Redefining the Shape of Numbers and Three Forms of Calculation
Abstract: This paper redefines the Shape of numbers, makes it more natural and concise, and the domain of definition is extended to ring. The inconvenient PCHG() and PH() are removed. The concept of subsets is also removed. The new definition can be used to calculate ∑n-0N-1Πi-1M (Ki+n×Di) ∑ni,j-0j-N-1Πi-1M (Ki+ni,j×Di), ni,j≤ni+1,j or ni,j=ni+1,j; Ki,Di∈ring. Three forms corresponding to three calculation methods are obtained. They can be used as a powerful tool for analysis. Some of the conclusions are: 1) Expressions and properties of two kinds of Stirling number, Lah number and Eulerian number; 2) Expression of power sum of natural numbers; 3) Vandermonde identity, Norlund identity; 4) New congruence and new proof of Wilson theorem; 5) ∑n-1P-1≡0 MOD P2, P>3; 6) ∑C-0C-M-1(-1)M-1-C∑PM(PS)-M,PB(PS)-CMIN(PS)=1. 1. Introduction

Peng, J. has introduced Shape of numbers in    : $\left({I}_{1},{I}_{2},\cdots ,{I}_{M}\right)$, ${I}_{i}\in Z$. There are $M-\text{1}$ intervals between adjacent numbers. ${I}_{i+1}-{I}_{i}\le 1$ means continuity; ${I}_{i+1}-{I}_{i}>1$ means discontinuity.

Shape of numbers: collect $\left({I}_{1},\cdots ,{I}_{M}\right)$ with the same continuity and discontinuity at the same position into a catalog, call it a Shape.

A Shape has a min Item: $\left({K}_{1},{K}_{2},\cdots \right)$. Use the symbol PS = [min Item] to represent it.

If ${K}_{i+1}-{K}_{i}=D>1$, only ${I}_{i+1}-{I}_{i}\ge D$ is allowed.

If ${K}_{i+1}-{K}_{i}=D\le 1$, only ${I}_{i+1}-{I}_{i}=D$ is allowed.

The single $\left({I}_{1},\cdots ,{I}_{M}\right)$ is an item. ${I}_{1}×\cdots ×{I}_{M}$ is the product. Ii is a factor.

Example 1.1:

$PS=\left[2,3\right]\to \left(2,3\right),\left(3,4\right),\left(1000,1001\right)\in PS$

$\begin{array}{l}PS=\left[-3,-1\right]\to \left(-3,-1\right),\left(-3,0\right),\left(-2,0\right),\left(-3,1\right),\\ \left(-2,1\right),\left(-1,1\right),\left(1000,2007\right)\in PS\end{array}$

$\begin{array}{l}PS=\left[1,4,4\right]\to \left(1,4,4\right),\left(1,5,5\right),\left(2,5,5\right),\left(1,6,6\right),\\ \left(2,6,6\right),\left(3,6,6\right)\in PS,\left(3,5,5\right)\notin PS\end{array}$

$\begin{array}{l}PS=\left[1,4,6,4\right]\to \left(1,4,7,5\right),\left(1,5,7,5\right),\left(2,5,7,5\right)\in PS,\\ \left(1,4,7,6\right),\left(3,5,7,5\right)\notin PS\end{array}$

PM(PS) = Count of factors.

PB(PS) = Count of discontinuities.

MIN(PS) = Min product: $MIN\left(\left[1,2,3\right]\right)=1×2×3$, $MIN\left(\left[1,2,4\right]\right)=1×2×4$

Basic Shape: K1 = 1 and intervals = 1 or 2

BASE(PS) = BS: if (1) PB(BS) = PB(PS), (2) PM(BS) = PM(PS), (3) BS is a Basic Shape, (4) BS has discontinuity intervals at the same positions of PS.

PH(PS) = (Max Factor)-1-PB(BS)

IDX(PS) = IDX of BS = {Max factor of BS} + 1 = PM(BS) + PB(BS) + 1

$PS=\left[{K}_{1},\cdots ,{K}_{M}\right]$, $BS=\left[{G}_{1},\cdots ,{G}_{M}\right]$ then ${K}_{i+1}-K\le 1\to {G}_{i+1}-{G}_{i}=1$ ; ${K}_{i+1}-K>1\to {G}_{i+1}-{G}_{i}=2$

Example 1.2:

$PS=\left[1,2\right],\left[1001,1002\right]\to BASE\left(PS\right)=\left[1,2\right]$

$PS=\left[1,3\right],\left[-9,4\right],\left[1,K>2\right]\to BASE\left(PS\right)=\left[1,3\right]$

$PS=\left[1,3,4\right],\left[-8,4,5\right],\left[1,K>2,X=K+1\right]\to BASE\left(PS\right)=\left[1,3,4\right]$

$PS=\left[1,3,5\right],\left[0,4,9\right],\left[1,K>2,X>K+1\right]\to BASE\left(PS\right)=\left[1,3,5\right]$

$PS=\left[1001,1002\right]\to PH\left(PS\right)=1002-1-0=1001$, $IDX\left(PS\right)=IDX\left(\left[1,2\right]\right)=3$

$PS=\left[0,4,9\right]\to PH\left(PS\right)=9-1-2=6$, $IDX\left(\left[0,4,9\right]\right)=IDX\left(\left[1,3,5\right]\right)=6$

SET(N, PS) = SET of items $\in$ PS in [K1, N-1], item’s max factor ≤ N-1

 introduced the subset: fix some interval of discontinuities.

SET(N,PS,PT) = Subset of SET(N, PS), $BASE\left(PS\right)=\left[{G}_{1},\cdots ,{G}_{M}\right]$, $PT=\left[{T}_{1},\cdots ,{T}_{M}\right]$

$=\left\{\begin{array}{l}{T}_{i+1}-{T}_{i}=1:{G}_{i+1}-{G}_{i}=1,\text{\hspace{0.17em}}\text{means}\text{\hspace{0.17em}}{I}_{i+1}-{I}_{i}={K}_{i+1}-{K}_{i}\\ {T}_{i+1}-{T}_{i}=1:{G}_{i+1}-{G}_{i}=2,\text{\hspace{0.17em}}\text{means}\text{\hspace{0.17em}}{I}_{i+1}-{I}_{i}={K}_{i+1}-{K}_{i}\\ {T}_{i+1}-{T}_{i}=2:{G}_{i+1}-{G}_{i}=2,\text{\hspace{0.17em}}\text{means}\text{\hspace{0.17em}}{I}_{i+1}-{I}_{i}>={K}_{i+1}-{K}_{i}\end{array}$ (*)

PT only has the change at (*). When a change happens, make the interval fixed.

PCHG(PS, PT) = Count of change from BASE(PS) to PT

Example 1.3:

$PCHG\left(\left[1,3,5\right],\left[1,3,5\right]\right)=0$

$PCHG\left(\left[1,3,5\right],\left[1,2,4\right]\right)=PCHG\left(\left[1,4,7\right],\left[1,2,4\right]\right)=1$, changed at T1

$PCHG\left(\left[1,3,5\right],\left[1,3,4\right]\right)=PCHG\left(\left[1,4,7\right],\left[1,3,4\right]\right)=1$, changed at T2

$PCHG\left(\left[1,3,5\right],\left[1,2,3\right]\right)=PCHG\left(\left[1,8,10\right],\left[1,2,3\right]\right)=2$, changed at T1, T2

|SET(N, PS, PT)| = Count of items in SET(N, PS, PT)

SUM(N, PS, PT) = Sum of all products in SET(N, PS, PT)

Example 1.4:

$SUM\left(6,\left[1,2,4\right]\right)=1×2×4+1×2×5+2×3×5$

$\begin{array}{c}SUM\left(9,\left[1,4,7\right]\right)=SUM\left(9,\left[1,4,7\right],\left[1,3,5\right]\right)\\ =1×4×7+1×4×8+1×5×8+2×5×8\end{array}$

$SUM\left(9,\left[1,4,7\right],\left[1,2,4\right]\right)=1×4×7+1×4×8+2×5×8$

    came to the following conclusion:

(1.1) $|SET\left(N,PS,PT\right)|=\left(\begin{array}{c}N-PH\left(PS\right)-PCHG\left(PS,PT\right)-1\\ PB\left(PT\right)+1\end{array}\right)$

The following uses count of $X\in K$ for count of $\left\{{X}_{1},{X}_{2},\cdots ,{X}_{M}\right\}\in \left\{{K}_{1},{K}_{2},\cdots ,{K}_{M}\right\}$

(1.2) Use the form $\left({T}_{1}+{K}_{1}\right)\left({T}_{2}+{K}_{2}\right)\cdots \left({T}_{M}+{K}_{M}\right)=\sum {X}_{1}{X}_{2}\cdots {X}_{M}$, ${X}_{i}={T}_{i}$ or ${K}_{i}$.

