Formal Calculation

1. Introduction

To calculate all products of M distinct integers in [1,N-1], items needs to be divided into 2^M-1 categories.

For example, calculate ∑3-products in [1,6]

$②1\times 2\times 4+\left(1\times 2+2\times 3\right)\times 5+\left(1\times 2+2\times 3+3\times 4\right)\times 6=1\times 2\times 4\left(\begin{array}{c}7\\ 5\end{array}\right)$

$③1\times 3\times 4+\left(1\times 4+2\times 4\right)\times 5+\left(1\times 5+2\times 5+3\times 5\right)\times 6=1\times 3\times 4\left(\begin{array}{c}7\\ 5\end{array}\right)$

$④1\times 3\times 5+\left(1\times 3+1\times 4+2\times 4\right)\times 6=1\times 3\times 5\left(\begin{array}{c}7\\ 6\end{array}\right)$

$\mathrm{An}\mathrm{item}\left({\mathrm{I}}_1,{\mathrm{I}}_2,{\mathrm{I}}_3\right),{\mathrm{I}}_{\mathrm{i}}+1={\mathrm{I}}_{\mathrm{i}+1}\mathrm{means}\mathrm{continuity}\left(\mathrm{A}\right),{\mathrm{I}}_{\mathrm{i}}+1<{\mathrm{I}}_{\mathrm{i}+1}\mathrm{means}\mathrm{discontinuity}\left(\mathrm{B}\right)$

So ①=AA,②=AB,③=BA,④=BB.Items of a category have the same discontinuities at the same positions.

Continuity means the distance between factors remains constant. Discontinuity assumes the distance between factors can be  arbitrarily expanded. As N increases, the number of items will also increase.In ②=AB:

N-1=6,products={1×2×6,2×3×6,3×4×6},1×2,2×3,3×4 keep  continuity. 2×6,3×6,4×6 keep discontinuity.

N-1=7,products={1×2×7,2×3×7,3×4×7,4×5×7}

  

Start here and going through several promotions, [1][2][3] introduced Formal Calculation:

M-series:Serie_i={K_i,K_i+D_i,K_i+  2D_i…K_i+(N-1)D_i}, i∈ [1,M], K_i,D_i∈ ring with identity elements.

Use PS=[K₁:D₁, K₂:D₂...K_M:D_M]  to represent the M-series.

[K₁:D, K₂:D...K_M:D]  is abbreviated as [K₁, K₂…K_M]:D. [K₁,K₂…K_M]:1  is abbreviated as [K₁,K₂…K_M]

  

$Recursivedefinieoperator{▽}^P,p\in \mathbb{Z}:$

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

${▽}^{1}f\left(n\right)=▽f\left(n\right)=f\left(n\right)-f\left(n-1\right),i{t}^{\prime }salittledifferentfrom\mathrm{Differential}\mathrm{calculation}.$

$Iff\left(n\right)=\sum A_i\left(\begin{array}{c}{N}_i\\ M_i\end{array}\right)and{M}_{i}isnotchangedwithN,then{▽}^{p}f\left(N\right)=\sum A_i\left(\begin{array}{c}{N}_i-p\\ M_i-p\end{array}\right)$

  

Use PT=[T₁,T₂...T_M]  to indicate some products in M-series.By default,the following uses:

PS=[K₁:D₁,K₂:D₂…K_M:D_M],PT=[T₁,T₂...T_M]

PS1=[K₁:D₁,K₂:D₂…K_M:D_M,K_M+1:D_M+1]=[PS,  K_M+1:D_M+1],PT1=[PT, T_M+1= T_M+2-p]

  

Recursive define SUN(N,PS,PT), abbreviated as  SUM(N):

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

$SUM\left(N,PS1,PT1\right)={{\displaystyle \sum }}_{n=0}^{N-1}\left(K_{M+1}+n\times D_{M+1}\right)\times {▽}^{p}SUM\left(n+1,PS,PT\right)$

This is actually nested summation. For example:

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

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

$SUM\left(N,PS,\left[1,4\right]\right)={{\displaystyle \sum }}_{n=0}^{N-1}\left(K_2+n{D}_2\right){{\displaystyle \sum }}_{n_2=0}^n{{\displaystyle \sum }}_{n_1=0}^{n_2}\left(K_1+n_1{D}_1\right)$

Formal  Computation was once named the shape of numbers because PS and PT specify the  shape of the product factors and extensively use the concepts of continuity,  discontinuity, and position. It has little to do with trigonometric numbers,  squares numbers, etc. To avoid confusion, it was renamed.

  

If T_i< T_i+1,T_i ∈ N,define PB(PT)=count of discontinuities of PT.  If compared to [1,2...M],PB(PT)=T_M-M.

Also define PB(PS)=count of discontinuities of  PS,it’s  compared to a certain benchmark.

If PB (PT)=PB (PS) and their discontinuities are at  the same positions, then say them have the same shape.

  

The following use K to represent the  set{K₁,K₂...K_M},  T to represent the set{T₁,T₂...T_M}

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

X(T)=Count of {X₁, X₂...X_M} ∈T , X_T-1=Count  of {X₁, X₂...X_i-1} ∈ T,X_K-1=Count of {X₁, X₂...X_i-1} ∈ K,X_T-1+X_K-1=i-1

Don't swap X_i,then each ∏ X_i corresponds to one expression in  the SUM(N).

$1.1)g=X\left(T\right),SUM\left(N,PS,PT\right)=$

$UseH\left(g\right)torepresentthethreeparameters{H}_{1,2,3}\left(g\right).$

  

$Define{F}_M^{K=\left\{K_1,K_2\dots K_M\right\}}=\sum \left(Mdistinctproducts,factors\in K\right),F_M^{N}isshortfor{F}_M^{\left\{1,2\dots N\right\}},F_0^K=0$

$Define{E}_M^K=\sum \left(Mproducts,factors\in K\right),E_M^{N}isshortfor{E}_M^{\left\{1,2\dots N\right\}},E_0^K=0$

$Easytoobtain:F_M^{\left\{1,1\dots 1\right\},countof1=N}=\left(\begin{array}{c}N\\ M\end{array}\right),E_M^N={\displaystyle \sum }_{{\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_N=M,{\lambda }_i\ge 0}{1}^{{\lambda }_1}{2}^{{\lambda }_2}\mathrm{...}{N}^{{\lambda }_N}$

$1.2)▽SUM\left(N,PS,\left[1,2\dots M\right]\right)={{\displaystyle \prod }}_{i=1}^M\left(K_i+\left(N-1\right)D_i\right)={{\displaystyle \prod }}_{i=1}^M\left(K_i+n{D}_i\right)$

$1.3)SUM\left(N,\left[\dots PS\dots \right],\left[\dots T+1,T+2\dots T+M\dots \right]\right),K_{i}canexchangeorder.$

$1.4)SUM\left(N,\left[L_1,L_2\dots L_P,PS\right],\left[L_1,L_2\dots L_P,PT\right]\right)={{\displaystyle \prod }}_{i=1}^P{L}_{i}SUM\left(N,PS,PT\right)\to {\mathrm{T}}_1\mathrm{can}\mathrm{be}\mathrm{greater}\mathrm{than}1.\mathrm{T}\in \mathbb{N}$

  

$1.5)SUM\left(N,\left[1,1\dots 1\right],\left[1,2\dots M\right]\right)=SUM\left(N,\left[1,1\dots 1\right],\left[2,3\dots M\right]\right)=1^M+2^M+\dots +N^M$

$1.6)SUM\left(N,\left[1,1\dots 1\right],\left[1,3\dots 2M-1\right]\right)=SUM\left(N,\left[1,1\dots 1\right],\left[3,5\dots 2M-1\right]\right)=S_2\left(N+M,N\right)=E_M^N$

$1.7)SUM\left(N,\left[1,2\dots M\right],\left[1,3\dots 2M-1\right]\right)=SUM\left(N,\left[2,3\dots M\right],\left[3,5\dots 2M-1\right]\right)=S_1\left(N+M,N\right)=F_M^{N+M-1}$ S₁(N,M) is unsigned stirling number of  the first kind. S₂(N,M) is stirling number of the second kind.

  

$1.8)SUM\left(N,PT,PT\right)={{\displaystyle \prod }}_{i=1}^M{T}_i\left(\begin{array}{c}N+T_M\\ T_M+1\end{array}\right).$

This is an important conclusion. It is the origin  of  Formal Computation and why ①②③④   came into being.

$Itgeneralizestheformula{{\displaystyle \sum }}_{n=0}^{N-1}\left(\begin{array}{c}n\\ M\end{array}\right)=\left(\begin{array}{c}N\\ M+1\end{array}\right)=\frac{1}{M!}SUM\left(N-M,\left[1,2\dots M\right],\left[1,2\dots M\right]\right)innestedsummation.$

$\mathrm{SUM}\left(4,\left[1,2,3\right],\left[1,2,3\right]\right)=①$

$\mathrm{SUM}\left(4-1,\left[1,2,4\right],\left[1,2,4\right]\right)=②,\mathrm{SUM}\left(4-1,\left[1,3,4\right],\left[1,3,4\right]\right)=③$

$\mathrm{SUM}\left(4-2,\left[1,3,5\right],\left[1,3,5\right]\right)=④$

$\mathrm{SUM}\left(\mathrm{N}-PB\left(PT\right),\mathrm{PT},\mathrm{PT}\right)=①②③④=1.8),\mathrm{PB}\left(\mathrm{PS}\right)compareswith\left[1,2\dots M\right]$

$\mathrm{SUM}\left(\mathrm{N},\left[1,2,3\right],\left[1,3,5\right]\right)=\mathrm{SUM}\left(\mathrm{N},\left[2,3\right],\left[3,5\right]\right)=①+②+③+④,{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}\mathrm{M}-\mathrm{Products}=1.7)$

  

$1.9)IfPT=\left[T_1=1,T_2\dots T_M\right],T_1<T_2<\mathrm{...}<T_M,then\left(CountofproductsinSUM\left(N\right)\right)=\left(\begin{array}{c}\mathrm{N}+PB\left(PT\right)\\ PB\left(PT\right)+1\end{array}\right)$

  

$\mathrm{SUM}\left(\mathrm{N},\left[2,3\right],\left[3,5\right]\right)\stackrel{Form}{\to }\left(2+T_1\right)\left(3+T_2\right),T_M-M=5-2=3\stackrel{For{m}_1}{\to }$

$=T_1\times T_2\left(\begin{array}{c}\mathrm{N}+3\\ 6\end{array}\right)+\left\{K_1\times \left(T_2-X_{K-1}\right)+T_1\times \left(K_2+X_{T-1}\right)\right\}\left(\begin{array}{c}\mathrm{N}+3\\ 5\end{array}\right)+K_1\times K_2\left(\begin{array}{c}\mathrm{N}+3\\ 4\end{array}\right)\stackrel{X_{K-1}=X_{T-1}=1}{\to }$

$=3\times 5\left(\begin{array}{c}\mathrm{N}+3\\ 6\end{array}\right)+\left\{2\times 4+3\times 4\right\}\left(\begin{array}{c}\mathrm{N}+3\\ 5\end{array}\right)+2\times 3\left(\begin{array}{c}\mathrm{N}+3\\ 4\end{array}\right)$

$\mathrm{SUM}\left(\mathrm{N},\left[1,2,3\right],\left[1,3,5\right]\right)\stackrel{Form}{\to }\left(1+T_1\right)\left(2+T_2\right)\left(3+T_3\right),T_M-M=5-3=2\stackrel{For{m}_1}{\to }$

$=1\times 3\times 5\left(\begin{array}{c}\mathrm{N}+2\\ 6\end{array}\right)+35\left(\begin{array}{c}\mathrm{N}+2\\ 5\end{array}\right)+26\left(\begin{array}{c}\mathrm{N}+2\\ 4\end{array}\right)+1\times 2\times 3\left(\begin{array}{c}\mathrm{N}+2\\ 3\end{array}\right)$

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

$=1\times 2\times \left(5-2\right)+1\times \left(3-1\right)\times \left(3+1\right)+1\times \left(2+1\right)\times \left(3+1\right)=26$

  

Formal Calculation can be calculated in the ring with identity elements.

$H_1\left(0\right)=\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}17& 13\\ 4& 5\end{array}\right)$

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

$H_1\left(2\right)=1\times \left(\begin{array}{cc}1& 3\\ 4& 2\end{array}\right)\times 3\times \left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}15& 27\\ 30& 48\end{array}\right)$

$\mathrm{SUM}\left(\mathrm{N},\mathrm{PS},\mathrm{PT}\right)=\left(\begin{array}{cc}15& 27\\ 30& 48\end{array}\right)\left(\begin{array}{c}\mathrm{N}+1\\ 4\end{array}\right)+\left(\begin{array}{cc}44& 70\\ 28& 38\end{array}\right)\left(\begin{array}{c}\mathrm{N}+1\\ 3\end{array}\right)+\left(\begin{array}{cc}17& 13\\ 4& 5\end{array}\right)\left(\begin{array}{c}\mathrm{N}+1\\ 2\end{array}\right)$

$For{m}_{1}isobtainedbyguessandprovedbyinduction.Therehavethreeformsbecause:{{\displaystyle \sum }}_{n=0}^{N-1}n\left(\begin{array}{c}n+K\\ M\end{array}\right)$

$=\left(M+1\right)\left(\begin{array}{c}N+K\\ M+2\end{array}\right)+\left(M-K\right)\left(\begin{array}{c}N+K\\ M+1\end{array}\right)=\left(M+1\right)\left(\begin{array}{c}N+K+1\\ M+2\end{array}\right)-\left(1+K\right)\left(\begin{array}{c}N+K\\ M+1\end{array}\right)=\left(M-K\right)\left(\begin{array}{c}N+K+1\\ M+2\end{array}\right)+\left(1+K\right)\left(\begin{array}{c}N+K\\ M+2\end{array}\right)$

Property of H(g)

$2.1)H_1\left(g,PT,PT\right)={{\displaystyle \prod }}_{i=1}^M{T}_i\left(\begin{array}{c}M\\ g\end{array}\right),H_2\left(g<M,PT,PT\right)=H_3\left(g>0,PT,PT\right)=0\to 1.8)$

  

The recursive relationship of H(g) can be obtained  by definition.

$H_1\left(g,PS1,PT1\right)=H_1\left(g-1\right)\left({\mathrm{T}}_{M+1}-\left[\mathrm{M}-\left(\mathrm{g}-1\right)\right]\right){\mathrm{D}}_{M+1}+H_1\left(g\right)\left({\mathrm{K}}_{M+1}+{\mathrm{gD}}_{M+1}\right)$

$H_2\left(g,PS1,PT1\right)=H_2\left(g-1\right)\left({\mathrm{T}}_{M+1}-\left[\mathrm{M}-\left(\mathrm{g}-1\right)\right]\right){\mathrm{D}}_{M+1}+H_2\left(g\right)\left({\mathrm{K}}_{M+1}+\left[-{\mathrm{T}}_{M+1}+\mathrm{M}-\mathrm{g}\right]{\mathrm{D}}_{M+1}\right)$

$H_3\left(g,PS1,PT1\right)=H_3\left(g-1\right)\left(-{\mathrm{K}}_{M+1}+\left({\mathrm{T}}_{M+1}-\left[\mathrm{g}-1\right]\right){\mathrm{D}}_{M+1}\right)+H_3\left(g\right)\left({\mathrm{K}}_{M+1}+{\mathrm{gD}}_{M+1}\right)$

Using the recursive relationships and induction to  prove →

$2.2)H_1\left(g\right)={{\displaystyle \sum }}_{k=g}^M{H}_2\left(k\right)\left(\begin{array}{c}k\\ g\end{array}\right)={{\displaystyle \sum }}_{k=0}^g{H}_3\left(k\right)\left(\begin{array}{c}M-k\\ M-g\end{array}\right),Inversion\to$

$2.3)H_2\left(g\right)={{\displaystyle \sum }}_{k=g}^M{\left(-1\right)}^{k+g}{H}_1\left(k\right)\left(\begin{array}{c}k\\ g\end{array}\right),H_3\left(g\right)={{\displaystyle \sum }}_{k=0}^g{\left(-1\right)}^{k+g}{H}_1\left(k\right)\left(\begin{array}{c}M-k\\ M-g\end{array}\right)$

  

Calculation with 2.2) →

$2.4){{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right)={{\displaystyle \sum }}_{g=0}^M{H}_2\left(g\right)2^g={{\displaystyle \sum }}_{g=0}^M{H}_3\left(g\right)2^{M-g}$

$2.5){{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right)\left(\begin{array}{c}A\\ B-g\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{H}_2\left(g\right)\left(\begin{array}{c}A+g\\ B\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{H}_3\left(g\right)\left(\begin{array}{c}A+M-g\\ B-g\end{array}\right)$

It  indicates Form₁=Form₂=Form₃. If  regardless of the actual meaning, PT’s domain can be extended to ℂ. A=B →

$2.6){{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right)\left(\begin{array}{c}A\\ g\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{H}_2\left(g\right)\left(\begin{array}{c}A+g\\ g\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{H}_3\left(g\right)\left(\begin{array}{c}A+M-g\\ M\end{array}\right)$

  

$2.7)Induction\to {{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right)g\left(\begin{array}{c}A+1\\ B-g\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{H}_2\left(g\right)g\left(\begin{array}{c}A+g\\ B-1\end{array}\right)={{\displaystyle \sum }}_{g=0}^M\left\{H_3\left(g\right)g\left(\begin{array}{c}A+M-g\\ B-g\end{array}\right)+M{H}_3\left(g\right)\left(\begin{array}{c}A+M-g\\ B-1-g\end{array}\right)\right\}$

$2.8)H_1\left(g,\left[AD:D,PS\right],\left[A,PT\right]\right)=AD\left\{H_1\left(g\right)+H_1\left(g-1\right)\right\}\to H_1\left(g,\left[1,PS\right],\left[1,PT\right]\right)=H_1\left(g\right)+H_1\left(g-1\right)$

  

$In1.1),H\left(g\right)=\prod B_i={\displaystyle \prod }_{X_i\in T}{B}_i{\displaystyle \prod }_{X_i\in K}{B}_i$

$DefineH\left(g,T\right)=H\left(g,T,PS,PT\right)={\displaystyle \prod }_{X_i\in T}{B}_i;H\left(g,\sum T\right)=\sum H\left(g,T\right),alsodefineH\left(g,K\right),H\left(g,\sum K\right)$

$2.9)If{D}_i=1and{T}_{i+1}-T_i=1then{H}_1\left(g,T\right)={{\displaystyle \prod }}_{i=1}^g{T}_i;H_1\left(g,\sum K,\left[1,1\dots 1\right],PT\right)=E_{M-g}^{g+1}$

  

Using induction to prove →

$2.10)If{D}_i=1,H_1\left(g,\sum K\right)=F_{M-g}^K{E}_0^g+F_{M-g-1}^K{E}_1^g+\mathrm{...}+F_0^K{E}_{M-g}^g$

$2.11)If{D}_i=1,H_1\left(g,\sum T\right)=F_g^T{E}_0^{M-g}-F_{g-1}^T{E}_1^{M-g}+\mathrm{...}+{\left(-1\right)}^g{F}_0^T{E}_g^{M-g}$

$2.12)If{D}_i=1and{T}_{i+1}-T_i=K_{i+1}-K_i=1then{H}_1\left(g\right)=\left(\begin{array}{c}\mathrm{M}\\ N\end{array}\right)T_1\dots T_g\times K_{g+1}\mathrm{...}{K}_M$

$\lambda isaprimitiveunitroot,{\lambda }^M=1,PS=\left[{\lambda }^1,{\lambda }^2\dots {\lambda }^M\right]\to F_M^K={\left(-1\right)}^{M+1},F_{0<x<M}^K=0$

$SUM\left(N,PS,\left[1,2\dots M\right]\right)\stackrel{2.10)\to H_1\left(g>0,\sum K\right)=E_{M-g}^g}{\to }M!\left(\begin{array}{c}N\\ M+1\end{array}\right)E_0^M+\dots +1!\left(\begin{array}{c}N\\ 2\end{array}\right)E_{M-1}^1+0!\left(\begin{array}{c}N\\ 1\end{array}\right){\left(-1\right)}^{M+1}$

$\left\{\begin{array}{c}▽SUM\left(N\right)={{\displaystyle \prod }}_{i=1}^M\left({\lambda }^i+n\right)={{\displaystyle \sum }}_{g=1}^{M}g!S_2\left(M,g\right)\left(\begin{array}{c}n\\ g\end{array}\right)+{\left(-1\right)}^{M+1}\stackrel{Definitionof{S}_2\left(M,g\right)}{\to }{n}^M+{\left(-1\right)}^{M+1}\\ {{\displaystyle \prod }}_{i=1}^M\left({\lambda }^i+n\right)\stackrel{\mathrm{Vieta}\mathrm{theorem}}{\to }{n}^M+{\left(-1\right)}^{M+1}\end{array}\right.$

  

$Define{E}_p^q\odot \left(\left[T_1,T_2\dots \right],C\right)$

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

$2.12)H_1\left(g,\left[1,1\dots 1\right],PT=\left[{\mathrm{T}}_i={\mathrm{T}}_1+\left(\mathrm{i}-1\right)\left(\mathrm{C}+1\right)\right]\right)=E_{M-g}^{g+1}\odot \left(PT,C\right)$

$2.13)H_3\left(g,\left[1,1\dots 1\right],PT=\left[{\mathrm{T}}_i={\mathrm{T}}_1+\left(\mathrm{i}-1\right)\left(\mathrm{C}+1\right)\right]\right)=E_{M-g}^{g+1}\odot \left(\left[{\mathrm{T}}_{\mathrm{i}}=\left({\mathrm{T}}_1-1\right)+\left(\mathrm{i}-1\right)\mathrm{C}\right],C+1\right)$

  

$PB\left(PS\right)compareswith\left[K_1,K_2\dots K_M\right]$

$DefineMI{N}_g\left(PS\right)=K_1\times {\displaystyle \sum }_{PB\left(PS\right)=g}{{\displaystyle \prod }}_{i=2}^M\left(K_i+{\lambda }_i{D}_i\right),{\lambda }_1=0,\left\{\begin{array}{c}{\lambda }_i={\lambda }_{i+1},continuity \\ {\lambda }_i+1={\lambda }_{i+1},discontinuity\end{array}\right.$

$MI{N}_g=MI{N}_g\left(M\right),0\le g\le M-1,isshortforMI{N}_g\left(\left[1,2\dots M\right]\right)$

It’s easy to understand MIN_g. Imagine  that discontinuity is a hole. Factors from small to large with a distance of 2  as an hole.MIN_g = ∑ products with  g holes.

For example M=3,MIN₀=1×2×3,MIN₁=1×2×4+1×3×4,MIN₂=1×3×5,they  are used to calcute ∑3-proudcts.

Due to historical reasons, they are called MIN,  which is actually the first item of SUM().D_i can be less than 0, at  this point, they are not the minimum terms.

$2.14)K_1\times H_1\left(g,\left[K_2:D_2,K_3:D_3\dots K_M:D_M\right],\left[\frac{K_2}{D_2}+1,\frac{K_3}{D_3}+2\dots \frac{K_M}{D_M}+\left(M-1\right)\right]\right)=MI{N}_g\left(\left[K_1:D_1\dots K_M:D_M\right]\right)$

$PS=\left[1,1\dots 1\right],PT=\left[2,3\dots M\right]\to \left(g+1\right)!E_{M-g-1}^{g+1}=\left(g+1\right)!S_2\left(M,g+1\right)=MI{N}_g\left(\left[1,1\dots 1\right]\right)$

Table 2. 1:H(g) of special functions.

Table 2. 1:H(g) of special functions.

$\langle \begin{array}{c}M\\ g\end{array}\rangle is Eulerian number,defined as N^M={{\displaystyle \sum }}_{g=0}^{M-1}\langle \begin{array}{c}M\\ g\end{array}\rangle \left(\begin{array}{c}N+g\\ M\end{array}\right);\langle \begin{array}{c}M\\ M-1-g\end{array}\rangle =\langle \begin{array}{c}M\\ g\end{array}\rangle ,\left(*\right) is its another formula$ [5].

S_1,2(M+1+g,g+1) and S_2,2(M+1+g,g+1) are described in the following section.

2. H(g) and Associated Stirling Numbers

$3.1)\mathrm{By}\mathrm{definition}:{\mathrm{MIN}}_{\mathrm{g}}\left(M\right)=\sum \frac{\left(\mathrm{M}+\mathrm{g}\right)!}{i_1{i}_2\dots i_g},2\le i_1<i_2<\mathrm{...}<i_g\le \mathrm{M}+\mathrm{g}-1,i_{j+1}-i_j\ge 2$

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 .

$\left(1\right){\mathrm{S}}_{1,\mathrm{r}}\left(\mathrm{n},\mathrm{k}\right)=\frac{\mathrm{n}!}{\mathrm{k}!}{\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_k=n,i_j\ge r}\frac{1}{i_1{i}_2\dots i_k}$

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

$\left(3\right){\mathrm{S}}_{1,\mathrm{r}}\left(\mathrm{n},\mathrm{k}\right)=\sum \frac{\left(\mathrm{n}-1\right)!}{i_1{i}_2\dots i_{k-1}},r\le i_1<i_2<\mathrm{...}<i_{K-1}\le n-r,i_{j+1}-i_j\ge r,\left[4\right]$

  

Derived from (2) or (3)→

$3.2){\mathrm{MIN}}_{\mathrm{g}}\left(\mathrm{M}\right)={\mathrm{S}}_{1,2}\left(\mathrm{M}+1+\mathrm{g},\mathrm{g}+1\right)=\frac{\left(\mathrm{M}+1+\mathrm{g}\right)!}{\left(\mathrm{g}+1\right)!}{\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_{g+1}=M+1+g}\frac{1}{i_1{i}_2\dots i_{g+1}},i_j\ge 2$

Table 3. 1:MINg(M)=S1,2(M+1+g,g+1).

