Formal Calculation of Binomial Coefficients

1. Calculation Formula

Definition 1.1. 

Recursively define ${\nabla }^p$, $p\in \mathbb{Z}$,

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

Definition 1.2. 

Recursively define $SUM\left(N\right)=SUM(N,PS,PT)$, $K_i,D_i\in \mathbb{C}$, $T_i\in \mathbb{N}$.

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

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

Abbreviations: $[{\mathit{K}}_{\mathit{1}}:\mathit{D},{\mathit{K}}_{\mathit{2}}:\mathit{D}...{\mathit{K}}_{\mathit{M}}:\mathit{D}]=[{\mathit{K}}_{\mathit{1}},{\mathit{K}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}}]:\mathit{D}$, $[{\mathit{K}}_{\mathit{1}},{\mathit{K}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}}]:\mathit{1}=[{\mathit{K}}_{\mathit{1}},{\mathit{K}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}}]$.

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

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

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

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

$SUM(N,PS,[1,4])={\sum }_{n_3=0}^{N-1}(K_2+n_3{D}_2){\sum }_{n_2=0}^{n_3}{\sum }_{n_1=0}^{n_2}(K_1+n_1{D}_1)$.

In this paper, the default $\mathit{PS}=[{\mathit{K}}_{\mathit{1}}:{\mathit{D}}_{\mathit{1}},{\mathit{K}}_{\mathit{2}}:{\mathit{D}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}}:{\mathit{D}}_{\mathit{M}}]$, $PT=[T_1,T_2...T_M],T_i<T_{i+1}$.

Use $\mathbb{K}$ to represent the set $\left\{K_i\right\}$, $\mathbb{T}$ to represent the set $\left\{K_i\right\}$.

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

Definition 1.3. 

$X\left(T\right)$=Number of $\{X_1,X_2...X_M\}\in \mathbb{T}$.

Definition 1.4. 

$X_{T-1}$=Number of $\{X_1,X_2...X_{i-1}\}\in \mathbb{T}$, $X_{K-1}$=Number of $\{X_1,X_2...X_{i-1}\}\in \mathbb{K}$.

Obviously, $X_{T-1}+X_{K-1}=i-1$. Use the auxiliary form and each $X_i$ cannot be exchanged:

Theorem 1.1. 

$H=T_M-M,SUM(N,PS,PT)=$

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

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

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

$H_i\left(g\right)=H_i(g,PS,PT)=H_i(g,M)$, is defined as ${\sum }_{X\left(T\right)=g}{\prod }_{i=1}^M{B}_i$.

Proof. 

$H_1\left(g\right)=H_1(g-1,M-1)(T_M-[(M-1)-(g-1)])D_M+H_1(g,M-1)(K_M+g{D}_M)$.

${\sum }_{n=0}^{N-1}n\left.{(}_M^{n+K}\right)={\sum }_{n=0}^{N-1}(n-K+M)\left.{(}_M^{n+K}\right)+(M-K)\left.{(}_M^{n+K}\right)=(M+1)\left.{(}_{M+2}^{N+K}\right)+(M-K)\left.{(}_{M+1}^{N+K}\right)$.

M = 1, it holds. Suppose that holds when M, $PS1=[PS,K_{M+1}:D_{M+1}]$, $PT1=[PT,T_{M+1}=T_M+2-p]$.

$SUM(N,PS1,PT1)={\sum }_{n=0}^{N-1}(K_{M+1}+n{D}_{M+1}){\nabla }^{p}SUM(n+1)$

$={\sum }_{n=0}^{N-1}(K_{M+1}+n{D}_{M+1}){\sum }_{g=0}^M{H}_1\left(g\right)\left.{(}_{T_M-M+1+g-p}^{n+1+T_M-M-p}\right)$

$={\sum }_{g=0}^M(K_{M+1}+g{D}_{M+1})H_1\left(g\right)\left.{(}_{T_M-M+2+g-p}^{N+1+T_M-M-p}\right)+{\sum }_{g=0}^M(T_M-M+2+g-p)D_{M+1}{H}_1\left(g\right)\left.{(}_{T_M-M+3+g-p}^{N+1+T_M-M-p}\right)$

$={\sum }_{g=0}^M(K_{M+1}+g{D}_{M+1})H_1\left(g\right)\left.{(}_{T_{M+1}-(M+1)+1+g}^{N+T_{M+1}-(M+1)}\right)+{\sum }_{g=0}^M(T_{M+1}-(M-g))D_{M+1}{H}_1\left(g\right)\left.{(}_{T_{M+1}-(M+1)+2+g}^{N+T_{M+1}-(M+1)}\right)$

$={\sum }_{g=0}^{M+1}{H}_1(g,M+1)\left.{(}_{T_{M+1}-(M+1)+1+g}^{N+T_{M+1}-(M+1)}\right)$. Proof of $For{m}_1$ completion.

$\sum _{n=0}^{N-1}n\left.{(}_M^{n+K}\right)$$=(M+1)\left.{(}_{M+2}^{N+K+1}\right)-(1-K)\left.{(}_{M+1}^{N+K}\right)=(M-K)\left.{(}_{M+2}^{N+K+1}\right)+(1+K)\left.{(}_{M+2}^{N+K}\right)$. Prove $For{m}_{2,3}$ through it. □

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

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

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

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

$H_1\left(g\right)=H_1(g-1,M-1)(T_M-[M-g])D_M+H_1(g,M-1)(K_M+g{D}_M)$.

$H_2\left(g\right)=H_2(g-1,M-1)(T_M-[M-g)])D_M+H_2(g,M-1)(K_M+[M-1-g-T_M]D_M)$.

$H_3\left(g\right)=H_3(g-1,M-1)(-K_M+(T_M-[g-1])D_M)+H_3(g,M-1)(K_M+g{D}_M)$.

They can be unified as: $H\left(g\right)=H(g-1,M-1)B+H(g,M-1)A$.

$H_1\left(g\right),B=B_M+(g-1)D_M,A=A_M+g{D}_M$, $A_M=K_M,B_M=T_M{D}_M-(M-1)D_M$.

$H_2\left(g\right),B=B_M+(g-1)D_M,A=A_M-g{D}_M$, $A_M=K_M+(M-1-T_M)D_M,B_M=T_M{D}_M-(M-1)D_M$.

$H_3\left(g\right),B=B_M-(g-1)D_M,A=A_M+g{D}_M$, $A_M=K_M,B_M=-K_M+T_M{D}_M$.

Consider the general two-dimensional second-order linear recursive equations:

$\mathit{R}(\mathit{M},\mathit{g})=\mathit{R}(\mathit{M}-\mathit{1},\mathit{g}-\mathit{1})({\mathit{B}}_{\mathit{M}}+(\mathit{g}-\mathit{1}){\mathit{E}}_{\mathit{M}})+\mathit{R}(\mathit{M}-\mathit{1},\mathit{g})({\mathit{A}}_{\mathit{M}}+{\mathit{gD}}_{\mathit{M}})$. It can be calculated in a similar way to $\mathit{H}\left(\mathit{g}\right)$. $\mathit{H}\left(\mathit{g}\right)$ itself requires $|{\mathit{D}}_{\mathit{i}}|=|{\mathit{E}}_{\mathit{i}}|$ and can’t change the sign, so that 3-forms exist.

2. Property

Theorem 2.1.

(1). $H_1\left(g\right)={\sum }_{k=g}^M{H}_2\left(k\right)\left.{(}_g^k\right)={\sum }_{k=0}^g{H}_3\left(k\right)\left.{(}_{M-g}^{M-k}\right)$.

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

(3). $H_2\left(g\right)={(-1)}^{M-g}{\sum }_{k=M-g}^M{H}_3\left(k\right)\left.{(}_{M-g}^k\right)$, $H_3\left(g\right)={(-1)}^g{\sum }_{k=g}^M{H}_2(M-k)\left.{(}_g^k\right)$.

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

(5). ${\sum }_{g=0}^M{H}_1\left(g\right)\left.{(}_{Y-g}^X\right)={\sum }_{g=0}^M{H}_2\left(g\right)\left.{(}_Y^{X+g}\right)={\sum }_{g=0}^M{H}_3\left(g\right)\left.{(}_{Y-g}^{X+M-g}\right),Y\in \mathbb{N}$.

