Formal Calculation

1. Introduction

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

Definition 1. 

Recursive define ${\nabla }^p,p\in \mathbb{Z},{\nabla }^{0}f\left(n\right)=f\left(n\right),{\displaystyle \sum _{n=0}^{N-1}}{\nabla }^{1}f(n+1)=f\left(N\right),{\displaystyle \sum _{n=0}^{N-1}}f(n+1)={\nabla }^{-1}f\left(N\right)$, ${\nabla }^1=\nabla$

Definition 2. 

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

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

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

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

$[K_1:D,K_2:D...K_M:D]$ 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 auxiliary 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 $\{X_1,X_2...X_{i-1}\}\in K$

$X_T$=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)=

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

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

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

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

The theorem is proved by induction.There have three forms because:${\displaystyle \sum _{n=0}^{N-1}}n\left({}_M^{n+K}\right)$

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

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. 

$E_M^{K=\{K_1,K_2...K_M\}}$=$\sum K_{I_1}{K}_{I_2}...K_{I_M},a<b,I_a\le I_b$,$E_M^N$ is short for $E_M^{\{1,2...N\}}$,$E_0^K=0$

Theorem 2. 

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

Theorem 3. 

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

Theorem 4. 

$SUM(N,[L_1,L_2...L_q,PS],[L_1,L_2...L_q,PT])={\displaystyle \prod _{i=1}^q}{L}_i\times SUM(N,PS,PT),so$$T_1$ can great than 1,$T_i\in \mathbb{N}$

Theorem 5. 

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

Theorem 6. 

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

$=\sum _{{\lambda }_1+...+{\lambda }_N=M,{\lambda }_i\ge 0}{1}^{{\lambda }_1}{2}^{{\lambda }_2}...N^{{\lambda }_N}=E_M^N=S_2(N+M,N)$. $S_2$ is Stirling numbers of the second kind.

Theorem 7. 

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

$=\sum _{1\le i_1<i_2<...<i_M\le N+M-1}{i}_1{i}_2...i_M=F_M^{N+M-1}=S_1(N+M,N)$. $S_1$ is unsigned Stirling numbers of the first kind.

Example 1.1:

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

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

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

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

It also can be calculated in the Ring with identity elements.$K_i,D_i$ 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 $H\left(g\right)$

By definition:

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

Using these relationships and induction can prove:

Theorem 9. 

$H_1\left(g\right)={\displaystyle \sum _{k=g}^M}{H}_2\left(k\right)\left({}_g^k\right)={\displaystyle \sum _{k=0}^g}{H}_3\left(k\right)\left({}_{M-g}^{M-k}\right)$ [2]

Inversion→

Theorem 10. 

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

Calculation with 9 →

Theorem 11. 

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

Theorem 12. 

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

This indicates $For{m}_1=For{m}_2=For{m}_3\to {\displaystyle \sum _{g=0}^M}{H}_1\left(g\right)\left({}_g^A\right)={\displaystyle \sum _{g=0}^M}{H}_2\left(g\right)\left({}_g^{A+g}\right)={\displaystyle \sum _{g=0}^M}{H}_3\left(g\right)\left({}_M^{A+M-g}\right)$

Induction [2]→

Theorem 13. 

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

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. 

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

If $D_i=1$ and $T_i+1=T_{i+1}$,$H_1(g,T)={\displaystyle \prod _{i=1}^g}{T}_i$. $H_1(g,T,PS,[1,2...M])=g!$.$H_1(g,\sum K,[1,1...1],PT)=E_{M-g}^{g+1}$

Theorem 14. 

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 15. 

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

Theorem 16. 

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

Theorem 17. 

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

Theorem 18. 

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

Definition 8. 

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

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

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

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

$eg:〈{}_2^5〉=E_2^3\odot ([1,1...],1)=\sum _{{\lambda }_1+{\lambda }_2+{\lambda }_3=2}{1}^{{\lambda }_1}{2}^{{\lambda }_2}{3}^{{\lambda }_3}(1+{\lambda }_1)(1+{\lambda }_1+{\lambda }_2)=66$

By simple calculation:

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

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

2.3. Shape of numbers

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

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

There are M-1 intervals between factors.If the interval=1,define 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:

$\sum _{1\le K_1<K_2<K_3\le N}{K}_1{K}_2{K}_3=SUM(N,[1,2,3],[1,2,3])+SUM(N-1,[1,2,4],[1,2,4])+SUM(N-1,[1,3,4],[1,3,4])+SUM(N-2,[1,3,5],[1,3,5])$

Definition 9. 

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

$(*)=\sum _{AlloftheShapeswithfactors=M}SUM(N-PB(PT),PT,PT)$.From 17 we can obtain a simple formula:

Theorem 19. 

$SUM(N,PT,PT)={\displaystyle \prod _{i=1}^M}{T}_i\left({}_{T_M-M+1}^{N+T_M-M}\right)$,PT has no restrictions.

This 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 (*) which led to the birth of Formal Calculation.

Theorem 20. 

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

Definition 10. 

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

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

Theorem 21. 

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

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

$eg:MI{N}_2\left(4\right)=\left(12357\right)+\left(12457\right)+\left(13457\right)+\left(12467\right)+\left(13467\right)+\left(13567\right)$.Here (...) is products.

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

Theorem 22. 

$SUM(N,[K_1:D_1,K_2:D_2...K_M:D_M],[T_1,T_2...T_M])=SUM(N,PS,[T_1...T_i,T_{i+1}-1...T_M-1])+SUM(N-1,[K_1:D_1...K_i:D_i,K_{i+1}+D_{i+1}:D_{i+1}...K_M+D_M:D_M],PT)$

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

$(*)=SUM(N,[1,2...M],[1,3...2M-1])={\displaystyle \sum _{g=0}^{M-1}}MI{N}_g\left(M\right)\left({}_{g+1}^N\right).I{t}^{\prime }sexactly.$

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.

• $S_{1,r}(n,k)=\frac{n!}{k!}\sum _{i_1+i_2+...+i_k=n,i_j\ge r}\frac{1}{i_1{i}_2...i_k}$
• $S_{1,r}(n+1,k)=n{S}_{1,r}(n,k)+{\left(n\right)}_{r-1}{S}_{1,r}(n-r+1,k-1),n\ge kr$
• $\sum \frac{(n-1)!}{i_1{i}_2...i_{k-1}},r\le i_1<i_2<...<i_{k-1}\le n-r,i_{j+1}-i_j\ge r$ [5]

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

Theorem 23. 

