Symmetry in Formal Calculation

1. Introduction

The notion of formal computation was introduced in [1].

Definition 1.

The definition of ${\nabla }^{\mathit{p}}$ is recursive. $\mathit{p}\in \mathbb{Z}$,

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

Definition 2.

The definition of $\mathit{SUM}\left(\mathit{N}\right)=\mathit{SUM}(\mathit{N},\mathit{PS},\mathit{PT})$ is recursive. ${\mathit{K}}_{\mathit{i}},{\mathit{D}}_{\mathit{i}}\in \mathbb{C};{\mathit{T}}_{\mathit{i}}\in \mathbb{N}$.

$\mathit{SUM}(\mathit{N},[{\mathit{K}}_{\mathit{1}}:{\mathit{D}}_{\mathit{1}}],[{\mathit{T}}_{\mathit{1}}=\mathit{1}])=\sum _{\mathit{n}=\mathit{0}}^{\mathit{N}-\mathit{1}}({\mathit{K}}_{\mathit{1}}+{\mathit{nD}}_{\mathit{1}})$.

$\mathit{SUM}(\mathit{N},[{\mathit{K}}_{\mathit{1}}:{\mathit{D}}_{\mathit{1}},{\mathit{K}}_{\mathit{2}}:{\mathit{D}}_{\mathit{2}}],[{\mathit{T}}_{\mathit{1}},{\mathit{T}}_{\mathit{2}}={\mathit{T}}_{\mathit{1}}+\mathit{2}-\mathit{p}])=\sum _{\mathit{n}=\mathit{0}}^{\mathit{N}-\mathit{1}}({\mathit{K}}_{\mathit{2}}+{\mathit{nD}}_{\mathit{2}}){\nabla }^{\mathit{p}}\mathit{SUM}(\mathit{n}+\mathit{1},[{\mathit{K}}_{\mathit{1}}:{\mathit{D}}_{\mathit{1}}],\left[{\mathit{T}}_{\mathit{1}}\right]).$

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

Use $\mathbb{K}$ to represent the set $\{{\mathit{K}}_{\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}$, $\mathbb{T}$ to represent the set $\{{\mathit{T}}_{\mathit{1}}...{\mathit{T}}_{\mathit{M}}\}$.

Use the auxiliary form: $({\mathit{K}}_{\mathit{1}}+{\mathit{T}}_{\mathit{1}})({\mathit{K}}_{\mathit{2}}+{\mathit{T}}_{\mathit{2}})...({\mathit{K}}_{\mathit{M}}+{\mathit{T}}_{\mathit{M}})=\sum {\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{X}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}={\mathit{T}}_{\mathit{i}}\mathit{or}{\mathit{K}}_{\mathit{i}}$.

Definition 3.

$\mathit{X}\left(\mathit{T}\right)$=Number of $\{{\mathit{X}}_{\mathit{1}},{\mathit{X}}_{\mathit{2}}...{\mathit{X}}_{\mathit{M}}\}\in \mathbb{T}$.

Definition 4.

${\mathit{X}}_{\mathit{T}-\mathit{1}}$=Number of $\{{\mathit{X}}_{\mathit{1}},{\mathit{X}}_{\mathit{2}}...{\mathit{X}}_{\mathit{i}-\mathit{1}}\}\in \mathbb{T}$, ${\mathit{X}}_{\mathit{K}-\mathit{1}}$=Number of $\{{\mathit{X}}_{\mathit{1}},{\mathit{X}}_{\mathit{2}}...{\mathit{X}}_{\mathit{i}-\mathit{1}}\}\in \mathbb{K}$.

Obviously:${\mathit{X}}_{\mathit{T}-\mathit{1}}+{\mathit{X}}_{\mathit{K}-\mathit{1}}=\mathit{i}-\mathit{1}$.

Use the auxiliary form and each ${\mathit{X}}_{\mathit{i}}$ cannot be exchanged, [1] draws conclusions:

$\mathit{SUM}(\mathit{N},\mathit{PS},\mathit{PT})=$

${\mathit{Form}}_{\mathit{1}}\to {\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)\left({ }_{\mathit{N}-\mathit{1}-\mathit{g}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{M}}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)\left({ }_{{\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{1}+\mathit{g}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{M}}\right),{\mathit{B}}_{\mathit{i}}=\left\{{ }_{{\mathit{K}}_{\mathit{i}}+{\mathit{X}}_{\mathit{T}-\mathit{1}}{\mathit{D}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}={\mathit{K}}_{\mathit{i}}}^{({\mathit{T}}_{\mathit{i}}-{\mathit{X}}_{\mathit{K}-\mathit{1}}){\mathit{D}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}={\mathit{T}}_{\mathit{i}}}\right.$,

${\mathit{Form}}_{\mathit{2}}\to {\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)\left({ }_{\mathit{N}-\mathit{1}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{g}}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)\left({ }_{{\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{1}+\mathit{g}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{g}}\right),{\mathit{B}}_{\mathit{i}}=\left\{{ }_{{\mathit{K}}_{\mathit{i}}+({\mathit{X}}_{\mathit{K}-\mathit{1}}-{\mathit{T}}_{\mathit{i}}){\mathit{D}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}={\mathit{K}}_{\mathit{i}}}^{({\mathit{T}}_{\mathit{i}}-{\mathit{X}}_{\mathit{K}-\mathit{1}}){\mathit{D}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}={\mathit{T}}_{\mathit{i}}}\right.$,

${\mathit{Form}}_{\mathit{3}}\to {\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)\left({ }_{\mathit{N}-\mathit{1}-\mathit{g}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{g}}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)\left({ }_{{\mathit{T}}_{\mathit{M}}+\mathit{1}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{g}}\right),{\mathit{B}}_{\mathit{i}}=\left\{{ }_{{\mathit{K}}_{\mathit{i}}+{\mathit{X}}_{\mathit{T}-\mathit{1}}{\mathit{D}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}={\mathit{K}}_{\mathit{i}}}^{-{\mathit{K}}_{\mathit{i}}+({\mathit{T}}_{\mathit{i}}-{\mathit{X}}_{\mathit{T}-\mathit{1}}){\mathit{D}}_{\mathit{i}},{\mathit{X}}_{\mathit{i}}={\mathit{T}}_{\mathit{i}}}\right.$.

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

For example:

$\mathit{SUM}(\mathit{N},\mathit{PS},[\mathit{1},\mathit{2},\mathit{3}...\mathit{M}])={\sum }_{\mathit{n}=\mathit{0}}^{\mathit{N}-\mathit{1}}{\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}({\mathit{K}}_{\mathit{i}}+{\mathit{nD}}_{\mathit{i}})$.

$\mathit{SUM}(\mathit{N},\mathit{PS},[\mathit{1},\mathit{3}...\mathit{2}\mathit{M}-\mathit{1}])={\sum }_{{\mathit{n}}_{\mathit{M}}=\mathit{0}}^{\mathit{N}-\mathit{1}}({\mathit{K}}_{\mathit{M}}+{\mathit{n}}_{\mathit{M}}{\mathit{D}}_{\mathit{M}})...{\sum }_{{\mathit{n}}_{\mathit{2}}=\mathit{0}}^{{\mathit{n}}_{\mathit{3}}}({\mathit{K}}_{\mathit{2}}+{\mathit{n}}_{\mathit{2}}{\mathit{D}}_{\mathit{2}}){\sum }_{{\mathit{n}}_{\mathit{1}}=\mathit{0}}^{{\mathit{n}}_{\mathit{2}}}({\mathit{K}}_{\mathit{1}}+{\mathit{n}}_{\mathit{1}}{\mathit{D}}_{\mathit{1}})$.

$\mathit{SUM}(\mathit{N},\mathit{PS},[\mathit{1},\mathit{2},\mathit{4}])={\sum }_{{\mathit{n}}_{\mathit{3}}=\mathit{0}}^{\mathit{N}-\mathit{1}}({\mathit{K}}_{\mathit{3}}+{\mathit{n}}_{\mathit{3}}{\mathit{D}}_{\mathit{3}}){\sum }_{\mathit{n}=\mathit{0}}^{{\mathit{n}}_{\mathit{3}}}({\mathit{K}}_{\mathit{1}}+\mathit{n}{\mathit{D}}_{\mathit{1}})({\mathit{K}}_{\mathit{2}}+\mathit{n}{\mathit{D}}_{\mathit{2}})$.

$\mathit{SUM}(\mathit{N},\mathit{PS},[\mathit{1},\mathit{3},\mathit{4}])={\sum }_{{\mathit{n}}_{\mathit{3}}=\mathit{0}}^{\mathit{N}-\mathit{1}}({\mathit{K}}_{\mathit{3}}+{\mathit{n}}_{\mathit{3}}{\mathit{D}}_{\mathit{3}})({\mathit{K}}_{\mathit{2}}+{\mathit{n}}_{\mathit{3}}{\mathit{D}}_{\mathit{2}}){\sum }_{\mathit{n}=\mathit{0}}^{{\mathit{n}}_{\mathit{3}}}({\mathit{K}}_{\mathit{1}}+{\mathit{nD}}_{\mathit{1}})$.