Recursive relations yields (1), inversion yields (2), (4) and (5) can be obtained from (1). (5) shows $For{m}_1=For{m}_2=For{m}_3$. If we ignore the practical significance, $T_i\in \mathbb{C}$. $q^X={\sum }_{k=0}^X{(-1)}^k{(1-q)}^k\left.{(}_k^X\right),{(1-q)}^X={\sum }_{k=0}^X{(-1)}^k{q}^k\left.{(}_k^X\right)$ and (4) yields (3).

Theorem 2.2.

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

(2). $SUM(N,[L_1,L_2...L_q,PS],[L_1,L_2...L_q,PT])={\prod }_{i=1}^q{L}_i\times SUM(N,PS,PT)$.

(3). $SUM(N,PT,PT)=\prod T_i\left.{(}_{T_M+1}^{N+T_M}\right),H_1\left(g\right)=\prod T_i\left.{(}_g^M\right)$.

(4). In $SUM(N,[...PS...],[...T,T+1..T+M-1...\left]\right)$, $K_i:D_i$ can exchange order.

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

(1) is obvious, (2) and (3) is derived from (1), so $T_1$ can great than 1. (3) is the promotion of ${\sum }_{\mathit{n}=\mathit{0}}^{\mathit{N}-\mathit{1}}\left.{(}_{\mathit{M}}^{\mathit{n}+\mathit{M}}\right)=\left.{(}_{\mathit{M}+\mathit{1}}^{\mathit{N}+\mathit{M}}\right)$. (4) comes from the definition. (5) can be proved by induction.

Definition 2.1. 

$F_g^{\mathbb{K}}={\sum }_{1\le {\lambda }_1<{\lambda }_2<...<{\lambda }_g\le M}{K}_{{\lambda }_1}{K}_{{\lambda }_2}...K_{{\lambda }_g},F_0^{\mathbb{K}}=1$, $F_g^N=F_g^{\{1,2...N\}}$.

Definition 2.2. 

$E_g^{\mathbb{K}}={\sum }_{1\le {\lambda }_1\le {\lambda }_2\le ...\le {\lambda }_g\le M}{K}_{{\lambda }_1}{K}_{{\lambda }_2}...K_{{\lambda }_g},E_0^{\mathbb{K}}=1$, $E_g^N=E_g^{\{1,2...N\}}$.

${\mathit{F}}_{\mathit{M}}^{\mathit{N}+\mathit{M}-\mathit{1}}={\mathit{S}}_{\mathit{1}}(\mathit{N}+\mathit{M},\mathit{N})$, $S_1$ is unsigned stirling number of the first kind.

${\mathit{E}}_{\mathit{M}}^{\mathit{N}}={\mathit{S}}_{\mathit{2}}(\mathit{N}+\mathit{M},\mathit{N})$, $S_2$ is stirling number of the second kind.

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

$\mathit{SUM}(\mathit{N},[\mathit{1},\mathit{1}...\mathit{1}],[\mathit{1},\mathit{3}...\mathit{2}\mathit{M}-\mathit{1}])=\mathit{SUM}(\mathit{N},[\mathit{1},\mathit{1}...\mathit{1}],[\mathit{3},\mathit{5}...\mathit{2}\mathit{M}-\mathit{1}])={\mathit{S}}_{\mathit{2}}(\mathit{N}+\mathit{M},\mathit{N})$,

$\mathit{SUM}(\mathit{N},[\mathit{1},\mathit{2}...\mathit{M}],[\mathit{1},\mathit{3}...\mathit{2}\mathit{M}-\mathit{1}])=\mathit{SUM}(\mathit{N},[\mathit{2},\mathit{3}...\mathit{M}],[\mathit{3},\mathit{5}...\mathit{2}\mathit{M}-\mathit{1}])={\mathit{S}}_{\mathit{1}}(\mathit{N}+\mathit{M},\mathit{N})$.

They can be used to calculate ${\mathit{S}}_{\mathit{1}}(\mathit{N},\mathit{N}-\mathit{M})$ and ${\mathit{S}}_{\mathit{2}}(\mathit{N},\mathit{N}-\mathit{M})$.

In the calculation of $\mathit{H}\left(\mathit{g}\right)$, $\prod {\mathit{X}}_{\mathit{i}}=(\prod {\mathit{X}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}\in \mathbb{T})(\prod {\mathit{X}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}\in \mathbb{K})$.

Definition 2.3. 

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

Obviously, $H_1(g,\sum K,[1,1..1],PT)=E_{M-g}^{g+1}$. $PT1=[T,T+1...T+M-1],PS1=[K+M-1,K+M-2...K]$,

$D_i=1,H_1(g,PS,PT1)={\left[T\right]}^{g}H(g,\sum K),H_1(g,PS1,PT)={[K+M-1]}_{g}H(g,\sum T)$.

Theorem 2.3. 

$\mathbb{S}=\{T_i-K_i-i+1\}$, $D_i=1$

(1). $H_1(g,\sum K)=H_3(g,\sum K)=F_{M-g}^{\mathbb{K}}{E}_0^g+F_{M-g-1}^{\mathbb{K}}{E}_1^g+...+F_0^{\mathbb{K}}{E}_{M-g}^g$.

(2). $H_1(g,\sum T)=H_2(g,\sum T)=F_g^{\mathbb{T}}{E}_0^{M-g}-F_{g-1}^{\mathbb{T}}{E}_1^{M-g}+...+{(-1)}^g{F}_0^{\mathbb{T}}{E}_g^{M-g}$.

(3). $H_2(g,\sum K)={(-1)}^{M-g}(F_{M-g}^{\mathbb{S}}{E}_0^g+...+F_0^{\mathbb{S}}{E}_{M-g}^g)$.

(4). $H_3(g,\sum T)={(-1)}^g{F}_g^{\mathbb{S}}{E}_0^{M-g}+{(-1)}^{g-1}{F}_{g-1}^{\mathbb{S}}{E}_1^{M-g}...+F_0^{\mathbb{S}}{E}_g^{M-g}$.

(5). $H_1(g,\sum K)=\sum {\prod }_{i=1}^{M-g}(K_{i+{\lambda }_i}+{\lambda }_i),0\le {\lambda }_1\le {\lambda }_2\le ...\le {\lambda }_{M-g}\le g$.

(6). $H_1(g,\sum T)=\sum {\prod }_{i=1}^g(T_{i+{\lambda }_i}-{\lambda }_i),0\le {\lambda }_1\le {\lambda }_2\le ...\le {\lambda }_g\le M-g$.

Proof. 

$\mathit{PS}\mathit{1}=[\mathit{PS},{\mathit{K}}_{\mathit{M}+\mathit{1}}],\mathit{PT}\mathit{1}=[\mathit{PT},{\mathit{T}}_{\mathit{M}+\mathit{1}}]$. Using induction to prove (2).

${\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{T},\mathit{M}+\mathit{1})={\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{T})+{\mathit{H}}_{\mathit{1}}(\mathit{g}-\mathit{1},\sum \mathit{T})({\mathit{T}}_{\mathit{M}+\mathit{1}}-(\mathit{M}-\mathit{g}+\mathit{1}))$.

${\mathit{F}}_{\mathit{g}-\mathit{x}}^{\left\{\mathit{PT}\mathit{1}\right\}}$ in ${\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{T},\mathit{M}+\mathit{1})$ has three sources.

