Subdivide the Shape of Numbers and a Theorem of Ring
Abstract: This article is based on the concept of Shape of numbers, introduce more shapes, obtain the calculation formulas and find an explanation of the formulas. By observing and associating, show a law about the symmetry of Ring. 1. Introduction

Peng, J. introduced the definition of Shape of numbers in :

$\left({K}_{1},{K}_{2},\cdots ,{K}_{M}\right),{K}_{i}\in N,{K}_{1}<{K}_{2}<\cdots <{K}_{M}$

there are M-1 intervals between adjacent numbers. Use A for continuity and B for discontinuity, record as a string of M-1 characters (e.g. AABB…) to represents a catalog.

Define collection of a catalog as Shape of numbers. Use the symbol PX to represent a catelog (if M = 1 then PX = 1).

The single $\left({K}_{1},{K}_{2},\cdots ,{K}_{M}\right)$ is an Item, ${K}_{1}{K}_{2}\cdots {K}_{M}$ is the product of an item.

For example:

$\left(1,2,4\right),\left(1,2,8\right),\left(2,3,6\right)\in PX=AB$, $\left(1,3,5\right),\left(1,3,6\right),\left(2,4,6\right)\in PX=BB$,

$\left(1,2,5,6\right),\left(2,3,6,7\right),\left(2,3,7,8\right),\left(1002,1003,6789,6790\right)\in PX=ABA$

$PM\left(PX\right)$ = Count of numbers of PX, $PA\left(PX\right)$ = Count of A, $PB\left(PX\right)$ = Count of B

$\to$ $PM\left(PX\right)=PA\left(PX\right)+PB\left(PX\right)+1$

$|PX|$ = Count of items belonging to PX

$MIN\left(PX\right)$ = Minimum product of PX: $MIN\left(AA\right)=1×2×3$, $MIN\left(AB\right)=1×2×4$

$IDX\left(PX\right)=2+PA\left(PX\right)+2×PB\left(PX\right)=PM\left(PX\right)+PB\left(PX\right)+1$ : $IDX\left(AA\right)=4$, $IDX\left(AB\right)=5$

$SUM\left(N,PX\right)$ = Sum of all product of items belonging to PX in $\left[1,N-1\right]$

For example: $SUM\left(6,AB\right)=1×2×4+1×2×5+2×3×5$

$END\left(N,PX\right)$ = Set of items belonging to PX with the maximum factor = N-1

For example: $END\left(6,B\right)=\left\{\left(1,5\right),\left(2,5\right),\left(3,5\right)\right\}$

 obtain the conclusion:

$|\text{Itemsin}SUM\left(N,PX\right)|=\left(\begin{array}{c}N-PM\left(PX\right)\\ PB\left(PX\right)+1\end{array}\right)=\left(\begin{array}{c}N-IDX\left(PX\right)+PB\left(PX\right)+1\\ PB\left(PX\right)+1\end{array}\right)$ (1)

$SUM\left(N,PX\right)=MIN\left(PX\right)\left(\begin{array}{c}N\\ IDX\left(PX\right)\end{array}\right)$ (2)

Definition: Subdividethe $\left({K}_{1},{K}_{2},\cdots ,{K}_{M}\right)$ by interval of adjacent numbers. If the discontinuity interval is D > 1, the interval of adjacent numbers ≥ D is classified into a same catelog. Use the symbol PY represent a catelog and represented by [min item]

For example:

$PY=\left[1,3\right]$, $\left(1,3\right),\left(1,4\right),\left(2,4\right),\left(1,5\right),\left(2,5\right),\left(3,5\right)\in PY$. Same as PX = B

$PY=\left[1,4\right]$, $\left(1,4\right),\left(1,5\right),\left(2,5\right),\left(1,6\right),\left(2,6\right),\left(3,6\right)\in PY$, $\left(3,5\right),\left(4,6\right)\notin PY$

$PY=\left[1,4,6\right]$, $\left(1,4,7\right),\left(1,5,7\right),\left(2,5,7\right)\in PY$, $\left(3,5,7\right)\notin PY$

Redefinition: PB(PY) = Count of discontinuity intervals in PY, compatible with PX

Redefinition: IDX(PY) = The maximum factor of MIN(PY) + 1, compatible with PX

Definition: BASE(PY) = PX, If PB(PX) = PB(PY), PM(PX) = PM(PY), PX has discontinuity intervals at the same positions of PY.

For example:

$PY=\left[1,3\right],\left[1,4\right],\left[1,K>2\right]$ ; $BASE\left(PY\right)=\left[1,3\right]=B$

$PY=\left[1,3,4\right],\left[1,4,5\right],\left[1,K>2,X=K+1\right]$ ; $BASE\left(PY\right)=\left[1,3,4\right]=BA$

$PY=\left[1,3,5\right],\left[1,4,9\right],\left[1,K>2,X>K+1\right]$ ; $BASE\left(PY\right)=\left[1,3,5\right]=BB$

Definition: if $f\left(n\right)=\sum {K}_{i}\left(\begin{array}{c}N-{n}_{i}\\ {m}_{i}\end{array}\right)$, then ${D}^{1}f\left(n\right)=\sum {K}_{i}\left(\begin{array}{c}N-{n}_{i}-1\\ {m}_{i}-1\end{array}\right)$

Definition: $PH\left(PY\right)=IDX\left(PY\right)-PB\left(PY\right)-2$ = Maximum factor of $MIN\left(PY\right)-PB\left(PY\right)-1$

$PY=\left[1,{K}_{1}\cdots {K}_{M}\right]$, $BS=BASE\left(PY\right)=\left[1,{G}_{1}\cdots {G}_{M}\right]$ $\to$

$PH\left(PY\right)={K}_{M}-IDX\left(BS\right)+PM\left(BS\right)$ (1.0)

[Proof]

$\begin{array}{c}PH\left(PY\right)={K}_{M}-PB\left(PY\right)-1={K}_{M}-PB\left(BS\right)-1={K}_{M}-{G}_{M}+M\\ ={K}_{M}-\left({G}_{M}+1\right)+\left(M+1\right)=\text{Right}\end{array}$

${\sum }_{n=K}^{N-1}n\left(\begin{array}{c}n-K\\ M\end{array}\right)=\left(M+1\right)\left(\begin{array}{c}N-K\\ M+2\end{array}\right)+\left(M+K\right)\left(\begin{array}{c}N-K\\ M+1\end{array}\right)$ (1.1)

[Proof]

$\begin{array}{l}\text{Left}={\sum }_{n=K}^{N-1}\left(n-K+1\right)\left(\begin{array}{c}n-K\\ M\end{array}\right)+\left(K-1\right){\sum }_{n=K}^{N-1}\left(\begin{array}{c}n-K\\ M\end{array}\right)\\ ={\sum }_{n=K}^{N-1}\left(M+1\right)\left(\begin{array}{c}n-K+1\\ M+1\end{array}\right)+\left(K-1\right){\sum }_{n=K}^{N-1}\left(\begin{array}{c}n-K\\ M\end{array}\right)\\ =\left(M+1\right)\left(\begin{array}{c}N-K+1\\ M+2\end{array}\right)+\left(K-1\right)\left(\begin{array}{c}N-K\\ M+1\end{array}\right)\\ =\left(M+1\right)\left(\begin{array}{c}N-K\\ M+2\end{array}\right)+\left(M+1\right)\left(\begin{array}{c}N-K\\ M+1\end{array}\right)+\left(K-1\right)\left(\begin{array}{c}N-K\\ M+1\end{array}\right)=\text{Right}\end{array}$

