Formal Calculation

1. Introduction

Formal Calculation has gone through 3 years of development, and this article contains its summary and latest achievements. The concept is introduced in [1,2,3,4] .

Definition 1. 

Recursive define the operator ${\nabla }^p$,p is an integer.

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

Definition 2. 

Recursive define SUM(N,PS,PT),abbreviated as SUM(N).$K_i,D_i\in$ Ring with identity elements.

$$SUM(N,[K_1:D_1],[T_1=1])={\displaystyle \sum _{n=0}^{N-1}}(K_1+n{D}_1)$$

$$SUM(N,[K_1:D_1,K_2:D_2],[T_1,T_2=T_1+2-p])={\displaystyle \sum _{n=0}^{N-1}}(K_2+n{D}_2){\nabla }^{p}SUM(n+1,[K_1:D_1],[T_1])$$

$$Iff(N)=\sum A_i\left({}_{M_i}^{N_i}\right)and{M}_{i}isnotchangedwithN,then{\nabla }^{p}f(N)=\sum A_i\left({}_{M_i-p}^{N_i-p}\right)$$

$[K_1:D,K_2:D...K_M:D]$ 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:

$$SUM(N,PS,[1,2,3...M])={\displaystyle \sum _{n=0}^{N-1}}{\displaystyle \prod _{i=1}^M}(K_i+n{D}_i)$$

$$SUM(N,PS,[1,3,5...2M-1])={\displaystyle \sum _{n_M=0}^{N-1}}(K_M+n_M{D}_M)...{\displaystyle \sum _{n_2=0}^{n_3}}(K_2+n_2{D}_2){\displaystyle \sum _{n_1=0}^{n_2}}(K_1+n_1{D}_1)$$

$$SUM(N,PS,[1,2,4])={\displaystyle \sum _{n_3=0}^{N-1}}(K_3+n_3{D}_3){\displaystyle \sum _{n=0}^{n_3}}(K_1+n{D}_1)(K_2+n{D}_2)$$

$$SUM(N,PS,[1,3,4])={\displaystyle \sum _{n_3=0}^{N-1}}(K_3+n_3{D}_3)(K_2+n_3{D}_2){\displaystyle \sum _{n=0}^{n_3}}(K_1+n{D}_1)$$

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 Form: $(K_1+T_1)(K_2+T_2)...(K_M+T_M)=\sum {\displaystyle \prod _{i=1}^M}{X}_i,X_i=T_{i}or{K}_i$

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 $\{X_1,X_2...X_{i-1}\}\in K$

Obviously:$X_{T-1}+X_{K-1}=i-1$

Theorem 1.1. 

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

$$For{m}_1\to {\displaystyle \sum _{g=0}^M}{H}_1(g)\left({}_{N-1-g}^{N+T_M-M}\right)={\displaystyle \sum _{g=0}^M}{H}_1(g)\left({}_{T_M-M+1+g}^{N+T_M-M}\right),B_i=\left\{{}_{K_i+X_{T-1}{D}_i,X_i=K_i}^{(T_i-X_{K-1})D_i,X_i=T_i}\right.$$

$$For{m}_2\to {\displaystyle \sum _{g=0}^M}{H}_2(g)\left({}_{N-1}^{N+T_M-M+g}\right)={\displaystyle \sum _{g=0}^M}{H}_2(g)\left({}_{T_M-M+1+g}^{N+T_M-M+g}\right),B_i=\left\{{}_{K_i+(X_{K-1}-T_i)D_i,X_i=K_i}^{(T_i-X_{K-1})D_i,X_i=T_i}\right.$$

$$For{m}_3\to {\displaystyle \sum _{g=0}^M}{H}_3(g)\left({}_{N-1-g}^{N+T_M-g}\right)={\displaystyle \sum _{g=0}^M}{H}_3(g)\left({}_{T_M+1}^{N+T_M-g}\right),B_i=\left\{{}_{K_i+X_{T-1}{D}_i,X_i=K_i}^{-K_i+(T_i-X_{T-1})D_i,X_i=T_i}\right.$$

The factors of $\prod X_i$ cannot be exchanged.$H_i(g)$, short for $H_i(g,PS,PT)$,is also defined above=${\displaystyle \sum _{X(T)=g}}{\displaystyle \prod _{i=1}^M}{B}_i$

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

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{N-1}}n\left({}_M^{n+K}\right)=(M+1)\left({}_{M+2}^{N+K}\right)+(M-K)\left({}_{M+1}^{N+K}\right)\hfill \\ =(M+1)\left({}_{M+2}^{N+K+1}\right)-(1-K)\left({}_{M+1}^{N+K}\right)=(M-K)\left({}_{M+2}^{N+K+1}\right)+(1+K)\left({}_{M+2}^{N+K}\right)\hfill \end{array}$$

Definition 5. 

$F_M^{K=\{K_1,K_2...K_M\}}$=∑(M-dictinct products,factors∈ K),$F_M^N$ is short for $F_M^{\{1,2...N\}}$,$F_0^K=0$

Definition 6. 

$E_M^{K=\{K_1,K_2...K_M\}}$=∑(M-products,factors∈ K),$E_M^N$ is short for $E_M^{\{1,2...N\}}$,$E_0^K=0$

Theorem 1.2. 

$\nabla SUM(N,PS,[1,2...M])={\displaystyle \prod _{i=1}^M}(K_i+(N-1)D_i)={\displaystyle \prod _{i=1}^M}(K_i+n{D}_i)$

Theorem 1.3. 

In SUM(N,[...PS...],[...T+1,T+2...T+M...]),$K_i$ can exchange orders.

Theorem 1.4. 

$SUM(N,[L_1,L_2...L_Q,PS],[L_1,L_2...L_Q,PT])={\displaystyle \prod _{i=1}^Q}{L}_{i}SUM(N,PS,PT)\to$$T_1$ can great than 1,$T_i\in \mathbb{N}$

Theorem 1.5. 

$SUM(N,[1,1...1],[1,2...M])=SUM(N,[1,1...1],[2,3...M])=1^M+2^M+...+N^M$

Theorem 1.6. 

$SUM(N,[1,1...1],[1,3...2M-1])=SUM(N,[1,1...1],[3,5...2M-1])$

$$={\displaystyle \sum _{{\lambda }_1+..+{\lambda }_N=M,{\lambda }_i\ge 0}}{1}^{{\lambda }_1}{2}^{{\lambda }_2}...N^{{\lambda }_N}={\displaystyle \sum _{1\le i_1\le i_2\le ...\le i_M\le N}}{i}_1{i}_2...i_M=E_M^N=S_2(N+M,N)$$

$S_2$ is Stirling numbers of the second kind.

Theorem 1.7. 

$SUM(N,[1,2...M],[1,3...2M-1])=SUM(N,[2,3...M],[3,5...2M-1])$

$$={\displaystyle \sum _{1\le i_1<i_2<...<i_M\le N+M-1}}{i}_1{i}_2...i_M=F_M^{N+M-1}=S_1(N+M,N)$$

$S_1$ is unsigned Stirling numbers of the first kind.

Example 1.1:

$$SUM(N,[1,2,3],[1,3,5])\to Form=(1+T_1)(2+T_2)(3+T_3)$$

$${\displaystyle \sum _{X(T)=1}}\prod X_i=1\times 2\times T_3+1\times T_2\times 3+T_1\times 2\times 3$$

$H_1(1)=1\times 2\times (T_3-X_{K-1})+1\times (T_2-X_{K-1})\times (3+X_{T-1})+T_1\times (2+X_{T-1})\times (3+X_{T-1})=1\times 2\times (5-2)+1\times (3-1)\times (3+1)+1\times (2+1)\times (3+1)=26$

$$SUM(N,[1,2,3],[1,3,5])=1\times 3\times 5\left({}_6^{N+2}\right)+35\left({}_5^{N+2}\right)+26\left({}_4^{N+2}\right)+1\times 2\times 3\left({}_3^{N+2}\right)$$

$$SUM(N,[2,3],[3,5])=3\times 5\left({}_6^{N+3}\right)+(2\times 4+3\times 4)\left({}_5^{N+3}\right)+2\times 3\left({}_4^{N+3}\right)$$

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

$SUM(N,[\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right):\left(\begin{array}{cc}1& 3\\ 4& 2\end{array}\right),\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right):\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)],[1,3])=\left(\begin{array}{cc}15& 27\\ 30& 48\end{array}\right)\left({}_4^{N+1}\right)+\left(\begin{array}{cc}44& 70\\ 28& 38\end{array}\right)\left({}_3^{N+1}\right)+\left(\begin{array}{cc}17& 13\\ 4& 5\end{array}\right)\left({}_2^{N+1}\right)$

$H_1(1)=1\times \left(\begin{array}{cc}1& 3\\ 4& 2\end{array}\right)[\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)+\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)]+\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right)(3-1)\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}44& 70\\ 28& 38\end{array}\right)$

2. Property

2.1. Equivalence of three forms

By definition:

• $H_1(g,PS1,PT1)=H_1(g-1)(T_{M+1}-[M-(g-1)])D_{M+1}+H_1(g)(K_{M+1}+g{D}_{M+1})$
• $H_2(g,PS1,PT1)=H_2(g-1)(T_{M+1}-[M-(g-1)])D_{M+1}+H_2(g)(K_{M+1}+[M-g-T_{M+1}]D_{M+1})$
• $H_3(g,PS1,PT1)=H_3(g-1)(-K_{M+1}+(T_{M+1}-[g-1])D_{M+1})+H_3(g)(K_{M+1}+g{D}_{M+1})$

Using these relationships and induction can prove:

Theorem 2.1. 

$H_1(g)={\displaystyle \sum _{k=g}^M}{H}_2(k)\left({}_g^k\right)={\displaystyle \sum _{k=0}^g}{H}_3(k)\left({}_{M-g}^{M-k}\right)$ [3]

Inversion→

Theorem 2.2. 

$H_2(g)={\displaystyle \sum _{k=g}^M}{(-1)}^{k+g}{H}_1(k)\left({}_g^k\right),H_3(g)={\displaystyle \sum _{k=0}^g}{(-1)}^{k+g}{H}_1(k)\left({}_{M-g}^{M-k}\right)$

Calculation with 2.1 →

Theorem 2.3. 

${\displaystyle \sum _{g=0}^M}{H}_1(g)={\displaystyle \sum _{g=0}^M}{H}_2(g)2^g={\displaystyle \sum _{g=0}^M}{H}_3(g)2^{M-g}$

Theorem 2.4. 

${\displaystyle \sum _{g=0}^M}{H}_1(g)\left({}_{B-g}^A\right)={\displaystyle \sum _{g=0}^M}{H}_2(g)\left({}_B^{A+g}\right)={\displaystyle \sum _{g=0}^M}{H}_3(g)\left({}_{B-g}^{A+M-g}\right),A,B\in \mathbb{N}$

This $\to For{m}_1=For{m}_2=For{m}_3.A=B\to$

Theorem 2.5. 

${\displaystyle \sum _{g=0}^M}{H}_1(g)\left({}_g^A\right)={\displaystyle \sum _{g=0}^M}{H}_2(g)\left({}_g^{A+g}\right)={\displaystyle \sum _{g=0}^M}{H}_3(g)\left({}_M^{A+M-g}\right),A,B\in \mathbb{N}$

Induction [3]→

Theorem 2.6. 

${\displaystyle \sum _{g=0}^M}{H}_1(g)g\left({}_{B-g}^{A+1}\right)={\displaystyle \sum _{g=0}^M}{H}_2(g)g\left({}_{B-1}^{A+g}\right)={\displaystyle \sum _{g=0}^M}\{H_3(g)g\left({}_{B-g}^{A+M-g}\right)+M\times H_3(g)\left({}_{B-1-g}^{A+M-g}\right)\}$

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. 

$H(g,T),shortforH(g,T,PS,PT)={\displaystyle \prod _{X_i\in T}}{B}_i,H(g,\sum T)=\sum {\displaystyle \prod _{X_i\in T}}{B}_i$

Also define $H(g,K),H(g,\sum K)$

Theorem 2.7. 

If $D_i=1$ and $T_i+1=T_{i+1}$,$H_1(g,T)={\displaystyle \prod _{i=1}^g}{T}_i$. $H_1(g,T,[K_1,K_2...K_M],[1,2...M])=g!$

Theorem 2.8. 