Table 3. 1:MINg(M)=S1,2(M+1+g,g+1).

$MI{N}_0=M!, MI{N}_1=\left(M+1\right)!\left(\frac{1}{2}+\frac{1}{3}+\mathrm{...}+\frac{1}{M}\right),MI{N}_{M-1}=\left(2M-1\right)!!$

$MI{N}_{M-2}=\frac{2\left(M-1\right)}{3}\left(2M-1\right)!! ,M\ge 2;MI{N}_{M-3}=\left(M-1\right)\left(M-2\right)\left(2M-3\right)!!\frac{4M-3}{9},M\ge 3$

  

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 .

$\left(4\right){\mathrm{S}}_{2,\mathrm{r}}\left(\mathrm{n},\mathrm{k}\right)=\frac{\mathrm{n}!}{\mathrm{k}!}{\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_k=n,i_j\ge r}\frac{1}{i_1!i_2!\dots i_k!}$

$\left(5\right){\mathrm{S}}_{2,\mathrm{r}}\left(\mathrm{n}+1,\mathrm{k}\right)=k{\mathrm{S}}_{2,\mathrm{r}}\left(\mathrm{n},\mathrm{k}\right)+\left(\begin{array}{c}n\\ r-1\end{array}\right){\mathrm{S}}_{2,\mathrm{r}}\left(\mathrm{n}-\mathrm{r}+1,\mathrm{k}-1\right),n\ge kr$

  

Derived from (5)→

$3.3){\mathrm{H}}_1\left(\mathrm{g},\left[1,1\dots 1\right],\left[3,5\dots 2\mathrm{M}-1\right]\right)=E_{M-1-g}^{g+1}\odot \left(PT,1\right)={\mathrm{S}}_{2,2}\left(\mathrm{M}+1+\mathrm{g},\mathrm{g}+1\right)$

Table 3. 2: H1(g,[1,1…1],[3,5…2M-1])=S2,2(M+1+g,g+1).

Table 3. 2: H1(g,[1,1…1],[3,5…2M-1])=S2,2(M+1+g,g+1).

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

$={{\displaystyle \sum }}_{g=0}^{M-1}{S}_{1,2}\left(M+1+g,g+1\right)\left(\begin{array}{c}N+M\\ M+1+g\end{array}\right)={{\displaystyle \sum }}_{g=1}^M{S}_{1,2}\left(M+g,g\right)\left(\begin{array}{c}N+M\\ M+g\end{array}\right)=S_1\left(N+M,N\right)=S_{1,1}\left(N+M,N\right)$

$=\frac{\left(N+M\right)!}{N!}{\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_N=N+M,i_j\ge 1}\frac{1}{i_1{i}_2\dots i_N}=\frac{\left(N+M\right)!}{N!}{{\displaystyle \sum }}_{g=1}^M{\displaystyle \sum }_{i_1+i_2+\dots +i_N=N+M,i_j\ge 1,(countof{i}_j>1)=g}\frac{1}{i_1{i}_2\dots i_N}\to$

$\frac{\left(N+M\right)!}{N!}{\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_N=N+M,i_j\ge 1,(countof{i}_j>1)=g}\frac{1}{i_1{i}_2\dots i_N}\stackrel{\left(\begin{array}{c}N\\ N-g\end{array}\right)=slecctN-gquantitiesof1fromN}{\to }$

$=\frac{\left(N+M\right)!}{N!}\left(\begin{array}{c}N\\ N-g\end{array}\right){\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_g=g+M,i_j>1,}\frac{1}{i_1{i}_2\dots i_g}=\left(\begin{array}{c}N+M\\ M+g\end{array}\right)S_{1,2}\left(M+g,g\right).$

  

Use the same way:

$S_{1,r}\left(rN+M,N\right)=\frac{\left(rN+M\right)!}{N!}{\displaystyle \sum }_{i_1+i_2+\dots +i_N=rN+M,i_j\ge r}\frac{1}{i_1{i}_2\dots i_N}$

$=\frac{\left(rN+M\right)!}{N!}{{\displaystyle \sum }}_{g=1}^M{\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_N=rN+M,i_j\ge r,\left(g=countof{i}_j\right)>r}\frac{1}{i_1{i}_2\dots i_N}$

$={{\displaystyle \sum }}_{g=1}^M\frac{\left(rN+M\right)!}{N!}\left(\begin{array}{c}N\\ g\end{array}\right)\frac{1}{r^{N-g}}{\displaystyle \sum }_{i_1+i_2+\mathrm{...}+i_g=rg+M,i_j\ge r+1}\frac{1}{i_1{i}_2\dots i_g}={{\displaystyle \sum }}_{g=1}^M\frac{1}{r^{N-g}}\frac{\left(rN+M\right)!}{N!}\left(\begin{array}{c}N\\ g\end{array}\right)\frac{g!}{\left(rg+M\right)!}{S}_{1,r+1}\left(rg+M,g\right)$

$3.4)S_{1,r}\left(rN+M,N\right)={{\displaystyle \sum }}_{g=1}^M\frac{1}{r^{N-g}}\frac{\left(rN-rg\right)!}{\left(N-g\right)!}\left(\begin{array}{c}rN+M\\ rg+M\end{array}\right)S_{1,r+1}\left(rg+M,g\right)$

$3.5)S_{2,r}\left(rN+M,N\right)={{\displaystyle \sum }}_{g=1}^M\frac{1}{r{!}^{N-g}}\frac{\left(rN-rg\right)!}{\left(N-g\right)!}\left(\begin{array}{c}rN+M\\ rg+M\end{array}\right)S_{2,r+1}\left(rg+M,g\right)$

  

P is a prime number.(1)(4)→

$\mathrm{Max}\left(\mathrm{g}+1,\mathrm{M}-\mathrm{g}+3\right)\le \mathrm{P}\le \mathrm{M}+\mathrm{g},S_{1,2}\left(M+g,g\right)\equiv S_{2,2}\left(M+g,g\right)\equiv 0\mathrm{MOD}\mathrm{P},2\le g\le M\to$

$3.6)S_1\left(P,N\right)\equiv S_2\left(P,N\right)\equiv 0\mathrm{MOD}\mathrm{P},2\le \mathrm{N}\le \mathrm{P}-1$

$3.7)S_1\left(N+M,N\right)\equiv S_2\left(N+M,N\right)\equiv \left\{\begin{array}{c}1 MOD P,N\equiv 1 MOD P\\ 0 MOD P,N\not\equiv 1 MOD P\end{array}\right.,\mathrm{P}=\mathrm{M}+2$

  

Proof by induction and recurrence formula of S₁(N,M) and S₂(N,M)

$3.8){\mathrm{S}}_1\left(\mathrm{P}-1,\mathrm{g}\right)\equiv 1\mathrm{MOD}\mathrm{P},\mathrm{g}!{\mathrm{S}}_2\left(\mathrm{P}-1,\mathrm{g}\right)\equiv {\left(-1\right)}^{g+1}\mathrm{MOD}\mathrm{P},1\le \mathrm{g}\le \mathrm{P}-1$

3. Application of SUM(N) and H(g)

$4.1){{\displaystyle \prod }}_{i=1}^M\left(K_i+n{D}_i\right)canbedecomposedintothreeformsby1.2)$

$4.2)SUM\left(N,PS,PT\right)=SUM\left(N,\left[1,1\dots 1,PS\right],\left[1,1\dots 1,PT\right]\right),expand{{\displaystyle \sum }}_{g=0}^M\left(\dots \right)to{{\displaystyle \sum }}_{g=0}^{M+countof1added}\left(\dots \right)$

  

$PS=\left[1,1\dots 1\right],PT=\left[1,2\dots M\right]\stackrel{2.2)}{\to }{H}_1\left(M\right)={{\displaystyle \sum }}_{g=0}^M{H}_3\left(g\right),H_1\left(1\right)={{\displaystyle \sum }}_{g=1}^M{H}_2\left(g\right)g\to$

$4.3){{\displaystyle \sum }}_{g=0}^M\begin{array}{c}M\\ g\end{array}=M!;{{\displaystyle \sum }}_{g=1}^M{\left(-1\right)}^{M-g}g\times g!S_2\left(M,g\right)=2^M-1$

$PS=\left[1,1\dots 1\right],PT=\left[2,3\dots M\right]\stackrel{2.2)}{\to }$

$4.4)g!S_2\left(M,g\right)={{\displaystyle \sum }}_{k=g}^M{\left(-1\right)}^{M-K}k!S_2\left(M,k\right)\left(\begin{array}{c}k-1\\ g-1\end{array}\right)={{\displaystyle \sum }}_{k=0}^{g-1}\begin{array}{c}\mathrm{M}\\ \mathrm{k}\end{array}\left(\begin{array}{c}M-1-k\\ M-g\end{array}\right),1\le g\le M$

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

  

$SUM\left(g,\left[2,3\dots M\right],\left[3,5\dots 2M-1\right]\right),SUM\left(g,\left[1,1\dots 1\right],\left[3,5\dots 2M-1\right]\right)\stackrel{2.2)}{\to }$

$4.5){\mathrm{S}}_{1,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)={{\displaystyle \sum }}_{k=g}^M{\left(-1\right)}^{M-k}{\mathrm{S}}_{2,2}\left(\mathrm{M}+\mathrm{k},\mathrm{k}\right)\left(\begin{array}{c}k-1\\ g-1\end{array}\right),{\mathrm{S}}_{2,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)={{\displaystyle \sum }}_{k=g}^M{\left(-1\right)}^{M-k}{\mathrm{S}}_{1,2}\left(\mathrm{M}+\mathrm{k},\mathrm{k}\right)\left(\begin{array}{c}k-1\\ g-1\end{array}\right)$

  

$SUM\left(N,\left[1,1\dots 1\right],\left[2,3\dots M\right]\right)\stackrel{2.4)}{\to }$

$4.6){{\displaystyle \sum }}_{g=1}^{M}g!S_2\left(M,g\right)={{\displaystyle \sum }}_{g=1}^M{\left(-1\right)}^{M-g}g!S_2\left(M,g\right)2^{g-1}={{\displaystyle \sum }}_{g=1}^M\begin{array}{c}M\\ M-g\end{array}{2}^{M-g}$

$SUM\left(N,\left[1,1\dots 1\right],\left[3,5\dots 2M-1\right]\right),SUM\left(N,\left[2,3\dots M\right],\left[3,5\dots 2M-1\right]\right)\stackrel{2.4)}{\to }$

$4.7){{\displaystyle \sum }}_{g=1}^M{\mathrm{S}}_{2,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)={{\displaystyle \sum }}_{g=1}^M{\left(-1\right)}^{M-g}{\mathrm{S}}_{1,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)2^{g-1};{{\displaystyle \sum }}_{g=1}^M{\mathrm{S}}_{1,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)={{\displaystyle \sum }}_{g=1}^M{\left(-1\right)}^{M-g}{\mathrm{S}}_{2,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)2^{g-1}$

  

$SUM\left(N,\left[1,1\dots 1\right],\left[2,3\dots M\right]\right)\stackrel{2.7)}{\to }$

$4.8){{\displaystyle \sum }}_{g=1}^{M}g!S_2\left(M,g\right)\left(g-1\right)\left(\begin{array}{c}A+1\\ g\end{array}\right)={{\displaystyle \sum }}_{g=1}^M{\left(-1\right)}^{M-g}g!S_2\left(M,g\right)\left(g-1\right)\left(\begin{array}{c}A+g-1\\ g\end{array}\right)$

  

$H_1\left(g,\left[1,2\dots M\right],\left[1,2\dots M\right]\right)=H_1\left(M-g\right)=M!\left(\begin{array}{c}M\\ g\end{array}\right)\stackrel{2.10)}{\to }g!\left(F_{M-g}^M{E}_0^g+\mathrm{...}+F_0^M{E}_{M-g}^g\right)\to$

$4.9)M!\left(\begin{array}{c}M\\ g\end{array}\right)=g!{{\displaystyle \sum }}_{i=0}^{M-g}{S}_1\left(M+1,g+1+i\right)S_2\left(g+i,g\right)=\left(M-g\right)!{{\displaystyle \sum }}_{i=0}^g{S}_1\left(M+1,M+1-i\right)S_2\left(M-i,M-g\right)$

$H_1\left(g,\left[1,1\dots 1\right],\left[1,2\dots M\right]\right)=g!S_2\left(M+1,g+1\right)\stackrel{2.10)}{\to }g!\left(F_{M-g}^{\left\{1,\mathrm{1...1}\right\}}{E}_0^g+\mathrm{...}+F_0^{\left\{1,\mathrm{1...1}\right\}}{E}_{M-g}^g\right)\to$

$4.10)S_2\left(M+1,g+1\right)={{\displaystyle \sum }}_{i=0}^{M-g}\left(\begin{array}{c}M\\ g+i\end{array}\right)S_2\left(g+i,g\right)={{\displaystyle \sum }}_{i=0}^{M-g}\left(\begin{array}{c}M\\ i\end{array}\right)S_2\left(M-i,g\right)$

  

$H_1\left(g,\left[1,2\dots M\right],\left[1,2\dots M\right]\right)=H_1\left(g,\left[M\dots 2,1\right],\left[1,2\dots M\right]\right)=M!\left(\begin{array}{c}M\\ g\end{array}\right)\stackrel{2.11)}{\to }\frac{M!}{g!}\left(F_g^M{E}_0^{M-g}+\mathrm{...}+{\left(-1\right)}^g{F}_0^M{E}_g^{M-g}\right)\to$

$4.11)g!\left(\begin{array}{c}M\\ g\end{array}\right)={{\displaystyle \sum }}_{i=0}^g{S}_1\left(M+1,M+1-g+i\right)S_2\left(M-g+i,M-g\right){\left(-1\right)}^g$

  

$H_1\left(g,\left[K+i\right],\left[T+i\right]\right)\stackrel{2.12)}{\to }\left(\begin{array}{c}\mathrm{M}\\ g\end{array}\right)T_1\dots T_g\times K_{g+1}\dots K_M\stackrel{2.10)}{\to }{T}_1\dots T_g\left(\dots \right)\stackrel{2.11)}{\to }{K}_{g+1}\dots K_M\left(\dots \right)$

$4.12){{\displaystyle \prod }}_{i=g+1}^M\left(\mathrm{K}+\mathrm{i}\right)\left(\begin{array}{c}M\\ g\end{array}\right)=F_{M-g}^{\left\{K+i\right\}}{E}_0^g+\dots +F_0^{\left\{K+i\right\}}{E}_{M-g}^g;{{\displaystyle \prod }}_{i=1}^g\left(\mathrm{T}+\mathrm{i}\right)\left(\begin{array}{c}M\\ g\end{array}\right)=F_g^{\left\{\mathrm{T}+i\right\}}{E}_0^{M-g}+\mathrm{...}{\left(-1\right)}^g{F}_0^{\left\{\mathrm{T}+i\right\}}{E}_g^{M-g}$

  

$SUM\left(1,\left[1,1\dots 1\right],\left[3,5\dots 2M-1\right]\right)=1,SUM\left(1,\left[2,3\dots M\right],\left[3,5\dots 2M-1\right]\right)=M!\stackrel{For{m}_2}{\to }{{\displaystyle \sum }}_{g=0}^{M-1}{H}_2\left(g\right)\left(\begin{array}{c}1+M+g\\ M+1+g\end{array}\right)\to$

$4.13){{\displaystyle \sum }}_{g=0}^{M-1}{\left(-1\right)}^{M-1-g}MI{N}_g={{\displaystyle \sum }}_{g=1}^M{\left(-1\right)}^{M-g}{\mathrm{S}}_{1,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)=1;{{\displaystyle \sum }}_{g=1}^M{\left(-1\right)}^{M-g}{\mathrm{S}}_{2,2}\left(\mathrm{M}+\mathrm{g},\mathrm{g}\right)=M!$

$\mathrm{PS}1=\left[{\mathrm{D}}_2:{\mathrm{D}}_2\mathrm{...}{\mathrm{D}}_{\mathrm{M}}:{\mathrm{D}}_{\mathrm{M}}\right],\mathrm{PT}1=\left[\frac{{\mathrm{K}}_2}{{\mathrm{D}}_2}+1\dots \frac{{\mathrm{K}}_{\mathrm{M}}}{{\mathrm{D}}_{\mathrm{M}}}+\mathrm{M}-1\right]\to {\mathrm{K}}_1\times H_2\left(g\right)={\left(-1\right)}^{M-1-g}MI{N}_g\left[{\mathrm{K}}_1:{\mathrm{D}}_1,{\mathrm{K}}_2:{\mathrm{D}}_2\dots \right]\to$

$4.14){{\displaystyle \sum }}_{g=0}^{M-1}{\left(-1\right)}^{M-1-g}MI{N}_g\left(PS\right)=K_1{{\displaystyle \prod }}_{i=2}^M{D}_i$

  

$SUM\left(2,\left[1,1\dots 1\right],\left[3,5\dots 2M-1\right]\right)=S_2\left(M+2,2\right)=2^{M+1}-1={{\displaystyle \sum }}_{g=0}^{M-1}{H}_2\left(g\right)\left(\begin{array}{c}2+M+g\\ 1+M+g\end{array}\right),combination4.13)\to$

$4.15){{\displaystyle \sum }}_{g=0}^{M-1}{\left(-1\right)}^{M-1-g}gMI{N}_g=2^{M+1}-M-3$

  

$SUM\left(N,\left[T,T\dots T\right],\left[T,T\dots T\right]\right)\stackrel{1.8)}{\to }{T}^M\left(\begin{array}{c}N+T\\ T+1\end{array}\right)\stackrel{2.1)}{\to }{{\displaystyle \sum }}_{g=0}^M{T}^M\left(\begin{array}{c}M\\ g\end{array}\right)\left(\begin{array}{c}N+T-M\\ T-M+1+g\end{array}\right)\to$

$4.16){{\displaystyle \sum }}_{g=0}^M\left(\begin{array}{c}M\\ g\end{array}\right)\left(\begin{array}{c}N+T-M\\ T-M+1+g\end{array}\right)=\left(\begin{array}{c}N+T\\ T+1\end{array}\right)$

  

$4.17)Pisprime,SUM\left(P,\left[\dots \right]:D,\left[1,2\dots M\right]\right)={{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right)\left(\begin{array}{c}P\\ 1+g\end{array}\right)\to \left\{\begin{array}{c}\equiv 0MODP,M<P-1\\ \equiv -1MODP,M=P-1\end{array}\right.,\left(P,D\right)=1$ Exclude products≡0 MOD P , then a series of congruence can be obtained.For Example:

$1^M+2^M+\mathrm{...}+{\left(P-1\right)}^M\equiv \left\{\begin{array}{c}0MODP,0<M<P-1\\ -1MODP,M=P-1\end{array}\right.$

$1^A{2}^B+2^A{3}^B+\mathrm{...}+{\left(P-2\right)}^A{\left(P-1\right)}^B\equiv \left\{\begin{array}{c}0MODP,0<A+B<P-1,0\le A,B\in N\\ -1MODP,A+B=P-1,0\le A,B\in N\end{array}\right.$

$1^A{3}^B+2^A{4}^B+\mathrm{...}+{\left(P-3\right)}^A{\left(P-1\right)}^B+{\left(P-1\right)}^A{1}^B\equiv \left\{\begin{array}{c}0MODP,0<A+B<P-1,0\le A,B\in N\\ -1MODP,A+B=P-1,0\le A,B\in N\end{array}\right.$

$1^A{3}^B{5}^C+2^A{4}^B{6}^C+\dots +{\left(P-5\right)}^A{\left(P-3\right)}^B{\left(P-1\right)}^C+{\left(P-3\right)}^A{\left(P-1\right)}^B{1}^C+{\left(P-1\right)}^A{1}^B{3}^C$

$\equiv \left\{\begin{array}{c}0MODP,0<A+B+C<P-1,0\le A,B,C\in N\\ -1MODP,A+B+C=P-1,0\le A,B,C\in N\end{array}\right.$

  

$\mathrm{Derived}\mathrm{from}4.17)\to$

$4.18)P is prime,{\displaystyle \sum }_{,{\mathsf{\lambda }}_1+{\mathsf{\lambda }}_2+\mathrm{...}+{\mathsf{\lambda }}_{\mathrm{k}}=q,1\le q\le P-2,{\mathsf{\lambda }}_{\mathrm{j}}\ge 0}{1}^{{\mathsf{\lambda }}_1}{2}^{{\mathsf{\lambda }}_2}\dots {\left(\mathrm{P}-1\right)}^{{\mathsf{\lambda }}_k}\equiv 0\mathrm{MOD}\mathrm{P}$

  

$PS1=\left[K_1:D_1\dots K_i:D_i,K_{i+1}+D_{i+1}:D_{i+1},K_{i+2}+D_{i+2}:D_{i+2}\dots \right],PT1=\left[T_1\dots T_i,T_{i+1}-1,T_{i+2}-1\dots T_M-1\right]$

$4.19)Classificationprinciple:SUM\left(N\right)=SUM\left(N,PS,PT1\right)+SUM\left(N-1,PS1,PT\right)$ Simply put, PS and PS have the same shape.

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

$=\left\{SUM\left(N,PT,\left[1,2,3\right]\right)+SUM\left(N-1,PT,\left[1,2,4\right]\right)\right\}+\left\{SUM\left(N-1,PT,\left[1,3,4\right]\right)+SUM\left(N-2,PT,\left[1,3,5\right]\right)\right\}$

If T_i= T_i+1, it is also established.

$SUM\left(N,\left[1,2\right],\left[1,2\right]\right)=2\left(\begin{array}{c}N+2\\ 3\end{array}\right)=SUM\left(N,\left[1,2\right],\left[1,1\right]\right)+SUM\left(N-1,\left[1,3\right],\left[1,2\right]\right)$

$=SUM\left(N,\left[2\right],\left[1\right]\right)+SUM\left(N-1,\left[3\right],\left[2\right]\right)=\left(\begin{array}{c}N\\ 2\end{array}\right)+2\left(\begin{array}{c}N\\ 1\end{array}\right)+2\left(\begin{array}{c}N\\ 3\end{array}\right)+3\left(\begin{array}{c}N\\ 2\end{array}\right)$

The classification principle also applies to the following Formal Calculation.

  

PT’s domain can be extended to ℂ. The Recursive Formula of H (g) can be used for.

$4.20)SolvingtheEquation:F\left(M+1,g\right)=F\left(M,g\right)\left(aM+bg+c\right)+F\left(M,g-1\right)\left(xM+bg+z\right),b\ne 0$

$H_1\left(g,PS1,PT1\right)=H_1\left(g-1\right)\left({\mathrm{T}}_{M+1}-\left[\mathrm{M}-\left(\mathrm{g}-1\right)\right]\right){\mathrm{D}}_{M+1}+H_1\left(g\right)\left({\mathrm{K}}_{M+1}+{\mathrm{gD}}_{M+1}\right)$

$\left\{\begin{array}{c}aM+bg+c={\mathrm{K}}_{M+1}+g{\mathrm{D}}_{M+1}\\ xM+bg+z=\left({\mathrm{T}}_{M+1}-\left[\mathrm{M}-\left(\mathrm{g}-1\right)\right]\right){\mathrm{D}}_{M+1}\end{array}\right.\stackrel{\mathrm{eliminate}g}{\to }\left\{\begin{array}{c}{\mathrm{D}}_{M+1}=b,{\mathrm{K}}_{M+1}=aM+c,PS=\left[a\left(i-1\right)+c:b\right]\\ {\mathrm{T}}_{M+1}=M+1+\frac{xM+z}{b},PT=\left[\frac{x\left(i-1\right)+z}{b}+i\right]\end{array}\right.$

$H_2\left(g\right),H_3\left(g\right)alsohassimilarproperties.\mathrm{This}\mathrm{way},\mathrm{we}\mathrm{can}\mathrm{use}\mathrm{the}\mathrm{attributes}\mathrm{of}\mathrm{H}\left(\mathrm{g}\right)\mathrm{for}\mathrm{analysis}.$

$eg:F\left(M+1,g\right)=\left(g+1\right)F\left(M,g\right)+\left(g+1\right)F\left(M,g-1\right)\to PS=\left[1,1\dots 1\left],PT=\right[2,3\dots M+1\right]$

$I{t}^{\prime }s\left(g+1\right)!S_2\left(M+1,g+1\right)=\left(g+1\right)\left(g+1\right)!S_2\left(M,g+1\right)+\left(g+1\right)g!S_2\left(M,g\right)$

  

$SUM\left(N,PS1,PT1\right)=SUM\left(N-1,PS1,PT1\right)+\left[{\mathrm{K}}_{M+1}+\left(N-1\right){\mathrm{D}}_{M+1}\right]{▽}^{P}SUM\left(N\right)candothesame.$

$SUM\left(N,\left[1,\mathrm{2...}M\right],\left[1,3\dots 2M-1\right]\right)\to S_1\left(N+M,N\right)=S_1\left(N+M-1,N-1\right)+\left(N+M-1\right)S_1\left(N+M-1,N\right)$

  

4. Combinatorial identity of SUM(N)

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

$5.1){\displaystyle \sum }_{\left(r\right)}^N{▽}^{P}SUM\left(k,PS,PT\right)={▽}^{P-r}SUM\left(N,PS,PT\right)$

  

$5.2){{\displaystyle \sum }}_{k_r=1}^N\dots {{\displaystyle \sum }}_{k_2=1}^{k_3}{{\displaystyle \sum }}_{k_1=1}^{k_2}▽SUM\left(k_r,PS,\left[1,2\dots M\right]\right)=SUM\left(N,PS,\left[1+\left(r-1\right),2+\left(r-1\right)\dots M+\left(r-1\right)\right]\right)$

[Proof]

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

$PT=\left[1,2\dots M\right],PTx=\left[1+\left(r-1\right),2+\left(r-1\right)\dots M+\left(r-1\right)\right]$

$f\left(N\right)={{\displaystyle \sum }}_{k_r=1}^N\dots {{\displaystyle \sum }}_{k_2=1}^{k_3}{{\displaystyle \sum }}_{k_1=1}^{k_2}▽SUM\left(k_r,PS,PT\right)={{\displaystyle \sum }}_{k_r=1}^N▽SUM\left(k_r,PS,PT\right)\dots {{\displaystyle \sum }}_{k_2=1}^{k_3}{{\displaystyle \sum }}_{k_1=1}^{k_2}1$

$PT=\left[1,2\dots M\right].AccordingtodefinitionofSUM\left(N\right)\to$

$f\left(N\right)=SUM\left(N,PS1,PT1\right)\stackrel{M+2\left(r-1\right)-\left(r-1+M\right)=r-1}{\to }{{\displaystyle \sum }}_{g=0}^{M+r-1}{H}_1\left(g,PS1,PT1\right)\left(\begin{array}{c}N+r-1\\ r+g\end{array}\right)$

$In{H}_1\left(g>M,PS1,PT1\right),i<r,X_i=\left\{\begin{array}{c}\left(T_i-X_{K-1}\right)D_i=0;X_i=T_i\\ K_i+X_{T-1}{D}_i=1;X_i=K_i \end{array}\right.=0$

$In{H}_1\left(g\le M,PS1,PT1\right),{{\displaystyle \prod }}_{i<r}{X}_i=1;i\ge r,X_i=\left\{\begin{array}{c}\left(T_i-X_{K-1}\right)D_i;X_i=T_i\\ K_i+X_{T-1}{D}_i;X_i=K_i \end{array}\right.=H_1\left(g,PS,PTx\right)$

$\to f\left(N\right)={{\displaystyle \sum }}_{g=0}^M{H}_1\left(g,PS1,PT1\right)\left(\begin{array}{c}N+r-1\\ r+g\end{array}\right)=SUM\left(N,PS,PTx\right)$

q.e.d.