By definition:

$SUM\left(N,PY\right)={\sum }_{n=0}^{N}\sum END\left(n,PY\right)$ (1.2)

Derived from (1.2)

$\sum END\left(N,PY\right)={D}^{1}SUM\left(N,PY\right)$ (1.3)

By definition:

$SUM\left(N,\left[1,\cdots ,K,K+1\right]\right)={\sum }_{n=0}^{N-1}n×\sum END\left(n,\left[1,\cdots ,K\right]\right)$ (1.4)

$SUM\left(N,\left[1,\cdots ,K,X>K+1\right]\right)={\sum }_{n=0}^{N-1}n×SUM\left(n-X+K+1,\left[1,\cdots ,K\right]\right)$ (1.5)

According to the method in :

$\begin{array}{c}|\text{Itemsin}SUM\left(N,PY\right)|=\left(\begin{array}{c}N-IDX\left(PY\right)+PB\left(PY\right)+1\\ PB\left(PY\right)+1\end{array}\right)\\ =\left(\begin{array}{c}N-PH\left(PY\right)-1\\ PB\left(PY\right)+1\end{array}\right),\end{array}$ (1.6)

compatible with (1).

2. Calculation Formula

If $PY=BASE\left(PY\right)$, the calculation formula has been given by (2).

Otherwise, it can be deduced from (1.1)-(1.5).

$SUM\left(N,\left[1,K\ge 3\right]\right)=3\left(\begin{array}{c}N-K+2\\ 4\end{array}\right)+K\left(\begin{array}{c}N-K+2\\ 3\end{array}\right)$ (2.1)

[Proof]

$\begin{array}{l}\text{Left}\stackrel{1.2\right)}{\to }{\sum }_{n=0}^{N}\sum END\left(n,\left[1,K\right]\right)={\sum }_{n=K+1}^{N}\left(n-1\right)\left(1+2+\cdots +\left(n-K\right)\right)\\ ={\sum }_{n=K+1}^{N}\left(n-1\right)\left(\begin{array}{c}n-K+1\\ 2\end{array}\right)={\sum }_{n=K}^{N-1}n\left(\begin{array}{c}n-K+2\\ 2\end{array}\right)\stackrel{1.1\right)}{\to }\text{Right}\end{array}$

$SUM\left(N,\left[1,2,K>3\right]\right)=2×4\left(\begin{array}{c}N-K+3\\ 5\end{array}\right)+2×K\left(\begin{array}{c}N-K+3\\ 4\end{array}\right)$ (2.2)

[Proof]

$\begin{array}{l}\stackrel{1.5\right)}{\to }{\sum }_{n=0}^{N-1}n×SUM\left(n-K+3,\left[1,2\right]\right)\\ \stackrel{\left(2\right)}{\to }{\sum }_{n=0}^{N-1}n×2\left(\begin{array}{c}n-K+3\\ 3\end{array}\right)\stackrel{1.1\right)}{\to }\text{Right}\end{array}$

$\begin{array}{l}SUM\left(N,\left[1,K\ge 3,X=K+1\right]\right)\\ =12\left(\begin{array}{c}N-X+2\\ 5\end{array}\right)+3\left(X+1\right)\left(\begin{array}{c}N-X+2\\ 4\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+3K\left(\begin{array}{c}N-X+2\\ 4\end{array}\right)+KX\left(\begin{array}{c}N-X+2\\ 3\end{array}\right)\end{array}$ (2.3)

[Proof]

$\begin{array}{l}\stackrel{1.2\right)\text{\hspace{0.17em}}1.4\right)\text{\hspace{0.17em}}2.1\right)}{\to }{\sum }_{n=0}^{N-1}n×{D}^{1}\left[3\left(\begin{array}{c}n-K+2\\ 4\end{array}\right)+K\left(\begin{array}{c}n-K+2\\ 3\end{array}\right)\right]\\ ={\sum }_{n=0}^{N-1}n×\left[3\left(\begin{array}{c}n-K+1\\ 3\end{array}\right)+K\left(\begin{array}{c}n-K+1\\ 2\end{array}\right)\right]\stackrel{1.1\right)}{\to }\\ =12\left(\begin{array}{c}N-K+1\\ 5\end{array}\right)+3\left(K+2\right)\left(\begin{array}{c}N-K+1\\ 4\end{array}\right)+3K\left(\begin{array}{c}N-K+1\\ 4\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+K\left(K+1\right)\left(\begin{array}{c}N-K+1\\ 3\end{array}\right)\\ =\text{Right}\end{array}$

$\begin{array}{l}SUM\left(N,\left[1,K\ge 3,\text{}X>K+1\right]\right)\\ =15\left(\begin{array}{c}N-X+3\\ 6\end{array}\right)+3\left(X+1\right)\left(\begin{array}{c}N-X+3\\ 5\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+4K\left(\begin{array}{c}N-X+3\\ 5\end{array}\right)+KX\left(\begin{array}{c}N-X+3\\ 4\end{array}\right)\end{array}$ (2.4)

[Proof]

$\stackrel{1.5\right)\text{\hspace{0.17em}}2.1\right)}{\to }{\sum }_{n=0}^{N-1}n×\left[3\left(\begin{array}{c}n-X+3\\ 4\end{array}\right)+K\left(\begin{array}{c}n-X+3\\ 3\end{array}\right)\right]\stackrel{1.1\right)}{\to }\text{Right}$

2.1. $SUM\left(N,\left[1,{K}_{1}\ge 3,\cdots ,{K}_{M}\right]\right)$

$PY=\left[1,{K}_{1}\ge 3,\cdots ,{K}_{M}\right]$, $BASE\left(PY\right)=BS=\left[1,{G}_{1},\cdots ,{G}_{M}\right]$

1*) $SUM\left(N,PY\right)=\sum {A}_{i}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$, ${2}^{M}$ items in total.

2*) ${M}_{i}=IDX\left(BS\right),IDX\left(BS\right)-1,\cdots ,IDX\left(BS\right)-M$

3*) $P=N-PH\left( P Y \right)$

Use the form $\left({G}_{1}+{K}_{1}\right)\left({G}_{2}+{K}_{2}\right)\cdots \left({G}_{M}+{K}_{M}\right)=\sum {X}_{1}{X}_{2}\cdots {X}_{M}$. The expansion function has 2M items in total.