$MI{N}_g\left(M\right)=S_{1,2}(M+g+1,g+1)$

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

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

	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

$S_{2,r}(n,k)=\frac{n!}{k!}\sum _{i_1+i_2+...+i_k=n,i_j\ge r}\frac{1}{i_1!i_2!...i_k!}$

2

$S_{2,r}(n+1,k)=k{S}_{2,r}(n,k)+\left({}_{r-1}^n\right)S_{2,r}(n-r+1,k-1),n\ge kr$

Derived from 2:

Theorem 24. 

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

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

$SUM(N,[2,3...M],[3,5...2M-1])=S_{1,1}(N+M,N)=\frac{(N+M)!}{N!}\sum _{i_1+i_2+...+i_N=N+M,i_j\ge 1}\frac{1}{i_1{i}_2...i_N}$

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

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

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

Use the same way:

Theorem 25. 

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

Theorem 26. 

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

2.5. Table of H(g)

Table 3. Table of H(g).

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

Table 3. Table of H(g).

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

3. Application

3.1. Number analysis

Theorem 27. 

9, 11, 13 →

• ${\displaystyle \sum _{g=0}^M}〈{}_g^M〉=M!,{\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g!S_2(M,g)=1,{\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g\times g!S_2(M,g)=2^M-1$
• $g!S_2(M,g)={\displaystyle \sum _{k=g}^M}{(-1)}^{M-k}k!S_2(M,k)\left({}_{g-1}^{K-1}\right)={\displaystyle \sum _{k=0}^{g-1}}〈{}_k^M〉\left({}_{M-g}^{M-1-k}\right),1\le g\le M$
• $S_{1,2}(M+g,g)={\displaystyle \sum _{k=g}^M}{(-1)}^{M-k}{S}_{2,2}(M+k,k)\left({}_{g-1}^{K-1}\right),S_{2,2}(M+g,g)={\displaystyle \sum _{k=g}^M}{(-1)}^{M-k}{S}_{1,2}(M+k,k)\left({}_{g-1}^{K-1}\right)$
• ${\displaystyle \sum _{g=1}^M}g!S_2(M,g)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g!S_2(M,g)2^{g-1}={\displaystyle \sum _{g=1}^M}〈{}_{M-g}^M〉2^{M-g}$
• ${\displaystyle \sum _{g=1}^M}{S}_{2,2}(M+g,g)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{1,2}(M+g,g)2^{g-1},{\displaystyle \sum _{g=1}^M}{S}_{1,2}(M+g,g)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{2,2}(M+g,g)2^{g-1}$
• ${\displaystyle \sum _{g=1}^M}g!S_2(M,g)(g-1)\left({}_g^{A+1}\right)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}g!S_2(M,g)(g-1)\left({}_g^{A+g-1}\right)$
• ${\displaystyle \sum _{g=0}^{M-1}}{(-1)}^{M-1-g}MI{N}_g\left(M\right)={\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{1,2}(M+g,g)=1,{\displaystyle \sum _{g=1}^M}{(-1)}^{M-g}{S}_{2,2}(M+g,g)=M!$

PS=PT=[1,2...M],$H_1\left(g\right)=H_1(M-g)=M!\left({}_g^M\right)$,14→

Theorem 28. 

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

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

Theorem 29. 

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

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

Theorem 30. 

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

$H_1(g,[K+i],[T+i])$,16$\to \left({}_g^M\right)T_1...T_g\times K_{g+1}...K_M$,14$\to T_1...T_g(...)$,15$\to K_{g+1}...K_M(...)$

Theorem 31. 

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

3.2. Merge and Expand

Theorem 32. 

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

Any ${\displaystyle \sum _{g=0}^M}{a}_g\left({}_{Y+g}^X\right)$ can be converted to $\frac{a_M}{M!}{\nabla }^{q}SUM(N+P,[K_1,K_2...K_M],[1,2...M])$

10 provides the necessary and sufficient condition for ${\displaystyle \sum _{g=0}^M}H\left(g\right)\left({}_{Y+g}^X\right)$ to be merged into ${\displaystyle \sum _{g=0}^{M-K}}(...)\left({}_{Y+K+g}^{X+K}\right):$

$H_2\left(g\right)={\displaystyle \sum _{x=g}^M}{(-1)}^{x}H\left(x\right)\left({}_g^x\right)=0,g<Kor{H}_3\left(g\right)={\displaystyle \sum _{x=0}^g}{(-1)}^{x}H\left(x\right)\left({}_{M-g}^{M-x}\right)=0,M-g<K$

For example:

$SUM(N,[1,2...M],[1,2...M]),\to {\displaystyle \sum _{x=g}^M}{(-1)}^x\left({}_x^M\right)\left({}_g^x\right)=0,0\le g<M$${\displaystyle \sum _{n=0}^{N-1}}\left({}_M^{M+dn}\right)=\frac{1}{M!}SUM(N,[1,2...M]:d,[1,2...M])=\frac{1}{M!}{\displaystyle \sum _{g=0}^M}{H}_1\left(g\right)\left({}_{1+g}^N\right)$

$IfM\ge kd,B_i(X_i=K_i)=i+(X_{K-1}-i)d\to H_2(g<k)=0\to {\displaystyle \sum _{n=0}^{N-1}}\left({}_M^{M+dn}\right)={\displaystyle \sum _{g=0}^{M-k}}(...)\left({}_{1+g+k}^{N+k}\right)$

After a simple calculation,it can be written as:

Theorem 33. 

Necessary and sufficient conditions for merging,$0\le g<K\le M$:

• ${\displaystyle \sum _{n=0}^M}H\left(n\right)\left({}_{Y+n}^X\right)\to {\displaystyle \sum _{n=0}^{M-K}}(...)\left({}_{Y+K+n}^{X+K}\right):{\displaystyle \sum _{x=0}^M}{(-1)}^{x}H\left(x\right)\left({}_g^{P\pm x}\right)=0$
• ${\displaystyle \sum _{n=0}^M}H\left(n\right)\left({}_{Y+n}^{X+n}\right)\to {\displaystyle \sum _{n=0}^{M-K}}(...)\left({}_{Y+K+n}^{X+K+n}\right):{\displaystyle \sum _{x=0}^M}H\left(x\right)\left({}_g^{P\pm x}\right)=0$
• ${\displaystyle \sum _{n=0}^M}H\left(n\right)\left({}_Y^{X-n}\right)\to {\displaystyle \sum _{n=0}^{M-K}}(...)\left({}_{Y-K}^{X-K-n}\right):{\displaystyle \sum _{x=0}^M}H\left(x\right)\left({}_g^{P\pm x}\right)=0$

Theorem 34. 