$\mathit{PS}=[{\mathit{K}}_{\mathit{1}}:{\mathit{D}}_{\mathit{1}},{\mathit{K}}_{\mathit{2}}:{\mathit{D}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}}:{\mathit{D}}_{\mathit{M}}],\mathit{PT}=[{\mathit{T}}_{\mathit{1}},{\mathit{T}}_{\mathit{2}}...{\mathit{T}}_{\mathit{M}}]$. There are recursive relationships:

• ${\mathit{H}}_{\mathit{1}}(\mathit{g},\mathit{M})=({\mathit{A}}_{\mathit{M}}+{\mathit{gD}}_{\mathit{M}}){\mathit{H}}_{\mathit{1}}(\mathit{g},\mathit{M}-\mathit{1})+({\mathit{B}}_{\mathit{M}}+(\mathit{g}-\mathit{1}){\mathit{D}}_{\mathit{M}}){\mathit{H}}_{\mathit{1}}(\mathit{g}-\mathit{1},\mathit{M}-\mathit{1})$,
  ${\mathit{A}}_{\mathit{M}}={\mathit{K}}_{\mathit{M}},{\mathit{B}}_{\mathit{M}}={\mathit{D}}_{\mathit{M}}({\mathit{T}}_{\mathit{M}}-((\mathit{i}-\mathit{1})-(\mathit{g}-\mathit{1})))-(\mathit{g}-\mathit{1}){\mathit{D}}_{\mathit{M}}={\mathit{T}}_{\mathit{M}}{\mathit{D}}_{\mathit{M}}-(\mathit{i}-\mathit{1}){\mathit{D}}_{\mathit{M}}$.
• ${\mathit{H}}_{\mathit{2}}(\mathit{g},\mathit{M})=({\mathit{A}}_{\mathit{M}}-{\mathit{gD}}_{\mathit{M}}){\mathit{H}}_{\mathit{2}}(\mathit{g},\mathit{M}-\mathit{1})+({\mathit{B}}_{\mathit{M}}+(\mathit{g}-\mathit{1}){\mathit{D}}_{\mathit{M}}){\mathit{H}}_{\mathit{2}}(\mathit{g}-\mathit{1},\mathit{M}-\mathit{1})$,
  ${\mathit{A}}_{\mathit{M}}={\mathit{K}}_{\mathit{M}}+(\mathit{i}-\mathit{1}-\mathit{g}-{\mathit{T}}_{\mathit{M}}){\mathit{D}}_{\mathit{M}}+{\mathit{gD}}_{\mathit{M}}={\mathit{K}}_{\mathit{M}}+(\mathit{i}-\mathit{1}-{\mathit{T}}_{\mathit{M}}){\mathit{D}}_{\mathit{M}},{\mathit{B}}_{\mathit{M}}={\mathit{T}}_{\mathit{M}}{\mathit{D}}_{\mathit{M}}-(\mathit{i}-\mathit{1}){\mathit{D}}_{\mathit{M}}$.
• ${\mathit{H}}_{\mathit{3}}(\mathit{g},\mathit{M})=({\mathit{A}}_{\mathit{M}}+{\mathit{gD}}_{\mathit{M}}){\mathit{H}}_{\mathit{3}}(\mathit{g},\mathit{M}-\mathit{1})+({\mathit{B}}_{\mathit{M}}-(\mathit{g}-\mathit{1}){\mathit{D}}_{\mathit{M}}){\mathit{H}}_{\mathit{3}}(\mathit{g}-\mathit{1},\mathit{M}-\mathit{1})$,
  ${\mathit{A}}_{\mathit{M}}={\mathit{K}}_{\mathit{M}},{\mathit{B}}_{\mathit{M}}=-{\mathit{K}}_{\mathit{M}}+({\mathit{T}}_{\mathit{M}}-(\mathit{g}-\mathit{1})){\mathit{D}}_{\mathit{M}}+(\mathit{g}-\mathit{1}){\mathit{D}}_{\mathit{M}}=-{\mathit{K}}_{\mathit{M}}+{\mathit{T}}_{\mathit{M}}{\mathit{D}}_{\mathit{M}}$.

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

$\mathit{R}(\mathit{M},\mathit{g})=({\mathit{A}}_{\mathit{M}}+{\mathit{gD}}_{\mathit{M}})\mathit{R}(\mathit{M}-\mathit{1},\mathit{g})+({\mathit{B}}_{\mathit{M}}+(\mathit{g}-\mathit{1}){\mathit{E}}_{\mathit{M}})\mathit{R}(\mathit{M}-\mathit{1},\mathit{g}-\mathit{1})$.

It can also be calculated in a similar way to $\mathit{H}\left(\mathit{g}\right)$, which is easier to understand. $\mathit{H}\left(\mathit{g}\right)$ itself requires $|{\mathit{D}}_{\mathit{i}}|=|{\mathit{E}}_{\mathit{i}}|$ and can’t change the sign, so that three forms exist. Many conclusions have been drawn [1]:

1.

${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={\sum }_{\mathit{k}=\mathit{g}}^{\mathit{M}}{\mathit{H}}_{\mathit{2}}\left(\mathit{k}\right)\left({ }_{\mathit{g}}^{\mathit{k}}\right)={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{g}}{\mathit{H}}_{\mathit{3}}\left(\mathit{k}\right)\left({ }_{\mathit{M}-\mathit{g}}^{\mathit{M}-\mathit{k}}\right)$.

2.

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={\sum }_{\mathit{k}=\mathit{g}}^{\mathit{M}}{(-\mathit{1})}^{\mathit{k}+\mathit{g}}{\mathit{H}}_{\mathit{1}}\left(\mathit{k}\right)\left({ }_{\mathit{g}}^{\mathit{k}}\right)$. ${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{g}}{(-\mathit{1})}^{\mathit{k}+\mathit{g}}{\mathit{H}}_{\mathit{1}}\left(\mathit{k}\right)\left({ }_{\mathit{M}-\mathit{g}}^{\mathit{M}-\mathit{k}}\right)$.

3.

${\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right){\mathit{q}}^{\mathit{g}}{(\mathit{1}-\mathit{q})}^{\mathit{M}-\mathit{g}}={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right){(\mathit{1}-\mathit{q})}^{\mathit{M}-\mathit{g}}={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right){\mathit{q}}^{\mathit{g}}$

4.

${\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right){\mathit{2}}^{\mathit{g}}={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right){\mathit{2}}^{\mathit{M}-\mathit{g}}$

5.

${\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)\left({ }_{\mathit{Y}-\mathit{g}}^{\mathit{X}}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)\left({ }_{\mathit{Y}}^{\mathit{X}+\mathit{g}}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)\left({ }_{\mathit{Y}-\mathit{g}}^{\mathit{X}+\mathit{M}-\mathit{g}}\right),\mathit{Y}\in \mathbb{N},\mathit{X}\in \mathbb{C}$.

6.

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

Recurrence relations → 1), inversion → 2). 1) → 3)4)5). 5) is the basis of $\mathit{SUM}\left(\mathit{N}\right)$.

6) show that ${\mathit{T}}_{\mathit{1}}$ can be great than 1. Regardless of the practical implications, we can make the definition domain of PT extend to $\mathbb{C}$.

When ${\mathit{D}}_{\mathit{i}}\ne \mathit{0},{\mathit{K}}_{\mathit{i}}+{\mathit{D}}_{\mathit{i}}\mathit{n}={\mathit{D}}_{\mathit{i}}({\displaystyle \frac{{\mathit{K}}_{\mathit{i}}}{{\mathit{D}}_{\mathit{i}}}}+\mathit{1})$, so only the case ${\mathit{D}}_{\mathit{i}}=\mathit{1}$ needs to be dealt with. In this paper, if not specified, the default is ${\mathit{D}}_{\mathit{i}}=\mathit{1}$. $\mathit{PS}=[{\mathit{K}}_{\mathit{1}},{\mathit{K}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}}],\mathit{PT}=[{\mathit{T}}_{\mathit{1}},{\mathit{T}}_{\mathit{2}}...{\mathit{T}}_{\mathit{M}}]$.

Definition 5.

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

Definition 6.

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

${\mathit{F}}_{\mathit{M}}^{\mathit{N}+\mathit{M}-\mathit{1}}={\mathit{S}}_{\mathit{1}}(\mathit{N}+\mathit{M},\mathit{N})$, $S_1$ 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.

2. Properties of H(g)

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 7.

$\mathit{H}(\mathit{g},\mathit{T})=\mathit{H}(\mathit{g},\mathit{T},\mathit{PS},\mathit{PT})={\prod }_{{\mathit{X}}_{\mathit{i}}\in \mathbb{T}}{\mathit{B}}_{\mathit{i}},\mathit{H}(\mathit{g},\sum \mathit{T})=\sum {\prod }_{{\mathit{X}}_{\mathit{i}}\in \mathbb{T}}{\mathit{B}}_{\mathit{i}}$.

Also define $\mathit{H}(\mathit{g},\mathit{K}),\mathit{H}(\mathit{g},\sum \mathit{K})$.

Theorem 1.

1.

${\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{K})={\mathit{H}}_{\mathit{3}}(\mathit{g},\sum \mathit{K})={\mathit{F}}_{\mathit{M}-\mathit{g}}^{\mathbb{K}}{\mathit{E}}_{\mathit{0}}^{\mathit{g}}+{\mathit{F}}_{\mathit{M}-\mathit{g}-\mathit{1}}^{\mathbb{K}}{\mathit{E}}_{\mathit{1}}^{\mathit{g}}+...+{\mathit{F}}_{\mathit{0}}^{\mathbb{K}}{\mathit{E}}_{\mathit{M}-\mathit{g}}^{\mathit{g}}$.

2.

${\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{T})={\mathit{H}}_{\mathit{2}}(\mathit{g},\sum \mathit{T})={\mathit{F}}_{\mathit{g}}^{\mathbb{T}}{\mathit{E}}_{\mathit{0}}^{\mathit{M}-\mathit{g}}-{\mathit{F}}_{\mathit{g}-\mathit{1}}^{\mathbb{T}}{\mathit{E}}_{\mathit{1}}^{\mathit{M}-\mathit{g}}+...+{(-\mathit{1})}^{\mathit{g}}{\mathit{F}}_{\mathit{0}}^{\mathbb{T}}{\mathit{E}}_{\mathit{g}}^{\mathit{M}-\mathit{g}}$.

3.

${\mathit{H}}_{\mathit{2}}(\mathit{g},\sum \mathit{K})={(-\mathit{1})}^{\mathit{M}-\mathit{g}}({\mathit{F}}_{\mathit{M}-\mathit{g}}^{\mathbb{S}}{\mathit{E}}_{\mathit{0}}^{\mathit{g}}+...+{\mathit{F}}_{\mathit{0}}^{\mathbb{S}}{\mathit{E}}_{\mathit{M}-\mathit{g}}^{\mathit{g}}),\mathbb{S}=\{{\mathit{T}}_{\mathit{i}}-{\mathit{K}}_{\mathit{i}}-\mathit{i}+\mathit{1}\}$.

4.