Don’t swap the factors of ${X}_{1}{X}_{2}\cdots {X}_{M}$, then each ${X}_{1}{X}_{2}\cdots {X}_{M}$ corresponds to a expression = ${A}_{q}\left(\begin{array}{c}N\text{\hspace{0.17em}}-\text{\hspace{0.17em}}PH\left(PS\right)\text{\hspace{0.17em}}-\text{\hspace{0.17em}}PCHG\left(PS,PT\right)-1\\ IDX\left(PT\right)-q\end{array}\right)$, q = count of $X\in K$.

$SUM\left(N,PS,PT\right)=\sum {A}_{q}\left(\begin{array}{c}N-PH\left(PS\right)-PCHG\left(PS,PT\right)-1\\ IDX\left(PT\right)-q\end{array}\right)$

${A}_{q}={\prod }_{i=1}^{M}\left({X}_{i}+{D}_{i}\right)$, ${D}_{i}=\left\{\begin{array}{l}-m:{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in K\\ +m:{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in T\end{array}$

Example 1.5:

$PS=\left[{K}_{1},{K}_{2}\ge {K}_{1}+2,{K}_{3}\ge {K}_{2}+2\right]$, $BS=BASE\left(PS\right)=\left[1,3,5\right]$,

$\to IDX\left(BS\right)=6$,

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

$\begin{array}{c}P=N-PH\left(PS\right)-PCHG\left(PS,BS\right)-1\\ =N-\left\{{K}_{3}-1-2\right\}-0-1=N-{K}_{3}+2\end{array}$

$\begin{array}{l}\to SUM\left(N,PS\right)\\ =1×3×5\left(\begin{array}{c}P\\ 6\end{array}\right)+1×3×\left({K}_{3}+2\right)\left(\begin{array}{c}P\\ 5\end{array}\right)+1×\left({K}_{2}+1\right)×\left(5-1\right)\left(\begin{array}{c}P\\ 5\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+1×\left({K}_{2}+1\right)×\left({K}_{3}+1\right)\left(\begin{array}{c}P\\ 4\end{array}\right)+{K}_{1}×\left(3-1\right)×\left(5-1\right)\left(\begin{array}{c}P\\ 5\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{K}_{1}×\left(3-1\right)×\left({K}_{3}+1\right)\left(\begin{array}{c}P\\ 4\end{array}\right)+{K}_{1}×\text{}{K}_{2}×\left(5-2\right)\left(\begin{array}{c}P\\ 4\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{K}_{1}×{K}_{2}×{K}_{3}\left( P 3 \right)\end{array}$

An item = $\left\{{K}_{1}+{E}_{1},\cdots ,{K}_{M}+{E}_{M}\right\}$, K is fixed, E is variable.

A product = $\left({K}_{1}+{E}_{1}\right)\cdots \left({K}_{M}+{E}_{M}\right)=\sum {F}_{1}{F}_{2}\cdots {F}_{M}$, ${F}_{i}={E}_{i}$ or ${K}_{i}$

That is, a product can be broken down into 2M parts.

Define $SUM\text{_}K\left(N,PS,PT,PF={F}_{1}\cdots {F}_{M}\right)$ = Sum of one part, PF indicates the part.

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

${X}_{i}+{D}_{i}=\left\{\begin{array}{l}\left\{{T}_{i}-{D}_{i}\right\}:{X}_{i}={T}_{i},{D}_{i}=\text{count of}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in K\\ \left\{{K}_{i}\right\}+\left\{{D}_{i}\right\}:{X}_{i}={K}_{i},{D}_{i}=\text{count of}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in T\end{array}$

Expand SUM() by {braces}:

(1.3) $SUM\text{_}K\left(N,PS,PT,PF\right)$

$=\sum \text{Expansion}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}SUM\left(\text{ }\right)\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\left\{{K}_{i}\right\}\in PF=\sum {\prod }_{i=1}^{M}{Y}_{i}\left(\begin{array}{c}A\\ {M}_{q}\end{array}\right)$

${Y}_{i}=\left\{\begin{array}{l}0:{F}_{i}={K}_{i},{X}_{i}={T}_{i}\\ {K}_{i}:{F}_{i}={K}_{i},{X}_{i}={K}_{i}\\ {T}_{i}-{D}_{i}:{F}_{i}={E}_{i},{X}_{i}={T}_{i},{D}_{i}=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\\ {D}_{i}:{F}_{i}={E}_{i},{X}_{i}={K}_{i},{D}_{i}=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\end{array}$

Example 1.6:

Use the example above

$SUM\text{_}K\left(\cdots {E}_{1}{E}_{2}{E}_{3}\right)=1×3×5\left(\begin{array}{c}P\\ 6\end{array}\right)+\left[1×3×2+1×1×\left(5-1\right)\right]\left(\begin{array}{c}P\\ 5\end{array}\right)+{1}^{3}\left( P 4 \right)$

$SUM\text{_}K\left(\cdots {E}_{1}{E}_{2}{K}_{3}\right)=1×3×{K}_{3}\left(\begin{array}{c}P\\ 5\end{array}\right)+1×1×{K}_{3}\left( P 4 \right)$

$SUM\text{_}K\left(\cdots {E}_{1}{K}_{2}{E}_{3}\right)=1×{K}_{2}×\left(5-1\right)\left(\begin{array}{c}P\\ 5\end{array}\right)+1×{K}_{2}×1\left( P 4 \right)$

$SUM\text{_}K\left(\cdots {K}_{1}{E}_{2}{E}_{3}\right)={K}_{1}×\left(3-1\right)×\left(5-1\right)\left(\begin{array}{c}P\\ 5\end{array}\right)+{K}_{1}×\left(3-1\right)×1\left( P 4 \right)$

$SUM\text{_}K\left(\cdots {E}_{1}{K}_{2}{K}_{3}\right)=1×{K}_{2}×{K}_{3}\left( P 4 \right)$

$SUM\text{_}K\left(\cdots {K}_{1}{E}_{2}{K}_{3}\right)={K}_{1}×\left(3-1\right)×{K}_{3}\left( P 4 \right)$

$SUM\text{_}K\left(\cdots {K}_{1}{K}_{2}{E}_{3}\right)={K}_{1}×{K}_{2}×\left(5-2\right)\left( P 4 \right)$

$SUM\text{_}K\left(\cdots {K}_{1}{K}_{2}{K}_{3}\right)={K}_{1}×{K}_{2}×{K}_{3}\left( P 3 \right)$

SUM_K() can explain why SUM() has that strange form:

We can calculate every part of SUM() by some way without the form. There may be complex relationships between the parts, but their sum match a simple form.

If understand this: In 1.2), when Ti and Di all increase L times

(1.4) ${K}_{1}×\cdots ×{K}_{M}+\left(L+{K}_{1}\right)×\cdots ×\left(L+{K}_{M}\right)+\cdots$

$+\left(\left(N-1\right)L+{K}_{1}\right)×\cdots ×\left(\left(N-1\right)L+{K}_{M}\right)$

$PT=\left[1×L,2×L,\cdots ,M×L\right]$, can use the form $\left({T}_{1}+{K}_{1}\right)\cdots \left({T}_{M}+{K}_{M}\right)=\sum {A}_{q}\left(\begin{array}{c}N\\ M+1-q\end{array}\right)$, q = count of $X\in K$, 2M items in total.

${A}_{q}={\prod }_{i=1}^{M}\left({X}_{i}+{D}_{i}\right)$, ${D}_{i}=\left\{\begin{array}{l}-mL:{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\\ +mL:{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\end{array}$

This paper starts from (1.4), tries to calculate

(1*) ${K}_{1}×\cdots ×{K}_{M}+\left({D}_{1}+{K}_{1}\right)×\cdots ×\left({D}_{M}+{K}_{M}\right)+\cdots$

$+\left(\left(N-1\right){D}_{1}+{K}_{1}\right)×\cdots ×\left(\left(N-1\right){D}_{M}+{K}_{M}\right)$

In the process, the concept of Shape is greatly expanded.

The form of 1.2) is obtained by guess. If the correct form is found, the rest is mainly inductive proof. With the forms, we can analyze the expression and coefficient and get a lot of results.

2. Redefinition

Change domain from Z to Ring, ${K}_{i},{D}_{i}\in \text{Ring}$. ${K}_{i}$ no longer compares big or small.

M-series:

$\left\{{K}_{1},{D}_{1}+{K}_{1},2{D}_{1}+{K}_{1},3{D}_{1}+{K}_{1},\cdots ,\left(N-1\right){D}_{1}+{K}_{1}\right\}=\left\{{K}_{1}+n×{D}_{1}\right\}$

$\left\{{K}_{2},{D}_{2}+{K}_{2},2{D}_{2}+{K}_{2},3{D}_{2}+{K}_{2},\cdots ,\left(N-1\right){D}_{2}+{K}_{2}\right\}=\left\{{K}_{2}+n×{D}_{2}\right\}$

$⋮$

$\begin{array}{l}\left\{{K}_{M},{D}_{M}+{K}_{M},2{D}_{M}+{K}_{M},3{D}_{M}+{K}_{M},\cdots ,\left(N-1\right){D}_{M}+{K}_{M}\right\}\\ =\left\{{K}_{M}+n×{D}_{M}\right\}\end{array}$

An item = $\left({I}_{1},{I}_{2},\cdots ,{I}_{M}\right)$, I1 come from serie1, I2 come from serie2…

Use $PS=\left[{K}_{1}:{D}_{1},{K}_{2}:{D}_{2},\cdots ,{K}_{M}:{D}_{M}\right]$ to represent the Shape.