$={(-\mathit{1})}^{\mathit{x}}{\mathit{F}}_{\mathit{g}-\mathit{x}}^{\mathbb{T}}{\mathit{E}}_{\mathit{x}}^{\mathit{M}-\mathit{g}}+{(-\mathit{1})}^{\mathit{x}}{\mathit{F}}_{(\mathit{g}-\mathit{1})-\mathit{x}}^{\mathbb{T}}{\mathit{E}}_{\mathit{x}}^{\mathit{M}-(\mathit{g}-\mathit{1})}{\mathit{T}}_{\mathit{M}+\mathit{1}}+{(-\mathit{1})}^{\mathit{x}-\mathit{1}}{\mathit{F}}_{(\mathit{g}-\mathit{1})-(\mathit{x}-\mathit{1})}^{\mathbb{T}}{\mathit{E}}_{\mathit{x}-\mathit{1}}^{\mathit{M}-(\mathit{g}-\mathit{1})}(-(\mathit{M}-\mathit{g}+\mathit{1}))$

$={(-\mathit{1})}^{\mathit{x}}{\mathit{F}}_{\mathit{g}-\mathit{x}}^{\mathbb{T}}({\mathit{E}}_{\mathit{x}}^{\mathit{M}-\mathit{g}}+{\mathit{E}}_{\mathit{x}-\mathit{1}}^{\mathit{M}+\mathit{1}-\mathit{g}}(\mathit{M}+\mathit{1}-\mathit{g}))+{(-\mathit{1})}^{\mathit{x}}{\mathit{F}}_{\mathit{g}-\mathit{x}-\mathit{1}}^{\mathbb{T}}{\mathit{E}}_{\mathit{x}}^{\mathit{M}+\mathit{1}-\mathit{g}}{\mathit{T}}_{\mathit{M}+\mathit{1}}$ (*)

$={(-\mathit{1})}^{\mathit{x}}{\mathit{F}}_{\mathit{g}-\mathit{x}}^{\mathbb{T}}{\mathit{E}}_{\mathit{x}}^{\mathit{M}+\mathit{1}-\mathit{g}}+{(-\mathit{1})}^{\mathit{x}}{\mathit{F}}_{\mathit{g}-\mathit{x}-\mathit{1}}^{\mathbb{T}}{\mathit{E}}_{\mathit{x}}^{\mathit{M}+\mathit{1}-\mathit{g}}{\mathit{T}}_{\mathit{M}+\mathit{1}}$$={(-\mathit{1})}^{\mathit{x}}{\mathit{E}}_{\mathit{x}}^{\mathit{M}+\mathit{1}-\mathit{g}}{\mathit{F}}_{\mathit{g}-\mathit{x}}^{\left\{\mathit{PT}\mathit{1}\right\}}$.

${\mathit{E}}_{\mathit{x}}^{\mathit{M}-\mathit{g}}+{\mathit{E}}_{\mathit{x}-\mathit{1}}^{\mathit{M}+\mathit{1}-\mathit{g}}(\mathit{M}+\mathit{1}-\mathit{g})$

$={\mathit{S}}_{\mathit{2}}(\mathit{M}-\mathit{g}+\mathit{x},\mathit{M}-\mathit{g})+(\mathit{M}+\mathit{1}-\mathit{g}){\mathit{S}}_{\mathit{2}}(\mathit{M}-\mathit{g}+\mathit{x},\mathit{M}+\mathit{1}-\mathit{g})$

$={\mathit{S}}_{\mathit{2}}(\mathit{M}-\mathit{g}+\mathit{x}+\mathit{1},\mathit{M}+\mathit{1}-\mathit{g})={\mathit{E}}_{\mathit{x}}^{\mathit{M}+\mathit{1}-\mathit{g}}\to$ (*).

(5) and (6) are definitions. □

Definition 2.4. 

${\mathit{F}}_{\mathit{T}}(\mathit{N}+\mathit{M},\mathit{N})={\mathit{F}}_{\mathit{M}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{N}+\mathit{M}-\mathit{1}\}},{\mathit{E}}_{\mathit{T}}(\mathit{N}+\mathit{M},\mathit{N})={\mathit{E}}_{\mathit{M}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{N}-\mathit{1}\}}$.

Obviously, $F_0(N+M,N)=S_1(N+M,N),E_1(N+M,N)=S_2(N+M,N)$, $F_T(M+1,g+1)=F_{M-g}^{\{T,T+1...T+M\}}=(T+M)F_T(M,g+1)+F_T(M,g)$, $E_T(M+1,g+1)=E_{M-g}^{\{T,T+1...T+g\}}=(T+g)E_T(M,g+1)+E_T(M,g)$.

Definition 2.5. 

${\left.{\langle }_j^n\right)}_T=(T-1+n-j){\left.{\langle }_{j-1}^{n-1}\right)}_T+(j+1){\left.{\langle }_j^{n-1}\right)}_T,{\left.{\langle }_0^1\right)}_T=T,{\left.{\langle }_1^1\right)}_T=0,{\left.{\langle }_j^n\right)}_T=0,j<0,j>n$.

Obviously: ${\left.{\langle }_0^n\right)}_1=T,{\left.{\langle }_n^n\right)}_T=0$. $n>0,{\left.{\langle }_j^n\right)}_1=\left.{\langle }_j^n\right)$ is Eulerian number.

Definition 2.6. 

${\left.{\langle }_{\mathit{g}}^{\mathit{M}}\right)}_{\mathit{T}}^{\mathit{j}}={\sum }_{\mathit{i}=\mathit{0}}^{\mathit{j}}{(-\mathit{1})}^{\mathit{i}}{\left.{\langle }_{\mathit{g}-\mathit{1}-\mathit{i}}^{\mathit{M}-\mathit{j}}\right)}_{\mathit{T}}\left.{(}_{\mathit{i}}^{\mathit{j}}\right),\mathit{0}\le \mathit{j}<\mathit{g},\mathit{0}<\mathit{g}<\mathit{M}$.

Theorem 2.4.

(1). $PT=[T,T+1...T+M-1],H_2(g,\sum K)={\sum }_{j=0}^{M-g}{(-1)}^{M-g-j}{F}_j^{\mathbb{K}}{E}_{M-g-j}^{\{T,T+1...T+g\}}$.

(2). ${\sum }_{j=0}^{M-g}{F}_j^{\{T-K_i\}}{E}_{M-g-j}^g={\sum }_{j=0}^{M-g}{(-1)}^j{F}_j^{\mathbb{K}}{E}_{M-g-j}^{\{T,T+1...T+g\}}$.

(3). $PT=[T,T+1...T+M-1],H_3(0<g<M,\sum K)={\sum }_{j=0}^{M-1}{\left.{\langle }_g^M\right)}_T^j{F}_j^{\mathbb{K}}+{(-1)}^g\left.{(}_g^M\right)F_M^{\mathbb{K}}$.

(4). $PS=[K,K-1...K-M+1]$, $H_2(0<g<M,\sum T)={\sum }_{j=0}^{M-1}-{\left.{\langle }_{M-g}^M\right)}_{-K}^j{F}_j^{\mathbb{T}}+{(-1)}^{M-g}\left.{(}_g^M\right)F_M^{\mathbb{T}}$.

(5). $PS=[K,K-1...K-M+1]$, $H_3(g,\sum T)={\sum }_{j=0}^g{(-1)}^{g-j}{F}_j^{\mathbb{T}}{E}_{g-j}^{\{K,K-1...K-M+g\}}$.

(6). ${\sum }_{j=0}^{M-g}{F}_j^{\{K_i-T\}}{E}_{M-g-j}^g={(-1)}^{M-g}{\sum }_{j=0}^{M-g}{(-1)}^j{F}_j^{\mathbb{K}}{E}_{M-g-j}^{\{T,T-1...T-g\}}$.

Proof. 

$PT=[T,T+1...T+M-1],H_2(g,\sum K)$ must be is a symmetric function $={\sum }_{j=0}^M{A}_M^g\left(j\right)F_j^{\mathbb{K}}$.

$H_2(0,1)=K_1-T,H_2(1,1)=T$. It’s holds when $M=1$.

$H_2(g,M)=(K_M-T-g)H_2(g,M-1)+(T+g-1)H_2(g-1,M-1)$.

$A_1^0\left(0\right)=-T,A_1^1\left(0\right)=T$. $A_M^g\left(0\right)=-(T+g)A_{M-1}^g\left(0\right)+(T+g-1)A_{M-1}^{g-1}\left(0\right)\to$

$A_M^g\left(0\right)={(-1)}^{M-g}{[T+g-1]}_g{E}_T(M+1,g+1)={(-1)}^{M-g}{[T+g-1]}_g{E}_{M-g}^{\{T,T+1...T+g\}}$.