${\mathit{H}}_{\mathit{3}}(\mathit{g},\sum \mathit{T})={(-\mathit{1})}^{\mathit{g}}{\mathit{F}}_{\mathit{g}}^{\mathbb{S}}{\mathit{E}}_{\mathit{0}}^{\mathit{M}-\mathit{g}}+{(-\mathit{1})}^{\mathit{g}-\mathit{1}}{\mathit{F}}_{\mathit{g}-\mathit{1}}^{\mathbb{S}}{\mathit{E}}_{\mathit{1}}^{\mathit{M}-\mathit{g}}...+{\mathit{F}}_{\mathit{0}}^{\mathbb{S}}{\mathit{E}}_{\mathit{g}}^{\mathit{M}-\mathit{g}},\mathbb{S}=\{-({\mathit{T}}_{\mathit{i}}-{\mathit{K}}_{\mathit{i}}-\mathit{i}+\mathit{1})\}$.

5.

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

6.

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

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).

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

${\mathit{F}}_{\mathit{g}-\mathit{x}}^{\left\{\mathit{PT}\mathit{1}\right\}}$ in ${\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{T},\mathit{PS}\mathit{1},\mathit{PT}\mathit{1})$ 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.

□

Theorem 2.

$\mathit{PS}=[{\mathit{K}}_{\mathit{i}}:{\mathit{D}}_{\mathit{i}}],{\mathit{D}}_{\mathit{i}}\ne \mathit{0}$,

${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{M}-\mathit{g}}{\mathit{H}}_{\mathit{2}}(\mathit{g},[-{\mathit{K}}_{\mathit{i}}+{\mathit{T}}_{\mathit{i}}-(\mathit{i}-\mathit{1}):{\mathit{D}}_{\mathit{i}}],\mathit{PT})={(-\mathit{1})}^{\mathit{g}}{\mathit{H}}_{\mathit{3}}(\mathit{g},\mathit{PS},[{\displaystyle \frac{{\mathit{K}}_{\mathit{i}}}{{\mathit{D}}_{\mathit{i}}}}-{\mathit{T}}_{\mathit{i}}+\mathit{i}-\mathit{1}])$.

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{M}-\mathit{g}}{\mathit{H}}_{\mathit{1}}(\mathit{g},[-{\mathit{K}}_{\mathit{i}}+{\mathit{T}}_{\mathit{i}}-(\mathit{i}-\mathit{1}):{\mathit{D}}_{\mathit{i}}],\mathit{PT}),{\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{g}}{\mathit{H}}_{\mathit{1}}(\mathit{g},\mathit{PS},[{\displaystyle \frac{{\mathit{K}}_{\mathit{i}}}{{\mathit{D}}_{\mathit{i}}}}-{\mathit{T}}_{\mathit{i}}+\mathit{i}-\mathit{1}])$.

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{M}}{\mathit{H}}_{\mathit{3}}(\mathit{g},[-{\mathit{T}}_{\mathit{i}}:{\mathit{D}}_{\mathit{i}}],[{\displaystyle \frac{{\mathit{K}}_{\mathit{i}}}{{\mathit{D}}_{\mathit{i}}}}-{\mathit{T}}_{\mathit{i}}+\mathit{i}-\mathit{1}])$.

${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{M}}{\mathit{H}}_{\mathit{2}}(\mathit{g},[-{\mathit{K}}_{\mathit{i}}+{\mathit{T}}_{\mathit{i}}-(\mathit{i}-\mathit{1}):{\mathit{D}}_{\mathit{i}}],[-{\displaystyle \frac{{\mathit{K}}_{\mathit{i}}}{{\mathit{D}}_{\mathit{i}}}}])$.

3. Symmetric Expressions in H(g)

$\mathit{PT}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}]$. It can be inferred from the definition of $\mathit{SUM}\left(\mathit{N}\right)$ that ${\mathit{K}}_{\mathit{i}}$ can exchange orders. ${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={[\mathit{T}+\mathit{g}-\mathit{1}]}_{\mathit{g}}{\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{K})$. It is clearly a symmetric function of $\mathbb{K}$, we also reached the same conclusion. ${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right),{\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)$ are also symmetric functions, and there’s no ${\mathit{K}}_{\mathit{i}}^{\mathit{2}}$ factors.

Definition 8.

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

Obviously:

• ${\mathit{F}}_{\mathit{T}}(\mathit{0},\mathit{0})={\mathit{E}}_{\mathit{T}}(\mathit{0},\mathit{0})=\mathit{1},{\mathit{F}}_{\mathit{T}}(\mathit{1},\mathit{0})={\mathit{E}}_{\mathit{T}}(\mathit{1},\mathit{0})=\mathit{0},{\mathit{F}}_{\mathit{T}}(\mathit{1},\mathit{1})={\mathit{E}}_{\mathit{T}}(\mathit{1},\mathit{1})=\mathit{1}$.
• ${\mathit{F}}_{\mathit{0}}(\mathit{N}+\mathit{M},\mathit{N})={\mathit{S}}_{\mathit{1}}(\mathit{N}+\mathit{M},\mathit{N}),{\mathit{E}}_{\mathit{1}}(\mathit{N}+\mathit{M},\mathit{N})={\mathit{S}}_{\mathit{2}}(\mathit{N}+\mathit{M},\mathit{N})$.
• ${\mathit{F}}_{\mathit{T}}(\mathit{M}+\mathit{1},\mathit{g}+\mathit{1})={\mathit{F}}_{\mathit{M}-\mathit{g}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}\}}=(\mathit{T}+\mathit{M}){\mathit{F}}_{\mathit{T}}(\mathit{M},\mathit{g}+\mathit{1})+{\mathit{F}}_{\mathit{T}}(\mathit{M},\mathit{g})$.
• ${\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1},\mathit{g}+\mathit{1})={\mathit{E}}_{\mathit{M}-\mathit{g}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{g}\}}=(\mathit{T}+\mathit{g}){\mathit{E}}_{\mathit{T}}(\mathit{M},\mathit{g}+\mathit{1})+{\mathit{E}}_{\mathit{T}}(\mathit{M},\mathit{g})$.
• ${\mathit{F}}_{\mathit{T}}(\mathit{N}+\mathit{M},\mathit{N})=\mathit{SUM}(\mathit{N},[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],[\mathit{1},\mathit{3}...\mathit{2}\mathit{M}-\mathit{1}])$.
• ${\mathit{E}}_{\mathit{T}}(\mathit{N}+\mathit{M},\mathit{N})=\mathit{SUM}(\mathit{N},[\mathit{T},\mathit{T}...\mathit{T}],[\mathit{1},\mathit{3}...\mathit{2}\mathit{M}-\mathit{1}])$.

Theorem 3.

$\mathit{PT}=[\mathit{T}...\mathit{T}+\mathit{M}-\mathit{1}]$, ${\mathit{H}}_{\mathit{2}}(\mathit{g},\sum \mathit{K})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{(-\mathit{1})}^{\mathit{M}-\mathit{g}-\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{K}}{\mathit{E}}_{\mathit{M}-\mathit{g}-\mathit{j}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{g}\}}$.

Proof. 

${\mathit{H}}_{\mathit{2}}(\mathit{0},\mathit{1})={\mathit{K}}_{\mathit{1}}-\mathit{T},{\mathit{H}}_{\mathit{2}}(\mathit{1},\mathit{1})=\mathit{T}$. It’s holds when $\mathit{M}=\mathit{1}$.

${\mathit{H}}_{\mathit{2}}(\mathit{g},\mathit{M})=({\mathit{K}}_{\mathit{M}}-\mathit{T}-\mathit{g}){\mathit{H}}_{\mathit{2}}(\mathit{g},\mathit{M}-\mathit{1})+(\mathit{T}+\mathit{g}-\mathit{1}){\mathit{H}}_{\mathit{2}}(\mathit{g}-\mathit{1},\mathit{M}-\mathit{1})$.

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)$ is a symmetric function $={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}}{\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{j}\right){\mathit{F}}_{\mathit{j}}^{\mathbb{K}}$

${\mathit{A}}_{\mathit{1}}^{\mathit{0}}\left(\mathit{0}\right)=-\mathit{T},{\mathit{A}}_{\mathit{1}}^{\mathit{1}}\left(\mathit{0}\right)=\mathit{T}$.

${\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{0}\right)=-(\mathit{T}+\mathit{g}){\mathit{A}}_{\mathit{M}-\mathit{1}}^{\mathit{g}}\left(\mathit{0}\right)+(\mathit{T}+\mathit{g}-\mathit{1}){\mathit{A}}_{\mathit{M}-\mathit{1}}^{\mathit{g}-\mathit{1}}\left(\mathit{0}\right)\to$

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

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

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

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

${\mathit{H}}_{\mathit{2}}(\mathit{g},\mathit{T})={[\mathit{T}+\mathit{g}-\mathit{1}]}_{\mathit{g}}\to$ the conclusion. □

Definition 9.

${〈{ }_{\mathit{j}}^{\mathit{n}}〉}_{\mathit{T}}=(\mathit{T}-\mathit{1}+\mathit{n}-\mathit{j}){〈{ }_{\mathit{j}-\mathit{1}}^{\mathit{n}-\mathit{1}}〉}_{\mathit{T}}+(\mathit{j}+\mathit{1}){〈{ }_{\mathit{j}}^{\mathit{n}-\mathit{1}}〉}_{\mathit{T}},{〈{ }_{\mathit{0}}^{\mathit{1}}〉}_{\mathit{T}}=\mathit{T},{〈{ }_{\mathit{1}}^{\mathit{1}}〉}_{\mathit{T}}=\mathit{0},{〈{ }_{\mathit{j}>\mathit{n},\mathit{j}<\mathit{0}}^{\mathit{n}}〉}_{\mathit{T}}=\mathit{0}$.

Obviously: ${〈{ }_{\mathit{0}}^{\mathit{n}}〉}_{\mathit{1}}=\mathit{T},{〈{ }_{\mathit{n}}^{\mathit{n}}〉}_{\mathit{T}}=\mathit{0}$. $\mathit{n}>\mathit{0},{〈{ }_{\mathit{j}}^{\mathit{n}}〉}_{\mathit{1}}=〈{ }_{\mathit{j}}^{\mathit{n}}〉$ is Eulerian number.