$H_1(g,\sum K,[1,1...1],PT)=E_{M-g}^{g+1}$

Theorem 2.9. 

If $D_i=1$,$H_1(g,\sum K)=F_{M-g}^K{E}_0^g+F_{M-g-1}^K{E}_1^g+...+F_0^K{E}_{M-g}^g$

Theorem 2.10. 

If $D_i=1$,$H_1(g,\sum T)=F_g^T{E}_0^{M-g}-F_{g-1}^T{E}_1^{M-g}+...+{(-1)}^g{F}_0^T{E}_g^{M-g}$

Theorem 2.11. 

If $D_i=1$ and $K_{i+1}-K_i=T_{i+1}-T_i=1$,$H_1(g)$=$\left({}_g^M\right)T_1...T_g\times K_{g+1}...K_M$

Theorem 2.12. 

If PS=PT,$H_1(g)={\displaystyle \prod _{i=1}^M}{T}_i\left({}_N^M\right),H_2(M)=H_3(0)={\displaystyle \prod _{i=1}^M}{T}_i,H_2(g<M)=H_3(g>0)=0$

Theorem 2.13. 

$H_1(g,[AD:D,PS],[A,PT])=AD(H_1(g-1)+H_1(g))\to H_1(g,[1,PS],[1,PT])=H_1(g-1)+H_1(g)$

Definition 8. 

$E_p^q\odot ([T_1,T_2....T_M],C)$

$={\displaystyle \sum _{{\lambda }_1+{\lambda }_2+...+{\lambda }_q=p,{\lambda }_i\ge 0}}{1}^{{\lambda }_1}{2}^{{\lambda }_2}...q^{{\lambda }_q}(T_1+{\lambda }_{1}C)(T_2+{\lambda }_{1}C+{\lambda }_{2}C)...(T_{q-1}+{\lambda }_{1}C+{\lambda }_{2}C+...+{\lambda }_{q-1}C)$

[5] has proved:$〈{}_g^M〉=〈{}_{M-1-g}^M〉=E_{M-g-1}^{g+1}\odot ([1,1...1],1])$

$〈{}_g^M〉$ is Eulerian numbers.Worpitzky identity:$N^M={\displaystyle \sum _{g=0}^{M-1}}〈{}_g^M〉\left({}_M^{N+g}\right)$

$$〈{}_2^5〉=E_2^3\odot ([1,1...],1)={\displaystyle \sum _{{\lambda }_1+{\lambda }_2+{\lambda }_3=2}}{1}^{{\lambda }_1}{2}^{{\lambda }_2}{3}^{{\lambda }_3}(1+{\lambda }_1)(1+{\lambda }_1+{\lambda }_2)$$

$$=1^2{2}^0{3}^0(1+2)(1+2)+1^0{2}^2{3}^0(1+0)(1+0+2)+1^0{2}^0{3}^2(1+0)(1+0+0)$$

$$+1^1{2}^1{3}^0(1+1)(1+1+1)+1^1{2}^0{3}^1(1+1)(1+1+0)+1^0{2}^1{3}^1(1+0)(1+0+1)=66$$

Theorem 2.14. 

$H_1(g,[1,1...1],PT=[T_i=T_1+(C+1)(i-1)])=E_{M-g}^{g+1}\odot (PT,C)$

Theorem 2.15. 

$H_3(g,[1,1...1],PT=[T_i=T_1+(C+1)(i-1)])=E_{M-g}^{g+1}\odot ([T_i=T_1-1+C(i-1)],C+1)$

2.3. Shape of numbers

In this section,if not specifically mentioned,$T_1=1,T_{i+1}-T_i=1or2$.

To calculate ${\displaystyle \sum _{1\le K_1<K_2<...<K_M\le N}}{K}_1{K}_2...K_M$ (*),products needs to be divided into $2^{M-1}$ categories.

There are M-1 intervals between factors.If the interval=1,define as continuity. If the interval>1,define as discontinuity. Continuities, Discontinuities and their Positions, is defined as Shape. So there have $2^{M-1}$ Shapes.From the definition of nested sum:

$${\displaystyle \sum _{1\le K_1<K_2\le N}}{K}_1{K}_2=SUM(N,[1,2],[1,2])+SUM(N-1,[1,3],[1,3])$$

$${\displaystyle \sum _{1\le K_1<K_2<K_3\le N}}{K}_1{K}_2{K}_3=$$

$$SUM(N,[1,2,3],[1,2,3])+SUM(N-1,[1,2,4],[1,2,4])+SUM(N-1,[1,3,4],[1,3,4])+SUM(N-2,[1,3,5],[1,3,5])$$

Definition 9. 

PB(PT)=Number of $T_i+1<T_{i+1}$=Number of discontinuities

$(*)={\displaystyle \sum _{AlloftheShapes}}SUM(N-PB(PT),PT,PT)$.From 2.12 we can obtain a simple formula:

Theorem 2.16. 

$SUM(N,PT,PT)={\displaystyle \prod _{i=1}^M}{T}_i\left({}_{M+1}^{N+T_M}\right),T_i\in \mathbb{N}$

This is an important conclusion,generalizes the famous formula ${\displaystyle \sum _{n=0}^{N-1}}\left({}_M^n\right)=\left({}_{M+1}^N\right)$. It was discovered during the calculation of (*) and leads to the birth of Formal Calculation. There are two ways to understand SUM(N,PT,PT),one is nested sum,another is the intervals between factors.

Theorem 2.17. 

Number of Products in SUM(N,PT,PT) =$\left({}_{PB(PT)+1}^{N+PB(PT)}\right)$

Definition 10. 

$MI{N}_g(M)={\displaystyle \sum _{PB(PT)=g}}{\displaystyle \prod _{i=1}^M}{T}_i=1\times {\displaystyle \sum _{PB(PT)=g}}{\displaystyle \prod _{i=2}^M}{T}_i$

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

$$(*)={\displaystyle \sum _{g=0}^{M-1}}MI{N}_g(M)\left({}_{g+1}^N\right)$$

By definition:

Theorem 2.18. 

$MI{N}_g(M)=\sum \frac{(M+g)!}{i_1{i}_2...i_g},2\le i_1<i_2<...<i_g\le M+g-1,i_{j+1}-i_j\ge 2$

Based on the concept of Shape rather than 2.18, it is easier to understand.eg:

$$MI{N}_1(3)=(124)+(134),MI{N}_2(4)=(12357)+(12457)+(13457)+(12467)+(13467)+(13567)$$

Here (...) is products.

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

Theorem 2.19. 

$SUM(N,[K_1:D_1,K_2:D_2...K_M:D_M],[T_1,T_2...T_M])$

$$=SUM(N,PS,[T_1...T_i,T_{i+1}-1...T_M-1])+SUM(N-1,[K_1:D_1...K_i:D_i,K_{i+1}+D_{i+1}:D_{i+1}...K_M+D_M:D_M],PT)$$

$$eg:SUM(N,[1,2,3],[1,3,5])=SUM(N,[1,2,3],[1,2,4])+SUM(N-1,[1,3,4],[1,3,5])$$

$$=SUM(N,[1,2,3],[1,2,3])+SUM(N-1,[1,2,4],[1,2,4])+SUM(N-1,[1,3,4],[1,3,4])+SUM(N-2,[1,3,5],[1,3,5])$$

$$(*)=SUM(N,[1,2...M],[1,3...2M-1]).I{t}^{\prime }sexactly1.7.$$

We can also define Shape for general PS:

$$\prod (K_i+{\lambda }_i{D}_i),\left\{{}_{{\lambda }_i<{\lambda }_{i+1},discontinuity}^{{\lambda }_i={\lambda }_{i+1},continuity}\right.$$

Definition 11. 

$MI{N}_g(PS)=K_1\times {\displaystyle \sum _{PB(\prod (K_i+{\lambda }_i{D}_i))=g}}{\displaystyle \prod _{i=2}^M}(K_i+{\lambda }_i{D}_i),{\lambda }_1=0,{\lambda }_{i+1}-{\lambda }_i=0or1$

By definition of H(g):

Theorem 2.20. 

$K_1\times H_1(g,[K_2:D_2...K_M:D_M],[\frac{K_2}{D_2}+1,\frac{K_3}{D_3}+2...\frac{K_M}{D_M}+M-1])=MI{N}_g(PS)$

Theorem 2.21. 

$K_1\times H_2(g,[D_2:D_2...D_M:D_M],[\frac{K_2}{D_2}+1,\frac{K_3}{D_3}+2...\frac{K_M}{D_M}+M-1])={(-1)}^{M-1-g}MI{N}_g(PS)$

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.

• $S_{1,r}(n,k)=\frac{n!}{k!}{\displaystyle \sum _{i_1+i_2+...+i_k=n,i_j\ge r}}\frac{1}{i_1{i}_2...i_k}$
• $S_{1,r}(n+1,k)=n{S}_{1,r}(n,k)+{(n)}_{r-1}{S}_{1,r}(n-r+1,k-1),n\ge kr$
• $\sum \frac{(n-1)!}{i_1{i}_2...i_{k-1}},r\le i_1<i_2<...<i_{k-1}\le n-r,i_{j+1}-i_j\ge r$ [6]

Derived from 2 and definition of H(g) or 3 and 2.18:

Theorem 2.22. 

$MI{N}_g(M)=S_{1,2}(M+g+1,g+1)=\frac{(M+g+1)!}{(g+1)!}{\displaystyle \sum _{i_1+i_2+...+i_{g+1}=M+g+1,i_j\ge 2}}\frac{1}{i_1{i}_2...i_{g+1}}$

Table 1. Table of $MI{N}_g(M)=S_{1,2}(M+g+1,g+1)$.

	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

Table 1. Table of $MI{N}_g(M)=S_{1,2}(M+g+1,g+1)$.

	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.

• $S_{2,r}(n,k)=\frac{n!}{k!}{\displaystyle \sum _{i_1+i_2+...+i_k=n,i_j\ge r}}\frac{1}{i_1!i_2!...i_k!}$
• $S_{2,r}(n+1,k)=k{S}_{2,r}(n,k)+\left({}_{r-1}^n\right)S_{2,r}(n-r+1,k-1),n\ge kr$

Derived from 2:

Theorem 2.23. 

$H_2(g,[1,1...1],[3,5...2M-1])=S_{2,2}(M+g+1,g+1)=\frac{(M+g+1)!}{(g+1)!}{\displaystyle \sum _{i_1+...+i_{g+1}=M+g+1,i_j\ge 2}}\frac{1}{i_1!...i_{g+1}!}$

Table 2. Table of $H_2(g,[1,1...1],[3,5...2M-1])=S_{2,2}(M+g+1,g+1)$.

	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

Table 2. Table of $H_2(g,[1,1...1],[3,5...2M-1])=S_{2,2}(M+g+1,g+1)$.

	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

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

$$={\displaystyle \sum _{g=0}^{M-1}}{S}_{1,2}(M+g+1,g+1)\left({}_{M+1+g}^{N+M}\right)={\displaystyle \sum _{g=1}^M}{S}_{1,2}(M+g,g)\left({}_{M+g}^{N+M}\right)$$

$$S_{1,1}(N+M,N)\to =\frac{(N+M)!}{N!}{\displaystyle \sum _{i_1+i_2+...+i_N=N+M,i_j\ge 1}}\frac{1}{i_1{i}_2...i_N}$$

$$=\frac{(N+M)!}{N!}{\displaystyle \sum _{g=1}^M}[{\displaystyle \sum _{i_1+...+i_N=N+M,i_j\ge 1,Numberof{i}_j>1=g}}\frac{1}{i_1{i}_2...i_N}]={\displaystyle \sum _{g=1}^M}f(*)$$

$$f(*)=\frac{(N+M)!}{N!}\left({}_{N-g}^N\right){\displaystyle \sum _{i_1+...+i_g=g+M,i_j\ge 1}}\frac{1}{i_1{i}_2...i_g}=\left({}_{M+g}^{N+M}\right)S_{1,2}(M+g,g)$$

Use the same way:

Theorem 2.24. 

$S_{1,r}(rN+M,N)={\displaystyle \sum _{g=1}^M}\frac{1}{r^{N-g}}\frac{(rN-rg)!}{(N-g)!}\left({}_{rg+M}^{rN+M}\right)S_{1,r+1}(rg+M,g)$

Theorem 2.25. 