The rest only need to consider terms that multiply with ${\mathit{K}}_{\mathit{M}}$.

${\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{j}\right)={\mathit{A}}_{\mathit{M}-\mathit{1}}^{\mathit{g}}(\mathit{j}-\mathit{1})={\mathit{A}}_{\mathit{M}-\mathit{j}}^{\mathit{g}}\left(\mathit{0}\right)={(-\mathit{1})}^{\mathit{M}-\mathit{g}-\mathit{j}}{[\mathit{T}+\mathit{g}-\mathit{1}]}_{\mathit{g}}{\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1}-\mathit{j},\mathit{g}+\mathit{1})$.

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={[\mathit{T}+\mathit{g}-\mathit{1}]}_{\mathit{g}}{\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{(-\mathit{1})}^{\mathit{M}-\mathit{g}-\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{K}}{\mathit{E}}_{\mathit{M}-\mathit{g}-\mathit{j}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{g}\}}\to \left(1\right)$. [Theorem 2.3(3)] and (1) $\to \left(2\right)$.

Using the same method to prove (3), (4), (5), (6). □

Theorem 2.5.

(1). $F_T(N+M,N)=SUM(N,[T,T+1...T+M-1],[1,3...2M-1])$, $E_T(N+M,N)=SUM(N,[T,T...T],[1,3...2M-1])$. (2). $(x+T)(x+T+1)...(x+T+M-1)={\sum }_{k=0}^M{F}_T(M,k)x^k,{(x+T)}^M={\sum }_{k=0}^M{E}_T(M+1,k+1){\left[x\right]}_k$.

(3) ${\sum }_{g=0}^M{(-1)}^{g}g!E_T(M+1,g+1)={(T-1)}^M$, ${\sum }_{g=0}^M{\left.{\langle }_g^M\right)}_T={\left[T\right]}^M$.

(4). ${\left.{\langle }_g^M\right)}_T=H_3(g,[T,1...1],[T...T+M-1])=T\times H_3(g,[1...1],[T+1...T+M-1])$.

Proof. (1) is obvious. $\nabla SUM(N,[T,T...T],[1,2...M])={(T+n)}^M$ $={\sum }_{g=0}^{M}g!E_T(M+1,g+1)\left.{(}_g^n\right)\to$ the latter half of (2). [Theorem 2.1] → (3). (4) is derived from the definition of $H_3\left(g\right)$. □

Definition 2.7. 

$E_p^q\odot ([I_1,I_2...I_M],C)$

$={\sum }_{{\lambda }_1+{\lambda }_2+...+{\lambda }_q=p,{\lambda }_i\ge 0}{1}^{{\lambda }_1}{2}^{{\lambda }_2}...q^{{\lambda }_q}(I_1+{\lambda }_{1}C)(I_2+{\lambda }_{1}C+{\lambda }_{2}C)...(I_{q-1}+{\lambda }_{1}C+{\lambda }_{2}C+...+{\lambda }_{q-1}C)$.

From the calculation formula: $H_3(g,[1,1...1],[1,2...M])=E_{M-g-1}^{g+1}\odot ([1,1...1],1)$. From Worpitzky identity: $N^M=\sum _{g=0}^{M-1}\left.{\langle }_g^M\right)\left.{(}_M^{N+g}\right)$, so $H_3\left(g\right)=\left.{\langle }_g^M\right)$. ${\langle }_g^M〉$ is Eulerian number, this is a new expression. It is the result of [1]. For example: $\left.{\langle }_2^5\right)={\sum }_{{\lambda }_1+{\lambda }_2+{\lambda }_3=2}{1}^{{\lambda }_1}{2}^{{\lambda }_2}{3}^{{\lambda }_3}(1+{\lambda }_1)(1+{\lambda }_1+{\lambda }_2)=66$.

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 $\ge r$. $S_{1,r}(n,k)={\displaystyle \frac{n!}{k!}}{\sum }_{i_1+...+i_k=n,i_j\ge r}{\displaystyle \frac{1}{i_1...i_k}}$. $S_{1,r}(n+1,k)=n{S}_{1,r}(n,k)+{\left(n\right)}_{r-1}{S}_{1,r}(n-r+1,k-1),n\ge kr$. From the recurrence relation, $H_1(g,[2,3..M],[3,5...2M-1])=S_{1,2}(M+g+1,g+1),H_2(g,[1,1..1],[3,5...2M-1])={(-1)}^{M-1-g}{S}_{1,2}(M+g+1,g+1)$.

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 $\ge r$. $S_{2,r}(n,k)={\displaystyle \frac{n!}{k!}}{\sum }_{i_1+...+i_k=n,i_j\ge r}{\displaystyle \frac{1}{i_1!...i_k!}}$. $S_{2,r}(n+1,k)=k{S}_{2,r}(n,k)+\left.{(}_{r-1}^n\right)S_{2,r}(n-r+1,k-1),n\ge kr$. From the recurrence relation, $H_1(g,[1,1..1],[3,5...2M-1])=S_{2,2}(M+g+1,g+1),H_2(g,[2,3..M],[3,5...2M-1])={(-1)}^{M-1-g}{S}_{2,2}(M+g+1,g+1)$.

Definition 2.8. 

${\mathit{I}}_{\mathit{1}}=\mathit{1}$, ${\mathit{I}}_{\mathit{i}+\mathit{1}}-{\mathit{I}}_{\mathit{i}}=$ 1 or 2, ${\mathit{I}}_{\mathit{i}+\mathit{1}}-{\mathit{I}}_{\mathit{i}}=\mathit{2}$ is defined as continuity, ${\mathit{I}}_{\mathit{i}+\mathit{1}}-{\mathit{I}}_{\mathit{i}}=\mathit{2}$ is defined as discontinuity.

${\mathit{MIN}}_{\mathit{g}}\left(\mathit{M}\right)={\mathit{I}}_{\mathit{1}}\times \sum {\mathit{I}}_{\mathit{2}}\times {\mathit{I}}_{\mathit{3}}...\times {\mathit{I}}_{\mathit{M}}$, Number of discontinuities=g.

It is easily understood by adding g holes in consecutive numbers. Easy to know: $MI{N}_g\left(M\right)=\sum {\displaystyle \frac{(M+g)!}{i_1{i}_2...i_g}},2\le i_1<i_2<...<i_g\le M+g-1,i_{j+1}-i_j\ge 2$. This is the conclusion of [2]. From the calculation formula: $H_1(g,[2,3...M],[3,5...2M-1])=MI{N}_g\left(M\right)$. So ${\mathit{MIN}}_{\mathit{g}}\left(\mathit{M}\right)={\mathit{S}}_{\mathit{1},\mathit{2}}(\mathit{M}+\mathit{g}+\mathit{1},\mathit{g}+\mathit{1})$.

From the definition of H(g):

Theorem 2.6. 

${\lambda }_1+...+{\lambda }_{g+1}=M-g,{\lambda }_i\ge 0$, ${\left.{\langle }_g^M\right)}_T=T\sum 1^{{\lambda }_1}{2}^{{\lambda }_2}...{(g+1)}^{{\lambda }_{g+1}}(T+{\lambda }_1)(T+{\lambda }_1+{\lambda }_2)...(T+{\lambda }_1+...+{\lambda }_g)$.

$PS=[T,T...T],PT=[1,3...2M-1]$.

$H_1\left(g\right)=\sum T^{{\lambda }_1}{(T+1)}^{{\lambda }_2}...{(T+g)}^{{\lambda }_{g+1}}(1+{\lambda }_1)(3+{\lambda }_1+{\lambda }_2)...(2g-1+{\lambda }_1+...+{\lambda }_g)$.

$H_2\left(g\right)$=Changed $MI{N}_{g-1}\left(M\right)+MI{N}_g\left(M\right)$. Select $X_i\in \mathbb{K}$ from M factors, change i to (T-i).

$PS=[T,T+1...T+M-1],PT=[1,3...2M-1]$.