Definition 10.

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

Theorem 4.

$\mathit{PT}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}]$, ${\mathit{H}}_{\mathit{3}}\left(\mathit{0}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{K}}_{\mathit{i}},{\mathit{H}}_{\mathit{3}}\left(\mathit{M}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}(\mathit{T}-{\mathit{K}}_{\mathit{i}})$,

${\mathit{H}}_{\mathit{3}}(\mathit{0}<\mathit{g}<\mathit{M})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{K}}+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{K}}$.

Proof. 

The coefficient before ${\mathit{F}}_{\mathit{M}}^{\mathbb{K}}$ is obvious, so ${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)$ can be written in that form.

${\mathit{H}}_{\mathit{3}}(\mathit{0}<\mathit{g}<\mathit{M})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{j}\right){\mathit{F}}_{\mathit{j}}^{\mathbb{K}}+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{K}}$.

${\mathit{H}}_{\mathit{3}}(\mathit{0},\mathit{1})={\mathit{K}}_{\mathit{1}},{\mathit{H}}_{\mathit{3}}(\mathit{1},\mathit{1})=\mathit{T}-{\mathit{K}}_{\mathit{1}}\to {\mathit{A}}_{\mathit{1}}^{\mathit{0}}\left(\mathit{0}\right)=\mathit{0},{\mathit{A}}_{\mathit{1}}^{\mathit{1}}\left(\mathit{0}\right)=\mathit{T}$.

${\mathit{H}}_{\mathit{3}}(\mathit{g},\mathit{M})=({\mathit{K}}_{\mathit{M}}+\mathit{g}){\mathit{H}}_{\mathit{3}}(\mathit{g},\mathit{M}-\mathit{1})+(\mathit{T}+\mathit{M}-\mathit{g}-{\mathit{K}}_{\mathit{M}}){\mathit{H}}_{\mathit{3}}(\mathit{g}-\mathit{1},\mathit{M}-\mathit{1})\to$

${\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{0}\right)={\mathit{gA}}_{\mathit{M}-\mathit{1}}^{\mathit{g}}\left(\mathit{0}\right)+(\mathit{T}+\mathit{M}-\mathit{g}){\mathit{A}}_{\mathit{M}-\mathit{1}}^{\mathit{g}-\mathit{1}}\left(\mathit{0}\right)\to {\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{0}\right)={〈{ }_{\mathit{g}-\mathit{1}}^{\mathit{M}}〉}_{\mathit{T}}$.

${\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{1}}^{\mathit{g}-\mathit{1}}(\mathit{j}-\mathit{1})$$\to A_M^g\left(j\right)={〈{ }_g^M〉}_T^j$ .

□

Similarly, $\mathit{PS}=[\mathit{K},\mathit{K}-\mathit{1}...\mathit{K}-\mathit{M}+\mathit{1}]$, then $\mathit{H}\left(\mathit{g}\right)$ are symmetric functions of $\mathbb{T}$.

Theorem 5.

$\mathit{PS}=[\mathit{K}...\mathit{K}-\mathit{M}+\mathit{1}]$, ${\mathit{H}}_{\mathit{3}}(\mathit{g},\sum \mathit{T})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{g}}{(-\mathit{1})}^{\mathit{g}-\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{T}}{\mathit{E}}_{\mathit{g}-\mathit{j}}^{\{\mathit{K},\mathit{K}-\mathit{1}...\mathit{K}-\mathit{M}+\mathit{g}\}}$.

Theorem 6.

$\mathit{PS}=[\mathit{K},\mathit{K}-\mathit{1}...\mathit{K}-\mathit{M}+\mathit{1}]$, ${\mathit{H}}_{\mathit{2}}\left(\mathit{M}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}},{\mathit{H}}_{\mathit{2}}\left(\mathit{0}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}(\mathit{K}-{\mathit{T}}_{\mathit{i}})$,

${\mathit{H}}_{\mathit{2}}(\mathit{0}<\mathit{g}<\mathit{M})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}-{〈{ }_{\mathit{M}-\mathit{g}}^{\mathit{M}}〉}_{-\mathit{K}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{T}}+{(-\mathit{1})}^{\mathit{M}-\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{T}}$.

Proof. 

${\mathit{H}}_{\mathit{2}}(\mathit{0}<\mathit{g}<\mathit{M})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{j}\right){\mathit{F}}_{\mathit{j}}^{\mathbb{T}}+{(-\mathit{1})}^{\mathit{M}-\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{T}}$.

${\mathit{H}}_{\mathit{2}}(\mathit{0},\mathit{1})=\mathit{K}-{\mathit{T}}_{\mathit{1}},{\mathit{H}}_{\mathit{2}}(\mathit{1},\mathit{1})={\mathit{T}}_{\mathit{1}}\to {\mathit{A}}_{\mathit{1}}^{\mathit{0}}\left(\mathit{0}\right)=\mathit{K},{\mathit{A}}_{\mathit{1}}^{\mathit{1}}\left(\mathit{0}\right)=\mathit{0}$.

${\mathit{H}}_{\mathit{2}}(\mathit{g},\mathit{M})=(\mathit{K}-\mathit{g}-{\mathit{T}}_{\mathit{M}}){\mathit{H}}_{\mathit{2}}(\mathit{g},\mathit{M}-\mathit{1})+({\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{g}){\mathit{H}}_{\mathit{2}}(\mathit{g}-\mathit{1},\mathit{M}-\mathit{1})\to$

${\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{0}\right)=(\mathit{K}-\mathit{g}){\mathit{A}}_{\mathit{M}-\mathit{1}}^{\mathit{g}}\left(\mathit{0}\right)-(\mathit{M}-\mathit{g}){\mathit{A}}_{\mathit{M}-\mathit{1}}^{\mathit{g}-\mathit{1}}\left(\mathit{0}\right)\to {\mathit{A}}_{\mathit{M}}^{\mathit{g}}\left(\mathit{0}\right)=-{〈{ }_{\mathit{M}-\mathit{g}-\mathit{1}}^{\mathit{M}}〉}_{-\mathit{K}}$.

${\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{1}}^{\mathit{g}-\mathit{1}}(\mathit{j}-\mathit{1})$$\to A_M^g\left(j\right)=-{〈{ }_{M-g}^M〉}_{-K}^j$ .

□

4. Properties of Extended Numbers

Theorem 7.

$(\mathit{x}+\mathit{T})(\mathit{x}+\mathit{T}+\mathit{1})...(\mathit{x}+\mathit{T}+\mathit{M}-\mathit{1})={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}}{\mathit{F}}_{\mathit{T}}(\mathit{M},\mathit{k}){\mathit{x}}^{\mathit{k}}$.

${(\mathit{x}+\mathit{T})}^{\mathit{M}}={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}}{\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1},\mathit{k}+\mathit{1}){\left[\mathit{x}\right]}_{\mathit{k}}$.

Proof. 

$\nabla \mathit{SUM}(\mathit{N},[\mathit{T},\mathit{T}...\mathit{T}],[\mathit{1},\mathit{2}...\mathit{M}])={(\mathit{T}+\mathit{n})}^{\mathit{M}}={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)\left({ }_{\mathit{g}}^{\mathit{n}}\right)$

$={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}\mathit{g}!{\mathit{E}}_{\mathit{M}-\mathit{g}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{g}\}}\left({ }_{\mathit{g}}^{\mathit{n}}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}\mathit{g}!{\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1},\mathit{g}+\mathit{1})\left({ }_{\mathit{g}}^{\mathit{n}}\right)\to$ 2).

□

${\mathit{x}}^{\mathit{M}}={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}}{\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{k}){\mathit{x}}^{\mathit{k}}$ because ${\mathit{E}}_{\mathit{0}}(\mathit{M}+\mathit{1},\mathit{k}+\mathit{1})={\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{k})$.

Theorem 8.

${\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1},\mathit{g}+\mathit{1})$

$={\displaystyle \frac{\mathit{1}}{\mathit{g}!}}{\sum }_{\mathit{k}=\mathit{0}}^{\mathit{g}}{(-\mathit{1})}^{\mathit{g}+\mathit{k}}\left({ }_{\mathit{k}}^{\mathit{g}}\right){(\mathit{T}+\mathit{k})}^{\mathit{M}}={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}-\mathit{g}}{\mathit{T}}^{\mathit{k}}\left({ }_{\mathit{k}}^{\mathit{M}}\right){\mathit{E}}_{\mathit{M}-\mathit{k}}^{\mathit{g}}$.

Proof. 

${\mathit{H}}_{\mathit{1}}(\mathit{g},[\mathit{T},\mathit{T}...\mathit{T}],[\mathit{1},\mathit{2}...\mathit{M}])=\mathit{g}!{\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1},\mathit{g}+\mathit{1})$.

Based on Crammer’s law, ${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={\sum }_{\mathit{k}=\mathit{1}}^{\mathit{g}+\mathit{1}}{(-\mathit{1})}^{\mathit{g}+\mathit{1}+\mathit{k}}\left({ }_{{\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{k}-\mathit{1}}^{{\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{g}}\right)\nabla \mathit{SUM}\left(\mathit{k}\right)$. [1]

${\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1},\mathit{g}+\mathit{1})={\displaystyle \frac{\mathit{1}}{\mathit{g}!}}{\sum }_{\mathit{k}=\mathit{1}}^{\mathit{g}+\mathit{1}}{(-\mathit{1})}^{\mathit{g}+\mathit{1}+\mathit{k}}\left({ }_{\mathit{k}-\mathit{1}}^{\mathit{g}}\right){(\mathit{T}+\mathit{k}-\mathit{1})}^{\mathit{M}}\to$ 1), $\to$ 2).

□

Theorem 9.

${〈{ }_{\mathit{j}}^{\mathit{n}}〉}_{-\mathit{1}}={(-\mathit{1})}^{\mathit{j}-\mathit{1}}\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{1}}\right)$. ${〈{ }_{\mathit{j}}^{\mathit{n}}〉}_{-\mathit{2}}={(-\mathit{2})}^{\mathit{j}-\mathit{1}}(\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{2}}\right)+{\mathit{2}}^{\mathit{n}-\mathit{1}}\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{2}}\right))$.