$\left[{K}_{1}:1,{K}_{2}:1,\cdots ,{K}_{M}:1\right]$ is abbreviated as $\left[{K}_{1},{K}_{2},\cdots ,{K}_{M}\right]$

If ${K}_{i}\in Z$, the new definition is similar to the old definition and allows ${D}_{i}\le 0$.

SET(N, PS, PT) = Set of some Items come from M-series.

${I}_{i}={K}_{i}+a×{D}_{i}$, ${I}_{i+1}={K}_{i+1}+b×{D}_{i+1}$, $PT=\left[{T}_{1}=1,{T}_{2},\cdots ,{T}_{M}\right]=\left\{\begin{array}{l}{T}_{i+1}-{T}_{i}=1:\text{means}\text{\hspace{0.17em}}a=b\\ {T}_{i+1}-{T}_{i}=2,\text{\hspace{0.17em}}\text{means}\text{\hspace{0.17em}}a\le b\end{array}$

There is no longer the idea of subsets.

We have tried to extend the domain of PT, but when M > 2, no rules have been found yet.

Basic Shape: K1 = 1 and intervals = 1 or 2, ${K}_{i}\in N$, Di = 1

PT is always a Basic Shape.

MIN(PS) = Min product of a Basic Shape = $\prod {K}_{i}$

|SET(N, PS, PT)| = Count of items $\in$ SET(N, PS, PT)

END(N, PS, PT) = Set of Items $\in$ SUM(N, PS, PT) and ${I}_{M}\text{ }=\left(N\text{\hspace{0.17em}}-1\right){D}_{M}\text{ }+\text{\hspace{0.17em}}{K}_{M}$.

SUM(N, PS, PT) = Sum of products in SET(N, PS, PT)

$SUM\left(N,PS,\left[1,2,\cdots ,M\right]\right)$ is abbreviated as SUM (N, PS)

$SUM\left(N,\cdots \right)=\text{Old Sum}\left(Max\left({K}_{i}\right)+N,\cdots \right)$, PH() and PCHG() are no longer needed.

The old does not allow $SUM\left(\left[2,3\right],\left[1,3\right]\right)$, $SUM\left(\left[3,2\right],\left[1,3\right]\right)$. The new it is allowed.

$SUM\left(N,PS\right)=\left(1*\right)={\sum }_{n=0}^{N-1}{\prod }_{i=1}^{M}\left({K}_{i}+n×{D}_{i}\right)$

$SUM\left(N,PS,\left[1,3,5,\cdots ,2M-1\right]\right)=\sum {\prod }_{i=1}^{M}\left({K}_{i}+{J}_{i,j}×{D}_{i}\right)$, ${J}_{1.i}\le {J}_{2,i}\le \cdots \le {J}_{M,i}$, ${J}_{i,j}\in \left[0,N-1\right]$

PM(PS) = M; PB(PT) = Count of discontinuities in PT

$IDX\left(PT\right)={T}_{M}+1=PB\left(PT\right)+PM\left(PT\right)+1$

2.1) $|SET\left(N,PS,PT\right)|=\left(\begin{array}{c}N+PB\left(PT\right)\\ PB\left(PT\right)+1\end{array}\right)$

3. Form1 of Calculation

Similar to , key points are:

Define $\nabla f\left(n\right)=f\left(n\right)-f\left(n-1\right)$ : if $f\left(n\right)=\sum {A}_{i}\left(\begin{array}{c}{N}_{i}\\ {m}_{i}\end{array}\right)$, then $\nabla f\left(n\right)=\sum {A}_{i}\left(\begin{array}{c}{N}_{i}-1\\ {m}_{i}-1\end{array}\right)$

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

By definition:

2) $\sum END\left(N,PS,PT\right)=\nabla SUM\left(N,PS,PT\right)$

3) $SUM\left(N,\left[PS,{K}_{M+1}:{D}_{M+1}\right],\left[PT,{T}_{M}={T}_{M}+1\right]\right)$

$={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×\sum END\left(n+1,PS,PT\right)$

4) $SUM\left(N,\left[PS,{K}_{M+1}:{D}_{M+1}\right],\left[PT,{T}_{M}={T}_{M}+2\right]\right)$

$={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×SUM\left(n+1,PS,PT\right)$

$PS=\left[{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right]$, $PT=\left[{T}_{1},\cdots ,{T}_{M}\right]$, $PS1=\left[PS,{K}_{M+1}:{D}_{M+1}\right]$

3.1) Use the Form1 = $\left({T}_{1}+{K}_{1}\right)\cdots \left({T}_{M}+{K}_{M}\right)=\sum {X}_{1}\cdots {X}_{M}$,

$SUM\left(N,PS,PT\right)=\sum {A}_{q}\left(\begin{array}{c}N+PB\left(PT\right)\\ IDX\left(PT\right)-q\end{array}\right)$, q = count of $X\in K$.

${A}_{q}={\prod }_{i=1}^{M}{B}_{i}$, ${B}_{i}=\left\{\begin{array}{l}\left({T}_{i}-m\right){D}_{i};{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in K\\ {K}_{i}+m{D}_{i};{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{{X}_{1},\cdots ,{X}_{i-1}\right\}\in T\end{array}$

[Proof]

Suppose $SUM\left(N,PS,PT\right)=\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}N+PB\left(PT\right)\\ IDX\left(PT\right)-q\end{array}\right)$

When $PT1=\left[PT,{T}_{M+1}=1+{T}_{M}\right]$

$\begin{array}{l}SUM\left(N,PS1,PT1\right)\\ ={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×\sum END\left(n+1,PS,PT\right)\\ ={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×\nabla SUM\left(n+1,PS,PT\right)\\ ={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}n+PB\left(PT\right)\\ IDX\left(PT\right)-q-1\end{array}\right)\stackrel{\left( 1 \right)}{\to }\end{array}$

$\begin{array}{l}=\sum {X}_{1}\cdots {X}_{M}{D}_{M+1}\left(IDX\left(PT\right)-q\right)\left(\begin{array}{c}N+PB\left(PT\right)\\ IDX\left(PT\right)-q+1\end{array}\right)\\ \text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left\{{K}_{M+1}+\left(IDX\left(PT\right)-q-1-PB\left(PT\right)\right){D}_{M+1}\right\}\left(\begin{array}{c}N+PB\left(PT\right)\\ IDX\left(PT\right)-q\end{array}\right)\\ \stackrel{PB\left(PT1\right)=PB\left(PT\right),IDX\left(PT1\right)=IDX\left(PT\right)+1,IDX\left(PT\right)=1+{T}_{M}={T}_{M+1},IDX\left(PT\right)-PB\left(PT\right)-1=M}{\to }\\ =\sum {X}_{1}\cdots {X}_{M}{D}_{M+1}\left({T}_{M+1}-q\right)\left(\begin{array}{c}N+PB\left(PT1\right)\\ IDX\left(PT1\right)-q\end{array}\right)\\ \text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left\{{K}_{M+1}+\left(M-q\right){D}_{M+1}\right\}\left(\begin{array}{c}N+PB\left(PT1\right)\\ IDX\left(PT1\right)-\left(q+1\right)\end{array}\right)\end{array}$

The previous expression means ${X}_{M+1}={T}_{M+1}$

M ? q = Count of $\left\{{X}_{1}\cdots {X}_{M}\right\}\in T$. The following expression means ${X}_{M+1}={K}_{M+1}$

à Match the Form $\left({T}_{1}+{K}_{1}\right)\cdots \left({T}_{M}+{K}_{M}\right)\left\{{T}_{M+1}+{K}_{M+1}\right\}$.

