)}^{g} G_{g}^{M}, a_{g}^{*} = q^{g (g + 1)} \sum_{k = 0}^{M} b_{k} G_{g}^{k}, [3.5] will lead to some special identities . \\ b_{g} = q^{g}, \sum_{k = 0}^{M} G_{g}^{k} q^{k - g} = G_{g + 1}^{M + 1} \to a_{g}^{*} = q^{g (g + 1)} \sum_{k = g}^{M} q^{k} G_{g}^{k} = q^{g (g + 2)} G_{g + 1}^{M + 1} . \\ calculate b_{g} using a_{g}^{*} \to \sum_{k = 0}^{M} {(- 1)}^{k + g} G_{g}^{k} q^{(؜_{2}^{k - g})} q^{k} G_{k + 1}^{M + 1} = q^{g} . \\ \sum_{g = 0}^{M} q^{g} {(q^{1 + g}; q)}_{M - g} = \sum_{g = 0}^{M} q^{g (g + 1)} G_{g + 1}^{M + 1} {(q^{1 + g}; q)}_{M - g} = {(q; q)}_{M + 1} \sum_{g = 0}^{M} \frac{q^{g (g + 1)} G_{g}^{M}}{{(q; q)}_{g + 1}} . \end{matrix}

\begin{matrix} From [(1)], G_{M + 1}^{N + M} = \sum_{g = 0}^{M} q^{(g + 1) g} G_{g}^{M} G_{1 + g}^{N} \to \sum_{g = 0}^{M} \frac{q^{g (g + 1)} G_{g}^{M}}{{(q; q)}_{g + 1}} = \frac{1}{{(q; q)}_{M + 1}} \\ \to \sum_{g = 0}^{M} q^{g} {(q^{1 + g}; q)}_{M - g} = 1 \to \sum_{g = 0}^{M} \frac{q^{g}}{{(q; q)}_{g}} = \frac{1}{{(q; q)}_{M}} (*) . It is a known formula [2] \\ It can be proven through induction that : \sum_{k = 0}^{M} G_{g}^{M - k} q^{(g + 1) k} = G_{g + 1}^{M + 1} \to c_{g}^{*} = {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} G_{g + 1}^{M + 1} \\ \to \sum_{g = 0}^{M} c_{g}^{*} q^{- g} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} G_{g + 1}^{M + 1} = - \sum_{g = 1}^{M + 1} {(- 1)}^{g} q^{\frac{g (g - 1)}{2}} G_{g}^{M + 1} = 1 - {(q^{0}; q)}_{M + 1} = 1, get (*) again . \\ \sum_{g = 0}^{M} c_{g}^{*} q^{- g} a^{g} = a^{- 1} - a^{- 1} {(a; q)}_{M + 1}, [(2)] \to 1 + \sum_{g = 0}^{M} \frac{a q^{g}}{{(a; q)}_{g + 1}} = \frac{1}{{(a; q)}_{M + 1}}, generalization of (*) [2] pp 113 \\ a_{g}^{*} and c_{g}^{*} \to G_{g + 1}^{M + 1} = \sum_{k = 0}^{M} {(- 1)}^{k} G_{M - g}^{M - k} G_{k + 1}^{M + 1} q^{\frac{k (k + 1)}{2}} = \sum_{k = 0}^{M} {(- 1)}^{k} G_{M - g}^{M - k} G_{k + 1}^{M + 1} q^{\frac{k (k + 1)}{2} - g (k + 1)} . \\ [(2)] \to \sum_{g = 0}^{M} b_{g} q^{- g} = M + 1 = q^{- M} \sum_{g = 0}^{M} c_{g}^{*} q^{- (M + 1) g} {(q; q)}_{g} = \sum_{g = 0}^{M} {(- 1)}^{g} G_{g + 1}^{M + 1} q^{\frac{(g + 1) g}{2} - M (g + 1)} {(q; q)}_{g} . \\ If we replace q by q^{- 1}, we will still get M + 1, which is Euler ’ s identity : \sum_{g = 1}^{M} G_{g}^{M} {(q; q)}_{g - 1} = M [2] pp 83 \\ a_{g}^{*} = G_{g}^{M} q^{g^{2} + g} \to c_{g}^{*} = {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} \sum_{k = 0}^{M} {(- 1)}^{k} G_{M - g}^{M - k} G_{k}^{M} q^{\frac{k (k - 1)}{2}}, [3.2] \to c_{0}^{*} = 1, c_{g > 0}^{*} = 