Formal Calculation

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Formal Calculation

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Ji Peng^*

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Submitted:

05 October 2023

Posted:

07 October 2023

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Abstract

Formal Calculation uses an auxiliary form to calculate various nested sums and provides results in three forms. In addition to computation, it is also a powerful tool for analysis, allowing one to study various numbers in a unified way. This article contains many results of two types of Stirling numbers, associated Stirling numbers, and Eulerian numbers, making a great generalization of Euler polynomials, Wilson's theorem, and Wolstenholme's theorem, showing that they are just special cases. Formal Calculation provides a novel method for obtaining combinatorial identities and analyzing q-binomial.This article has obtained a large number of results in q-analogues, including inversion formulas for q-binomial coefficients. This article also introduces a theorem on symmetry.

Keywords:

Subject: Computer Science and Mathematics - Discrete Mathematics and Combinatorics

1. Introduction

Formal Calculation is introduced in [1,2,3] , this article contains its summary and latest achievements.

Definition 1.

Recursive define

\nabla^{p}, p \in Z, \nabla^{0} f (n) = f (n), \sum_{n = 0}^{N - 1} \nabla^{1} f (n + 1) = f (N), \sum_{n = 0}^{N - 1} f (n + 1) = \nabla^{- 1} f (N)

\nabla^{1} = \nabla

Definition 2.

Recursive define SUM(N)=SUM(N,PS,PT).

K_{i}, D_{i} \in

Ring with identity elements.

S U M (N, [K_{1} : D_{1}], [T_{1} = 1]) = \sum_{n = 0}^{N - 1} (K_{1} + n D_{1})

S U M (N, [K_{1} : D_{1}, K_{2} : D_{2}], [T_{1}, T_{2} = T_{1} + 2 - p]) = \sum_{n = 0}^{N - 1} (K_{2} + n D_{2}) \nabla^{p} S U M (n + 1, [K_{1} : D_{1}], [T_{1}])

I f f (N) = \sum A_{i} (_{M_{i}}^{N_{i}}) a n d M_{i} i s n o t c h a n g e d w i t h N, t h e n \nabla^{p} f (N) = \sum A_{i} (_{M_{i} - p}^{N_{i} - p})

[K_{1} : D, K_{2} : D . . . K_{M} : D]

is abbreviated as

[K_{1}, K_{2} . . . K_{M}] : D

[K_{1}, K_{2} . . . K_{M}] : 1

is abbreviated as

[K_{1}, K_{2} . . . K_{M}]

By default,this paper use:

PS=

[K_{1} : D_{1}, K_{2} : D_{2} . . . K_{M} : D_{M}]

,PT=

[T_{1}, T_{2} . . . T_{M}]

,PS1=[PS,

K_{M + 1} : D_{M + 1}

],PT1=[PT,

T_{M + 1}

]

This is actually nested summation. For example:

S U M (N, P S, [1, 2, 3 . . . M]) = \sum_{n = 0}^{N - 1} \prod_{i = 1}^{M} (K_{i} + n D_{i})

S U M (N, P S, [1, 3, 5 . . . 2 M - 1]) = \sum_{n_{M} = 0}^{N - 1} (K_{M} + n_{M} D_{M}) . . . \sum_{n_{2} = 0}^{n_{3}} (K_{2} + n_{2} D_{2}) \sum_{n_{1} = 0}^{n_{2}} (K_{1} + n_{1} D_{1})

S U M (N, P S, [1, 2, 4]) = \sum_{n_{3} = 0}^{N - 1} (K_{3} + n_{3} D_{3}) \sum_{n = 0}^{n_{3}} (K_{1} + n D_{1}) (K_{2} + n D_{2})

S U M (N, P S, [1, 3, 4]) = \sum_{n_{3} = 0}^{N - 1} (K_{3} + n_{3} D_{3}) (K_{2} + n_{3} D_{2}) \sum_{n = 0}^{n_{3}} (K_{1} + n D_{1})

The following use K to represent the set

[K_{1}, K_{2} . . . K_{M}]

, T to represent the set

[T_{1}, T_{2} . . . T_{M}]

Use the auxiliary form:

(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}

Definition 3.

X(T)=Number of

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

Definition 4.

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

,and also define

X_{K}

Obviously:

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

Theorem 1.

[1] SUM(N,PS,PT)=

F o r m_{1} \to \sum_{g = 0}^{M} H_{1} (g) (_{N - 1 - g}^{N + T_{M} - M}) = \sum_{g = 0}^{M} H_{1} (g) (_{T_{M} - M + 1 + g}^{N + T_{M} - M}), B_{i} = \{_{K_{i} + X_{T - 1} D_{i}, X_{i} = K_{i}}^{(T_{i} - X_{K - 1}) D_{i}, X_{i} = T_{i}}

F o r m_{2} \to \sum_{g = 0}^{M} H_{2} (g) (_{N - 1}^{N + T_{M} - M + g}) = \sum_{g = 0}^{M} H_{2} (g) (_{T_{M} - M + 1 + g}^{N + T_{M} - M + g}), B_{i} = \{_{K_{i} + (X_{K - 1} - T_{i}) D_{i}, X_{i} = K_{i}}^{(T_{i} - X_{K - 1}) D_{i}, X_{i} = T_{i}}

F o r m_{3} \to \sum_{g = 0}^{M} H_{3} (g) (_{N - 1 - g}^{N + T_{M} - g}) = \sum_{g = 0}^{M} H_{3} (g) (_{T_{M} + 1}^{N + T_{M} - g}), B_{i} = \{_{K_{i} + X_{T - 1} D_{i}, X_{i} = K_{i}}^{- K_{i} + (T_{i} - X_{T - 1}) D_{i}, X_{i} = T_{i}}

The factors of

\prod X_{i}

cannot be exchanged.

H_{i} (g)

,short for

H_{i} (g, P S, P T)

, is also defined above as

\sum_{X (T) = g} \prod_{i = 1}^{M} B_{i}

The theorem is proved by induction.There have three forms because:

\sum_{n = 0}^{N - 1} n (_{M}^{n + K})

= (M + 1) (_{M + 2}^{N + K}) + (M - K) (_{M + 1}^{N + K})

= (M + 1) (_{M + 2}^{N + K + 1}) - (1 - K) (_{M + 1}^{N + K}) = (M - K) (_{M + 2}^{N + K + 1}) + (1 + K) (_{M + 2}^{N + K})

Definition 5.

F_{M}^{K = {K_{1}, K_{2} . . . K_{M}}}

\sum K_{I_{1}} K_{I_{2}} . . . K_{I_{M}}, a < b, I_{a} < I_{b}

F_{M}^{N}

is short for

F_{M}^{{1, 2 . . . N}}

F_{0}^{K} = 0

Definition 6.

E_{M}^{K = {K_{1}, K_{2} . . . K_{M}}}

\sum K_{I_{1}} K_{I_{2}} . . . K_{I_{M}}, a < b, I_{a} \leq I_{b}

E_{M}^{N}

is short for

E_{M}^{{1, 2 . . . N}}

E_{0}^{K} = 0

Theorem 2.

\nabla S U M (N, P S, [1, 2 . . . M]) = \prod_{i = 1}^{M} (K_{i} + n D_{i})

Theorem 3.

In SUM(N,[...PS...],[...T+1,T+2...T+M...]),

K_{i}

can exchange orders.

Theorem 4.

S U M (N, [L_{1}, L_{2} . . . L_{q}, P S], [L_{1}, L_{2} . . . L_{q}, P T]) = \prod_{i = 1}^{q} L_{i} \times S U M (N, P S, P T), s o

T_{1}

can great than 1,

T_{i} \in N

Theorem 5.

S U M (N, [1, 1 . . . 1], [1, 2 . . . M]) = S U M (N, [1, 1 . . . 1], [2, 3 . . . M]) = 1^{M} + 2^{M} + . . . + N^{M}

Theorem 6.

S U M (N, [1, 1 . . . 1], [1, 3 . . . 2 M - 1]) = S U M (N, [1, 1 . . . 1], [3, 5 . . . 2 M - 1])

= \sum_{λ_{1} + . . . + λ_{N} = M, λ_{i} \geq 0} 1^{λ_{1}} 2^{λ_{2}} . . . N^{λ_{N}} = E_{M}^{N} = S_{2} (N + M, N)

S_{2}

is Stirling numbers of the second kind.

Theorem 7.

S U M (N, [1, 2 . . . M], [1, 3 . . . 2 M - 1]) = S U M (N, [2, 3 . . . M], [3, 5 . . . 2 M - 1])

= \sum_{1 \leq i_{1} < i_{2} < . . . < i_{M} \leq N + M - 1} i_{1} i_{2} . . . i_{M} = F_{M}^{N + M - 1} = S_{1} (N + M, N)

S_{1}

is unsigned Stirling numbers of the first kind.

Example 1.1:

F o r m = (1 + T_{1}) (2 + T_{2}) (3 + T_{3}), \sum_{X (T) = 1} \prod X_{i} = 1 \times 2 \times T_{3} + 1 \times T_{2} \times 3 + T_{1} \times 2 \times 3

H_{1} (1) = 1 \times 2 \times (T_{3} - X_{K - 1}) + 1 \times (T_{2} - X_{K - 1}) \times (3 + X_{T - 1}) + T_{1} \times (2 + X_{T - 1}) \times (3 + X_{T - 1}) = 1 \times 2 \times (5 - 2) + 1 \times (3 - 1) \times (3 + 1) + 1 \times (2 + 1) \times (3 + 1) = 26

S U M (N, [1, 2, 3], [1, 3, 5]) = 1 \times 3 \times 5 (_{6}^{N + 2}) + 35 (_{5}^{N + 2}) + 26 (_{4}^{N + 2}) + 1 \times 2 \times 3 (_{3}^{N + 2})

S U M (N, [2, 3], [3, 5]) = 3 \times 5 (_{6}^{N + 3}) + (2 \times 4 + 3 \times 4) (_{5}^{N + 3}) + 2 \times 3 (_{4}^{N + 3})

It also can be calculated in the Ring with identity elements.

K_{i}, D_{i}

can be a matrix.

Theorem 8.

\prod (K_{i} + n D_{i})

can be decomposed into three forms by 1 and ∇.

2. Property

2.1. Relationships between $H (g)$

By definition:

$H_{1} (g, P S 1, P T 1) = H_{1} (g - 1) (T_{M + 1} - [M - (g - 1)]) D_{M + 1} + H_{1} (g) (K_{M + 1} + g D_{M + 1})$
$H_{2} (g, P S 1, P T 1) = H_{2} (g - 1) (T_{M + 1} - [M - (g - 1)]) D_{M + 1} + H_{2} (g) (K_{M + 1} + [M - g - T_{M + 1}] D_{M + 1})$
$H_{3} (g, P S 1, P T 1) = H_{3} (g - 1) (- K_{M + 1} + (T_{M + 1} - [g - 1]) D_{M + 1}) + H_{3} (g) (K_{M + 1} + g D_{M + 1})$

Using these relationships and induction can prove:

Theorem 9.

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

[2]

Inversion→

Theorem 10.

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

Calculation with 9 →

Theorem 11.

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

Theorem 12.

\sum_{g = 0}^{M} H_{1} (g) (_{B - g}^{A}) = \sum_{g = 0}^{M} H_{2} (g) (_{B}^{A + g}) = \sum_{g = 0}^{M} H_{3} (g) (_{B - g}^{A + M - g}), A, B \in N

This indicates

F o r m_{1} = F o r m_{2} = F o r m_{3} \to \sum_{g = 0}^{M} H_{1} (g) (_{g}^{A}) = \sum_{g = 0}^{M} H_{2} (g) (_{g}^{A + g}) = \sum_{g = 0}^{M} H_{3} (g) (_{M}^{A + M - g})

Induction [2]→

Theorem 13.

\sum_{g = 0}^{M} H_{1} (g) g (_{B - g}^{A + 1}) = \sum_{g = 0}^{M} H_{2} (g) g (_{B - 1}^{A + g}) = \sum_{g = 0}^{M} {H_{3} (g) g (_{B - g}^{A + M - g}) + M \times H_{3} (g) (_{B - 1 - g}^{A + M - g})}

2.2. Property of H(g)

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

.In some cases, H(g) is easy to calculate.

Definition 7.

H (g, T) = H (g, T, P S, P T) = \prod_{X_{i} \in T} B_{i}, H (g, \sum T) = \sum \prod_{X_{i} \in T} B_{i} . A l s o d e f i n e H (g, K), H (g, \sum K)

D_{i} = 1

and

T_{i} + 1 = T_{i + 1}

H_{1} (g, T) = \prod_{i = 1}^{g} T_{i}

H_{1} (g, T, P S, [1, 2 . . . M]) = g!

H_{1} (g, \sum K, [1, 1 . . . 1], P T) = E_{M - g}^{g + 1}

Theorem 14.

D_{i} = 1

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

Theorem 15.