4*) ${M}_{i}=IDX\left(BS\right)-\left(\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}X\in K\right)$

5*) ${A}_{i}={\prod }_{i=1}^{M}\left({X}_{i}+{D}_{i}\right)$,

${D}_{i}=\left\{\begin{array}{l}-m,{X}_{i}\in G,m=\text{countof}\text{\hspace{0.17em}}\left\{{X}_{i}\cdots {X}_{i-1}\right\}\in K\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5.1*\right)\\ +m,{X}_{i}\in K,m=\text{countof}\text{\hspace{0.17em}}\left\{{X}_{i}\cdots {X}_{i-1}\right\}\in G\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5.2*\right)\end{array}$

[Proof]

$PY=\left[1,{K}_{1}\ge 3,\cdots ,{K}_{M}\right]$, $BS=BASE\left( P Y \right)$

Suppose $SUM\left(N,PY\right)=\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$, $P=N-PH\left(PY\right)$.

According to inductive hypothesis:

$\begin{array}{l}\left({G}_{1}+{K}_{1}\right)\left({G}_{2}+{K}_{2}\right)\cdots \left({G}_{M}+{K}_{M}\right)\left\{\left({G}_{M}+1\right)+\left({K}_{M}+1\right)\right\}\\ =\sum {X}_{1}{X}_{2}\cdots {X}_{M}\left\{\left({G}_{M}+1\right)+\left({K}_{M}+1\right)\right\}\end{array}$

$C=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}X\in K$, $M-C=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}X\in G$, ${M}_{i}=IDX\left(BS\right)-C$

$PY1=\left[PY,{K}_{M}+1\right]$, $BS1=BASE\left(PY1\right)$ (2.1.1)

$\begin{array}{l}SUM\left(N,PY1\right)\stackrel{1,2\right)\text{\hspace{0.17em}}1.3\right)}{\to }{\sum }_{n=0}^{N-1}n×{D}^{1}SUM\left(n,PY\right)\\ ={\sum }_{n=0}^{N-1}n×\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}P-1\\ {M}_{i}-1\end{array}\right)\stackrel{1.1\right)}{\to }\\ ={\sum }^{\text{​}}\left({X}_{1}\cdots {X}_{M}{M}_{i}\left(\begin{array}{c}P-1\\ {M}_{i}+1\end{array}\right)+{X}_{1}\cdots {X}_{M}\left(PH\left(PY\right)+{M}_{i}\right)\left(\begin{array}{c}P-1\\ {M}_{i}\end{array}\right)\right)\\ =\sum {X}_{1}\cdots {X}_{M}\left(IDX\left(BS\right)-C\right)\left(\begin{array}{c}P-1\\ {M}_{i}+1\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left(PH\left(PY\right)+IDX\left(BS\right)-C\right)\left(\begin{array}{c}P-1\\ {M}_{i}\end{array}\right)\end{array}$

$\begin{array}{l}=\sum {X}_{1}\cdots {X}_{M}\left(\left({G}_{M}+1\right)-C\right)\left(\begin{array}{c}P-1\\ {M}_{i}+1\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left(\left({K}_{M}+1\right)+\left(PM\left(BS\right)-1\right)-C\right)\left(\begin{array}{c}P-1\\ {M}_{i}\end{array}\right)\\ =\sum {X}_{1}\cdots {X}_{M}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS\right)-C+1\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left({K}_{M+1}+M-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS\right)-C\end{array}\right)\end{array}$

$\begin{array}{l}=\sum {X}_{1}\cdots {X}_{M}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-C\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left({K}_{M+1}+M-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)\end{array}$

1*) is obvious.

Mi change form $IDX\left(BS\right),\cdots ,IDX\left(BS\right)-M$ to $IDX\left(BS\right)+1,\cdots ,IDX\left(BS\right)-M$ $=IDX\left(BS1\right),\cdots ,IDX\left(BS1\right)-\left(M+1\right)$

2*) proved

$\begin{array}{c}P-1=N-PH\left(PY\right)-1=N-\left({K}_{M}-IDX\left(BS\right)+PM\left(BS\right)\right)-1\\ =N-\left\{\left({K}_{M}+1\right)-\left(IDX\left(BS\right)+1\right)+\left(PM\left(BS\right)+1\right)\right\}\\ =N-\left({K}_{M+1}-IDX\left(BS1\right)+PM\left(BS1\right)\right)=N-PH\left(PY1\right)\end{array}$

3*) proved

${X}_{1}\cdots {X}_{M}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-C\end{array}\right)$ $\to$ 4*) 5.1*)

${X}_{1}\cdots {X}_{M}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)$ $\to$ 4*) 5.2*)

4*) 5*) proved

$PY1=\left[PY,{K}_{M+1}>{K}_{M}+1\right]$, $BS1=BASE\left(PY1\right)$ (2.1.2)

$\begin{array}{l}SUM\left(N,PY1\right)\stackrel{1.5\right)}{\to }{\sum }_{n=0}^{N-1}n×SUM\left(n-{K}_{M+1}+{K}_{M}+1,PY\right)\\ ={\sum }_{n=0}^{N-1}n×\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}n-{K}_{M+1}+{K}_{M}+1-PH\left(PY\right)\\ {M}_{i}\end{array}\right)\\ ={\sum }_{n=0}^{N-1}n×\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}n-{K}_{M+1}+{K}_{M}+1-\left({K}_{M}-IDX\left(BS\right)+PM\left(BS\right)\right)\\ {M}_{i}\end{array}\right)\\ ={\sum }_{n=0}^{N-1}n×\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}n-\left({K}_{M+1}-IDX\left(BS\right)+PM\left(BS\right)-1\right)\\ {M}_{i}\end{array}\right)\end{array}$

$\begin{array}{l}=\sum {X}_{1}\cdots {X}_{M}\left({M}_{i}+1\right)\left(\begin{array}{c}P1\\ {M}_{i}+2\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left({K}_{M+1}-IDX\left(BS\right)+M+{M}_{i}\right)\left(\begin{array}{c}P1\\ {M}_{i}+1\end{array}\right)\\ =\sum {X}_{1}\cdots {X}_{M}\left(IDX\left(BS\right)-C+1\right)\left(\begin{array}{c}P1\\ {M}_{i}+2\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left({K}_{M+1}+M-C\right)\left(\begin{array}{c}P1\\ {M}_{i}+1\end{array}\right)\end{array}$

$\begin{array}{l}=\sum {X}_{1}\cdots {X}_{M}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS\right)-C+2\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left({K}_{M+1}+M-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS\right)-C+1\end{array}\right)\\ =\sum {X}_{1}\cdots {X}_{M}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-C\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left({K}_{M+1}+M-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)\end{array}$

1*) is obvious.

Mi change form $IDX\left(BS\right),\cdots ,IDX\left(BS\right)-M$ to $IDX\left(BS\right)+2,\cdots ,IDX\left(BS\right)-M+1$ $=IDX\left(BS1\right),\cdots ,IDX\left(BS1\right)-\left(M+1\right)$

2*) proved

$P1=N-\left\{{K}_{M+1}-\left(IDX\left(BS\right)+2\right)+\left(PM\left(BS\right)+1\right)\right\}=N-PH\left(PY1\right)$

3*) proved

${X}_{1}\cdots {X}_{M}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-C\end{array}\right)$ $\to$ 4*) 5.1*)

${X}_{1}\cdots {X}_{M}\left({K}_{M+1}+M-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)$ $\to$ 4*) 5.2*)

4*) 5*) proved

q.e.d.