$H_1\left(g\right)$=Changed $MI{N}_{g-1}\left(M\right)+MI{N}_g\left(M\right)$. Select $X_i\in \mathbb{K}$ from M factors, change i to (T+i-1).

$H_2\left(g\right)=\sum {(T-1)}^{{\lambda }_1}{(T-2)}^{{\lambda }_2}...{(T-g-1)}^{{\lambda }_{g+1}}(1+{\lambda }_1)(3+{\lambda }_1+{\lambda }_2)...(2g-1+{\lambda }_1+...+{\lambda }_g)$.

For example, express the product in terms of ():

${\mathit{MIN}}_{\mathit{0}}\left(\mathit{3}\right)+{\mathit{MIN}}_{\mathit{1}}\left(\mathit{3}\right)=\left(\mathit{123}\right)+\left(\mathit{124}\right)+\left(\mathit{134}\right)$, ${\mathit{H}}_{\mathit{2}}\left(\mathit{1}\right)=(\mathit{T}-\mathit{1},\mathit{T}-\mathit{2},\mathit{3})+(\mathit{T}-\mathit{1},\mathit{2},\mathit{T}-\mathit{4})+(\mathit{1},\mathit{T}-\mathit{3},\mathit{T}-\mathit{4})$.

${\mathit{MIN}}_{\mathit{1}}\left(\mathit{3}\right)+{\mathit{MIN}}_{\mathit{2}}\left(\mathit{3}\right)=\left(\mathit{124}\right)+\left(\mathit{134}\right)+\left(\mathit{135}\right)$, ${\mathit{H}}_{\mathit{1}}\left(\mathit{2}\right)=(\mathit{T},\mathit{2},\mathit{4})+(\mathit{1},\mathit{T}+\mathit{2},\mathit{3})+(\mathit{1},\mathit{3},\mathit{T}+\mathit{4})$.

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

Theorem 2.7. 

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

Table 1. Table of H(g)

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

Table 1. Table of H(g)

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

3. Number Analysis

From the relations between H(g) [Theorem 2.1], it’s easy to get.:

$\sum _{g=0}^M\left.{\langle }_g^M\right)=M!,\sum _{g=1}^M{(-1)}^{M-g}g!S_2(M,g)=1$. ${\sum }_{g=1}^M{(-1)}^{M-g}{S}_{1,2}(M+g,g)=1,{\sum }_{g=1}^M{(-1)}^{M-g}{S}_{2,2}(M+g,g)=M!$.

$g!S_2(M,g)={\sum }_{k=g}^M{(-1)}^{M-k}k!S_2(M,k)\left.{(}_{g-1}^{k-1}\right)={\sum }_{k=0}^{g-1}\left.{\langle }_k^M\right)\left.{(}_{M-g}^{M-1-k}\right),1\le g\le M$.

From $H(g,\sum K)$, $H(g,\sum T)$ [Theorem 2.3], it’s easy to get:

Theorem 3.1.

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

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

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

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

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

Proof. 

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

$PS=[1,1...1],PT=[1,2...M],F_i^{\{1,1...1\}}=\left.{(}_i^M\right)\to \left(2\right)$.

$PS=[M,M-1...1],PT=[1,2...M],H_1\left(g\right)=M!\left.{(}_g^M\right)\to \left(3\right)$.

$PS=[K,K+1...K+M-1],PT=[T,T+1...T+M-1]\to \left(4\right),\left(5\right)$. □

Theorem 3.2. 

$SUM(N,[1,1..1,PS],[1,1..1,PT\left]\right)=SUM\left(N\right)$. Number of 1 added=X,

$T_M-M-X\ge 0,SUM(N,[1,1...1,PS],[1,1...1,PT])$ expands ${\sum }_{g=0}^M(...)$ to ${\sum }_{g=0}^{M+X}(...)$.

Any ${\sum }_{g=0}^M{a}_g\left.{(}_{Y+g}^X\right)$ can be converted to ${\displaystyle \frac{a_M}{M!}}{\nabla }^{q}SUM(N+p,[K_1,K_2...K_M],[1,2...M])$, relations between H(g) [Theorem 2.1] provides necessary and sufficient conditions for:

Theorem 3.3. 

$\mathit{f}\left(\mathit{g}\right)$ is an integral polynomial of degree equals X, $0\le X<K$.

${\sum }_{g=0}^M{a}_g\left.{(}_{Y+n}^X\right)={\sum }_{n=0}^{M-K}(...)\left.{(}_{Y+K+n}^{X+K}\right)\leftrightarrow {\sum }_{g=0}^M{(-1)}^x{a}_{g}f\left(g\right)=0$.

Proof. 

The condition is $H_2\left(g\right)=H_3(M-g)=0,0\le g<K$. □

Define $(M+1)\times (M+1)$ matrix, $A_{1,2,3}(P,Q,M)=$

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

.

$P=N+T_M-M,Q=N-1$, List SUM(N)...SUM(N+M) from top to bottom, g increases from left to right, they correspond to $For{m}_{1,2,3}$. It is easy to prove by Gaussian elimination:

Theorem 3.4. 

$\Vert A_1(P,Q,M)\Vert =\Vert A_2(P,Q,M)\Vert =\Vert A_3(P,Q,M)\Vert ,\Vert A(P,Q,M)\Vert =\Vert A(P,P-Q,M)\Vert$.

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

If $SUM\left(N\right)$ or $\nabla SUM\left(N\right)$ is easy to obtained, consider H(g) as variables, use Cramer’s law:

Theorem 3.5.

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

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

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

$\nabla SUM(N,[1,1...1],[2,3...M])=N^M$, the following classical formula can be obtained.

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

(3) $\to \left.{\langle }_g^M\right)={\sum }_{k=1}^{g+1}{(-1)}^{1+g+k}\left.{(}_{g+1-k}^{M+1}\right)k^M={\sum }_{k=0}^g{(-1)}^k\left.{(}_k^{M+1}\right){(g+1-k)}^M$.

$T\in \mathbb{N},E_T(M+1,g+1)={\displaystyle \frac{1}{g!}}{\sum }_{k=0}^g{(-1)}^{g+k}\left.{(}_k^g\right){(T+k)}^M={\sum }_{k=0}^{M-g}{T}^k\left.{(}_k^M\right)E_{M-k}^g$. The latter equation comes from [Theorem 2.3(1)].

$T\in \mathbb{N},{\left.{\langle }_g^M\right)}_T$$={\displaystyle \frac{1}{(T-1)!}}{\sum }_{k=0}^g{(-1)}^{g+k}\left.{(}_{g-k}^{T+M}\right){[T+k]}_T{(k+1)}^{M-1}$

$=T\times ({\sum }_{k=0}^{M-2}{\left.{\langle }_g^{M-1}\right)}_{T+1}^k\left.{(}_k^{M-1}\right)+{(-1)}^g\left.{(}_g^{M-1}\right)),0<g<M-1$. The latter equation comes from [Theorem 2.4].

Theorem 3.6. 

${\left.{\langle }_g^M\right)}_T^j={\displaystyle \frac{1}{(T-1)!}}{\sum }_{k=0}^g{(-1)}^{g+k}\left.{(}_{g-k}^{T+M}\right){[T+k]}_T{(k+1)}^{M-1-j},T\in \mathbb{N}$.

Proof. 

From ${\prod }_{i=1}^M(x+K_i)={\sum }_{j=0}^M{F}_j^{\mathbb{K}}{x}^{M-j}$, it’s easy to get: ${\sum }_{j=0}^M{(-1)}^j{k}^{M-j}{F}_j^M=0,M\ge k\ge 1$.

$PS=[0,-1...-(M-1)],PT=[T,T+1...T+M-1],H_3(g<M)=0\to {\sum }_{j=0}^{M-1}{(-1)}^j{\left.{\langle }_g^M\right)}_T^j{F}_j^{M-1}=0,0<g<M$.