$5.3){{\displaystyle \sum }}_{k_r=1}^N\dots {{\displaystyle \sum }}_{k_2=1}^{k_3}{{\displaystyle \sum }}_{k_1=1}^{k_2}▽SUM\left(k_i,PS,\left[1,2\dots M\right]\right)={▽}^{i-r}SUM\left(N,PS,\left[1+\left(i-1\right),2+\left(i-1\right)\dots M+\left(i-1\right)\right]\right)$

${{\displaystyle \sum }}_{k_r=0}^N\dots {{\displaystyle \sum }}_{k_2=0}^{k_3}{{\displaystyle \sum }}_{k_1=0}^{k_2}\left(\begin{array}{c}{k}_i\\ j\end{array}\right)=\frac{1}{j!}{▽}^{i-r}SUM\left(N+1,\left[0,-1,-2\dots -j+1\left],\right[1+\left(i-1\right),2+\left(i-1\right)\dots j+\left(i-1\right)\right]\right)$

$\stackrel{H_1\left(g<j\right)=0,T_M-M=i-1}{\to }\frac{1}{j!}{▽}^{i-r}\frac{\left(j+i-1\right)!}{\left(i-1\right)!}\left(\begin{array}{c}N+1+i-1\\ i+j\end{array}\right)=\left(\begin{array}{c}j+i-1\\ i-1\end{array}\right)\left(\begin{array}{c}N+i-\left(i-r\right)\\ i+j-\left(i-r\right)\end{array}\right)=\left(\begin{array}{c}j+i-1\\ i-1\end{array}\right)\left(\begin{array}{c}N+r\\ r+j\end{array}\right)\to recordat\left[6\right]:Lemma1$

  

$PS=\left[1,2\dots T_1-P_1,\frac{K_1\left(T_1+1-P_1\right)}{T_1{D}_1},1,2\dots T_2-P_2,\frac{K_2\left(T_2+1-P_2\right)}{T_2{D}_2}\right],PT=\left[1,2,3\dots T_1+T_2+2-P_1-P_2\right]$

$5.4){{\displaystyle \sum }}_{n=0}^{N-1}{▽}^{P_1}SUM\left(n+1,\left[K_1:D_1\right],\left[T_1\right]\right){▽}^{P_2}SUM\left(n+1,\left[K_2:D_2\right],\left[T_2\right]\right)=\frac{T_1{T}_2{D}_1{D}_2}{\left(T_1+1-P_1\right)!\left(T_2+1-P_2\right)!}SUM\left(N,PS,PT\right)$

  

$5.5){\displaystyle \sum }_{0\le n_1\le n_2\le \mathrm{...}\le n_M\le N-1}\left(K+n_1+n_2+\dots +n_M\right)={▽}^{\left(\begin{array}{c}M\\ 2\end{array}\right)}SUM\left(N,\left[K\right],\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right)\right]\right)=\left(\begin{array}{c}M+1\\ 2\end{array}\right)\left(\begin{array}{c}N+M-1\\ M+1\end{array}\right)+K\left(\begin{array}{c}N+M-1\\ M\end{array}\right)$  ${\displaystyle \sum }_{0\le n_1\le \mathrm{...}\le n_p\le n}\left(n_1+\dots +n_p\right)={{\displaystyle \sum }}_{n_p=0}^n\dots {{\displaystyle \sum }}_{n_1=0}^{n_2}\left(n_1+\mathrm{...}+n_p\right)=\left(\begin{array}{c}p+1\\ 2\end{array}\right)\left(\begin{array}{c}n+p\\ p+1\end{array}\right)=\frac{pn}{2}\left(\begin{array}{c}n+p\\ p\end{array}\right)recordat\left[6\right]:Theorem2$

  

$5.6){\displaystyle \sum }_{0\le n_1\le \mathrm{...}\le n_M\le N-1}\left(K+n_1{D}_1+\dots +n_M{D}_M\right)={▽}^{\left(\begin{array}{c}M\\ 2\end{array}\right)}SUM\left(N,\left[K:\left(D_1+2D_2+\mathrm{...}+M{D}_M\right){\left(\begin{array}{c}M+1\\ 2\end{array}\right)}^{-1}\right],\left[\left(\begin{array}{c}M+1\\ 2\end{array}\right)\right]\right)$

$=\frac{D_1+2D_2+\mathrm{...}+M{D}_M}{\left(\begin{array}{c}M+1\\ 2\end{array}\right)}\left(\begin{array}{c}M+1\\ 2\end{array}\right)\left(\begin{array}{c}N+M-1\\ M+1\end{array}\right)+K\left(\begin{array}{c}N+M-1\\ M\end{array}\right)$

  

Use 1.2)-1.4) and properties of H(g) can be used to derive combinatorial identities.For Example:

$\left(\begin{array}{c}n+A\\ A\end{array}\right)\left(\begin{array}{c}n+M+B\\ M\end{array}\right)=\frac{1}{\mathrm{A}!\mathrm{M}!}▽SUM\left(N,\left[1,2\dots \mathrm{A},\mathrm{B}+1,\mathrm{B}+2\dots \mathrm{B}+M\right],\left[1,2\dots \mathrm{A},\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right]\right)$

$=\frac{1}{\mathrm{M}!}▽SUM\left(N,\left[\mathrm{B}+1\dots \mathrm{B}+2,\mathrm{B}+\mathrm{M}\right],\left[\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right]\right)\stackrel{2.12)}{\to }$

$H_1\left(g\right)=\left(\begin{array}{c}M\\ g\end{array}\right){\left[B+M\right]}_{M-g}{\left[A+1\right]}^g=\left(\begin{array}{c}M\\ g\end{array}\right)\left(\mathrm{M}-\mathrm{g}\right)!\left(\begin{array}{c}M+B\\ M-g\end{array}\right)g!\left(\begin{array}{c}A+g\\ g\end{array}\right)=M!\left(\begin{array}{c}\mathrm{A}+\mathrm{g}\\ \mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{M}+\mathrm{B}\\ \mathrm{M}-\mathrm{g}\end{array}\right)$

  

$5.7)\left(\begin{array}{c}\mathrm{n}+\mathrm{A}\\ \mathrm{A}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{M}+\mathrm{B}\\ \mathrm{M}\end{array}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}\left(\begin{array}{c}\mathrm{A}+\mathrm{g}\\ \mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{M}+\mathrm{B}\\ \mathrm{M}-\mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{A}\\ \mathrm{A}+\mathrm{g}\end{array}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^{\mathrm{M}-\mathrm{g}}\left(\begin{array}{c}\mathrm{A}-\mathrm{B}\\ \mathrm{M}-\mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{A}+\mathrm{g}\\ \mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{A}+\mathrm{g}\\ \mathrm{A}+\mathrm{g}\end{array}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}\left(\begin{array}{c}\mathrm{A}-\mathrm{B}\\ \mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{M}+\mathrm{B}\\ \mathrm{M}-\mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{A}+\mathrm{M}-\mathrm{g}\\ \mathrm{A}+\mathrm{M}\end{array}\right)$

  

$5.8)\left(\begin{array}{c}\mathrm{n}+\mathrm{X}\\ \mathrm{A}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{Y}\\ \mathrm{M}\end{array}\right)={{\displaystyle \sum }}_{\mathrm{q}=0}^{\mathrm{A}}\left(\begin{array}{c}\mathrm{M}+\mathrm{q}\\ \mathrm{q}\end{array}\right)\left(\begin{array}{c}\mathrm{M}+\mathrm{X}-\mathrm{Y}\\ \mathrm{A}-\mathrm{q}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{Y}\\ \mathrm{M}+\mathrm{q}\end{array}\right),0\le \mathrm{Y}\le \mathrm{M}$

[Proof]

$=\frac{1}{\mathrm{A}!\mathrm{M}!}▽SUM\left(N,\left[X,X-1\dots X-A+1,Y,Y-1\dots Y-M+1\right],\left[1,2\dots \mathrm{A}+\mathrm{M}\right]\right)$

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

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

$If{\mathrm{H}}_1\left(\mathrm{g}\right)\ne 0,X_1,X_2\dots X_{\mathrm{M}-\mathrm{Y}}mustbelongtoT.letC=A+M-Y$

${\mathrm{H}}_1\left(\mathrm{g}\ge \mathrm{M}-\mathrm{Y},\sum \mathrm{K}\right)\ne 0\stackrel{\mathrm{count}\mathrm{of}\mathrm{K}=\left(\begin{array}{c}C-\left(M-Y\right)\\ \mathrm{C}-\mathrm{g}\end{array}\right)}{\to }\left(\begin{array}{c}C-M+Y\\ \mathrm{C}-\mathrm{g}\end{array}\right){\left[\mathrm{X}+\left(\mathrm{M}-\mathrm{Y}\right)\right]}_{\mathrm{C}-\mathrm{g}}=\left(\begin{array}{c}A\\ \mathrm{C}-\mathrm{g}\end{array}\right){\left[\mathrm{X}+\left(\mathrm{M}-\mathrm{Y}\right)\right]}_{\mathrm{C}-\mathrm{g}}$

$\frac{\mathrm{Y}!}{\mathrm{A}!\mathrm{M}!}{H}_1\left(g\ge \mathrm{M}-\mathrm{Y}\right)=\frac{\mathrm{Y}!}{\mathrm{A}!\mathrm{M}!}\left(\begin{array}{c}\mathrm{A}\\ M+A-Y-\mathrm{g}\end{array}\right){\left[\mathrm{X}+\mathrm{M}-\mathrm{Y}\right]}_{A+M-Y-\mathrm{g}}{\left[Y+1\right]}^g\stackrel{\mathrm{q}≔-\left(\mathrm{M}-\mathrm{Y}-\mathrm{g}\right)}{\to }$

$\frac{\mathrm{Y}!}{\mathrm{A}!\mathrm{M}!}{H}_1\left(\mathrm{M}-\mathrm{Y}+\mathrm{q}\right)=\frac{\mathrm{Y}!}{\mathrm{A}!\mathrm{M}!}\left(\begin{array}{c}\mathrm{A}\\ A-q\end{array}\right)\left(\begin{array}{c}\mathrm{X}+\mathrm{M}-\mathrm{Y}\\ A-q\end{array}\right)\left(A-q\right)!\left(\begin{array}{c}\mathrm{M}+\mathrm{q}\\ M-y+q\end{array}\right)\left(M-y+q\right)!=\left(\begin{array}{c}\mathrm{M}+\mathrm{q}\\ \mathrm{q}\end{array}\right)\left(\begin{array}{c}\mathrm{M}+\mathrm{X}-\mathrm{Y}\\ \mathrm{A}-\mathrm{q}\end{array}\right)$

$\left(\begin{array}{c}\mathrm{n}+\mathrm{X}\\ \mathrm{A}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{Y}\\ \mathrm{M}\end{array}\right)=\frac{\mathrm{Y}!}{\mathrm{A}!\mathrm{M}!}▽{{\displaystyle \sum }}_{g=M-Y}^C{H}_1\left(g\right)\left(\begin{array}{c}N+\left(A+M\right)-C\\ \left(A+M\right)-C+1+q\end{array}\right)=\frac{\mathrm{Y}!}{\mathrm{A}!\mathrm{M}!}{{\displaystyle \sum }}_{g=M-Y}^{A+M-Y}{H}_1\left(g\right)\left(\begin{array}{c}n+Y\\ \mathrm{Y}+g\end{array}\right)$

q.e.d.

$Comparewith5.7),therehasnoformulafor\left(\begin{array}{c}\mathrm{n}+\mathrm{X}\\ \mathrm{A}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{Y}\\ \mathrm{M}\end{array}\right),Y>M,X\ne A$

  

$\mathrm{PS}=\left[0,-1\dots -\left(B-1\right),A-\left(M-B-1\right):-1\dots A:-1\right],\mathrm{PT}=\left[1,\mathrm{2...}\mathrm{M}\right]\to$

$5.9)\left(\begin{array}{c}\mathrm{n}\\ \mathrm{B}\end{array}\right)\left(\begin{array}{c}\mathrm{A}-\mathrm{n}\\ \mathrm{M}-\mathrm{B}\end{array}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}-\mathrm{B}}{\left(-1\right)}^{M-B-g}\left(\begin{array}{c}\mathrm{A}-\mathrm{M}+\mathrm{g}\\ \mathrm{g}\end{array}\right)\left(\begin{array}{c}M-g\\ \mathrm{B}\end{array}\right)\left(\begin{array}{c}n\\ \mathrm{M}-\mathrm{g}\end{array}\right),n\ge 0$

  

$\mathrm{PS}=\left[0,-1\dots -A+1,0,-1\dots -\mathrm{B}+1\right],\mathrm{PT}=\left[1,\mathrm{2...}\mathrm{A}+\mathrm{B}\right]\to$

$5.10)\left(\begin{array}{c}\mathrm{n}\\ \mathrm{A}\end{array}\right)\left(\begin{array}{c}\mathrm{n}\\ \mathrm{B}\end{array}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{B}}\left(\begin{array}{c}\mathrm{B}\\ \mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{A}+\mathrm{B}-\mathrm{g}\\ \mathrm{B}-\mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{n}\\ \mathrm{A}+\mathrm{B}-\mathrm{g}\end{array}\right),\mathrm{record}\mathrm{at}\left[7\right]:\left(6.44\right)$

  

$5.11){\displaystyle \prod }_{i=1}^M\left(A+2i+n\right)={\left(\begin{array}{c}n+A\\ A\end{array}\right)}^{-1}{{\displaystyle \sum }}_{g=0}^M\left(2\left[M-g\right]-1\right)!!\left(\begin{array}{c}2M-g\\ g\end{array}\right){\left[A+1\right]}^g\left(\begin{array}{c}n+A+g\\ A+g\end{array}\right),A\in \mathbb{N} orA=0$

[Proof]

PS=[A+2,A+4…A+2M],PT=[A+1,A+2…A+M]

$H_2\left(g,\sum K\right)=\mathrm{SUM}\left(\mathrm{g}+1,\left[1,3\dots 2\left(\mathrm{M}-\mathrm{g}\right)-1\right],\left[1,3\dots 2\left(\mathrm{M}-\mathrm{g}\right)-1\right]\right)\stackrel{1.8)}{\to }\left(2\left(\mathrm{M}-\mathrm{g}\right)-1\right)!!\left(\begin{array}{c}2\mathrm{M}-\mathrm{g}\\ \mathrm{g}\end{array}\right)$

$H_2\left(g,T\right)={\left[A+1\right]}^g$

$\left(\begin{array}{c}\mathrm{n}+\mathrm{A}\\ \mathrm{A}\end{array}\right){\displaystyle \prod }_{i=1}^M\left(A+2i+n\right)=\frac{1}{\mathrm{A}!}▽\mathrm{SUM}\left(\mathrm{N},\left[1,2\dots \mathrm{A},\mathrm{PS}\right],\left[1,2\dots \mathrm{A},\mathrm{PT}\right]\right)=▽\mathrm{SUM}\left(\mathrm{N}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{H}_2\left(g\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{A}+\mathrm{g}\\ \mathrm{A}+\mathrm{g}\end{array}\right)$

q.e.d.

$5.12){\displaystyle \prod }_{i=1}^M\left(2i+n\right)={{\displaystyle \sum }}_{g=0}^M\left(2\left[M-g\right]-1\right)!!\left(\begin{array}{c}2M-g\\ g\end{array}\right)g!\left(\begin{array}{c}n+A+g\\ A+g\end{array}\right)$

  

SUM(N,[1,3…2M-1],[1,2…M])= SUM(N,[3,5…2M-1],[2,3…M])→

$5.13){\displaystyle \prod }_{i=1}^M\left(2i-1+n\right)={{\displaystyle \sum }}_{g=0}^{M-1}\left(2\left[\mathrm{M}-1-\mathrm{g}\right]-1\right)!!\left(\begin{array}{c}2\left(M-1\right)-g\\ g\end{array}\right)\left(g+1\right)!\left(\begin{array}{c}\mathrm{n}+1+\mathrm{g}\\ \mathrm{g}+1\end{array}\right)$

  

$5.14)SUM\left(N,\left[\mathrm{A}+1,\mathrm{A}+3\dots \mathrm{A}+2\mathrm{M}-1\right],\left[1,3\dots 2\mathrm{M}-1\right]\right)={{\displaystyle \sum }}_{g=0}^M{\left[\mathrm{A}\right]}^{M-g}\left(2g-1\right)!!\left(\begin{array}{c}M+g\\ 2g\end{array}\right)\left(\begin{array}{c}\mathrm{N}+\mathrm{M}-1+\mathrm{g}\\ \mathrm{M}+\mathrm{g}\end{array}\right)$

[Proof]

$H_2\left(g,K\right)={\left[A\right]}^{M-g},H_2\left(g,T\right)=\mathrm{SUM}\left(\mathrm{M}-\mathrm{g}+1,\left[1,3\dots 2g-1\right],\left[1,3\dots 2\mathrm{g}-1\right]\right)\stackrel{1.8)}{\to }\left(2\mathrm{g}-1\right)!!\left(\begin{array}{c}\mathrm{M}+\mathrm{g}\\ 2\mathrm{g}\end{array}\right)$

q.e.d.

$5.15){{\displaystyle \sum }}_{n_M=0}^{N-1}\left(2M+n_M\right)\dots {{\displaystyle \sum }}_{n_2=0}^{n_3}\left(4+n_2\right){{\displaystyle \sum }}_{n_1=0}^{n_2}\left(2+n_1\right)={{\displaystyle \sum }}_{g=0}^M\frac{\left(M+g\right)!}{\mathrm{g}!2^g}\left(\begin{array}{c}\mathrm{N}+\mathrm{M}-1+\mathrm{g}\\ \mathrm{M}+\mathrm{g}\end{array}\right)={{\displaystyle \sum }}_{q=0}^M\left(\begin{array}{c}\mathrm{N}-1+\mathrm{g}\\ \mathrm{g}\end{array}\right)\frac{{\left[N+g\right]}^M}{2^g}$

$5.16)SUM\left(\mathrm{N},\left[\mathrm{A},\mathrm{A}+1\dots \mathrm{A}+\mathrm{M}-1\right]:2,\left[1,3\dots 2\mathrm{M}-1\right]\right)=\left(\begin{array}{c}M+N-1\\ M\end{array}\right){\left[A+M+N-2\right]}_M$

[Proof]

${\mathrm{H}}_1\left(\mathrm{g},\sum \mathrm{K},\left[A,A+1\dots \mathrm{A}+\left(\mathrm{M}-1\right)\right],\left[1,2\dots \mathrm{M}\right]\right)$

$=SUM\left(g+1,\left[A,A+1\dots \mathrm{A}+\left(\mathrm{M}-1-\mathrm{g}\right)\right]:2,\left[1,3\dots 2\left(M-\mathrm{g}\right)-1\right]\right)$

$={\mathrm{H}}_1\left(\mathrm{g},\sum \mathrm{K},\left[\mathrm{A}+\left(\mathrm{M}-1\right)\dots A+1,A\right],\left[1,2\dots \mathrm{M}\right]\right)=\left(\begin{array}{c}M\\ g\end{array}\right){\left[\mathrm{A}+\left(\mathrm{M}-1\right)\right]}_{\mathrm{M}-\mathrm{g}}$

  

$SUM\left(g+1,\left[A,A+1\dots \mathrm{A}+\left(\mathrm{M}-1-\mathrm{g}\right)\right]:2,\left[1,3\dots 2\left(M-\mathrm{g}\right)-1\right]\right)=\left(\begin{array}{c}M\\ g\end{array}\right){\left[\mathrm{A}+\left(\mathrm{M}-1\right)\right]}_{\mathrm{M}-\mathrm{g}}\stackrel{M≔M-g}{\to }$

$SUM\left(g+1\right)=\left(\begin{array}{c}M+g\\ g\end{array}\right){\left[\mathrm{A}+\left(\mathrm{M}+\mathrm{g}-1\right)\right]}_{\mathrm{M}}\stackrel{N≔g+1}{\to }SUM\left(N\right)=\left(\begin{array}{c}M+N-1\\ M\end{array}\right){\left[A+M+N-2\right]}_M$

q.e.d.