D_{i} = 1

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

Theorem 16.

D_{i} = 1

and

K_{i + 1} - K_{i} = T_{i + 1} - T_{i} = 1

H_{1} (g)

(_{g}^{M}) T_{1} . . . T_{g} \times K_{g + 1} . . . K_{M}

Theorem 17.

If PS=PT,

H_{1} (g) = \prod_{i = 1}^{M} T_{i} (_{N}^{M}), H_{2} (M) = H_{3} (0) = \prod_{i = 1}^{M} T_{i}, H_{2} (g < M) = H_{3} (g > 0) = 0

Theorem 18.

H_{1} (g, [A D : D, P S], [A, P T]) = A D (H_{1} (g - 1) + H_{1} (g)) \to H_{1} (g, [1, P S], [1, P T]) = H_{1} (g - 1) + H_{1} (g)

Definition 8.

E_{p}^{q} ⊙ ([T_{1}, T_{2} . . . T_{M}], C)

= \sum_{λ_{1} + λ_{2} + . . . + λ_{q} = p, λ_{i} \geq 0} 1^{λ_{1}} 2^{λ_{2}} . . . q^{λ_{q}} (T_{1} + λ_{1} C) (T_{2} + λ_{1} C + λ_{2} C) . . . (T_{q - 1} + λ_{1} C + λ_{2} C + . . . + λ_{q - 1} C)

[4] has proved:

〈_{g}^{M}〉 = 〈_{M - g - 1}^{M}〉 = E_{M - g - 1}^{g + 1} ⊙ ([1, 1 . . . 1], 1])

〈_{g}^{M}〉

is Eulerian numbers.It is known that there exists Worpitzky identity:

N^{M} = \sum_{g = 0}^{M - 1} 〈_{g}^{M}〉 (_{M}^{N + g})

e g : 〈_{2}^{5}〉 = E_{2}^{3} ⊙ ([1, 1 . . .], 1) = \sum_{λ_{1} + λ_{2} + λ_{3} = 2} 1^{λ_{1}} 2^{λ_{2}} 3^{λ_{3}} (1 + λ_{1}) (1 + λ_{1} + λ_{2}) = 66

By simple calculation:

H_{1} (g, [1, 1 . . . 1], P T = [T_{i} = T_{1} + (C + 1) (i - 1)]) = E_{M - g}^{g + 1} ⊙ (P T, C)

H_{3} (g, [1, 1 . . . 1], P T = [T_{i} = T_{1} + (C + 1) (i - 1)]) = E_{M - g}^{g + 1} ⊙ ([T_{i} = T_{1} - 1 + C (i - 1)], C + 1)

2.3. Shape of numbers

In this section,if not specifically mentioned,

T_{1} = 1, T_{i + 1} - T_{i} = 1 o r 2

To calculate

\sum_{1 \leq K_{1} < K_{2} < . . . < K_{M} \leq N} K_{1} K_{2} . . . K_{M}

(*),products needs to be divided into

2^{M - 1}

categories.

There are M-1 intervals between factors.If the interval=1,define it as Continuity.If the interval>1,define it as Discontinuity. Continuities,Discontinuities and their Positions are defined as Shape.So there have

2^{M - 1}

Shapes.

From the definition of nested sum:

\sum_{1 \leq K_{1} < K_{2} < K_{3} \leq N} K_{1} K_{2} K_{3} = S U M (N, [1, 2, 3], [1, 2, 3]) + S U M (N - 1, [1, 2, 4], [1, 2, 4]) + S U M (N - 1, [1, 3, 4], [1, 3, 4]) + S U M (N - 2, [1, 3, 5], [1, 3, 5])

Definition 9.

PB(PT)=Number of

T_{i} + 1 < T_{i + 1}

=Number of discontinuities

(*) = \sum_{A l l o f t h e S h a p e s w i t h f a c t o r s = M} S U M (N - P B (P T), P T, P T)

.From 17 we can obtain a simple formula:

Theorem 19.

S U M (N, P T, P T) = \prod_{i = 1}^{M} T_{i} (_{T_{M} - M + 1}^{N + T_{M} - M})

,PT has no restrictions.

This generalizes the famous formula

\sum_{n = 0}^{N - 1} (_{M}^{n}) = (_{M + 1}^{N})

. It was discovered during the calculation of (*) which led to the birth of Formal Calculation.

Theorem 20.

Number of Products in SUM(N,PT,PT) =

(_{P B (P T) + 1}^{N + P B (P T)})

Definition 10.

M I N_{g} (M) = \sum_{P B (P T) = g} \prod_{i = 1}^{M} T_{i} = 1 \times \sum_{P B (P T) = g} \prod_{i = 2}^{M} T_{i}

This is the sum of the products of PTs with the same number of discontinuities.By definition:

Theorem 21.

M I N_{g} (M) = \sum \frac{(M + g)!}{i_{1} i_{2} . . . i_{g}}, 2 \leq i_{1} < i_{2} < . . . < i_{g} \leq M + g - 1, i_{j + 1} - i_{j} \geq 2

Based on the concept of Shape rather than 21, it is easier to understand.

e g : M I N_{2} (4) = (12357) + (12457) + (13457) + (12467) + (13467) + (13567)

.Here (...) is products.

From the definition of nested sum,there exists general classification principles:

Theorem 22.

S U M (N, [K_{1} : D_{1}, K_{2} : D_{2} . . . K_{M} : D_{M}], [T_{1}, T_{2} . . . T_{M}]) = S U M (N, P S, [T_{1} . . . T_{i}, T_{i + 1} - 1 . . . T_{M} - 1]) + S U M (N - 1, [K_{1} : D_{1} . . . K_{i} : D_{i}, K_{i + 1} + D_{i + 1} : D_{i + 1} . . . K_{M} + D_{M} : D_{M}], P T)

e g : S U M (N, [1, 2, 3], [1, 3, 5]) = S U M (N, [1, 2, 3], [1, 2, 4]) + S U M (N - 1, [1, 3, 4], [1, 3, 5]) = S U M (N, [1, 2, 3], [1, 2, 3]) + S U M (N - 1, [1, 2, 4], [1, 2, 4]) + S U M (N - 1, [1, 3, 4], [1, 3, 4]) + S U M (N - 2, [1, 3, 5], [1, 3, 5])

(*) = S U M (N, [1, 2 . . . M], [1, 3 . . . 2 M - 1]) = \sum_{g = 0}^{M - 1} M I N_{g} (M) (_{g + 1}^{N}) . I t^{'} s e x a c t l y .

2.4. H(g) and Associated Stirling Numbers

Associated Stirling Numbers of the first kind

S_{1, r} (n, k)

is defined as the number of permutations of a set of n elements having exactly k cycles, all length >= r.

$S_{1, r} (n, k) = \frac{n!}{k!} \sum_{i_{1} + i_{2} + . . . + i_{k} = n, i_{j} \geq r} \frac{1}{i_{1} i_{2} . . . i_{k}}$
$S_{1, r} (n + 1, k) = n S_{1, r} (n, k) + {(n)}_{r - 1} S_{1, r} (n - r + 1, k - 1), n \geq k r$
$\sum \frac{(n - 1)!}{i_{1} i_{2} . . . i_{k - 1}}, r \leq i_{1} < i_{2} < . . . < i_{k - 1} \leq n - r, i_{j + 1} - i_{j} \geq r$ [5]

Derived from 2 and definition of H(g) or 3 and 21:

Theorem 23.

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

Table 1. Table of

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

Table 1. Table of

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

	g=0	g=1	g=2	g=3	g=4	g=5	g=6
M=1	1
M=2	2	3
M=3	6	20	15
M=4	24	130	210	105
M=5	120	924	2380	2520	945
M=6	720	7308	26432	44100	34650	10395
M=7	5040	64224	303660	705320	866250	540540	135135

Associated Stirling Numbers of the second kind

S_{2, r} (n, k)

is defined as the number of permutations of a set of n elements having exactly k blocks, all length >= r.

2: $S_{2, r} (n, k) = \frac{n!}{k!} \sum_{i_{1} + i_{2} + . . . + i_{k} = n, i_{j} \geq r} \frac{1}{i_{1}! i_{2}! . . . i_{k}!}$
2: $S_{2, r} (n + 1, k) = k S_{2, r} (n, k) + (_{r - 1}^{n}) S_{2, r} (n - r + 1, k - 1), n \geq k r$

Derived from 2:

Theorem 24.

H_{2} (g, [1, 1 . . . 1], [3, 5 . . . 2 M - 1]) = S_{2, 2} (M + g + 1, g + 1)

Table 2. Table of

H_{2} (g, [1, 1 . . . 1], [3, 5 . . . 2 M - 1]) = S_{2, 2} (M + g + 1, g + 1)

Table 2. Table of

H_{2} (g, [1, 1 . . . 1], [3, 5 . . . 2 M - 1]) = S_{2, 2} (M + g + 1, g + 1)

	g=0	g=1	g=2	g=3	g=4	g=5	g=6
M=1	1
M=2	1	3
M=3	1	10	15
M=4	1	25	105	105
M=5	1	56	490	1260	945
M=6	1	119	1918	9450	17325	10395
M=7	1	246	6825	56980	190575	270270	135135

S U M (N, [2, 3 . . . M], [3, 5 . . . 2 M - 1]) = S_{1, 1} (N + M, N) = \frac{(N + M)!}{N!} \sum_{i_{1} + i_{2} + . . . + i_{N} = N + M, i_{j} \geq 1} \frac{1}{i_{1} i_{2} . . . i_{N}}

= \sum_{g = 0}^{M - 1} S_{1, 2} (M + g + 1, g + 1) (_{M + 1 + g}^{N + M}) = \sum_{g = 1}^{M} S_{1, 2} (M + g, g) (_{M + g}^{N + M})

= \frac{(N + M)!}{N!} \sum_{g = 1}^{M} [\sum_{i_{1} + . . . + i_{N} = N + M, i_{j} \geq 1, N u m b e r o f i_{j} > 1 = g} \frac{1}{i_{1} i_{2} . . . i_{N}}] = \sum_{g = 1}^{M} f (*)

f (*) = \frac{(N + M)!}{N!} (_{N - g}^{N}) \sum_{i_{1} + . . . + i_{g} = g + M, i_{j} \geq 1} \frac{1}{i_{1} i_{2} . . . i_{g}} = (_{M + g}^{N + M}) S_{1, 2} (M + g, g)

Use the same way:

Theorem 25.

S_{1, r} (r N + M, N) = \sum_{g = 1}^{M} \frac{1}{r^{N - g}} \frac{(r N - r g)!}{(N - g)!} (_{r g + M}^{r N + M}) S_{1, r + 1} (r g + M, g)

Theorem 26.

S_{2, r} (r N + M, N) = \sum_{g = 1}^{M} \frac{1}{{(r!)}^{N - g}} \frac{(r N - r g)!}{(N - g)!} (_{r g + M}^{r N + M}) S_{2, r + 1} (r g + M, g)

2.5. Table of H(g)

Table 3. Table of H(g).

PS	PT	$H_{1} (g)$	$H_{2} (g)$	$H_{3} (g)$
$[1, 1 . . . 1]$	$[1, 2 . . . M]$	$g! E_{M - g}^{g + 1} = g! S_{2} (M + 1, g + 1)$	${(- 1)}^{M - g} g! S_{2} (M, g)$	$〈_{g}^{M}〉$
$[1, 1 . . . 1]$	$[2, 3 . . . M]$	$(g + 1)! S_{2} (M, g + 1)$	${(- 1)}^{M - 1 - g} (g + 1)! S_{2} (M, g + 1)$	$〈_{g}^{M}〉$
$[1, 1 . . . 1]$	$[1, 3 . . . 2 M - 1]$	$E_{M - g}^{g + 1} ⊙ (P T, 1)$	${(- 1)}^{M - g} M I N_{g - 1} (M)$	$E_{M - g}^{g + 1} ⊙ ([0, 1 . . .], 2)$
$[1, 1 . . . 1]$	$[3, 5 . . . 2 M - 1]$	$S_{2, 2} (M + 1 + g, g + 1)$	${(- 1)}^{M - 1 - g} M I N_{g} (M)$	$E_{M - 1 - g}^{g + 1} ⊙ ([2, 3 . . .], 2)$
$[1, 2 . . . M]$	$[1, 3 . . . 2 M - 1]$	$M I N_{g - 1} (M) + M I N_{g} (M)$	$1 \times {(- 1)}^{M - g} E_{M - g}^{g} ⊙ ([3, 5 . . .], 1)$	$1 \times E_{g}^{M - g} ⊙ ([2, 3 . . .], 2)$
$[2, 3 . . . M]$	$[3, 5 . . . 2 M - 1]$	$M I N_{g} (M)$	${(- 1)}^{M - 1 - g} S_{2, 2} (M + 1 + g, g + 1)$	$E_{g}^{M - g} ⊙ ([2, 3 . . .], 2)$

3. Application

3.1. Number analysis

Theorem 27.