From these two equations it is clear that ${\left.{\langle }_g^M\right)}_T^j$ and ${\left.{\langle }_g^M\right)}_T^0={\left.{\langle }_{g-1}^M\right)}_T$ have the same thing:

${\left.{\langle }_g^M\right)}_T^0=a_1{1}^x+a_2{2}^y+...$, ${\left.{\langle }_{\mathit{g}}^{\mathit{M}}\right)}_{\mathit{T}}^{\mathit{j}}={\mathit{a}}_{\mathit{1}}{\mathit{1}}^{\mathit{x}-\mathit{j}}+{\mathit{a}}_{\mathit{2}}{\mathit{2}}^{\mathit{y}-\mathit{j}}+...$. Complete the proof from the expression of ${\left.{\langle }_{g-1}^M\right)}_T$. □

4. Combinatorial Identities

Formal Calculation provides many ways to obtain an identity, and here are some of the methods.

Theorem 4.1. 

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

Proof. 

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

It can be obtained: $H_1(g>M,PS1,PT1)=0,H_1(g\le M,PS1,PT1)=H_1\left(g\right)$.

$SUM(N,PS1,PT1)={\sum }_{g=0}^M{H}_1\left(g\right)\left.{(}_{j+g}^{N+j-1}\right)={\sum }_{k_j=1}^N\nabla SUM\left(k_j\right)...{\sum }_{k_2=1}^{k_3}\sum _{k_1=1}^{k_2}1$. The remaining step is obvious. □

For example: ${\sum }_{k_r=1}^N...{\sum }_{k_2=1}^{k_3}{\sum }_{k_1=1}^{k_2}\left.{(}_M^{k_j}\right)={\displaystyle \frac{1}{M!}}{\nabla }^{j-r}SUM(N+1,[0,-1...-(M-1)],[T_i=i+j-1])$.

$H_1(g<M)=0,H_1\left(M\right)={\displaystyle \frac{(M+j-1)!}{(j-1)!}}\to$${\displaystyle \frac{1}{M!}}{\nabla }^{j-r}{\displaystyle \frac{(M+j-1)!}{(j-1)!}}\left.{(}_{j+1+(M-1)}^{(N+1)+(j-1)}\right)=\left.{(}_{j-1}^{M+j-1}\right)\left.{(}_{j+M-(j-r)}^{N+j-(j-r)}\right)=\left.{(}_{j-1}^{M+j-1}\right)\left.{(}_{M+r}^{N+r}\right)$. [3].

Theorem 4.2. 

${\sum }_{g=0}^M\left.{(}_g^M\right)\left.{(}_A^{A+T+g}\right)\left.{(}_{Y+g}^X\right)={\sum }_{g=0}^A\left.{(}_{g+T}^{A+T}\right)\left.{(}_{M+T}^{M+T+g}\right)\left.{(}_{Y+M-A+g}^{X+M-A}\right),A\ge 0,M\ge A+1$.

Proof. 

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

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

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

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

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

$Y=T+A+1,X=N+A+T$ then proves completion. The two sides correspond to merging and unfolding. □

Theorem 4.3. 

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

Proof. 

$PS=[A,A+1...A+M-1]$.

Expand by definition : $H_1(g,\sum K)=SUM(g+1,[A,A+1...A+M-1-g]:2,[1,3...2(M-g)-1])$.

Form [Theorem 2.2(5)], $H_1(g,\sum K)=\left.{(}_g^M\right){[A+M-1]}_{M-g}$. Just compare the results. □

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

5. Congruence

In this section, P is prime, $K_i,D_i$ is any integer, $(P,D_i)=1$.

The following congruences are based on P. $D_i{D}_i^{-1}\equiv 1,K_i+D_{i}n\equiv D_i(K_i{D}_i^{-1}+n)$.

Definition 5.1. 

If $K_i{D}_i^{-1}\equiv T_i,0\le i\le X$, define $PTS=X$, $0\le PTS\le M$.

Theorem 5.1. 

$0\le Y\le T_M-M+PTS$,

$T_M<P-1$, $SUM(P-Y)\equiv 0(modP{\prod }_{i=1}^{PTS}{T}_i{D}_i)$.

$T_M=P-1$, $SUM(P-Y)\equiv \prod T_i{D}_i$.

$T_M=P$, $SUM(P-Y)\equiv H_1(M-1)\equiv H_2(M-1)$. $PT=[1,2...P]$, $SUM\left(P\right)\equiv -\prod D_i\sum K_i{D}_i^{-1}$.

Proof. 

$H=T_M-M,0\le H<T_M\le P.0\le Y\le H$.

$SUM\left(P\right)=H_1\left(M\right)\left.{(}_{T_M+1}^{P+H}\right)+...+H_1\left(0\right)\left.{(}_{H+1}^{P+H}\right)={\sum }_{g=0}^M{H}_1\left(g\right)\left.{(}_{g+1+H}^{P+H}\right)$.

$T_M<P-1\to SUM(P-Y)\equiv 0$. $T_M=P-1\to SUM(P-Y)\equiv H_1\left(M\right)\equiv \prod T_i{D}_i$.

$T_M=P\to H_1\left(M\right)\equiv 0\to SUM(P-Y)\equiv SUM\left(P\right)\equiv H_1(M-1)$.

Prove the part of PTS through [Theorem 2.2(2)].

$PT=[1,2..P]$, [Theorem 2.3](1) $\to H_1(M-1)=(P-1)!\prod D_i(F_1^{\left\{K_i{D}_i^{-1}\right\}}+E_1^{P-1})=-\prod D_i{F}_1^{\left\{K_i{D}_i^{-1}\right\}}\equiv -\prod D_i\sum K_i{D}_i^{-1}$. □

Remove products$\equiv 0$, it can derive many congruence equations. Wilson’s Theorem $(\mathit{P}-\mathit{1})!\equiv -\mathit{1}$ is a special case.

Theorem 5.2.

(1). $P>3$, $S_1(P+M-Z,P-Z)\equiv S_2(P+M-Z,P-Z)\equiv 0,0\le Z\le M,1\le M\le P-2$.

(2). $S_1(P,K)\equiv S_2(P,K)\equiv 0,2\le K\le P-1$.

(3). $S_1(N+P-1,N)\equiv -A,S_2(N+P-1,N)\equiv A,A\equiv [(N-1)/P]+1,N>0$.

(4). $S_1(N+P-2,N)\equiv S_2(N+P-2,N)\equiv \left.{\{}_{0,N\neg \equiv 1}^{1,N\equiv 1}\right),N>0$.

(5). $S_1(P-1,N)\equiv 1,S_2(P-1,K)\equiv (P-1-K)!,K!S_2(P-1,K)\equiv {(-1)}^{K+1},K>0$.

(6). $S_1(P-2,P-2-N)\equiv 2^{N+1}-1$.

(7). $P>3,S_1(P,2)\equiv 0(mod{P}^2)$.

Proof. 

$PT=[3,5...2M-1],PS1=[2,3...M],PS2=[1,1...1]$,

$H_1(g,PS1,PT)=S_{1,2}(M+1+g,g+1)$, $H_1(g,PS2,PT)=S_{2,2}(M+1+g,g+1)$.

If $2M-1\ge P$, $SUM\left(P\right)=H_1(M-1)\left.{(}_{2M}^{P+M}\right)+...H_1(P-M-1)\left.{(}_P^{P+M}\right)+...={\sum }_{g=0}^{M-1}{H}_1\left(g\right)\left.{(}_{g+1+M}^{P+M}\right)$.

If $g\ge P-M-1$, in the express of $S_{1,2}(n,g+1),S_{2,2}(n,g+1)$, $n=M+g+1\ge P$, when the largest ${\mathit{i}}_{\mathit{j}}$ is encountered where the number of 2 is g. $M+g+1-2g=M+1-g$, $Max\left(i_j\right)=M+1-(P-M-1)<P$, so $H_1\left(g\right)\equiv 0\to \left(1\right),\left(2\right)$.

$M=P-1,SUM(N,PS1,PT),H_1(g>0)\equiv 0,SUM\left(N\right)\equiv -\left.{(}_P^{N+P-1}\right)$.

$M=P-1,SUM(N,PS2,PT),H_1(g>0)\equiv 0,SUM\left(N\right)\equiv \left.{(}_P^{N+P-1}\right)$.