$S_{2,r}(rN+M,N)={\displaystyle \sum _{g=1}^M}\frac{1}{{(r!)}^{N-g}}\frac{(rN-rg)!}{(N-g)!}\left({}_{rg+M}^{rN+M}\right)S_{2,r+1}(rg+M,g)$

2.5. Table of H(g)

Table 3. Table of H(g).

PS	PT	${\mathit{H}}_1(\mathit{g})$	${\mathit{H}}_2(\mathit{g})$	${\mathit{H}}_3(\mathit{g})$
$[1,1...1]$	$[1,2...M]$	$g!E_{M-g}^{g+1}=g!S_2(M+1,g+1)$	${(-1)}^{M-g}g!S_2(M,g)$	$〈{}_g^M〉$
$[1,1...1]$	$[2,3...M]$	$(g+1)!S_2(M,g+1)$	${(-1)}^{M-1-g}(g+1)!S_2(M,g+1)$	$〈{}_g^M〉$
$[1,1...1]$	$[1,3...2M-1]$	$E_{M-g}^{g+1}\odot (PT,1)$	${(-1)}^{M-g}MI{N}_{g-1}(M)$	$E_{M-g}^{g+1}\odot ([0,1...],2)$
$[1,1...1]$	$[3,5...2M-1]$	$S_{2,2}(M+1+g,g+1)$	${(-1)}^{M-1-g}MI{N}_g(M)$	$E_{M-1-g}^{g+1}\odot ([2,3...],2)$
$[1,2...M]$	$[1,3...2M-1]$	$MI{N}_{g-1}(M)+MI{N}_g(M)$	$1\times {(-1)}^{M-g}{E}_{M-g}^g\odot ([3,5...],1)$	$1\times E_g^{M-g}\odot ([2,3...],2)$
$[2,3...M]$	$[3,5...2M-1]$	$MI{N}_g(M)$	${(-1)}^{M-1-g}{S}_{2,2}(M+1+g,g+1)$	$E_g^{M-g}\odot ([2,3...],2)$

Table 3. Table of H(g).

PS	PT	${\mathit{H}}_1(\mathit{g})$	${\mathit{H}}_2(\mathit{g})$	${\mathit{H}}_3(\mathit{g})$
$[1,1...1]$	$[1,2...M]$	$g!E_{M-g}^{g+1}=g!S_2(M+1,g+1)$	${(-1)}^{M-g}g!S_2(M,g)$	$〈{}_g^M〉$
$[1,1...1]$	$[2,3...M]$	$(g+1)!S_2(M,g+1)$	${(-1)}^{M-1-g}(g+1)!S_2(M,g+1)$	$〈{}_g^M〉$
$[1,1...1]$	$[1,3...2M-1]$	$E_{M-g}^{g+1}\odot (PT,1)$	${(-1)}^{M-g}MI{N}_{g-1}(M)$	$E_{M-g}^{g+1}\odot ([0,1...],2)$
$[1,1...1]$	$[3,5...2M-1]$	$S_{2,2}(M+1+g,g+1)$	${(-1)}^{M-1-g}MI{N}_g(M)$	$E_{M-1-g}^{g+1}\odot ([2,3...],2)$
$[1,2...M]$	$[1,3...2M-1]$	$MI{N}_{g-1}(M)+MI{N}_g(M)$	$1\times {(-1)}^{M-g}{E}_{M-g}^g\odot ([3,5...],1)$	$1\times E_g^{M-g}\odot ([2,3...],2)$
$[2,3...M]$	$[3,5...2M-1]$	$MI{N}_g(M)$	${(-1)}^{M-1-g}{S}_{2,2}(M+1+g,g+1)$	$E_g^{M-g}\odot ([2,3...],2)$

3. Application

3.1. Number analysis

Theorem 3.1. 

$\prod (K_1+n{D}_i)$ can be decomposed into three forms by 1.1 and ∇.

Theorem 3.2. 

SUM(N,PS,PT)=SUM(N,[1,1...1,PS],[1,1...1,PT]) expand ${\displaystyle \sum _{g=0}^M}(...)={\displaystyle \sum _{g=0}^{M+(numberof1added)}}(...)$

PS=[1,1...1],PT=[1,2...M], 2.1$\to H_1(M)={\displaystyle \sum _{g=0}^M}{H}_3(g),H_1(1)={\displaystyle \sum _{g=1}^M}{H}_2(g)g$

Theorem 3.3. 

${\displaystyle \sum _{g=0}^M}〈{}_g^M〉=M!,{\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g\times g!S_2(M,g)=2^M-1$

PS=[1,1...1],PT=[2,3...M], 2.1$\to {\displaystyle \sum _{k=0}^M}{(-1)}^{M-k}k!S_2(M,k)=1$

Theorem 3.4. 

$g!S_2(M,g)={\displaystyle \sum _{k=g}^M}{(-1)}^{M-k}k!S_2(M,k)\left({}_{g-1}^{K-1}\right)={\displaystyle \sum _{k=0}^{g-1}}〈{}_k^M〉\left({}_{M-g}^{M-1-k}\right),1\le g\le M$

SUM(N,[2,3...M],[3,5...2M-1]),SUM(N,[1,1...1],[3,5...2M-1]), 2.1 →

Theorem 3.5. 

$S_{1,2}(M+g,g)={\displaystyle \sum _{k=g}^M}{(-1)}^{M-k}{S}_{2,2}(M+k,k)\left({}_{g-1}^{K-1}\right),S_{2,2}(M+g,g)={\displaystyle \sum _{k=g}^M}{(-1)}^{M-k}{S}_{1,2}(M+k,k)\left({}_{g-1}^{K-1}\right)$

SUM(N,[1,1...1],[2,3...M]), 2.3 →

Theorem 3.6. 

${\displaystyle \sum _{g=1}^M}g!S_2(M,g)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g!S_2(M,g)2^{g-1}={\displaystyle \sum _{g=1}^M}〈{}_{M-g}^M〉2^{M-g}$

SUM(N,[1,1...1],[3,5...2M-1]),SUM(N,[2,3...M],[3,5...2M-1]), 2.3 →

Theorem 3.7. 

${\displaystyle \sum _{g=1}^M}{S}_{2,2}(M+g,g)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{1,2}(M+g,g)2^{g-1},{\displaystyle \sum _{g=1}^M}{S}_{1,2}(M+g,g)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{2,2}(M+g,g)2^{g-1}$

SUM(N,[1,1...1],[1,2...M]), 2.6 →

Theorem 3.8. 

${\displaystyle \sum _{g=1}^M}g!S_2(M,g)(g-1)\left({}_g^{A+1}\right)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g!S_2(M,g)(g-1)\left({}_g^{A+g-1}\right)$

PS=PT=[1,2...M],$H_1(g)=H_1(M-g)=M!\left({}_g^M\right)$, 2.9$\to g!(F_{M-g}^K{E}_0^g+F_{M-g-1}^K{E}_1^g+...+F_0^K{E}_{M-g}^g)$

Theorem 3.9. 

$M!\left({}_g^M\right)=g!{\displaystyle \sum _{i=0}^{M-g}}{S}_1(M+1,g+1+i)S_2(g+i,g)=(M-g)!{\displaystyle \sum _{i=0}^g}{S}_1(M+1,M+1-i)S_2(M-i,M-g)$

$H_1(g,[1,1...1],[1,2...M])=g!S_2(M+1,g+1),F_i^{\{1,1...1\}}=\left({}_i^M\right),2.9\to$

Theorem 3.10. 

$S_2(M+1,g+1)={\displaystyle \sum _{i=0}^{M-g}}{S}_2(g+i,g)\left({}_{g+i}^M\right)={\displaystyle \sum _{i=0}^{M-g}}{S}_2(M-i,g)\left({}_i^M\right)$

$H_1(g,[1,2...M],[1,2...M])=H_1(g,[M,M-1...1],[1,2...M])=M!\left({}_g^M\right)2.10\to$

$H_1(g)=\frac{M!}{g!}(F_g^M{E}_0^{M-g}-F_{g-1}^M{E}_1^{M-g}+...+{(-1)}^g{F}_0^M{E}_g^{M-g})$

Theorem 3.11. 

$g!\left({}_g^M\right)={\displaystyle \sum _{i=0}^g}{S}_1(M+1,M+1-g+i)S_2(M-g+i,M-g){(-1)}^i$

$H_1(g,[K+i],[T+i])$, 2.11$\to \left({}_g^M\right)T_1...T_g\times K_{g+1}...K_M$, 2.9$\to T_1...T_g(...)$, 2.10$\to K_{g+1}...K_M(...)$

Theorem 3.12. 

${\displaystyle \prod _{i=g+1}^M}(K+i)\left({}_g^M\right)=F_{M-g}^{\{K+i\}}{E}_0^g+...+F_0^{\{K+i\}}{E}_{M-g}^g,{\displaystyle \prod _{i=1}^g}(T+i)\left({}_g^M\right)=F_g^{\{T+i\}}{E}_0^{M-g}+...+{(-1)}^g{F}_0^{\{T+i\}}{E}_g^{M-g}$

SUM(1,[1,1…1],[3,5…2M-1])=1,SUM(1,[2,3…M],[3,5…2M-1])=M!,$=SUM(1)={\displaystyle \sum _{g=0}^{M-1}}{H}_2(g)\left({}_{M+g+1}^{M+g+1}\right)\to$

Theorem 3.13. 

${\displaystyle \sum _{g=0}^{M-1}}{(-1)}^{M-1-g}MI{N}_g(M)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{1,2}(M+g,g)=1,{\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{2,2}(M+g,g)=M!$

2.20 and 2.21 →

Theorem 3.14. 

${\displaystyle \sum _{g=0}^{M-1}}{(-1)}^{M-1-g}MI{N}_g(PS)=K_1{\displaystyle \prod _{i=2}^M}{D}_i$

$SUM(N,[T,T...T],[T,T...T])=T^M\left({}_{T+1}^{N+T}\right)={\displaystyle \sum _{g=0}^M}{T}^M\left({}_g^M\right)\left({}_{T-M+1+g}^{N+T-M}\right)\to$

Theorem 3.15. 

${\displaystyle \sum _{g=0}^M}\left({}_g^M\right)\left({}_{T-M+1+g}^{N+T-M}\right)=\left({}_{T+1}^{N+T}\right),T\in \mathbb{N}$

3.2. Congruences

P is prime.$K_i$ is any integer.

Theorem 3.16. 

$(P,D)=1,SUM(P,[K_1,K_2...K_M]:D,[1,2...M])={\displaystyle \sum _{g=0}^M}{H}_1(g)\left({}_{1+g}^P\right)\equiv \left\{{}_{-1(modP),M=P-1}^{0(modP),M<P-1}\right.$

Proof. 

If M=P-1,$SUM(P)\equiv H_1(P-1)\left({}_P^P\right)\equiv H_1(P-1)\equiv (P-1)!D^{P-1}\equiv -1(modP)$ □

If a product has a factor $\equiv 0(modP)$,ignore it;change factors to its minimum positive residue,we can obtain many congruences.Wilson’s Theorem is just a special case . $A,B,C\in \mathbb{N}$

$$1^A{2}^B+2^A{3}^B+...+{(P-2)}^A{(P-1)}^B\equiv 1^A{3}^B+2^A{4}^B+...+{(P-3)}^A{(P-1)}^B+{(P-1)}^A{1}^B\equiv \left\{{}_{-1(modP),A+B=P-1}^{0(modP),A+B<P-1}\right.$$

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

Theorem 3.17. 

${\displaystyle \sum _{0<K_i,K_j<P,K_i\ne K_j}}{K_1}^{{\lambda }_1}{K_2}^{{\lambda }_2}...{K_q}^{{\lambda }_q}\equiv \left\{{}_{0\left(modP\right),{\lambda }_1+{\lambda }_2+...+{\lambda }_q<P-1}^{-1\left(modP\right),{\lambda }_1+{\lambda }_2+...+{\lambda }_q=P-1}\right.,{\lambda }_i\in \mathbb{N}$

Theorem 3.18. 