Proof. 

1) is to prove: $\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{1}}\right)=-(\mathit{n}-\mathit{j}-\mathit{2})\left({ }_{\mathit{j}-\mathit{1}}^{\mathit{n}-\mathit{2}}\right)+(\mathit{j}+\mathit{1})\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{2}}\right)$.

$\mathit{Right}=-(\mathit{n}-\mathit{1})\left({ }_{\mathit{j}-\mathit{1}}^{\mathit{n}-\mathit{2}}\right)+(\mathit{j}+\mathit{1})\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{1}}\right)=-\mathit{j}\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{1}}\right)+(\mathit{j}+\mathit{1})\left({ }_{\mathit{j}}^{\mathit{n}-\mathit{1}}\right)=\mathit{Left}$.

Prove 2) in a similar way.

□

By recurrence relation:

Theorem 10.

${〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}={\mathit{H}}_{\mathit{3}}(\mathit{g},[\mathit{T},\mathit{1}...\mathit{1}],[\mathit{T}...\mathit{T}+\mathit{M}-\mathit{1}])=\mathit{T}\times {\mathit{H}}_{\mathit{3}}(\mathit{g},[\mathit{1}...\mathit{1}],[\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}])$.

Theorem 11.

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

$=\mathit{T}\times ({\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}-\mathit{2}}{〈{ }_{\mathit{g}}^{\mathit{M}-\mathit{1}}〉}_{\mathit{T}+\mathit{1}}^{\mathit{k}}\left({ }_{\mathit{k}}^{\mathit{M}-\mathit{1}}\right)+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}-\mathit{1}}\right)),\mathit{0}<\mathit{g}<\mathit{M}-\mathit{1}$.

Proof. 

Based on Crammer’s law, ${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)={\sum }_{\mathit{n}=\mathit{1}}^{\mathit{g}+\mathit{1}}{(-\mathit{1})}^{\mathit{g}+\mathit{n}+\mathit{1}}\left({ }_{\mathit{g}+\mathit{1}-\mathit{n}}^{{\mathit{T}}_{\mathit{M}}+\mathit{1}}\right)\nabla \mathit{SUM}\left(\mathit{n}\right)$. [1]

$\mathit{PS}=[\mathit{1}...\mathit{1}],\mathit{PT}=[\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],\mathit{PS}\mathit{1}=[\mathit{1},\mathit{2}...\mathit{T},\mathit{1}...\mathit{1}],\mathit{PT}\mathit{1}=[\mathit{1},\mathit{2}...\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}]$.

${〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}=\mathit{T}\times {\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)={\displaystyle \frac{\mathit{1}}{(\mathit{T}-\mathit{1})!}}{\mathit{H}}_{\mathit{3}}(\mathit{g},\mathit{PS}\mathit{1},\mathit{PT}\mathit{1})$.

$={\displaystyle \frac{\mathit{1}}{(\mathit{T}-\mathit{1})!}}{\sum }_{\mathit{n}=\mathit{1}}^{\mathit{g}+\mathit{1}}{(-\mathit{1})}^{\mathit{g}+\mathit{n}+\mathit{1}}\left({ }_{\mathit{g}+\mathit{1}-\mathit{n}}^{\mathit{T}+\mathit{M}}\right){[\mathit{T}+\mathit{n}-\mathit{1}]}_{\mathit{T}}\times {\mathit{n}}^{\mathit{M}-\mathit{1}}\to$ 1), $\to$ 2).

□

${〈{ }_{\mathit{1}}^{\mathit{M}}〉}_{\mathit{T}}=\mathit{T}(\mathit{T}+\mathit{1}){\mathit{2}}^{\mathit{M}-\mathit{1}}-\mathit{T}(\mathit{T}+\mathit{M})$.

Theorem 12.

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

Proof. 

${\mathit{H}}_{\mathit{2}}\left(\mathit{0}\right)={(\mathit{T}-\mathit{1})}^{\mathit{M}}={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}}{(-\mathit{1})}^{\mathit{k}}{\mathit{H}}_{\mathit{1}}\left(\mathit{k}\right)={\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}}{(-\mathit{1})}^{\mathit{k}}\mathit{k}!{\mathit{E}}_{\mathit{M}-\mathit{k}}^{\{\mathit{T}...\mathit{T}+\mathit{k}\}}\to$ 1).

$\mathit{PS}=[\mathit{1}...\mathit{1}],\mathit{PT}=[\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],\mathit{T}\times {\mathit{H}}_{\mathit{1}}\left(\mathit{M}\right)=\mathit{T}\times {\sum }_{\mathit{k}=\mathit{0}}^{\mathit{M}}{\mathit{H}}_{\mathit{3}}\left(\mathit{k}\right)\to$2).

□

${\mathit{K}}_{\mathit{1}}$ is a constant, ${\mathit{K}}_{\mathit{1}}{\mathit{K}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}},{\mathit{K}}_{\mathit{i}+\mathit{1}}-{\mathit{K}}_{\mathit{i}}=$ 1 or 2, there are M-1 intervals between factors. ${\mathit{K}}_{\mathit{i}+\mathit{1}}-{\mathit{K}}_{\mathit{i}}=\mathit{1}$ is defined as continuity, ${\mathit{K}}_{\mathit{i}+\mathit{1}}-{\mathit{K}}_{\mathit{i}}=\mathit{2}$ is defined as discontinuity.

Definition 11.

${\mathit{MIN}}_{\mathit{g}}^{\mathit{K}}\left(\mathit{M}\right)=\mathit{K}\sum {\mathit{K}}_{\mathit{2}}...{\mathit{K}}_{\mathit{M}},\mathit{K}={\mathit{K}}_{\mathit{1}},{\mathit{MIN}}_{\mathit{g}}^{\mathit{1}}\left(\mathit{M}\right)={\mathit{MIN}}_{\mathit{g}}\left(\mathit{M}\right)$, count of discontinuities=g.

Obviously, $\mathit{0}\le \mathit{g}\le \mathit{M}-\mathit{1}$, ${\mathit{MIN}}_{\mathit{g}}^{\mathit{K}}\left(\mathit{M}\right)$ have $\left({ }_{\mathit{g}}^{\mathit{M}-\mathit{1}}\right)$ items.

$\mathit{PS}=[\mathit{T}+\mathit{1},\mathit{T}+\mathit{2}...\mathit{T}+\mathit{M}-\mathit{1}],\mathit{PT}=[\mathit{T}+\mathit{2},\mathit{T}+\mathit{4}...\mathit{T}+\mathit{2}\mathit{M}-\mathit{2}]$, $\mathit{T}\times {\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={\mathit{MIN}}_{\mathit{g}}^{\mathit{T}}\left(\mathit{M}\right)$

$\mathit{PS}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],\mathit{PT}=[\mathit{T},\mathit{T}+\mathit{2}...\mathit{T}+\mathit{2}\mathit{M}-\mathit{2}]$, ${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={\mathit{MIN}}_{\mathit{g}}^{\mathit{T}}\left(\mathit{M}\right)+{\mathit{MIN}}_{\mathit{g}-\mathit{1}}^{\mathit{T}}\left(\mathit{M}\right)$

By definition:

Theorem 13.

${\lambda }_1+{\lambda }_2+...+{\lambda }_{g+1}=M-g,{\lambda }_i\ge 0$,

${〈{ }_g^M〉}_T=T\sum 1^{{\lambda }_1}{2}^{{\lambda }_2}...{(g+1)}^{{\lambda }_{g+1}}(T+{\lambda }_1)(T+{\lambda }_1+{\lambda }_2)...(T+{\lambda }_1+...+{\lambda }_g)$.

$\mathit{PS}=[\mathit{T},\mathit{T}...\mathit{T}],\mathit{PT}=[\mathit{1},\mathit{3}...\mathit{2}\mathit{M}-\mathit{1}]$.

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

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)$=Changed ${\mathit{MIN}}_{\mathit{g}-\mathit{1}}\left(\mathit{M}\right)+{\mathit{MIN}}_{\mathit{g}}\left(\mathit{M}\right)$. Select M-g from M factors, change i to (T-i).

$\mathit{PS}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],\mathit{PT}=[\mathit{1},\mathit{3}...\mathit{2}\mathit{M}-\mathit{1}]$.

${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)$=Changed ${\mathit{MIN}}_{\mathit{g}-\mathit{1}}\left(\mathit{M}\right)+{\mathit{MIN}}_{\mathit{g}}\left(\mathit{M}\right)$. Select M-g 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 ():

${\mathit{MIN}}_{\mathit{0}}\left(\mathit{3}\right)+{\mathit{MIN}}_{\mathit{1}}\left(\mathit{3}\right)=\left(\mathit{123}\right)+\left(\mathit{124}\right)+\left(\mathit{134}\right)$,

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

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

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

5. The Formulas for Symmetric Functions

If $\mathit{1}+{\mathit{K}}_{\mathit{i}}={\mathit{K}}_{\mathit{i}+\mathit{1}}$ and $\mathit{1}+{\mathit{T}}_{\mathit{i}}={\mathit{T}}_{\mathit{i}+\mathit{1}}$, ${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)=\left({ }_{\mathit{g}}^{\mathit{M}}\right){\prod }_{\mathit{i}=\mathit{1}}^{\mathit{g}}{\mathit{T}}_{\mathit{i}}{\prod }_{\mathit{i}=\mathit{g}+\mathit{1}}^{\mathit{M}}{\mathit{K}}_{\mathit{i}}$. 1 → [1]:

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{K},\mathit{K}+\mathit{1}...\mathit{K}+\mathit{M}-\mathit{1}\}}{\mathit{E}}_{\mathit{M}-\mathit{g}-\mathit{j}}^{\mathit{g}}={\prod }_{\mathit{i}=\mathit{g}+\mathit{1}}^{\mathit{M}}(\mathit{K}+\mathit{i}-\mathit{1})\left({ }_{\mathit{g}}^{\mathit{M}}\right)$.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{g}}{(-\mathit{1})}^{\mathit{j}}{\mathit{F}}_{\mathit{g}-\mathit{j}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}\}}{\mathit{E}}_{\mathit{j}}^{\mathit{M}-\mathit{g}}={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{g}}(\mathit{T}+\mathit{i}-\mathit{1})\left({ }_{\mathit{g}}^{\mathit{M}}\right)$.