9, 11, 13 →

$\sum_{g = 0}^{M} 〈_{g}^{M}〉 = M!, \sum_{g = 1}^{M} {(- 1)}^{M - g} g! S_{2} (M, g) = 1, \sum_{g = 1}^{M} {(- 1)}^{M - g} g \times g! S_{2} (M, g) = 2^{M} - 1$
$g! S_{2} (M, g) = \sum_{k = g}^{M} {(- 1)}^{M - k} k! S_{2} (M, k) (_{g - 1}^{K - 1}) = \sum_{k = 0}^{g - 1} 〈_{k}^{M}〉 (_{M - g}^{M - 1 - k}), 1 \leq g \leq M$
$S_{1, 2} (M + g, g) = \sum_{k = g}^{M} {(- 1)}^{M - k} S_{2, 2} (M + k, k) (_{g - 1}^{K - 1}), S_{2, 2} (M + g, g) = \sum_{k = g}^{M} {(- 1)}^{M - k} S_{1, 2} (M + k, k) (_{g - 1}^{K - 1})$
$\sum_{g = 1}^{M} g! S_{2} (M, g) = \sum_{g = 1}^{M} {(- 1)}^{M - g} g! S_{2} (M, g) 2^{g - 1} = \sum_{g = 1}^{M} 〈_{M - g}^{M}〉 2^{M - g}$
$\sum_{g = 1}^{M} S_{2, 2} (M + g, g) = \sum_{g = 1}^{M} {(- 1)}^{M - g} S_{1, 2} (M + g, g) 2^{g - 1}, \sum_{g = 1}^{M} S_{1, 2} (M + g, g) = \sum_{g = 1}^{M} {(- 1)}^{M - g} S_{2, 2} (M + g, g) 2^{g - 1}$
$\sum_{g = 1}^{M} g! S_{2} (M, g) (g - 1) (_{g}^{A + 1}) = \sum_{g = 1}^{M} {(- 1)}^{M - g} g! S_{2} (M, g) (g - 1) (_{g}^{A + g - 1})$
$\sum_{g = 0}^{M - 1} {(- 1)}^{M - 1 - g} M I N_{g} (M) = \sum_{g = 1}^{M} {(- 1)}^{M - g} S_{1, 2} (M + g, g) = 1, \sum_{g = 1}^{M} {(- 1)}^{M - g} S_{2, 2} (M + g, g) = M!$

PS=PT=[1,2...M],

H_{1} (g) = H_{1} (M - g) = M! (_{g}^{M})

,14→

Theorem 28.

M! (_{g}^{M}) = g! \sum_{i = 0}^{M - g} S_{1} (M + 1, g + 1 + i) S_{2} (g + i, g) = (M - g)! \sum_{i = 0}^{g} S_{1} (M + 1, M + 1 - i) S_{2} (M - i, M - g)

H_{1} (g, [1, 1 . . . 1], [1, 2 . . . M]) = g! S_{2} (M + 1, g + 1), F_{i}^{{1, 1 . . . 1}} = (_{i}^{M}), \to

Theorem 29.

S_{2} (M + 1, g + 1) = \sum_{i = 0}^{M - g} S_{2} (M - i, g) (_{i}^{M})

H_{1} (g, [1, 2 . . . M], [1, 2 . . . M]) = H_{1} (g, [M, M - 1 . . . 1], [1, 2 . . . M]) = M! (_{g}^{M}) \to

Theorem 30.

g! (_{g}^{M}) = \sum_{i = 0}^{g} S_{1} (M + 1, M + 1 - g + i) S_{2} (M - g + i, M - g) {(- 1)}^{i}

H_{1} (g, [K + i], [T + i])

,16

\to (_{g}^{M}) T_{1} . . . T_{g} \times K_{g + 1} . . . K_{M}

,14

\to T_{1} . . . T_{g} (. . .)

,15

\to K_{g + 1} . . . K_{M} (. . .)

Theorem 31.

\prod_{i = g + 1}^{M} (K + i) (_{g}^{M}) = F_{M - g}^{{K + i}} E_{0}^{g} + . . . + F_{0}^{{K + i}} E_{M - g}^{g}, \prod_{i = 1}^{g} (T + i) (_{g}^{M}) = F_{g}^{{T + i}} E_{0}^{M - g} + . . . + {(- 1)}^{g} F_{0}^{{T + i}} E_{g}^{M - g}

3.2. Merge and Expand

Theorem 32.

SUM(N,PS,PT)=SUM(N,[1,1...1,PS],[1,1...1,PT]) expand

\sum_{g = 0}^{M} (. . .) = \sum_{g = 0}^{M + (n u m b e r o f 1 a d d e d)} (. . .)

Any

\sum_{g = 0}^{M} a_{g} (_{Y + g}^{X})

can be converted to

\frac{a_{M}}{M!} \nabla^{q} S U M (N + P, [K_{1}, K_{2} . . . K_{M}], [1, 2 . . . M])

10 provides the necessary and sufficient condition for

\sum_{g = 0}^{M} H (g) (_{Y + g}^{X})

to be merged into

\sum_{g = 0}^{M - K} (. . .) (_{Y + K + g}^{X + K}) :

H_{2} (g) = \sum_{x = g}^{M} {(- 1)}^{x} H (x) (_{g}^{x}) = 0, g < K o r H_{3} (g) = \sum_{x = 0}^{g} {(- 1)}^{x} H (x) (_{M - g}^{M - x}) = 0, M - g < K

For example:

S U M (N, [1, 2 . . . M], [1, 2 . . . M]), \to \sum_{x = g}^{M} {(- 1)}^{x} (_{x}^{M}) (_{g}^{x}) = 0, 0 \leq g < M

\sum_{n = 0}^{N - 1} (_{M}^{M + d n}) = \frac{1}{M!} S U M (N, [1, 2 . . . M] : d, [1, 2 . . . M]) = \frac{1}{M!} \sum_{g = 0}^{M} H_{1} (g) (_{1 + g}^{N})

I f M \geq k d, B_{i} (X_{i} = K_{i}) = i + (X_{K - 1} - i) d \to H_{2} (g < k) = 0 \to \sum_{n = 0}^{N - 1} (_{M}^{M + d n}) = \sum_{g = 0}^{M - k} (. . .) (_{1 + g + k}^{N + k})

After a simple calculation,it can be written as:

Theorem 33.

Necessary and sufficient conditions for merging,

0 \leq g < K \leq M

$\sum_{n = 0}^{M} H (n) (_{Y + n}^{X}) \to \sum_{n = 0}^{M - K} (. . .) (_{Y + K + n}^{X + K}) : \sum_{x = 0}^{M} {(- 1)}^{x} H (x) (_{g}^{P \pm x}) = 0$
$\sum_{n = 0}^{M} H (n) (_{Y + n}^{X + n}) \to \sum_{n = 0}^{M - K} (. . .) (_{Y + K + n}^{X + K + n}) : \sum_{x = 0}^{M} H (x) (_{g}^{P \pm x}) = 0$
$\sum_{n = 0}^{M} H (n) (_{Y}^{X - n}) \to \sum_{n = 0}^{M - K} (. . .) (_{Y - K}^{X - K - n}) : \sum_{x = 0}^{M} H (x) (_{g}^{P \pm x}) = 0$

Theorem 34.

\sum_{g = 0}^{M} (_{g}^{M}) (_{A}^{A + T + g}) (_{Y + g}^{X}) = \sum_{g = 0}^{A} (_{g + T}^{A + T}) (_{M + T}^{M + T + g}) (_{Y + M - A + g}^{X + M - A}), A \geq 0

Proof.

It can be proved by induction, but it is cumbersome.

SUM(N,[T+1,T+2...T+M],[T+A+1,T+A+2...T+A+M])

= \sum_{g = 0}^{M} (_{g}^{M}) {[T + A + g]}_{g} {[T + M]}_{M - g} (_{T + A + 1 + g}^{N + T + A})

= \sum_{g = 0}^{M} (_{g}^{M}) \frac{(T + A + g)!}{(T + A)!} \frac{(T + M)!}{(T + g)!} (_{T + A + 1 + g}^{N + T + A}) = \frac{A! (T + M)!}{(T + A)!} \sum_{g = 0}^{M} (_{g}^{M}) (_{A}^{A + T + g}) (_{T + A + 1 + g}^{N + T + A})

= \frac{(T + M)!}{(T + A)!} S U M (N, [T + 1, T + 2 . . . T + A], [T + M + 1, T + M + 2 . . . T + M + A])

= \frac{(T + M)!}{(T + A)!} \sum_{g = 0}^{A} (_{g}^{A}) \frac{(T + M + g)!}{(T + M)!} \frac{(T + A)!}{(T + g)!} (_{T + M + 1 + g}^{N + T + M})

\sum_{g = 0}^{M} (_{g}^{M}) (_{A}^{A + T + g}) (_{T + A + 1 + g}^{N + T + A}) = \sum_{g = 0}^{A} \frac{(T + M + g)!}{(T + M)!} \frac{(T + A)!}{(T + g)!} \frac{A!}{(A - g)! g!} \frac{1}{A!} (_{T + M + 1 + g}^{N + T + M})

□

g : = M - g \to \sum_{g = 0}^{M} (_{g}^{M}) (_{A}^{A + T + M - g}) (_{Y + g}^{X}) = \sum_{g = 0}^{A} (_{g}^{A + T}) (_{M + T}^{A + M + T - g}) (_{Y + M - A + g}^{X + M - A}), A \geq 0

When A>M, it is an expansion; When A<M, it is a merge.Combining with 33:

Theorem 35.

\sum_{g = 0}^{M} {(- 1)}^{g} (_{g}^{M}) (_{A}^{X_{1} \pm g}) (_{B}^{X_{2} \pm g}) = 0, M > A + B, A, B \geq 0

I f f (n) = \sum_{i = 0}^{B} a_{i} n^{i}, B < M \to \sum_{g = 0}^{M} {(- 1)}^{g} (_{g}^{M}) f (g) = 0 \to \sum_{g = 0}^{M} {(- 1)}^{g} (_{g}^{M}) (_{Y_{1}}^{X_{1} \pm g}) (_{Y_{2}}^{X_{2} \pm g}) . . . = 0, \sum Y_{i} < M

It can be proved by induction:

Theorem 36.

\sum_{g = 0}^{M} {(- 1)}^{g} (_{g}^{M}) (_{M + K}^{T + g}) = \{_{{(- 1)}^{M} (_{K}^{T}), K \geq 0}^{0, K < 0}

;

\sum_{g = 0}^{M} {(- 1)}^{g} (_{g}^{M}) (_{M + K}^{T - g}) = \{_{(_{K}^{T}), K \geq 0}^{0, K < 0} T \pm g \in Z

This helps to understand the differential sequence.

3.3. Congruences

P is prime.

K_{i}, D

is any integer,D ≠0.

Theorem 37.

(P, D) = 1, S U M (P, [K_{1}, K_{2} . . . K_{M}] : D, [1, 2 . . . M]) \equiv \{_{- 1 (mod P), M = P - 1}^{0 (mod P), M < P - 1}

Proof.

If M=P-1,

S U M (P) \equiv H_{1} (P - 1) (_{P}^{P}) \equiv H_{1} (P - 1) \equiv (P - 1)! D^{P - 1} \equiv - 1 (mod P)

□

If a product has a factor that is divisible by P then ignore it and change the factor to its minimum positive residue, then we can obtain many congruences. Wilson’s Theorem is a special case.eg:

A, B, C \in N

1^{A} 2^{B} + 2^{A} 3^{B} + . . . + {(P - 2)}^{A} {(P - 1)}^{B} \equiv 1^{A} 3^{B} + . . . + {(P - 3)}^{A} {(P - 1)}^{B} + {(P - 1)}^{A} 1^{B} \equiv \{_{- 1 (mod P), A + B = P - 1}^{0 (mod P), A + B < P - 1}

1^{A} 2^{B} 3^{C} + 2^{A} 3^{B} 4^{C} + . . . + {(P - 3)}^{A} {(P - 2)}^{B} {(P - 1)}^{C} \equiv \{_{- 1 (mod P), A + B + C = P - 1}^{0 (mod P), A + B + C < P - 1}

Theorem 38.

\sum_{0 < K_{i}, K_{j} < P, K_{i} \neq K_{j}} {K_{1}}^{λ_{1}} {K_{2}}^{λ_{2}} . . . {K_{q}}^{λ_{q}} \equiv \{_{0 (mod P), λ_{1} + λ_{2} + . . . + λ_{q} < P - 1}^{- 1 (mod P), λ_{1} + λ_{2} + . . . + λ_{q} = P - 1}, λ_{i} \in N

Wolstenholme’s Theorem is also a special case.P>3.