0 . \\ \to \frac{\sum_{g = 0}^{M} G_{g}^{M} q^{g^{2} + g - y g} {(a q)}^{g} {(a q \times q^{2 - y + g}; q)}_{M - g}}{{(a q; q)}_{M + 2 - y}} = \frac{1}{{(a q; q)}_{M + 2 - y}} \to \frac{\sum_{g = 0}^{M} G_{g}^{M} q^{g^{2}} a^{g} {(a q^{1 + g}; q)}_{M - g}}{{(a q; q)}_{M}} = \frac{1}{{(a q; q)}_{M}} . \\ It ’ s a finite form of Jacobi ’ s Durfee square identity [2] pp 158 - 159 . \sum_{g = 0}^{\infty} \frac{q^{g^{2} + g r} a^{g}}{{(q; q)}_{g} {(a q; q)}_{g + r}} = \frac{1}{{(a q; q)}_{\infty}}, r = 2 - y \geq 0 . \\ [(4)], g > x, \sum_{k = 0}^{g} {(- 1)}^{k} G_{M - g}^{M - k} G_{k}^{M} q^{(؜_{2}^{k}) - x k} = 0, further promotion can be done . Calculating b_{g} is also acceptable . \\ [4.4] is flexible . We can consider A, a, z as an independent variable or as a part of a_{g}, b_{g}, c_{g}, for example : \\ a_{g}^{*, 1} = q^{g (g + 1) + g x} G_{g}^{M}, a_{g}^{*, 2} = a_{g}^{*, 1} z^{g} [3.4] \to \\ b_{g}^{*, 1} = q^{x g} G_{g}^{M} {(q^{x}; q)}_{M - g}, b_{g}^{*, 2} = q^{x g} G_{g}^{M} {(z q^{x}; q)}_{M - g} z^{g}, c_{g}^{*, 1} = {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} G_{g}^{M} {(q^{x}; q)}_{g}, c_{g}^{*, 2} = {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} G_{g}^{M} {(z q^{x}; q)}_{g} . \\ \sum_{g = 0}^{M} a_{g}^{*, 1} q^{- \frac{g (g + 1)}{2}} z^{g} = \sum_{g = 0}^{M} a_{g}^{*, 2} q^{- \frac{g (g + 1)}{2}} = {(- z q^{1 + x}; q)}_{M} \\ = \sum_{g = 0}^{M} q^{x g} G_{g}^{M} {(q^{x}; q)}_{M - g} {(- z q; q)}_{g} (*) = \sum_{g = 0}^{M} q^{x g} z^{g} G_{g}^{M} {(z q^{x}; q)}_{M - g} {(- q; q)}_{g} \\ = \sum_{g = 0}^{M} {(- 1)}^{g} q^{g} z^{g} G_{g}^{M} {(q^{x}; q)}_{g} {(- z q; q)}_{M - g} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{g} G_{g}^{M} {(z q^{x}; q)}_{g} {(- q; q)}_{M - g} . \\ x = 1, replace z q by z, (*) \to \frac{{(- z q; q)}_{\infty}}{{(q; q)}_{\infty}} = \sum_{g = 0}^{\infty} \frac{q^{g} {(- z; q)}_{g}}{{(q; q)}_{g}} . \\ [(1)] a^{M} \sum_{g = 0}^{M} b_{g} {(- q^{A + 1} \frac{z}{a}; q)}_{g} = \sum_{g = 0}^{M} c_{g}^{*} q^{- \frac{g (g + 1)}{2} + A g} a^{M - g} z^{g} {(- q^{A + 1} \frac{z}{a}; q)}_{M - g} \\ = \sum_{g = 0}^{M} q^{- \frac{g (g + 1)}{2} + A g} a^{M - g} z^{g} {(- q^{A + 1} \frac{z}{a}; q)}_{M - g} {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} \sum_{k = 0}^{M} b_{k} G_{g}^{M - k} q^{g k}, let b_{g \neq k} = 0 \to \\ a^{M} {(- q^{A + 1} \frac{z}{a}; q)}_{k} = \sum_{g = 0}^{M} q^{g + A g} a^{M - g} {(- q^{A + 1} \frac{z}{a}; q)}_{M - g} {(- z)}^{g} G_{g}^{M - k} q^{g k}; {(z; q)}_{k} = \sum_{g = 0}^{M} {(z; q)}_{M - g} z^{g} G_{g}^{M - k} q^{g k} . \\ (3) \to \sum_{g = 0}^{M} c_{g}^{*} = \sum_{g = 0}^{M} a_{g}^{*} {(q^{2 + g}; q)}_{M - g} = \sum_{g = 0}^{M} b_{g} {(q^{2 + g}; q)}_{M - g} . Let a_{g}^{*} = b_{g} = x_{g}, calculate c_{g} by a_{g}^{*} and b_{g} \to \\ For any x_{g}, \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} \sum_{k = 0}^{M} {(- 1)}^{k} G_{M - g}^{M - k} q^{\frac{- k (k + 3)}{2}} x_{k} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} \sum_{k = 0}^{M} {(- 1)}^{k} G_{g}^{M - k} q^{g k} x_{k} . \\ More basically, x_{g \neq k} = 0 \to q^{\frac{- k (k + 3)}{2}} \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} G_{M - g}^{M - k} = \sum_{g = 0}^{M} {(- 1)}^{g} q^{\frac{g (g + 3)}{2}} G_{g}^{M - k} q^{g k} . \\ In the previous text, T_{i} \in N, but excluding the actual meaning of S U M_{q} (N), T_{i} can be any number . \\ P S = [b q^{- T}, b q^{- T} . . . b q^{- T}] : b q^{- T} (q - 1), P T = [T, T + 1 . . . . T + M - 1], a = b q^{- T}, due to T, a and b are independent . \\ At H_{1}^{q} (g), B_{i} = \{\begin{matrix} q^{(T_{i} - T_{i - 1} - 1) X_{T - 1}} (K_{i} + G_{1}^{X_{T - 1}} D_{i}) = a q^{X_{T}}, X_{i} = K_{i} \\ q^{X_{T}} G_{1}^{T + i - 1 - X_{K - 1}} (q - 1) b q^{- T} = q^{X_{T}} (b q^{X_{T} - 1} - a), X_{i} = T_{i} \end{matrix}, H_{1}^{q} (g) = G_{g}^{M} q^{\frac{g (g + 1)}{2}} (b - a) (b q - a) . . . (b q^{g - 1} - a) a^{M - g} . \\ At H_{3}^{q} (g), B_{i} = \{\begin{matrix} (K_{i} + G_{1}^{X_{T - 1}} D_{i}) = a q^{X_{T}}, X_{i} = K_{i} \\ q^{1} {(q^{X_{T - 1}} G_{1}^{T_{i}} - q^{T_{i}} G_{1}^{X_{T - 1}}) D_{i} - K_{i} q^{T_{i}}} = - a q^{X_{T}}, X_{i} = T_{i} \end{matrix}, H_{3}^{q} (g) = G_{g}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} a^{M} . \\ \frac{\sum_{g = 0}^{M} q^{- y g} G_{g}^{M} q^{\frac{g (g + 1)}{2}} (b - a) (b q - a) . . . (b q^{g - 1} - a) {(b q^{- T})}^{M - g} {(b x)}^{g} {(b x \times q^{2 + T_{M} - M - y + g}; q)}_{M - g}}{{(b x; q)}_{T_{M} + 2 - y}}, T_{M} = T + M - 1, \\ T_{M} + 2 - y = M, y = T + 1 \to \frac{b^{M} q^{- T M} \sum_{g = 0}^{M} q^{- g} G_{g}^{M} q^{\frac{g (g + 1)}{2}} (b - a) (b q - a) . . . (b q^{g - 1} - a) x^{g} {(b x \times q^{g}; q)}_{M - g}}{{(b x; q)}_{M}} \\ = \frac{\sum_{g = 0}^{M} H_{3}^{q} (g) {(b x)}^{g} q^{- y g}}{{(b x; q)}_{M}} = \frac{a^{M} \sum_{g = 0}^{M} G_{g}^{M} {(- 1)}^{g} q^{\frac{g (g + 1)}{2}} {(b x)}^{g} q^{- y g}}{{(b x; q)}_{M}} = \frac{a^{M} \sum_{g = 0}^{M} G_{g}^{M} {(- 1)}^{g} q^{\frac{g (g - 1)}{2}} {(b q^{- T} x)}^{g}}{{(b x; q)}_{M}} . \end{matrix}