Example 2.1:

$\begin{array}{l}\left(3+{K}_{1}\right)\left(5+{K}_{2}\right)\left(7+{K}_{3}\right)\\ =3×5×7+3×5×{K}_{3}+3×{K}_{2}×7+3×{K}_{2}×{K}_{3}+{K}_{1}×5×7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+{K}_{1}×5×{K}_{3}+{K}_{1}×{K}_{2}×7+{K}_{1}×\text{}{K}_{2}×{K}_{3}\end{array}$

$P=N-{K}_{3}+IDX\left(\left[1,3,5,7\right]\right)-PM\left(\left[1,3,5,7\right]\right)=N-{K}_{3}+8-4=N-{K}_{3}+4$

$\begin{array}{l}SUM\left(N,\left[1,{K}_{1}\ge 3,{K}_{2}\ge {K}_{1}+2,{K}_{3}\ge {K}_{2}+2\right]\right)\\ =3×5×7\left(\begin{array}{c}P\\ 8\end{array}\right)+3×5×\left({K}_{3}+2\right)\left(\begin{array}{c}P\\ 7\end{array}\right)+3×\left({K}_{2}+1\right)×\left(7-1\right)\left(\begin{array}{c}P\\ 7\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+3×\left({K}_{2}+1\right)×\left({K}_{3}+1\right)\left(\begin{array}{c}P\\ 7\end{array}\right)+{K}_{1}×\left(5-1\right)×\left(7-1\right)\left(\begin{array}{c}P\\ 7\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{K}_{1}×\left(5-1\right)×\left({K}_{3}+1\right)\left(\begin{array}{c}P\\ 6\end{array}\right)+{K}_{1}×\text{}{K}_{2}×\left(7-2\right)\left(\begin{array}{c}P\\ 6\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+{K}_{1}×{K}_{2}×{K}_{3}\left( P 5 \right)\end{array}$

2.2. $SUM\left(N,\left[1,2,\cdots ,n,{K}_{1}\ge n+2,\cdots ,{K}_{M}\right]\right)$

$PY=\left[1,2,3,\cdots ,n,{K}_{1}\ge n+2,\cdots ,{K}_{M}\right]$,

$BASE\left(PY\right)=BS=\left[1,2,3,\cdots ,n,{G}_{1},\cdots ,{G}_{M}\right]$

1*) $SUM\left(N,PY\right)=\sum {A}_{i}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$, ${2}^{M}$ items in total.

2*) ${M}_{i}=IDX\left(BS\right),IDX\left(BS\right)-1,\cdots ,IDX\left(BS\right)-M$

3*) $P=N\text{\hspace{0.17em}}-\text{\hspace{0.17em}}{K}_{M}\text{ }+\text{\hspace{0.17em}}IDX\left(BS\right)-\left(M\text{\hspace{0.17em}}+1\right)=N\text{\hspace{0.17em}}-\left({K}_{M}\text{ }-\text{\hspace{0.17em}}PB\left(PY\right)-1\right)=N\text{\hspace{0.17em}}-\text{\hspace{0.17em}}PH\left( P Y \right)$

Use the form $\left({G}_{1}+{K}_{1}\right)\left({G}_{2}+{K}_{2}\right)\cdots \left({G}_{M}+{K}_{M}\right)=\sum {X}_{1}{X}_{2}\cdots {X}_{M}$.

4*) ${M}_{i}=IDX\left(BS\right)-\left(\text{Countof}\text{\hspace{0.17em}}X\in K\right)$

5*) ${A}_{i}=n\text{!}×$ (Same as 2.1)

Example 2.2:

$\left(5+{K}_{1}\right)\left(7+{K}_{2}\right)=5×7+5×{K}_{2}+{K}_{1}×7+{K}_{1}×{K}_{2}$, $P=N-{K}_{2}+IDX\left(\left[1,2,3,5,7\right]\right)-3=N-{K}_{2}+5$

$\begin{array}{l}SUM\left(N,\left[1,2,3,{K}_{1}\ge 5,{K}_{2}\ge {K}_{1}+2\right]\right)\\ =3!\left\{5×7\left(\begin{array}{c}P\\ 8\end{array}\right)+5×\left({K}_{2}+1\right)\left(\begin{array}{c}P\\ 7\end{array}\right)+{K}_{1}×\left(7-1\right)\left(\begin{array}{c}P\\ 7\end{array}\right)+{K}_{1}{K}_{2}\left(\begin{array}{c}P\\ 6\end{array}\right)\right\}\end{array}$

2.3. The Meaning of the Expansion of SUM (N, PY)

$PY=\left[1,{K}_{1},\cdots ,{K}_{M}\right]$, $\text{anItem}=\left\{\text{begin},{K}_{1}+{E}_{1},\cdots ,{K}_{M}+{E}_{M}\right\}$

$\left(\text{productofanitem}\right)=\text{begin}×\left({K}_{1}+{E}_{1}\right)\cdots \left({K}_{M}+{E}_{M}\right)=\text{begin}×\sum {F}_{1}\cdots {F}_{M}$

${F}_{i}=E$ (means ${F}_{i}={E}_{i}$ ) or ${F}_{i}={K}_{i}$,

$SUM\left(N,PY\right)=\sum \text{product}=\sum \sum \text{begin}×{F}_{1}\cdots {F}_{M}$ (2.3.1*)

Define 2.3. $SUM\text{_}K\left(N,PY,PF={F}_{1}{F}_{2}\cdots {F}_{M}\right)=\sum \text{Items}$ in (2.3.1*) with the same PF

Example 2.3:

$\begin{array}{l}SUM\left(N,\left[1,{K}_{1}\ge 3,{K}_{2}\ge {K}_{1}+2\right]\right)\\ =15\left(\begin{array}{c}N-{K}_{2}+3\\ 6\end{array}\right)+3\left({K}_{2}+1\right)\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+4{K}_{1}\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)+{K}_{1}{K}_{2}\left(\begin{array}{c}N-{K}_{2}+3\\ 4\end{array}\right)\\ =15\left(\begin{array}{c}N-{K}_{2}+3\\ 6\end{array}\right)+3\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)+3{K}_{2}\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+4{K}_{1}\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)+{K}_{1}{K}_{2}\left(\begin{array}{c}N-{K}_{2}+3\\ 4\end{array}\right)\\ ={\sum }_{n=0}^{N-{K}_{2}}\sum n×\left({K}_{1}+{E}_{1,i}\right)\left({K}_{2}+{E}_{2,i}\right)\end{array}$

then can prove

$\begin{array}{l}SUM\text{_}K\left(N,PY,EE\right)\\ ={\sum }_{\text{allitems}}\text{begin}\ast {E}_{1,i}{E}_{2,i}=15\left(\begin{array}{c}N-{K}_{2}+3\\ 6\end{array}\right)+3\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)\end{array}$