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

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])

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

$={\displaystyle \sum _{g=0}^M}\left({}_g^M\right)\frac{(T+A+g)!}{(T+A)!}\frac{(T+M)!}{(T+g)!}\left({}_{T+A+1+g}^{N+T+A}\right)=\frac{A!(T+M)!}{(T+A)!}{\displaystyle \sum _{g=0}^M}\left({}_g^M\right)\left({}_A^{A+T+g}\right)\left({}_{T+A+1+g}^{N+T+A}\right)$

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

$=\frac{(T+M)!}{(T+A)!}{\displaystyle \sum _{g=0}^A}\left({}_g^A\right)\frac{(T+M+g)!}{(T+M)!}\frac{(T+A)!}{(T+g)!}\left({}_{T+M+1+g}^{N+T+M}\right)$

${\displaystyle \sum _{g=0}^M}\left({}_g^M\right)\left({}_A^{A+T+g}\right)\left({}_{T+A+1+g}^{N+T+A}\right)={\displaystyle \sum _{g=0}^A}\frac{(T+M+g)!}{(T+M)!}\frac{(T+A)!}{(T+g)!}\frac{A!}{(A-g)!g!}\frac{1}{A!}\left({}_{T+M+1+g}^{N+T+M}\right)$□

$g:=M-g\to {\displaystyle \sum _{g=0}^M}\left({}_g^M\right)\left({}_A^{A+T+M-g}\right)\left({}_{Y+g}^X\right)={\displaystyle \sum _{g=0}^A}\left({}_g^{A+T}\right)\left({}_{M+T}^{A+M+T-g}\right)\left({}_{Y+M-A+g}^{X+M-A}\right),A\ge 0$

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

Theorem 35. 

${\displaystyle \sum _{g=0}^M}{(-1)}^g\left({}_g^M\right)\left({}_A^{X_1\pm g}\right)\left({}_B^{X_2\pm g}\right)=0,M>A+B,A,B\ge 0$

$Iff\left(n\right)={\displaystyle \sum _{i=0}^B}{a}_i{n}^i,B<M\to {\displaystyle \sum _{g=0}^M}{(-1)}^g\left({}_g^M\right)f\left(g\right)=0\to {\displaystyle \sum _{g=0}^M}{(-1)}^g\left({}_g^M\right)\left({}_{Y_1}^{X_1\pm g}\right)\left({}_{Y_2}^{X_2\pm g}\right)...=0,\sum Y_i<M$

It can be proved by induction:

Theorem 36. 

${\displaystyle \sum _{g=0}^M}{(-1)}^g\left({}_g^M\right)\left({}_{M+K}^{T+g}\right)=\left\{{}_{{(-1)}^M\left({}_K^T\right),K\ge 0}^{0,K<0}\right.$; ${\displaystyle \sum _{g=0}^M}{(-1)}^g\left({}_g^M\right)\left({}_{M+K}^{T-g}\right)=\left\{{}_{\left({}_K^T\right),K\ge 0}^{0,K<0}\right.T\pm g\in \mathbb{Z}$

This helps to understand the differential sequence.

3.3. Congruences

P is prime.$K_i,D$ is any integer,D ≠0.

Theorem 37.$(P,D)=1,SUM(P,[K_1,K_2...K_M]:D,[1,2...M])\equiv \left\{{}_{-1(modP),M=P-1}^{0(modP),M<P-1}\right.$

Proof. 

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

If a product has a factor 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: $A,B,C\in \mathbb{N}$

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

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

Theorem 38. 

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

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

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

They are two extremes.In fact,there have:

Theorem 39. 

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

Proof. 

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

For $\sum A{B}^{P-3},A\ne B$,if $P-A\ne B$ then add $A{B}^{P-3}$ with $(P-A)B^{P-3}$ to $P{B}^{P-3}$,if $P-A=B$ then add $A{B}^{P-3}$ with $B{B}^{P-3}$ to $P{B}^{P-3}$.so $\sum A{B}^{P-3}+\sum B^{P-2}=xP\sum B^{P-3},0<x<P\to \sum A{B}^{P-3}\equiv 0(mod{P}^2)$.

Similarly:

For $\sum AB{C}^{P-4},A\ne B\ne C$,treat P-A=B,P-A=C,$P-A\ne B,C$ separately.

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

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

$\sum AB{C}^{P-4}+\sum B^2{C}^{P-4}=\sum (\left({}_2^P\right)-B)B{C}^{P-4}+\sum B^2{C}^{P-4}=\left({}_2^P\right)\sum B{C}^{P-4}\equiv 0\left(mod{P}^2\right)$

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

Prove the conclusion in a similar way... □

Theorem 40. 

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

$eg:S_2(7,4)=350,S_2(8,5)=1050\equiv 0(mod25),S_2(11,6)=179487,S_2(12,7)=627396\equiv 0(mod49)$

4. Combinatorial Identities

Definition 11. 

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

By nested sum:

Theorem 41. 

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

Theorem 42. 

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

Proof. 

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

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

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

Theorem 43. 

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

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

$H_1(g<j)=0,H_1\left(j\right)=\frac{(j+i-1)!}{(i-1)!},T_j-j=i-1\to$

$(*)=\frac{1}{j!}{\nabla }^{i-r}\frac{(j+i-1)!}{(i-1)!}\left({}_{j+1+(i-1)}^{(N+1)+(i-1)}\right)=\left({}_{i-1}^{j+i-1}\right)\left({}_{j+i-(i-r)}^{N+i-(i-r)}\right)=\left({}_{i-1}^{j+i-1}\right)\left({}_{j+r}^{N+r}\right)$ [6]

Using induction to prove:

Theorem 44. 

$\sum _{0\le n_1\le ...\le n_M\le N-1}(K+n_1{D}_1+...+n_M{D}_M)=(D_1+2D_2+...+M{D}_M)\left({}_{M+1}^{N+M-1}\right)+K\left({}_M^{N+M-1}\right)$

$eg:\sum _{0\le n_1\le ...\le n_p\le n}(n_1+...+n_p)={\displaystyle \sum _{n_p=0}^n}{\displaystyle \sum _{n_{p-1}=0}^{n_p}}...{\displaystyle \sum _{n_1=0}^{n_2}}(n_1+...+n_p)=\left({}_2^{p+1}\right)\left({}_{p+1}^{n+p}\right)=\frac{pn}{2}\left({}_p^{n+p}\right)$. [6]

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

Theorem 45. 