1 →${\mathit{H}}_{\mathit{2}}(\mathit{g},\sum \mathit{K})={(-\mathit{1})}^{\mathit{M}-\mathit{g}}{\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{T}-{\mathit{K}}_{\mathit{i}}\}}{\mathit{E}}_{\mathit{M}-\mathit{g}-\mathit{j}}^{\mathit{g}}$, 3→

Theorem 14.

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

${\mathit{K}}_{\mathit{i}}=\mathit{T}\to {\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{g})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{(-\mathit{T})}^{\mathit{j}}\left({ }_{\mathit{j}}^{\mathit{M}}\right){\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1}-\mathit{j},\mathit{g}+\mathit{1})$.

${\mathit{K}}_{\mathit{i}}=\mathit{T}=\mathit{1}\to {\mathit{S}}_{\mathit{2}}(\mathit{M},\mathit{g})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{(-\mathit{1})}^{\mathit{j}}\left({ }_{\mathit{j}}^{\mathit{M}}\right){\mathit{S}}_{\mathit{2}}(\mathit{M}+\mathit{1}-\mathit{j},\mathit{g}+\mathit{1})$.

${\mathit{K}}_{\mathit{i}}=-\mathit{q}\to {\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{(\mathit{T}+\mathit{q})}^{\mathit{j}}\left({ }_{\mathit{j}}^{\mathit{M}}\right){\mathit{S}}_{\mathit{2}}(\mathit{M}-\mathit{j},\mathit{g})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{\mathit{q}}^{\mathit{j}}\left({ }_{\mathit{j}}^{\mathit{M}}\right){\mathit{E}}_{\mathit{T}}(\mathit{M}+\mathit{1}-\mathit{j},\mathit{g}+\mathit{1})$.

${\mathit{K}}_{\mathit{i}}=\mathit{q},\mathit{T}=\mathit{1}\to {\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{(\mathit{1}-\mathit{q})}^{\mathit{j}}\left({ }_{\mathit{j}}^{\mathit{M}}\right){\mathit{S}}_{\mathit{2}}(\mathit{M}-\mathit{j},\mathit{g})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{(-\mathit{q})}^{\mathit{j}}\left({ }_{\mathit{j}}^{\mathit{M}}\right){\mathit{S}}_{\mathit{2}}(\mathit{M}+\mathit{1}-\mathit{j},\mathit{g}+\mathit{1})$.

1 →$\mathit{PS}=[\mathit{K}...\mathit{K}-\mathit{M}+\mathit{1}],{\mathit{H}}_{\mathit{3}}(\mathit{g},\sum \mathit{T})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{g}}{\mathit{F}}_{\mathit{j}}^{\{{\mathit{T}}_{\mathit{i}}-\mathit{K}\}}{\mathit{E}}_{\mathit{g}-\mathit{j}}^{\mathit{M}-\mathit{g}}$, 5→

Theorem 15.

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

Compared to 14, this is a little different.

Theorem 16.

$\mathit{PS}\mathit{1}=[\mathit{0},-\mathit{1}...-(\mathit{p}-\mathit{1}),{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}],\mathit{PT}\mathit{1}=[{\mathit{L}}_{\mathit{1}},{\mathit{L}}_{\mathit{2}}...{\mathit{L}}_{\mathit{p}},{\mathit{T}}_{\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}]$,

$\mathit{SUM}(\mathit{N},\mathit{PS}\mathit{1},\mathit{PT}\mathit{1})=\prod {\mathit{L}}_{\mathit{i}}\times \mathit{SUM}(\mathit{N}-\mathit{p},[{\mathit{K}}_{\mathit{p}+\mathit{1}}+\mathit{p}...{\mathit{K}}_{\mathit{M}}+\mathit{p}],[{\mathit{T}}_{\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}])$.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{0},-\mathit{1}...-(\mathit{p}-\mathit{1}),{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}}{\mathit{E}}_{\mathit{M}-\mathit{g}-\mathit{j}}^{\mathit{g}}=\mathit{0},\mathit{g}<\mathit{p}$.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}-\mathit{p}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{0}...-(\mathit{p}-\mathit{1}),{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}}{\mathit{E}}_{\mathit{M}-\mathit{g}-\mathit{p}-\mathit{j}}^{\mathit{g}+\mathit{p}}={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{p}-\mathit{g}}{\mathit{F}}_{\mathit{j}}^{\{{\mathit{K}}_{\mathit{p}+\mathit{1}}+\mathit{p}...{\mathit{K}}_{\mathit{M}}+\mathit{p}\}}{\mathit{E}}_{\mathit{M}-\mathit{p}-\mathit{g}-\mathit{j}}^{\mathit{g}},\mathit{0}\le \mathit{g}\le \mathit{M}-\mathit{p}$.

Proof. 

By the definition of ${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)$, there clearly is:

${\mathit{H}}_{\mathit{1}}(\mathit{g}+\mathit{p},\mathit{PS}\mathit{1},\mathit{PT}\mathit{1})=\prod {\mathit{L}}_{\mathit{i}}\times {\mathit{H}}_{\mathit{1}}(\mathit{g},[{\mathit{K}}_{\mathit{p}+\mathit{1}}+\mathit{p}...{\mathit{K}}_{\mathit{M}}+\mathit{p}],[{\mathit{T}}_{\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}])\to \mathit{SUM}\left(\mathit{N}\right)$.

Let $\mathit{PT}=[\mathit{1},\mathit{2}...\mathit{M}]$, 1→ the rest.

□

Special:

${\mathit{H}}_{\mathit{1}}(\mathit{0},\sum \mathit{K},[{\mathit{K}}_{\mathit{i}}+\mathit{p}],\mathit{PT})\to {\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}(\mathit{p}+{\mathit{K}}_{\mathit{i}})={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{0},-\mathit{1}...-(\mathit{p}-\mathit{1}),{\mathit{K}}_{\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}}{\mathit{E}}_{\mathit{M}-\mathit{j}}^{\mathit{p}}$.

${\mathit{p}}^{\mathit{M}}={\sum }_{\mathit{j}=\mathit{0}}^{\mathit{p}-\mathit{1}}{(-\mathit{1})}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathit{p}-\mathit{1}}{\mathit{E}}_{\mathit{M}-\mathit{j}}^{\mathit{p}},\mathit{p}>\mathit{1}$.

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

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}}{\mathit{k}}^{\mathit{M}-\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathit{M}}={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}(\mathit{k}+\mathit{i})$. ${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}}{(-\mathit{1})}^{\mathit{j}}{\mathit{k}}^{\mathit{M}-\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathit{M}}=\mathit{0},\mathit{M}\ge \mathit{k}\ge \mathit{1}$.

The extension can be obtained with ${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)$.

$\mathit{PS}=[\mathit{0},-\mathit{1}...-(\mathit{M}-\mathit{1})],\mathit{PT}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],{\mathit{H}}_{\mathit{3}}(\mathit{g}<\mathit{M})=\mathit{0}$,

$\mathit{PS}\mathit{1}=\mathit{PT}\mathit{1}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],{\mathit{H}}_{\mathit{3}}(\mathit{g}>\mathit{0})=\mathit{0}\to$

Theorem 17.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{(-\mathit{1})}^{\mathit{j}}{〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathit{M}-\mathit{1}}=\mathit{0},\mathit{0}<\mathit{g}<\mathit{M}$.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{K}}+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{K}}=\mathit{0},\mathit{0}<\mathit{g}<\mathit{M},\mathbb{K}=\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}\}$.

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

Proof. 

Proof of the third equation. From ${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}}{(-\mathit{1})}^{\mathit{j}}{\mathit{k}}^{\mathit{M}-\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathit{M}}=\mathit{0}$ and the first equation,

it can be seen that ${〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}$ and ${〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{0}}={〈{ }_{\mathit{g}-\mathit{1}}^{\mathit{M}}〉}_{\mathit{T}}$ have the same thing:

11$\to {〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{0}}={\mathit{a}}_{\mathit{1}}{\mathit{1}}^{\mathit{x}}+{\mathit{a}}_{\mathit{2}}{\mathit{2}}^{\mathit{y}}+...\to$${〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}={\mathit{a}}_{\mathit{1}}{\mathit{1}}^{\mathit{x}-\mathit{j}}+{\mathit{a}}_{\mathit{2}}{\mathit{2}}^{\mathit{y}-\mathit{j}}+...\to$ the expression.

The same conclusion can be obtained by combining definition of ${〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}$ and 11.

□

$\mathit{PS}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{p}-\mathit{1},{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}],\mathit{PT}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}]$,

${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{p}}(\mathit{T}+\mathit{i}-\mathit{1}){\mathit{H}}_{\mathit{3}}(\mathit{g},[{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}],[\mathit{T}+\mathit{p}...\mathit{T}+\mathit{M}-\mathit{1}])\to$

Theorem 18.

$\mathbb{K}=\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{p}-\mathit{1},{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\},{\mathbb{K}}_{\mathbb{1}}=\{{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}$,

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{K}}+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{K}}=\mathit{0},\mathit{M}>\mathit{g}>\mathit{M}-\mathit{p}$.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{〈{ }_{\mathit{g}}^{\mathit{M}}〉}_{\mathit{T}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{K}}+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{K}}=$

${\prod }_{\mathit{i}=\mathit{1}}^{\mathit{p}}(\mathit{T}+\mathit{i}-\mathit{1})\left({\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{p}-\mathit{1}}{〈{ }_{\mathit{g}}^{\mathit{M}-\mathit{p}}〉}_{\mathit{T}+\mathit{p}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{{\mathbb{K}}_{\mathbb{1}}}+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}-\mathit{p}}\right){\mathit{F}}_{\mathit{M}-\mathit{p}}^{{\mathbb{K}}_{\mathbb{1}}}\right),\mathit{0}<\mathit{g}<\mathit{M}-\mathit{p}$.