When $PT1=\left[PT,{T}_{M+1}=2+{T}_{M}\right]$

$\begin{array}{l}SUM\left(N,PS1,PT1\right)\\ ={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×SUM\left(n+1,PS,PT\right)\\ ={\sum }_{n=0}^{N-1}\left({K}_{M+1}+n×{D}_{M+1}\right)×\sum {X}_{1}\cdots {X}_{M}\left(\begin{array}{c}n+1+PB\left(PT\right)\\ IDX\left(PT\right)-q\end{array}\right)\stackrel{\left(1\right)}{\to }\\ =\sum {X}_{1}\cdots {X}_{M}{D}_{M+1}\left(IDX\left(PT\right)-q+1\right)\left(\begin{array}{c}N+1+PB\left(PT\right)\\ IDX\left(PT\right)-q+2\end{array}\right)\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left\{{K}_{M+1}+\left(IDX\left(PT\right)-q-1-PB\left(PT\right)\right){D}_{M+1}\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left(\begin{array}{c}N+1+PB\left(PT\right)\\ IDX\left(PT\right)-q+1\end{array}\right)\stackrel{PB\left(PT1\right)=PB\left(PT\right)+1,IDX\left(PT1\right)=IDX\left(PT\right)+2}{\to }\end{array}$

$\begin{array}{l}=\sum {X}_{1}\cdots {X}_{M}{D}_{M+1}\left({T}_{M+1}-q\right)\left(\begin{array}{c}N+PB\left(PT1\right)\\ IDX\left(PT1\right)-q\end{array}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\sum {X}_{1}\cdots {X}_{M}\left\{{K}_{M+1}+\left(M-q\right){D}_{M+1}\right\}\left(\begin{array}{c}N+PB\left(PT1\right)\\ IDX\left(PT1\right)-\left(q+1\right)\end{array}\right)\end{array}$

à Match the Form $\left({T}_{1}+{K}_{1}\right)\cdots \left({T}_{M}+{K}_{M}\right)\left\{{T}_{M+1}+{K}_{M+1}\right\}$.

q.e.d.

The proof process can be extended to ring.

Example 3.1:

$PS=\left[7:13,-7.7:2.2,15:-23\right]$, $PT\text{\hspace{0.17em}}=\left[1,3,5\right]$, ${\text{Form}}_{1}\text{ }=\left(1+7\right)\left(3-7.7\right)\left(5+15\right)$

$\begin{array}{l}SUM\left(N,PS,PT\right)\\ =-9867.0\left(\begin{array}{c}N+2\\ 6\end{array}\right)+1084.6\left(\begin{array}{c}N+2\\ 5\end{array}\right)+4044.7\left(\begin{array}{c}N+2\\ 4\end{array}\right)-808.5\left(\begin{array}{c}N+2\\ 3\end{array}\right)\end{array}$

$-808.5=7×\left(-7.7\right)×15$

$\begin{array}{c}4044.7=\left[\left(1-0\right)×13\right]×\left(-7.7+2.2×1\right)×\left(15-23×1\right)\\ \text{\hspace{0.17em}}+7×\left[\left(3-1\right)×2.2\right]×\left(15-23×1\right)\\ \text{\hspace{0.17em}}+7×\left(-7.7\right)×\left[\left(5-2\right)×\left(-23\right)\right]\end{array}$

$\begin{array}{c}1084.6=\left[\left(1-0\right)×13\right]×\left[\left(3-0\right)×2.2\right]×\left(15-23×2\right)\\ \text{\hspace{0.17em}}+\left[\left(1-0\right)×13\right]×\left(-7.7+2.2×1\right)×\left[\left(5-1\right)×\left(-23\right)\right]\\ \text{\hspace{0.17em}}+7×\left[\left(3-1\right)×2.2\right]×\left[\left(5-1\right)×\left(-23\right)\right]\end{array}$

$-9867.0=\left[\left(1-0\right)×13\right]×\left[\left(3-0\right)×2.2\right]×\left[\left(5-0\right)×\left(-23\right)\right]$

$\begin{array}{l}SUM\left(2,PS,PT\right)\\ =7×\left(-7.7\right)×15+7×\left(-7.7-5.5\right)×\left(-8\right)+20×\left(-5.5\right)×\left(-8\right)\\ =4044.7\left(\begin{array}{c}4\\ 4\end{array}\right)-808.5×\left(\begin{array}{c}4\\ 3\end{array}\right)=810.7\end{array}$

$\begin{array}{l}SUM\left(3,PS,PT\right)\\ =SUM\left(2,PS,PT\right)+7×\left(-7.7-5.5-3.3\right)×\left(-31\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+20×\left(-5.5-3.3\right)×\left(-31\right)+33×\left(-3.3\right)×\left(-31\right)\\ =1084.6\left(\begin{array}{c}5\\ 5\end{array}\right)+4044.7\left(\begin{array}{c}5\\ 4\end{array}\right)-808.5\left(\begin{array}{c}5\\ 3\end{array}\right)=13223.1\end{array}$

$\begin{array}{l}SUM\left(4,PS,PT\right)\\ =SUM\left(3,PS,PT\right)+7×\left(-7.7-5.5-3.3-1.1\right)×{\left(-54\right)}_{{}_{}}^{{}^{}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+20×\left(-5.5-3.3-1.1\right)×\left(-54\right)+33×\left(-3.3-1.1\right)×{\left(-54\right)}_{{}_{}}^{{}^{}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+46×\left(-1.1\right)×{\left(-54\right)}_{{}_{}}\\ =-9867.0\left(\begin{array}{c}6\\ 6\end{array}\right)+1084.6\left(\begin{array}{c}6\\ 5\end{array}\right)+4044.7\left(\begin{array}{c}6\\ 4\end{array}\right)-808.5\left(\begin{array}{c}6\\ 3\end{array}\right)=41141.1\end{array}$

Example 3.2:

$PS=\left[\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right):\left(\begin{array}{cc}1& 3\\ 4& 2\end{array}\right),\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right):\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)\right]$, $PT=\left[1,3\right]$

${\text{Form}}_{\text{1}}=\left(1+\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right)\right)\left(3+\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\right)$

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

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

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

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

$\begin{array}{c}SUM\left(2,PS,PT\right)=\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)+\left[\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right)+\left(\begin{array}{cc}8& 6\\ 5& 4\end{array}\right)\right]\left(\begin{array}{cc}4& 4\\ 2& 4\end{array}\right)\\ =\left(\begin{array}{c}2+1\\ 3\end{array}\right)\left(\begin{array}{cc}44& 70\\ 28& 38\end{array}\right)+\left(\begin{array}{c}2+1\\ 2\end{array}\right)\left(\begin{array}{cc}17& 13\\ 4& 5\end{array}\right)\\ =\left(\begin{array}{cc}95& 109\\ 40& 53\end{array}\right)\end{array}$

$\begin{array}{c}SUM\left(3,PS,PT\right)=SUM\left(2,PS,PT\right)+\left[\left(\begin{array}{cc}7& 3\\ 1& 2\end{array}\right)+\left(\begin{array}{cc}8& 6\\ 5& 4\end{array}\right)+\left(\begin{array}{cc}9& 9\\ 9& 6\end{array}\right)\right]\left(\begin{array}{cc}6& 7\\ 3& 6\end{array}\right)\\ =\left(\begin{array}{c}3+1\\ 4\end{array}\right)\left(\begin{array}{cc}15& 27\\ 30& 48\end{array}\right)+\left(\begin{array}{c}3+1\\ 3\end{array}\right)\left(\begin{array}{cc}44& 70\\ 28& 38\end{array}\right)+\left(\begin{array}{c}3+1\\ 2\end{array}\right)\left(\begin{array}{cc}17& 13\\ 4& 5\end{array}\right)\\ =\left(\begin{array}{cc}293& 385\\ 166& 230\end{array}\right)\end{array}$

A product = $\left({K}_{1}+{E}_{1}\right)\cdots \left({K}_{M}+{E}_{M}\right)=\sum {F}_{1}\cdots {F}_{M}$, ${F}_{i}={E}_{i}$ or ${K}_{i}$

Here’s another extension: Let ${F}_{i}={E}_{i}$ or ${K}_{i}$ or ${R}_{i}$, ${R}_{i}$ means $\left({K}_{i}+{E}_{i}\right)$

So a product can be broken down into ${2}^{0},{2}^{1},{2}^{2},\cdots ,{2}^{M}$ parts.

$SUM\text{_}K\left(N,PS,PT,PF={F}_{1}{F}_{2}\cdots {F}_{M}\right)$ = Sum of one part, PF indicates the part.

$SUM\left(N,PS,PT\right)=\sum {\prod }_{i=1}^{M}{A}_{q}\left(\begin{array}{c}A\\ {M}_{q}\end{array}\right)$, ${A}_{q}={\prod }_{i=1}^{M}{B}_{i}$

${B}_{i}=\left\{\begin{array}{l}\left\{\left({T}_{i}-m\right){D}_{i}\right\};{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\\ \left\{{K}_{i}\right\}+\left\{m{D}_{i}\right\};{F}_{i}\ne {R}_{i},{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\\ \left\{{K}_{i}+m{D}_{i}\right\};{F}_{i}={R}_{i},{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\end{array}$

Expand SUM() by {braces}:

3.2) $SUM\text{_}K\left(N,PS,PT,PF\right)=\sum \text{Expansion}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\left\{{K}_{i},{R}_{i}\right\}\in PF$

Use the similar method of  to prove.

4. Coefficient Analysis

Define ${H}_{1}\left(PS,PT,C\right)={A}_{M-C}$ of 3.1), C = Count of $X\in T$, $T\in \text{Ring}$.

It’s the coefficient in the expression. Here $T\in Ring$ is just for analysis.

${F}_{M}^{K}=\sum M-\text{productwithdifferentfactors}\in K$, the sum traverse all combinations.

${E}_{M}^{K}=\sum M-\text{productwithfactors}\in K$, the sum traverse all combinations.