$M=P-2,SUM(N,PS1,PT),H_1(g>0)\equiv 0,SUM\left(N\right)\equiv \left.{(}_{P-1}^{N+P-2}\right)$.

$M=P-2,SUM(N,PS2,PT),H_1(g>0)\equiv 0,SUM\left(N\right)\equiv \left.{(}_{P-1}^{N+P-2}\right)\to \left(3\right),\left(4\right)$.

$H_2(g,PS1,PT)={(-1)}^{M-1-g}{S}_{2,2}(M+1+g,g+1)$, $H_2(g,PS2,PT)={(-1)}^{M-1-g}{S}_{1,2}(M+1+g,g+1)$.

$SUM(P-M-1)=H_2(M-1)\left.{(}_{2M}^{P+M-2}\right)+...+H_2(P-M-1)\left.{(}_P^{...}\right)+...+H_2\left(1\right)\left.{(}_{2+M}^P\right)+H_2\left(0\right)\left.{(}_{1+M}^{P-1}\right)$.

$H_2(g\ge P-M-1)\equiv 0\to SUM(P-M-1)\equiv H_2\left(0\right){(-1)}^{M+1}$.

$H_2(0,PS1,PT)={(-1)}^{M-1}$, $H_2(0,PS2,PT)={(-1)}^{M-1}M!\to \left(5\right)$.

$S_1(N,M)=S_1(N-1,M-1)+(N-1)S_1(N-1,M)$ and $S_1(P-1,N)\equiv 1$, induction shows that (6).

$P>3,SUM(2,[2,3..P-2],[3,5...2P-5])=H_1\left(1\right)\left.{(}_P^P\right)+H_1\left(0\right)\left.{(}_{P-1}^P\right)=S_{1,2}(P,2)+P(P-2)!$.

$H_1\left(1\right)/P=(P-1)!({\displaystyle \frac{1}{2(P-2)}}+{\displaystyle \frac{1}{3(P-3)}}+...+{\displaystyle \frac{1}{\frac{P-1}{2}\frac{P+1}{2}}})$$\equiv -1(-2^{-2}-3^{-2}...-{\left({\displaystyle \frac{P-1}{2}}\right)}^{-2})\equiv -1\to \left(7\right)$.

This is Wolstenholme’s Theory: ${\sum }_{\mathit{i}=\mathit{1}}^{\mathit{P}-\mathit{1}}{\displaystyle \frac{(\mathit{P}-\mathit{1})!}{\mathit{i}}}(mod{\mathit{P}}^{\mathit{2}})\}$. □

A Direct Proof of Wilson’s Theorem:

$PS=PT=[1,2...P-1],\nabla SUM\left(2\right)=H_1\left(1\right)\left.{(}_{2-1}^{N-1}\right)+H_1\left(0\right)\left.{(}_{1-1}^{N-1}\right)=P!/1\equiv 0$.

$\nabla SUM\left(2\right)=1!(F_{P-2}^{P-1}{E}_0^1+...+F_0^{P-1}{E}_{P-2}^1)+(P-1)!\equiv 1+(P-1)!$

$\mathit{PS}=\mathit{PT}=[\mathit{1},\mathit{2}...\mathit{P}-\mathit{2}],\mathit{2}\le \mathit{K}\le \mathit{P}-\mathit{2},\nabla \mathit{SUM}(\mathit{K}+\mathit{1})={\mathit{H}}_{\mathit{1}}\left(\mathit{K}\right)\left.{(}_{\mathit{K}}^{\mathit{K}}\right)+...+{\mathit{H}}_{\mathit{1}}\left(\mathit{0}\right)\left.{(}_{\mathit{0}}^{\mathit{K}}\right)\equiv \mathit{0}$.

${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)=\mathit{g}!({\mathit{F}}_{\mathit{P}-\mathit{2}-\mathit{g}}^{\mathit{P}-\mathit{2}}{\mathit{E}}_{\mathit{0}}^{\mathit{g}}+...+{\mathit{F}}_{\mathit{0}}^{\mathit{P}-\mathit{2}}{\mathit{E}}_{\mathit{P}-\mathit{2}-\mathit{g}}^{\mathit{g}})\equiv \mathit{g}!({\mathit{E}}_{\mathit{0}}^{\mathit{g}}+...+{\mathit{E}}_{\mathit{P}-\mathit{2}-\mathit{g}}^{\mathit{g}})$.

Let $\mathit{f}\left(\mathit{g}\right)=\mathit{g}!({\mathit{E}}_{\mathit{0}}^{\mathit{g}}+...+{\mathit{E}}_{\mathit{P}-\mathit{2}-\mathit{g}}^{\mathit{g}}),\mathit{f}\left(\mathit{0}\right)\equiv \mathit{1}$. It can be proved by induction: $\mathit{f}\left(\mathit{g}\right)\equiv {(-\mathit{1})}^{\mathit{g}}(\mathit{g}+\mathit{1})$.

$PS=[P-2...2,1]$. Use $H_1(g,\sum T)$, similar conclusions were reached.

Theorem 5.3. 

${\sum }_{\mathit{M}=\mathit{0}}^{\mathit{P}-\mathit{2}}{\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{K})\equiv {(-\mathit{1})}^{\mathit{K}}(\mathit{K}+\mathit{1}){(\mathit{K}!)}^{-\mathit{1}}$. ${\sum }_{\mathit{M}=\mathit{0}}^{\mathit{P}}{\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{K})\equiv {(-\mathit{1})}^{\mathit{K}}\mathit{K}{(\mathit{K}!)}^{-\mathit{1}},\mathit{0}<\mathit{K}<\mathit{P}$.

${\sum }_{\mathit{M}=\mathit{0}}^{\mathit{P}-\mathit{2}}{(-\mathit{1})}^{\mathit{M}-\mathit{K}}{\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{P}-\mathit{2}-\mathit{K})\equiv {(-\mathit{1})}^{\mathit{K}}(\mathit{K}+\mathit{1})!$.

${\sum }_{\mathit{M}=\mathit{0}}^{\mathit{P}}{(-\mathit{1})}^{\mathit{M}-\mathit{K}}{\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{P}-\mathit{2}-\mathit{K})\equiv ({(-\mathit{1})}^{\mathit{K}}-\mathit{1})(\mathit{K}+\mathit{1})!,\mathit{0}\le \mathit{K}<\mathit{P}-\mathit{2}$.

Theorem 5.4.

(1). ${\sum }_{0<K_i,K_j<P,K_i\ne K_j}{K}_1^{C_1}{K}_2^{C_2}...K_q^{C_q}\equiv 0,1\le \sum C_i\le P-2,C_i>0$.

(2). $0\le K_{i+1}-K_i\le 1,\prod K_i=1^{C_1}{2}^{C_2}..q^{C_q},\sum C_i\le P-2,C_i>0;T_{i+1}-T_i=K_{i+1}-K_i+1,H=T_M-M$,

$\sum SUM(P-1-H,[K_1=1,K_2...K_M],[T_1=1,T_2...T_M])\equiv 0$. For $\left\{{\mathit{C}}_{\mathit{i}}\right\}$, sum over all PTs meet the condition.

(3). In (1) and (2),$\sum C_i=P-2,P>3$ then $0\equiv (mod{P}^2)$.

Proof. 

Obviously, $\mathit{q}=\mathit{1}$ holds for (1).

$q=2$, If $K_1+K_2\ne p,K_1^{C_1}{(P-K_2)}^{C_2}$ in the summation term; If $K_1+K_2=p,K_1^{C_1}{(P-K_2)}^{C_2}=K_1^{C_1+C_2}$.

$\sum K_1^{C_1+C_2}\equiv 0,\sum K_1^{C_1}{K}_2^{C_2}=yP\sum K_1^{C_1}-\sum K_1^{C_1+C_2},y\in \left(N\right)\to q=2$ holds. Prove by induction that (1) holds.

Removing the repeated products, it still holds due to symmetry. That’s (2). It can be proven that: $\sum n^{P-2}\equiv 0(mod{P}^2),P>3$, applying the same method → (3), they’re generalization of Wolstenholme’s Theory. □

Special: $E_M^P=S_2(P+M,P),E_M^{P-1}=S_2(P+M-1,P-1)\equiv 0,1\le M\le P-2$. $E_{P-2}^P,E_{P-2}^{P-1}\equiv 0(mod{P}^2)$.