$SUM\text{_}K\left(N,PY,{K}_{1}E\right)={\sum }_{\text{allitems}}\text{begin}\ast {K}_{1}{E}_{2,i}=4{K}_{1}\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)$

$SUM\text{_}K\left(N,PY,E{K}_{2}\right)={\sum }_{\text{allitems}}\text{begin}\ast {E}_{1,i}{K}_{2}=3{K}_{2}\left(\begin{array}{c}N-{K}_{2}+3\\ 5\end{array}\right)$

$SUM\text{_}K\left(N,PY,{K}_{1}{K}_{2}\right)={\sum }_{\text{allitems}}\text{begin}\ast {K}_{1}{K}_{2}={K}_{1}{K}_{2}\left(\begin{array}{c}N-{K}_{2}+3\\ 4\end{array}\right)$

$SUM\left(N,PY\right)=\sum {\prod }_{i=1}^{M}\left({X}_{i}+{D}_{i}\right)\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$,

${X}_{i}+{D}_{i}=\left\{\begin{array}{l}\left\{{G}_{i}-{D}_{i}\right\},{X}_{i}\in G,{D}_{i}=\text{countof}\text{\hspace{0.17em}}\left\{{X}_{1}\cdots {X}_{i-1}\right\}\in K\\ \left\{{K}_{i}\right\}+\left\{{D}_{i}\right\},{X}_{i}\in K,{D}_{i}=\text{countof}\text{\hspace{0.17em}}\left\{{X}_{1}\cdots {X}_{i-1}\right\}\in G\end{array}$ (2.3.2*)

Theorem 2.3 $SUM\text{_}K\left(N,PY,PF\right)=\sum \text{Items}$ in (2.3.2*) expand by {}, factors has the same $\left\{{K}_{i}\right\}\in F$

$=\sum {\prod }_{i=1}^{M}{Y}_{i}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$,

${Y}_{i}=\left\{\begin{array}{l}0,{F}_{i}\in K,{X}_{i}\in G\\ {K}_{i},{F}_{i}\in K,{X}_{i}\in K\\ {G}_{i}-{D}_{i},{F}_{i}=E,{X}_{i}\in G,{D}_{i}=\text{countof}\text{\hspace{0.17em}}\left\{\cdots {X}_{i-1}\right\}\in K\\ {D}_{i},{F}_{i}=E,{X}_{i}\in K,{D}_{i}=\text{countof}\text{\hspace{0.17em}}\left\{\cdots {X}_{i-1}\right\}\in G\end{array}$ (2.3.3*)

$P=N-PH\left(PY\right)$, ${M}_{i}=\text{}IDX\left(BS\right)-\left(\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}X\in K\right)$

[Proof]

Suppose $SUM\text{_}K\left(N,PY,PZ\right)=\sum {\prod }_{i=1}^{M}{Y}_{i}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$, $P=N-PH\left(PY\right)$.

$PY1=\left[PY,{K}_{M+1}\right]$, $C=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}X\in K$,

$M-C=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}X\in G$, ${M}_{i}=IDX\left(BS\right)-C$

When ${K}_{M+1}={K}_{M}+1$

$\begin{array}{l}SUM\left(N,PY1\right)\stackrel{1,2\right)\text{\hspace{0.17em}}1.3\right)}{\to }{\sum }_{n=0}^{N-1}n×{D}^{1}SUM\left(n,PY\right)\\ \to {\sum }_{n=0}^{N-1}n×{D}^{1}SUM\text{_}K\left(n,PY,PF\right)\\ ={\sum }_{n=0}^{N-1}n×\sum {\prod }_{i=1}^{M}{Y}_{i}\left(\begin{array}{c}P-1\\ {M}_{i}-1\end{array}\right)\stackrel{\text{reference}\text{\hspace{0.17em}}\left(2.1.1\right)}{\to }\end{array}$

$\begin{array}{l}=\sum {\prod }_{i=1}^{M}{Y}_{i}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-C\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {\prod }_{i=1}^{M}{Y}_{i}\left(M-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {\prod }_{i=1}^{M}{Y}_{i}{K}_{M+1}\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)\\ =SUM\text{_}K\left(N,PY1,\left[PF,E\right]\right)+SUM\text{_}K\left(N,PY1,\left[PF,{K}_{M+1}\right]\right)\end{array}$

$P-1=N-PH\left(PY1\right)$ $\to$ (2.3.3*)

$SUM\text{_}K\left(N,PY1,\left[PF,{K}_{M+1}\right]\right)=\sum {\prod }_{i=1}^{M}{Y}_{i}{K}_{M+1}\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)$ $\to$ (2.3.3*)

$\begin{array}{l}SUM\text{_}K\left(N,PY1,\left[PF,E\right]\right)\\ =\sum {\prod }_{i=1}^{M}{Y}_{i}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-C\end{array}\right)\end{array}$

$\text{\hspace{0.17em}}\text{ }+\sum {\prod }_{i=1}^{M}{Y}_{i}\left(M-C\right)\left(\begin{array}{c}P-1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)$ $\to$ (2.3.3*)

When ${K}_{M+1}>{K}_{M}+1$

$\begin{array}{l}SUM\left(N,PY1\right)\stackrel{1,5\right)}{\to }{\sum }_{n=0}^{N-1}n×SUM\left(n-{K}_{M+1}+{K}_{M}+1,PY\right)\\ \to {\sum }_{n=0}^{N-1}n×SUM\text{_}K\left(n-{K}_{M+1}+{K}_{M}+1,PY,PF\right)\\ ={\sum }_{n=0}^{N-1}n×\sum {\prod }_{i=1}^{M}{Y}_{i}\left(\begin{array}{c}P1\\ {M}_{i}\end{array}\right)\stackrel{\text{reference}\text{\hspace{0.17em}}\left(2.1.2\right)}{\to }\end{array}$

$\begin{array}{l}=\sum {\prod }_{i=1}^{M}{Y}_{i}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-C\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {\prod }_{i=1}^{M}{Y}_{i}\left(M-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {\prod }_{i=1}^{M}{Y}_{i}{K}_{M+1}\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)\\ =SUM\text{_}K\left(N,PY1,\left[PF,E\right]\right)+SUM\text{_}K\left(N,PY1,\left[PF,{K}_{M+1}\right]\right)\end{array}$

$P1=N-PH\left(PY1\right)$ $\to$ (2.3.3*)

$SUM\text{_}K\left(N,PY1,\left[PF,{K}_{M+1}\right]\right)=\sum {\prod }_{i=1}^{M}{Y}_{i}{K}_{M+1}\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)$ $\to$ (2.3.3*)

$\begin{array}{l}SUM\text{_}K\left(N,PY1,\left[PF,E\right]\right)\\ =\sum {\prod }_{i=1}^{M}{Y}_{i}\left({G}_{M+1}-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-C\end{array}\right)\end{array}$

$\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\sum {\prod }_{i=1}^{M}{Y}_{i}\left(M-C\right)\left(\begin{array}{c}P1\\ IDX\left(BS1\right)-\left(C+1\right)\end{array}\right)$ $\to$ (2.3.3*)

q.e.d.