${F}_{M}^{\left\{1,2,\cdots ,N\right\}}$ is abbreviated as ${F}_{M}^{N}$, ${E}_{M}^{\left\{1,2,\cdots ,N\right\}}$ is abbreviated as ${E}_{M}^{N}$ ;

By definition:

${E}_{M}^{N}=\underset{{N}_{1}+\cdots +{N}_{C}=M}{\sum }{1}^{{N}_{1}}{2}^{{N}_{2}}\cdots {N}^{{N}_{C}}$

${F}_{0}^{K}={E}_{0}^{K}=1;{F}_{M>|K|}^{K}=0$

${E}_{M}^{N+1}=\left(N+1\right){E}_{M-1}^{N+1}+{E}_{M}^{N};{F}_{M}^{\left[K,{K}_{M+1}\right]}={K}_{M+1}{F}_{M-1}^{K}+{F}_{N}^{K}$

4.1) ${H}_{1}\left(K,T,0\right)=\prod {K}_{i}$, ${H}_{1}\left(K,T,M\right)=\prod {T}_{i}\prod {D}_{i}$

 has proved:

4.2) ${H}_{1}\left(\left[D×A:D,{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right],\left[A,{T}_{1},\cdots ,{T}_{M}\right],C\right)$

$\begin{array}{l}=D×A×{H}_{1}\left(\left[{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right],\left[{T}_{1},\cdots ,{T}_{M}\right],C\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+D×A×{H}_{1}\left(\left[{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right],\left[{T}_{1},\cdots ,{T}_{M}\right],C-1\right)\end{array}$

this à

4.3) $SUM\left(N,\left[1,2,\cdots ,n,{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right],\left[1,2,\cdots ,n,{T}_{1},\cdots ,{T}_{M}\right]\right)$, $PT=\left[1,\cdots ,n,{T}_{1},\cdots ,{T}_{M}\right]$ can use the form:

$\left({T}_{1}+{K}_{1}\right)\cdots \left({T}_{M}+{K}_{M}\right)=n!\sum {A}_{q}\left(\begin{array}{c}N+PB\left(PT\right)+n\\ IDX\left(PT\right)-q\end{array}\right)$

 has proved:

4.4) ${H}_{1}\left(\left[D×{T}_{1}:D,\cdots ,D×{T}_{M}:D\right],\left[{T}_{1},\cdots ,{T}_{M}\right],C\right)={D}^{M}\left(\begin{array}{c}M\\ C\end{array}\right)\prod {T}_{i}$

D = 1, this à

4.5) $SUM\left(N,PT,PT\right)=MIN\left(PT\right)\left(\begin{array}{c}N+PB\left(PT\right)+PM\left(PT\right)\\ IDX\left(PT\right)\end{array}\right)$

This is a generalization of ${\sum }_{n=0}^{N}\left(\begin{array}{c}n\\ M\end{array}\right)=\left(\begin{array}{c}N+1\\ M+1\end{array}\right)$

Eg: $SUM\left(2,\left[1,2,4\right]\right)=1×2×4+1×2×5+2×3×5=1×2×4\left(\begin{array}{c}2+1+3\\ 5\end{array}\right)$

4.6) ${H}_{1}\left(PS,PT,C\right)={\nabla }^{C}SUM\left(C,PS,PT\right)$, ${\nabla }^{IDX\left(PT\right)}SUM\left(N>M,\cdots \right)=\prod {T}_{i}\prod {D}_{i}$

The last row value of the difference sequence is not arbitrary.

Comparison with 4.4),  has proved:

4.7) if ${T}_{i}+1={T}_{i+1}$, then

$\left\{\begin{array}{l}{K}_{i}+1={K}_{i+1},{H}_{1}\left(\left[D×{K}_{1}:D,\cdots \right],PT,C\right)={D}^{M}\left(\begin{array}{c}M\\ C\end{array}\right){T}_{1}\cdots {T}_{C}{K}_{C+1}\cdots {K}_{M}\\ {K}_{i}-1={K}_{i+1},{H}_{1}\left(\left[D×{K}_{1}:D,\cdots \right],PT,C\right)={D}^{M}\left(\begin{array}{c}M\\ C\end{array}\right){T}_{1}\cdots {T}_{C}{K}_{1}\cdots {K}_{M-C}\\ {K}_{i}+1={K}_{i+1},{H}_{1}\left(\left[D×{K}_{1}:-D,\cdots \right],PT,C\right)={\left(-1\right)}^{C}{D}^{M}\left(\begin{array}{c}M\\ C\end{array}\right){T}_{1}\cdots {T}_{C}{K}_{1}\cdots {K}_{M-C}\\ {K}_{i}-1={K}_{i+1},{H}_{1}\left(\left[D×{K}_{1}:-D,\cdots \right],PT,C\right)={\left(-1\right)}^{C}{D}^{M}\left(\begin{array}{c}M\\ C\end{array}\right){T}_{1}\cdots {T}_{C}{K}_{C+1}\cdots {K}_{M}\end{array}$

$\begin{array}{c}{\left[x+y\right]}_{n}=\nabla SUM\left(y+1,\left[x-n+1,x-n+2,\cdots ,x\right]\right)\\ =\left(\begin{array}{c}y\\ n\end{array}\right)\left(\begin{array}{c}n\\ n\end{array}\right)n!{\left[x\right]}_{0}+\left(\begin{array}{c}y\\ n-1\end{array}\right)\left(\begin{array}{c}n\\ n-1\end{array}\right)\left(n-1\right)!{\left[x\right]}_{1}+\cdots +\left(\begin{array}{c}y\\ 0\end{array}\right)\left(\begin{array}{c}n\\ 0\end{array}\right)0!{\left[x\right]}_{n}\\ =\left(\begin{array}{c}n\\ n\end{array}\right){\left[y\right]}_{n}{\left[x\right]}_{0}+\left(\begin{array}{c}n\\ n-1\end{array}\right){\left[y\right]}_{n-1}{\left[x\right]}_{1}+\cdots +\left(\begin{array}{c}n\\ 0\end{array}\right){\left[y\right]}_{0}{\left[x\right]}_{n}\end{array}$

à Vandermonde identity : ${\left[x+y\right]}_{n}={\sum }_{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){\left[x\right]}_{k}{\left[y\right]}_{n-k}$

$\begin{array}{c}{\left[x+y\right]}^{n}=\nabla SUM\left(-y+1,\left[x:-1,x+1:-1,\cdots ,x+n-1:-1\right]\right)\\ ={\left(-1\right)}^{n}\left(\begin{array}{c}-y\\ n\end{array}\right)\left(\begin{array}{c}n\\ n\end{array}\right)n!{\left[x\right]}^{0}+{\left(-1\right)}^{n-1}\left(\begin{array}{c}-y\\ n-1\end{array}\right)\left(\begin{array}{c}n\\ n-1\end{array}\right)\left(n-1\right)!{\left[x\right]}^{1}+\cdots \\ =\left(\begin{array}{c}n\\ n\end{array}\right){\left[y\right]}^{n}{\left[x\right]}^{0}+\left(\begin{array}{c}n\\ n-1\end{array}\right){\left[y\right]}^{n-1}{\left[x\right]}^{1}+\cdots +\left(\begin{array}{c}n\\ 0\end{array}\right){\left[y\right]}^{0}{\left[x\right]}^{n}\end{array}$

à Norlund identity : ${\left[x+y\right]}^{n}={\sum }_{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){\left[x\right]}^{k}{\left[y\right]}^{n-k}$

$\begin{array}{l}\left({K}_{1}+N×{D}_{1}\right)\cdots \left({K}_{M}+N×{D}_{M}\right)\\ =\nabla SUM\left(N+1,\left[{K}_{1}:{D}_{1}..{K}_{M}:{D}_{M}\right]\right)={\sum }_{q=0}^{M}{A}_{q}\left( N q \right)\end{array}$

4.8) $\left({K}_{1}+N×{D}_{1}\right)\cdots \left({K}_{M}+N×{D}_{M}\right)$ can be decomposed to ${\sum }_{q=0}^{M}{C}_{q}\left(\begin{array}{c}N\\ q\end{array}\right)$ by 3.1)

$N=x$, ${K}_{i}=i-1$, $PS=\left[0,1,2,\cdots ,M-1\right]$ à

$\begin{array}{l}{\left[x\right]}^{M}=\left(\begin{array}{c}x\\ M\end{array}\right)\left(\begin{array}{c}M\\ M\end{array}\right)M!{\left[M-1\right]}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\begin{array}{c}x\\ M-1\end{array}\right)\left(\begin{array}{c}M\\ M-1\end{array}\right)\left(M-1\right)!{\left[M-1\right]}_{1}\cdots \left(\begin{array}{c}x\\ 0\end{array}\right)\left(\begin{array}{c}M\\ 0\end{array}\right)0!{\left[M-1\right]}_{M}\end{array}$

4.9) ${\left[x\right]}^{M}={\sum }_{k=0}^{M}\left(\begin{array}{c}M\\ k\end{array}\right){\left[x\right]}_{k}{\left[M-1\right]}_{M-k}$