Wolstenholme’s Theorem: $(P - 1)! \sum_{n = 1}^{P - 1} \frac{1}{n} = \sum_{0 < K_{i}, K_{j} < P, K_{i} \neq K_{j}} K_{1} K_{2} . . . K_{P - 2} \equiv 0 (mod P^{2})$
$\sum_{n = 1}^{P - 1} n^{P - 2} \equiv 0 (mod P^{2})$

They are two extremes.In fact,there have:

Theorem 39.

\sum_{0 < K_{i}, K_{j} < P, K_{i} \neq K_{j}} {K_{1}}^{C_{1}} {K_{2}}^{C_{2}} . . . {K_{q}}^{C q} \equiv 0 (mod P^{2}), C_{1} + C_{2} + . . . + C_{q} = P - 2, C_{i} > 0

Proof.

X (mod P^{2})

and

X + Y (mod P^{2})

then

Y (mod P^{2})

.The Sum has symmetry.

For

\sum A B^{P - 3}, A \neq B

,if

P - A \neq B

then add

A B^{P - 3}

with

(P - A) B^{P - 3}

P B^{P - 3}

,if

P - A = B

then add

A B^{P - 3}

with

B B^{P - 3}

P B^{P - 3}

.so

\sum A B^{P - 3} + \sum B^{P - 2} = x P \sum B^{P - 3}, 0 < x < P \to \sum A B^{P - 3} \equiv 0 (mod P^{2})

Similarly:

For

\sum A B C^{P - 4}, A \neq B \neq C

,treat P-A=B,P-A=C,

P - A \neq B, C

separately.

\sum A B C^{P - 4} + x \sum B^{2} C^{P - 4} + y \sum B C^{P - 3} \equiv 0 mod P^{2}, 0 < x, y < P

\to \sum A B C^{P - 4} + x \sum B^{2} C^{P - 4} \equiv 0 (mod P^{2}), 0 < x < P

\sum A B C^{P - 4} + \sum B^{2} C^{P - 4} = \sum ((_{2}^{P}) - B) B C^{P - 4} + \sum B^{2} C^{P - 4} = (_{2}^{P}) \sum B C^{P - 4} \equiv 0 (mod P^{2})

\to \sum B^{2} C^{P - 4} \equiv 0 mod P^{2} \to \sum A B C^{P - 4} \equiv 0 mod P^{2}

Prove the conclusion in a similar way... □

Theorem 40.

E_{P - 2}^{P - 1} = S_{2} (2 P - 3, P - 1) \equiv 0 (mod P^{2}); E_{P - 2}^{P} = S_{2} (2 P - 2, P) \equiv 0 (mod P^{2})

e g : S_{2} (7, 4) = 350, S_{2} (8, 5) = 1050 \equiv 0 (mod 25), S_{2} (11, 6) = 179487, S_{2} (12, 7) = 627396 \equiv 0 (mod 49)

4. Combinatorial Identities

Definition 11.

R-FOLD SUM:

\sum_{(r)}^{N} f (k) = \sum_{k_{r} = 1}^{N} . . . \sum_{k_{2} = 1}^{k_{3}} \sum_{k_{1} = 1}^{k_{2}} f (k_{1}) = \sum_{k_{r} = 0}^{N - 1} . . . \sum_{k_{2} = 0}^{k_{3}} \sum_{k_{1} = 0}^{k_{2}} f (k_{1} + 1)

By nested sum:

Theorem 41.

\sum_{(r)}^{N} \nabla^{p} S U M (k, P S, P T) = \nabla^{p - r} S U M (N, P S, P T)

Theorem 42.

\sum_{k_{r} = 1}^{N} . . . \sum_{k_{2} = 1}^{k_{3}} \sum_{k_{1} = 1}^{k_{2}} \nabla S U M (k_{r}, P S, [1, 2 . . . M]) = S U M (N, P S, P T = [T_{i} = i + r - 1])

Proof.

P S 1 = [1 : 0, 1 : 0 . . . 1 : 0, P S], P T 1 = [1, 3 . . . 2 (r - 1) - 1, 1 + 2 (r - 1), 2 + 2 (r - 1) . . . M + 2 (r - 1)]

B_{i} = \{_{K_{i} + X_{K - 1} D_{i} = 1, X_{i} \in K}^{(T_{i} - X_{T - 1}) D_{i} = 0, X_{i} \in T} \to H_{1} (g > M, P S 1, P T 1) = 0, H_{1} (g \leq M, P S 1, P T 1) = H_{1} (g, P S, P T)

S U M (N, P S 1, P T 1) = \sum_{g = 0}^{M} H_{1} (g, P S, P T) (_{r + g}^{N + r - 1}) = \sum_{k_{r} = 1}^{N} \nabla S U M (k_{r}, P S, [1, 2 . . . M]) . . . \sum_{k_{2} = 1}^{k_{3}} \sum_{k_{1} = 1}^{k_{2}} 1

□

Theorem 43.

\sum_{k_{x} = 1}^{N} . . . \sum_{k_{2} = 1}^{k_{3}} \sum_{k_{1} = 1}^{k_{2}} \nabla S U M (k_{r}, P S, [1, 2 . . . M]) = \nabla^{r - x} S U M (N, P S, [T_{i} = i + r - 1])

e g : (*) \sum_{k_{r} = 1}^{N} . . . \sum_{k_{2} = 1}^{k_{3}} \sum_{k_{1} = 1}^{k_{2}} (_{j}^{k_{i}}) = \frac{1}{j!} \nabla^{i - r} S U M (N + 1, [0, - 1, - 2 . . . - (j - 1)], [T_{x} = x + (i - 1)])

H_{1} (g < j) = 0, H_{1} (j) = \frac{(j + i - 1)!}{(i - 1)!}, T_{j} - j = i - 1 \to

(*) = \frac{1}{j!} \nabla^{i - r} \frac{(j + i - 1)!}{(i - 1)!} (_{j + 1 + (i - 1)}^{(N + 1) + (i - 1)}) = (_{i - 1}^{j + i - 1}) (_{j + i - (i - r)}^{N + i - (i - r)}) = (_{i - 1}^{j + i - 1}) (_{j + r}^{N + r})

[6]

Using induction to prove:

Theorem 44.

\sum_{0 \leq n_{1} \leq . . . \leq n_{M} \leq N - 1} (K + n_{1} D_{1} + . . . + n_{M} D_{M}) = (D_{1} + 2 D_{2} + . . . + M D_{M}) (_{M + 1}^{N + M - 1}) + K (_{M}^{N + M - 1})

e g : \sum_{0 \leq n_{1} \leq . . . \leq n_{p} \leq n} (n_{1} + . . . + n_{p}) = \sum_{n_{p} = 0}^{n} \sum_{n_{p - 1} = 0}^{n_{p}} . . . \sum_{n_{1} = 0}^{n_{2}} (n_{1} + . . . + n_{p}) = (_{2}^{p + 1}) (_{p + 1}^{n + p}) = \frac{p n}{2} (_{p}^{n + p})

. [6]

2 , 3 , 4 can be used to derive combinatorial identities.

Theorem 45.

(_{A}^{n + A}) (_{M}^{n + M + B}) =

$\sum_{g = 0}^{M} (_{g}^{A + g}) (_{M - g}^{M + B}) (_{A + g}^{n + A})$
$\sum_{g = 0}^{M} {(- 1)}^{M - g} (_{g}^{A + g}) (_{M - g}^{A - B}) (_{A + g}^{n + A + g})$
$\sum_{g = 0}^{M} (_{g}^{A - B}) (_{M - g}^{M + B}) (_{A + M}^{n + A + M - g})$

Proof.

= \frac{1}{A! M!} \nabla S U M (N, [1, 2 . . . A, B + 1 . . . B + M], [1, 2 . . . A + M]) = \frac{1}{M!} \nabla S U M (N, [B + 1 . . . B + M], [A + 1 . . . A + M])

H_{1} (g) = (_{g}^{M}) (A + 1) . . . (A + g) (B + g + 1) . . . (B + M) = (_{g}^{M}) g! (_{g}^{A + g}) (M - g)! (_{M - g}^{B + M})

Using a similar method to obtain

H_{2} (g), H_{3} (g)

□

Theorem 46.

(_{A}^{n + X}) (_{M}^{n + Y}) = \sum_{x = 0}^{A} (_{x}^{M + x}) (_{A - x}^{M + X - Y}) (_{M + x}^{n + Y}), 0 \leq Y \leq M

Proof.

= \frac{1}{A! M!} \nabla S U M (N, [X, X - 1 . . . X - A + 1, Y, Y - 1 . . . Y - M + 1], [1, 2 . . . A + M])

= \frac{1}{A! M!} \nabla S U M (N, [1, 2 . . . Y, 0, - 1, - 2 . . . - (M - Y) + 1, X, X - 1 . . . X - A + 1], [1, 2 . . . A + M])

= \frac{Y!}{A! M!} \nabla S U M (N, [0, - 1, - 2 . . . - (M - Y) + 1, X, X - 1 . . . X - A + 1], [Y + 1, Y + 2 . . . A + M])

I f H_{1} (g) \neq 0 t h e n X_{1}, X_{2} . . . X_{M - Y} \in T \to H_{1} (g < M - Y) = 0, L e t C = A + M - Y

I f H_{1} (g \geq M - Y) \neq 0, N u m b e r o f X \in K = (_{C - g}^{C - M + Y}) \to H_{1} (g) = (_{C - g}^{C - M + Y}) {[X + M - Y]}_{C - g} {[Y + 1]}^{g}

x:=-(M-Y-g)

\to H_{1} (M - Y + x) = \frac{A!}{(A - x)! x!} \frac{(M + X - Y)!}{(M + X - Y - A + x)!} \frac{(M + x)!}{Y!}

□

Theorem 47.

\prod_{i = 1}^{M} (A + 2 i + n) = {(_{A}^{n + A})}^{- 1} \sum_{g = 0}^{M} (2 (M - g) - 1)!! (_{g}^{2 M - g}) {[A + 1]}^{g} (_{A + g}^{n + A + g}), A \geq 0

Proof.

P S = [A + 2, A + 4 . . . A + 2 M], P T = [A + 1, A + 2 . . . A + M], P T 1 = [1, 3 . . . 2 (M - g) - 1]

H_{2} (g, \sum K) = S U M (g + 1, P T 1, P T 1)) = (2 (M - g) - 1)!! (_{g}^{2 M - g}), H_{2} (g, T) = {[A + 1]}^{g}

(_{A}^{n + A}) \prod_{i = 1}^{M} (A + 2 i + n) = \frac{1}{A!} \nabla S U M (N, [1, 2 . . . A, P S], [1, 2 . . . A, P T]) = \nabla S U M (N, P S, P T)

□

Theorem 48.

SUM(N,[A+1,A+3...A+2M-1],[1,3...2M-1])=

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

Proof.

H_{2} (g, \sum T) = S U M (M - g + 1, [1, 3 . . . 2 g - 1], [1, 3 . . . 2 g - 1])) = (2 g - 1)!! (_{2 g}^{M + g}), H_{2} (g, K) = {[A]}^{M - g}

□

Theorem 49.

SUM(N,[A,A+1...A+M-1]:2,[1,3...2M-1])=

(_{M}^{M + N - 1}) {[A + M + N - 2]}_{M}

Proof.

S U M (g + 1, [A, A + 1 . . . A + M - 1 - g] : 2, [1, 3 . . . 2 (M - g) - 1])

= H_{1} (g, \sum K, [A, A + 1 . . . A + M - 1], [1, 2 . . . M]) = (_{g}^{M}) {[A + M - 1]}_{M - g}

□

S U M (N, [1, 2 . . . M] : 2, [1, 3 . . . 2 M - 1]) = M! {(_{M}^{N + M - 1})}^{2}, 1 + 3 + . . . + (2 N - 1) = N^{2}

49 and 45

\to \frac{1}{M!} H (g, [A + 1, A + 2 . . . A + M] : 2, [1, 3 . . . 2 M - 1]) = H (g, [A + 1, A + 2 . . . A + M], [M + 1, M + 2 . . . 2 M])

Theorem 50.

S U M (N, [A + 2, A + 4 . . . A + 2 M] : 3, [1, 3 . . . 2 M - 1]) = \sum_{g = 0}^{M} (_{g}^{A + N - 1 + g}) (_{M - g}^{N + M - 1 - g}) \frac{{[M + N - g]}^{M}}{2^{M - g}}

Proof.

P S 1 = [A + 2, A + 4 . . . A + 2 M], P T 1 = [A + 1, A + 2 . . . A + M]

H_{1} (g, P S 1, P T 1) = {[A + 1]}^{g} S U M (g + 1, [A + 2, A + 4 . . . A + 2 (M - g)] : 3, [1, 3 . . . 2 (M - g) - 1]))

= \sum_{k = g}^{M} H_{2} (g, P S 1, P T 1) (_{g}^{k}), \to = \sum_{k = g}^{M} (2 (M - k) - 1)!! (_{k}^{2 M - k}) {[A + 1]}^{k} (_{g}^{k})

□

5. Matrix of SUM(N)

Consider H(g) as variables,list SUM(N),SUM(N+1)…SUM(N+M),we can obtain a (M+1) ×(M+1) matrix.