$MI{N}_g(M)\equiv 0(modP),0<g<M-1,g+M=P-1$

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

• Wolstenholme’s Theorem:$(P-1)!{\displaystyle \sum _{n=1}^{P-1}}\frac{1}{n}={\displaystyle \sum _{0<K_i,K_j<P,K_i\ne K_j}}{K}_1{K}_2...K_{P-2}\equiv 0(mod{P}^2)$
• ${\displaystyle \sum _{n=1}^{P-1}}{n}^{P-2}\equiv 0(mod{P}^2)$

They are two extremes.In fact,there have:

Theorem 3.19. 

${\displaystyle \sum _{0<K_i,K_j<P,K_i\ne K_j}}{K_1}^{C_1}{K_2}^{C_2}...{K_q}^{Cq}\equiv 0(mod{P}^2),C_1+C_2+...+C_q=P-2,C_i>0$

Proof. 

If $X(mod{P}^2)$ and $X+Y(mod{P}^2)$ then $Y(mod{P}^2)$.The Sum has symmetry.

For $\sum A{B}^{P-3}$,pair each $A{B}^{P-3}$ with $(P-A)B^{P-3}$ to $P{B}^{P-3}$,$(P-A)B^{P-3}\in \sum A{B}^{P-3}or\sum B^{P-2}$.

$\sum (PairedProdcuts)=x\sum A{B}^{P-3}+y\sum B^{P-2}=zP\sum B^{P-3},0<x,y<P$.

$2\to y\sum B^{P-2}\equiv 0(mod{P}^2);P\sum B^{P-3}\equiv 0(mod{P}^2)$.So $\sum A{B}^{P-3}\equiv 0(mod{P}^2)$.

Similarly:

$$\sum (PairedProdcutsofAB{C}^{P-4}+B^2{C}^{P-4})=x\sum AB{C}^{P-4}+y\sum B^2{C}^{P-4}\equiv 0(mod{P}^2),0<x,y<P$$

$$x\sum AB{C}^{P-4}+y\sum B^2{C}^{P-4},symmetry\to$$

$$\equiv z\sum (A+B)B{C}^{P-4}\equiv z\sum AB{C}^{P-4}(modP),0<z<P\to$$

$$\sum AB{C}^{P-4}\equiv \sum B^2{C}^{P-4}\equiv 0(mod{P}^2)$$

Prove the conclusion in a similar way... □

Theorem 3.20. 

$E_{P-2}^{P-1}=S_2(2P-3,P-1)\equiv 0(mod{P}^2);E_{P-2}^P=S_2(2P-2,P)\equiv 0(mod{P}^2)$

$$eg:S_2(7,4)=350\equiv 0(mod25),S_2(8,5)=1050\equiv 0(mod25)$$

$$eg:S_2(11,6)=179487\equiv 0(mod49),S_2(12,7)=627396\equiv 0(mod49)$$

4. Combinatorial Identities

Definition 12. 

R-FOLD SUM:${\sum }_{(r)}^{N}f(k)={\displaystyle \sum _{k_r=1}^N}...{\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}f(k_1)={\displaystyle \sum _{k_r=0}^{N-1}}...{\displaystyle \sum _{k_2=0}^{k_3}}{\displaystyle \sum _{k_1=0}^{k_2}}f(k_1+1)$

By nested sum:

Theorem 4.1. 

${\sum }_{(r)}^N{\nabla }^{p}SUM(k,PS,PT)={\nabla }^{p-r}SUM(N,PS,PT)$

Theorem 4.2. 

${\displaystyle \sum _{k_r=1}^N}...{\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}\nabla SUM(k_r,PS,[1,2...M])=SUM(N,PS,[T_i=i+r-1])$

Proof. 

$$PT=[T_i=i+r-1]$$

$$PS1=[1:0,1:0...1:0,PS],PT1=[1,3...2(r-1)-1,1+2(r-1),2+2(r-1)...M+2(r-1)]$$

$$B_{i<r}=\left\{{}_{K_i+X_{K-1}{D}_i=1,X_i\in K}^{(T_i-X_{T-1})D_i=0,X_i\in T}\right.\to H_1(g>M,PS1,PT1)=0,H_1(g\le M,PS1,PT1)=H_1(g,PS,PT)$$

$$SUM(N,PS1,PT1)={\displaystyle \sum _{g=0}^M}{H}_1(g,PS,PT)\left({}_{r+g}^{N+r-1}\right)={\displaystyle \sum _{k_r=1}^N}\nabla SUM(k_r,PS,[1,2...M])...{\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}1$$

□

Theorem 4.3. 

${\displaystyle \sum _{k_x=1}^N}...{\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}\nabla SUM(k_r,PS,[1,2...M])={\nabla }^{r-x}SUM(N,PS,[T_i=i+(r-1)])$

$$eg:(*){\displaystyle \sum _{k_r=1}^N}...{\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}\left({}_j^{k_i}\right)=\frac{1}{j!}{\nabla }^{i-r}SUM(N,[0,-1,-2...-(j-1)],[T_n=n+(i-1)])$$

$$H_1(g<j)=0,T_j-j=i-1\to$$

$(*)=\frac{1}{j!}{\nabla }^{i-r}{H}_1(j)\left({}_{j+1+(i-1)}^{(N+1)+(i-1)}\right)=\frac{1}{j!}{\nabla }^{i-r}\frac{(j+i-1)!}{(i-1)!}\left({}_{j+i}^{N+i}\right)=\left({}_{i-1}^{j+i-1}\right)\left({}_{j+i-(i-r)}^{N+i-(i-r)}\right)=\left({}_{i-1}^{j+i-1}\right)\left({}_{j+r}^{N+r}\right)$ [7]

Using induction to prove:

Theorem 4.4. 

${\displaystyle \sum _{0\le n_1\le ....\le n_M\le N-1}}(K+n_1{D}_1+....+n_M{D}_M)={\nabla }^{\left({}_2^M\right)}SUM(N,[K:\frac{D_1+2D_2+....+M{D}_M}{\left({}_2^{M+1}\right)},[\left({}_2^{M+1}\right)])$

$$=(D_1+2D_2+....+M{D}_M)\left({}_{M+1}^{N+M-1}\right)+K\left({}_M^{N+M-1}\right)$$

$${\displaystyle \sum _{0\le n_1\le n_2\le ....\le n_M\le N-1}}(K+n_1+n_2+....+n_M)=\left({}_2^{M+1}\right)\left({}_{M+1}^{N+M-1}\right)+K\left({}_M^{N+M-1}\right)$$

${\displaystyle \sum _{0\le n_1\le ....\le n_p\le n}}(n_1+....+n_p)={\displaystyle \sum _{n_p=0}^n}{\displaystyle \sum _{n_{p-1}=0}^{n_p}}...{\displaystyle \sum _{n_1=0}^{n_2}}(n_1+....+n_p)=\left({}_2^{p+1}\right)\left({}_{p+1}^{n+p}\right)=\frac{pn}{2}\left({}_p^{n+p}\right)$. [7]

1.2, 1.3, 1.4 can be used to derive combinatorial identities.

Theorem 4.5. 

$\left({}_A^{n+A}\right)\left({}_M^{n+M+B}\right)=$

$${\displaystyle \sum _{g=0}^M}\left({}_g^{A+g}\right)\left({}_{M-g}^{M+B}\right)\left({}_{A+g}^{n+A}\right)={\displaystyle \sum _{g=0}^M}{(-1)}^{M-g}\left({}_g^{A+g}\right)\left({}_{M-g}^{A-B}\right)\left({}_{A+g}^{n+A+g}\right)={\displaystyle \sum _{g=0}^M}\left({}_g^{A-B}\right)\left({}_{M-g}^{M+B}\right)\left({}_{A+M}^{n+A+M-g}\right)$$

Proof. 

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

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

$$H_1(g)=\left({}_g^M\right)(A+1)....(A+g)(B+g+1)....(B+M)=\left({}_g^M\right)g!\left({}_g^{A+g}\right)(M-g)!\left({}_{M-g}^{B+M}\right)$$

Using a similar method to obtain $H_2(g),H_3(g)$

□

Theorem 4.6. 

$\left({}_A^{n+X}\right)\left({}_M^{n+Y}\right)={\displaystyle \sum _{g=0}^A}\left({}_g^{M+g}\right)\left({}_{A-g}^{M+X-Y}\right)\left({}_{M+g}^{n+Y}\right),0\le Y\le M$

Proof. 

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

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

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

$$If{H}_1(g)\ne 0then{X}_1,X_2...X_{M-Y}\in T\to H_1(g<M-Y)=0,LetC=A+M-Y$$

$$H_1(g\ge M-Y)\ne 0\to NumberofK=\left({}_{C-g}^{C-(M-Y)}\right)\to H_1(g\ge M-Y,\sum K)=\left({}_{C-g}^{C-(M-Y)}\right){[X+M-Y]}_{C-g}$$

$$H_1(g\ge M-Y)=\left({}_{C-g}^{C-(M-Y)}\right){[X+M-Y]}_{C-g}{[Y+1]}^g$$

$$original=\frac{Y!}{M!A!}{\displaystyle \sum _{g=M-Y}^C}\left({}_{C-g}^{C-(M-Y)}\right){[X+M-Y]}_{C-g}{[Y+1]}^g\left({}_{Y+g}^{n+Y}\right),-q=M-Y-g\to$$

$$=\frac{Y!}{M!A!}{\displaystyle \sum _{q=0}^A}\left({}_{A-q}^A\right)\left({}_{A-q}^{X+M-Y}\right)(A-q)!\left({}_{M-Y+q}^{M+q}\right)(M-Y+q)!\left({}_{M+q}^{n+Y}\right)$$

□

PS=[0,-1,-2...-(B-1),A-(M-B-1):-1,A-(M-B-1)+1:-1...A:-1],PT=[1,2...M]→

Theorem 4.7. 

$\left({}_B^n\right)\left({}_{M-B}^{A-n}\right)={\displaystyle \sum _{g=0}^{M-B}}{(-1)}^{M-B-g}\left({}_g^{A-M+g}\right)\left({}_B^{M-g}\right)\left({}_{M-g}^n\right),n\ge 0$

PS=[0,-1,-2...-A+1,0,-1,-2...-B+1],PT=[1,2...A+B]→

Theorem 4.8. 

$\left({}_A^n\right)\left({}_B^n\right)={\displaystyle \sum _{g=0}^B}\left({}_g^B\right)\left({}_{B-g}^{A+B-g}\right)\left({}_{A+B-g}^n\right)$. record at [8]: (6.44)

Theorem 4.9. 

${\displaystyle \prod _{i=1}^M}(A+2i+n)={\left({}_A^{n+A}\right)}^{-1}{\displaystyle \sum _{g=0}^M}(2(M-g)-1)!!\left({}_g^{2M-g}\right){[A+1]}^g\left({}_{A+g}^{n+A+g}\right)$ A ∈$\mathbb{N}$ or A = 0

Proof. 

PS=[A+2,A+4...A+2M],PT=[A+1,A+2...A+M]

$$H_2(g,\sum K)=SUM(g+1,[1,3...2(M-g)-1],[1,3...2(M-g)-1]))=(2(M-g)-1)!!\left({}_g^{2M-g}\right),H_2(g,T)={[A+1]}^g$$

$$\left({}_A^{n+A}\right){\displaystyle \prod _{i=1}^M}(A+2i+n)=\frac{1}{A!}\nabla SUM(N,[1,2...A,PS],[1,2...A,PT])=\nabla SUM(N,PS,PT)$$

□

Theorem 4.10. 

SUM(N,[A+1,A+3...A+2M-1],[1,3...2M-1])=${\displaystyle \sum _{g=0}^M}{[A]}^{M-g}(2g-1)!!\left({}_{2g}^{M+g}\right)\left({}_{M+g}^{N+M-1+g}\right)$

Proof. 

$$H_2(g,\sum T)=SUM(M-g+1,[1,3...2g-1],[1,3...2g-1]))=(2g-1)!!\left({}_{2g}^{M+g}\right),H_2(g,K)={[A]}^{M-g}$$

□

Theorem 4.11. 