$N=x$, ${K}_{i}=1-i$, $PS=\left[0:-1,-1:-1,\cdots ,-M+1:-1\right]$ à

$\begin{array}{l}{\left[-x\right]}_{M}={\left(-1\right)}^{M+0}\left(\begin{array}{c}X\\ M\end{array}\right)\left(\begin{array}{c}M\\ M\end{array}\right)M!{\left[M-1\right]}_{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(-1\right)}^{\left(M-1\right)+1}\left(\begin{array}{c}X\\ M-1\end{array}\right)\left(\begin{array}{c}M\\ M-1\end{array}\right)\left(M-1\right)!{\left[M-1\right]}_{1}\end{array}$

4.10) ${\left[-x\right]}_{M}={\sum }_{k=0}^{M}{\left(-1\right)}^{M}\left(\begin{array}{c}M\\ k\end{array}\right){\left[x\right]}_{k}{\left[M-1\right]}_{M-K}$

 has proved:

4.11) if ${T}_{i}+1={T}_{i+1}$, then

$\begin{array}{l}{H}_{1}\left(\left[D×{K}_{1}:D,\cdots ,D×{K}_{M}:D\right],\left[{T}_{1},\cdots ,{T}_{M}\right],C\right)\\ ={D}^{C}{T}_{1}\cdots {T}_{C}\left[{F}_{M-C}^{D×K}{E}_{0}^{D×\left\{1,2,\cdots ,C\right\}}+{F}_{M-C-1}^{D×K}{E}_{1}^{D×\left\{1,2,\cdots ,C\right\}}+\cdots +{F}_{0}^{D×K}{E}_{M-C}^{D×\left\{1,2,\cdots ,C\right\}}\right]\end{array}$

This à Ki can exchange order in $SUM\left(N,\left[{K}_{1},\cdots ,{K}_{M}\right]\right)$.

In fact, Ki can exchange order in $SUM\left(N,\left[{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right]\right)$ by definition.

This à

4.12) if $\lambda$ is a primitive unit root, ${\lambda }^{M}=1$, then

$\begin{array}{l}SUM\left(N,\left[{\lambda }^{1},{\lambda }^{2},\cdots ,{\lambda }^{M}\right]\right)\\ =M!\left(\begin{array}{c}N\\ M+1\end{array}\right){E}_{0}^{M}+\left(M-1\right)!\left(\begin{array}{c}N\\ M\end{array}\right){E}_{1}^{M-1}+\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+1!\left(\begin{array}{c}N\\ 2\end{array}\right){E}_{M-1}^{1}+0!\left(\begin{array}{c}N\\ 1\end{array}\right){\left(-1\right)}^{M+1}\end{array}$

$\left({\lambda }^{1}+1\right)\cdots \left({\lambda }^{M}+1\right)=SUM\left(2,\left[\cdots \right]\right)-SUM\left(1,\left[\cdots \right]\right)=1+{\left(-1\right)}^{M+1}$

It’s obvious when M is even; if M is odd then $\left({\lambda }^{1}+1\right)\cdots \left({\lambda }^{M-1}+1\right)=1$

$\left({\lambda }^{1}+2\right)\cdots \left({\lambda }^{M}+2\right)=SUM\left(3,\left[\cdots \right]\right)-SUM\left(2,\left[\cdots \right]\right)={2}^{M}+{\left(-1\right)}^{M+1}$

It can be concluded from the definition:

4.13) 1) ${H}_{1}\left(\left[1,\cdots ,M\right],\left[1,\cdots ,2M-1\right],C\right)$

$={\sum }_{PM\left(\text{ }\right)=M,PB\left(\text{ }\right)=C}MIN\left(PS\right)+{\sum }_{PM\left(\text{ }\right)=M,PB\left(\text{ }\right)=C-1}MIN\left( P S \right)$

2) ${H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],C\right)={\sum }_{PM\left(PS\right)=M,PB\left(PS\right)=C}MIN\left( P S \right)$

PS are Basic Shapes.

5. Special Functions

5.1) $SUM\left(N,\left[a:0\right]\right)=a\left( N 1 \right)$

5.2) $SUM\left(N,\left[a:d\right]\right)={\sum }_{n=0}^{N-1}\left(a+nd\right)=d\left(\begin{array}{c}N\\ 2\end{array}\right)+a\left( N 1 \right)$

5.3) $SUM\left(N,\left[1,2,\cdots ,M\right]\right)=M!\left(\begin{array}{c}N+M\\ M+1\end{array}\right)$

5.4) $SUM\left(N,\left[1,2,\cdots ,M\right],\left[1,3,\cdots ,2M-1\right]\right)=\sum {I}_{1}{I}_{2}\cdots {I}_{M}$, $1\le {I}_{1}<{I}_{2}<\cdots <{I}_{M}\le N+M-1$

5.5) $SUM\left(N,\left[1,1,\cdots ,1\right],\left[1,2,\cdots ,M\right]\right)={1}^{M}+{2}^{M}+\cdots +{N}^{M}$

5.6) $SUM\left(N,\left[0:{D}_{1},\cdots ,0:{D}_{M}\right],PT\right)=SUM\text{_}K\left(N,PS,PT,{E}_{1}\cdots {E}_{M}\right)$

5.7) $SUM\left(N,\left[1,1,\cdots ,1\right],\left[1,3,\cdots ,2M-1\right]\right)={E}_{M}^{N}$

6. Stirling, Lah Number

S1(M, K), S2(M, K) is unsigned Stirling number. LM,K is Lah number.

 use 4.5) to calculate

${S}_{1}\left(N,N-M\right)={F}_{M}^{N-1}=\sum MIN\left(PS\right)\left(\begin{array}{c}N+PB\left(PT\right)+PM\left(PT\right)\\ IDX\left(PT\right)\end{array}\right)$, PS traverses all Basic Shapes, $PM\left(PS\right)=M$

This conforms to 4.13). In this paper, it can be written as:

6.1) $SUM\left(N,\left[1,2,\cdots ,M\right],\left[1,3,\cdots ,2M-1\right]\right)={S}_{1}\left(N+M,N\right)$, this is 5.4)

${E}_{M}^{N}\stackrel{def}{\to }\underset{{N}_{1}+\cdots +{N}_{C}=M}{\sum }{1}^{{N}_{1}}{2}^{{N}_{2}}\cdots {N}^{{N}_{C}}$, It’s a known property of ${S}_{2}\left(N+M,N\right)$

6.2) $SUM\left(N,\left[1,\cdots ,1\right],\left[1,3,\cdots ,2M-1\right]\right)={E}_{M}^{N}={S}_{2}\left(N+M,N\right)$

6.3) 1) ${H}_{1}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,M\right],K\right)$

$=K!{S}_{2}\left(M+1,K+1\right)=K!{\sum }_{i=0}^{M-K}\left(\begin{array}{c}M\\ i\end{array}\right){S}_{2}\left(M-i,K\right)$

2) ${H}_{1}\left(\left[1,\cdots ,1\right],\left[2,\cdots ,M\right],K\right)=\left(K+1\right)!{S}_{2}\left(M,K+1\right)$

[Proof]

Definition of S2(M, K) is

${N}^{M}={\sum }_{K=0}^{M}{S}_{2}\left(M,K\right){\left[N\right]}_{k}={\sum }_{K=0}^{M}K!{S}_{2}\left(M,K\right)\left( N K \right)$

$\begin{array}{c}{\left(N+1\right)}^{M}=\nabla SUM\left(N+1,\left[1,\cdots ,1\right],\left[1,\cdots ,M\right]\right)\\ ={\sum }_{q=0}^{M}{A}_{q}\left(\begin{array}{c}N\\ q\end{array}\right)={\sum }_{K=0}^{M}K!{S}_{2}\left(M,K\right)\left(\begin{array}{c}N+1\\ K\end{array}\right)\\ ={\sum }_{K=0}^{M}K!{S}_{2}\left(M,K\right)\left(\begin{array}{c}N\\ K+1\end{array}\right)+{\sum }_{K=0}^{M}K!{S}_{2}\left(M,K\right)\left(\begin{array}{c}N\\ K\end{array}\right)\end{array}$ à

${H}_{1}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,M\right],K\right)=K!{S}_{2}\left(M,K\right)+\left(K+1\right)!{S}_{2}\left(M,K+1\right)$ à the first equation

$\stackrel{4.11\right)}{\to }{H}_{1}\left(\cdots ,K\right)=K!\left[{F}_{M-K}^{\left[1,\cdots ,1\right]}{E}_{0}^{K}+\cdots +{F}_{0}^{\left[1,\cdots ,1\right]}{E}_{M-K}^{K}\right]$ $\stackrel{{F}_{M-K}^{\left[1,\cdots ,1\right]}=\left(\begin{array}{c}M\\ M-K\end{array}\right)}{\to }$ the second equation

$\begin{array}{l}\stackrel{4.2\right)}{\to }{H}_{1}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,M\right],K\right)\\ ={H}_{1}\left(\left[1,\cdots ,1\right],\left[2,\cdots ,M\right],K\right)+{H}_{1}\left(\left[2,\cdots ,M\right],K-1\right)\end{array}$ à 2)

q.e.d.