Form classification [Theorem 2.7], $H=T_M-M,SUM(P-1,[1,1...1],[1,3...2M-1])$ $={\sum }_{X=0}^{M-1}{\sum }_{H=X}SUM(P-1-X,...)$.

We can get ${\sum }_{H=X}SUM(P-1-X,...)\equiv 0$. For Example: $\mathit{P}=\mathit{7},\mathit{M}=\mathit{P}-\mathit{2}$,

$\mathit{SUM}(\mathit{6},[\mathit{1},\mathit{1},\mathit{1},\mathit{1},\mathit{1}],[\mathit{1},\mathit{2},\mathit{3},\mathit{4},\mathit{5}])={\sum }_{\mathit{n}=\mathit{1}}^{\mathit{6}}{\mathit{n}}^{\mathit{5}}$,

$\mathit{SUM}(\mathit{5},[\mathit{1},\mathit{1},\mathit{1},\mathit{1},\mathit{2}],[\mathit{1},\mathit{2},\mathit{3},\mathit{4},\mathit{6}\left]\right)+\mathit{SUM}(\mathit{5},[\mathit{1},\mathit{1},\mathit{1},\mathit{2},\mathit{2}],[\mathit{1},\mathit{2},\mathit{3},\mathit{5},\mathit{6}\left]\right)$

$+\mathit{SUM}(\mathit{5},[\mathit{1},\mathit{1},\mathit{2},\mathit{2},\mathit{2}],[\mathit{1},\mathit{2},\mathit{4},\mathit{5},\mathit{6}\left]\right)+\mathit{SUM}(\mathit{5},[\mathit{1},\mathit{2},\mathit{2},\mathit{2},\mathit{2}],[\mathit{1},\mathit{3},\mathit{4},\mathit{5},\mathit{6}\left]\right)$,

$\mathit{SUM}(\mathit{4},[\mathit{1},\mathit{1},\mathit{1},\mathit{2},\mathit{3}],[\mathit{1},\mathit{2},\mathit{3},\mathit{5},\mathit{7}\left]\right)+\mathit{SUM}(\mathit{4},[\mathit{1},\mathit{1},\mathit{2},\mathit{2},\mathit{3}],[\mathit{1},\mathit{2},\mathit{4},\mathit{5},\mathit{7}\left]\right)$

$+\mathit{SUM}(\mathit{4},[\mathit{1},\mathit{2},\mathit{2},\mathit{2},\mathit{3}],[\mathit{1},\mathit{3},\mathit{4},\mathit{5},\mathit{7}\left]\right)+\mathit{SUM}(\mathit{4},[\mathit{1},\mathit{2},\mathit{3},\mathit{3},\mathit{3}],[\mathit{1},\mathit{3},\mathit{5},\mathit{6},\mathit{7}\left]\right)$

$+\mathit{SUM}(\mathit{4},[\mathit{1},\mathit{2},\mathit{2},\mathit{3},\mathit{3}],[\mathit{1},\mathit{3},\mathit{4},\mathit{6},\mathit{7}\left]\right)+\mathit{SUM}(\mathit{4},[\mathit{1},\mathit{1},\mathit{2},\mathit{3},\mathit{3}],[\mathit{1},\mathit{2},\mathit{4},\mathit{6},\mathit{7}\left]\right)$,

$\mathit{SUM}(\mathit{3},[\mathit{1},\mathit{1},\mathit{2},\mathit{3},\mathit{4}],[\mathit{1},\mathit{2},\mathit{4},\mathit{6},\mathit{8}\left]\right)+\mathit{SUM}(\mathit{3},[\mathit{1},\mathit{2},\mathit{2},\mathit{3},\mathit{4}],[\mathit{1},\mathit{3},\mathit{4},\mathit{6},\mathit{8}\left]\right)$

$+\mathit{SUM}(\mathit{3},[\mathit{1},\mathit{2},\mathit{3},\mathit{3},\mathit{4}],[\mathit{1},\mathit{3},\mathit{5},\mathit{6},\mathit{8}\left]\right)+\mathit{SUM}(\mathit{3},[\mathit{1},\mathit{2},\mathit{3},\mathit{4},\mathit{4}],[\mathit{1},\mathit{3},\mathit{5},\mathit{7},\mathit{8}\left]\right)$,

$\mathit{SUM}(\mathit{2},[\mathit{1},\mathit{2},\mathit{3},\mathit{4},\mathit{5}],[\mathit{1},\mathit{3},\mathit{5},\mathit{7},\mathit{9}])={\sum }_{\mathit{i}=\mathit{1}}^{\mathit{P}-\mathit{1}}{\displaystyle \frac{(\mathit{P}-\mathit{1})!}{\mathit{i}}}$. All of them $\equiv 0(mod{P}^2)$

In fact,$SUM(5,[1,1,1,1,2],[1,2,3,4,6])+SUM(5,[1,2,2,2,2],[1,3,4,5,6])\equiv 0(mod{P}^2)$. (*)

6. Generalization of Eulerian polynomials

In this section, $q\ne 0,q\ne 1$. By induction it can be shown that:

${\sum }_{n=0}^{N-1}{q}^n\left.{(}_M^{n+A}\right)=q^N{\sum }_{k=0}^M{(-1)}^k{\displaystyle \frac{\left.{(}_{M-k}^{N+A-1-k}\right)}{{(q-1)}^{k+1}}}+{\displaystyle \frac{q^{M-A}}{{(1-q)}^{M+1}}},0\le A\le M$.

$X=T_M-M-p\ge 0,0\le Y\le 1,{\sum }_{n=0}^{N-1}{q}^n{\nabla }^{p}SUM(n+Y)={\sum }_{g=0}^M{H}_1\left(g\right){\sum }_{n=0}^{N-1}{q}^n\left.{(}_{1+X+g}^{n+Y+X}\right)$

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

From [Theorem 2.1 (5)], ${\sum }_{g=0}^M{H}_1\left(g\right)q^g{(1-q)}^{M-g}={\sum }_{g=0}^M{H}_2\left(g\right){(1-q)}^{M-g}={\sum }_{g=0}^M{H}_3\left(g\right)q^g$. Define it as ${\mathbf{A}}_{\mathbf{q}}(\mathbf{PS},\mathbf{PT})$.

${\sum }_{n=0}^{N-1}{q}^n{\nabla }^{p}SUM(n+Y)=q^N{\sum }_{g=0}^M{H}_1\left(g\right){\sum }_{k=0}^{X+1+g}{\displaystyle \frac{{(-1)}^k}{{(q-1)}^{k+1}}}\left.{(}_{1+X+g-k}^{N+X+Y-1-k}\right)+{\displaystyle \frac{A_q(PS,PT)q^{1-Y}}{{(1-q)}^{T_M+2-p}}}$.

$=q^N{\sum }_{k=0}^M{\displaystyle \frac{{(-1)}^k}{{(q-1)}^{k+1}}}{\sum }_{g=0}^M{H}_1\left(g\right)\left.{(}_{1+X+g-k}^{N+X+Y-1-k}\right)+{\displaystyle \frac{A_q(PS,PT)q^{1-Y}}{{(1-q)}^{T_M+2-p}}}$.

Theorem 6.1. 

$T_M-M-p\ge 0,0\le Y\le 1$,

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

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

Special: $PS=[1,1...1],PT=[1,2...M]$, define ${\mathbf{A}}_{\mathbf{q}}(\mathbf{PS},\mathbf{PT})={\mathbf{A}}_{\mathbf{q}}\left(\mathbf{M}\right)$, $A_q\left(M\right)$ is Eulerian polynomial.

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

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

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

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

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

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

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

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

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

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

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

Theorem 6.2. 

$A_q\left(M\right)=q^{-1}{\sum }_{k=0}^M{(q-1)}^{M-k}{S}_2(M+1,k+1)k!$. Together with $A_q(PS,PT)$, there are four expressions.

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