SUM(N,[A,A+1...A+M-1]:2,[1,3...2M-1])=$\left({}_M^{M+N-1}\right){[A+M+N-2]}_M$

Proof. 

$$SUM(g+1,[A,A+1...A+M-1-g]:2,[1,3...2(M-g)-1])$$

$$=H_1(g,\sum K,[A,A+1...A+M-1],[1,2...M])=\left({}_g^M\right){[A+M-1]}_{M-g}$$

□

$$SUM(N,[1,2...M]:2,[1,3...2M-1])=M!{\left({}_M^{N+M-1}\right)}^2,M=1\to 1+3+...+(2N-1)=N^2$$

4.11 and 4.5 →

Theorem 4.12. 

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

$$This\to H_1(g,[A+1,A+2...A+M]:2,[1,3...2M-1])=M!\left({}_g^{M+g}\right)\left({}_{M-g}^{M+A}\right)$$

$$Definitionof{H}_1(g)\to$$

$$H_1(1)=2\{(A+1)...(A+M-1)\times M+(A+1)...(A+M-2)\times (M-1)\times (A+M+2)\times (M-1)+...\}\to$$

Theorem 4.13. 

${\displaystyle \sum _{n=0}^{M-1}}(M-n){\displaystyle \prod _{k=1}^{M-1}}(A+k+3[\frac{k+n}{M}])=\frac{(M+1)!}{2}\left({}_{M-1}^{M+A}\right)$.[n] is truncates integer.

Theorem 4.14. 

$SUM(N,[A+2,A+4...A+2M]:3,[1,3...2M-1])={\displaystyle \sum _{g=0}^M}\left({}_g^{A+N-1+g}\right)\left({}_{M-g}^{N+M-1-g}\right)\frac{{[M+N-g]}^M}{2^{M-g}}$

Proof. 

PS1=[A+2,A+4...A+2M],PT1=[A+1,A+2...A+M]

$$H_1(g,PS1,PT1)={[A+1]}^{g}SUM(g+1,[A+2,A+4...A+2(M-g)]:3,[1,3...2(M-g)-1]))$$

$$={\displaystyle \sum _{k=g}^M}{H}_2(g,PS1,PT1)\left({}_g^k\right),4.9\to ={\displaystyle \sum _{k=g}^M}(2(M-k)-1)!!\left({}_k^{2M-k}\right){[A+1]}^k\left({}_g^k\right)$$

□

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 $P=N+T_M-M,Q=N-1$,define A(P,Q,M), respectively corresponding to the three forms.

$$A_1(P,Q,M)=\left(\begin{array}{ccc}\left({}_Q^P\right)& \dots & \left({}_{Q-M}^P\right)\\ ⋮& \ddots & ⋮\\ \left({}_{Q+M}^{P+M}\right)& \cdots & \left({}_Q^{P+M}\right)\end{array}\right)$$

$$A_2(P,Q,M)=\left(\begin{array}{ccc}\left({}_Q^P\right)& \dots & \left({}_Q^{P+M}\right)\\ ⋮& \ddots & ⋮\\ \left({}_{Q+M}^{P+M}\right)& \cdots & \left({}_{Q+M}^{P+2M}\right)\end{array}\right)$$

$$A_3(P,Q,M)=\left(\begin{array}{ccc}\left({}_Q^{P+M}\right)& \dots & \left({}_{Q-M}^P\right)\\ ⋮& \ddots & ⋮\\ \left({}_{Q+M}^{P+2M}\right)& \cdots & \left({}_Q^{P+M}\right)\end{array}\right)$$

Theorem 5.1. 

$\Vert A_1(P,Q,M)\Vert =\Vert A_2(P,Q,M)\Vert =\Vert A_3(P,Q,M)\Vert ,\Vert A(P,Q,M)\Vert =\Vert A(P,P-Q,M)\Vert$ [3]

Theorem 5.2. 

$\Vert A(P,0,M)\Vert =1,\Vert A(P,1,M)\Vert =\left({}_{1+M}^{P+M}\right),\Vert A(P,Q>1,M)\Vert ={\displaystyle \prod _{g=0}^{Q-1}}\frac{\left({}_{1+M}^{P+M-g}\right)}{\left({}_{1+M}^{1+M-g}\right)}$ [3]

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

Theorem 5.3. 

$H_1(g)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({}_{T_M-M+k}^{T_M-M+1+g}\right)SUM(k)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({}_{T_M-M+k-1}^{T_M-M+g}\right)\nabla SUM(k)$

$$SUM(N,[1,2...M],[1,2...M])=M!\left({}_{M+1}^{N+M}\right)\to \left({}_g^M\right)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({}_k^{1+g}\right)\left({}_{M+1}^{k+M}\right)$$

$$SUM(N,[2,3...M],[3,5...2M-1])=S_1(N+M,N)\to MI{N}_g(M)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({}_{M+k}^{M+1+g}\right)S_1(M+k,k)$$

$$(1)SUM(N,[1,1...1],[2,3...M])=N^M\to S_2(M,g)=\frac{1}{g!}{\displaystyle \sum _{k=0}^g}{(-1)}^{g+k}\left({}_k^g\right)k^M=\frac{1}{g!}{\displaystyle \sum _{k=0}^g}{(-1)}^k\left({}_k^g\right){(g-k)}^M$$

Theorem 5.4. 

$z(k)={\displaystyle \sum _{i=1}^k}{(-1)}^{i+k}\left({}_i^k\right)\nabla SUM(k),H_2(g)={\displaystyle \sum _{k=g+1}^{M+1}}{(-1)}^{g+k-1}\left({}_g^{k-1}\right)z(k)$

$$SUM(N,[1,1...1],[2,3...M])=N^M\to z(k)={\displaystyle \sum _{i=1}^k}{(-1)}^{i+k}\left({}_i^k\right)i^M=k!S_2(M,k)$$

$$\to g!S_2(M,g)={\displaystyle \sum _{k=g}^M}{(-1)}^{M+k}\left({}_{g-1}^{k-1}\right)k!S_2(M,k)3.4$$

Theorem 5.5. 

$H_3(g)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({}_{g+1-k}^{2+T_M}\right)SUM(k)={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{g+1+k}\left({}_{g+1-k}^{1+T_M}\right)\nabla SUM(k)$

$$(2)SUM(N,[1,1...1],[2,3...M])=N^M\to 〈{}_g^M〉={\displaystyle \sum _{k=1}^{g+1}}{(-1)}^{1+g+k}\left({}_{g+1-k}^{M+1}\right)k^M={\displaystyle \sum _{k=0}^g}{(-1)}^k\left({}_k^{M+1}\right){(g+1-k)}^M$$

(1)(2) are already known formula.

6. Eulerian polynomials and Beyond

In this section,$q\ne 0,q\ne 1$.

Inductive proof:

Theorem 6.1. 

${\displaystyle \sum _{n=0}^{N-1}}{q}^n\left({}_M^{n+K}\right)=q^N{\displaystyle \sum _{g=0}^M}{(-1)}^g\frac{\left({}_{M-g}^{N+K-1-g}\right)}{{(q-1)}^{g+1}}+\frac{q^{M-K}}{{(1-q)}^{M+1}}$

Definition 13. 

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

Table 4. Table of $A_q^M$.

	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

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

Use 3.4 $\to$

Theorem 6.2. 

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

$$n^M=\nabla SUM(N,[1,1...1],[2,3...M])={\displaystyle \sum _{g=0}^M}{S}_2(M,g)g!\left({}_g^n\right)={\displaystyle \sum _{g=0}^M}{(-1)}^{M-g}{S}_2(M,g)g!\left({}_g^{n+g}\right)={\displaystyle \sum _{g=0}^M}〈{}_g^M〉\left({}_M^{n+g}\right)$$

$$(1){\displaystyle \sum _{n=0}^{N-1}}{q}^n{n}^M={\displaystyle \sum _{g=0}^M}{S}_2(M,g)g!{\displaystyle \sum _{n=0}^{N-1}}{q}^n\left({}_g^n\right)$$

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

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

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

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

Use the $Fro{m}_2$ of $n^M$ in (1),the first part of (*) keep same, we can obtain 6.2. Use the $Fro{m}_3$:

Theorem 6.3. 

$A_q^M={\displaystyle \sum _{g=0}^M}〈{}_g^M〉q^{M-g}={\displaystyle \sum _{g=0}^M}〈{}_g^M〉q^{1+g}$

$n^M=\nabla SUM(N,[1,1...1],[1,2...M])={\displaystyle \sum _{g=0}^M}{S}_2(M+1,g+1)g!\left({}_g^{n-1}\right)\to$

Theorem 6.4. 

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

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

$${\nabla }^k{(N-1)}^M={\displaystyle \sum _{j=0}^k}{(-1)}^j\left({}_j^k\right){(N-1-j)}^M={\displaystyle \sum _{j=0}^k}{(-1)}^j\left({}_j^k\right){\displaystyle \sum _{g=0}^M}\left({}_g^M\right)N^g{(j+1)}^{M-g}{(-1)}^{M-g}$$

$$={\displaystyle \sum _{g=0}^M}{(-1)}^{M-g-k}\left({}_g^M\right)N^g{\displaystyle \sum _{j=0}^k}{(-1)}^{j+k}\left({}_j^k\right){(j+1)}^{M-g}={\displaystyle \sum _{g=0}^M}{(-1)}^{M-g-k}\left({}_g^M\right)N^g{S}_2(M-g+1,k+1)k!\to$$

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

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

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

Theorem 6.5. 

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

$$A_q^{M-g}={\displaystyle \sum _{k=0}^M}{(q-1)}^{M-k-g}{S}_2(M-g+1,k+1)k!From(**)\to$$

Theorem 6.6. 

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

Table 5. Table of ${\displaystyle \sum _{n=0}^{N-1}}{q}^n{n}^M$.

${\displaystyle \sum _{\mathit{n}=0}^{\mathit{N}-1}}{\mathit{q}}^{\mathit{n}}\mathit{n}=\frac{{\mathit{q}}^{\mathit{N}}((\mathit{q}-1)\mathit{N}-\mathit{q})+\mathit{q}}{{(\mathit{q}-1)}^2}$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n=2^N-1$

${\displaystyle \sum _{n=0}^{N-1}}{2}^{n}n=2^N(N-2)+2$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^2=2^N(N^2-4N+6)-6$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^3=2^N(N^3-6N^2+18N-26)+26$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^4=2^N(N^4-8N^3+36N^2-104N+150)-150$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^5=2^N(N^5-10N^4+60N^3-260N^2+750N-1082)+1082$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n=\frac{3^N-1}{2}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^{n}n=\frac{3^N(2N-3)+3}{4}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n{n}^2=\frac{3^N(4N^2-12N+12)-12}{8}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n{n}^3=\frac{3^N(8N^3-36N^2+72N-66)+66}{16}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n{n}^4=\frac{3^N(16N^4-96N^3+288N^2-528N+480)-480}{32}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^0=\frac{4^N-1}{3}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^1=\frac{4^N(3N-4)+4}{9}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^2=\frac{4^N(9N^2-24N+20)-20}{27}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^3=\frac{4^N(27N^3-108N^2+180N-132)+132}{81}$

Table 5. Table of ${\displaystyle \sum _{n=0}^{N-1}}{q}^n{n}^M$.

${\displaystyle \sum _{\mathit{n}=0}^{\mathit{N}-1}}{\mathit{q}}^{\mathit{n}}\mathit{n}=\frac{{\mathit{q}}^{\mathit{N}}((\mathit{q}-1)\mathit{N}-\mathit{q})+\mathit{q}}{{(\mathit{q}-1)}^2}$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n=2^N-1$

${\displaystyle \sum _{n=0}^{N-1}}{2}^{n}n=2^N(N-2)+2$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^2=2^N(N^2-4N+6)-6$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^3=2^N(N^3-6N^2+18N-26)+26$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^4=2^N(N^4-8N^3+36N^2-104N+150)-150$

${\displaystyle \sum _{n=0}^{N-1}}{2}^n{n}^5=2^N(N^5-10N^4+60N^3-260N^2+750N-1082)+1082$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n=\frac{3^N-1}{2}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^{n}n=\frac{3^N(2N-3)+3}{4}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n{n}^2=\frac{3^N(4N^2-12N+12)-12}{8}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n{n}^3=\frac{3^N(8N^3-36N^2+72N-66)+66}{16}$

${\displaystyle \sum _{n=0}^{N-1}}{3}^n{n}^4=\frac{3^N(16N^4-96N^3+288N^2-528N+480)-480}{32}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^0=\frac{4^N-1}{3}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^1=\frac{4^N(3N-4)+4}{9}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^2=\frac{4^N(9N^2-24N+20)-20}{27}$

${\displaystyle \sum _{n=0}^{N-1}}{4}^n{n}^3=\frac{4^N(27N^3-108N^2+180N-132)+132}{81}$

We know that Eulerian polynomials: $A_M(t):{\displaystyle \sum _{i=0}^{\infty }}{t}^i{i}^M=\frac{t{A}_M(t)}{{(1-t)}^{M+1}},A_M(t)={\displaystyle \sum _{g=0}^{M-1}}〈{}_g^M〉t^g$

$\left|q\right|<1,{\displaystyle \underset{N\to \infty }{lim}}{\displaystyle \sum _{n=0}^{N-1}}{q}^n{n}^M=\frac{A_q^M}{{(1-t)}^{M+1}}\to A_t^M=t{A}_M(t)$. There have 5 expressions for $A_M(t)$.