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

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

Proof. 

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

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

Using a similar method to obtain $H_2\left(g\right),H_3\left(g\right)$ □

Theorem 46. 

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

Proof. 

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

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

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

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

$If{H}_1(g\ge M-Y)\ne 0,NumberofX\in K=\left({}_{C-g}^{C-M+Y}\right)\to H_1\left(g\right)=\left({}_{C-g}^{C-M+Y}\right){[X+M-Y]}_{C-g}{[Y+1]}^g$

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. 

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

Proof. 

$PS=[A+2,A+4...A+2M],PT=[A+1,A+2...A+M],PT1=[1,3...2(M-g)-1]$

$H_2(g,\sum K)=SUM(g+1,PT1,PT1))=(2(M-g)-1)!!\left({}_g^{2M-g}\right),H_2(g,T)={[A+1]}^g$

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

Theorem 48. 

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

Proof. 

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

Theorem 49. 

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. 

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

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

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

49 and 45 $\to \frac{1}{M!}H(g,[A+1,A+2...A+M]:2,[1,3...2M-1])=H(g,[A+1,A+2...A+M],[M+1,M+2...2M])$

Theorem 50. 

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

Proof. 

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

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

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

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 $P=N+T_M-M,Q=N-1$, corresponding to the three forms.define $A_{1,2,3}(P,Q,M)=$

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

Theorem 51. 

$\Vert A_1(P,Q,M)\Vert =\Vert A_2(P,Q,M)\Vert =\Vert A_3(P,Q,M)\Vert ,\Vert A(P,Q,M)\Vert =\Vert A(P,P-Q,M)\Vert$ [2]

Theorem 52. 

$\Vert A(P,0,M)\Vert =1,\Vert A(P,1,M)\Vert =\left({}_{1+M}^{P+M}\right),\Vert A(P,Q>1,M)\Vert ={\displaystyle \prod _{g=0}^{Q-1}}\frac{\left({}_{1+M}^{P+M-g}\right)}{\left({}_{1+M}^{1+M-g}\right)}$ [2]

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

Theorem 53. 

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

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

Theorem 54. 

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

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

Theorem 55. 

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

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

(1)(2) are already known formula.

6. Eulerian polynomials and Beyond

In this section,$q\ne 0,q\ne 1$.Inductive proof:

Theorem 56. 

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

Definition 12. 

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

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

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

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

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

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

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

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

Theorem 57. 

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

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

Theorem 58. 

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

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

By definition of difference:

${\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 59. 

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

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

Theorem 60. 

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

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

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

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

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

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

Theorem 61. 

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

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

Definition 13. 

$A_q(PS,PT)=$

Theorem 62. 

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

Theorem 63. 

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

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

• $G_0^N=1,G_{M<0,M>N}^N=0,G_M^N=G_{N-M}^N$
• $G_M^N=q^M{G}_M^{N-1}+G_{M-1}^{N-1}=G_M^{N-1}+q^{N-M}{G}_{M-1}^{N-1}$
• ${\displaystyle \sum _{n=0}^{N-1}}{q}^n{G}_M^{n+K}=q^{M-K}{G}_{M+1}^{N+K}$
• $G_K^M=\sum _{w\in \Omega (0^{M-K},1^K)}{q}^{inv\left(w\right)}$[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 ${\nabla }_q^p,p\in \mathbb{N},{\nabla }_q^{0}f\left(n\right)=f\left(n\right),{\displaystyle \sum _{n=0}^{N-1}}{q}^n{\nabla }_q^{1}f(n+1)=f\left(N\right),{\displaystyle \sum _{n=0}^{N-1}}{q}^{n}f(n+1)={\nabla }_q^{-1}f\left(N\right)$

Definition 15. 

Recursive define $SU{M}_q\left(N\right)=SU{M}_q(N,PS,PT)$.

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

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

Theorem 64. 

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

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

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

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

Use this to prove:

Theorem 65. 

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

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

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

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

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

7.2. Property

Theorem 66. 

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

Theorem 67. 

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

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

Theorem 68. 

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

Definition 16. 

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

$For{m}_2\to$

Theorem 69. 

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

Theorem 70. 

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

$SU{M}_q(N,[1_{q-},2_{q-}...M_{q-}],[1,2...M])\to$${\displaystyle \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. 

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

7.3. Application

Theorem 72. 

$\sum _{0\le {\lambda }_1\le {\lambda }_2\le ...\le {\lambda }_M\le N}{q}^{{\lambda }_1+{\lambda }_2+...+{\lambda }_M}=G_M^{N+M}=G_N^{N+M}$ [6]

Proof. 

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

Theorem 73. 

$\sum _{1\le {\lambda }_1<{\lambda }_2<...<{\lambda }_M\le N+M}{q}^{{\lambda }_1+{\lambda }_2+...+{\lambda }_M}=q^{\left({}_2^{M+1}\right)}{G}_M^{N+M}$

Proof. 

${\displaystyle \sum _{{\lambda }_M=0}^N}{q}^{M+{\lambda }_M}...{\displaystyle \sum _{{\lambda }_2=0}^{{\lambda }_3}}{q}^{2+{\lambda }_2}{\displaystyle \sum _{{\lambda }_1=0}^{{\lambda }_2}}{q}^{1+{\lambda }_1}=q^{1+2+...+M}{\displaystyle \sum _{{\lambda }_M=0}^N}{q}^{{\lambda }_M}...{\displaystyle \sum _{{\lambda }_2=0}^{{\lambda }_3}}{q}^{{\lambda }_2}{\displaystyle \sum _{{\lambda }_1=0}^{{\lambda }_2}}{q}^{{\lambda }_1}$□

Theorem 74. 

$\sum _{A\le {\lambda }_1<{\lambda }_2<...<{\lambda }_M\le B}{q}^{{\lambda }_1+{\lambda }_2+...+{\lambda }_M}=q^{\left({}_2^{M+1}\right)+(A-1)M}{G}_M^{B-A+1},A,B\in \mathbb{Z}$

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

Theorem 75. 