$\mathit{PS}=[\mathit{K},\mathit{K}-\mathit{1}..\mathit{K}-\mathit{M}+\mathit{1}],\mathit{PT}=[\mathit{0},\mathit{1}...\mathit{M}-\mathit{1}],{\mathit{H}}_{\mathit{2}}(\mathit{g}>\mathit{0})=\mathit{0}$,

$\mathit{PS}\mathit{1}=\mathit{PT}\mathit{1}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}],{\mathit{H}}_{\mathit{2}}(\mathit{g}<\mathit{M})=\mathit{0}\to$

Theorem 19.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}-{〈{ }_{\mathit{M}-\mathit{g}}^{\mathit{M}}〉}_{-\mathit{K}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathit{M}-\mathit{1}}=\mathit{0},\mathit{0}<\mathit{g}<\mathit{M}$.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}-{〈{ }_{\mathit{M}-\mathit{g}}^{\mathit{M}}〉}_{\mathit{1}-\mathit{M}-\mathit{T}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{K}}+{(-\mathit{1})}^{\mathit{M}-\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{K}}=\mathit{0},\mathit{0}<\mathit{g}<\mathit{M},\mathbb{K}=\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}\}$.

$\mathit{PS}=[\mathit{K},\mathit{K}-\mathit{1}...\mathit{K}-\mathit{M}+\mathit{1}],\mathit{PT}=[\mathit{K},\mathit{K}-\mathit{1}...\mathit{K}-\mathit{p}+\mathit{1},{\mathit{T}}_{\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}]$,

${\mathit{H}}_{\mathit{2}}(\mathit{g}+\mathit{p})={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{p}}(\mathit{K}-\mathit{i}+\mathit{1}){\mathit{H}}_{\mathit{2}}(\mathit{g},[\mathit{K}-\mathit{p}...\mathit{K}-\mathit{M}+\mathit{1}],[{\mathit{T}}_{\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}])\to$

Theorem 20.

$\mathbb{T}=\{\mathit{K},\mathit{K}-\mathit{1}...\mathit{K}-\mathit{p}+\mathit{1},{\mathit{T}}_{\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}\},{\mathbb{T}}_{\mathbb{1}}=\{{\mathit{T}}_{\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}\}$,

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}-{〈{ }_{\mathit{M}-\mathit{g}}^{\mathit{M}}〉}_{-\mathit{K}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{T}}+{(-\mathit{1})}^{\mathit{M}-\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{T}}=\mathit{0},\mathit{0}<\mathit{g}<\mathit{p}$.

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}-{〈{ }_{\mathit{M}-\mathit{p}-\mathit{g}}^{\mathit{M}}〉}_{-\mathit{K}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{\mathbb{T}}+{(-\mathit{1})}^{\mathit{M}-\mathit{p}-\mathit{g}}\left({ }_{\mathit{g}+\mathit{p}}^{\mathit{M}}\right){\mathit{F}}_{\mathit{M}}^{\mathbb{T}}=$

${\prod }_{\mathit{i}=\mathit{1}}^{\mathit{p}}(\mathit{K}-\mathit{i}+\mathit{1})\left({\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{p}-\mathit{1}}-{〈{ }_{\mathit{M}-\mathit{p}-\mathit{g}}^{\mathit{M}-\mathit{p}}〉}_{\mathit{p}-\mathit{K}}^{\mathit{j}}{\mathit{F}}_{\mathit{j}}^{{\mathbb{T}}_{\mathbb{1}}}+{(-\mathit{1})}^{\mathit{g}}\left({ }_{\mathit{g}}^{\mathit{M}-\mathit{p}}\right){\mathit{F}}_{\mathit{M}-\mathit{p}}^{{\mathbb{T}}_{\mathbb{1}}}\right),\mathit{0}<\mathit{g}<\mathit{M}-\mathit{p}$.

Theorem 21.

$\mathit{T}\in \mathbb{Z},\mathit{p}\in \mathbb{N},\mathit{0}\le \mathit{g}\le \mathit{M},\mathit{p}\le \mathit{M}$,

${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{g}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{p}-\mathit{1},{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}}{\mathit{E}}_{\mathit{M}-\mathit{g}-\mathit{j}}^{\mathit{g}}=$

${\sum }_{\mathit{i}=\mathit{0}}^{\mathit{p}}\left({ }_{\mathit{i}}^{\mathit{p}}\right){(\mathit{T}+\mathit{p}+\mathit{g}-\mathit{i}-\mathit{1})}_{\mathit{p}-\mathit{i}}{\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{p}-\mathit{g}+\mathit{i}}{\mathit{F}}_{\mathit{j}}^{\{{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}}{\mathit{E}}_{\mathit{M}-\mathit{p}-\mathit{g}+\mathit{i}-\mathit{j}}^{\mathit{g}-\mathit{i}}$.

Proof. 

$\mathit{PS}\mathit{1}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+(\mathit{p}-\mathit{1}),{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}],\mathit{PT}\mathit{1}=[\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}]$,

$\mathit{PS}\mathit{2}=[{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}],\mathit{PT}\mathit{2}=[\mathit{T}+\mathit{p}-\mathit{1}...\mathit{T}+\mathit{M}-\mathit{1}]$.

${\mathit{H}}_{\mathit{1}}(\mathit{g},[\mathit{T},\mathit{PS}],[\mathit{T},\mathit{PT}])=\mathit{T}\times {\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)+\mathit{T}\times {\mathit{H}}_{\mathit{1}}(\mathit{g}-\mathit{1})\to$

${\mathit{H}}_{\mathit{1}}(\mathit{g},\mathit{PS}\mathit{1},\mathit{PT}\mathit{1})={[\mathit{T}+\mathit{g}-\mathit{1}]}_{\mathit{g}}{\mathit{H}}_{\mathit{1}}(\mathit{g},\sum \mathit{K},\mathit{PS}\mathit{1},\mathit{PT}\mathit{1})$

$={[\mathit{T}+\mathit{p}-\mathit{1}]}_{\mathit{p}}{\sum }_{\mathit{i}=\mathit{0}}^{\mathit{p}}\left({ }_{\mathit{i}}^{\mathit{p}}\right){\mathit{H}}_{\mathit{1}}(\mathit{g}-\mathit{i},\mathit{PS}\mathit{2},\mathit{PT}\mathit{2})$.

$={[\mathit{T}+\mathit{p}-\mathit{1}]}_{\mathit{p}}{\sum }_{\mathit{i}=\mathit{0}}^{\mathit{p}}\left({ }_{\mathit{i}}^{\mathit{p}}\right){[\mathit{T}+\mathit{p}-\mathit{1}+\mathit{g}-\mathit{i}]}_{\mathit{g}-\mathit{i}}{\mathit{H}}_{\mathit{1}}(\mathit{g}-\mathit{i},\sum \mathit{K},\mathit{PS}\mathit{2},\mathit{PT}\mathit{2})$.

□

${\mathit{K}}_{\mathit{i}}=\mathit{0},\mathit{g}=\mathit{M}-\mathit{p}\to$${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{p}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{T}...\mathit{T}+\mathit{p}-\mathit{1}\}}{\mathit{E}}_{\mathit{p}-\mathit{j}}^{\mathit{M}-\mathit{p}}={\sum }_{\mathit{i}=\mathit{0}}^{\mathit{p}}\left({ }_{\mathit{i}}^{\mathit{p}}\right){[\mathit{T}-\mathit{1}+\mathit{M}-\mathit{i}]}_{\mathit{p}-\mathit{i}}{\mathit{E}}_{\mathit{i}}^{\mathit{M}-\mathit{p}-\mathit{i}}$.

$\mathit{g}=\mathit{1}\to$${\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{1}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{T}...\mathit{T}+\mathit{p}-\mathit{1},{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}}={\sum }_{\mathit{i}=\mathit{0}}^{\mathit{p}}\left({ }_{\mathit{i}}^{\mathit{p}}\right){[\mathit{T}-\mathit{1}+\mathit{p}-\mathit{i}]}_{\mathit{p}-\mathit{i}}{\sum }_{\mathit{j}=\mathit{0}}^{\mathit{M}-\mathit{p}-\mathit{1}+\mathit{i}}{\mathit{F}}_{\mathit{j}}^{\{{\mathit{K}}_{\mathit{p}+\mathit{1}}...{\mathit{K}}_{\mathit{M}}\}}$.

$\mathit{p}=\mathit{M}-\mathit{1}\to {\sum }_{\mathit{j}=\mathit{0}}^{\mathit{p}}{\mathit{F}}_{\mathit{j}}^{\{\mathit{T},\mathit{T}+\mathit{1}...\mathit{T}+\mathit{p}-\mathit{1}\}}={[\mathit{T}+\mathit{p}]}_{\mathit{p}}$.

6. PT=PS and Its Promotion

${\mathit{H}}_{\mathit{1}}(\mathit{g},\mathit{PT},\mathit{PT})={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{g}}^{\mathit{M}}\right)$, promoting it:

Theorem 22.

$\mathit{PS}=[{\mathit{T}}_{\mathit{1}},{\mathit{T}}_{\mathit{2}}...{\mathit{T}}_{\mathit{M}-\mathit{p}},\mathit{0},-\mathit{1}...-(\mathit{p}-\mathit{1})]$, then

${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{g}-\mathit{p}}^{\mathit{M}-\mathit{p}}\right)$,${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{M}-\mathit{g}}{\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{M}-\mathit{g}}^{\mathit{p}}\right)$,

${\mathit{H}}_{\mathit{3}}\left(\mathit{p}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}},{\mathit{H}}_{\mathit{3}}(\mathit{g}\ne \mathit{p})=\mathit{0}$, $\mathit{SUM}\left(\mathit{N}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{{\mathit{T}}_{\mathit{M}}+\mathit{1}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{p}}\right)$.

Proof. 

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

Using induction and the recurrence relationship one can obtain ${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)$.

$\mathit{PT}\mathit{1}=[{\mathit{T}}_{\mathit{M}-\mathit{p}+\mathit{1}}...{\mathit{T}}_{\mathit{M}}]$, 2→

${\mathit{H}}_{\mathit{2}}(\mathit{g},[\mathit{0}...-(\mathit{p}-\mathit{1})],\mathit{PT}\mathit{1})={(-\mathit{1})}^{\mathit{M}-\mathit{g}}{\mathit{H}}_{\mathit{1}}(\mathit{g},\mathit{PT}\mathit{1},\mathit{PT}\mathit{1})={(-\mathit{1})}^{\mathit{M}-\mathit{g}}{\prod }_{\mathit{i}=\mathit{M}-\mathit{p}+\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{g}}^{\mathit{P}}\right)$.