$5.17)\mathrm{SUM}\left(\mathrm{N},\left[1,2\dots \mathrm{M}\right]:2,\left[1,3\dots 2\mathrm{M}-1\right]\right)=\mathrm{M}!{\left(\begin{array}{c}\mathrm{N}+\mathrm{M}-1\\ \mathrm{M}\end{array}\right)}^2\stackrel{M=1}{\to }1+3+\dots +\left(2\mathrm{N}-1\right)={\mathrm{N}}^2$

  

$\frac{1}{M!}SUM\left(\mathrm{N},\left[\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right]:2,\left[1,3\dots 2\mathrm{M}-1\right]\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{H}}_1\left(\mathrm{g}\right)\left(\begin{array}{c}\mathrm{N}+\mathrm{M}-1\\ \mathrm{M}+\mathrm{g}\end{array}\right)=\left(\begin{array}{c}M+N-1\\ M\end{array}\right)\left(\begin{array}{c}A+M+N-1\\ M\end{array}\right)\stackrel{5.7),A=M,B=A}{\to }$

$=▽SUM\left(\mathrm{N},\mathrm{PS}1=\left[\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right],\mathrm{PT}1=\left[\mathrm{M}+1,\mathrm{M}+2\dots 2\mathrm{M}\right]\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{H}}_1\left(\mathrm{g},\mathrm{PS}1,\mathrm{PT}1\right)\left(\begin{array}{c}\mathrm{N}+\mathrm{M}-1\\ \mathrm{M}+\mathrm{g}\end{array}\right)\to$

$5.18)\frac{1}{M!}\mathrm{H}\left(\mathrm{g},\left[\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right]:2,\left[1,3\dots 2\mathrm{M}-1\right]\right)=\mathrm{H}\left(\mathrm{g},\left[\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right],\left[\mathrm{M}+1,\mathrm{M}+2\dots 2\mathrm{M}\right]\right)$

${\mathrm{H}}_1\left(\mathrm{g},\left[\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right]:2,\left[1,3\dots 2\mathrm{M}-1\right]\right)=M!\left(\begin{array}{c}\mathrm{M}+\mathrm{g}\\ \mathrm{g}\end{array}\right)\left(\begin{array}{c}\mathrm{M}+\mathrm{A}\\ \mathrm{M}-\mathrm{g}\end{array}\right).ThedefinitionofH\left(g\right)iscomplex.eg:$

${\mathrm{H}}_1\left(1\right)=2\left\{\left(\mathrm{A}+1\right)\dots \left(\mathrm{A}+\mathrm{M}-1\right)\mathrm{ⅹ}\mathrm{M}+\left(\mathrm{A}+1\right)\dots \left(\mathrm{A}+\mathrm{M}-2\right)\left(\mathrm{A}+\mathrm{M}+2\right)\mathrm{ⅹ}\left(M-1\right)+\dots +\left(A+4\right)\left(A+5\right)\dots \left(A+M+2\right)ⅹ1\right\}\to$

$5.19){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{M}-1}\left(\mathrm{M}-\mathrm{n}\right){{\displaystyle \prod }}_{k=1}^{M-1}\left(A+k+3\left[\frac{\mathrm{k}+\mathrm{n}}{M}\right]\right)=\frac{\left(M+1\right)!}{2}\left(\begin{array}{c}\mathrm{M}+\mathrm{A}\\ \mathrm{M}-1\end{array}\right),\left[\mathrm{n}\right]istruncatesinteger.$

  

$5.20)SUM\left(\mathrm{N},\left[\mathrm{A}+2,\mathrm{A}+4\dots \mathrm{A}+2\mathrm{M}\right]:3,\left[1,3\dots 2\mathrm{M}-1\right]\right)={{\displaystyle \sum }}_{\mathrm{y}=0}^{\mathrm{M}}\left(\begin{array}{c}A+N-1+y\\ y\end{array}\right)\left(\begin{array}{c}N+M-1-y\\ M-y\end{array}\right)\frac{{\left[M+N-y\right]}^M}{2^{M-y}}$

[Proof]

$PS1=\left[\mathrm{A}+2,\mathrm{A}+4\dots \mathrm{A}+2\mathrm{M}\right],PT1=\left[\mathrm{A}+1,\mathrm{A}+2\dots \mathrm{A}+\mathrm{M}\right]$

${\mathrm{H}}_1\left(\mathrm{g},\mathrm{T},PS1,\mathrm{PT}1\right)={\left[\mathrm{A}+1\right]}^g$

${\mathrm{H}}_1\left(\mathrm{g},\sum \mathrm{K},PS1,\mathrm{PT}1\right)=SUM\left(g+1,\left[\mathrm{A}+2,\mathrm{A}+4\dots \mathrm{A}+2\left(\mathrm{M}-\mathrm{g}\right)\right]:3,\left[1,3\dots 2\left(M-g\right)-1\right]\right)$

${\mathrm{H}}_1\left(\mathrm{g},\mathrm{PS}1,\mathrm{PT}1\right)\stackrel{2.2)}{\to }{{\displaystyle \sum }}_{\mathrm{x}=\mathrm{g}}^{\mathrm{M}}{\mathrm{H}}_2\left(\mathrm{x},\mathrm{PS}1,\mathrm{PT}1\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{g}\end{array}\right)\stackrel{5.11)}{\to }{{\displaystyle \sum }}_{\mathrm{x}=\mathrm{g}}^{\mathrm{M}}\left(2\left[\mathrm{M}-\mathrm{x}\right]-1\right)!!\left(\begin{array}{c}2M-x\\ x\end{array}\right){\left[\mathrm{A}+1\right]}^x\left(\begin{array}{c}\mathrm{x}\\ \mathrm{g}\end{array}\right)$

  

$SUM(g+1,\left[\mathrm{A}+2,\mathrm{A}+4\dots \mathrm{A}+2\left(\mathrm{M}-\mathrm{g}\right)\right]:3,\left[1,3\dots 2\left(M-g\right)-1\right]$

$={{\displaystyle \sum }}_{\mathrm{x}=\mathrm{g}}^{\mathrm{M}}\left(2\left[\mathrm{M}-\mathrm{x}\right]-1\right)!!\left(\begin{array}{c}2M-x\\ x\end{array}\right){\left[\mathrm{A}+1\right]}^x\left(\begin{array}{c}\mathrm{x}\\ \mathrm{g}\end{array}\right){\left[\mathrm{A}+1\right]}^{-g}$

$={{\displaystyle \sum }}_{\mathrm{x}=\mathrm{g}}^{\mathrm{M}}\left(2\left[\mathrm{M}-\mathrm{x}\right]-1\right)!!\frac{\left(2M-x\right)!}{\left(2m-2x\right)!\mathrm{x}!}\frac{\left(\mathrm{A}+\mathrm{x}\right)!}{A!}\frac{\mathrm{x}!}{g!\left(x-g\right)!}\frac{\mathrm{A}!}{\left(A+g\right)!}={{\displaystyle \sum }}_{\mathrm{x}=\mathrm{g}}^{\mathrm{M}}\frac{\left(2M-x\right)!}{\left(m-x\right)!2^{M-x}}\frac{\left(A+x\right)!}{\left(A+g\right)!\mathrm{g}!\left(x-g\right)!}$

$\stackrel{\mathrm{y}≔\mathrm{x}-\mathrm{g}}{\to }{{\displaystyle \sum }}_{\mathrm{y}=0}^{\mathrm{M}-\mathrm{y}}\frac{\left(2M-g-y\right)!}{\left(M-g-y\right)!2^{\left(M-g-y\right)}}\frac{\left(A+y+g\right)!}{\left(A+g\right)!y!g!}\stackrel{M≔M-g}{\to }$

$\mathrm{SUM}\left(\mathrm{g}+1,\left[\mathrm{A}+2,\mathrm{A}+4\dots \mathrm{A}+2\mathrm{M}\right]:3,\left[1,3\dots 2\mathrm{M}-1\right]\right)={{\displaystyle \sum }}_{\mathrm{y}=0}^{\mathrm{M}}\frac{\left(2M+g-y\right)!}{\left(M-y\right)!2^{M-y}}\frac{\left(A+y+g\right)!}{g!y!\left(\mathrm{A}+\mathrm{g}\right)!}\stackrel{N≔g+1}{\to }conclusion$

q.e.d.

The identity of PS=[K₁,K₂…K_M]:D similar to 5.7)-5.20) can also be obtained.

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

$6.1)\left|\left|{\mathrm{A}}_1\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)\left|\left|=\right|\right|{\mathrm{A}}_2\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)\left|\left|=\right|\right|{\mathrm{A}}_3\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)\right|\right|$

$6.2)\left|\left|\mathrm{A}\left(\mathrm{P},0,\mathrm{M}\right)\right|\right|=1,\left|\left|\mathrm{A}\left(\mathrm{P},1,\mathrm{M}\right)\right|\right|=\left(\begin{array}{c}\mathrm{P}+\mathrm{M}\\ 1+\mathrm{M}\end{array}\right),\left|\left|\mathrm{A}\left(\mathrm{P},\mathrm{Q}>0,\mathrm{M}\right)\right|\right|={{\displaystyle \prod }}_{g=0}^{Q-1}\frac{\left(\begin{array}{c}\mathrm{P}+\mathrm{M}-\mathrm{g}\\ 1+\mathrm{M}\end{array}\right)}{\left(\begin{array}{c}1+\mathrm{M}+\mathrm{g}\\ 1+\mathrm{M}\end{array}\right)}$

$6.3)\left|\left|\mathrm{A}\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)\right|\right|=\left|\left|\mathrm{A}\left(\mathrm{P},\mathrm{P}-\mathrm{Q},\mathrm{M}\right)\right|\right|$

${\mathrm{A}}_1\left(\mathrm{P},0,\mathrm{M}\right)=\left[\begin{array}{ccc}\left(\begin{array}{c}\mathrm{P}\\ 0\end{array}\right)& \cdots & 0\\ ⋮& \ddots & ⋮\\ \left(\begin{array}{c}\mathrm{P}+\mathrm{M}\\ \mathrm{M}\end{array}\right)& \cdots & \left(\begin{array}{c}\mathrm{P}+\mathrm{M}\\ 0\end{array}\right)\end{array}\right]=\left[\begin{array}{ccc}1& \cdots & 0\\ ⋮& \ddots & ⋮\\ \left(\begin{array}{c}\mathrm{P}+\mathrm{M}\\ \mathrm{M}\end{array}\right)& \cdots & 1\end{array}\right]$

${\mathrm{A}}_3\left(\mathrm{P},0,\mathrm{M}\right)=\left[\begin{array}{ccc}\left(\begin{array}{c}\mathrm{P}+\mathrm{M}\\ 0\end{array}\right)& \cdots & 0\\ ⋮& \ddots & ⋮\\ \left(\begin{array}{c}\mathrm{P}+2\mathrm{M}\\ \mathrm{M}\end{array}\right)& \cdots & \left(\begin{array}{c}\mathrm{P}+\mathrm{M}\\ 0\end{array}\right)\end{array}\right]=\left[\begin{array}{ccc}1& \cdots & 0\\ ⋮& \ddots & ⋮\\ \left(\begin{array}{c}\mathrm{P}+2\mathrm{M}\\ \mathrm{M}\end{array}\right)& \cdots & 1\end{array}\right]$

${\mathrm{A}}_2\left(1,0,\mathrm{M}\right)\mathrm{can}\mathrm{be}\mathrm{changed}\mathrm{to}\left[\begin{array}{ccc}\left(\begin{array}{c}0\\ 0\end{array}\right)& \cdots & \left(\begin{array}{c}\mathrm{M}\\ 0\end{array}\right)\\ ⋮& \ddots & ⋮\\ 0& \cdots & \left(\begin{array}{c}\mathrm{M}\\ \mathrm{M}\end{array}\right)\end{array}\right]$

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

Table 6. 1: Calculate H(g) with matrix,TM>=M.

Many conclusions of [8] can be obtained through the matrix. Use 5.16)-5.18) we can get some new identities.

Table 6. 1: Calculate H(g) with matrix,TM>=M.

Many conclusions of [8] can be obtained through the matrix. Use 5.16)-5.18) we can get some new identities.

6. Formal Calculation of Gaussian Coefficients

$\mathrm{Gaussian}\mathrm{Coefficients}:{\mathrm{G}}_M^N={\left[\begin{array}{c}N\\ M\end{array}\right]}_q=\frac{\left({\mathrm{q}}^N-1\right)\left({\mathrm{q}}^{N-1}-1\right)\dots \left({\mathrm{q}}^{N-\mathrm{M}+1}-1\right)}{\left({\mathrm{q}}^M-1\right)\left({\mathrm{q}}^{M-1}-1\right)\dots \left(\mathrm{q}-1\right)},q\ne 0,1$

$\left(1\right){\mathrm{G}}_0^N=1,{\mathrm{G}}_{M〈0orM〉N}^N=0,{\mathrm{G}}_M^N={\mathrm{G}}_{N-M}^N$

$\left(2\right){\mathrm{G}}_M^N=q^M{\mathrm{G}}_M^{N-1}+{\mathrm{G}}_{M-1}^{N-1}={\mathrm{G}}_M^{N-1}+q^{N-M}{\mathrm{G}}_{M-1}^{N-1}$

$\left(3\right){\mathrm{G}}_M^N={{\displaystyle \sum }}_{n=0}^{N-1}{q}^n{\mathrm{G}}_M^{n+K}=q^{M-K}{\mathrm{G}}_{M+1}^{N+K}$

  

Formal Calculation of Gaussian Coefficients has been obtained by [3].

$\mathrm{It}\mathrm{uses}{q}^n\left({\mathrm{K}}_{\mathrm{i}}+{\mathrm{G}}_1^{\mathrm{n}}\right){\mathrm{instead}\mathrm{of}\mathrm{K}}_{\mathrm{i}}+{\mathrm{q}}^{\mathrm{n}}andnestedsummationcanalsobeperformed.$

${▽}_q^{0}f\left(N\right)=f\left(N\right),{{\displaystyle \sum }}_{n=0}^{N-1}{q}^n{▽}_q^{1}f\left(n+1\right)=f\left(N\right),{{\displaystyle \sum }}_{n=0}^{N-1}{q}^{n}f\left(n+1\right)={▽}_q^{-1}f\left(N\right).Recursivedefine{▽}_q^P$

  

${\mathrm{SUM}}_q\left(\mathrm{N},\left[{\mathrm{K}}_1:{\mathrm{D}}_1\right],\left[{\mathrm{T}}_1=1\right]\right)={{\displaystyle \sum }}_{n=0}^{N-1}{q}^n\left({\mathrm{K}}_1+{\mathrm{D}}_1{\mathrm{G}}_1^n\right)$

${\mathrm{SUM}}_q\left(\mathrm{N},\left[{\mathrm{K}}_1:{\mathrm{D}}_1,{\mathrm{K}}_2:{\mathrm{D}}_2\right],\left[{\mathrm{T}}_1=1,{\mathrm{T}}_2={\mathrm{T}}_1+2-\mathrm{P}\right]\right)={{\displaystyle \sum }}_{n=0}^{N-1}{q}^n\left({\mathrm{K}}_2+{\mathrm{D}}_2{\mathrm{G}}_1^n\right){▽}_q^P{\mathrm{SUM}}_q\left(\mathrm{n}+1,\left[{\mathrm{K}}_1:{\mathrm{D}}_1\right],\left[1\right]\right)$

$Recursivedefine{\mathrm{SUM}}_q\left(\mathrm{N},PS,PT\right)$

  

$TheForm:\left({\mathrm{T}}_1+{\mathrm{K}}_1\right)\left({\mathrm{T}}_2+{\mathrm{K}}_2\right)\dots \left({\mathrm{T}}_M+{\mathrm{K}}_M\right)=\sum {{\displaystyle \prod }}_{i=1}^M{\mathrm{X}}_i$

$7.1){\mathrm{H}}^q\left(\mathrm{g}\right)={\mathrm{H}}^q\left(\mathrm{g},\mathrm{PS},\mathrm{PT}\right)={{\displaystyle \sum }}_{\mathrm{X}\left(\mathrm{T}\right)=g}{{\displaystyle \prod }}_{i=1}^M{B}_i,{\mathrm{SUM}}_q\left(N\right)=$

$\underset{q\to 1}{\lim }{H}^q\left(g\right)=H\left(g\right),\underset{q\to 1}{\lim }SU{M}_q\left(N\right)=SUM\left(N\right)$

$7.2){▽}_q^1{\mathrm{SUM}}_q\left(\mathrm{N},PS,\left[1,2\dots \mathrm{M}\right]\right)={{\displaystyle \prod }}_{i=1}^M\left({\mathrm{K}}_{\mathrm{i}}+{\mathrm{D}}_{\mathrm{i}}{\mathrm{G}}_1^n\right)$

$7.3)SU{M}_q\left(N,\left[\dots PS\dots \right],\left[\dots T+1,T+2\dots T+M\dots \right]\right),K_{i}canexchangeorder.$

  

Derived from (3), Form₂ is the simplest,Let X=T_M-M-p:

$7.4){▽}_q^{P}SUM\left(N\right)\stackrel{{\mathrm{Form}}_1}{\to }{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{H}}_1^{\mathrm{q}}\left(\mathrm{g}\right){\mathrm{G}}_{\mathrm{X}+1+\mathrm{g}}^{\mathrm{N}+\mathrm{X}}{\mathrm{q}}^{-\mathrm{gp}}\stackrel{{\mathrm{Form}}_2}{\to }{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{H}}_2^{\mathrm{q}}\left(\mathrm{g}\right){\mathrm{G}}_{\mathrm{X}+1+\mathrm{g}}^{\mathrm{N}+\mathrm{X}+\mathrm{g}}\stackrel{{\mathrm{Form}}_3}{\to }{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{H}}_3^{\mathrm{q}}\left(\mathrm{g}\right){\mathrm{G}}_{X+M+1}^{\mathrm{N}+\mathrm{X}+\mathrm{M}-\mathrm{g}}{\mathrm{q}}^{-\mathrm{gp}}$

  

$\mathrm{PS}=\left[q^{-T_1}{G}_1^{T_1},q^{-T_2}{G}_1^{T_2}\dots q^{-T_{\mathrm{M}}}{G}_1^{T_{\mathrm{M}}}\right],\to {\mathrm{H}}_2^{\mathrm{q}}\left(\mathrm{g}<\mathrm{M}\right)=0\to SU{M}_q\left(N\right)={\displaystyle \prod }_{i=1}^M{q}^{-T_i}{G}_1^{T_i}{G}_{T_M+1}^{N+T_M}$

${\mathrm{define}\mathrm{n}}_{q-}=q^{-\mathrm{n}}{G}_1^{\mathrm{n}},{\mathrm{n}}_{q+}=q^{\mathrm{n}}{G}_1^{\mathrm{n}},{\mathrm{n}}_{q-orq+}!=1_q{2}_q\dots {\mathrm{n}}_q,0_{q-}!=0_{q+}!=1,{\mathrm{Form}}_2\to$

$7.5)SU{M}_q\left(N,\left[{\mathrm{L}}_1{}_{q-},{\mathrm{L}}_2{}_{q-}\dots {\mathrm{L}}_{\mathrm{p}}{}_{q-},PS\right],\left[L_1,L_2\dots L_P,PT\right]\right)={\displaystyle \prod }_{i=1}^P{\mathrm{L}}_{\mathrm{i}}{}_{q-}SU{M}_q\left(N,PS,PT\right),similarto1.4)$

This indicates T₁ can be greater than 1, T is defined in ℕ.

$7.6)SU{M}_q\left(N,\left[{\mathrm{T}}_1{}_{q-},{\mathrm{T}}_2{}_{q-}\dots {\mathrm{T}}_{\mathrm{M}}{}_{q-}\right],\mathrm{PT}\right)={\displaystyle \prod }_{i=1}^M{\mathrm{T}}_{\mathrm{i}}{}_{q-}{G}_{T_M+1}^{N+T_M},similar to 1.8)$

  

$7.7)G_{\mathrm{M}+1}^{N+\mathrm{M}}=q^{\left(\begin{array}{c}\mathrm{M}+1\\ 2\end{array}\right)}{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}\left\{q^{\left(\begin{array}{c}\mathrm{g}+1\\ 2\end{array}\right)-\left(\mathrm{M}+1\right)\left(\mathrm{M}-\mathrm{g}\right)}{{\displaystyle \sum }}_{1\le i_1<i_2<\mathrm{...}<i_{M-g}\le M}{q}^{i_1}{q}^{i_2}\dots q^{i_{M-g}}\right\}{\mathrm{G}}_{1+\mathrm{g}}^{\mathrm{N}}$

[Proof]

It is difficult to derive directly from (2).

$SUM\left(N,\left[1_{q-},2_{q-}\dots {\mathrm{M}}_{q-}\right],\left[1,2\dots \mathrm{M}\right]\right)\stackrel{7.6)}{\to }{M}_{q-}!G_{\mathrm{M}+1}^{N+\mathrm{M}}=SUM\left(N,\left[{\mathrm{M}}_{q-}\dots 1_{q-}\right],\left[1,2\dots \mathrm{M}\right]\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{H}}_1^{\mathrm{q}}\left(\mathrm{g}\right){\mathrm{G}}_{1+\mathrm{g}}^{\mathrm{N}}$

$B_i\stackrel{For{m}_1}{\to }\left\{\begin{array}{c}{q}^{X_{T-1}+1}{G}_1^{T_i-X_{K-1}}=q^{X_{T-1}+1}{G}_1^{i-X_{K-1}}=q^{X_{T-1}+1}{G}_1^{X_{T-1}+1}=q^{X_T}{G}_1^{X_T};X_i=T_i\\ K_i+G_1^{X_{T-1}}=q^{-\left(M+1-i\right)}{G}_1^{M+1-i}+G_1^{X_{T-1}}=\frac{q^{\mathrm{M}+1-\mathrm{i}+{\mathrm{X}}_{\mathrm{T}-1}}-1}{q^{\mathrm{M}+1-\mathrm{i}}\left(q-1\right)}=\frac{q^{\mathrm{M}-{\mathrm{X}}_{\mathrm{K}-1}}-1}{q^{\mathrm{M}+1-\mathrm{i}}\left(q-1\right)};X_i=K_i \end{array}\right.$

$H_1^q\left(g,T\right)=g_{q+}!;H_1^q\left(g,\sum K\right)=G_1^M{G}_1^{M-1}\dots G_1^{g+1}{q}^{-\left(\mathrm{M}+1\right)\left(\mathrm{M}-\mathrm{g}\right)}{{\displaystyle \sum }}_{1\le i_1<i_2<\mathrm{...}<i_{M-g}\le M}{q}^{i_1}{q}^{i_2}\dots q^{i_{M-g}}$

$\frac{{\mathrm{H}}_1^q\left(g,\left[{\mathrm{M}}_{q-}\dots 2_{q-},1_{q-}\right],\left[1,2\dots \mathrm{M}\right]\right)}{M_{q-}!}=\frac{q^{1+2+\dots g}{q}^{1+2+\dots M}{{\displaystyle \sum }}_{1\le i_1<i_2<\mathrm{...}<i_{M-g}\le M}{q}^{i_1}{q}^{i_2}\dots q^{i_{M-g}}}{q^{\left(\mathrm{M}+1\right)\left(\mathrm{M}-\mathrm{g}\right)}}$

q.e.d.

$G_{\mathrm{M}+1}^{N+\mathrm{M}}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}f\left(g\right){\mathrm{G}}_{1+\mathrm{g}}^{\mathrm{N}}\to f\left(0\right)=1, f\left(M\right)=q^{\mathrm{M}\left(\mathrm{M}+1\right)}$

$G_3^{N+2}=q^6{\mathrm{G}}_3^{\mathrm{N}}+q\left(q^1+q^2\right){\mathrm{G}}_2^{\mathrm{N}}+{\mathrm{G}}_1^{\mathrm{N}}$  $G_4^{N+3}=q^{12}{\mathrm{G}}_4^{\mathrm{N}}+q^5\left(q^1+q^2+q^3\right){\mathrm{G}}_3^{\mathrm{N}}+q^{-1}\left(q^1{q}^2+q^1{q}^3+q^2{q}^3\right){\mathrm{G}}_2^{\mathrm{N}}+{\mathrm{G}}_1^{\mathrm{N}}$

$=q^{12}{\mathrm{G}}_4^{\mathrm{N}}+\left(q^6+q^7+q^8\right){\mathrm{G}}_3^{\mathrm{N}}+\left(q^2+q^3+q^4\right){\mathrm{G}}_2^{\mathrm{N}}+{\mathrm{G}}_1^{\mathrm{N}}=Equationderivedfrom\left(2\right)$

  

$E_M^N={\displaystyle \sum }_{{\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_N=M,{\lambda }_{\mathrm{i}}\ge 0}{1}^{{\lambda }_1}{2}^{{\lambda }_2}\mathrm{..}{N}^{{\lambda }_N}=S_2\left(\mathrm{N}+\mathrm{M},\mathrm{N}\right)$

$define{E}_q{}_M^N={\displaystyle \sum }_{{\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_N=M,{\lambda }_{\mathrm{i}}\ge 0}{G}_1^1{}^{{\lambda }_1}{G}_1^2{}^{{\lambda }_2}\dots G_1^{\mathrm{N}}{}^{{\lambda }_N}=S_2^q\left(\mathrm{N}+\mathrm{M},\mathrm{N}\right)$

$define{E}_{q-}{}_M^N={\displaystyle \sum }_{{\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_N=M,{\lambda }_{\mathrm{i}}\ge 0}{G}_1^1{}^{{\lambda }_1}{G}_1^2{}^{{\lambda }_2}\dots G_1^{\mathrm{N}}{}^{{\lambda }_N}{q}^{-\left({\lambda }_1+2{\lambda }_2+\mathrm{...}+N{\lambda }_{\mathrm{N}}\right)}$

$define{E}_q{}_M^N\odot \left(\left[T_1,T_2\dots \right],K\right)={\displaystyle \sum }_{{\lambda }_1\mathrm{..}+{\lambda }_N=M,{\lambda }_i\ge 0}{G}_1^1{}^{{\lambda }_1}\dots G_1^{\mathrm{N}}{}^{{\lambda }_N}{G}_1^{T_1+{\lambda }_{1}K}{G}_1^{T_2+{\lambda }_{1}K+{\lambda }_{2}K}\dots G_1^{T_{q-1}+{\lambda }_{1}K+{\lambda }_{2}K+\mathrm{..}+{\lambda }_{N-1}K}$

Table 7. 1: $S_2^q$(N, M).

Table 7. 1: $S_2^q$(N, M).

$7.8){\left(G_1^N\right)}^M={{\displaystyle \sum }}_{g=1}^M{g}_{q+}!S_2^q\left(M,g\right)G_g^N{q}^{-\mathrm{g}},{\mathrm{similar}\mathrm{to}\mathrm{N}}^{\mathrm{M}}={{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{M}}\mathrm{g}!{\mathrm{S}}_2\left(\mathrm{M},\mathrm{g}\right)\left(\begin{array}{c}\mathrm{N}\\ \mathrm{g}\end{array}\right)$

[Proof]

$PS=\left[1_{q-},1_{q-}\dots 1_{q-}\right],PT=\left[2,3\dots M\right],D_i=1,B_i\stackrel{For{m}_1}{\to }\left\{\begin{array}{c}{q}^{X_{T-1}+1}{G}_1^{T_i-X_{K-1}};X_i=T_i\\ 1_{q-}+G_1^{X_{T-1}}=q^{-1}{G}_1^{X_{T-1}+1};X_i=K_i \end{array}\right.$

$H_1^q\left(g,T\right)=\frac{{\left(g+1\right)}_{q+}!}{q^{\mathrm{g}+1}},H_1^q\left(g,\sum K\right)=q^{-\left(\mathrm{M}-1-\mathrm{g}\right)}{E}_q{}_{M-1-g}^{g+1}\to H_1^q\left(g\right)=q^{-\mathrm{M}}{\left(g+1\right)}_{q+}!E_q{}_{M-1-g}^{g+1}$

$1_{q-}+G_1^n=\frac{1}{\mathrm{q}}+\frac{q^{\mathrm{n}}-1}{\mathrm{q}-1}=q^{-1}{G}_1^{n+1}\to$

$q^{-1}{▽}_q^{1}SU{M}_q\left(N,\left[1_{q-},1_{q-}\dots 1_{q-}\right],\left[2,\mathrm{3...}M\right]\right)\stackrel{7.5)}{\to }{▽}_q^{1}SU{M}_q\left(N,\left[1_{q-},1_{q-}\dots 1_{q-}\right],\left[1,2\dots M\right]\right)={\left(q^{-1}{G}_1^N\right)}^M$

$=q^{-1}{{\displaystyle \sum }}_{g=0}^{M-1}{H}_1^q\left(g\right)G_{1+g}^N{q}^{-g}={{\displaystyle \sum }}_{g=0}^{M-1}{q}^{-\mathrm{M}}{\left(g+1\right)}_{q+}!S_2^q\left(M,g+1\right)G_{1+g}^N{q}^{-g}$

${\left(G_1^N\right)}^M=q^{\mathrm{M}-1}{{\displaystyle \sum }}_{g=0}^{M-1}{q}^{-\mathrm{M}}{\left(g+1\right)}_{q+}!S_2^q\left(M,g+1\right)G_{1+g}^N{q}^{-g}$

q.e.d.