Let

P = N + T_{M} - M, Q = N - 1

, corresponding to the three forms.define

A_{1, 2, 3} (P, Q, M) =

(\begin{matrix} (_{Q}^{P}) & \dots & (_{Q - M}^{P}) \\ ⋮ & ⋱ & ⋮ \\ (_{Q + M}^{P + M}) & \dots & (_{Q}^{P + M}) \end{matrix}), (\begin{matrix} (_{Q}^{P}) & \dots & (_{Q}^{P + M}) \\ ⋮ & ⋱ & ⋮ \\ (_{Q + M}^{P + M}) & \dots & (_{Q + M}^{P + 2 M}) \end{matrix}), (\begin{matrix} (_{Q}^{P + M}) & \dots & (_{Q - M}^{P}) \\ ⋮ & ⋱ & ⋮ \\ (_{Q + M}^{P + 2 M}) & \dots & (_{Q}^{P + M}) \end{matrix})

Theorem 51.

‖ A_{1} (P, Q, M) ‖ = ‖ A_{2} (P, Q, M) ‖ = ‖ A_{3} (P, Q, M) ‖, ‖ A (P, Q, M) ‖ = ‖ A (P, P - Q, M) ‖

[2]

Theorem 52.

‖ A (P, 0, M) ‖ = 1, ‖ A (P, 1, M) ‖ = (_{1 + M}^{P + M}), ‖ A (P, Q > 1, M) ‖ = \prod_{g = 0}^{Q - 1} \frac{(_{1 + M}^{P + M - g})}{(_{1 + M}^{1 + M - g})}

[2]

If SUM(N) or ∇ SUM(N) is easy to obtained,then H(g) can be calculated with the Cramer’s law.Below,

T_{M} \geq M

Theorem 53.

H_{1} (g) = \sum_{k = 1}^{g + 1} {(- 1)}^{g + 1 + k} (_{T_{M} - M + k}^{T_{M} - M + 1 + g}) S U M (k) = \sum_{k = 1}^{g + 1} {(- 1)}^{g + 1 + k} (_{T_{M} - M + k - 1}^{T_{M} - M + g}) \nabla S U M (k)

(1) \nabla S U M (N, [1, 1 . . . 1], [2, 3 . . . M]) = N^{M} \to S_{2} (M, g) = \frac{1}{g!} \sum_{k = 0}^{g} {(- 1)}^{g + k} (_{k}^{g}) k^{M} = \frac{1}{g!} \sum_{k = 0}^{g} {(- 1)}^{k} (_{k}^{g}) {(g - k)}^{M}

Theorem 54.

z (k) = \sum_{i = 1}^{k} {(- 1)}^{i + k} (_{i}^{k}) \nabla S U M (k), H_{2} (g) = \sum_{k = g + 1}^{M + 1} {(- 1)}^{g + k - 1} (_{g}^{k - 1}) z (k)

\nabla S U M (N, [1, 1 . . . 1], [2, 3 . . . M]) = N^{M} \to z (k) = \sum_{i = 1}^{k} {(- 1)}^{i + k} (_{i}^{k}) i^{M} = k! S_{2} (M, k) \to . 2

Theorem 55.

H_{3} (g) = \sum_{k = 1}^{g + 1} {(- 1)}^{g + 1 + k} (_{g + 1 - k}^{2 + T_{M}}) S U M (k) = \sum_{k = 1}^{g + 1} {(- 1)}^{g + 1 + k} (_{g + 1 - k}^{1 + T_{M}}) \nabla S U M (k)

(2) \nabla S U M (N, [1, 1 . . . 1], [2, 3 . . . M]) = N^{M} \to 〈_{g}^{M}〉 = \sum_{k = 1}^{g + 1} {(- 1)}^{1 + g + k} (_{g + 1 - k}^{M + 1}) k^{M} = \sum_{k = 0}^{g} {(- 1)}^{k} (_{k}^{M + 1}) {(g + 1 - k)}^{M}

(1)(2) are already known formula.

6. Eulerian polynomials and Beyond

In this section,

q \neq 0, q \neq 1

.Inductive proof:

Theorem 56.

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

Definition 12.

A_{q}^{M} = \sum_{k = 0}^{M} {(1 - q)}^{M - k} q^{k} S_{2} (M, k) k!, A_{q}^{0} = 1, A_{q}^{1} = q

Table 4. Table of

A_{q}^{M}

Table 4. Table of

A_{q}^{M}

	M=0	M=1	M=2	M=3	M=4	M=5	M=6	OEIS
$A_{2}^{M}$	1	2	6	26	150	1082	9366	A000629
$A_{3}^{M}$	1	3	12	66	480	4368	47712	A123227
$A_{4}^{M}$	1	4	20	132	1140	12324	160020	A201355

n^{M} = \nabla S U M (N, [1, 1 . . . 1], [2, 3 . . . M]) = \sum_{g = 0}^{M} S_{2} (M, g) g! (_{g}^{n}) = \sum_{g = 0}^{M} {(- 1)}^{M - g} S_{2} (M, g) g! (_{g}^{n + g}) = \sum_{g = 0}^{M} 〈_{g}^{M}〉 (_{M}^{n + g})

\sum_{n = 0}^{N - 1} q^{n} n^{M} = \sum_{g = 0}^{M} S_{2} (M, g) g! \sum_{n = 0}^{N - 1} q^{n} (_{g}^{n}) = \sum_{g = 0}^{M} S_{2} (M, g) g! {q^{N} \sum_{k = 0}^{g} {(- 1)}^{k} \frac{(_{g - k}^{N - 1 - k})}{{(q - 1)}^{k + 1}} + \frac{q^{g}}{{(1 - q)}^{g + 1}}}

= \frac{q^{N}}{{(q - 1)}^{M + 1}} \sum_{g = 0}^{M} S_{2} (M, g) g! \sum_{k = 0}^{M} {(- 1)}^{k} (_{g - k}^{N - 1 - k}) {(q - 1)}^{M - k} + \frac{\sum_{g = 0}^{M} S_{2} (M, g) g! {(1 - q)}^{M - g} q^{g}}{{(1 - q)}^{M + 1}}

= \frac{q^{N}}{{(q - 1)}^{M + 1}} \sum_{k = 0}^{M} {(- 1)}^{k} {(q - 1)}^{M - k} \sum_{g = 0}^{M} S_{2} (M, g) g! (_{g - k}^{N - 1 - k}) + \frac{A_{q}^{M}}{{(1 - q)}^{M + 1}}

= \frac{q^{N}}{{(q - 1)}^{M + 1}} \sum_{k = 0}^{M} {(- 1)}^{k} {(q - 1)}^{M - k} \nabla^{k} {(N - 1)}^{M} + \frac{A_{q}^{M}}{{(1 - q)}^{M + 1}} (*)

Use the

F r o m_{2}

and

F r o m_{3}

n^{M}

,the first part of (*) keep same, we can obtain:

Theorem 57.

A_{q}^{M} = q \sum_{k = 0}^{M} {(q - 1)}^{M - k} S_{2} (M, k) k! = \sum_{g = 0}^{M} 〈_{g}^{M}〉 q^{M - g} = \sum_{g = 0}^{M} 〈_{g}^{M}〉 q^{1 + g}

n^{M} = \nabla S U M (N, [1, 1 . . . 1], [1, 2 . . . M]) = \sum_{g = 0}^{M} S_{2} (M + 1, g + 1) g! (_{g}^{n - 1}) \to

Theorem 58.

A_{q}^{M} = \sum_{k = 0}^{M} {(1 - q)}^{M - k} q^{k + 1} S_{2} (M + 1, k + 1) k!, M > 0

S_{2} (M + 1, k + 1) = \frac{1}{(k + 1)!} \sum_{j = 0}^{k + 1} {(- 1)}^{j + k + 1} (_{j}^{k + 1}) j^{M + 1} = \frac{1}{k!} \sum_{j = 0}^{k} {(- 1)}^{j + k} (_{j}^{k}) {(j + 1)}^{M}

By definition of difference:

\nabla^{k} {(N - 1)}^{M} = \sum_{j = 0}^{k} {(- 1)}^{j} (_{j}^{k}) {(N - 1 - j)}^{M}

= \sum_{j = 0}^{k} {(- 1)}^{j} (_{j}^{k}) \sum_{g = 0}^{M} (_{g}^{M}) N^{g} {(j + 1)}^{M - g} {(- 1)}^{M - g}

= \sum_{g = 0}^{M} {(- 1)}^{M - g - k} (_{g}^{M}) N^{g} \sum_{j = 0}^{k} {(- 1)}^{j + k} (_{j}^{k}) {(j + 1)}^{M - g} = \sum_{g = 0}^{M} {(- 1)}^{M - g - k} (_{g}^{M}) N^{g} S_{2} (M - g + 1, k + 1) k! \to

(*) = \frac{q^{N}}{{(q - 1)}^{M + 1}} \sum_{k = 0}^{M} {(- 1)}^{k} {(q - 1)}^{M - k} {\sum_{g = 0}^{M} {(- 1)}^{M - g - k} (_{g}^{M}) N^{g} S_{2} (M - g + 1, k + 1) k!} + \frac{A_{q}^{M}}{{(1 - q)}^{M + 1}}

= \frac{q^{N}}{{(q - 1)}^{M + 1}} \sum_{g = 0}^{M} {(- 1)}^{M - g} {(q - 1)}^{g} (_{g}^{M}) N^{g} \sum_{k = 0}^{M} {(q - 1)}^{M - k - g} S_{2} (M - g + 1, k + 1) k! + \frac{A_{q}^{M}}{{(1 - q)}^{M + 1}} (* *)

N = 0, (*) = 0 \to \frac{q^{N}}{{(q - 1)}^{M + 1}} (. . .) + \frac{A_{q}^{M}}{{(1 - q)}^{M + 1}} = 0 \to

Theorem 59.

A_{q}^{M} = \sum_{k = 0}^{M} {(q - 1)}^{M - k} S_{2} (M + 1, k + 1) k!

A_{q}^{M - g} = \sum_{k = 0}^{M} {(q - 1)}^{M - k - g} S_{2} (M - g + 1, k + 1) k!, (* *) \to

Theorem 60.

\sum_{n = 0}^{N - 1} q^{n} n^{M} = \frac{q^{N}}{{(q - 1)}^{M + 1}} \sum_{g = 0}^{M} {(- 1)}^{M - g} {(q - 1)}^{g} (_{g}^{M}) A_{q}^{M - g} N^{g} + \frac{A_{q}^{M}}{{(1 - q)}^{M + 1}}

The Eulerian polynomials:

A_{M} (t) : \sum_{i = 0}^{\infty} t^{i} i^{M} = \frac{t A_{M} (t)}{{(1 - t)}^{M + 1}}, A_{M} (t) = \sum_{g = 0}^{M - 1} 〈_{g}^{M}〉 t^{g}

| q | < 1, lim_{N \to \infty} \sum_{n = 0}^{N - 1} q^{n} n^{M} = \frac{A_{q}^{M}}{{(1 - t)}^{M + 1}} \to A_{t}^{M} = t A_{M} (t)

. There have 5 expressions for

A_{M} (t)

Eulerian Numbers and Polynomials is just a special case,we can handler:

\sum_{n = 0}^{N - 1} q^{n} \nabla^{p} S U M (n + Y, P S, P T), X = T_{M} - M - p

= \{\begin{matrix} \sum_{g = 0}^{M} H_{1} (g) \sum_{n = 0}^{N - 1} q^{n} (_{X + 1 + g}^{n + Y + X}) = \sum_{g = 0}^{M} H_{1} (g) {q^{N} \sum_{k = 0}^{X + 1 + g} {(- 1)}^{k} \frac{(_{X + 1 + g - k}^{N + Y + X - 1 - k})}{{(q - 1)}^{k + 1}} + \frac{q^{1 + g - Y}}{{(1 - q)}^{X + 2 + g}}} \\ \sum_{g = 0}^{M} H_{2} (g) \sum_{n = 0}^{N - 1} q^{n} (_{X + 1 + g}^{n + Y + X + g}) = \sum_{g = 0}^{M} H_{2} (g) {q^{N} \sum_{k = 0}^{X + 1 + g} {(- 1)}^{k} \frac{(_{X + 1 + g - k}^{N + Y + X + g - 1 - k})}{{(q - 1)}^{k + 1}} + \frac{q^{1 - Y}}{{(1 - q)}^{X + 2 + g}}} \\ \sum_{g = 0}^{M} H_{3} (g) \sum_{n = 0}^{N - 1} q^{n} (_{X + 1 + M}^{n + Y + X + M - g}) = \sum_{g = 0}^{M} H_{3} (g) {q^{N} \sum_{k = 0}^{X + 1 + M} {(- 1)}^{k} \frac{(_{X + 1 + M - k}^{N + Y + X + M - g - 1 - k})}{{(q - 1)}^{k + 1}} + \frac{q^{1 + g - Y}}{{(1 - q)}^{X + 2 + M}}} \end{matrix}

Theorem 61.