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

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{\nabla }^{p}SUM(n+Y,PS,PT),X=T_M-M-p$$

$$=\left\{\begin{array}{c}{\displaystyle \sum _{g=0}^M}{H}_1(g){\displaystyle \sum _{n=0}^{N-1}}{q}^n\left({}_{X+1+g}^{n+Y+X}\right)={\displaystyle \sum _{g=0}^M}{H}_1(g)\{q^N{\displaystyle \sum _{k=0}^{X+1+g}}{(-1)}^k\frac{\left({}_{X+1+g-k}^{N+Y+X-1-k}\right)}{{(q-1)}^{k+1}}+\frac{q^{1+g-Y}}{{(1-q)}^{X+2+g}}\}\\ {\displaystyle \sum _{g=0}^M}{H}_2(g){\displaystyle \sum _{n=0}^{N-1}}{q}^n\left({}_{X+1+g}^{n+Y+X+g}\right)={\displaystyle \sum _{g=0}^M}{H}_2(g)\{q^N{\displaystyle \sum _{k=0}^{X+1+g}}{(-1)}^k\frac{\left({}_{X+1+g-k}^{N+Y+X+g-1-k}\right)}{{(q-1)}^{k+1}}+\frac{q^{1-Y}}{{(1-q)}^{X+2+g}}\}\\ {\displaystyle \sum _{g=0}^M}{H}_3(g){\displaystyle \sum _{n=0}^{N-1}}{q}^n\left({}_{X+1+M}^{n+Y+X+M-g}\right)={\displaystyle \sum _{g=0}^M}{H}_3(g)\{q^N{\displaystyle \sum _{k=0}^{X+1+M}}{(-1)}^k\frac{\left({}_{X+1+M-k}^{N+Y+X+M-g-1-k}\right)}{{(q-1)}^{k+1}}+\frac{q^{1+g-Y}}{{(1-q)}^{X+2+M}}\}\end{array}\right.$$

$$Y=0\to$$

Theorem 6.7. 

${\displaystyle \sum _{n=0}^M}{H}_1(g){(1-q)}^{M-g}{q}^{g+1}=q{\displaystyle \sum _{n=0}^M}{H}_2(g){(1-q)}^{M-g}={\displaystyle \sum _{n=0}^M}{H}_3(g)q^{g+1}$

Here q can take any value,which is magical.$q=0.5\to$ 2.3

Definition 14. 

$A_q(PS,PT)=6.7$

Theorem 6.8. 

${\displaystyle \sum _{n=0}^{N-1}}{q}^n{\nabla }^{p}SUM(n+Y,PS,PT)=\frac{q^N}{{(q-1)}^{M+1}}{\displaystyle \sum _{k=0}^M}{(q-1)}^{M-k}{(-1)}^k{\nabla }^{p+K-1}SUM(N+Y-2)+\frac{A_q(PS,PT)q^{-Y}}{{(1-q)}^{T_M-p+2}}$

Theorem 6.9. 

${\displaystyle \sum _{n=0}^{\infty }}{q}^n{\nabla }^{p}SUM(n+Y,PS,PT)=\frac{A_q(PS,PT)q^{-Y}}{{(1-q)}^{T_M+2-p}}$

$$eg:A_q^M([d,d...d]:d,[1,2...M])=d^M{A}_q^M$$

$${\displaystyle \sum _{n=0}^{N-1}}{q}^n{(nd)}^M=\frac{q^N}{{(q-1)}^{M+1}}{\displaystyle \sum _{g=0}^M}{(-1)}^{M-g}{(q-1)}^g\left({}_g^M\right)A_q^{M-g}([d,d...d]:d,[1,2...M]){(nd)}^g+\frac{d^M{A}_q^M}{{(1-q)}^{M+1}}$$

Many results of [9,10] can be obtained by this.

7. Formal Calculation of Gaussian Coefficients

7.1. Basic concepts

Gaussian Coefficients:${\left[{}_M^N\right]}_q=\frac{(q^N-1)(q^{N-1}-1)...(q^{N-M+1}-1)}{(q^M-1)(q^{M-1}-1)...(q^1-1)},q\ne 0,1$,abbreviated as $G_M^N$

• $G_0^N=1,G_{M<0,M>N}^N=0,G_M^N=G_{N-M}^N$
• $G_M^N=q^M{G}_M^{N-1}+G_{M-1}^{N-1}=G_M^{N-1}+q^{N-M}{G}_{M-1}^{N-1}$
• ${\displaystyle \sum _{n=0}^{N-1}}{q}^n{G}_M^{n+K}=q^{M-K}{G}_{M+1}^{N+K}$
• $G_K^M={\displaystyle \sum _{w\in \Omega (0^{M-K},1^K)}}{q}^{inv(w)}$,that is,all words $w=w_1{w}_2\dots w_M$ with M-K zeroes and K ones, and inv(·) denotes the inversion statistic defined

The Formal Calculation was obtained by [3]. It use $q^n(K_i+G_1^n{D}_i)$ instead of $K_i+q^n{D}_i$ .

Definition 15. 

Recursive define the operator ${\nabla }_q^p$,p is an integer.

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

Definition 16. 

Recursive define $SU{M}_q(N,PS,PT)$,abbreviated as $SU{M}_q(N)$.

$$SU{M}_q(N,[K_1:D_1],[T_1=1])={\displaystyle \sum _{n=0}^{N-1}}{q}^n(K_1+G_1^n{D}_1)$$

$$SU{M}_q(N,[K_1:D_1,K_2:D_2],[T_1,T_2=T_1+2-p])={\displaystyle \sum _{n=0}^{N-1}}{q}^n(K_2+G_1^n{D}_2){\nabla }^{p}SU{M}_q(n+1,[K_1:D_1],[T_1])$$

Theorem 7.1. 

${\displaystyle \sum _{n=0}^{N-1}}{q}^n{G}_1^n{G}_M^{n+K},M>0,M\ge K$

$=q^{2(M-K)+1}{G}_1^{M+1}{G}_{M+2}^{N+K}+q^{M-K}{G}_1^{M-K}{G}_{M+1}^{N+K}$

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

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

Use this and induction to prove:

Theorem 7.2. 

$SU{M}_q(N,PS,PT)$=

$$For{m}_1\to {\displaystyle \sum _{g=0}^M}{H}_1^q(g)G_{N-1-g}^{N+T_M-M}={\displaystyle \sum _{g=0}^M}{H}_1^q(g)G_{T_M-M+1+g}^{N+T_M-M},B_i=\left\{{}_{q^{(T_i-T_{i-1}-1)X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i),X_i=K_i}^{q^{1+(T_i-T_{i-1})X_{T-1}}{G}_1^{T_i-X_{K-1}}{D}_i,X_i=T_i}\right.$$

$$For{m}_2\to {\displaystyle \sum _{g=0}^M}{H}_2^q(g)G_{N-1}^{N+T_M-M+g}={\displaystyle \sum _{g=0}^M}{H}_2^q(g)G_{T_M-M+1+g}^{N+T_M-M+g},B_i=\left\{{}_{K_i-q^{-(T_i-X_{K-1})}{G}_1^{T_i-X_{K-1}}{D}_i,X_i=K_i}^{q^{-(T_i-X_{K-1})}{G}_1^{T_i-X_{K-1}}{D}_i,X_i=T_i}\right.$$

$$For{m}_3\to {\displaystyle \sum _{g=0}^M}{H}_3^q(g)G_{N-1-g}^{N+T_M-g}={\displaystyle \sum _{g=0}^M}{H}_3^q(g)G_{T_M+1}^{N+T_M-g},B_i=\left\{{}_{q^{(T_i-T_{i-1}-1)X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i),X_i=K_i}^{q^{1+(T_i-T_{i-1}-1)X_{T-1}}\{(q^{X_{T-1}}{G}_1^{T_i}-q^{T_i}{G}_1^{X_{T-1}})D_i-K_i{q}^{T_i}\},X_i=T_i}\right.$$

$$H^q(g)={\displaystyle \sum _{X(T)=g}}{\displaystyle \prod _{i=1}^M}{B}_i,{\displaystyle \underset{q\to 1}{lim}}SU{M}_q(N)=SUM(q),{\displaystyle \underset{q\to 1}{lim}}{H}^q(g)=H(g)$$

7.2. Property

Theorem 7.3. 

${\nabla }_q^{1}SU{M}_q(N,PS,[1,2...M])={\displaystyle \prod _{i=1}^M}(K_i+D_i{G}_1^n)$

Theorem 7.4. 

In $SU{M}_q(N,[...PS...],[...T+1,T+2...T+M...])$,$K_i$ can exchange orders.

$For{m}_2$ is simplest,$X=T_M-M-p$, 3→

Theorem 7.5. 

${\nabla }_q^{p}SU{M}_q(N)={\displaystyle \sum _{g=0}^M}{H}_1^q(g)G_{X+1+g}^{N+X}{q}^{-gp}={\displaystyle \sum _{g=0}^M}{H}_2^q(g)G_{X+1+g}^{N+X+g}={\displaystyle \sum _{g=0}^M}{H}_3^q(g)G_{X+M+1}^{N+X+M-g}{q}^{-gp}$

Definition 17. 

$n_{q-}=q^{-n}{G}_1^n,n_{q+}=q^n{G}_1^n,n_{q-}!=n_{q-}...2_{q-}{1}_{q-},0_{q-}!=0,n_{q+}!issimilar$

$For{m}_2\to$

Theorem 7.6. 

$SU{M}_q(N,[L_{1_{q-}},L_{2_{q-}}...L_{Q_{q-}},PS],[L_1,L_2...L_Q,PT])={\displaystyle \prod _{i=1}^Q}{L}_{i_{q-}}SU{M}_q(N,PS,PT)$,$T_1$ can great than 1,$T_i\in \mathbb{N}$

Theorem 7.7. 

$SU{M}_q(N,[T_{1_{q-}},T_{2_{q-}}...T_{M_{q-}}],[T_1,T_2...T_M])={\displaystyle \prod _{i=1}^M}{T}_{i_{q-}}{G}_{T_M+1}^{N+T_M}$

By definition and 4

Theorem 7.8. 

$In{H}^q(g),{\displaystyle \sum _{X_i\in K}}\prod q^{X_T}=G_{M-g}^M=G_g^M,X_T=numberof\{X_1,X_2...X_i\}\in T$

Induction →:

Theorem 7.9. 

$PT=[1,2....M],q^{-\left({}_2^{g+1}\right)}{H}_1^q(g)=q^{\left({}_2^{g+1}\right)}{\displaystyle \sum _{k=g}^M}{H}_2^q(k)G_g^k$

Theorem 7.10. 

$H_1^q(0)={\displaystyle \sum _{k=0}^M}{H}_2^q(k)$

Except for 7.9 and 7.10, other situations are relatively complex and no conclusions similar to 2.1 have been drawn.

7.3. Application

Theorem 7.11. 

${\displaystyle \sum _{0\le {\lambda }_1\le {\lambda }_2\le ...\le {\lambda }_M\le N}}{q}^{{\lambda }_1+{\lambda }_2+...+{\lambda }_M}=G_M^{N+M}=G_N^{N+M}$ [7]

Proof. 