Table 7. 2 H(g) of PS=[1,1…1],PT=[1,2…M].

Table 7. 2 H(g) of PS=[1,1…1],PT=[1,2…M].

[Proof of (*)]

$H_3^q\left(g\right)\to B_i\stackrel{For{m}_3}{\to }\left\{\begin{array}{c}q\left\{\left(q^{X_{T-1}}{G}_1^{T_i}-q^{T_i}{G}_1^{X_{T-1}}\right)-1_{q-}{q}^{T_i}\right\};X_i=T_i\\ 1_{q-}+G_1^{X_{T-1}}=q^{-1}{G}_1^{X_{T-1}+1};X_i=K_i\end{array}\right.$

$H_3^q\left(g,\sum K\right)={\mathrm{q}}^{-\left(\mathrm{M}-\mathrm{g}\right)}{E}_q{}_{M-g}^{g+1}$

$H_3^q\left(g,T\right)=\prod q\left\{\left(q^{X_{T-1}}{G}_1^i-q^i{G}_1^{X_{T-1}}\right)-{\mathrm{q}}^{-1}{q}^i\right\}=\prod {\left(\mathrm{q}-1\right)}^{-1}\mathrm{q}\left\{{\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}-1}}\left({\mathrm{q}}^{\mathrm{i}}-1\right)-{\mathrm{q}}^{\mathrm{i}}\left({\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}-1}}-1\right)-{\mathrm{q}}^{\mathrm{i}-1}\left(\mathrm{q}-1\right)\right\}$

$=\prod \frac{q^i-q^{1+X_{T-1}}}{\mathrm{q}-1}=\prod q^{1+X_{T-1}}\frac{q^{X_{K-1}}-1}{\mathrm{q}-1}=\prod q^{X_T}\frac{q^{X_{K-1}}-1}{\mathrm{q}-1}=q^{1+2+\dots g}\prod G_1^{X_{K-1}}$

$\prod G_1^{X_{K-1}}={\mathrm{G}}_1^{{\lambda }_1}{\mathrm{G}}_1^{{\lambda }_1+{\lambda }_2}\dots {\mathrm{G}}_1^{{\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_{g-1}}$

q.e.d.

$7.9)\mathrm{Induction}\to \mathrm{PT}=\left[1,2\dots \mathrm{M}\right],q^{-\left(\begin{array}{c}g+1\\ 2\end{array}\right)}{H}_1^q\left(\mathrm{g}\right)=q^{\left(\begin{array}{c}g+1\\ 2\end{array}\right)}{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{H}_2^q\left(\mathrm{k}\right){\mathrm{G}}_{\mathrm{g}}^{\mathrm{k}}$

$7.10)\mathrm{Induction}\to H_1^q\left(0\right)={{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{H}_2^q\left(\mathrm{k}\right)$

Other situations are relatively complex and no conclusions similar to 2.2) have been drawn.

  

$SU{M}_q\left(N,\left[1_{q-},1_{q-}\dots 1_{q-}\right],\left[1,3\dots 2M-1\right]\right)$

$={{\displaystyle \sum }}_{n_M=0}^{N-1}{q}^{n_M-1}{G}_1^{n_M+1}\dots {{\displaystyle \sum }}_{n_2=0}^{n_3}{q}^{n_2-1}{G}_1^{n_2+1}{{\displaystyle \sum }}_{n_1=0}^{n_2}{q}^{n_1-1}{G}_1^{n_1+1}$

$=q^{-2M}{{\displaystyle \sum }}_{1\le I_1\le I_2\le \mathrm{...}\le I_M\le N}{G}_1^{I_1}{G}_1^{I_2}\dots G_1^{I_M}{q}^{I_1+I_2+\mathrm{...}+I_M}=q^{-2M}{\displaystyle \sum }_{{\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_N=M,{\lambda }_i\ge 0}{G}_1^1{}^{{\lambda }_1}{G}_1^2{}^{{\lambda }_2}\dots G_1^{\mathrm{N}}{}^{{\lambda }_N}{q}^{{\lambda }_1+2{\lambda }_2+\mathrm{...}+N{\lambda }_N}$

  

$SU{M}_q\left(N,\left[1_{q-},2_{q-}\dots M_{q-}\right],\left[1,3\dots 2M-1\right]\right)$

$={{\displaystyle \sum }}_{n_M=0}^{N-1}{q}^{n_M-M}{G}_1^{n_M+M}\dots {{\displaystyle \sum }}_{n_2=0}^{n_3}{q}^{n_2-2}{G}_1^{n_2+2}{{\displaystyle \sum }}_{n_1=0}^{n_2}{q}^{n_1-1}{G}_1^{n_1+1}$

$=q^{-\left(M+1\right)M}{{\displaystyle \sum }}_{1\le I_1<I_2<\mathrm{...}<I_M\le N+M-1}{G}_1^{I_1}{G}_1^{I_2}\dots G_1^{I_M}{q}^{I_1+I_2+\mathrm{...}+I_M}$

  

$7.11){\mathrm{P}}_1〈{\mathrm{T}}_1,{\mathrm{P}}_2〈{\mathrm{T}}_2,{\mathrm{T}}_1〉0,{\mathrm{T}}_2〉0,\mathrm{PT}=\left[1,2,3\dots {\mathrm{T}}_1+{\mathrm{T}}_2+2-{\mathrm{P}}_1-{\mathrm{P}}_2\right]$

$\mathrm{PS}=\left[1_{q-},2_{q-}\dots {\left({\mathrm{T}}_1-{\mathrm{P}}_1\right)}_{q-},\frac{{\mathrm{K}}_1{\mathrm{G}}_1^{{\mathrm{T}}_1+1-{\mathrm{P}}_1}}{{\mathrm{q}}^{1-{\mathrm{P}}_1}{\mathrm{G}}_1^{{\mathrm{T}}_1}{\mathrm{D}}_1},1_{q-},2_{q-}\dots {\left({\mathrm{T}}_2-{\mathrm{P}}_2\right)}_{q-},\frac{{\mathrm{K}}_2{\mathrm{G}}_1^{{\mathrm{T}}_2+1-{\mathrm{P}}_2}}{{\mathrm{q}}^{1-{\mathrm{P}}_2}{\mathrm{G}}_1^{{\mathrm{T}}_2}{\mathrm{D}}_2}\right]$

${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}{▽}^{{\mathrm{P}}_1}{\mathrm{SUM}}_{\mathrm{q}}\left(\mathrm{n}+1,\left[{\mathrm{K}}_1:{\mathrm{D}}_1\right],\left[{\mathrm{T}}_1\right]\right){▽}^{{\mathrm{P}}_2}{\mathrm{SUM}}_{\mathrm{q}}\left(\mathrm{n}+1,\left[{\mathrm{K}}_2:{\mathrm{D}}_2\right],\left[{\mathrm{T}}_2\right]\right)$

$=\frac{1}{{\left({\mathrm{T}}_1-{\mathrm{P}}_1\right)}_{q-}!}\frac{{\mathrm{q}}^{1-{\mathrm{P}}_1}{\mathrm{G}}_1^{{\mathrm{T}}_1}{\mathrm{D}}_1}{ {\mathrm{G}}_1^{{\mathrm{T}}_1-{\mathrm{P}}_1+1}}\frac{1}{{\left({\mathrm{T}}_2-{\mathrm{P}}_2\right)}_{q-}!}\frac{{\mathrm{q}}^{1-{\mathrm{P}}_2}{\mathrm{G}}_1^{{\mathrm{T}}_2}{\mathrm{D}}_2}{{\mathrm{G}}_1^{{\mathrm{T}}_2-{\mathrm{P}}_2+1}}{\mathrm{SUM}}_{\mathrm{q}}\left(\mathrm{N},\mathrm{PS},\mathrm{PT}\right)$

  

$7.12){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}{{\displaystyle \prod }}_{i=1}^M\left({\mathrm{K}}_{\mathrm{i}}+{\mathrm{q}}^{\mathrm{n}}{\mathrm{D}}_{\mathrm{i}}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{H}_1^q\left(g\right)G_{\mathrm{g}+1}^{\mathrm{N}},{\mathrm{B}}_{\mathrm{i}}=\left\{\begin{array}{c}{q}^{X_T}\left(q^{X_T}-1\right)D_i;X\in T\\ K_i+q^{X_T}{D}_i;X\in K \end{array}\right.$

$7.13)\mathrm{Induction}\to {{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{{\displaystyle \prod }}_{i=1}^M\left({\mathrm{K}}_{\mathrm{i}}+{\mathrm{q}}^{\mathrm{n}}{\mathrm{D}}_{\mathrm{i}}\right)={{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{M}}f\left(g\right)G_{\mathrm{g}}^{\mathrm{N}}+N{{\displaystyle \prod }}_{i=1}^M{\mathrm{K}}_{\mathrm{i}}$

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

  

Similar to Section 6, P=N+T_M-M,Q=N-1,define A^q(P,Q,M)

$A_1^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)=\left[\begin{array}{ccc}{G}_Q^P& \cdots & G_{Q-M}^P\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+M}& \cdots & G_Q^{P+M}\end{array}\right],A_2^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)=\left[\begin{array}{ccc}{G}_Q^P& \cdots & G_Q^{P+M}\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+M}& \cdots & G_{Q+M}^{P+2M}\end{array}\right],A_3^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)=\left[\begin{array}{ccc}{G}_Q^{P+M}& \cdots & G_{Q-M}^P\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+2M}& \cdots & G_Q^{P+M}\end{array}\right]$

  

Turn it into an upper triangular matrix:

$7.14)\left|\left|A_2^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)\right|\right|=\frac{G_Q^P}{G_0^P}\frac{G_{Q+1}^{P+2}}{G_1^{P+2}}\frac{G_{Q+2}^{P+4}}{G_2^{P+4}}\dots \frac{G_{Q+M}^{P+2M}}{G_M^{P+2M}}{\mathrm{q}}^{\left(P+1\right)+2\left(P+2\right)+\mathrm{...}+M\left(P+M\right)}=\left|\left|A_2^q\left(\mathrm{P},\mathrm{P}-\mathrm{Q},\mathrm{M}\right)\right|\right|$

$7.15)\left|\left|A_2^q\left(\mathrm{P},0,\mathrm{M}\right)\right|\right|={\mathrm{q}}^{\left(P+1\right)+2\left(P+2\right)+\mathrm{...}+M\left(P+M\right)};\left|\left|A_2^q\left(\mathrm{P},1,\mathrm{M}\right)\right|\right|=G_{1+M}^{P+M}{\mathrm{q}}^{\left(P+1\right)+2\left(P+2\right)+\mathrm{...}+M\left(P+M\right)}$

  

$A_2^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)\stackrel{{\mathrm{Row}}_{\mathrm{i}+1}:={\mathrm{Row}}_{\mathrm{i}+1}-{\mathrm{Row}}_{\mathrm{i}}\mathrm{and}\mathrm{repeat}}{\to }\mathrm{Change}\mathrm{the}\mathrm{first}\mathrm{column}\mathrm{to}\mathrm{to}{\left[G_Q^P,q^{Q+1}{G}_{Q+1}^P,q^{2\left(Q+2\right)}{G}_{Q+2}^P\dots \right]}^T$

${\mathrm{q}}^{\left(Q+1\right)+2\left(Q+2\right)\dots +M\left(Q+M\right)}\left[\begin{array}{ccc}{G}_Q^P& \cdots & G_Q^{P+M}\\ ⋮& \ddots & ⋮\\ G_{Q+M}^P& \cdots & G_{Q+M}^{P+M}\end{array}\right]\stackrel{\mathrm{transpose}}{\to }{\mathrm{q}}^{\left\{\dots \right\}}\left[\begin{array}{ccc}{G}_Q^P& \cdots & G_{Q+M}^P\\ ⋮& \ddots & ⋮\\ G_Q^{P+M}& \cdots & G_{Q+M}^{P+M}\end{array}\right]={\mathrm{q}}^{\left\{\dots \right\}}{\mathrm{A}}_1\left(\mathrm{P},\mathrm{P}-\mathrm{Q},\mathrm{M}\right)\to$

$\left|\left|A_1^q\left(\mathrm{P},\mathrm{P}-\mathrm{Q},\mathrm{M}\right)\right|\right|=\frac{G_Q^P}{G_0^P}\frac{G_{Q+1}^{P+2}}{G_1^{P+2}}\frac{G_{Q+2}^{P+4}}{G_2^{P+4}}\dots \frac{G_{Q+M}^{P+2M}}{G_M^{P+2M}}{\mathrm{q}}^{\left(P-Q\right)+2\left(P-Q\right)+\mathrm{...}+M\left(P-Q\right)}$

  

$A_1^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)=\left[\begin{array}{ccc}{G}_Q^P& \cdots & G_{Q-M}^P\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+M}& \cdots & G_Q^{P+M}\end{array}\right]\stackrel{use\left(2\right),co{l}_{\mathrm{i}}=co{l}_{\mathrm{i}}+{\mathrm{q}}^{x}co{l}_{\mathrm{i}+1}\mathrm{and}\mathrm{repeat}}{\to }\left[\begin{array}{ccc}{G}_Q^{P+M}& \cdots & G_{Q-M}^P\\ ⋮& \ddots & ⋮\\ G_{Q+M}^{P+2M}& \cdots & G_Q^{P+M}\end{array}\right]=A_3^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)$

$7.16)\left|\left|A_{1,3}^q\left(\mathrm{P},\mathrm{Q},\mathrm{M}\right)\right|\right|=\frac{G_Q^P}{G_0^P}\frac{G_{Q+1}^{P+2}}{G_1^{P+2}}\frac{G_{Q+2}^{P+4}}{G_2^{P+4}}\dots \frac{G_{Q+M}^{P+2M}}{G_M^{P+2M}}{\mathrm{q}}^{Q\left(\begin{array}{c}M+1\\ 2\end{array}\right)};\left|\left|A_{1,3}^q\left(\mathrm{P},0,\mathrm{M}\right)\right|\right|=1;\left|\left|A_{1,3}^q\left(\mathrm{P},1,\mathrm{M}\right)\right|\right|={\mathrm{q}}^{\left(\begin{array}{c}M+1\\ 2\end{array}\right)}{G}_{1+M}^{P+M}$

$So H^q\left(\mathrm{g}\right)\mathrm{can}\mathrm{be}\mathrm{calculated}\mathrm{by}\mathrm{the}\mathrm{matrix}.$

7. Application of SUM_q(N) and H^q(g)

$8.1)q^{Mn}={{\displaystyle \sum }}_{g=0}^M{q}^{-\left(M-g\right)}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^{-1}}{{\displaystyle \prod }}_{i=1}^g\left(q^n-q^{-i}\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{{\displaystyle \prod }}_{i=0}^{g-1}\left(q^n-q^i\right)$

$={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{\left(-1\right)}^g{q}^{\left(\begin{array}{c}\mathrm{g}\\ 2\end{array}\right)}{G}_{\mathrm{M}}^{\mathrm{n}+\mathrm{M}-\mathrm{g}}={\mathrm{q}}^{-\left(\begin{array}{c}M+1\\ 2\end{array}\right)}{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{q}}^{\left(\begin{array}{c}M-g\\ 2\end{array}\right)}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{{\displaystyle \prod }}_{i=0}^{g-1}\left({\mathrm{q}}^{n+g-i}-1\right)$

[Proof]

$q^{Mn}=▽SU{M}_q\left(N,\left[1,1\dots 1\right]:q-1,\left[1,2\dots \mathrm{M}\right]\right)$

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

$H_2^q\left(g,T\right)={\left(q-1\right)}^{\mathrm{g}}{\mathrm{g}}_{\mathrm{q}-}!$

$H_2^q\left(g,\sum K\right)=\sum \prod \left\{1-\frac{q^{i-{\mathrm{X}}_{\mathrm{K}-1}}-1}{q^{i-{\mathrm{X}}_{\mathrm{K}-1}}\left(q-1\right)}\left(q-1\right)\right\}=\sum \prod \frac{1}{q^{i-{\mathrm{X}}_{\mathrm{K}-1}}}=\sum \prod \frac{1}{q^{1+{\mathrm{X}}_{\mathrm{T}-1}}}=q^{-\left(M-g\right)}\sum \prod \frac{1}{q^{{\mathrm{X}}_{\mathrm{T}}}}$

MacMahon [9] ${\mathrm{observes}\mathrm{G}}_{\mathrm{K}}^{\mathrm{M}}={\displaystyle \sum }_{w\in \Omega \left(0^{M-K},1^K\right)}{q}^{inv\left(w\right)},\mathrm{that}\mathrm{is},\mathrm{all}\mathrm{words}\mathrm{w}={\mathrm{w}}_1{\mathrm{w}}_2\dots {\mathrm{w}}_M\mathrm{with}\mathrm{M}-\mathrm{K}\mathrm{zeroes}\mathrm{and}\mathrm{K}\mathrm{ones},\mathrm{and}\mathrm{inv}\left(·\right)\mathrm{denotes}\mathrm{the}\mathrm{inversion}\mathrm{statistic}\mathrm{defined}$

$\to H_2^q\left(g,\sum K\right)=q^{-\left(M-g\right)}{\left[\begin{array}{c}M\\ M-g\end{array}\right]}_{q^{-1}}=q^{-\left(M-g\right)}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^{-1}}\to \mathrm{The}\mathrm{first}\mathrm{equation}.$

$H_1^q\left(g,T\right)={\left(q-1\right)}^{\mathrm{g}}{\mathrm{g}}_{\mathrm{q}+}!;H_1^q\left(g,\sum K\right)=\sum \left\{1+\frac{q^{{\mathrm{X}}_{\mathrm{T}-1}}-1}{\left(q-1\right)}\left(q-1\right)\right\}=\sum \prod {\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}-1}}=\sum \prod {\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}}}={\mathrm{G}}_{\mathrm{M}-\mathrm{g}}^{\mathrm{M}}={\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}$

$q^{Mn}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(q-1\right)}^{\mathrm{g}}{\mathrm{g}}_{\mathrm{q}+}!{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{q}^{-\mathrm{g}}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{n}}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{q}^{-\mathrm{g}}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{q}^{\left(\begin{array}{c}\mathrm{g}+1\\ 2\end{array}\right)}{{\displaystyle \prod }}_{i=0}^{g-1}\left(q^{n-i}-1\right)\to The \sec \mathrm{ond}\mathrm{equation}$

$H_3^q\left(g,\sum K\right)={\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}},H_3^q\left(g,T\right)={\left(-1\right)}^g{q}^{\left(\begin{array}{c}\mathrm{g}+1\\ 2\end{array}\right)},q^{Mn}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{\left(-1\right)}^g{q}^{\left(\begin{array}{c}\mathrm{g}+1\\ 2\end{array}\right)}{q}^{-\mathrm{g}}{G}_{\mathrm{M}}^{\mathrm{n}+\mathrm{M}-\mathrm{g}}\to Thethird\mathrm{equation}$

  

$▽SU{M}_q\left(N,\left[{\mathrm{q}}^1:\left(q-1\right){\mathrm{q}}^1,{\mathrm{q}}^2:\left(q-1\right){\mathrm{q}}^2\dots {\mathrm{q}}^M:\left(q-1\right){\mathrm{q}}^M\right],\left[1,2\dots M\right]\right)=q^{n+1}{q}^{n+2}\mathrm{..}{q}^{n+M}=q^{Mn+\left(\begin{array}{c}M+1\\ 2\end{array}\right)}$

$B_i\stackrel{For{m}_2}{\to }\left\{\begin{array}{c}{\mathrm{q}}^{-\left({\mathrm{T}}_{\mathrm{i}}-{\mathrm{X}}_{\mathrm{K}-1}\right)}\left\{{\mathrm{q}}^{\left({\mathrm{T}}_{\mathrm{i}}-{\mathrm{X}}_{\mathrm{K}-1}\right)}-1\right\}{\mathrm{q}}^{T_i}={\mathrm{q}}^{T_i}-{\mathrm{q}}^{{\mathrm{X}}_{\mathrm{K}-1}}={\mathrm{q}}^{{\mathrm{X}}_{\mathrm{K}-1}}\left({\mathrm{q}}^{T_i-{\mathrm{X}}_{\mathrm{K}-1}}-1\right);X_i=T_i\\ {\mathrm{K}}_{\mathrm{i}}-{\mathrm{q}}^{-\left({\mathrm{T}}_{\mathrm{i}}-{\mathrm{X}}_{\mathrm{K}-1}\right)}{\mathrm{G}}_1^{{\mathrm{T}}_{\mathrm{i}}-{\mathrm{X}}_{\mathrm{K}-1}}{\mathrm{D}}_{\mathrm{i}}={\mathrm{q}}^{{\mathrm{X}}_{\mathrm{K}-1}};X_i=K_i \end{array}\right.$

$H_2^q\left(g,K\right)={\mathrm{q}}^0{\mathrm{q}}^1\dots {\mathrm{q}}^{\mathrm{M}-\mathrm{g}-1}={\mathrm{q}}^{\left(\begin{array}{c}M-g\\ 2\end{array}\right)};H_2^q\left(g,\sum T\right)={{\displaystyle \prod }}_{i=0}^g\left({\mathrm{q}}^i-1\right){\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}\to Thefouth\mathrm{equation}$

  

$B_i\stackrel{For{m}_3}{\to }\left\{\begin{array}{c}{\mathrm{q}}^1\left\{\left({\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}-1}}{\mathrm{G}}_1^{{\mathrm{T}}_{\mathrm{i}}}-{\mathrm{q}}^{{\mathrm{T}}_{\mathrm{i}}}{\mathrm{G}}_1^{{\mathrm{X}}_{\mathrm{T}-1}}\right){\mathrm{D}}_{\mathrm{i}}-{\mathrm{K}}_{\mathrm{i}}{\mathrm{q}}^{{\mathrm{T}}_{\mathrm{i}}}\right\}=-{\mathrm{q}}^{i+{\mathrm{X}}_{\mathrm{T}}};X_i=T_i\\ {\mathrm{K}}_{\mathrm{i}}+{\mathrm{G}}_1^{{\mathrm{X}}_{\mathrm{T}-1}}{\mathrm{D}}_{\mathrm{i}}={\mathrm{q}}^{i+{\mathrm{X}}_{\mathrm{T}-1}}={\mathrm{q}}^{i+{\mathrm{X}}_{\mathrm{T}}};X_i=K_i \end{array},Every item of {\mathrm{H}}_3^{\mathrm{q}}\left(\mathrm{g}\right) has {\mathrm{q}}^{1+2+\dots +M}\right.$

$\prod \left(X\in T\right)of{\mathrm{H}}_3^{\mathrm{q}}\left(\mathrm{g}\right)q^{-\left(\begin{array}{c}M+1\\ 2\end{array}\right)}={\left(-1\right)}^g{\mathrm{q}}^{1+2+\mathrm{...}+g},\sum \prod \left(X\in K\right)of{\mathrm{H}}_3^{\mathrm{q}}\left(\mathrm{g}\right)q^{-\left(\begin{array}{c}M+1\\ 2\end{array}\right)}=\sum \prod {\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}}}={\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}$

$q^{Mn}=q^{-\left(\begin{array}{c}M+1\\ 2\end{array}\right)}{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^g{q}^{\left(\begin{array}{c}M+1\\ 2\end{array}\right)}{q}^{\left(\begin{array}{c}\mathrm{g}+1\\ 2\end{array}\right)}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{G}_{\mathrm{M}}^{\mathrm{n}+\mathrm{M}-\mathrm{g}}{q}^{-g}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{G}}_{\mathrm{g}}^{\mathrm{M}}{\left(-1\right)}^g{q}^{\left(\begin{array}{c}\mathrm{g}\\ 2\end{array}\right)}{G}_{\mathrm{M}}^{\mathrm{n}+\mathrm{M}-\mathrm{g}}=Thethird\mathrm{equation}$

q.e.d.