${\displaystyle \prod _{i=1}^M}(1+q^{A+i}z)={\displaystyle \sum _{g=0}^M}{q}^{\left({}_2^{g+1}\right)+Ag}{G}_g^M{z}^g$

A=-1 or 1,it’s q-Binomial Theorem:${\displaystyle \prod _{i=1}^M}(1+q^{i-1}z)={\displaystyle \sum _{g=0}^M}{q}^{\left({}_2^g\right)}{G}_g^M{z}^g$ , ${\displaystyle \prod _{i=1}^M}(1+q^{i}z)={\displaystyle \sum _{g=0}^M}{q}^{\left({}_2^{g+1}\right)}{G}_g^M{z}^g$

If $T_i+1=T_{i+1},D_i=1,K_i=K_{i_{q-}}$:

$B_{i}of{H}_{1,2,3}^q\left(g\right)=\left\{{}_{X_{{}_i}=K_{{}_{i_{q-}}}}^{X_{{}_i}=T}\right.=\left\{{}_{q^{-K_i}{G}^{K_i+X_{T-1}}}^{q^{X_T}{G}^{T_i-X_{K-1}}}\right.,=\left\{{}_{q^{-K_i}{G}^{K_i-T_i+X_{K-1}}}^{q^{-(T_i-X_{K-1})}{G}^{T_i-X_{K-1}}}\right.,=\left\{{}_{q^{-K_i}{G}^{K_i+X_{T-1}}}^{q^{1+X_T}{G}^{T_i-K_i-X_{T-1}}}\right.$

It’s similar to 1.Replace each $B_i$ to $G_1^{B_i}$ in $H(g,[K_1,K_2...K_M],PT)$ and multiply by $q^{?}$ to obtain $H^q\left(g\right)$.

If there is another $K_i+1=K_{i+1}$,74 can be used to obtain general formulas.

Theorem 76. 

Expansion of ii:$G_{M+1}^{N+M}={\displaystyle \sum _{g=0}^M}{q}^{(g+1)g}{G}_g^M{G}_{1+g}^N$,$G_{M+1+P}^{N+M+P}={\displaystyle \sum _{g=0}^M}{q}^{(g+1+P)g}{G}_g^M{G}_{P+1+g}^{N+P}$

Proof. 

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

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

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

Theorem 77. 

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

Proof. 

$SU{M}_q(N+1,[q^1:(q-1)q^1,q^2:(q-1)q^2...q^M:(q-1)q^M],[1,3...2M-1])$

$={\displaystyle \sum _{{\lambda }_M=0}^N}{q}^{2{\lambda }_M+M}...{\displaystyle \sum _{{\lambda }_2=0}^{{\lambda }_3}}{q}^{2{\lambda }_2+2}{\displaystyle \sum _{{\lambda }_1=0}^{{\lambda }_2}}{q}^{2{\lambda }_1+1}=q^{\left({}_2^{M+1}\right)}\sum _{0\le {\lambda }_1\le ...\le {\lambda }_M\le N}{q}^{2({\lambda }_1+...+{\lambda }_M)}=q^{\left({}_2^{M+1}\right)}{\left[{}_M^{N+M}\right]}_{q^2}$

$B_{i}of{H}_3^q\left(g\right)=\left\{{}_{q^{X_{T-1}}(K_i+G_1^{X_{T-1}}{D}_i)=q^{X_{T-1}}(q^i+G_1^{X_{T-1}}{q}^i(q-1))=q^{2X_{T-1}+i}=q^{i+2X_T},X_i=K_i}^{q^{1+X_{T-1}}\{(q^{X_{T-1}}{G}_1^{T_i}-q^{T_i}{G}_1^{X_{T-1}})D_i-K_i{q}^{T_i}\}=-q^{i+1+2X_{T-1}}=-q^{i-1+2X_T},X_i=T_i}\right.$

Factor $q^i$ is present in all of them here, so $q^{(1+2+...+M)}$ can be extracted.

$H_3^q\left(g\right)={(-1)}^g{q}^{(1+2+...+M)-g+2(1+2+...+g)}\sum _{X_i\in K}\prod q^{2X_T}={(-1)}^g{q}^{\left({}_2^{M+1}\right)+g^2}{\left[{}_g^M\right]}_{q^2}$

$q^{\left({}_2^{M+1}\right)}{\left[{}_M^{N+M}\right]}_{q^2}={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)G_{2M}^{N+2M-g}=q^{\left({}_2^{M+1}\right)}{\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{g^2}{\left[{}_g^M\right]}_{q^2}{G}_{2M}^{N+2M-g}$

□

Definition 17. 

${}_q{E}_M^N=\sum _{1\le {\lambda }_1\le {\lambda }_2\le ..\le {\lambda }_M\le N}{G}_1^{{\lambda }_1}{G}_1^{{\lambda }_2}...G_1^{{\lambda }_M}=S_2^q(N+M,N)$

Theorem 78.${\left(G_1^N\right)}^M={(1+q+...+q^{N-1})}^M={\displaystyle \sum _{g=1}^M}{g}_{q+}!S_2^q(M,g)G_g^N{q}^{-g}$

Proof. 

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

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

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

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

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

Use $PS=[K_i+1:D_i(q-1)],PT=[1,2...M]\to$

Theorem 79. 

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

$K_i=1,D_i=q^{-i}z\to \nabla SUM(M+1)={\displaystyle \prod _{i=1}^M}(1+q^{M-i}z)={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)G_g^M{q}^{-g}(*)$

$B_i=\left\{{}_{1+q^{X_T}{q}^{-i}z=1+q^{-1-X_{K-1}}z=1+q^{-X_K}z,X_i=K_i}^{q^{X_T}(q^{X_T}-1)q^{-i}z,X_i=T_i}\right.$

$H_1^q\left(g\right)=q^{1+2+...+g}{z}^g(q^1-1)...(q^g-1)\left[\sum _{-M\le {\lambda }_1<...<{\lambda }_g\le -1}{q}^{{\lambda }_1+...+{\lambda }_g}\right](1+q^{-1}z)...(1+q^{-(M-g)}z)$

$(*)={\displaystyle \sum _{g=0}^M}{q}^{\left({}_2^{g+1}\right)}{z}^g(q^1-1)...(q^g-1)q^{\left({}_2^{g+1}\right)-g(M+1)}{G}_g^M(1+q^{-1}z)...(1+q^{-(M-g)}z)G_g^M{q}^{-g}$

$={\displaystyle \sum _{g=0}^M}{q}^{g^2-g(M+1)-\left({}_2^{M-g+1}\right)}{z}^g(q^M-1)...(q^{M-g+1}-1)(q^1+z)...(q^{(M-g)}+z)G_g^M$

Combining q-Binomial Theorem →

Theorem 80. 