\sum_{n = 0}^{M} H_{1} (g) {(1 - q)}^{M - g} q^{g + 1} = q \sum_{n = 0}^{M} H_{2} (g) {(1 - q)}^{M - g} = \sum_{n = 0}^{M} H_{3} (g) q^{g + 1}

Here q can take any value,which is magical.

q = 0.5 \to

Definition 13.

A_{q} (P S, P T) =

Theorem 62.

X = T_{M} - p, \sum_{n = 0}^{N - 1} q^{n} \nabla^{p} S U M (n + Y) = \frac{q^{N}}{{(q - 1)}^{X + 2}} \sum_{k = 0}^{M} {(q - 1)}^{X - k} {(- 1)}^{k} \nabla^{p + K - 1} S U M (n + Y - 2) + \frac{A_{q} (P S, P T) q^{- Y}}{{(1 - q)}^{X + 2}}

Theorem 63.

| q | < 1, \sum_{n = 0}^{\infty} q^{n} \nabla^{p} S U M (n + Y, P S, P T) = \frac{A_{q} (P S, P T) q^{- Y}}{{(1 - q)}^{T_{M} + 2 - p}}

We can handler 63 of SUM(N,[a,a...a]:d,[1,2...M]),SUM(N,[1,1...1,2,2...2...k,k...k],[1,2...kM]). Many results of [7,8] can be obtained by this.

7. Formal Calculation of q-Binomial

7.1. Concept

q-Binomial:

{[_{M}^{N}]}_{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}

$G_{0}^{N} = 1, G_{M < 0, M > N}^{N} = 0, G_{M}^{N} = G_{N - M}^{N}$
$G_{M}^{N} = q^{M} G_{M}^{N - 1} + G_{M - 1}^{N - 1} = G_{M}^{N - 1} + q^{N - M} G_{M - 1}^{N - 1}$
$\sum_{n = 0}^{N - 1} q^{n} G_{M}^{n + K} = q^{M - K} G_{M + 1}^{N + K}$
$G_{K}^{M} = \sum_{w \in Ω (0^{M - K}, 1^{K})} q^{i n v (w)}$ [9]. $w_{1} \dots w_{M}$ with M-K(zeros) and K(ones),inv(·) denotes the inversion statistic.

The Formal Calculation use

q^{n} (K_{i} + G_{1}^{n} D_{i})

instead of

K_{i} + q^{n} D_{i}

Definition 14.

Recurssive define

\nabla_{q}^{p}, p \in N, \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)

Definition 15.

Recursive define

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

S U M_{q} (N, [K_{1} : D_{1}], [T_{1} = 1]) = \sum_{n = 0}^{N - 1} q^{n} (K_{1} + G_{1}^{n} 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} + G_{1}^{n} D_{2}) \nabla^{p} S U M_{q} (n + 1, [K_{1} : D_{1}], [T_{1}])

Theorem 64.

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

= 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}

= (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}

Use this to prove:

Theorem 65.

[2]

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

F o r m_{1} \to \sum_{g = 0}^{M} H_{1}^{q} (g) G_{N - 1 - g}^{N + T_{M} - M} = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{T_{M} - M + 1 + g}^{N + T_{M} - M}, 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_{N - 1}^{N + T_{M} - M + g} = \sum_{g = 0}^{M} H_{2}^{q} (g) G_{T_{M} - M + 1 + g}^{N + T_{M} - M + 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_{N - 1 - g}^{N + T_{M} - g} = \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^{q} (g) = \sum_{X (T) = g} \prod_{i = 1}^{M} B_{i}, lim_{q \to 1} S U M_{q} (N) = S U M (q), lim_{q \to 1} H^{q} (g) = H (g)

7.2. Property

Theorem 66.

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

Theorem 67.

S U M_{q} (N, [. . . P S . . .], [. . . T + 1, T + 2 . . . T + M . . .])

K_{i}

can exchange orders.

F o r m_{2}

is simplest,

X = T_{M} - M - p

, iii→

Theorem 68.

\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}

Definition 16.

n_{q -} = q^{- n} G_{1}^{n}, n_{q +} = q^{n} G_{1}^{n}, n_{q -}! = n_{q -} . . . 2_{q -} 1_{q -}, 0_{q -}! = 0, n_{q +}! i s s i m i l a r .

F o r m_{2} \to

Theorem 69.

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, P S, P T)

T_{1}

can great than 1,

T_{i} \in N

Theorem 70.

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}}

S U M_{q} (N, [1_{q -}, 2_{q -} . . . M_{q -}], [1, 2 . . . M]) \to

\sum_{n = 0}^{N - 1} q^{n} G_{1}^{n + 1} G_{1}^{n + 2} . . . G_{1}^{n + M} = G_{1}^{1} G_{1}^{2} . . . G_{1}^{M} G_{M + 1}^{N + M}

By definition and iv

Theorem 71.

I n H^{q} (g), \sum_{X_{i} \in K} \prod q^{X_{T}} = G_{M - g}^{M} = G_{g}^{M}

7.3. Application

Theorem 72.

\sum_{0 \leq λ_{1} \leq λ_{2} \leq . . . \leq λ_{M} \leq N} q^{λ_{1} + λ_{2} + . . . + λ_{M}} = G_{M}^{N + M} = G_{N}^{N + M}

[6]

Proof.

S U M_{q} (N + 1, [1, 1 . . . 1] : 0, [1, 3 . . . 2 M - 1]) \to H_{2}^{q} (g > 0) = 0, H_{2}^{q} (0) = 1 \to \sum_{λ_{M} = 0}^{N} q^{λ_{M}} . . . \sum_{λ_{1} = 0}^{λ_{2}} q^{λ_{1}} = G_{M}^{N + M}

□

Theorem 73.

\sum_{1 \leq λ_{1} < λ_{2} < . . . < λ_{M} \leq N + M} q^{λ_{1} + λ_{2} + . . . + λ_{M}} = q^{(_{2}^{M + 1})} G_{M}^{N + M}

Proof.

\sum_{λ_{M} = 0}^{N} q^{M + λ_{M}} . . . \sum_{λ_{2} = 0}^{λ_{3}} q^{2 + λ_{2}} \sum_{λ_{1} = 0}^{λ_{2}} q^{1 + λ_{1}} = q^{1 + 2 + . . . + M} \sum_{λ_{M} = 0}^{N} q^{λ_{M}} . . . \sum_{λ_{2} = 0}^{λ_{3}} q^{λ_{2}} \sum_{λ_{1} = 0}^{λ_{2}} q^{λ_{1}}

□

Theorem 74.

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

By simply following the definition of product, we can obtain:

Theorem 75.

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

A=-1 or 1,it’s q-Binomial Theorem:

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

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

T_{i} + 1 = T_{i + 1}, D_{i} = 1, K_{i} = K_{i_{q -}}

B_{i} o f H_{1, 2, 3}^{q} (g) = \{_{X_{_{i}} = K_{_{i_{q -}}}}^{X_{_{i}} = T} = \{_{q^{- K_{i}} G^{K_{i} + X_{T - 1}}}^{q^{X_{T}} G^{T_{i} - X_{K - 1}}}, = \{_{q^{- K_{i}} G^{K_{i} - T_{i} + X_{K - 1}}}^{q^{- (T_{i} - X_{K - 1})} G^{T_{i} - X_{K - 1}}}, = \{_{q^{- K_{i}} G^{K_{i} + X_{T - 1}}}^{q^{1 + X_{T}} G^{T_{i} - K_{i} - X_{T - 1}}}

It’s similar to 1.Replace each

B_{i}

G_{1}^{B_{i}}

H (g, [K_{1}, K_{2} . . . K_{M}], P T)

and multiply by

q^{?}

to obtain

H^{q} (g)

If there is another

K_{i} + 1 = K_{i + 1}

,74 can be used to obtain general formulas.

Theorem 76.

Expansion of ii:

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}

Proof.

S U M_{q} (N, [1_{q -}, 2_{q -} . . . M_{q -}], [1, 2 . . . M]) = M_{q -}! G_{M + 1}^{N + M} = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{1 + g}^{N} = S U M_{q} (N, [M_{q -}, . . ., 1_{q -}], P T)

H_{1}^{q} (g, T) = g_{q +}!, 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}}

\frac{H_{1}^{q} (g)}{M_{q -}!} = \frac{q^{(_{2}^{g + 1})} q^{- (M + 1) (M - g)} \sum_{1 \leq λ_{1} < . . . < λ_{M - g} \leq M} q^{λ_{1} + . . . + λ_{M - g}}}{q^{- (1 + 2 + . . . + M)}} = \frac{q^{(_{2}^{g + 1})} q^{- (M + 1) (M - g)} q^{(_{2}^{M - g + 1})} G_{M - g}^{M}}{q^{- (_{2}^{M + 1})}} = q^{(g + 1) g} G_{g}^{M}

□

Theorem 77.

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

Proof.

S U M_{q} (N + 1, [q^{1} : (q - 1) q^{1}, q^{2} : (q - 1) q^{2} . . . q^{M} : (q - 1) q^{M}], [1, 3 . . . 2 M - 1])

= \sum_{λ_{M} = 0}^{N} q^{2 λ_{M} + M} . . . \sum_{λ_{2} = 0}^{λ_{3}} q^{2 λ_{2} + 2} \sum_{λ_{1} = 0}^{λ_{2}} q^{2 λ_{1} + 1} = q^{(_{2}^{M + 1})} \sum_{0 \leq λ_{1} \leq . . . \leq λ_{M} \leq N} q^{2 (λ_{1} + . . . + λ_{M})} = q^{(_{2}^{M + 1})} {[_{M}^{N + M}]}_{q^{2}}

B_{i} o f H_{3}^{q} (g) = \{_{q^{X_{T - 1}} (K_{i} + G_{1}^{X_{T - 1}} D_{i}) = q^{X_{T - 1}} (q^{i} + G_{1}^{X_{T - 1}} q^{i} (q - 1)) = q^{2 X_{T - 1} + i} = q^{i + 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^{i + 1 + 2 X_{T - 1}} = - q^{i - 1 + 2 X_{T}}, X_{i} = T_{i}}

Factor

q^{i}

is present in all of them here, so

q^{(1 + 2 + . . . + M)}

can be extracted.

H_{3}^{q} (g) = {(- 1)}^{g} q^{(1 + 2 + . . . + M) - g + 2 (1 + 2 + . . . + g)} \sum_{X_{i} \in K} \prod q^{2 X_{T}} = {(- 1)}^{g} q^{(_{2}^{M + 1}) + g^{2}} {[_{g}^{M}]}_{q^{2}}

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

□

Definition 17.

_{q} E_{M}^{N} = \sum_{1 \leq λ_{1} \leq λ_{2} \leq . . \leq λ_{M} \leq N} G_{1}^{λ_{1}} G_{1}^{λ_{2}} . . . G_{1}^{λ_{M}} = S_{2}^{q} (N + M, N)

Theorem 78.

{(G_{1}^{N})}^{M} = {(1 + q + . . . + q^{N - 1})}^{M} = \sum_{g = 1}^{M} g_{q +}! S_{2}^{q} (M, g) G_{g}^{N} q^{- g}

Proof.

\nabla_{q}^{1} S U M (N, [1_{q -}, 1_{q -} . . . 1_{q -}], [1, 2 . . . M]) = \prod (\frac{1}{q} + \frac{q^{n} - 1}{q - 1}) = q^{- M} {(G_{1}^{n + 1})}^{M} = q^{- M} {(G_{1}^{N})}^{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}

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} \in K}^{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} \in T}

H_{1}^{q} (g, T) = q^{- (g + 1)} {(g + 1)}_{q +}!, H_{1}^{q} (g, \sum K) = {q^{- (M - 1 - g)}}_{q} E_{M - 1 - g}^{g + 1}

{(G_{1}^{N})}^{M} = q^{M - 1} \sum_{g = 0}^{M - 1} q^{- (g + 1)} {(g + 1)}_{q +} q^{- (M - 1 - g)} S_{2}^{q} (M, g + 1)! G_{1 + g}^{N} q^{- g}

□

Use

P S = [K_{i} + 1 : D_{i} (q - 1)], P T = [1, 2 . . . M] \to

Theorem 79.