This à ${S}_{2}\left(N+1,M\right)={\sum }_{k=M-1}^{N}\left(\begin{array}{c}N\\ k\end{array}\right){S}_{2}\left(K,M-1\right)$, which is recorded in 

Example 6.1:

Directly according to the definition of H1()

$\begin{array}{l}{H}_{1}\left(\cdots ,1\right)=\left(1×2×2×\cdots ×2+1×1×2×\cdots ×2+\cdots +1×1×\cdots ×1×1\right)\\ \to {S}_{2}\left(M+1,2\right)={2}^{M}-1\end{array}$

${H}_{1}\left(\cdots ,M-1\right)=\left(M-1\right)!\left(1+2+\cdots +M\right)\to {S}_{2}\left(M+1,\text{}M\right)=\left(\begin{array}{c}M+1\\ 2\end{array}\right)$

6.4) $\left(\begin{array}{c}M\\ K\end{array}\right)\frac{M!}{K!}={S}_{1}\left(M+1,K+1\right){S}_{2}\left(K,K\right)+\cdots +{S}_{1}\left(M+1,M+1\right){S}_{2}\left(M,K\right)$

[Proof]

$\begin{array}{l}{H}_{1}\left(\left[1,\cdots ,M\right],\left[1,\cdots ,M\right],K\right)\stackrel{4.4\right)}{\to }=\left(\begin{array}{c}M\\ K\end{array}\right)M!\\ \stackrel{4.11\right)}{\to }=K!\left[{F}_{M-K}^{M}{E}_{0}^{K}+\cdots +{F}_{0}^{M}{E}_{M-K}^{K}\right]\\ =K!\left[{S}_{1}\left(M+1,K+1\right){S}_{2}\left(K,K\right)+\cdots +{S}_{1}\left(M+1,M+1\right){S}_{2}\left(M,K\right)\right]\end{array}$

q.e.d.

Definition of Lah number  is ${\left[-X\right]}_{M}={\sum }_{k=0}^{M}{L}_{M,K}{\left[x\right]}_{k}\stackrel{4.10\right)}{\to }$

6.5) ${L}_{M,K}={\left(-1\right)}^{M}\frac{M!}{K!}\left(\begin{array}{c}M-1\\ K-1\end{array}\right)$, this is recorded in .

7. Congruence Analysis

P is a prime number, we already know:

(7*) ${S}_{1}\left(P,P-K\right)={F}_{K}^{P-1}\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P,0

 has proved, it is easy to infer from 4.5):

7.1) $SUM\left(P-PB\left(PT\right)-PM\left(PT\right),PT,PT\right)$

$=MIN\left(PT\right)\left(\begin{array}{c}P\\ IDX\left(PT\right)\end{array}\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P,\text{\hspace{0.17em}}IDX\left(PT\right)

E.g.: $1×2+2×3+3×4\equiv 1×3+1×4+2×4\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}5$

This is the promotion of (7*).

 has proved, it is easy to infer from 3.1):

7.2) For arbitrary ${K}_{i}\in Z$

1) $M, $SUM\left(P,\left[{K}_{1}:D,\cdots ,{K}_{M}:D\right]\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

2) $M=P-1$, $\left(D,P\right)=1$, $SUM\left(P,\left[{K}_{1}:D,\cdots ,{K}_{M}:D\right]\right)\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

$SUM\left(P,\left[{K}_{1},\cdots ,{K}_{M}\right]\right)=\prod {K}_{i}+\prod \left(1+{K}_{i}\right)+\cdots +\prod \left(P-1+{K}_{i}\right)$. Exclude products≡0 MOD P:

Example 7.1:

${1}^{Q}+{2}^{Q}+\cdots +{\left(P-1\right)}^{Q}\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $Q ; $\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $Q=P-1$

${1}^{2}×2×3+{2}^{2}×3×4+\cdots +{\left(P-3\right)}^{2}×\left(P-2\right)×\left(P-1\right)\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P=5$ ; $\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P>5$

${1}^{3}×2+{2}^{3}×3+\cdots +{\left(P-2\right)}^{3}×\left(P-1\right)\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P=5$ ; $\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P>5$

$1×{2}^{3}+2×{3}^{3}+\cdots +\left(P-2\right)×{\left(P-1\right)}^{3}\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P=5$ ; $\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P>5$

${1}^{2}\text{\hspace{0.17em}}×{2}^{2}\text{\hspace{0.17em}}+{2}^{2}\text{\hspace{0.17em}}×{3}^{2}+\cdots +{\left(P-2\right)}^{2}\text{\hspace{0.17em}}×{\left(P-1\right)}^{2}\text{\hspace{0.17em}}\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P=5$ ; $\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $P>5$

In 1), let $PS=\left[{K}_{1}=1,\cdots ,{K}_{M}\right]$ is a Basic Shape and ${K}_{M}=P-1$

Rewrite $PS=〚{L}_{1}{L}_{2}\cdots {L}_{Q}〛$, ${L}_{i}$ = count of continuity. $\left({L}_{i},{L}_{i+1}\right)$ means a discontinuity.

$〚{L}_{1}\cdots {L}_{Q}〛$ can been slided to $\left[{L}_{Q},{L}_{1},{L}_{2},\cdots \right]$, $\left[{L}_{Q-1},{L}_{Q},{L}_{1},{L}_{2},\cdots \right]$ by $SUM\left(P,PS\right)\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

 has proved:

7.3) $\sum MIN\left(PS\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

{PS} = All of the Basic Shapes $\left[1,\cdots ,{K}_{M}=P-1\right]\ne \left[1,2,\cdots ,P-1\right]$ can slide to.

Example 7.2:

$\begin{array}{l}1×\left(3×4×5\right)×\left(7×8\right)×10+1×3×\left(5×6×7\right)×\left(9×10\right)\\ +\left(1×2\right)×4×6×\left(8×9×10\right)+\left(1×2×3\right)×\left(5×6\right)×8×10\\ =139260\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}11\end{array}$

This à  has proved:

7.4) $\sum MIN\left(PS\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, {PS} = All of the Basic Shapes with the same PM() and the same PB(), PB() > 0 and ${K}_{M}=P-1$

In Example 7.1, the SUM() is not symmetrical, it’s part of some symmetrical express.

E.g.: $\left({1}^{2}×{2}^{2}+{2}^{2}×{3}^{2}+{3}^{3}×{4}^{2}\right)+\left({1}^{2}×{3}^{2}+{2}^{2}×{4}^{2}+{4}^{2}×{1}^{2}\right)$

$\equiv SUM\left(5,\left[1,1,2,2\right]\right)+SUM\left(5,\left[1,1,3,3\right]\right)\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}5$

Use F(PS) = The symmetrical express. Obviously: $F\left(\left[{K}_{1}\cdots {K}_{M}\right]\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$, $M ;

7.5) $F\left(\lambda \left(PS\right)\right)\equiv -\frac{\left(\begin{array}{c}P-1\\ LEN\left(PS\right)\end{array}\right)\frac{\left({\mu }_{1}+{\mu }_{2}+\cdots \right)!}{{\mu }_{1}!{\mu }_{2}!\cdots }}{P-LEN\left(PS\right)}MOD\text{\hspace{0.17em}}P,M=P-1$

[Proof]

Use CNT(PS) = Count of $PS\in F\left(PS\right)$ à $F\left(PS\right)\equiv -CNT\left(PS\right)\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

Use LEN(PS) =Count of different ${K}_{i}\in PS$

Obviously:

$LEN\left(PS\in F\left(PS\right)\right)$ is same, count of products $\in$ F(PS) = P-LEN(PS)

CNT(PS) = [Count of products $\in$ F(PS)]/(P-LEN(PS))

Use $\lambda \left(PS\right)$ to classify PS.