\begin{matrix} That is to say : \frac{\sum_{g = 0}^{M} G_{g}^{M} q^{\frac{g (g - 1)}{2}} (b - a) (b q - a) . . . (b q^{g - 1} - a) x^{g} {(b x \times q^{g}; q)}_{M - g}}{{(b x; q)}_{M}} = \frac{{(a x; q)}_{M}}{{(b x; q)}_{M}} . \\ This proves Cauchy ’ s identity [2] : \frac{{(a x; q)}_{\infty}}{{(b x; q)}_{\infty}} = \sum_{g = 0}^{\infty} \frac{q^{\frac{g (g - 1)}{2}} x^{g} (b - a) (b q - a) . . . (b q^{g - 1} - a)}{{(q; q)}_{g} {(b x; q)}_{g}} . \end{matrix}

5. An Example

\begin{matrix} We can verify various propositions and theorems . PS = [A, B, C], PT = [1, 3, 5], M = 3, p = 5 - M = 2. \\ \sum_{n_{3} = 0}^{N - 1} \sum_{n_{2} = 0}^{n_{3}} \sum_{n_{1} = 0}^{n_{2}} q^{n_{1} + n_{2} + n_{3}} (A + {[n_{1}]}_{q}) (B + {[n_{2}]}_{q}) (C + {[n_{3}]}_{q}) \\ = \sum_{g = 0}^{3} H_{1}^{q} (g) G_{3 + g}^{N + 2} = \sum_{g = 0}^{3} H_{2}^{q} (g) G_{3 + g}^{N + 2 + g} = \sum_{g = 0}^{3} H_{3}^{q} (g) G_{6}^{N + 5 - g} . \\ H_{1}^{q} (0) = A B C . \\ H_{1}^{q} (1) = A B 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) . \\ H_{1}^{q} (2) = 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} . \\ H_{1}^{q} (3) = q^{1} G_{1}^{1} q^{3} G_{1}^{3} q^{5} G_{1}^{5} . \\ H_{2}^{q} (0) = (A - q^{- 1} G_{1}^{1}) (B - q^{- 2} G_{1}^{2}) (C - q^{- 3} G_{1}^{3}) . \\ H_{2}^{q} (1) = (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}) . \\ H_{2}^{q} (2) = 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} . \\ H_{2}^{q} (3) = q^{- 1} G_{1}^{1} q^{- 3} G_{1}^{3} q^{- 5} G_{1}^{5} . \\ H_{3}^{q} (0) = A B C . \\ H_{3}^{q} (1) = A B 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) . \\ H_{3}^{q} (2) = 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) . \\ H_{3}^{q} (3) = 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) . \\ P S = [A, B], P T = [1, 3] . \\ \sum_{n_{2} = 0}^{N - 1} \sum_{n_{1} = 0}^{n_{2}} (A + n_{1}) (B + n_{2}) = \sum_{g = 0}^{2} H_{1} (g) (\begin{matrix} 2 + g \\ N + 1 \end{matrix}) = \sum_{g = 0}^{2} H_{2} (g) (\begin{matrix} 2 + g \\ N + 1 + g \end{matrix}) = \sum_{g = 0}^{2} H_{3} (g) (\begin{matrix} 4 \\ N + 3 - g \end{matrix}) . \\ H_{1} (0) = A B, H_{1} (1) = A \times 2 + 1 \times (B + 1), H_{1} (2) = 1 \times 3 . \\ H_{2} (0) = (A - 1) (B - 2), H_{2} (1) = (A - 1) \times 2 + 1 \times (B - 3), H_{2} (2) = 1 \times 3 . \\ H_{3} (0) = A B, H_{3} (1) = A (3 - B) + (1 - A) (B + 1), H_{3} (2) = (1 - A) (2 - B) . \end{matrix}

References

P.A. MacMahon. The Indices of Permutations and the Derivation Therefrom of Functions of a Single Variable Associated with the Permutations of Any Assemblage of Objects, American Journal of Mathematics. 35 (1913) 281-322. [CrossRef]
Warren P. Johnson , An Introduction to q-analysis. American Mathematical Society. (2020).
QI Deng-Ji. A New Explicit Expression for the Eulerian Numbers, Journal of Qingdao University of Science and Technology: Natural Science Edition. 04 (2012) 33. [CrossRef]

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Formal Calculation of Q-Binomial

Abstract

Keywords:

Subject:

1. Calculation Formula

2. Property

3. Application

4. The Extension of q-Euler Polynomials and the Relationship Between Three Forms

5. An Example

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