  

$SU{M}_q\left(N,\left[1,1\dots 1\right]:0,\left[1,3\dots 2\mathrm{M}-1\right]\right)\stackrel{For{m}_2}{\to }{\mathrm{G}}_{\mathrm{M}}^{\mathrm{N}+\mathrm{M}-1}\to {\displaystyle \sum }_{0\le {\lambda }_1\le \mathrm{...}\le {\lambda }_M\le N}{\mathrm{q}}^{{\lambda }_1+\mathrm{...}+{\lambda }_M}={\mathrm{G}}_{\mathrm{M}}^{\mathrm{N}+\mathrm{M}}\to record at$ [6]: 3.1

$8.2){\displaystyle \sum }_{0\le {\lambda }_1\le {\lambda }_2\le \mathrm{...}\le {\lambda }_N\le M-N}{\mathrm{q}}^{{\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_N}={\mathrm{G}}_{\mathrm{M}-\mathrm{N}}^{\mathrm{M}}={\mathrm{G}}_{\mathrm{N}}^{\mathrm{M}}={\displaystyle \sum }_{w\in \Omega \left(0^N,1^{M-N}\right)}{q}^{inv\left(w\right)}$

$8.3){\left[\begin{array}{c}N+M-1\\ M\end{array}\right]}_{q^2}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^g{q}^{g^2}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^2}{G}_{2\mathrm{M}}^{\mathrm{N}+2\mathrm{M}-1-\mathrm{g}}$

[Proof]

$SU{M}_q\left(N,\left[{\mathrm{q}}^1:\left(q-1\right){\mathrm{q}}^1,{\mathrm{q}}^3:\left(q-1\right){\mathrm{q}}^3\dots {\mathrm{q}}^{T_M}:\left(q-1\right){\mathrm{q}}^{T_M}\right],\left[1,3\dots 2M-1\right]\right)$

$={{\displaystyle \sum }}_{n_M=0}^{N-1}{q}^{n_M}{q}^{n_M+2M-1}\dots {{\displaystyle \sum }}_{n_1=0}^{n_2}{q}^{n_1}{q}^{n_1+1}=q^{M^2}{{\displaystyle \sum }}_{n_M=0}^{N-1}{q}^{2n_M}\dots {{\displaystyle \sum }}_{n_1=0}^{n_2}{q}^{2n_1}$

$=q^{M^2}{\displaystyle \sum }_{0\le {\lambda }_1\le {\lambda }_2\le \mathrm{...}\le {\lambda }_M\le N-1}{\mathrm{q}}^{2\left({\lambda }_1+{\lambda }_2+\mathrm{...}+{\lambda }_M\right)}=q^{M^2}{\left[\begin{array}{c}N+M-1\\ M\end{array}\right]}_{q^2}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\mathrm{H}}_3^{\mathrm{q}}\left(\mathrm{g}\right)G_{2\mathrm{M}}^{\mathrm{N}+2\mathrm{M}-1-\mathrm{g}}$

$B_i\stackrel{For{m}_3}{\to }\left\{\begin{array}{c}{\mathrm{q}}^{1+{\mathrm{X}}_{\mathrm{T}-1}}\left\{\left({\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}-1}}{\mathrm{G}}_1^{{\mathrm{T}}_{\mathrm{i}}}-{\mathrm{q}}^{{\mathrm{T}}_{\mathrm{i}}}{\mathrm{G}}_1^{{\mathrm{X}}_{\mathrm{T}-1}}\right){\mathrm{D}}_{\mathrm{i}}-{\mathrm{K}}_{\mathrm{i}}{\mathrm{q}}^{{\mathrm{T}}_{\mathrm{i}}}\right\}=-{\mathrm{q}}^{{\mathrm{T}}_{\mathrm{i}}-1+2{\mathrm{X}}_{\mathrm{T}}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{T}}_{\mathrm{i}}\\ {\mathrm{q}}^{{\mathrm{X}}_{\mathrm{T}-1}}\left({\mathrm{K}}_{\mathrm{i}}+{\mathrm{G}}_1^{{\mathrm{X}}_{\mathrm{T}-1}}{\mathrm{D}}_{\mathrm{i}}\right)={\mathrm{q}}^{{\mathrm{T}}_{\mathrm{i}}+2{\mathrm{X}}_{\mathrm{T}-1}}={\mathrm{q}}^{{\mathrm{T}}_{\mathrm{i}}+2{\mathrm{X}}_{\mathrm{T}}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{K}}_{\mathrm{i}}\end{array}\right.,Every item of {\mathrm{H}}_3^{\mathrm{q}}\left(\mathrm{g}\right) has {\mathrm{q}}^{1+3+\dots +2M-1}$

$H_3^q\left(g\right)={\left(-1\right)}^g{q}^{\left(1+3+\dots +\left(2M-1\right)\right)-g+2\left(\begin{array}{c}\mathrm{g}+1\\ 2\end{array}\right)}\sum {{\displaystyle \prod }}_{X\in K}{\mathrm{q}}^{2{\mathrm{X}}_{\mathrm{T}}}={\left(-1\right)}^g{q}^{M^2+g^2}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^2}$

q.e.d.

  

$8.4){{\displaystyle \sum }}_{1\le i_1<i_2<\mathrm{...}<i_M\le N+M-1}{q}^{i_1}{q}^{i_2}\dots q^{i_M}=q^{\frac{M\left(M+1\right)}{2}}{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^g{q}^{\frac{g^2}{2}}{\left[\begin{array}{c}M\\ g\end{array}\right]}_q{\left[\begin{array}{c}\mathrm{N}+2\mathrm{M}-1-\mathrm{g}\\ 2M\end{array}\right]}_{q^{\frac{1}{2}}}$

[Proof]

$SU{M}_q\left(N,\left[{\mathrm{q}}^2:\left(q-1\right){\mathrm{q}}^2,{\mathrm{q}}^4:\left(q-1\right){\mathrm{q}}^4\dots {\mathrm{q}}^{2M}:\left(q-1\right){\mathrm{q}}^{2M}\right],\left[1,3\dots 2M-1\right]\right)$

$={{\displaystyle \sum }}_{n_M=0}^{N-1}{q}^{n_M}{q}^{n_M+2M}\dots {{\displaystyle \sum }}_{n_2=0}^{n_3}{q}^{n_2}{q}^{n_2+4}{{\displaystyle \sum }}_{n_1=0}^{n_2}{q}^{n_1}{q}^{n_1+2}={{\displaystyle \sum }}_{n_M=0}^{N-1}{q}^{2\left(n_M+M\right)}\dots {{\displaystyle \sum }}_{n_2=0}^{n_3}{q}^{2\left(n_2+2\right)}{{\displaystyle \sum }}_{n_1=0}^{n_2}{q}^{2(n_1+1)}$

$B_i\stackrel{For{m}_3}{\to }\left\{\begin{array}{c}-{\mathrm{q}}^{2i-1+2{\mathrm{X}}_{\mathrm{T}}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{T}}_{\mathrm{i}}\\ {\mathrm{q}}^{2i+2{\mathrm{X}}_{\mathrm{T}}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{K}}_{\mathrm{i}}\end{array},Everyitemof{\mathrm{H}}_3^{\mathrm{q}}\left(\mathrm{g}\right)has{\mathrm{q}}^{2+4+\dots +2M}\right.$

$H_3^q\left(g\right)={\left(-1\right)}^g{q}^{2\left(1+2+\mathrm{...}+M\right)-g+2\left(\begin{array}{c}\mathrm{g}+1\\ 2\end{array}\right)}\sum {{\displaystyle \prod }}_{X\in K}{\mathrm{q}}^{2{\mathrm{X}}_{\mathrm{T}}}={\left(-1\right)}^g{q}^{M\left(M+1\right)+g^2}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^2}$

$SU{M}_q\left(N\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^g{q}^{M\left(M+1\right)+g^2}{\left[\begin{array}{c}M\\ g\end{array}\right]}_{q^2}{G}_{2\mathrm{M}}^{\mathrm{N}+2\mathrm{M}-1-\mathrm{g}}={{\displaystyle \sum }}_{1\le i_1<i_2<\mathrm{...}<i_M\le N+M-1}{q}^{2i_1}{q}^{2i_2}\dots q^{2i_M}$

q.e.d.

$8.5){{\displaystyle \sum }}_{k_r=1}^N\dots {{\displaystyle \sum }}_{k_2=1}^{k_3}{{\displaystyle \sum }}_{k_1=1}^{k_2}{\mathrm{q}}^{k_1+k_2+\mathrm{...}+k_r}▽SU{M}_q\left(k_i,PS,\left[1,2\dots M\right]\right)={▽}^{i-r}SU{M}_q\left(N,PS,\left[T_j=j+\left(i-1\right)\right]\right)$

  

$8.6){\left(K+qD\right)}^M=\left(q-1\right)D{{\displaystyle \sum }}_{g=0}^{M-1}{\left(K+D\right)}^g{\left(K+qD\right)}^{M-1-g}+{\left(K+D\right)}^M$

[Proof]

$PS=\left[K+D,K+D\dots K+D\right]:\left(q-1\right)D,PT=\left[1,\mathrm{2...}M\right],{\mathrm{B}}_{\mathrm{i}}\stackrel{For{m}_1}{\to }\left\{\begin{array}{c}{q}^{X_T}\left(q^{X_T}-1\right)D;{\mathrm{X}}_{\mathrm{i}}={\mathrm{T}}_{\mathrm{i}}\\ K+q^{X_{T-1}}D;{\mathrm{X}}_{\mathrm{i}}={\mathrm{K}}_{\mathrm{i}} \end{array}\right.$

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

$SU{M}_q\left(2\right)-SU{M}_q\left(1\right)=q{{\displaystyle \prod }}_{\mathrm{i}=1}^{\mathrm{M}}\left(K+D+{\mathrm{G}}_1^1\left(q-1\right)D\right)=q{{\displaystyle \prod }}_{\mathrm{i}=1}^{\mathrm{M}}\left(K+qD\right)=H_1^q\left(1\right)+H_1^q\left(0\right){\mathrm{G}}_1^2-H_1^q\left(0\right){\mathrm{G}}_1^1$

$=H_1^q\left(1\right)+q{H}_1^q\left(0\right)=H_1^q\left(1\right)+q{\left(K+D\right)}^M$

q.e.d.

8. Multiparameter Formal Calcution

$2-\mathrm{parameters}\mathrm{Formal}\mathrm{Calculation}\mathrm{calculate}nestedsummationof{\mathrm{K}}_{\mathrm{i}}+\left(\begin{array}{c}\mathrm{n}\\ 1\end{array}\right){\mathrm{D}}_{1,\mathrm{i}}+\left(\begin{array}{c}\mathrm{n}\\ 2\end{array}\right){\mathrm{D}}_{2,\mathrm{i}}$

SUM(N, PS, PT) changed to

SUM(N, [K₁,K₂...K_M], [T_1,1:D_1,1...T_1,M:D_1,M], [T_2,1: D_2,1...T_2,M: D_2,M])=SUM(N,PS,PT₁,PT₂), always T_i=T_1,i=T_2,i

$UsetheForm\left(K_1+T_{1,1}+T_{2,1}\right)\left(K_2+T_{1,2}+T_{2,2}\right)\dots \left(K_M+T_{1,M}+T_{2,M}\right)=\sum {{\displaystyle \prod }}_{i=1}^M{X}_i$

$X\left(P{T}_1\right)=countofX\in P{T}_1,X\left(P{T}_2\right)=countofX\in P{T}_2;X\left(PT\right)=X\left(P{T}_1\right)+2X\left(P{T}_2\right)$

$X_{P{T}_1}=countof\left\{X_1,X_2\dots X_i\right\}\in P{T}_1,X_{P{T}_2}=countof\left\{X_1,X_2\dots X_i\right\}\in P{T}_2,X_{PT}=X_{P{T}_1}+2X_{P{T}_2}$

$9.1)g=X\left(T\right),\mathrm{H}\left(\mathrm{g}\right)={{\displaystyle \sum }}_{\prod X_{i}withX\left(T\right)=g}{{\displaystyle \prod }}_{i=1}^M{B}_i,SUM\left(N,PS,P{T}_1,P{T}_2\right)=$

$\stackrel{{\mathrm{Form}}_1}{\to }{{\displaystyle \sum }}_{\mathrm{g}=0}^{2\mathrm{M}}{\mathrm{H}}_1\left(\mathrm{g}\right)\left(\begin{array}{c}\mathrm{N}+{\mathrm{T}}_{\mathrm{M}}-\mathrm{M}\\ {\mathrm{T}}_{\mathrm{M}}-\mathrm{M}+1+\mathrm{g}\end{array}\right),{\mathrm{B}}_{\mathrm{i}}=\left\{\begin{array}{c}{\mathrm{K}}_{\mathrm{i}}+{\mathrm{X}}_{\mathrm{PT}}{\mathrm{D}}_{1,\mathrm{i}}+\left(\begin{array}{c}{\mathrm{X}}_{\mathrm{PT}}\\ 2\end{array}\right){\mathrm{D}}_{2,\mathrm{i}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{K}}_{\mathrm{i}}\\ \left({\mathrm{T}}_{\mathrm{i}}-i+{\mathrm{X}}_{\mathrm{PT}}\right){\mathrm{D}}_{1,\mathrm{i}}+\left({\mathrm{T}}_{\mathrm{i}}-i+{\mathrm{X}}_{\mathrm{PT}}\right)\left({\mathrm{X}}_{\mathrm{PT}}-1\right){\mathrm{D}}_{2.\mathrm{i}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{PT}}_{1.\mathrm{i}}\\ \left(\begin{array}{c}{\mathrm{T}}_{\mathrm{i}}-\mathrm{i}+{\mathrm{X}}_{\mathrm{PT}}\\ 2\end{array}\right){\mathrm{D}}_{2,\mathrm{i}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{PT}}_{2.\mathrm{i}}\end{array}\right.$

$\stackrel{{\mathrm{Form}}_2}{\to }{{\displaystyle \sum }}_{\mathrm{g}=0}^{2\mathrm{M}}{\mathrm{H}}_2\left(\mathrm{g}\right)\left(\begin{array}{c}\mathrm{N}+{\mathrm{T}}_{\mathrm{M}}-\mathrm{M}+\mathrm{g}\\ {\mathrm{T}}_{\mathrm{M}}-\mathrm{M}+1+\mathrm{g}\end{array}\right),{\mathrm{B}}_{\mathrm{i}}=\left\{\begin{array}{c}{\mathrm{K}}_{\mathrm{i}}-\left({\mathrm{T}}_{\mathrm{i}}-i+{\mathrm{X}}_{\mathrm{PT}}+1\right){\mathrm{D}}_{1,\mathrm{i}}+\left(\begin{array}{c}{\mathrm{T}}_{\mathrm{i}}-\mathrm{i}+{\mathrm{X}}_{\mathrm{PT}}+2\\ 2\end{array}\right){\mathrm{D}}_{2,\mathrm{i}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{K}}_{\mathrm{i}}\\ \left({\mathrm{T}}_{\mathrm{i}}-i+{\mathrm{X}}_{\mathrm{PT}}\right){\mathrm{D}}_{1,\mathrm{i}}-\left({\mathrm{T}}_{\mathrm{i}}-i+{\mathrm{X}}_{\mathrm{PT}}\right)\left({\mathrm{T}}_{\mathrm{i}}-i+{\mathrm{X}}_{\mathrm{PT}}+1\right){\mathrm{D}}_{2.\mathrm{i}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{PT}}_{1.\mathrm{i}}\\ \left(\begin{array}{c}{\mathrm{T}}_{\mathrm{i}}-\mathrm{i}+{\mathrm{X}}_{\mathrm{PT}}\\ 2\end{array}\right){\mathrm{D}}_{2,\mathrm{i}};{\mathrm{X}}_{\mathrm{i}}={\mathrm{PT}}_{2.\mathrm{i}}\end{array}\right.$

$\stackrel{{\mathrm{Form}}_3}{\to }{{\displaystyle \sum }}_{\mathrm{g}=0}^{2\mathrm{M}}{\mathrm{H}}_3\left(\mathrm{g}\right)\left(\begin{array}{c}N+T_M+M-g\\ T_M+M+1\end{array}\right){\mathrm{B}}_{\mathrm{i}}=\left\{\begin{array}{c}{K}_i+{\mathrm{X}}_{\mathrm{PT}-1}{D}_{1,i}+\left(\begin{array}{c}{\mathrm{X}}_{\mathrm{PT}-1}\\ 2\end{array}\right)D_{2,i};{\mathrm{X}}_{\mathrm{i}}={\mathrm{K}}_{\mathrm{i}}\\ -2K_i+\left({\mathrm{T}}_{\mathrm{i}}+i-1-2{\mathrm{X}}_{\mathrm{PT}-1}\right)D_{1,i}+\left({\mathrm{T}}_{\mathrm{i}}+i-{\mathrm{X}}_{\mathrm{PT}-1}\right){\mathrm{X}}_{\mathrm{PT}-1}{D}_{2,i};{\mathrm{X}}_{\mathrm{i}}={\mathrm{PT}}_{1.\mathrm{i}}\\ K_i-\left({\mathrm{T}}_{\mathrm{i}}+\mathrm{i}-1-{\mathrm{X}}_{\mathrm{PT}-1}\right)D_{1,i}+\left(\begin{array}{c}{\mathrm{T}}_{\mathrm{i}}+\mathrm{i}-{\mathrm{X}}_{\mathrm{PT}-1}\\ 2\end{array}\right)D_{2,i};{\mathrm{X}}_{\mathrm{i}}={\mathrm{PT}}_{2.\mathrm{i}}\end{array}\right.$

[Proof]

[3] has obtained Form₁ and ignores Form_2,3. All that remains is some technical work.Here proves Form₂.

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

$\left(**\right){{\displaystyle \sum }}_{n=0}^{N-1}\left(\begin{array}{c}n\\ 2\end{array}\right)\left(\begin{array}{c}n+K\\ M\end{array}\right)=\left(\begin{array}{c}M+2\\ 2\end{array}\right)\left(\begin{array}{c}N+K+2\\ M+3\end{array}\right)-\left(M+1\right)\left(K+2\right)\left(\begin{array}{c}N+K+1\\ M+2\end{array}\right)+\left(\begin{array}{c}K+2\\ 2\end{array}\right)\left(\begin{array}{c}N+K\\ M+1\end{array}\right)，M>K$

$PS1=\left[PS,K_{M+1}\right],PT{1}_1=\left[P{T}_1,T_{M+1}=T_M+2-p:D_{1,M+1}\right],PT{1}_2=\left[P{T}_2,T_{M+1}=T_M+2-p:D_{2,M+1}\right]$

$Useinduction,\mathrm{it}canbeverifiedwhenM=1,supposeSUM\left(N\right)={{\displaystyle \sum }}_{g=0}^{2M}{H}_2\left(g\right)\left(\begin{array}{c}N+T_M-M+g\\ T_M-M+1+g\end{array}\right)$

$LetA=T_M-M+1-p=T_{M+1}-\left(M+1\right)$

$SUM\left(N,PS1,PT{1}_1,PT{1}_2\right)$

$={{\displaystyle \sum }}_{n=0}^{N-1}\left(K_{M+1}+n{D}_{1,M+1}+\left(\begin{array}{c}n\\ 2\end{array}\right)D_{2,M+1}\right){▽}^p{{\displaystyle \sum }}_{g=0}^{2M}{H}_2\left(g\right)\left(\begin{array}{c}n+1+T_M-M+g\\ T_M-M+1+g\end{array}\right)$  $={{\displaystyle \sum }}_{n=0}^{N-1}{{\displaystyle \sum }}_{g=0}^{2M}\left(K_{M+1}+n{D}_{1,M+1}+\left(\begin{array}{c}n\\ 2\end{array}\right)D_{2,M+1}\right)H_2\left(g\right)\left(\begin{array}{c}n+A+g\\ A+g\end{array}\right)$

$={{\displaystyle \sum }}_{g=0}^{2M}\left(K_{M+1}-\left(A+g+1\right)D_{1,M+1}+\left(\begin{array}{c}A+g+2\\ 2\end{array}\right)D_{M+1,2}\right)H_2\left(g\right)\left(\begin{array}{c}N+A+g\\ A+g+1\end{array}\right)①$

$+{{\displaystyle \sum }}_{g=0}^{2M}\left(\left(A+g+1\right)D_{1,M+1}-\left(A+g+1\right)\left(A+g+2\right)D_{2,M+1}\right)H_2\left(g\right)\left(\begin{array}{c}N+A+g+1\\ A+g+2\end{array}\right)②$

$+{{\displaystyle \sum }}_{g=0}^{2M}\left(\begin{array}{c}A+g+2\\ 2\end{array}\right)D_{2,M+1}{H}_2\left(g\right)\left(\begin{array}{c}N+A+g+2\\ A+g+3\end{array}\right)③$

$①\left(\begin{array}{c}N+A+g\\ A+g+1\end{array}\right)=\left(\begin{array}{c}N+T_{M+1}-\left(M+1\right)+g\\ T_{M+1}-\left(M+1\right)+1+g\end{array}\right),X_i=K_i,X_{PT}unchanged,X_{PT}=g,T_{M+1}=T_i,M+1=i$

$②\left(\begin{array}{c}N+A+g+1\\ A+g+2\end{array}\right)=\left(\begin{array}{c}N+T_{M+1}-\left(M+1\right)+\left(g+1\right)\\ T_{M+1}-\left(M+1\right)+1+\left(g+1\right)\end{array}\right),X_i=T_{1,i},X_{PT}=g+1,T_{M+1}=T_i,M+1=i$

$③\left(\begin{array}{c}N+A+g+2\\ A+g+3\end{array}\right)=\left(\begin{array}{c}N+T_{M+1}-\left(M+1\right)+\left(g+2\right)\\ T_{M+1}-\left(M+1\right)+1+\left(g+2\right)\end{array}\right),X_i=T_{2,i},X_{PT}=g+2,T_{M+1}=T_i,M+1=i$

Form₂ has been proved. Use the following two formulas to prove Form₃.

${{\displaystyle \sum }}_{n=0}^{N-1}n\left(\begin{array}{c}n+K\\ M\end{array}\right)=-\left(K-M\right)\left(\begin{array}{c}N+K+2\\ M+3\end{array}\right)+\left(2K-M+1\right)\left(\begin{array}{c}N+K+1\\ M+3\end{array}\right)-\left(K+1\right)\left(\begin{array}{c}N+K\\ M+3\end{array}\right)$

${{\displaystyle \sum }}_{n=0}^{N-1}\left(\begin{array}{c}n\\ 2\end{array}\right)\left(\begin{array}{c}n+K\\ M\end{array}\right)=\left(\begin{array}{c}M-K\\ 2\end{array}\right)\left(\begin{array}{c}N+K+2\\ M+3\end{array}\right)+\left(K+2\right)\left(M-K\right)\left(\begin{array}{c}N+K+1\\ M+3\end{array}\right)+\left(\begin{array}{c}K+2\\ 2\end{array}\right)\left(\begin{array}{c}N+K\\ M+3\end{array}\right)$

q.e.d.

According to this method, it can be extended to multiparameter SUM(N) and multiparameter SUM_q(N).

$1.Seri{e}_i:{\mathrm{K}}_{\mathrm{i}}+\left(\begin{array}{c}\mathrm{n}\\ 1\end{array}\right){\mathrm{D}}_{1,\mathrm{i}}+\left(\begin{array}{c}\mathrm{n}\\ 2\end{array}\right){\mathrm{D}}_{2,\mathrm{i}}+\left(\begin{array}{c}\mathrm{n}\\ 3\end{array}\right){\mathrm{D}}_{3,\mathrm{i}}+\dots or {\mathrm{K}}_{\mathrm{i}}+{\mathrm{G}}_1^n{\mathrm{D}}_{1,\mathrm{i}}+{\mathrm{G}}_2^n{\mathrm{D}}_{2,\mathrm{i}}+{\mathrm{G}}_3^n{\mathrm{D}}_{3,\mathrm{i}}+\dots ;T_{1,\mathrm{i}}=T_{2,\mathrm{i}}=T_{3,\mathrm{i}}\dots$

$2.X\left(PT\right)=X\left(P{T}_1\right)+2X\left(P{T}_2\right)+3X\left(P{T}_3\right)\dots ;X_{PT}=X_{P{T}_1}+2X_{P{T}_2}+3X_{P{T}_3}\dots$

$3.Use{X}_{PT}or{X}_{PT-1}Indicatethepositions.$

9. A theorem of symmetry

$Therehave2.1)\to H_1\left(g,PT,\mathrm{PT}\right)={{\displaystyle \prod }}_{i=1}^M{T}_i\left(\begin{array}{c}M\\ g\end{array}\right)={{\displaystyle \prod }}_{i=1}^M{T}_i\left(\begin{array}{c}M\\ g,M-g\end{array}\right)$

$Promoteit,theSet\left\{T_1,T_2\dots T_M\right\}comeformpsource{S}_1,S_2\dots S_P,T_i\in S_{j}means{T}_{i}comefromsourcej$

$defineDiff\left(S_x,S_y\right)=Z_{x,y},x\ne y;Diff\left(S_y,S_x\right)=-Diff\left(S_x,S_y\right);Diff\left(S_x,S_x\right)=0$

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

$defineW\left(g_1,g_2\dots g_p,\left[T_1,T_2\dots T_M\right]\right)={\displaystyle \sum }_{g_1+g_2+\mathrm{...}+g_p=M,g_i\ge 0}{{\displaystyle \prod }}_{i=1}^M\{T_i+{\displaystyle \sum }_{j<i}Diff(T_j,T_i)\}$

$Obvious:H_1\left(g,PT,\mathrm{PT}\right)=W\left(g,M-g,\left[T_1,T_2\dots T_M\right]\right),Z_{x,y}=-1,x<y$

$\left[T_1,T_2\dots T_M\right]canbearrangedarbitrarily,Z_{x,y\left(x\ne y\right)}canbearbitrarilytoo.$

$10.1)Induction\to W\left(g_1,g_2\dots g_p,\left[T_1,T_2\dots T_M\right]\right)={{\displaystyle \prod }}_{i=1}^M{T}_i\left(\begin{array}{c}M\\ g_1,g_2\dots g_p\end{array}\right).Aconclusionof\left[10\right].$

  

$PromotetoGaussiancoefficient:G_{g>0}^0=0;G_g^{M<0}isstilldefined$

$defineDif{f}_q\left(S_x,S_y\right)=-1,x<y;Dif{f}_q\left(S_y,S_x\right)=-Dif{f}_q\left(S_x,S_y\right);Dif{f}_q\left(S_x,S_x\right)=0$

$define{W}_q\left(g_1,g_2\dots g_p,\left[T_1,T_2\dots T_M\right]\right)={\displaystyle \sum }_{g_1+g_2+\mathrm{...}+g_p=M,g_i\ge 0}{{\displaystyle \prod }}_{i=1}^M{G}_1^{T_i+{\displaystyle \sum }_{j<i}Dif{f}_q\left(T_j,T_i\right)}{q}^{{\displaystyle \sum }_{j<i,Dif{f}_q\left(T_j,T_i\right)=-1}1}$

$10.2)W_q\left(g_1,g_2,\left[T_1,T_2\dots T_M\right]\right)={{\displaystyle \prod }}_{i=1}^M{G}_1^{T_i}{G}_{g_1}^M$

[Proof]

$Obvious:ItholdswhenM=1or{g}_1=0or{g}_1=M;$

$Verifiedbycalculation:ItholdswhenM=2:G_1^{T_1}{G}_1^{T_2+1}+q{G}_1^{T_1}{G}_1^{T_2-1}=G_1^{T_1}{G}_1^{T_2}{G}_1^2$

$WhenM+1,provebyinduction:PT=\left[T_1,T_2\dots T_M\right],PT1=\left[T_1,T_2\dots T_M,T_{M+1}\right]$

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

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

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

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

$Justneedtoprove{G}_1^{T_{M+1}}\left(q^{{\mathrm{g}}_1}{G}_{g_1}^M+{\mathrm{G}}_{{\mathrm{g}}_1-1}^{\mathrm{M}}\right)=G_{g_1}^M{G}_1^{T_{M+1}+g_1}+G_{g_1-1}^M{G}_1^{T_{M+1}-\left(M+1-g_1\right)}{q}^{M+1-g_1}$

$Bothsidesmultiply\left(q^{g_1}-1\right)\left(q^{g_1-1}-1\right)\dots \left(q-1\right)anddivide\left(q^M-1\right)\left(q^{M-1}-1\right)\dots \left(q^{M-{\mathrm{g}}_1+2}-1\right)$

$Left=\left(q^{T_{M+1}+g-1}+\mathrm{...}+q^g\right)\left(q^{M-g_1+1}-1\right)+\left(q^{T_{M+1}-1}+\mathrm{...}+q^0\right)\left(q^g-1\right)$

$Right=\left(q^{T_{M+1}+g-1}+\mathrm{...}+q^0\right)\left(q^{M-g_1+1}-1\right)+\left(q^{T_{M+1}-1}+\mathrm{...}+q^{M+1-g_1}\right)\left(q^g-1\right)$

$Left-Right=\left(q^g-1\right)\left(q^{M-g}+\dots +1\right)-\left(q^{M-g_1+1}-1\right)\left(q^{g-1}+\dots +1\right)$

$=\left(q^M+\mathrm{...}+q^g\right)-\left(q^{M-g}+\mathrm{...}+1\right)+\left(q^{g-1}+\mathrm{...}+1\right)-\left(q^M+\mathrm{...}+q^{M-g_1+1}\right)=0$

q.e.d.