\sum_{n = 0}^{N - 1} q^{n} \prod_{i = 1}^{M} (K_{i} + D_{i} q^{n}) = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{1 + g}^{N}, B_{i} = \{_{K_{i} + q^{X_{T}} D_{i}, X_{i} \in K}^{q^{X_{T}} (q^{X_{T}} - 1) D_{i}, X_{i} \in T}

K_{i} = 1, D_{i} = q^{- i} z \to \nabla S U M (M + 1) = \prod_{i = 1}^{M} (1 + q^{M - i} z) = \sum_{g = 0}^{M} H_{1}^{q} (g) G_{g}^{M} q^{- g} (*)

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

H_{1}^{q} (g) = q^{1 + 2 + . . . + g} z^{g} (q^{1} - 1) . . . (q^{g} - 1) [\sum_{- M \leq λ_{1} < . . . < λ_{g} \leq - 1} q^{λ_{1} + . . . + λ_{g}}] (1 + q^{- 1} z) . . . (1 + q^{- (M - g)} z)

(*) = \sum_{g = 0}^{M} q^{(_{2}^{g + 1})} z^{g} (q^{1} - 1) . . . (q^{g} - 1) q^{(_{2}^{g + 1}) - g (M + 1)} G_{g}^{M} (1 + q^{- 1} z) . . . (1 + q^{- (M - g)} z) G_{g}^{M} q^{- g}

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

Combining q-Binomial Theorem →

Theorem 80.

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

Use 79 and induction →

Theorem 81.

\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} = \{\begin{matrix} 1, X_{i} \in T, X_{T - 1} = 0 \\ q^{X_{T - 1}} (q^{X_{T - 1}} - 1) D_{i}, X_{i} \in T, X_{T - 1} > 0 \\ K_{i}, X_{i} \in K, X_{T - 1} = 0 \\ K_{i} + q^{X_{T - 1} - 1} D_{i}, X_{i} \in K, X_{T - 1} > 0 \end{matrix}

Theorem 82.

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}

Proof.

q^{M n} = \nabla_{q}^{1} S U M (N, [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} \in K}^{q^{1 + X_{T - 1}} G_{1}^{i - X_{K - 1}} (q - 1) = q^{X_{T}} (q^{X_{T}} - 1), X_{i} \in T}, H_{1}^{q} (g, T) = {(q - 1)}^{g} g_{q +}!, H_{1}^{q} (g, \sum K) = G_{g}^{M}

B_{i} = \{_{q^{- (1 + X_{T - 1})} = q^{- 1} q^{- X_{T}}, X \in K}^{q^{- (1 + X_{T - 1)}} (q^{1 + X_{T - 1}} - 1) = q^{- X_{T}} (q^{X_{T}} - 1), X \in T}, H_{2}^{q} (g, T) = {(q - 1)}^{g} g_{q -}!, H_{2}^{q} (g, \sum K) = q^{- (M - g)} {[_{g}^{M}]}_{q^{- 1}}

B_{i} = \{_{q^{X_{T - 1}} = q^{X_{T}}, X \in K}^{- q^{1 + X_{T - 1}} = - q^{X_{T}}, X \in T}, H_{3}^{q} (g, T) = {(- 1)}^{g} q^{(_{2}^{g + 1})}, H_{3}^{q} (g, \sum K) = G_{g}^{M}

□

Theorem 83.

\sum_{k_{r} = 1}^{N} . . . \sum_{k_{2} = 1}^{k_{3}} \sum_{k_{1} = 1}^{k_{2}} q^{k_{1} + k_{2} + . . . + k_{r}} \nabla_{q}^{p} S U M_{q} (k_{1}, P S, P T) = \nabla_{q}^{p - r} S U M_{q} (N, P S, P T)

Theorem 84.

\sum_{k_{x} = 1}^{N} . . . \sum_{k_{2} = 1}^{k_{3}} \sum_{k_{1} = 1}^{k_{2}} q^{k_{1} + k_{2} + . . . + k_{x}} \nabla_{q}^{1} S U M_{q} (k_{r}, P S, [1, 2 . . . M]) = \nabla_{q}^{r - x} S U M_{q} (N, P S, [T_{i} = i + (r - 1)])

Theorem 85.

\sum_{0 \leq n_{1} \leq . . . \leq n_{M} \leq N - 1} q^{n_{1} + n_{2} + . . . + n_{M}} (K + G_{1}^{n_{1}} D_{1} + . . . + G_{1}^{n_{M}} D_{M}) = {q^{1} D_{M} + q^{2} (D_{M} + D_{M + 1}) + . . . + q^{M} (D_{M} + D_{M + 1} + . . . + D_{1})} G_{M + 1}^{N + M - 1} + K G_{M}^{N + M - 1}

Theorem 86.

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

Proof.

P S = [K + D, K + D . . . K + D] : (q - 1) D, P T = [1, 2 . . . M]

B_{i} = \{_{K + D q^{X_{T}}, X \in K}^{q^{X_{T}} (q^{X_{T}} - 1) D, X \in T}, 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}, H_{1}^{q} (0) = {(K + D)}^{M}

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}

□

7.4. Relationships between $H^{q} (g)$

Theorem 87.

P T = [1, 2 . . . M], H_{1}^{q} (g) = q^{g (g + 1)} \sum_{k = g}^{M} H_{2}^{q} (k) G_{g}^{k}

[3]

Theorem 88.

P T = [1, 2 . . . M], H_{1}^{q} (g) = \sum_{k = 0}^{g} H_{3}^{q} (k) G_{M - g}^{M - k} q^{(g + 1) (g - k)}

Proof.

Direct verification when M=1, assuming M holds.

(*) H_{1}^{q} (P S 1, P T 1, g) = q^{g} G_{1}^{g} H_{1}^{q} (g - 1) + (K_{M + 1} + G_{1}^{g} D_{M + 1}) H_{1}^{q} (g)

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

(* *) H_{1}^{q} (P S 1, P T 1, g) = \sum_{x = 0}^{M + 1} H_{3}^{q} (P S 1, P T 1, x) G_{M + 1 - g}^{M + 1 - x} q^{(g + 1) (g - x)}

= \sum_{x = 0}^{M} H_{3}^{q} (x) {\frac{q^{M + 2} - q^{X + 1}}{q - 1} D_{M + 1} - K_{M + 1} q^{M + 2}} G_{M + 1 - g}^{M - x} q^{(g + 1) (g - x - 1)} + \sum_{x = 0}^{M} H_{3}^{q} (x) (K_{M + 1} + G_{1}^{X} D_{M + 1}) G_{M + 1 - g}^{M + 1 - x} q^{(g + 1) (g - x)}

Items containing

K_{M + 1}

K_{M + 1} G_{M + 1 - g}^{M + 1 - x} q^{(g + 1) (g - x)} - q^{M + 2} K_{M + 1} G_{M + 1 - g}^{M - x} q^{(g + 1) (g - x - 1)} = K_{M + 1} G_{M - g}^{M - x} q^{(g + 1) (g - x)}

Items does not contain

K_{M + 1}

I n (*) = q^{g} G_{1}^{g} D_{M + 1} G_{M - g + 1}^{M - x} q^{g (g - 1 - x)} + G_{1}^{g} D_{M + 1} G_{M - g}^{M - x} q^{(g + 1) (g - x)}

D i v i d e b y D_{M + 1} q^{g} q^{g (g - 1 - x)} {(q - 1)}^{- 1} = (q^{g} - 1) G_{M + 1 - g}^{M - x} + (q^{g} - 1) G_{M - g}^{M - x} q^{g - x} = (q^{g} - 1) G_{M + 1 - g}^{M + 1 - x}

I n (* *) = \frac{q^{M + 2} - q^{x + 1}}{q - 1} D_{M + 1} G_{M + 1 - g}^{M - x} q^{(g + 1) (g - x - 1)} + G_{1}^{x} D_{M + 1} G_{M + 1 - g}^{M + 1 - x} q^{(g + 1) (g - x)}

D i v i d e b y D_{M + 1} q^{g} q^{g (g - 1 - x)} {(q - 1)}^{- 1} = (q^{M + 1 - x} - 1) G_{M + 1 - g}^{M - x} + (q^{g} - q^{g - x}) G_{M + 1 - g}^{M + 1 - x} = (q^{g} - 1) G_{M + 1 - g}^{M + 1 - x}

□

Theorem 89.

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

Proof.

88 and 82 can obtain the first equation, the second equation is derived from 75. □

Similarly, using induction to prove:

Theorem 90.

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

Theorem 91.

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

Theorem 92.

PT = [1,2...M] and

D_{i} = 1

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

This indicates any

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

can be converted to

\frac{a_{M}}{M_{q +}!} \nabla^{A} S U M_{q} (N + B, [K_{1}, K_{2} . . . K_{M}], [1, 2 . . . M])

Use 90 , 91 and 68 ,we can obtain the inversion formulas.

Theorem 93.

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

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

7.5. Merge and Expand

Theorem 94.

Necessary and sufficient conditions for merging,

0 < K \leq M

$\sum_{n = 0}^{M} H (n) G_{Y + n}^{X} \to \sum_{n = 0}^{M - K} (. . .) G_{Y + K + n}^{X + K} : \sum_{x = g}^{M} {(- 1)}^{x} G_{g}^{x} q^{- x (x + 1) + (_{2}^{x - g})} H (x) q^{(1 - Y) x} = 0, 0 \leq g < K$
$\sum_{n = 0}^{M} H (n) G_{Y + n}^{X} \to \sum_{n = 0}^{M - K} (. . .) G_{Y + K + n}^{X + K} : \sum_{x = 0}^{g} {(- 1)}^{x} G_{M - g}^{M - x} q^{(g + 1) (g - x) - (_{2}^{g - x})} H (x) q^{(1 - Y) x} = 0, 0 \leq M - g < K$
$\sum_{n = 0}^{M} H (n) G_{Y + n}^{X + n} \to \sum_{n = 0}^{M - K} (. . .) G_{Y + K + n}^{X + K + n} : \sum_{x = g}^{M} H (x) G_{g}^{x} = 0, 0 \leq g < K$
$\sum_{n = 0}^{M} H (n) G_{M + Y}^{X - n} \to \sum_{n = 0}^{M - K} (. . .) G_{M + Y - K}^{X - K - n} : \sum_{x = 0}^{g} H (x) G_{M - g}^{M - x} q^{(g + 1) (g - x)} q^{(1 - Y) x} = 0, 0 \leq M - g < K$

Y=1,76 →:

Theorem 95.

\sum_{x = 0}^{M} {(- 1)}^{x} G_{g}^{x} G_{x}^{M} q^{(_{2}^{x - g})} = 0, 0 \leq g < M

\sum_{x = 0}^{M} {(- 1)}^{x} G_{x}^{M} q^{(_{2}^{x})} = 0

Theorem 96.

P=A+T+1-Y,M > A ≥ 0,

\sum_{g = 0}^{M} G_{g}^{M} G_{A}^{A + T + g} G_{Y + g}^{X} q^{g (g + 1 + T - P)} = \sum_{x = 0}^{A} G_{x + T}^{A + T} G_{M + T}^{M + T + x} G_{Y + M - A + x}^{X + M - A} q^{x (x + 1 + T - P)}

Proof.

S U M_{q} (N, [{(T + 1)}_{q -}, {(T + 2)}_{q -} . . . {(T + M)}_{q -}], [T + A + 1 . . . T + A + M]), B_{i} o f H_{1}^{q} (g) = \{_{q^{- (T + i)} G_{1}^{T + i + X_{T}}, X_{i} = K_{i}}^{q^{X_{T}} G_{1}^{T + A + i - X_{T}}, X_{i} = T_{i}}

= \sum_{g = 0}^{M} G_{A + T + 1 + g}^{X} q^{\frac{g (1 + g)}{2}} G_{1}^{T + A + 1} . . . G_{1}^{T + A + g} \times G_{1}^{T + M} . . . G_{1}^{T + g + 1} q^{- (M - g) (T + M + 1)} \sum_{1 \leq λ_{1} < . . . < λ_{M - g} \leq M} q^{\sum λ_{i}}

= \frac{(q^{A} - 1) . . . (q - 1)}{{(q - 1)}^{M}} \sum_{g = 0}^{M} G_{A + T + 1 + g}^{X} q^{\frac{g (1 + g)}{2} - (M - g) (T + M + 1) + \frac{(M - g + 1) (M - g)}{2}} \times (q^{T + M} - 1) . . . (q^{T + A + 1} - 1) G_{A}^{A + T + g} G_{g}^{M}

= q^{- (T + A + 1)} G_{1}^{T + A + 1} . . . q^{- (T + M)} G_{1}^{T + M} S U M_{q} (N, [{(T + 1)}_{q -} . . . . {(T + A)}_{q -}], [T + M + 1 . . . T + M + A])

= q^{\frac{- (M - A) (T + A + 1 + T + M)}{2}} G_{1}^{T + A + 1} . . . G_{1}^{T + M} \sum_{x = 0}^{A} G_{M + T + 1 + x}^{X + M - A} q^{\frac{x (1 + x)}{2} - (A - x) (T + A + 1) + \frac{(A - x + 1) (A - x)}{2}} \times G_{1}^{T + M + 1} . . G_{1}^{T + M + x} \times G_{1}^{T + A} . . . G_{1}^{T + x + 1} G_{x}^{A}

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

\to \sum_{g = 0}^{M} G_{g}^{M} G_{A}^{A + T + g} G_{A + T + 1 + g}^{X} q^{g (g + 1 + T)} = \sum_{x = 0}^{A} G_{x + T}^{A + T} G_{M + T}^{M + T + x} G_{M + T + 1 + x}^{X + M - A} q^{x (x + 1 + T)}

□

The difference between this and 34 is that M>A is required.