${\displaystyle \sum _{g=0}^M}{q}^{\left({}_2^g\right)}{z}^g{G}_g^M(q^M-1)...(q^{M-g+1}-1)(q^{M-g}+z)...(q^1+z)=q^{\left({}_2^{M+1}\right)}{\displaystyle \sum _{g=0}^M}{q}^{\left({}_2^g\right)}{z}^g{G}_g^M$

Use 79 and induction →

Theorem 81. 

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

Theorem 82. 

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

Proof. 

$q^{Mn}={\nabla }_q^{1}SUM(N,[1,1...1]:q-1,[1,2...M])={\displaystyle \sum _{g=0}^M}{H}_1^q\left(g\right)G_g^n{q}^{-g}={\displaystyle \sum _{g=0}^M}{H}_2^q\left(g\right)G_g^{n+g}={\displaystyle \sum _{g=0}^M}{H}_3^q\left(g\right)G_M^{n+M-g}{q}^{-g}$

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

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

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

□

Theorem 83. 

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

Theorem 84. 

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

Theorem 85. 

$\sum _{0\le n_1\le ...\le n_M\le N-1}{q}^{n_1+n_2+...+n_M}(K+G_1^{n_1}{D}_1+...+G_1^{n_M}{D}_M)=\{q^1{D}_M+q^2(D_M+D_{M+1})+...+q^M(D_M+D_{M+1}+...+D_1)\}{G}_{M+1}^{N+M-1}+K{G}_M^{N+M-1}$

Theorem 86.${(K+qD)}^M=(q-1)D{\displaystyle \sum _{g=0}^{M-1}}{(K+D)}^g{(K+qD)}^{M-1-g}+{(K+D)}^M$

Proof. 

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

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

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

7.4. Relationships between $H^q\left(g\right)$

Theorem 87. 

$PT=[1,2...M],H_1^q\left(g\right)=q^{g(g+1)}{\displaystyle \sum _{k=g}^M}{H}_2^q\left(k\right)G_g^k$[3]

Theorem 88. 

$PT=[1,2...M],H_1^q\left(g\right)={\displaystyle \sum _{k=0}^g}{H}_3^q\left(k\right)G_{M-g}^{M-k}{q}^{(g+1)(g-k)}$

Proof. 

Direct verification when M=1, assuming M holds.

$(*)H_1^q(PS1,PT1,g)=q^g{G}_1^g{H}_1^q(g-1)+(K_{M+1}+G_1^g{D}_{M+1})H_1^q\left(g\right)$

$=q^g{G}_1^g{H}_1^q(g-1){\displaystyle \sum _{x=0}^M}{H}_3^q\left(x\right)G_{M-g+1}^{M-x}{q}^{g(g-1-x)}+(K_{M+1}+G_1^g{D}_{M+1}){\displaystyle \sum _{x=0}^M}{H}_3^q\left(x\right)G_{M-g}^{M-x}{q}^{(g+1)(g-x)}$

$(**)H_1^q(PS1,PT1,g)={\displaystyle \sum _{x=0}^{M+1}}{H}_3^q(PS1,PT1,x)G_{M+1-g}^{M+1-x}{q}^{(g+1)(g-x)}$

$={\displaystyle \sum _{x=0}^M}{H}_3^q\left(x\right)\{\frac{q^{M+2}-q^{X+1}}{q-1}{D}_{M+1}-K_{M+1}{q}^{M+2}\}{G}_{M+1-g}^{M-x}{q}^{(g+1)(g-x-1)}+{\displaystyle \sum _{x=0}^M}{H}_3^q\left(x\right)(K_{M+1}+G_1^X{D}_{M+1})G_{M+1-g}^{M+1-x}{q}^{(g+1)(g-x)}$

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 $K_{M+1}$ :

$In(*)=q^g{G}_1^g{D}_{M+1}{G}_{M-g+1}^{M-x}{q}^{g(g-1-x)}+G_1^g{D}_{M+1}{G}_{M-g}^{M-x}{q}^{(g+1)(g-x)}$

$Divideby{D}_{M+1}{q}^g{q}^{g(g-1-x)}{(q-1)}^{-1}=(q^g-1)G_{M+1-g}^{M-x}+(q^g-1)G_{M-g}^{M-x}{q}^{g-x}=(q^g-1)G_{M+1-g}^{M+1-x}$

$In(**)=\frac{q^{M+2}-q^{x+1}}{q-1}{D}_{M+1}{G}_{M+1-g}^{M-x}{q}^{(g+1)(g-x-1)}+G_1^x{D}_{M+1}{G}_{M+1-g}^{M+1-x}{q}^{(g+1)(g-x)}$

$Divideby{D}_{M+1}{q}^g{q}^{g(g-1-x)}{(q-1)}^{-1}=(q^{M+1-x}-1)G_{M+1-g}^{M-x}+(q^g-q^{g-x})G_{M+1-g}^{M+1-x}=(q^g-1)G_{M+1-g}^{M+1-x}$□

Theorem 89.$(q^M-1)...(q^{M-K+1}-1)={\displaystyle \sum _{g=0}^K}{(-1)}^g{q}^{\left({}_2^{K-g+1}\right)}{G}_g^M{G}_{M-K}^{M-g}={\displaystyle \sum _{g=0}^K}{(-1)}^g{q}^{\left({}_2^{K-g+1}\right)+(K-g)(M-K)}{G}_g^K$

Proof. 

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

Similarly, using induction to prove:

Theorem 90. 

$PT=[1,2...M],H_2^q\left(g\right)={\displaystyle \sum _{k=g}^M}{(-1)}^{k+g}{G}_g^k{q}^{-k(k+1)+\left({}_2^{k-g}\right)}{H}_1^q\left(k\right)$

Theorem 91. 

$PT=[1,2...M],H_3^q\left(g\right)={\displaystyle \sum _{k=0}^g}{(-1)}^{k+g}{G}_{M-g}^{M-k}{q}^{\left(g+1\right)(g-k)-\left({}_2^{g-k}\right)}{H}_1^q\left(k\right)$

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 ${\displaystyle \sum _{g=0}^M}{a}_g{G}_{Y+g}^X$ can be converted to $\frac{a_M}{M_{q+}!}{\nabla }^{A}SU{M}_q(N+B,[K_1,K_2...K_M],[1,2...M])$

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

Theorem 93. 