$define{\left[\begin{array}{c}M\\ g_1\dots g_p\end{array}\right]}_q=\frac{\left({\mathrm{q}}^M-1\right)\left({\mathrm{q}}^{M-1}-1\right)\dots \left({\mathrm{q}}^1-1\right)}{{{\displaystyle \prod }}_{i=1}^{g_1}\left({\mathrm{q}}^i-1\right){{\displaystyle \prod }}_{i=1}^{g_2}\left({\mathrm{q}}^i-1\right)\dots {{\displaystyle \prod }}_{i=1}^{g_p}\left({\mathrm{q}}^i-1\right)},g_1+g_2+\mathrm{...}+g_p=M$

$10.3)W_q\left(g_1,g_2\dots g_p,\left[T_1,T_2\dots T_M\right]\right)={{\displaystyle \prod }}_{i=1}^M{G}_1^{T_i}{\left[\begin{array}{c}M\\ g_1,g_2\dots g_p\end{array}\right]}_q$

[Proof]

$\mathrm{Only}\mathrm{needs}\mathrm{to}\mathrm{prove}\mathrm{that}\mathrm{it}\mathrm{holds}\mathrm{when}\mathrm{p}=3$

$W_q\left(g_1,g_2+g_3,\left[T_1,T_2\dots T_M\right]\right)={{\displaystyle \prod }}_{i=1}^M{G}_1^{T_i}{\left[\begin{array}{c}{g}_1+g_2+g_3\\ g_1,g_2+g_3\end{array}\right]}_q$

$\mathrm{Every}\mathrm{item}\mathrm{has}{g}_2+g_{3}factorscomefromSourc{e}_2,dividethemto{g}_2\mathrm{ⅹ}Sourc{e}_2+g_3\mathrm{ⅹ}Sourc{e}_3$

$g_1-factors\mathrm{are}\mathrm{invariant},\left(g_2+g_3\right)-factors\mathrm{are}\mathrm{variant}.$

$\sum \prod \left(\mathrm{variant}factors\right)=W_q\left(g_2,g_3,\left[X_1,X_2\dots X_{g_2+g_3}\right]\right){{\displaystyle \prod }}_{i=1}^{g_2+g_3}{G}_1^{X_i}{\left[\begin{array}{c}{g}_2+g_3\\ g_2,g_3\end{array}\right]}_q$

$W_q\left(g_1,g_2,g_3\right)={{\displaystyle \prod }}_{i=1}^M{G}_1^{T_i}{\left[\begin{array}{c}{g}_1+g_2+g_3\\ g_1,g_2+g_3\end{array}\right]}_q{\left[\begin{array}{c}{g}_2+g_3\\ g_2,g_3\end{array}\right]}_q={{\displaystyle \prod }}_{i=1}^M{G}_1^{T_i}{\left[\begin{array}{c}{g}_1+g_2+g_3\\ g_1,g_2,g_3\end{array}\right]}_q$

$\mathrm{The}\mathrm{rest}\mathrm{is}\mathrm{to}\mathrm{prove}\mathrm{that}\mathrm{the}\mathrm{change}\mathrm{of}Dif{f}_q\left(T_j,T_i\right)meetsthedefinition.$

$\mathrm{Obvious}:g_1-\left(\mathrm{invariant}factors\right)meetthedefinition.$

$When{W}_q\left(g_2,g_3\right)isembeddedin{W}_q\left(g_1,g_2,g_3\right),allDif{f}_q\left(T_j\in Sourc{e}_1,T_i\in Sourc{e}_{2,3}\right)willadd1.$

q.e.d.

10. Eulerian polynomials and Beyond

The main proof method of ${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n$n^M comes from [2]. Extensive promotion here.

$11.1){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n\left(\begin{array}{c}\mathrm{n}+\mathrm{M}\\ \mathrm{M}\end{array}\right)=q^N{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^g\frac{\left(\begin{array}{c}\mathrm{N}+\mathrm{M}-1-\mathrm{g}\\ \mathrm{M}-\mathrm{g}\end{array}\right)}{{\left(q-1\right)}^{g+1}}+\frac{{\left(-1\right)}^{M+1}}{{\left(q-1\right)}^{M+1}}$

[Proof]

$\mathrm{M}=0,{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}\left(\begin{array}{c}\mathrm{n}\\ 0\end{array}\right)=\frac{{\mathrm{q}}^{\mathrm{N}}-1}{\mathrm{q}-1}=\frac{{\mathrm{q}}^{\mathrm{N}}\left(\begin{array}{c}\mathrm{N}-1\\ 0\end{array}\right)}{\mathrm{q}-1}-\frac{1}{\mathrm{q}-1},\mathrm{holds}$

$\mathrm{M}=1,{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}\left(\begin{array}{c}\mathrm{n}+1\\ 1\end{array}\right)={{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}\left(\mathrm{n}+1\right)={{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}+{{\displaystyle \sum }}_{\mathrm{n}=1}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}+\mathrm{...}+{{\displaystyle \sum }}_{\mathrm{n}=\mathrm{N}-1}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}$

$=\mathrm{N}{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^{\mathrm{n}}-\left({{\displaystyle \sum }}_{\mathrm{n}=0}^0{\mathrm{q}}^{\mathrm{n}}+{{\displaystyle \sum }}_{\mathrm{n}=0}^1{\mathrm{q}}^{\mathrm{n}}+\mathrm{...}+{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-2}{\mathrm{q}}^{\mathrm{n}}\right)$

$=\mathrm{N}\frac{{\mathrm{q}}^{\mathrm{N}}-1}{\mathrm{q}-1}-\left(\frac{\mathrm{q}-1}{\mathrm{q}-1}+\frac{{\mathrm{q}}^2-1}{\mathrm{q}-1}+\mathrm{...}+\frac{{\mathrm{q}}^{\mathrm{N}-1}-1}{\mathrm{q}-1}\right)=\mathrm{N}\frac{{\mathrm{q}}^{\mathrm{N}}-1}{\mathrm{q}-1}-\frac{\left(1+\mathrm{q}+{\mathrm{q}}^2+\mathrm{...}+{\mathrm{q}}^{\mathrm{N}-1}\right)-\mathrm{N}}{\mathrm{q}-1}=\frac{{\mathrm{q}}^{\mathrm{N}}\left(\begin{array}{c}\mathrm{N}\\ 1\end{array}\right)}{\mathrm{q}-1}-\frac{{\mathrm{q}}^{\mathrm{N}}\left(\begin{array}{c}\mathrm{N}-1\\ 0\end{array}\right)}{{\left(\mathrm{q}-1\right)}^2}+\frac{1}{{\left(\mathrm{q}-1\right)}^2},\mathrm{holds}$

  

$\mathrm{When}\mathrm{N}=1,\mathrm{right}={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^{\mathrm{g}}\frac{1}{{\left(\mathrm{q}-1\right)}^{\mathrm{g}+1}}+\frac{{\left(-1\right)}^{\mathrm{M}+1}}{{\left(\mathrm{q}-1\right)}^{\mathrm{M}+1}}=\frac{\mathrm{q}}{\mathrm{q}-1}{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}\frac{{\left(-1\right)}^{\mathrm{g}}}{{\left(\mathrm{q}-1\right)}^{\mathrm{g}}}+\frac{{\left(-1\right)}^{\mathrm{M}+1}}{{\left(\mathrm{q}-1\right)}^{\mathrm{M}+1}}=1=left$

$Use\left(\begin{array}{c}\mathrm{n}+\mathrm{M}\\ \mathrm{M}\end{array}\right)=\left(\begin{array}{c}\mathrm{n}+\mathrm{M}-1\\ \mathrm{M}\end{array}\right)+\left(\begin{array}{c}\mathrm{n}+\mathrm{M}-1\\ \mathrm{M}-1\end{array}\right),inductiveproof.$

q.e.d.

$11.2){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n\left(\begin{array}{c}\mathrm{n}+\mathrm{K}\\ \mathrm{M}\end{array}\right)=q^N{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(-1\right)}^g\frac{\left(\begin{array}{c}\mathrm{N}+\mathrm{K}-1-\mathrm{g}\\ \mathrm{M}-\mathrm{g}\end{array}\right)}{{\left(q-1\right)}^{g+1}}+\frac{{\left(-1\right)}^{M+1}{\mathrm{q}}^{M-K}}{{\left(q-1\right)}^{M+1}}$

  

${\mathrm{Define}\mathrm{A}}_{\mathrm{q}}^{\mathrm{M}}={{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(1-\mathrm{q}\right)}^{\mathrm{M}-\mathrm{k}}{\mathrm{q}}^{\mathrm{k}}{\mathrm{S}}_2\left(\mathrm{M},\mathrm{k}\right)\mathrm{k}!,\mathrm{q}\ne 0,1,\mathrm{M}\in \mathbb{N},{\mathrm{A}}_{\mathrm{q}}^0=1,{\mathrm{A}}_{\mathrm{q}}^1=\mathrm{q}$

Table 11. 1: ${\mathrm{A}}_{\mathrm{q}}^{\mathrm{M}}$.

Table 11. 1: ${\mathrm{A}}_{\mathrm{q}}^{\mathrm{M}}$.

$11.3){\mathrm{A}}_{\mathrm{q}}^{\mathrm{M}}=\mathrm{q}{{\displaystyle \sum }}_{k=0}^M{\left(\mathrm{q}-1\right)}^{\mathrm{M}-\mathrm{k}}{S}_2\left(M,k\right)k!$

[Proof]

${\mathrm{A}}_{\mathrm{q}}^{\mathrm{M}}\stackrel{4.4)}{\to }{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(1-\mathrm{q}\right)}^{\mathrm{M}-\mathrm{k}}{\mathrm{q}}^{\mathrm{k}}{{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}g!S_2\left(M,g\right)\left(\begin{array}{c}g-1\\ k-1\end{array}\right)$

$={{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}g!S_2\left(M,g\right){{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(1-\mathrm{q}\right)}^{\mathrm{M}-\mathrm{k}}{\mathrm{q}}^{\mathrm{k}}\left(\begin{array}{c}g-1\\ k-1\end{array}\right)$

$={{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}g!S_2\left(M,g\right){\left(1-\mathrm{q}\right)}^{\mathrm{M}-\mathrm{g}}q{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{g}}{\left(1-\mathrm{q}\right)}^{\mathrm{g}-\mathrm{k}}{\mathrm{q}}^{\mathrm{k}-1}\left(\begin{array}{c}g-1\\ k-1\end{array}\right)\stackrel{{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{g}}{\left(1-\mathrm{q}\right)}^{\mathrm{g}-\mathrm{k}}{\mathrm{q}}^{\mathrm{k}-1}\left(\begin{array}{c}g-1\\ k-1\end{array}\right)=1}{\to }11.3)$

q.e.d.

  

${\mathrm{n}}^M=▽SUM\left(n,\left[1\dots 1\right],\left[2\dots M\right]\right)={{\displaystyle \sum }}_{g=0}^M{S}_2\left(M,g\right)g!\left(\begin{array}{c}\mathrm{n}\\ \mathrm{g}\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}{S}_2\left(M,g\right)g!\left(\begin{array}{c}\mathrm{n}+\mathrm{g}-1\\ \mathrm{g}\end{array}\right)={{\displaystyle \sum }}_{g=0}^M\langle \begin{array}{c}M\\ g\end{array}\rangle \left(\begin{array}{c}n+g\\ M\end{array}\right)$  ${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M={{\displaystyle \sum }}_{g=0}^M{S}_2\left(M,g\right)g!{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n\left(\begin{array}{c}\mathrm{n}\\ \mathrm{g}\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{S}_2\left(M,g\right)g!\left\{q^N{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{g}}{\left(-1\right)}^k\frac{\left(\begin{array}{c}\mathrm{N}-1-\mathrm{k}\\ \mathrm{g}-\mathrm{k}\end{array}\right)}{{\left(q-1\right)}^{k+1}}+\frac{{\left(-1\right)}^{g+1}{\mathrm{q}}^g}{{\left(q-1\right)}^{g+1}}\right\}$

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

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

$=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(q-1\right)}^{M-k}{\left(-1\right)}^k{▽}^k{(\mathrm{N}-1)}^M+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M}{{\left(q-1\right)}^{M+1}} \left(*\right)$

  

${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M={{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}{S}_2\left(M,g\right)g!{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n\left(\begin{array}{c}\mathrm{n}+\mathrm{g}-1\\ \mathrm{g}\end{array}\right)$

$=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}{S}_2\left(M,g\right)g!\left\{{{\displaystyle \sum }}_{\mathrm{k}=0}^g{\left(-1\right)}^k\left(\begin{array}{c}\mathrm{N}+\mathrm{g}-2-\mathrm{k}\\ \mathrm{g}-\mathrm{k}\end{array}\right){\left(q-1\right)}^{M-k}\right\}+\frac{{\left(-1\right)}^{M+1}{{\displaystyle \sum }}_{g=0}^M{S}_2\left(M,g\right)g!{\left(1-q\right)}^{M-g}q}{{\left(q-1\right)}^{M+1}}$

$=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{k}=0}^M{\left(q-1\right)}^{M-k}{\left(-1\right)}^k{▽}^k\left(\mathrm{N}-1{)}^M+\frac{{\left(-1\right)}^{M+1}\left\{\dots \right\}}{{\left(q-1\right)}^{M+1}}\stackrel{comparewith\left(*\right)}{\to }11.3\right)$

  

${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M={{\displaystyle \sum }}_{g=0}^M\langle \begin{array}{c}M\\ g\end{array}\rangle {{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n\left(\begin{array}{c}n+g\\ M\end{array}\right)\to$

$11.4)A_q^M={{\displaystyle \sum }}_{g=0}^M\langle \begin{array}{c}M\\ g\end{array}\rangle {\mathrm{q}}^{M-g}={{\displaystyle \sum }}_{g=0}^M\langle \begin{array}{c}M\\ g\end{array}\rangle {\mathrm{q}}^{1+g}$

  

${\mathrm{n}}^M=▽SUM\left(n,\left[1,1\dots 1\right],\left[1,2\dots M\right]\right)={{\displaystyle \sum }}_{g=0}^M{S}_2\left(M,g+1\right)\left(g+1\right)!\left(\begin{array}{c}\mathrm{n}-1\\ \mathrm{g}\end{array}\right)\to$

$11.5)A_q^M={{\displaystyle \sum }}_{k=0}^M{\left(1-\mathrm{q}\right)}^{M-k}{\mathrm{q}}^{K+1}{S}_2\left(M+1,k+1\right)k!,M>0$

  

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

${▽}^k{\left(\mathrm{N}-1\right)}^M\stackrel{Bydefinition}{\to }{{\displaystyle \sum }}_{j=0}^k{\left(-1\right)}^j\left(\begin{array}{c}k\\ \mathrm{j}\end{array}\right){\left(\mathrm{N}-1-\mathrm{j}\right)}^M={{\displaystyle \sum }}_{j=0}^k{\left(-1\right)}^j\left(\begin{array}{c}k\\ \mathrm{j}\end{array}\right){{\displaystyle \sum }}_{g=0}^M\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\mathrm{N}}^g{\left(j+1\right)}^{M-g}{\left(-1\right)}^{M-g}$ =${{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\mathrm{N}}^g\left({{\displaystyle \sum }}_{j=0}^k{\left(-1\right)}^j\left(\begin{array}{c}\mathrm{k}\\ \mathrm{j}\end{array}\right){\left(j+1\right)}^{M-g}\right)={{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g-k}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\mathrm{N}}^g\left({{\displaystyle \sum }}_{j=0}^k{\left(-1\right)}^{j+k}\left(\begin{array}{c}\mathrm{k}\\ \mathrm{j}\end{array}\right){\left(j+1\right)}^{M-g}\right)$ =${{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g-k}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\mathrm{N}}^g{S}_2\left(M-g+1,k+1\right)k!$

  

$f\left(N\right)={{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M=\left(*\right)=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(q-1\right)}^{M-k}{\left(-1\right)}^k{▽}^k{(\mathrm{N}-1)}^M+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M}{{\left(q-1\right)}^{M+1}}$ Gf $=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(q-1\right)}^{M-k}{\left(-1\right)}^k\left\{{{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g-k}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\mathrm{N}}^g{S}_2\left(M-g+1,k+1\right)k!\right\}+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M}{{\left(q-1\right)}^{M+1}}$

$=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(q-1\right)}^g{\left(-1\right)}^{M-g}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\mathrm{N}}^g\left\{{{\displaystyle \sum }}_{k=0}^M{\left(q-1\right)}^{M-k-g}{S}_2\left(M-g+1,k+1\right)k!\right\}+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M}{{\left(q-1\right)}^{M+1}}\left(**\right)$

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

$11.6)A_q^M={{\displaystyle \sum }}_{k=0}^M{\left(\mathrm{q}-1\right)}^{M-k}{S}_2\left(M+1,k+1\right)k!$

  

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

$11.7){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{M}}{\left(q-1\right)}^g{\left(-1\right)}^{M-g}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right)A_{\mathrm{q}}^{M-g}{\mathrm{N}}^g+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M}{{\left(q-1\right)}^{M+1}}$

Table 11. 1: ${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M$.

Table 11. 1: ${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M$.

${\mathrm{A}}_{\mathrm{q}}^{\mathrm{M}}={{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(1-\mathrm{q}\right)}^{\mathrm{M}-\mathrm{k}}{\mathrm{q}}^{\mathrm{k}}{\mathrm{S}}_2\left(\mathrm{M},\mathrm{k}\right)\mathrm{k}!=\mathrm{q}{{\displaystyle \sum }}_{k=0}^M{\left(\mathrm{q}-1\right)}^{\mathrm{M}-\mathrm{k}}{S}_2\left(M,k\right)k! ={{\displaystyle \sum }}_{g=0}^M\langle \begin{array}{c}M\\ g\end{array}\rangle {\mathrm{q}}^{1+g}=$

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

  

${\mathrm{A}}_2^{\mathrm{M}}={{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(-1\right)}^{\mathrm{M}-\mathrm{k}}{2}^{\mathrm{k}}{\mathrm{S}}_2\left(\mathrm{M},\mathrm{k}\right)\mathrm{k}!=2{{\displaystyle \sum }}_{k=0}^M{S}_2\left(M,k\right)k! ={{\displaystyle \sum }}_{g=0}^M\langle \begin{array}{c}M\\ g\end{array}\rangle 2^{1+g}=$

${{\displaystyle \sum }}_{k=0}^M{\left(-1\right)}^{M-k}{2}^{K+1}{S}_2\left(M+1,k+1\right)k!,M>0={{\displaystyle \sum }}_{k=0}^M{S}_2\left(M+1,k+1\right)k!$

These expressions can validate 2.2).2.3).2.4).2.8)

  

$WeknowthatEulerianpolynomials{A}_M\left(t\right):{{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i{i}^M=\frac{t{A}_M\left(t\right)}{{\left(1-t\right)}^{M+1}};A_M\left(t\right)={{\displaystyle \sum }}_{g=0}^{M-1}\langle \begin{array}{c}M\\ g\end{array}\rangle {\mathrm{t}}^g$

$0<\mathrm{q}<1,\underset{N\to \infty }{\lim }{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\mathrm{n}}^M=\frac{A_{\mathrm{q}}^M}{{\left(1-q\right)}^{M+1}}\to$

$11.8)A_{\mathrm{t}}^M=t{A}_M\left(t\right)$

$Sotherearefiveexpressionsfor{A}_M\left(t\right).Thiscanalsobeobtainedfrom11.4).$

  

$Tocalculate{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\left(a+\left(n-1\right)d\right)}^M,investigation{\left(a+\left(n-1\right)d\right)}^M=▽SUM\left(n,\left[a,a\dots a\right]:d,\left[1,2\dots M\right]\right)$

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

  

${{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\left(a+\left(n-1\right)d\right)}^M={{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n\left(\begin{array}{c}\mathrm{n}-1\\ \mathrm{g}\end{array}\right)={{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right)\left\{q^N{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{g}}{\left(-1\right)}^k\frac{\left(\begin{array}{c}\mathrm{N}-2-\mathrm{k}\\ \mathrm{g}-\mathrm{k}\end{array}\right)}{{\left(q-1\right)}^{k+1}}+\frac{{\left(-1\right)}^{g+1}{\mathrm{q}}^{g+1}}{{\left(q-1\right)}^{g+1}}\right\}$

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

$=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(q-1\right)}^{M-k}{\left(-1\right)}^k\left\{{{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right)\left(\begin{array}{c}\mathrm{N}-2-\mathrm{k}\\ \mathrm{g}-\mathrm{k}\end{array}\right)\right\}+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M\left(a,d\right)}{{\left(q-1\right)}^{M+1}}$

$=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(q-1\right)}^{M-k}{\left(-1\right)}^k{▽}^k{(a+\left(N-2\right)d)}^M+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M\left(a,d\right)}{{\left(q-1\right)}^{M+1}}\left(*\right)$

$A_{\mathrm{q}}^M\left(a,d\right)isdefinedas{{\displaystyle \sum }}_{g=0}^M{H}_1\left(g,\left[a,a\dots a\right]:d,\left[1,2\dots M\right]\right){\left(1-q\right)}^{M-g}{\mathrm{q}}^{g+1}.Similarly:$

$11.9)A_{\mathrm{q}}^M\left(a,d\right)={{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right){\left(1-q\right)}^{M-g}{\mathrm{q}}^{g+1}={{\displaystyle \sum }}_{g=0}^M{H}_2\left(g\right){\left(1-q\right)}^{M-g}q={{\displaystyle \sum }}_{g=0}^M{H}_3\left(g\right)q^{g+1}$

  

$▽SUM\left(n,\left[d,d\dots d\right]:d,\left[1,2\dots M\right]\right)={\left(nd\right)}^M={{\displaystyle \sum }}_{g=0}^M{H}_1\left(g,\left[d,d\dots d\right]:d,\left[1,2\dots M\right]\right)\left(\begin{array}{c}\mathrm{n}-1\\ \mathrm{g}\end{array}\right)$

${\mathrm{H}}_1\left(\mathrm{g},\left[d,d\dots d\right]:d,\left[1,2\dots M\right]\right)=g!d^{g}f\left(M,g\right)\stackrel{Matrixrule}{\to }{{\displaystyle \sum }}_{\mathrm{j}=1}^{\mathrm{g}+1}{\left(-1\right)}^{\mathrm{g}+1+\mathrm{j}}\left(\begin{array}{c}\mathrm{g}\\ \mathrm{j}-1\end{array}\right)▽\mathrm{SUM}\left(\mathrm{j}\right)$

$={{\displaystyle \sum }}_{\mathrm{j}=1}^{\mathrm{g}+1}{\left(-1\right)}^{\mathrm{g}+1+\mathrm{j}}\left(\begin{array}{c}\mathrm{g}\\ \mathrm{j}-1\end{array}\right){\left(jd\right)}^M={{\displaystyle \sum }}_{\mathrm{j}=0}^{\mathrm{g}}{\left(-1\right)}^{\mathrm{g}+\mathrm{j}}\left(\begin{array}{c}\mathrm{g}\\ \mathrm{j}\end{array}\right){\left[\left(j+1\right)d\right]}^M$