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

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

. It’s the q-Vandermorde Theorem.

0 \leq B + A < M, \to \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}^{X_{1} + g} G_{B}^{g} q^{(_{2}^{g}) - g (A + B)} {(- 1)}^{g} = 0

A and B have symmetry,ii→.

Theorem 97.

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

Theorem 98.

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

Proof.

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})} \times G_{1}^{T + K + M + 1} . . . G_{1}^{T + K + M + g} \times G_{1}^{T + M} . . . G_{1}^{T + 1 + g} \times G_{1}^{T + M + 1} G_{g}^{M}

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} = {(- 1)}^{M} q^{- M (T + 1 + T + M)} G_{1}^{K + M} G_{1}^{K + M - 1} . . . G_{1}^{K + 1}

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

\to {(- 1)}^{M} G_{K}^{T + K + M} = \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}

□

7.6. Matrix of $S U M_{q} (N)$

Let

P = N + T_{M} - M, Q = N - 1

, corresponding to the three forms,define

A_{1, 2, 3}^{q} (P, Q, M) =

(\begin{matrix} G_{Q}^{P} & \dots & G_{Q - M}^{P} \\ ⋮ & ⋱ & ⋮ \\ G_{Q + M}^{P + M} & \dots & G_{Q}^{P + M} \end{matrix}), (\begin{matrix} G_{Q}^{P} & \dots & G_{Q}^{P + M} \\ ⋮ & ⋱ & ⋮ \\ G_{Q + M}^{P + M} & \dots & G_{Q + M}^{P + 2 M} \end{matrix}), (\begin{matrix} G_{Q}^{P + M} & \dots & G_{Q - M}^{P} \\ ⋮ & ⋱ & ⋮ \\ G_{Q + M}^{P + 2 M} & \dots & G_{Q}^{P + M} \end{matrix})

Theorem 99.

‖ A_{2}^{q} (P, Q, M) ‖ = ‖ A_{2}^{q} (P, P - Q, M) ‖ = \frac{G_{Q}^{P}}{G_{0}^{P}} \frac{G_{Q + 1}^{P + 2}}{G_{1}^{P + 2}} . . . \frac{G_{Q + M}^{P + 2 M}}{G_{M}^{P + 2 M}} q^{(P + 1) + 2 (P + 2) + . . . + M (P + M)}

‖ A_{2}^{q} (P, 0, M) ‖ = q^{(P + 1) + 2 (P + 2) + . . . + M (P + M)}, ‖ A_{2}^{q} (P, 1, M) ‖ = G_{M + 1}^{P + M} q^{(P + 1) + 2 (P + 2) + . . . + M (P + M)}

Theorem 100.

‖ A_{1, 3}^{q} (P, Q, M) ‖ = \frac{G_{Q}^{P}}{G_{0}^{P}} \frac{G_{Q + 1}^{P + 2}}{G_{1}^{P + 2}} . . . \frac{G_{Q + M}^{P + 2 M}}{G_{M}^{P + 2 M}} q^{Q (_{2}^{M + 1})}

8. Multi-parameter Formal Calculation

2-parameters Formal Calculation calculate nested sum of

K_{i} + (_{1}^{n}) D_{1, i} + (_{2}^{n}) D_{2, i}

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

S U M (N, [K_{1}, K_{2}], [T_{1, 1} : D_{1, 1}, T_{1, 2} = T_{1, 1} + 2 - p : D_{1, 2}], [T_{2, 1} = T_{1, 1} : D_{2, 1}, T_{2, 2} = T_{1, 2} : D_{2, 2}])

= \sum_{n = 0}^{N - 1} (K_{2} + D_{1, 2} n + D_{2, 2} (_{2}^{n})) \nabla^{p} S U M (N, [K_{1}], [T_{1, 1} : D_{1, 1}], [T_{2, 1} : D_{2, 1}])

Recursive define

S U M (N, P S, P T_{1}, P T_{2})

. There is always

T_{i} = T_{1, i} = T_{2, i}

Use the Form:

(K_{1} + T_{1, 1} + T_{2, 1}) (K_{2} + T_{1, 2} + T_{2, 2}) . . . (K_{M} + T_{1, M} + T_{2, M})

Use

T_{1}

to represent the set {

T_{1, 1}, T_{1, 2} . . . T_{1, M}

T_{2}

to represent the set {

T_{2, 1}, T_{2, 2} . . . T_{2, M}

Definition 18.

P T_{1}

)=Number of

{X_{1} . . . X_{M}} \in T_{1}

, X(

P T_{2}

)=Number of

{X_{1} . . . X_{M}} \in T_{2}

,X(PT)=X(

P T_{1}

)+2X(

P T_{2}

)

Definition 19.

X_{P T 1}

=Number of

{X_{1} . . . X_{i}} \in T_{1}

X_{P T 2}

=Number of

{X_{1} . . . X_{i}} \in T_{2}

X_{P T} = X_{P T 1} + 2 X_{P T 2}

Theorem 101.

S U M (N, P S, P T_{1}, P T_{2})

= \sum_{g = 0}^{2 M} H_{1} (g) (_{T_{M} - M + 1 + g}^{N + T_{M} - M}), B_{i} = \{\begin{matrix} K_{i} + X_{P T} D_{1, i} + (_{2}^{X_{P T}}) D_{2, i}, X_{i} = K_{i} \\ (T_{i} - i + X_{P T}) D_{1, i} + (T_{i} - i + X_{P T}) (X_{P T} - 1) D_{2, i}, X_{i} = T_{1, i} \\ (_{2}^{T_{i} - i + X_{P T}}) D_{2, i}, X_{i} = T_{2, i} \end{matrix}

= \sum_{g = 0}^{2 M} H_{2} (g) (_{T_{M} - M + 1 + g}^{N + T_{M} - M + g}), B_{i} = \{\begin{matrix} K_{i} - (T_{i} - i + X_{P T} + 1) D_{1, i} + (_{2}^{T_{i} - i + X_{P T} + 2}) D_{2, i}, X_{i} = K_{i} \\ (T_{i} - i + X_{P T}) D_{1, i} - (T_{i} - i + X_{P T}) (T_{i} - i + X_{P T} + 1) D_{2, i}, X_{i} = T_{1, i} \\ (_{2}^{T_{i} - i + X_{P T}}) D_{2, i}, X_{i} = T_{2, i} \end{matrix}

= \sum_{g = 0}^{2 M} H_{3} (g) (_{T_{M} + M + 1}^{N + T_{M} + M - g}), B_{i} = \{\begin{matrix} K_{i} + X_{P T - 1} D_{1, i} + (_{2}^{X_{P T - 1}}) D_{2, i}, X_{i} = K_{i} \\ - 2 K_{i} + (T_{i} + i - 1 - 2 X_{P T - 1}) D_{1, i} + (T_{i} + i - X_{P T - 1}) D_{2, i}, X_{i} = T_{1, i} \\ K_{i} - (T_{i} + i - 1 - X_{P T - 1}) D_{1, i} + (_{2}^{T_{i} + i - X_{P T - 1}}) D_{2, i}, X_{i} = T_{2, i} \end{matrix}

According to this way, it can be extended to multi-parameter

S U M (N)

and

S U M_{q} (N)

. This formula is complex and has not been studied in terms of analysis yet.

9. A theorem of symmetry

In this section,

T_{i} \geq i

\to H_{1} (g, P T, P T) = \prod_{i = 1}^{M} T_{i} (_{g}^{M}) = \prod_{i = 1}^{M} T_{i} (_{g, M - g}^{M})

.Promoted it:Set T come from p Source:

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

Definition 20.

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

Definition 21.

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}

Definition 22.

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

In set T,

g_{1}

come from

S_{1}

g_{2}

comes from

S_{2}

...

g_{M}

comes from

S_{M}

Theorem 102.

W (g_{1}, g_{2} . . . g_{p}, [T_{1}, T_{2} . . . T_{M}]) = \prod_{i = 1}^{M} T_{i} (_{g_{1}, g_{2}, . . ., g_{M}}^{M})

Definition 23.

W_{q} (g_{1}, g_{2} . . . g_{p}, [T_{1}, T_{2} . . . T_{M}]) = \sum_{g_{i} + g_{2} + . . . + 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}

Definition 24.

G_{g_{1}, g_{2} . . . g_{p}}^{M} = \frac{(q^{M} - 1) (q^{M - 1} - 1) . . . (q^{1} - 1)}{\prod_{i = 0}^{p} (q^{g_{i}} - 1) (q^{g_{i} - 1} - 1) . . . (q^{i} - 1)}, g_{1} + g_{2} + . . . + g_{p} = M

Theorem 103.

W_{q} (g_{1}, g_{2} . . . g_{p}, [T_{1}, T_{2} . . . T_{M}]) = (\prod_{i = 1}^{M} G_{1}^{T_{i}}) G_{g_{1}, g_{2} . . . g_{p}}^{M}, T_{i} \geq i

Proof.

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}

.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 \frac{(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 \frac{(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→It’s holds when p=2.

W_{q} (g_{1}, g_{2} + g_{3}, [T_{1}, T_{2} . . . T_{M}]) = (\prod_{i = 1}^{M} G_{1}^{T_{i}}) G_{g_{1}, g_{2} + g_{3}}^{g_{1} + g_{2} + g_{3}}

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 (v a r i a n t f a c t o r s) = W_{q} (g_{2}, g_{3}, [X_{1}, X_{2} . . . X_{g_{2} + g_{3}}]) = \prod_{i = 1}^{g_{2} + g_{3}} G_{1}^{X_{i}} G_{g_{2}, g_{3}}^{g_{2} + g_{3}}

W_{q} (g_{1}, g_{2}, g_{3}, [T_{1}, T_{2} . . . T_{M}]) = (\prod_{i = 1}^{M} G_{1}^{T_{i}}) 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}}

□

e g : W_{q} (1, 2, [1, 2, 3]) = S_{1} S_{2} S_{2} + S_{2} S_{1} S_{2} + S_{2} S_{2} S_{1} = G_{1}^{1} G_{1}^{3} G_{1}^{4} + G_{1}^{1} G_{1}^{1} q G_{1}^{4} + G_{1}^{1} G_{1}^{2} G_{1}^{1} q^{2} = G_{1}^{1} G_{1}^{2} G_{1}^{3} G_{1, 2}^{3}

e g : W h e n W_{q} (1, 2, [1, 2, 3]) c h a n g e s t o W_{q} (1, 1, 1, [1, 2, 3])

= S_{1} S_{2} S_{3} + S_{1} S_{3} S_{2} + S_{2} S_{1} S_{3} + S_{3} S_{1} S_{2} + S_{2} S_{3} S_{1} + S_{3} S_{2} S_{1}

= G_{1}^{1} {G_{1}^{3} G_{1}^{5} + G_{1}^{3} G_{1}^{3} q} + G_{1}^{1} q {G_{1}^{1} G_{1}^{5} + G_{1}^{1} G_{1}^{3} q} + G_{1}^{1} q^{2} {G_{1}^{1} G_{1}^{3} + G_{1}^{1} G_{1}^{1} q}

= G_{1}^{1} {G_{1}^{3} G_{1}^{4} G_{1, 1}^{2}} + G_{1}^{1} q {G_{1}^{1} G_{1}^{4} G_{1, 1}^{2}} + G_{1}^{1} q^{2} {G_{1}^{1} G_{1}^{2} G_{1, 1}^{2}} = G_{1}^{1} G_{1}^{2} G_{1}^{3} G_{1, 2}^{3} \times G_{1, 1}^{2} = G_{1}^{1} G_{1}^{2} G_{1}^{3} G_{1, 1, 1}^{3}

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Formal Calculation

Abstract

1. Introduction

2. Property

2.1. Relationships between H ( g )

2.2. Property of H(g)

2.3. Shape of numbers

2.4. H(g) and Associated Stirling Numbers

2.5. Table of H(g)

3. Application

3.1. Number analysis

3.2. Merge and Expand

3.3. Congruences

4. Combinatorial Identities

5. Matrix of SUM(N)

6. Eulerian polynomials and Beyond

7. Formal Calculation of q-Binomial

7.1. Concept

7.2. Property

7.3. Application

7.4. Relationships between H q ( g )

7.5. Merge and Expand

7.6. Matrix of S U M q ( N )

8. Multi-parameter Formal Calculation

9. A theorem of symmetry

References

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2.1. Relationships between $H (g)$

7.4. Relationships between $H^{q} (g)$

7.6. Matrix of $S U M_{q} (N)$