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

• $a_g{q}^{(1-Y)g}=q^{g(g+1)}{\displaystyle \sum _{k=g}^M}{b}_k{G}_g^k={\displaystyle \sum _{k=0}^g}{c}_k{G}_{M-g}^{M-k}{q}^{(g+1)(g-k)}{q}^{(1-Y)k}$
• $b_g={\displaystyle \sum _{k=g}^M}{(-1)}^{k+g}{G}_g^k{q}^{-k(k+1)+\left({}_2^{k-g}\right)}{a}_k{q}^{(1-Y)k}$
• $c_g{q}^{(1-Y)g}={\displaystyle \sum _{k=0}^g}{(-1)}^{k+g}{G}_{M-g}^{M-k}{q}^{(g+1)(g-k)-\left({}_2^{g-k}\right)}{a}_k{q}^{(1-Y)k}$

7.5. Merge and Expand

Theorem 94. 

Necessary and sufficient conditions for merging,$0<K\le M$:

• ${\displaystyle \sum _{n=0}^M}H\left(n\right)G_{Y+n}^X\to {\displaystyle \sum _{n=0}^{M-K}}(...)G_{Y+K+n}^{X+K}:{\displaystyle \sum _{x=g}^M}{(-1)}^x{G}_g^x{q}^{-x(x+1)+\left({}_2^{x-g}\right)}H\left(x\right)q^{(1-Y)x}=0,0\le g<K$
• ${\displaystyle \sum _{n=0}^M}H\left(n\right)G_{Y+n}^X\to {\displaystyle \sum _{n=0}^{M-K}}(...)G_{Y+K+n}^{X+K}:{\displaystyle \sum _{x=0}^g}{(-1)}^x{G}_{M-g}^{M-x}{q}^{\left(g+1\right)(g-x)-\left({}_2^{g-x}\right)}H\left(x\right)q^{(1-Y)x}=0,0\le M-g<K$
• ${\displaystyle \sum _{n=0}^M}H\left(n\right)G_{Y+n}^{X+n}\to {\displaystyle \sum _{n=0}^{M-K}}(...)G_{Y+K+n}^{X+K+n}:{\displaystyle \sum _{x=g}^M}H\left(x\right)G_g^x=0,0\le g<K$
• ${\displaystyle \sum _{n=0}^M}H\left(n\right)G_{M+Y}^{X-n}\to {\displaystyle \sum _{n=0}^{M-K}}(...)G_{M+Y-K}^{X-K-n}:{\displaystyle \sum _{x=0}^g}H\left(x\right)G_{M-g}^{M-x}{q}^{(g+1)(g-x)}{q}^{(1-Y)x}=0,0\le M-g<K$

Y=1,76 →:

Theorem 95. 

${\displaystyle \sum _{x=0}^M}{(-1)}^x{G}_g^x{G}_x^M{q}^{\left({}_2^{x-g}\right)}=0,0\le g<M$,${\displaystyle \sum _{x=0}^M}{(-1)}^x{G}_x^M{q}^{\left({}_2^x\right)}=0$

Theorem 96. 

P=A+T+1-Y,M > A ≥ 0,

${\displaystyle \sum _{g=0}^M}{G}_g^M{G}_A^{A+T+g}{G}_{Y+g}^X{q}^{g(g+1+T-P)}={\displaystyle \sum _{x=0}^A}{G}_{x+T}^{A+T}{G}_{M+T}^{M+T+x}{G}_{Y+M-A+x}^{X+M-A}{q}^{x(x+1+T-P)}$

Proof. 

$SU{M}_q(N,[{(T+1)}_{q-},{(T+2)}_{q-}...{(T+M)}_{q-}],[T+A+1...T+A+M]),B_{i}of{H}_1^q\left(g\right)=\left\{{}_{q^{-(T+i)}{G}_1^{T+i+X_T},X_i=K_i}^{q^{X_T}{G}_1^{T+A+i-X_T},X_i=T_i}\right.$

$={\displaystyle \sum _{g=0}^M}{G}_{A+T+1+g}^X{q}^{\frac{g(1+g)}{2}}{G}_1^{T+A+1}...G_1^{T+A+g}\times G_1^{T+M}...G_1^{T+g+1}{q}^{-(M-g)(T+M+1)}\sum _{1\le {\lambda }_1<...<{\lambda }_{M-g}\le M}{q}^{\sum {\lambda }_i}$

$=\frac{(q^A-1)...(q-1)}{{(q-1)}^M}{\displaystyle \sum _{g=0}^M}{G}_{A+T+1+g}^X{q}^{\frac{g(1+g)}{2}-(M-g)(T+M+1)+\frac{(M-g+1)(M-g)}{2}}\times (q^{T+M}-1)...(q^{T+A+1}-1)G_A^{A+T+g}{G}_g^M$

$=q^{-(T+A+1)}{G}_1^{T+A+1}...q^{-(T+M)}{G}_1^{T+M}SU{M}_q(N,[{(T+1)}_{q-}....{(T+A)}_{q-}],[T+M+1...T+M+A])$

$=q^{\frac{-(M-A)(T+A+1+T+M)}{2}}{G}_1^{T+A+1}...G_1^{T+M}{\displaystyle \sum _{x=0}^A}{G}_{M+T+1+x}^{X+M-A}{q}^{\frac{x(1+x)}{2}-(A-x)(T+A+1)+\frac{(A-x+1)(A-x)}{2}}\times G_1^{T+M+1}..G_1^{T+M+x}\times G_1^{T+A}...G_1^{T+x+1}{G}_x^A$

$\to {\displaystyle \sum _{g=0}^M}{G}_{A+T+1+g}^X{q}^{\frac{g(1+g)-(M-g)(g+M+1+2T)}{2}}{G}_A^{A+T+g}{G}_g^M=q^{\frac{-(M-A)(T+A+1+T+M)}{2}}{\displaystyle \sum _{x=0}^A}{G}_{M+T+1+x}^{X+M-A}{q}^{\frac{x(1+x)-(A-x)(x+A+1+2T)}{2}}{G}_{M+T}^{M+T+x}{G}_{x+T}^{A+T}$

$\to {\displaystyle \sum _{g=0}^M}{G}_g^M{G}_A^{A+T+g}{G}_{A+T+1+g}^X{q}^{g(g+1+T)}={\displaystyle \sum _{x=0}^A}{G}_{x+T}^{A+T}{G}_{M+T}^{M+T+x}{G}_{M+T+1+x}^{X+M-A}{q}^{x(x+1+T)}$□

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

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

$K=T+1+M\to {\displaystyle \sum _{g=0}^K}{G}_g^M{G}_{K-g}^N{q}^{(M-g)(K-g)}=G_K^{N+M}$. It’s the q-Vandermorde Theorem.