$SU{M}_q(N,[1,1...1]:0,[1,3...2M-1])\to H_2^q(g>0)=0,H_2^q(0)=1\to {\displaystyle \sum _{{\lambda }_M=0}^{N-1}}{q}^{{\lambda }_M}...{\displaystyle \sum _{{\lambda }_1=0}^{{\lambda }_2}}{q}^{{\lambda }_1}=G_M^{N+M-1}$□

Theorem 7.12. 

${\displaystyle \sum _{1\le {\lambda }_1<{\lambda }_2<...<{\lambda }_M\le N}}{q}^{{\lambda }_1+{\lambda }_2+...+{\lambda }_M}=q^{\left({}_2^{M+1}\right)}{\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{\frac{g^2}{2}}{\left[{}_g^M\right]}_q{\left[{}_{2M}^{N+M-g}\right]}_{q^{\frac{1}{2}}}$

Proof. 

$$SU{M}_q(N,[q^2:(q-1)q^2,q^4:(q-1)q^4...q^{2M}:(q-1)q^{2M}],[1,3...2M-1])={\displaystyle \sum _{g=0}^M}{H}_3^q(g)G_{2M}^{N+2M-1-g}$$

$$={\displaystyle \sum _{{\lambda }_M=0}^{N-1}}{q}^{2({\lambda }_M+M)}...{\displaystyle \sum _{{\lambda }_2=0}^{{\lambda }_3}}{q}^{2({\lambda }_2+2)}{\displaystyle \sum _{{\lambda }_1=0}^{{\lambda }_2}}{q}^{2({\lambda }_1+1)}={\displaystyle \sum _{1\le {\lambda }_1<{\lambda }_2<...<{\lambda }_M\le N+M-1}}{q}^{2({\lambda }_1+{\lambda }_2+...+{\lambda }_M)}=*$$

$$B_i=\left\{{}_{q^{X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i)=q^{X_{T-1}}(q^{2i}+G_1^{X_{T-1}}{q}^{2i}(q-1))=q^{2X_{T-1}+2i}=q^{2i+2X_T},X_i=K_i}^{q^{1+X_{T-1}}\{(q^{X_{T-1}}{G}_1^{T_i}-q^{T_i}{G}_1^{X_{T-1}})D_i-K_i{q}^{T_i}\}=-q^{2i+1+2X_{T-1}}=-q^{-1+2i+2X_T},X_i=T_i}\right.$$

$$H_3^q(g)={(-1)}^g{q}^{2(1+2+...+M)+2(1+2+...+g)-g}\sum \prod q^{2X_T}={(-1)}^g{q}^{M(M+1)+g^2}{\left[{}_g^M\right]}_{q^2}$$

$$*=q^{M(M+1)}{\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{g^2}{\left[{}_g^M\right]}_{q^2}{G}_{2M}^{N+2M-1-g}$$

□

Theorem 7.13. 

$G_{M+1}^{N+M}=q^{-\left({}_2^{M+1}\right)}{\displaystyle \sum _{g=0}^M}[q^{\left({}_2^{g+1}\right)+g(M+1)}{\displaystyle \sum _{1\le {\lambda }_1<{\lambda }_2<...<{\lambda }_{{}_{M-g}}\le M}}{q}^{{\lambda }_1+{\lambda }_2+...+{\lambda }_{M-g}}]G_{1+g}^N$

Proof. 

$$SU{M}_q(N,[1_{q-},2_{q-}...M_{q-}],[1,2...M])=M_{q-}!G_{M+1}^{N+M}$$

$$=SU{M}_q(N,[M_{q-}...2_{q-},1_{q-}],[1,2...M])={\displaystyle \sum _{g=0}^M}{H}_1^q(g)G_{1+g}^N$$

$$B_i=\left\{{}_{K_i+G_1^{X_{T-1}}=q^{-(M+1-i)}{G}_1^{M+1-i}+G_1^{X_{T-1}}=\frac{q^{X_{T-1}}-q^{-(M+1-i)}}{q-1}=q^{-(M+1-i)}{G}_1^{M+1-i+X_{T-1}}=q^{-(M+1-i)}{G}_1^{M-X_{K-1}},X_i=K_i}^{q^{1+X_{T-1}}{G}_1^{T_i-X_{k-1}}=q^{1+X_{T-1}}{G}_1^{{}_i-X_{k-1}}=q^{1+X_{T-1}}{G}_1^{1+X_{T-1}},X_i=T_i}\right.$$

$$H_1^q(g,T)=g_{q+}!,H_1^q(g,\sum K)=G_1^M{G}_1^{M-1}...G_1^{g+1}{q}^{-(M+1)(M-g)}{\displaystyle \sum _{1\le {\lambda }_1<{\lambda }_2<...<{\lambda }_{M-g}\le M}}{q}^{{\lambda }_1+{\lambda }_2+...+{\lambda }_{M-g}}$$

$$\frac{H_1^q(g,T)}{M_{q-}!}=\frac{q^{\left({}_2^{g+1}\right)}{q}^{-(M+1)(M-g)}{\displaystyle \sum _{1\le {\lambda }_1<...<{\lambda }_{M-g}\le M}}{q}^{{\lambda }_1+...+{\lambda }_{M-g}}}{q^{-(1+2+...+M)}}$$

□

$$eg:G_4^{n+3}=q^{12}{G}_4^n+q^5(q^1+q^2+q^3)G_3^n+q^{-1}(q^1{q}^2+q^1{q}^3+q^2{q}^3)G_2^n+q^{-6}(q^1{q}^2{q}^3)G_1^n$$

Definition 18. 

${E_q}_M^N={\displaystyle \sum _{1\le {\lambda }_1<...<{\lambda }_M\le N}}{G}_1^{{\lambda }_1}...G_1^{{\lambda }_M}=S_2^q(N+M,N)$

Theorem 7.14. 

${(G_1^N)}^M={(1+q+...+q^{n-1})}^M={\displaystyle \sum _{g=1}^M}{g}_{q+}!S_2^q(M,g)G_g^N{q}^{-g}$

Proof. 

$${\nabla }_q^{1}SUM(N,[1_{q-},1_{q-}...1_{q-}],[1,2...M])=\prod (\frac{1}{q}+\frac{q^n-1}{q-1})=q^{-M}{(G_1^{n+1})}^M=q^{-M}{(G_1^N)}^M$$

$$=q^{-1}{\nabla }_q^{1}SUM(N,[1_{q-},1_{q-}...1_{q-}],[2,3...M])=q^{-1}{\displaystyle \sum _{g=0}^{M-1}}{H}_1^q(g)G_{1+g}^N{q}^{-g}$$

$$B_i=\left\{{}_{q^{-1}+G_1^{X_{T-1}}=q^{-1}{G}_1^{1+X_{T-1}},X_i\in K}^{q^{1+X_{T-1}}{G}_1^{(i+1)-X_{K-1}}=q^{X_T}{G}_1^{1+X_T},X_i\in T}\right.$$

$$H_1^q(g,T)=q^{-(g+1)}{(g+1)}_{q+}!,H_1^q(g,\sum K)=q^{-(M-1-g)}{E_q}_{M-1-g}^{g+1}$$

$${(G_1^N)}^M=q^{M-1}{\displaystyle \sum _{g=0}^{M-1}}{q}^{-(g+1)}{(g+1)}_{q+}{q}^{-(M-1-g)}{S}_2^q(M,g+1)!G_{1+g}^N{q}^{-g}$$

□

$SU{M}_q(N,[K_i+1:D_i(q-1)],[1,2...M])\to$

Theorem 7.15. 

${\displaystyle \sum _{n=0}^{N-1}}{q}^n{\displaystyle \prod _{i=1}^M}(K_i+D_1{q}^n)={\displaystyle \sum _{g=0}^M}{H}_1^q(g)G_{1+g}^N,B_i=\left\{{}_{K_i+q^{X_T}{D}_i,X_i\in K}^{q^{X_T}(q^{X_T}-1)D_i,X_i\in T}\right.$

induction →

Theorem 7.16. 

$${\displaystyle \sum _{n=0}^{N-1}}{\displaystyle \prod _{i=1}^M}(K_i+D_1{q}^n)={\displaystyle \sum _{g=1}^M}f(g)G_g^N+N\prod K_i,f(g)=\prod B_i=\left\{\begin{array}{c}1,X_i\in T,X_{T-1}=0\\ q^{X_{T-1}}(q^{X_{T-1}}-1)D_i,X_i\in T,X_{T-1}>0\\ K_i,X_i\in K,X_{T-1}=0\\ K_i+q^{X_{T-1}-1}{D}_i,X_i\in K,X_{T-1}>0\end{array}\right.$$

Theorem 7.17. 

$q^{Mn}=[1]{\displaystyle \sum _{g=0}^M}{G}_g^M{\displaystyle \prod _{i=0}^{g-1}}(q^n-q^i)=[2]q^{-M}{\displaystyle \sum _{g=0}^M}{q}^g{\left[{}_g^M\right]}_{q^{-1}}{\displaystyle \prod _{i=1}^g}(q^n-q^{-i})=[3]{\displaystyle \sum _{g=0}^M}{G}_g^M{(-1)}^g{q}^{\left({}_2^g\right)}{G}_M^{n+M-g}$

Proof. 

$$q^{Mn}={\nabla }_q^{1}SUM(N,[1,1...1]:q-1,[1,2...M])$$

$$={\displaystyle \sum _{g=0}^M}{H}_1^q(g)G_g^n{q}^{-g}={\displaystyle \sum _{g=0}^M}{H}_2^q(g)G_g^{n+g}={\displaystyle \sum _{g=0}^M}{H}_3^q(g)G_M^{n+M-g}{q}^{-g}$$

$$B_i=\left\{{}_{1+G_1^{X_{T-1}}(q-1)=q^{X_{T-1}}=q^{X_T},X_i\in K}^{q^{1+X_{T-1}}{G}_1^{i-X_{K-1}}(q-1)=q^{X_T}(q^{X_T}-1),X_i\in T}\right.,H_1^q(g,T)={(q-1)}^g{g}_{q+}!,H_1^q(g,\sum K)=G_g^M\to [1]$$

$$B_i=\left\{{}_{q^{-(1+X_{T-1})}=q^{-1}{q}^{-X_T},X\in K}^{q^{-(1+X_{T-1)}}(q^{1+X_{T-1}}-1),X\in T}\right.,H_2^q(g,T)={(q-1)}^g{g}_{q-}!,H_2^q(g,\sum K)=q^{-(M-g)}{\left[{}_g^M\right]}_{q^{-1}}\to [2]$$

$$B_i=\left\{{}_{q^{X_{T-1}}=q^{X_T},X\in K}^{-q^{1+X_{T-1}}=-q^{X_T},X\in T}\right.,H_3^q(g,T)={(-1)}^g\left({}_2^{g+1}\right),H_3^q(g,\sum K)=G_g^M\to [3]$$

□

Theorem 7.18. 

${\displaystyle \sum _{k_r=1}^N}...{\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}{q}^{k_1+k_2+...+k_r}{\nabla }_q^{p}SU{M}_q(k_1,PS,PT)={\nabla }_q^{p-r}SU{M}_q(N,PS,PT)$

Theorem 7.19. 

${\displaystyle \sum _{k_x=1}^N}...{\displaystyle \sum _{k_2=1}^{k_3}}{\displaystyle \sum _{k_1=1}^{k_2}}{q}^{k_1+k_2+...+k_x}{\nabla }_q^{1}SU{M}_q(k_r,PS,[1,2...M])={\nabla }_q^{x-r}SU{M}_q(N,PS,[T_i=i+(r-1)])$

Theorem 7.20. 