2.3.1) $SUM\text{_}K\left(N,PY=\left[1,2,\cdots ,n,{K}_{1}\ge n+2,\cdots ,{K}_{M}\right],PF\right)$ has the similar conclusions.

2.3.2) $SUM\text{_}K\left(N,PY=\left[1,2,\cdots ,n,{K}_{1}\ge n+2,\cdots ,{K}_{M}\right],{K}_{1}{K}_{2}\cdots {K}_{M}\right)$,

$\begin{array}{c}BS=BASE\left(PY\right)=MIN\left(PY\right)\left(\begin{array}{c}N-PH\left(PY\right)\\ IDX\left(BS\right)-M\end{array}\right)\\ =MIN\left(PY\right){\sum }_{n=0}^{N-1}\left(\begin{array}{c}n-PH\left(PY\right)-1\\ IDX\left(BS\right)-M-1\end{array}\right)\\ =MIN\left(PY\right){\sum }_{n=0}^{N-1}\left(\begin{array}{c}n-PH\left(PY\right)-1\\ PB\left(BS\right)+1\end{array}\right)\\ =MIN\left(PY\right){\sum }_{n=0}^{N-1}\left(\begin{array}{c}n-PH\left(PY\right)-1\\ PB\left(PY\right)+1\end{array}\right)\\ =MIN\left(PY\right){\sum }_{n=0}^{N-1}|\text{Itemsin}SUM\left(n+1,PY\right)|\end{array}$

3. A Theorem of Ring

$PY=\left[1,{K}_{1}\ge 3,\cdots ,{K}_{M}\right]$, $BASE\left(PY\right)=BS=\left[1,{G}_{1},\cdots ,{G}_{M}\right]$

$SUM\left(N,PY\right)=\sum {A}_{i}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$,

When PY = BS, $SUM\left(N,PY\right)=MIN\left(BS\right)\left(\begin{array}{c}N\\ IDX\left(BS\right)\end{array}\right)$

This inspired us:

In the form $\left({G}_{1}+{K}_{1}\right)\left({G}_{2}+{K}_{2}\right)\cdots \left({G}_{M}+{K}_{M}\right)$, ${K}_{i}={G}_{i}$ $\to$ $MIN\left(BS\right)|\sum {A}_{i}$ (Mi is same).

Definition 3.1:

$S,\left\{{K}_{i}\right\}\in \text{Ring}$ ; ${K}_{i}={K}_{j}$, ${K}_{i}<{K}_{j}$, ${K}_{i}>{K}_{j}$ are allowed.

Choice N from $\left({K}_{1},{K}_{2},\cdots ,{K}_{M}\right)$, $R={R}_{1}\cdots {R}_{M}$

${R}_{i}=+1$ Indicates that Ki was selected, ${R}_{i}=-1$ Indicates that Ki was unselected,

$F\left(N,M,R,K,S\right)={\prod }_{i=1}^{M}\left({K}_{i}+{D}_{i}\right)$,

${D}_{i}=\left\{\begin{array}{l}-mS,{R}_{i}=-1,m=\text{count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\left\{{K}_{1},\cdots ,{K}_{i-1}\right\}\text{\hspace{0.17em}}\text{selected}\\ +mS,{R}_{i}=+1,m=\text{count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\left\{{K}_{1},\cdots ,{K}_{i-1}\right\}\text{\hspace{0.17em}}\text{unselected}\end{array}$ (*)

When S = 1, abbreviated as $F\left(N,M,R,K\right)$

Definition 3.2:

$H\left({N}_{1},{N}_{2},K,S\right)=\sum F\left({N}_{1},{N}_{1}+{N}_{2},R,K,S\right)$, Sum traverses all N1-Choice of K

Theorem 3.1: $H\left({N}_{1},{N}_{2},K,S\right)=\left(\begin{array}{c}M\\ {N}_{1},{N}_{2}\end{array}\right){K}_{1}\cdots {K}_{M}$, $M={N}_{1}+{N}_{2}$

[Proof]

When ${N}_{1}=0$ or ${N}_{2}=0$, it is obvious.

$\begin{array}{c}H\left(1,1,K,S\right)=\text{select}\text{\hspace{0.17em}}{K}_{1}+\text{select}\text{\hspace{0.17em}}{K}_{2}\\ =\text{unselect}\text{\hspace{0.17em}}{K}_{2}+\text{select}\text{\hspace{0.17em}}{K}_{2}\\ ={K}_{1}\left({K}_{2}-S\right)+{K}_{1}\left({K}_{2}+S\right)\\ =2{K}_{1}{K}_{2}\end{array}$

Suppose $H\left(1,M-1,K,S\right)=\left(\begin{array}{c}M\\ 1\end{array}\right){K}_{1}\cdots {K}_{M}$

$\begin{array}{l}H\left(1,M,K,S\right)\\ =\text{select}\text{\hspace{0.17em}}{K}_{M+1}+\text{unselect}\text{\hspace{0.17em}}{K}_{M+1}\\ =H\left(0,M,K,S\right)\left({K}_{M+1}+MS\right)+H\left(1,M-1,K,S\right)\left({K}_{M+1}-S\right)\\ =\left(\begin{array}{c}M\\ 0\end{array}\right){K}_{1}\cdots {K}_{M}\left({K}_{M+1}+MS\right)+\left(\begin{array}{c}M\\ 1\end{array}\right){K}_{1}\cdots {K}_{M}\left({K}_{M+1}-S\right)\\ =\left\{\left(\begin{array}{c}M\\ 0\end{array}\right)+\left(\begin{array}{c}M\\ 1\end{array}\right)\right\}{K}_{1}\cdots {K}_{M}{K}_{M+1}+\left\{M\left(\begin{array}{c}M\\ 0\end{array}\right)-\left(\begin{array}{c}M\\ 1\end{array}\right)\right\}S×{K}_{1}\cdots {K}_{M}{K}_{M+1}\\ =\left(\begin{array}{c}M+1\\ 1\end{array}\right){K}_{1}\cdots {K}_{M}{K}_{M+1}\end{array}$

$\to$ $H\left(1,{N}_{2},K,S\right)$ holds $\to$ Symmetry $\to$ $H\left({N}_{2},1,K,S\right)$ holds

Suppose $H\left({N}_{1},M-{N}_{1},K,S\right)=\left(\begin{array}{c}M\\ {N}_{1}\end{array}\right){K}_{1}\cdots {K}_{M}$