$0\le B+A<M,\to {\displaystyle \sum _{g=0}^M}{G}_g^M{G}_A^{A+T+g}{G}_B^g{q}^{-gA+\left({}_2^{g-B}\right)}{(-1)}^g=0\to {\displaystyle \sum _{g=0}^M}{G}_g^M{G}_A^{X_1+g}{G}_B^g{q}^{\left({}_2^g\right)-g(A+B)}{(-1)}^g=0$

A and B have symmetry,ii→.

Theorem 97. 

${\displaystyle \sum _{g=0}^M}{G}_g^M{G}_A^{X_1+g}{G}_B^{X_2+g}{q}^{\left({}_2^g\right)-g(A+B)}{(-1)}^g=0,0\le B+A<M$

Theorem 98. 

${\displaystyle \sum _{g=0}^M}{(-1)}^g{G}_g^M{G}_{M+K}^{X+g}{q}^{\left({}_2^{M+1-g}\right)+(M-g)K}={(-1)}^M{G}_K^X,X+g\ge 0,K\ge 0$

Proof. 

$SU{M}_q(N,[{(T+1)}_{q-},{(T+2)}_{q-}...{(T+M)}_{q-}],[T+K+M+1,T+K+M+2...T+K+2M])$

$H_1^q\left(g\right)=q^{\frac{g(1+g)}{2}-(M-g)(T+M+1)+\left({}_2^{M-g+1}\right)}\times G_1^{T+K+M+1}...G_1^{T+K+M+g}\times G_1^{T+M}...G_1^{T+1+g}\times G_1^{T+M+1}{G}_g^M$

$H_2^q\left(0\right)={\displaystyle \sum _{g=0}^M}{(-1)}^g{H}_1^q\left(g\right)q^{-g(g+1)+\frac{g(g-1)}{2}-(T+K+M)g}={(-1)}^M{q}^{-M(T+1+T+M)}{G}_1^{K+M}{G}_1^{K+M-1}...G_1^{K+1}$

$\to {(-1)}^M{G}_1^{K+M}...G_1^{K+1}={\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{\left({}_2^{M-g+1}\right)+(M-g)K}\times G_1^{T+K+M+g}...G_1^{T+K+M+1}\times G_1^{T+M}...G_1^{T+1+g}\times G_g^M$

$\to {(-1)}^M{G}_K^{T+K+M}={\displaystyle \sum _{g=0}^M}{(-1)}^g{q}^{\left({}_2^{M-g+1}\right)+(M-g)K}\times G_{M+K}^{T+K+M+g}{G}_g^M$□

7.6. Matrix of $SU{M}_q\left(N\right)$

Let $P=N+T_M-M,Q=N-1$, corresponding to the three forms,define $A_{1,2,3}^q(P,Q,M)=$

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

Theorem 99. 

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

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

Theorem 100. 

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

8. Multi-parameter Formal Calculation

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

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

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

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

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 18. 

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 19. 

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

Theorem 101. 

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

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

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

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

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

9. A theorem of symmetry

In this section,$T_i\ge i$.

17$\to H_1(g,PT,PT)={\displaystyle \prod _{i=1}^M}{T}_i\left({}_g^M\right)={\displaystyle \prod _{i=1}^M}{T}_i\left({}_{g,M-g}^M\right)$.Promoted it:Set T come from p Source:$S_1,S_2...S_p$.

Definition 20. 

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

Definition 21. 

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

Definition 22. 

$W(g_1,g_2...g_p,[T_1,T_2...T_M])=\sum _{g_1+g_2+...+g_p=M,g_i=\left|S_i\right|}{\displaystyle \prod _{i=1}^M}(T_i+\sum _{j<i}Diff(T_j,T_i\left)\right)$

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

Theorem 102. 

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

Definition 23. 

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

Definition 24. 

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

Theorem 103. 

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

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 $W_q(g_1,g_2,PT)$ holds,$W_q(g_1,g_2+1,[PT,T_{M+1}])=T_{M+1}\in Sourc{e}_1+T_{M+1}\in Sourc{e}_2$

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

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

Just need to prove:$G_{g_1}^M{G}_1^{T_{M+1}+g_1}+G_{g_1-1}^M{G}_1^{T_{M+1}-(M-g_1+1)}{q}^{M-g_1+1}=G_1^{T_{M+1}}{G}_{g_1}^{M+1}$

$\left(Rightside\right)\times \frac{(q^{M-g_1+1}-1)}{G_{g_1}^M}=(q^{T_{M+1}-1}+...+q+1)(q^{M+1}-1)=\left(1\right)$

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

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

(1)-(2)=0→It’s holds when p=2.

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

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.

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

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

$eg:W_q(1,2,[1,2,3])=S_1{S}_2{S}_2+S_2{S}_1{S}_2+S_2{S}_2{S}_1=G_1^1{G}_1^3{G}_1^4+G_1^1{G}_1^{1}q{G}_1^4+G_1^1{G}_1^2{G}_1^1{q}^2=G_1^1{G}_1^2{G}_1^3{G}_{1,2}^3$

$eg:When{W}_q(1,2,[1,2,3])changesto{W}_q(1,1,1,[1,2,3])$

$=S_1{S}_2{S}_3+S_1{S}_3{S}_2+S_2{S}_1{S}_3+S_3{S}_1{S}_2+S_2{S}_3{S}_1+S_3{S}_2{S}_1$

$=G_1^1\{G_1^3{G}_1^5+G_1^3{G}_1^{3}q\}+G_1^{1}q\{G_1^1{G}_1^5+G_1^1{G}_1^{3}q\}+G_1^1{q}^2\{G_1^1{G}_1^3+G_1^1{G}_1^{1}q\}$

$=G_1^1\left\{G_1^3{G}_1^4{G}_{1,1}^2\right\}+G_1^{1}q\left\{G_1^1{G}_1^4{G}_{1,1}^2\right\}+G_1^1{q}^2\left\{G_1^1{G}_1^2{G}_{1,1}^2\right\}=G_1^1{G}_1^2{G}_1^3{G}_{1,2}^3\times G_{1,1}^2=G_1^1{G}_1^2{G}_1^3{G}_{1,1,1}^3$