${(K+qD)}^M=(q-1)D{\displaystyle \sum _{g=0}^{M-1}}{(K+D)}^g{(K+qD)}^{M-1-g}+{(K+D)}^M$

Proof. 

$$PS=[K+D,K+D...K+D]:(q-1)D,PT=[1,2...M]$$

$$B_i=\left\{{}_{K+D{q}^{X_T},X\in K}^{q^{X_T}(q^{X_T}-1)D,X\in T}\right.,H_1^q(1)=q^1(q^1-1)D{\displaystyle \sum _{a+b=M-1,a,b\ge 0}}{(K+D)}^a{(K+qD)}^b,H_1^q(0)={(K+D)}^M$$

$$SU{M}_q\left(2\right)-SU{M}_q\left(1\right)=H_1^q(1)+H_1^q(0)G_1^2-H_1^q(0)=H_1^q(1)+q{H}_1^q(0)$$

$$=q{\displaystyle \prod _{i=1}^M}(K+D+G_1^1(q-1)D)=q{(K+qD)}^M$$

□

7.4. Matrix of $SU{M}_q(N)$

Similar to Section "Matrix of SUM(N)",Let $P=N+T_M-M,Q=N-1$,define $A^q(P,Q,M)$, respectively corresponding to the three forms.

$$A_1^q(P,Q,M)=\left(\begin{array}{ccc}{G}_Q^P& \dots & G_{Q-M}^P\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+M}& \cdots & G_Q^{P+M}\end{array}\right)$$

$$A_2^q(P,Q,M)=\left(\begin{array}{ccc}{G}_Q^P& \dots & G_Q^{P+M}\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+M}& \cdots & G_{Q+M}^{P+2M}\end{array}\right)$$

$$A_3^q(P,Q,M)=\left(\begin{array}{ccc}{G}_Q^{P+M}& \dots & G_{Q-M}^P\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+2M}& \cdots & G_Q^{P+M}\end{array}\right)$$

Theorem 7.21. 

$\Vert A_2^q(P,Q,M)\Vert =\Vert A_2^q(P,P-Q,M)\Vert =\frac{G_Q^P}{G_0^P}\frac{G_{Q+1}^{P+2}}{G_1^{P+2}}...\frac{G_{Q+M}^{P+2M}}{G_M^{P+2M}}{q}^{(P+1)+2(P+2)+...+M(P+M)}$

Theorem 7.22. 

$\Vert A_2^q(P,0,M)\Vert =q^{(P+1)+2(P+2)+...+M(P+M)},\Vert A_2^q(P,1,M)\Vert =G_{M+1}^{P+M}{q}^{(P+1)+2(P+2)+...+M(P+M)}$

Theorem 7.23. 

$\Vert A_{1,3}^q(P,Q,M)\Vert =\frac{G_Q^P}{G_0^P}\frac{G_{Q+1}^{P+2}}{G_1^{P+2}}...\frac{G_{Q+M}^{P+2M}}{G_M^{P+2M}}{q}^{Q\left({}_2^{M+1}\right)},\Vert A_{1,3}^q(P,0,M)\Vert =1,\Vert A_{1,3}^q(P,1,M)\Vert =G_{M+1}^{P+M}{q}^{\left({}_2^{M+1}\right)}$

8. Multi-parameter Formal Calculation

2-parameters Formal Calculation calculate nested sum of $K_i+\left({}_1^n\right)D_{1,i}+\left({}_2^n\right)D_{2,i}$.

$$SUM(N,[K_1],[T_{1,1}=1:D_{1,1}],[T_{2,1}=1:D_{2,1}])={\displaystyle \sum _{n=0}^{N-1}}(K_1+D_{1,1}n+D_{2,1}\left({}_2^n\right))$$

$$SUM(N,[K_1,K_2],[T_{1,1}:D_{1,1},T_{1,2}=T_{1,1}+2-p:D_{1,2}],[T_{2,1}:D_{2,1},T_{2,2}=T_{1,2}:D_{2,2}])$$

$$={\displaystyle \sum _{n=0}^{N-1}}(K_2+D_{1,2}n+D_{2,2}\left({}_2^n\right)){\nabla }^{p}SUM(N,[K_1],[T_{1,1}:D_{1,1}],[T_{2,1}:D_{2,1}])$$

Recursive define $SUM(N,PS,P{T}_1,P{T}_2)$. 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 19. 

X($P{T}_1$)=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 20. 

$X_{PT1}$=Number of $\{X_1...X_i\}\in T_1$,$X_{PT2}$=Number of $\{X_1...X_i\}\in T_2$,$X_{PT}=X_{PT1}+2X_{PT2}$

Use induction to prove:

Theorem 8.1. 

$SUM(N,PS,P{T}_1,P{T}_2)$

$$={\displaystyle \sum _{g=0}^{2M}}{H}_1(g)\left({}_{T_M-M+1+g}^{N+T_M-M}\right),B_i=\left\{\begin{array}{c}{K}_i+X_{PT}{D}_{1,i}+\left({}_2^{X_{PT}}\right)D_{2,i},X_i=K_i\\ (T_i-i+X_{PT})D_{1,i}+(T_i-i+X_{PT})(X_{PT}-1)D_{2,i},X_i=T_{1,i}\\ \left({}_2^{T_i-i+X_{PT}}\right)D_{2,i},X_i=T_{2,i}\end{array}\right.$$

$$={\displaystyle \sum _{g=0}^{2M}}{H}_2(g)\left({}_{T_M-M+1+g}^{N+T_M-M+g}\right),B_i=\left\{\begin{array}{c}{K}_i-(T_i-i+X_{PT}+1)D_{1,i}+\left({}_2^{T_i-i+X_{PT}+2}\right)D_{2,i},X_i=K_i\\ (T_i-i+X_{PT})D_{1,i}-(T_i-i+X_{PT})(T_i-i+X_{PT}+1)D_{2,i},X_i=T_{1,i}\\ \left({}_2^{T_i-i+X_{PT}}\right)D_{2,i},X_i=T_{2,i}\end{array}\right.$$

$$={\displaystyle \sum _{g=0}^{2M}}{H}_3(g)\left({}_{T_M+M+1}^{N+T_M+M-g}\right),B_i=\left\{\begin{array}{c}{K}_i+X_{PT-1}{D}_{1,i}+\left({}_2^{X_{PT-1}}\right)D_{2,i},X_i=K_i\\ -2K_i+(T_i+i-1-2X_{PT-1})D_{1,i}+(T_i+i-X_{PT-1})D_{2,i},X_i=T_{1,i}\\ K_i-(T_i+i-1-X_{PT-1})D_{1,i}+\left({}_2^{T_i+i-X_{PT-1}}\right)D_{2,i},X_i=T_{2,i}\end{array}\right.$$

According to this way, it can be extended to multi-parameter $SUM(N)$ and $SU{M}_q(N)$.This formula is very complex and has not been studied in terms of analysis yet.

9. A theorem of symmetry

In this section,$T_i\ge i$.

From 2.12$\to H_1(g,PT,PT)={\displaystyle \prod _{i=1}^M}{T}_i\left({}_g^M\right)={\displaystyle \prod _{i=1}^M}{T}_i\left({}_{g,M-g}^M\right)$.Promoted it:

The Set $\{T_1,T_2...T_M\}$ come from p Source:$S_1,S_2...S_p$,$T_i\in S_{j}means{T}_{i}comefromsourej$

Definition 21. 

$Diff(S_x,S_x)=0,Diff(S_x,S_y)=-Diff(S_y,S_x)=1,x\ne y$

Definition 22. 

$Diff(T_i,T_j)=Diff(S_x,S_y),T_i\in S_x,T_j\in S_y$

Definition 23. 

$W(g_1,g_2...g_p,[T_1,T_2....T_M])={\displaystyle \sum _{g_1+g_2+...+g_p=M,g_{{}_i}\ge 0}}{\displaystyle \prod _{i=1}^M}(T_i+{\displaystyle \sum _{j<i}}Diff(T_j,T_i))$

In set T, $g_1$ come from $S_1$,$g_2$ comes from $S_2$...,$g_M$ comes from $S_M$

[1] has proved:

Theorem 9.1. 

$W(g_1,g_2...g_p,[T_1,T_2....T_M])={\displaystyle \prod _{i=1}^M}{T}_i\left({}_{g_1,g_2,...,g_M}^M\right)$

Promote to Gaussian coefficient:

Definition 24. 

$$W_q(g_1,g_2...g_p,[T_1,T_2....T_M])={\displaystyle \sum _{g_1+g_2+...+g_p=M,g_i\ge 0}}{\displaystyle \prod _{i=1}^M}{G}_1^{T_i+{\displaystyle \sum _{j<i}}Diff(T_j,T_i)}{q}^{{\displaystyle \sum _{j<i,Diff(T_j,T_i)=-1}}1}$$

Definition 25. 

$G_{g_1,g_2...g_p}^M=\frac{(q^M-1)(q^{M-1}-1)...(q^1-1)}{{\displaystyle \prod _{i=0}^p}(q^{g_i}-1)(q^{g_i-1}-1)...(q^i-1)},g_1+g_2+...+g_p=M$

Theorem 9.2. 

$W_q(g_1,g_2...g_p,[T_1,T_2....T_M])=({\displaystyle \prod _{i=1}^M}{G}_1^{T_i})G_{g_1,g_2...g_p}^M,T_i\ge i$

Proof. 

When g=2,it clearly holds when M=1 or $g_1$=0 or $g_1$=M.

when M=2:$G_1^{T_1}{G}_2^{T_2+1}+G_1^{T_1}{G}_2^{T_2-1}q=G_1^{T_1}{G}_1^{T_2}{G}_1^2$

When M+1,prove by induction,PT1=[PT,$T_{M+1}$]

$$W_q(g_1,g_2+1,PT1)=W_q(g_1,g_2,PT)G_1^{T_{M+1}+g_1}+W_q(g_1-1,g_2+1,PT)G_1^{T_{M+1}-(g_2+1)}{q}^{g_2+1}$$

$$=({\displaystyle \prod _{i=1}^M}{G}_1^{T_i})G_{g_1,g_2}^M{G}_1^{T_{M+1}+g_1}+({\displaystyle \prod _{i=1}^M}{G}_1^{T_i})G_{g_1-1,g_2+1}^M{G}_1^{T_{M+1}-(g_2+1)}{q}^{g_2+1}$$

Just need to prove:

$$G_{g_1}^M{G}_1^{T_{M+1}+g_1}+G_{g_1-1}^M{G}_1^{T_{M+1}-(M-g_1+1)}{q}^{M-g_1+1}=G_1^{T_{M+1}}{G}_{g_1}^{M+1}=G_1^{T_{M+1}}(q^{g_1}{G}_{g_1}^M+G_{g_1-1}^M)(*)$$

$$G_1^{T_{M+1}}(q^{g_1}{G}_{g_1}^M+G_{g_1-1}^M)\times \frac{(q^{M-g_1+1}-1)}{G_{g_1}^M}=G_1^{T_{M+1}}(q^{g_1}(q^{M-g_1+1}-1)+(q^{g_1}-1))$$

$$=(q^{T_{M+1}-1}+...+q+1)(q^{M+1}-1)(1)$$

$$Left\times \frac{(q^{M-g_1+1}-1)}{G_{g_1}^M}=(q^{M-g_1+1}-1)G_1^{T_{M+1}+g_1}+(q^{g_1}-1)G_1^{T_{M+1}-(M-g_1+1)}{q}^{M-g_1+1}$$

$$=(q^{M-g_1+1}-1)(q^{T_{M+1}+g_1-1}+...+q+1)+(q^{g_1}-1)(q^{T_{M+1}-1}+...+q^{M-g_1+2}+q^{M-g_1+1})=(2)$$

$(1)-(2)=0\to$It’s holds when g=2.

$$W_q(g_1,g_2+g_3,[T_1,T_2....T_M])=({\displaystyle \prod _{i=1}^M}{G}_1^{T_i})G_{g_1,g_2+g_3}^{g_1+g_2+g_3}$$

Every product has $g_2+g_3$ factors come from $Sourc{e}_2$,divide them to $g_2\times Sourc{e}_2+g_3\times Sourc{e}_3$

$g_1$-factors are invariant,$g_2+g_3$)-factors are variant.

$$\sum \prod (variantfactors)=W_q(g_2,g_3,[X_1,X_2....X_{g_2+g_3}])={\displaystyle \prod _{i=1}^{g_2+g_3}}{G}_1^{X_i}{G}_{g_2,g_3}^{g_2+g_3}$$

$$W_q(g_1,g_2,g_3,[T_1,T_2....T_M])=({\displaystyle \prod _{i=1}^M}{G}_1^{T_i})G_{g_1,g_2+g_3}^{g_1+g_2+g_3}{G}_{g_2,g_3}^{g_2+g_3}=({\displaystyle \prod _{i=1}^M}{G}_1^{T_i})G_{g_1,g_2,g_3}^{g_1+g_2+g_3}$$

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