$f\left(M,g\right)canbedefinedassomekindofGeneralStirlingNumbersofthesecondkind.$

  

${▽}^k{\left(a+\left(N-2\right)d\right)}^M\stackrel{Bydefinition}{\to }{{\displaystyle \sum }}_{j=0}^k{\left(-1\right)}^j\left(\begin{array}{c}k\\ \mathrm{j}\end{array}\right){\left(a+\left(N-2-j\right)d\right)}^M$

$={{\displaystyle \sum }}_{j=0}^k{\left(-1\right)}^j\left(\begin{array}{c}k\\ \mathrm{j}\end{array}\right){{\displaystyle \sum }}_{g=0}^M\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\left(a+\left(\mathrm{N}-1\right)\mathrm{d}\right)}^g{\left[\left(\mathrm{j}+1\right)\mathrm{d}\right]}^{M-g}{\left(-1\right)}^{M-g}$

$={{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g-k}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\left(a+\left(\mathrm{N}-1\right)\mathrm{d}\right)}^g\left({{\displaystyle \sum }}_{j=0}^k{\left(-1\right)}^{j+k}\left(\begin{array}{c}\mathrm{k}\\ \mathrm{j}\end{array}\right){\left[\left(\mathrm{j}+1\right)\mathrm{d}\right]}^{M-g}\right)$

$={{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g-k}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\left(a+\left(\mathrm{N}-1\right)\mathrm{d}\right)}^g{\mathrm{H}}_1\left(\mathrm{k},\left[d,d\dots d\right]:d,\left[1,2\dots M-g\right]\right)$

  

$11.10){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\left(a+\left(n-1\right)d\right)}^M=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{g=0}^M{\left(-1\right)}^{M-g}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right){\left(a+\left(\mathrm{N}-1\right)\mathrm{d}\right)}^g\{{{\displaystyle \sum }}_{\mathrm{k}=0}^{\mathrm{M}}{\left(q-1\right)}^{M-k}{\mathrm{H}}_1\left(k,\left[d,d\dots d\right]:d,\left[1,2\dots M-g\right]\right)\}+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M\left(a,d\right)}{{\left(q-1\right)}^{M+1}}$

$11.11){{\displaystyle \sum }}_{i=1}^{\infty }{\mathrm{t}}^i{\left(a+\left(i-1\right)d\right)}^M=\frac{A_{\mathrm{t}}^M\left(a,d\right)}{{\left(1-t\right)}^{M+1}}$

  

$11.12)A_{\mathrm{q}}^M\left(d,d\right)={{\displaystyle \sum }}_{k=0}^M{\left(q-1\right)}^{M-k}{\mathrm{H}}_1\left(k,\left[d,d\dots d\right]:d,\left[1,2\dots M\right]\right)=d^M{A}_{\mathrm{q}}^M$

$11.13){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{\left(nd\right)}^M=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{g=0}^M{\left(q-1\right)}^g{\left(-1\right)}^{M-g}\left(\begin{array}{c}\mathrm{M}\\ \mathrm{g}\end{array}\right)A_{\mathrm{q}}^{M-g}\left(d,d\right){\left(N\mathrm{d}\right)}^g+\frac{{\left(-1\right)}^{M+1}{A}_{\mathrm{q}}^M\left(d,d\right)}{{\left(q-1\right)}^{M+1}}$

$GeneralEulerianNumbersandPolynomials$ [11] $aresimilarto{H}_3\left(g\right)and{A}_{\mathrm{q}}^M\left(a,d\right)\mathrm{of}{H}_3\left(g\right)$

  

$\mathrm{We}\mathrm{can}\mathrm{handle}{{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n{▽}^{p}SUM\left(n+Y,\mathrm{PS},\mathrm{PT}\right),\mathrm{X}={\mathrm{T}}_M-\mathrm{M}-\mathrm{P}$

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

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

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

$11.14)\mathrm{Y}=0\to {{\displaystyle \sum }}_{g=0}^M{H}_1\left(g\right){\left(1-q\right)}^{M-g}{\mathrm{q}}^{g+1}=q{{\displaystyle \sum }}_{g=0}^M{H}_2\left(g\right){\left(1-q\right)}^{M-g}={{\displaystyle \sum }}_{g=0}^M{H}_3\left(g\right)q^{g+1}\stackrel{\mathrm{def}}{=}{\mathrm{A}}_q\left(PS,PT\right)$

$\mathrm{Here}\mathrm{q}\mathrm{can}\mathrm{take}\mathrm{any}\mathrm{value},\mathrm{which}\mathrm{is}\mathrm{magical}.\mathrm{q}=0.5\to 2.4);q=2\to 4.7)$

  

$11.15){{\displaystyle \sum }}_{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{q}}^n▽\mathrm{SUM}\left(\mathrm{n},\mathrm{PS},\left[1,2\dots \mathrm{M}\right]\right)=\frac{q^N}{{\left(q-1\right)}^{M+1}}{{\displaystyle \sum }}_{k=0}^M{\left(q-1\right)}^{M-k}{\left(-1\right)}^k{▽}^k\mathrm{SUM}\left(\mathrm{N}-1\right)+\frac{{\left(-1\right)}^{M+1}{\mathrm{A}}_q\left(PS,PT\right)}{{\left(q-1\right)}^{M+1}}$

$11.16){{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i{▽}^P\mathrm{SUM}\left(i+Y,\mathrm{PS},\mathrm{PT}\right)=\frac{{\mathrm{A}}_q\left(PS,PT\right)t^{-Y}}{{\left(1-t\right)}^{{\mathrm{T}}_M-\mathrm{P}+2}}$

  

${{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i{▽}^P\mathrm{SUM}\left(i+Y,\left[1\right],\left[1\right]\right)={{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i\left(\begin{array}{c}i+Y+1-p\\ 2-p\end{array}\right)=\frac{\mathrm{t}}{t^Y{\left(1-t\right)}^{3-\mathrm{P}}}\stackrel{M=2-p}{\to }$

$11.17){{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i\left(\begin{array}{c}i+Y+M-1\\ M\end{array}\right)=\frac{\mathrm{t}}{t^Y{\left(1-t\right)}^{\mathrm{M}+1}}\to {{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i\left(\begin{array}{c}i+K\\ M\end{array}\right)=\frac{t^{M-K}}{{\left(1-t\right)}^{\mathrm{M}+1}},match 11.2)$

$\frac{1}{M!}{{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i{▽}^1\mathrm{SUM}\left(i+Y,\left[1,2\dots \mathrm{M}\right],\left[1,2\dots \mathrm{M}\right]\right)={{\displaystyle \sum }}_{i=0}^{\infty }{\mathrm{t}}^i\left(\begin{array}{c}i+Y+\mathrm{M}-1\\ \mathrm{M}+1-1\end{array}\right)canalsoreachthesameconclusion.$

${{\displaystyle \sum }}_{\mathrm{n}=0}^{\infty }{\mathrm{q}}^n{▽}^1\mathrm{SUM}\left(\mathrm{i},\left[1,1\dots 1,2,2\dots 2,\mathrm{k},\mathrm{k}\dots \mathrm{k}\right],\left[1,2\dots .\mathrm{kM}\right]\right)={{\displaystyle \sum }}_{\mathrm{n}=0}^{\infty }{\mathrm{q}}^n{\left(n+1\right)}^M{\left(n+2\right)}^M\mathrm{...}{\left(n+k\right)}^M$

$H_3\left(g\right)solvedwithmatrixistheGeneralEulerianNumbersof$ [12]$,\frac{{\mathrm{A}}_q\left(PS,PT\right)}{{\left(1-q\right)}^{kM+1}}isitslimitvalue.$

11. Generalization of Wolstenholme Theorem

In this section, p is a prime number,p>3.

$\left(*\right)WolstenholmeTheorem:\left(\mathrm{p}-1\right)!{{\displaystyle \sum }}_{\mathrm{n}=1}^{\mathrm{p}-1}\frac{1}{\mathrm{n}}\equiv 0MOD{p}^2$

$\left(\mathrm{p}-1\right)\left(\mathrm{p}-2\right)\dots \left(\mathrm{p}-\mathrm{p}+1\right)=p^{p-1}-A_1{p}^{p-2}+\dots +A_{p-3}{p}^2-A_{p-2}{p}^1+A_{p-1}=\left(p-1\right)!$

$\stackrel{A_{p-1}=\left(p-1\right)!}{\to }{p}^{p-2}-A_1{p}^{p-3}+\mathrm{...}+A_{p-3}{p}^1-A_{p-2}=0\stackrel{p|A_1,p|A_2\dots p|A_{p-3}and{A}_{p-2}=\left(\mathrm{p}-1\right)!{{\displaystyle \sum }}_{\mathrm{n}=1}^{\mathrm{p}-1}\frac{1}{\mathrm{n}}}{\to }\left(*\right)$ [13]

  

$12.1){{\displaystyle \sum }}_{\mathrm{n}=1}^{\mathrm{p}-1}{n}^{p-2}\equiv 0{\mathrm{MOD}\mathrm{p}}^2$

[Proof]

${\mathrm{p}}^{\mathrm{p}-2}+{{\displaystyle \sum }}_{\mathrm{n}=1}^{\mathrm{p}-1}{\mathrm{n}}^{\mathrm{p}-2}={{\displaystyle \sum }}_{\mathrm{n}=1}^{\mathrm{p}}{\mathrm{n}}^{\mathrm{p}-2}=\mathrm{SUM}\left(\mathrm{P},\left[1,1\dots 1\right],\left[1,2\dots \mathrm{p}-2\right]\right)={{\displaystyle \sum }}_{\mathrm{g}=0}^{\mathrm{p}-2}\mathrm{g}!{\mathrm{S}}_2\left(\mathrm{p}-1,\mathrm{g}+1\right)\left(\begin{array}{c}\mathrm{p}\\ \mathrm{g}+1\end{array}\right)$

$={{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{p}-1}\left(\mathrm{g}-1\right)!{\mathrm{S}}_2\left(\mathrm{p}-1,\mathrm{g}\right)\left(\begin{array}{c}\mathrm{P}\\ \mathrm{g}\end{array}\right)=\mathrm{p}{{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{p}-1}\frac{\mathrm{g}!{\mathrm{S}}_2\left(\mathrm{p}-1,\mathrm{g}\right)}{{\mathrm{g}}^2}\left(\begin{array}{c}\mathrm{p}-1\\ \mathrm{g}-1\end{array}\right)\stackrel{\frac{\mathrm{g}!{\mathrm{S}}_2\left(\mathrm{p}-1,\mathrm{g}\right)}{{\mathrm{g}}^2}\left(\begin{array}{c}\mathrm{p}-1\\ \mathrm{g}-1\end{array}\right) is an integer}{\to }$

$\frac{{{\displaystyle \sum }}_{\mathrm{n}=1}^{\mathrm{p}}{\mathrm{n}}^{\mathrm{p}-2}}{\mathrm{p}}={{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{p}-1}\frac{\mathrm{g}!{\mathrm{S}}_2\left(\mathrm{p}-1,\mathrm{g}\right)}{{\mathrm{g}}^2}\left(\begin{array}{c}\mathrm{p}-1\\ \mathrm{g}-1\end{array}\right)\stackrel{3.8)}{\to }\equiv {{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{p}-1}\frac{{\left(-1\right)}^{\mathrm{g}+1}}{{\mathrm{g}}^2}\left(\begin{array}{c}\mathrm{p}-1\\ \mathrm{g}-1\end{array}\right)\stackrel{\left(\begin{array}{c}\mathrm{p}-1\\ \mathrm{g}-1\end{array}\right)\equiv {\left(-1\right)}^{\mathrm{g}-1}}{\to }\equiv {{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{p}-1}\frac{1}{{\mathrm{g}}^2}\equiv {{\displaystyle \sum }}_{\mathrm{g}=1}^{\mathrm{p}-1}{\mathrm{g}}^2\equiv 0\mathrm{MODp}$

q.e.d.

(*)=SUM(2,[1,2…p-2],[1,3…2(p-2)-1]); 12.1)=SUM(p-1,[1,1…1],[1,2…p-2])

They are two extremes and their proof methods are special.Reconsidering shape:

$\mathrm{PS}\mathrm{starts}\mathrm{from}\left[1,1\dots 1\right],\mathrm{ends}\mathrm{at}\left[1,2\dots \mathrm{p}-2\right],{\mathrm{K}}_1=1,\left\{\begin{array}{c}{\mathrm{K}}_i={\mathrm{K}}_{i+1},continuity\\ {\mathrm{K}}_i+1={\mathrm{K}}_{i+1},discontinuity\end{array}\right.$

$\mathrm{PT}\mathrm{starts}\mathrm{from}\left[1,2\dots \mathrm{p}-2\right],\mathrm{ends}\mathrm{at}\left[1,3\dots 2\mathrm{p}-5\right],{\mathrm{T}}_1=1,\left\{\begin{array}{c}{\mathrm{T}}_i+1={\mathrm{T}}_{i+1},continuity\\ {\mathrm{T}}_i+2={\mathrm{T}}_{i+1},discontinuity\end{array}\right.$

$12.2){{\displaystyle \sum }}_{c_1+c_2+\dots +c_q=P-2,c_i>0}{k}_1^{c_1}{k}_2^{c_2}\dots k_q^{c_q}\equiv 0{\mathrm{MOD}\mathrm{p}}^2,K_i\ne K_{j}1\le K_i\le P-1$

[Proof]

$\mathrm{If}\mathrm{A}\equiv 0{\mathrm{MOD}\mathrm{p}}^2 and \mathrm{A}+\mathrm{B}\equiv 0{\mathrm{MOD}\mathrm{p}}^2,then B\equiv 0{\mathrm{MOD}\mathrm{p}}^2.$

$The Sum has symmetry.For \sum {\mathrm{A}}^1{\mathrm{B}}^{P-3},{\mathrm{pair}\mathrm{each}\mathrm{product}\mathrm{A}}^1{\mathrm{B}}^{P-3}\mathrm{with}\mathrm{another}{\left(\mathrm{p}-\mathrm{A}\right)}^1{\mathrm{B}}^{P-3}\mathrm{to}\mathrm{p}\mathrm{ⅹ}{\mathrm{B}}^{P-3}$

${\left(\mathrm{p}-\mathrm{A}\right)}^1{\mathrm{B}}^{P-3}\in \sum {\mathrm{A}}^1{\mathrm{B}}^{P-3}or\sum {\mathrm{B}}^{P-2}\stackrel{symmetry}{\to }\sum \left(\mathrm{Paired}\mathrm{products}\right)=x\sum {\mathrm{A}}^1{\mathrm{B}}^{P-3}+y\sum {\mathrm{B}}^{P-2}=p\sum {\mathrm{B}}^{P-3}$

$\sum {\mathrm{B}}^{P-3}\equiv 0\mathrm{MOD}\mathrm{p}\to p\sum {\mathrm{B}}^{P-3}\equiv 0{\mathrm{MOD}\mathrm{p}}^2\stackrel{obviously:0<x,y<p}{\to }\sum {\mathrm{A}}^1{\mathrm{B}}^{P-3}\equiv 0{\mathrm{MOD}\mathrm{p}}^2$  $For \sum {\mathrm{A}}^2{\mathrm{B}}^{P-4},{\mathrm{pair}\mathrm{A}}^2{\mathrm{B}}^{P-4}\mathrm{with}{\left(\mathrm{p}-\mathrm{A}\right)}^{1}A{\mathrm{B}}^{P-4}\mathrm{to}\mathrm{p}\mathrm{ⅹ}{\mathrm{AB}}^{P-4} ,{\left(\mathrm{p}-\mathrm{A}\right)}^{1}A{\mathrm{B}}^{P-4}\in \sum {\mathrm{A}}^2{\mathrm{B}}^{P-4} or \sum {\mathrm{AB}}^{P-3}$

$\sum \left(\mathrm{Paired}\mathrm{products}\right)=x\sum {\mathrm{A}}^2{\mathrm{B}}^{P-4}+y\sum A{\mathrm{B}}^{P-3}=p\sum A{\mathrm{B}}^{P-4}\equiv 0{\mathrm{MOD}\mathrm{p}}^2\to \sum {\mathrm{A}}^2{\mathrm{B}}^{P-4}\equiv 0{\mathrm{MOD}\mathrm{p}}^2$

$Usethesameway\to {{\displaystyle \sum }}_{a+b=P-2,A\ne B,1\le A,B\le P-1}{\mathrm{A}}^a{\mathrm{B}}^b\equiv 0{\mathrm{MOD}\mathrm{p}}^2$

  

$\sum {\mathrm{A}}^1{\mathrm{B}}^b{C}^c,\mathrm{pair}\mathrm{with}{\left(\mathrm{p}-\mathrm{A}\right)}^1{\mathrm{B}}^b{C}^c\mathrm{to}\mathrm{p}\mathrm{ⅹ}{\mathrm{B}}^b{C}^c,{\left(\mathrm{p}-\mathrm{A}\right)}^1{\mathrm{B}}^b{C}^c\in \sum {\mathrm{A}}^1{\mathrm{B}}^b{C}^c or \sum {\mathrm{B}}^{b+1}{C}^c or \sum {\mathrm{B}}^b{C}^{c+1}$

$Usethesameway\to conclusion$

q.e.d.

  

$12.3){{\displaystyle \sum }}_{sameshape,M=p-2}SUM\left(p-1-PB\left(PT\right),PS,PT\right)\equiv 0{\mathrm{MOD}\mathrm{p}}^2,0\le PB\le p-3$

SUM(3,[1,1,2],[1,2,4])+SUM(3,[1,2,2],[1,3,4])

=2(1²)+3(1²+2²)+4(1²+2²+3² )+ 2² (1)+ 3² (1+2)+ 4² (1+2+3)=200≡0MOD25

  

(#) SUM(5,[1,1,1,1,2],[1,2,3,4,6]) + SUM(5,[1,2,2,2,2],[1,3,4,5,6])=

2(1⁴)+3(1⁴+2⁴)+4(1⁴+2⁴+3⁴)+ 5(1⁴+2⁴+3⁴+4⁴)+6(1⁴+2⁴+3⁴+4⁴+5⁴)+

2⁴(1)+ 3⁴ (1+2)+ 4⁴ (1+2+3)+ 5⁴ (1+2+3+4)+ 6⁴ (1+2+3+4+5)= 35574≡0MOD49

  

(##) SUM(5,[ 1,1,1,2,2],[ 1,2,3,5,6]) + SUM(5,[ 1,1,2,2,2],[ 1,2,4,5,6]) =

2²(1³)+ 3² (1³+2³)+ 4² (1³+2³+3³)+ 5² (1³+2³+3³+4³)+ 6² (1³+2³+3³+4³+5³)

2³(1²)+ 3³ (1²+2²)+ 4³ (1²+2²+3²)+ 5³ (1²+2²+3²+4²)+ 6³ (1²+2²+3²+4²+5²) =27930≡0MOD49

(#)+(##)=12.3)

  

$12.4)E_{P-2}^{P-1}=S_2\left(2p-3,p-1\right)\equiv 0{\mathrm{MOD}\mathrm{p}}^2;E_{p-2}^p=S_2\left(2p-2,p\right)\equiv 0{\mathrm{MOD}\mathrm{p}}^2$

$S_2\left(7,4\right)=350\mathrm{and}{S}_2\left(8,5\right)=1050\equiv 0\mathrm{MOD}{5}^2;S_2\left(11,6\right)=179487\mathrm{and}{S}_2\left(12,7\right)=627396\equiv 0\mathrm{MOD}{7}^2$

  

$\mathrm{Generally},\mathrm{PS}\mathrm{starts}\mathrm{from}\left[K_2,K_3\dots K_M\right],\mathrm{ends}\mathrm{at}\left[K_2+1,K_3+2\dots K_M+M-1\right],K_i\in \mathbb{N}$

$\mathrm{PS}=\left[K_2+I_2,K_3+I_3\dots K_M+I_M\right],{\mathrm{I}}_1=0,\left\{\begin{array}{c}{\mathrm{I}}_i={\mathrm{I}}_{i+1},continuity\\ {\mathrm{I}}_i+1={\mathrm{I}}_{i+1},discontinuity\end{array}\right.$

$\mathrm{PT}\mathrm{starts}\mathrm{from}\left[K_2+1,K_3+2\dots K_M+M-1\right],\mathrm{ends}\mathrm{at}\left[K_2+2,K_3+4\dots K_M+2\left(M-1\right)\right]$

$\mathrm{PT}=\left[K_2+J_2,K_3+J\dots K_M+J_M\right],{\mathrm{J}}_1=0,\left\{\begin{array}{c}{\mathrm{J}}_i+1={\mathrm{J}}_{i+1},continuity\\ {\mathrm{J}}_i+2={\mathrm{J}}_{i+1},discontinuity\end{array}\right.$

$F\left(PB\right)\stackrel{\mathrm{def}}{=}{{\displaystyle \sum }}_{sameshape,PB\left(PTx\right)=PB,0\le PB\le M-1}SUM\left(N,PSx,PTx\right)={{\displaystyle \sum }}_{sameshape}{{\displaystyle \sum }}_{g=0}^{M-1}{H}_1\left(g,PSx,PTx\right)\left(\begin{array}{c}N+X\\ X+1+g\end{array}\right)$

$={{\displaystyle \sum }}_{g=0}^{M-1}{{\displaystyle \sum }}_{sameshape}{H}_1\left(g,PSx,PTx\right)\left(\begin{array}{c}N+X\\ X+1+g\end{array}\right)\stackrel{\mathrm{def}}{=}{{\displaystyle \sum }}_{g=0}^{M-1}{H}_1\left(g,PB\right)\left(\begin{array}{c}N+X\\ X+1+g\end{array}\right)$

$12.5)H_1\left(g,PB\right)=H_1\left(PB,g\right)$

[Proof]

$K_1\times H_1\left(g,0\right)\stackrel{2.14)}{\to }{\mathrm{MIN}}_g\left(\left[K_1,K_2\dots K_M\right]\right)\stackrel{\mathrm{def}}{\to }{K}_1\times H_1\left(0,g\right)$

$K_1\times H_1\left(g,M-1\right)\stackrel{2.14)}{\to }{\mathrm{MIN}}_g\left(\left[K_1,K_2+1\dots K_M+M-1\right]\right)\stackrel{\mathrm{def}}{\to }{K}_1\times H_1\left(M-1,g\right)$

$K_1\times H_1\left(g,PB\right)={\displaystyle \sum }_{K_1\times H_1\left(g,0\right)=MI{N}_{PB}\left(\left[K_1,K_2\dots K_M\right]\right)}{H}_1\left(g\right)={\displaystyle \sum }_{K_1\times H_1\left(g,0\right)=MI{N}_{PB}\left(\left[K_1,K_2\dots K_M\right]\right)}MI{N}_g\left(PS\right)$

$K_1\times H_1\left(PB,g\right)={\displaystyle \sum }_{K_1\times H_1\left(g,0\right)=MI{N}_g\left(\left[K_1,K_2\dots K_M\right]\right)}MI{N}_{PB}\left(PS\right)$

$Theirquantityis\left(\begin{array}{c}M-1\\ PB\end{array}\right)\left(\begin{array}{c}M-1\\ g\end{array}\right)andtheyarepairedonebyone$

q.e.d.

  

g from M-1 to 0, list H₁(PB,g) of F(M-1),F(M-2)…F(0) row by row, the matrix will be symmetrical diagonally.

Table 12. 1:H1(PB,g) of PS=[1,1,1],PT=[2,3,4].

Table 12. 1:H1(PB,g) of PS=[1,1,1],PT=[2,3,4].

12. Possible future research directions

The proof process of formal calculation needs to meet f(a+b)=f(a)f(b), so it cannot be further generalized through the method of this article.

$1)CanFormalCalculationbeextendedtootherfunctions?$

  

Examining various products,SUM(3,[1,1],[1,4])=1×1+2×{(1+2)+1} +3×{(1+2+3)+(1+2)+1}

$2)PT=\left[1,4,7\dots .\right],PT=\left[1,5,9\dots \right]arerarelystudied.$

$3)FindsomecalculationstoextendPTto ℂ, especiallytorationalfractions.$

$4)Otherexpressionsfor{E}_p^q\odot \left(\left[T_1,T_2\dots \right],C\right).$

  

$From2.2)and2.3),thereexistsformulasbetween{S}_{1,2}\left(M,g\right)and{S}_{2,2}\left(M,g\right)$

$5)Arethereanyformulasbetween{S}_{1,r}\left(M,g\right)and{S}_{2,r}\left(M,g\right)$

$From3.4)and3.5),thereis\left(\begin{array}{c}rN+M\\ rg+M\end{array}\right)intheformulas,butthereisalso\frac{1}{r^{N-g}}\frac{\left(rN-rg\right)!}{\left(N-g\right)!}or\frac{1}{r{!}^{N-g}}\frac{\left(rN-rg\right)!}{\left(N-g\right)!}$

$6)CanFormalCalculationbeextendedto\left(\begin{array}{c}rN+Y\\ X\end{array}\right)?$

$7)Findarelationshipsimilarto2.2)ingeneral{H}^q\left(g\right)$

  

$From10.1),W\left(g_1,g_2\dots g_p,\left[T_1,T_2\dots T_M\right]\right)={{\displaystyle \prod }}_{i=1}^M{T}_i\left(\begin{array}{c}M\\ g_1,g_2\dots g_p\end{array}\right).$

$Wewanttouseit,whichleadstothegenerationofMultiparameterFormalCalcution.Butitwasstillnotused.$

$8)FindsomecalculationstouseW\left(g_1,g_2\dots g_p,\left[T_1,T_2\dots T_M\right]\right)$

  

Even 2-parameters Formal Calcution are complex.

$9)UsingMultiparameterFormalCalcutionforanalysis$