$\begin{array}{l}H\left({N}_{1}+1,M-{N}_{1},K,S\right)=\text{select}\text{\hspace{0.17em}}{K}_{M+1}+\text{unselect}\text{\hspace{0.17em}}{K}_{M+1}\\ =H\left({N}_{1},M-{N}_{1},K,S\right)\left({K}_{M+1}+\left(M-{N}_{1}\right)S\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+H\left({N}_{1}+1,M-{N}_{1}-1,K,S\right)\left({K}_{M+1}-\left({N}_{1}+1\right)S\right)\\ =\left(\begin{array}{c}M\\ {N}_{1}\end{array}\right){K}_{1}\cdots {K}_{M}\left({K}_{M+1}+\left(M-{N}_{1}\right)S\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\begin{array}{c}M\\ {N}_{1}+1\end{array}\right){K}_{1}\cdots {K}_{M}\left({K}_{M+1}-\left({N}_{1}+1\right)S\right)\end{array}$

$\begin{array}{l}=\left\{\left(\begin{array}{c}M\\ {N}_{1}\end{array}\right)+\left(\begin{array}{c}M\\ {N}_{1}+1\end{array}\right)\right\}{K}_{1}\cdots {K}_{M+1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\left\{\left(M-{N}_{1}\right)\left(\begin{array}{c}M\\ {N}_{1}\end{array}\right)-\left({N}_{1}+1\right)\left(\begin{array}{c}M\\ {N}_{1}+1\end{array}\right)\right\}S×{K}_{1}\cdots {K}_{M+1}\\ =\left(\begin{array}{c}M+1\\ {N}_{1}+1\end{array}\right){K}_{1}\cdots {K}_{M}{K}_{M+1}\end{array}$

$\to$ $H\left({N}_{1},{N}_{2},K,S\right)$ holds

q.e.d.

Example 3.1: $\left\{{K}_{1},{K}_{2},{K}_{3}\right\}=\left\{A,B,C\right\}$

$\begin{array}{c}H\left(1,2,K\right)=\text{select}\text{\hspace{0.17em}}A+\text{select}\text{\hspace{0.17em}}B+\text{select}\text{\hspace{0.17em}}C\\ =A\left(B-1\right)\left(C-1\right)+A\left(B+1\right)\left(C-1\right)+AB\left(C+2\right)\\ =3ABC\end{array}$

$\begin{array}{c}H\left(2,1,K\right)=\text{select}\text{\hspace{0.17em}}AB+\text{select}\text{\hspace{0.17em}}BC+\text{select}\text{\hspace{0.17em}}AC\\ =AB\left(C-2\right)+A\left(B+1\right)\left(C+1\right)+A\left(B-1\right)\left(C+1\right)\\ =3ABC\end{array}$

Definition 3.3:

Ki come from q sources: ${S}_{1},{S}_{2},\cdots ,{S}_{q}$, ${K}_{i}\in {S}_{j}$ indicates Ki come from Sj $diff\left({K}_{i},{K}_{j}\right)=diff\left({K}_{i}\in {S}_{a},{K}_{j}\in {S}_{b}\right)=diff\left(a,b\right)$, $diff\left(a,b\right)=S$, ( $a\ne b$ ); $diff\left(a,b\right)=-diff\left(b,a\right)$ ; $diff\left(a,a\right)=0$ then can change related parts of definition 3.1 to

${\prod }_{i=1}^{M}\left({K}_{i}+{D}_{i}\right),{D}_{i}=\underset{j

and define $H\left({N}_{1},{N}_{2},\cdots ,{N}_{q},K,S\right)$

Theorem 3.2 $H\left({N}_{1},{N}_{2},\cdots ,{N}_{q},K,S\right)=\left(\begin{array}{c}M\\ {N}_{1},{N}_{2},\cdots ,{N}_{q}\end{array}\right){K}_{1}\cdots {K}_{M}$, $M={N}_{1}+{N}_{2}+\cdots +{N}_{q}$

[Proof]

Only need to prove S=1, and specify $diff\left(a,b\right)=1$, ( $a ) $H\left({N}_{1},{N}_{2},K\right)=\sum \text{Item}$, ${K}_{i}\in \left\{{S}_{1},{S}_{2}\right\}$. An item has M factors,

Choice N1 factors, Ki in these factors, ${K}_{i}\in {S}_{1}$, It is called invariant factor{F}.

Others are called variable factors {V}

${L}_{2}+{L}_{3}={N}_{2}$, choice L2 factors in {V} to S2, L3 to S3

By definition, $H\left({L}_{2},{L}_{3},V\right)=\sum \left\{\left({L}_{2},{L}_{3}\right)-\text{Choice}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}V\right\}$

$\begin{array}{l}\sum \left(\prod \left\{F\right\}×\sum \left\{\left({L}_{2},{L}_{3}\right)-\text{Choice}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}V\right\}\right)\\ =\sum \left(\prod \left\{F\right\}×H\left({L}_{2},{L}_{3},V\right)\right)=\sum \left(\prod \left\{F\right\}×\left(\begin{array}{c}{N}_{2}\\ {L}_{1},{L}_{2}\end{array}\right)×\prod \left\{V\right\}\right)\\ =\left(\begin{array}{c}{N}_{2}\\ {L}_{2},{L}_{3}\end{array}\right)\sum \text{Item}=\left(\begin{array}{c}{N}_{2}\\ {L}_{2},{L}_{3}\end{array}\right)\left(\begin{array}{c}M\\ {N}_{1},{N}_{2}\end{array}\right){K}_{1}\cdots {K}_{M}\\ =\left(\begin{array}{c}M\\ {N}_{1},{L}_{2},{L}_{3}\end{array}\right){K}_{1}\cdots {K}_{M}=H\left({N}_{1},{L}_{2},{L}_{3},K\right)\end{array}$

and items gives all (N1, L2, L3)-Choice of K.

Prove $\sum \left(\prod \left\{F\right\}×\sum \left\{\left({L}_{2},{L}_{3}\right)-\text{Choice}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}V\right\}\right)$ satisfies the definition of $H\left({N}_{1},\cdots ,K\right)$

Let $\text{Item}={\prod }_{i=1}^{M}\left({K}_{i}+{D}_{i}\right)$, $\left(\text{anitemchanged}\right)={\prod }_{i=1}^{M}\left({H}_{i}+{E}_{i}\right)$, ${K}_{i}={H}_{i}$

$A1=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\left\{{K}_{1},\cdots ,{K}_{i-1}\right\}\in {S}_{1}$, $A2=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\left\{{K}_{1},\cdots ,{K}_{i-1}\right\}\in {S}_{2}$

$B1=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\left\{\cdots ,{H}_{i-1}\right\}\in {S}_{1}$, $B2=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\left\{\cdots ,{H}_{i-1}\right\}\in {S}_{2}$,

$B3=\text{Count}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\left\{\cdots ,{H}_{i-1}\right\}\in {S}_{3}$

$\to$ $A1=B1$, $A2=B2+B3$

${K}_{i+1}\in {S}_{1}\to {H}_{i+1}\in {S}_{1}\to {D}_{i+1}=A2=i-A1=i-B1=B2+B3={E}_{i+1}$

$\to$ The invariant factors match the definition

${K}_{i+1}\in {S}_{2},{H}_{i+1}\in {S}_{2}$ ${E}_{i}={D}_{i}+B3=-A1+B3=-B1+B3$

$\stackrel{\text{definitionof}H\left(N,{L}_{2},{L}_{3},V\right)}{\to }$ Match the definition

${K}_{i+1}\in {S}_{2},{H}_{i+1}\in {S}_{3}$ ${E}_{i}={D}_{i}–B2=-A1-B2=-B1-B2$

$\stackrel{\text{definitionof}H\left(N,{L}_{2},{L}_{3},V\right)}{\to }$ Match the definition

$\to$ Match the definition of $H\left({N}_{1},{N}_{2},{N}_{3},K\right)$ $\stackrel{\text{recursion}}{\to }$ $\cdots$

q.e.d.