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={\prod }_{\mathit{j}=\mathit{1}}^{\mathit{M}-\mathit{p}}{\mathit{T}}_{\mathit{i}}\times {\mathit{H}}_{\mathit{2}}(\mathit{g}-(\mathit{M}-\mathit{p}),[\mathit{0}...-(\mathit{p}-\mathit{1})],\mathit{PT}\mathit{1})\to {\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)$.

${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)$ can be obtained from the definition. ${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)\to \mathit{SUM}\left(\mathit{N}\right)$.

□

Theorem 23.

$\mathit{PS}=[\mathit{0},-\mathit{1}...-(\mathit{M}-\mathit{p}-\mathit{1}),{\mathit{T}}_{\mathit{M}-\mathit{p}+\mathit{1}}-(\mathit{M}-\mathit{p})...{\mathit{T}}_{\mathit{M}}-(\mathit{M}-\mathit{p})]$, then

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{M}-\mathit{g}}{\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{g}-\mathit{p}}^{\mathit{M}-\mathit{p}}\right)$, ${\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{M}-\mathit{g}}^{\mathit{p}}\right)$,

${\mathit{H}}_{\mathit{3}}(\mathit{M}-\mathit{p})={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}},{\mathit{H}}_{\mathit{3}}(\mathit{g}\ne \mathit{M}-\mathit{p})=\mathit{0},\mathit{SUM}\left(\mathit{N}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{{\mathit{T}}_{\mathit{M}}+\mathit{1}}^{\mathit{N}+{\mathit{T}}_{\mathit{M}}-\mathit{M}+\mathit{p}}\right)$.

Proof. 

2 & 22 $\to {\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)$.

$\mathit{p}=\mathit{0},{\mathit{H}}_{\mathit{1}}\left(\mathit{M}\right)=\prod _{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}$. Recurrence relation & induction on p $\to {\mathit{H}}_{\mathit{1}}\left(\mathit{g}\right)$.

${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)$ can be obtained from the definition. ${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)\to \mathit{SUM}\left(\mathit{N}\right)$.

□

Using similar methods:

Theorem 24.

$\mathit{PS}=[{\mathit{T}}_{\mathit{1}},{\mathit{T}}_{\mathit{2}}...{\mathit{T}}_{\mathit{M}-\mathit{p}},\mathit{0},-\mathit{1}...-(\mathit{p}-\mathit{1})]$,

$\mathit{PT}=[\mathit{0},\mathit{1}...(\mathit{M}-\mathit{p}-\mathit{1}),(\mathit{M}-\mathit{p})-{\mathit{T}}_{\mathit{M}-\mathit{p}+\mathit{1}}...(\mathit{M}-\mathit{p})-{\mathit{T}}_{\mathit{M}}]$, then

${\mathit{H}}_{\mathit{3}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{g}}{\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{g}-\mathit{p}}^{\mathit{M}-\mathit{p}}\right)$, ${\mathit{H}}_{\mathit{1}}\left(\mathit{p}\right)={(-\mathit{1})}^{\mathit{p}}{\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}},{\mathit{H}}_{\mathit{1}}(\mathit{g}\ne \mathit{p})=\mathit{0}$,

${\mathit{H}}_{\mathit{2}}\left(\mathit{g}\right)={(-\mathit{1})}^{\mathit{g}}{\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{g}}^{\mathit{M}-\mathit{p}}\right)$,

$\mathit{SUM}\left(\mathit{N}\right)={(-\mathit{1})}^{\mathit{p}}{\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}}\left({ }_{\mathit{1}-{\mathit{T}}_{\mathit{M}}}^{\mathit{N}-{\mathit{T}}_{\mathit{M}}-\mathit{p}}\right),\mathit{p}>\mathit{0};\mathit{SUM}\left(\mathit{N}\right)={\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}{\mathit{T}}_{\mathit{i}},\mathit{p}=\mathit{0}$.

7. Transformation of SUM(N)

${\prod }_{\mathit{i}=\mathit{1}}^{\mathit{M}}(\mathit{x}+{\mathit{K}}_{\mathit{i}})={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{a}}_{\mathit{g}}{\mathit{x}}^{\mathit{M}-\mathit{g}}={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{F}}_{\mathit{g}}^{\mathbb{K}}{\mathit{x}}^{\mathit{M}-\mathit{g}}=\nabla \mathit{SUM}(\mathit{x}+\mathit{1},\mathit{PS},[\mathit{1},\mathit{2}...\mathit{M}])$.

No need to know the value of ${\mathit{K}}_{\mathit{i}}$. Any polynomial can be converted to $\nabla \mathit{SUM}(\mathit{N},\mathit{PS},[\mathit{1},\mathit{2}...\left]\right)$.

By choosing $\mathit{c}$ and ${\mathit{K}}_{\mathit{1}}^{{ }^{\prime }},{\mathit{K}}_{\mathit{2}}^{{ }^{\prime }}...$ appropriately, $\mathit{SUM}\left(\mathit{N}\right)$ can be converted to $\mathit{c}\times {\nabla }^{\mathit{q}}\mathit{SUM}(\mathit{N},[{\mathit{K}}_{\mathit{1}}^{{ }^{\prime }},{\mathit{K}}_{\mathit{2}}^{{ }^{\prime }}...],\mathit{PT}\mathit{1})$. However, it is generally necessary to solve higher-order equations to solve for ${\mathbb{K}}^{{ }^{\prime }}$. But specifies that every nested sum can be flattened, converting to $\mathit{c}\times {\nabla }^{\mathit{q}}\mathit{SUM}(\mathit{N},\mathit{PS},[\mathit{1},\mathit{2}...])$.

$\left(\mathit{1}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{b}}_{\mathit{g}}\left({ }_{\mathit{g}+\mathit{X}+\mathit{1}}^{\mathit{N}+\mathit{Y}}\right)$,$\left(\mathit{2}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{c}}_{\mathit{g}}\left({ }_{\mathit{g}+\mathit{X}+\mathit{1}}^{\mathit{N}+\mathit{Y}+\mathit{g}}\right)$, $\left(\mathit{3}\right)={\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{d}}_{\mathit{g}}\left({ }_{\mathit{M}+\mathit{1}+\mathit{X}}^{\mathit{N}+\mathit{Y}+\mathit{M}+\mathit{X}-\mathit{g}}\right)$, can be converted to $\mathit{c}\times {\nabla }^{-\mathit{X}}\mathit{SUM}(\mathit{N}+\mathit{Y}-\mathit{X},\mathit{PS},[\mathit{1},\mathit{2}...\mathit{M}])$. $\mathit{c}={\displaystyle \frac{{\mathit{b}}_{\mathit{M}}}{\mathit{M}!}}={\displaystyle \frac{{\mathit{c}}_{\mathit{M}}}{\mathit{M}!}}={\displaystyle \frac{{\mathit{d}}_{\mathit{M}}}{\prod (\mathit{1}-{\mathit{K}}_{\mathit{i}})}}$. It is only necessary to solve the system of linear equations to find ${\mathit{F}}_{\mathit{j}}^{\mathbb{K}}$. so them can also be converted to ${\sum }_{\mathit{g}=\mathit{0}}^{\mathit{M}}{\mathit{a}}_{\mathit{g}}{\mathit{x}}^{\mathit{M}-\mathit{g}}$.

Special:

${\mathit{b}}_{\mathit{0}}={\mathit{b}}_{\mathit{1}}=...={\mathit{b}}_{\mathit{q}}=\mathit{0}\to {\mathit{K}}_{\mathit{1}}=\mathit{0},{\mathit{K}}_{\mathit{2}}=-\mathit{1}...{\mathit{K}}_{\mathit{q}}=-(\mathit{q}-\mathit{1})$.

${\mathit{c}}_{\mathit{0}}={\mathit{c}}_{\mathit{1}}=...={\mathit{c}}_{\mathit{q}}=\mathit{0}\to {\mathit{K}}_{\mathit{1}}=\mathit{1},{\mathit{K}}_{\mathit{2}}=\mathit{2}...{\mathit{K}}_{\mathit{q}}=\mathit{q}$.

${\mathit{d}}_{\mathit{0}}={\mathit{d}}_{\mathit{1}}=...={\mathit{d}}_{\mathit{q}}=\mathit{0}\to {\mathit{K}}_{\mathit{1}}=\mathit{0},{\mathit{K}}_{\mathit{2}}=-\mathit{1}...{\mathit{K}}_{\mathit{q}}=-(\mathit{q}-\mathit{1})$.

$\left(\mathit{1}\right),\left(\mathit{2}\right),\left(\mathit{3}\right)$ can also be converted to $\mathit{c}\times \nabla \mathit{SUM}(\mathit{N}+\mathit{Y}-\mathit{X},\mathit{PS},[\mathit{X}+\mathit{1},\mathit{X}+\mathit{2}...\mathit{X}+\mathit{M}\left]\right)$.

$\mathit{c}={\displaystyle \frac{{\mathit{b}}_{\mathit{M}}}{(\mathit{X}+\mathit{1})...(\mathit{X}+\mathit{M})}}={\displaystyle \frac{{\mathit{c}}_{\mathit{M}}}{(\mathit{X}+\mathit{1})...(\mathit{X}+\mathit{M})}}={\displaystyle \frac{{\mathit{d}}_{\mathit{M}}}{\prod (\mathit{X}+\mathit{1}-{\mathit{K}}_{\mathit{i}})}}$.

Special: ${\mathit{c}}_{\mathit{0}}={\mathit{c}}_{\mathit{1}}=...={\mathit{c}}_{\mathit{q}}=\mathit{0}\to {\mathit{K}}_{\mathit{1}}=\mathit{X}+\mathit{1},{\mathit{K}}_{\mathit{2}}=\mathit{X}+\mathit{2}...{\mathit{K}}_{\mathit{q}}=\mathit{X}+\mathit{q}$.