Example 3.2 $\left\{{K}_{1},{K}_{2},{K}_{3}\right\}=\left\{A,B,C\right\}$

$H\left(1,2,K\right)=A\left(B-1\right)\left(C-1\right)+A\left(B+1\right)\left(C-1\right)+AB\left(C+2\right)=3ABC$

$A\left(B-1\right)\left(C-1\right)\in {S}_{1}{S}_{2}{S}_{2}\to \left\{F\right\}=\left\{A\right\}\to A\left\{\left(B-1\right)\left(C-2\right)+\left(B-1\right)C\right\}$

$\to A\left(B-1\right)\left(C-2\right)+A\left(B-1\right)C\in {S}_{1}{S}_{2}{S}_{3}+{S}_{1}{S}_{3}{S}_{2}$

$A\left(B+1\right)\left(C-1\right)\in {S}_{2}{S}_{1}{S}_{2}\to \left\{F\right\}=\left\{B+1\right\}\to \left(B+1\right)\left\{A\left(C-2\right)+AC\right\}$

$\to A\left(B+1\right)\left(C-2\right)+A\left(B+1\right)C\in {S}_{2}{S}_{1}{S}_{3}+{S}_{3}{S}_{1}{S}_{2}$

$AB\left(C+2\right)\in {S}_{2}{S}_{2}{S}_{1}\to \left\{F\right\}=\left\{C+2\right\}\to \left(C+2\right)\left\{A\left(B-1\right)+A\left(B+1\right)\right\}$

$\to A\left(B-1\right)\left(C+2\right)+A\left(B+1\right)\left(C+2\right)\in {S}_{2}{S}_{3}{S}_{1}+{S}_{3}{S}_{2}{S}_{1}$

$H\left(2,1,K\right)=AB\left(C-2\right)+A\left(B+1\right)\left(C+1\right)+A\left(B-1\right)\left(C+1\right)=3ABC$

$AB\left(C-2\right)\in {S}_{1}{S}_{1}{S}_{2}\to \left\{F\right\}=\left\{C-2\right\}\to \left(C-2\right)\left\{A\left(B-1\right)+A\left(B+1\right)\right\}$

$\to A\left(B-1\right)\left(C-2\right)+A\left(B+1\right)\left(C-2\right)\in {S}_{1}{S}_{2}{S}_{3}+{S}_{2}{S}_{1}{S}_{3}$

$A\left(B+1\right)\left(C+1\right)\in {S}_{2}{S}_{1}{S}_{1}\to \left\{F\right\}=\left\{A\right\}\to A\left\{\left(B+1\right)C+\left(B+1\right)\left(C+2\right)\right\}$

$\to A\left(B+1\right)C+A\left(B+1\right)\left(C+2\right)\in {S}_{3}{S}_{1}{S}_{2}+{S}_{3}{S}_{2}{S}_{1}$

$A\left(B-1\right)\left(C+1\right)\in {S}_{1}{S}_{2}{S}_{1}\to \left\{F\right\}=\left\{B-1\right\}\to \left(B-1\right)\left\{AC+A\left(C+2\right)\right\}$

$\to A\left(B-1\right)C+A\left(B-1\right)\left(C+2\right)\in {S}_{1}{S}_{3}{S}_{2}+{S}_{2}{S}_{3}{S}_{1}$

$\begin{array}{c}H\left(1,1,1,K\right)=A\left(B-1\right)\left(C-2\right)+A\left(B-1\right)C+A\left(B+1\right)\left(C-2\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+A\left(B+1\right)C+A\left(B-1\right)\left(C+2\right)+A\left(B+1\right)\left(C+2\right)\\ =6ABC\end{array}$

3.1) $\left({K}_{1},{K}_{2},\cdots ,{K}_{M}\right)$ is permutable in $H\left({N}_{1},{N}_{2},\cdots ,{N}_{q},K,S\right)$

3.2) $H\left({N}_{1},{N}_{2},\cdots ,{N}_{q},K,S\right)={\prod }_{i=1}^{M}\left({K}_{i}+{D}_{i}\right)$

$\to$ $H\left({N}_{q},{N}_{q-1},\cdots ,{N}_{1},K,S\right)={\prod }_{i=1}^{M}\left({K}_{i}-{D}_{i}\right)$

3.3) In the section 2, $SUM\left(N,PY\right)=\sum {A}_{i}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$

${A}_{i}$ is generated by the form $\left({G}_{1}+{K}_{1}\right)\left({G}_{2}+{K}_{2}\right)\cdots \left({G}_{M}+{K}_{M}\right)$. It can be understood asgenerated by 2-SET: $\left\{{G}_{i}\right\}$, $\left\{{K}_{i}\right\}$.

If ${K}_{i}={J}_{1,i}+{J}_{2,i}+\cdots +{J}_{x,i}$, ( ${J}_{x,i}>0$, ${J}_{x,i}=0$, ${J}_{x,i}<0$ are allowed),

$SUM\left(N,PY\right)=\sum {B}_{i}\left(\begin{array}{c}P\\ {M}_{i}\end{array}\right)$

${B}_{i}$ can be understood asgenerated by (X + 1)-SET: $\left\{{G}_{i}\right\},\left\{{J}_{1}\right\},\left\{{J}_{2}\right\},\cdots ,\left\{{J}_{X}\right\}$

3.4) ${K}_{i}={J}_{1,i}+{J}_{2,i}+\cdots +{J}_{x,i}$, Theorem 2.3 has the similar promotion.

4. Conclusion

Review the whole process, $\left(\begin{array}{c}N\\ M\end{array}\right)+\left(\begin{array}{c}N\\ M+1\end{array}\right)=\left(\begin{array}{c}N+1\\ M+1\end{array}\right)$ $\to$ Basic Shapes in  $\to$ More Shapes in this article.

Acknowledgements

The author is very grateful to Mr. HanSan Z. (Department of Electronic Information, Nanjing University) for his suggestions on this paper.

Cite this paper: Peng, J. (2020) Subdivide the Shape of Numbers and a Theorem of Ring. Open Access Library Journal, 7, 1-14. doi: 10.4236/oalib.1106719.
References

   Peng, J. (2020) Shape of Numbers and Calculation Formula of Stirling Numbers. Open Access Library Journal, 7: e6081. https://doi.org/10.4236/oalib.1106081

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