$\lambda \left(PS\right):={1}^{{\lambda }_{1}}{2}^{{\lambda }_{2}}\cdots {\left(P-1\right)}^{{\lambda }_{p-1}},PM\left(PS\right)=1×{\lambda }_{1}+2×{\lambda }_{2}+\cdots +\left(P-1\right)×{\lambda }_{p-1}$

$\lambda \left(PS\right):={4}^{1}:{1}^{4}+{2}^{4}+{3}^{4}+\cdots$

$\lambda \left(PS\right):={1}^{4}:1×2×3×4+\cdots$

$\begin{array}{l}\lambda \left(PS\right):={1}^{1}{3}^{1}:{1}^{3}×2+{2}^{3}×3+\cdots ;{1}^{3}×3+{2}^{3}×4+\cdots ;\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }1×{2}^{3}+2×{3}^{3}+\cdots ;1×{3}^{3}+2×{4}^{3}+\cdots \end{array}$

$\lambda \left(PS\right):={2}^{2}:{1}^{2}×{2}^{2}+{2}^{2}×{3}^{2}+\cdots ;{1}^{2}×{3}^{2}+{2}^{2}×{4}^{2}+\cdots ;{1}^{2}×{4}^{2}+{2}^{2}×{5}^{2}+\cdots$

$\lambda \left(PS\right):={1}^{2}{2}^{1}:{1}^{2}×2×3+\cdots ;1×{2}^{2}×3+\cdots ;1×2×{3}^{2}+\cdots ;{1}^{2}×3×5+\cdots$

1) The combination number of LEN(PS) in $P-\text{1}=\left(\begin{array}{c}P-1\\ LEN\left(PS\right)\end{array}\right)$

2) Record $\lambda \left(PS\right)$ as $\left\{{\mu }_{1},{\mu }_{2},\cdots \right\}$

There are ${\mu }_{1}$ numbers in ${\lambda }_{1},{\lambda }_{2},\cdots$ equal to $x>0$

There are ${\mu }_{2}$ numbers in ${\lambda }_{1},{\lambda }_{2},\cdots$ equal to $y>0,y\ne x$

The combination number of ${\mu }_{1},{\mu }_{2},\cdots$ = $\frac{\left({\mu }_{1}+{\mu }_{2}+\cdots \right)!}{{\mu }_{1}!{\mu }_{2}!\cdots }$

Count of products $\in$ F(PS) = $\left(\begin{array}{c}P-1\\ LEN\left(PS\right)\end{array}\right)\frac{\left({\mu }_{1}+{\mu }_{2}+\cdots \right)!}{{\mu }_{1}!{\mu }_{2}!\cdots }$

q.e.d.

Example 7.3:

$\begin{array}{l}\left({1}^{3}×2+{2}^{3}×3+{3}^{3}×4\right)+\left({1}^{3}×3+{2}^{3}×4+{4}^{3}×1\right)+\left({1}^{3}×4+{3}^{3}×1+{4}^{3}×2\right)\\ +\left(1×{2}^{3}+2×{3}^{3}+3×{4}^{3}\right)\equiv -\frac{4!}{2!2!}×\frac{2!}{1!1!}/\left(5-2\right)\equiv -4\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\end{array}$

$\begin{array}{l}\left({1}^{2}×{2}^{2}+{2}^{2}×{3}^{2}+{3}^{2}×{4}^{2}\right)+\left({1}^{2}×{3}^{2}+{2}^{2}×{4}^{2}+{4}^{2}×{1}^{2}\right)\\ \equiv -\frac{4!}{2!2!}×\frac{2!}{2!}/\left(5-2\right)\equiv -2\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\end{array}$

$\begin{array}{l}\left({1}^{2}×2×3+{2}^{2}×3×4\right)+\left({1}^{2}×2×4+{3}^{2}×4×1\right)+\left({1}^{2}×3×4+{4}^{2}×1×2\right)\\ +\left(1×{2}^{2}×3+2×{3}^{2}×4\right)+\left(1×{2}^{2}×4+3×{4}^{2}×1\right)+\left(1×2×{3}^{2}+2×3×{4}^{2}\right)\\ \equiv -\frac{4!}{3!1!}×\frac{3!}{2!1!}/\left(5-3\right)\equiv -6\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\end{array}$

7.6) 1) $K!{S}_{2}\left(P-1,K\right)=K!{E}_{P-1-K}^{K}\equiv {\left(-1\right)}^{K+1}MOD\text{\hspace{0.17em}}P,1\le K\le P-1$

2) ${S}_{2}\left(P,\text{}K\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P,1

[Proof]

$\begin{array}{l}SUM\left(N,\left[1,2,\cdots ,P-1\right]\right)\stackrel{4.11\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(7*\right)}{\to }\\ \equiv \left(P-1\right)!\left(\begin{array}{c}N\\ P\end{array}\right){E}_{0}^{P-1}+\cdots +1!\left(\begin{array}{c}N\\ 2\end{array}\right){E}_{P-2}^{1}+0!\left(\begin{array}{c}N\\ 1\end{array}\right)\left(P-1\right)!\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\end{array}$

$\begin{array}{l}\nabla SUM\left(N,\left[1,\cdots ,P-1\right]\right)\\ \equiv \left(P-1\right)!\left(\begin{array}{c}N-1\\ P-1\end{array}\right){E}_{0}^{P-1}+\cdots +1!\left(\begin{array}{c}N-1\\ 1\end{array}\right){E}_{P-2}^{1}+0!\left(\begin{array}{c}N-1\\ 0\end{array}\right)\left(p-1\right)!\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\end{array}$

$\begin{array}{l}{\left[P\right]}_{P-1}=\nabla SUM\left(2,\left[1,\cdots ,P-1\right]\right)\equiv 1+\left(P-1\right)!\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\\ \to \left(P-1\right)!\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\end{array}$

This is a new proof of Wilson theorem.

$1!{E}_{P-2}^{1}=1\equiv 1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

${\left[P+1\right]}_{P-1}=\nabla SUM\left(3,\left[1,\cdots ,P-1\right]\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\to 2!{E}_{P-3}^{2}\equiv -1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

${\left[P+2\right]}_{P-1}=\nabla SUM\left(4,\left[1,\cdots ,P-1\right]\right)\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\to 3!{E}_{P-4}^{3}\equiv 1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P$

$\cdots$

à 1)

${S}_{2}\left(P,K\right)={S}_{2}\left(P-1,K-1\right)+K×{S}_{2}\left(P-1,K\right)$ à

$\begin{array}{l}\left(K-1\right)!{S}_{2}\left(P,\text{}K\right)=\left(K-1\right)!{S}_{2}\left(P-1,K-1\right)+K!{S}_{2}\left(P-1,K\right)\\ \equiv {\left(-1\right)}^{K}+{\left(-1\right)}^{K+1}\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P\end{array}$ à 2)

q.e.d.

${N}^{P-1}\equiv 1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P={\sum }_{K=0}^{P-1}K!{S}_{2}\left(P-1,K\right)\left(\begin{array}{c}N\\ K\end{array}\right)\stackrel{7.6\right),{S}_{2}\left(P-1,0\right)=0}{\to }$

7.7) $-\left(\begin{array}{c}N\\ P-1\end{array}\right)+\left(\begin{array}{c}N\\ P-2\end{array}\right)+\cdots -\left(\begin{array}{c}N\\ 2\end{array}\right)+\left(\begin{array}{c}N\\ 1\end{array}\right)\equiv 1\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}P,\left(N,P\right)=1$

7.8) ${\sum }_{n=1}^{P-1}{n}^{P-2}\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}{P}^{2},P>3$

[Proof]

$\begin{array}{l}{P}^{P-2}+{\sum }_{n=1}^{P-1}{n}^{P-2}={\sum }_{n=1}^{P}{n}^{P-2}=SUM\left(P,\left[1,\cdots ,1\right],\left[1,\cdots ,P-2\right]\right)\\ \stackrel{6.3\right)}{\to }={\sum }_{k=0}^{P-2}k!{S}_{2}\left(P-1,k+1\right)\left(\begin{array}{c}P\\ K+1\end{array}\right)\\ ={\sum }_{k=1}^{\frac{P-1}{2}}\left[\left(k-1\right)!{S}_{2}\left(P-1,k\right)\left(\begin{array}{c}P\\ k\end{array}\right)+\left(P-k-1\right)!{S}_{2}\left(P-1,P-k\right)\left(\begin{array}{c}P\\ P-k\end{array}\right)\right]\\ \stackrel{7.6\right)}{\to }\equiv {\sum }_{k=1}^{\frac{P-1}{2}}\left[{\left(-1\right)}^{K+1}+{\left(-1\right)}^{P-K+1}\right]\left(\begin{array}{c}P\\ K\end{array}\right)\\ \equiv {\sum }_{k=1}^{\frac{P-1}{2}}P×{\lambda }_{k}\left(\begin{array}{c}P\\ K\end{array}\right)MOD\text{\hspace{0.17em}}P\equiv 0\text{\hspace{0.17em}}MOD\text{\hspace{0.17em}}{P}^{2}\end{array}$

q.e.d.

8. Form2 and Analysis

Rewrite (1) of section 3 as

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

$PS=\left[{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right]$, $PT=\left[{T}_{1},\cdots ,{T}_{M}\right]$, use the same method of 3.1) to prove:

8.1) Use the ${\text{Form}}_{\text{2}}=\left({T}_{1}+{K}_{1}\right)\cdots \left({T}_{M}+{K}_{M}\right)=\sum {X}_{1}\cdots {X}_{M}$,

$SUM\left(N,PS,PT\right)=\sum {A}_{q}\left(\begin{array}{c}N+{T}_{M}-q\\ IDX\left(PT\right)-q\end{array}\right)$, q = count of $X\in K$. ${A}_{q}={\prod }_{i=1}^{M}{B}_{i}$

${B}_{i}=\left\{\begin{array}{l}\left({T}_{i}-m\right){D}_{i};{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\\ {K}_{i}+\left(m-{T}_{i}\right){D}_{i};{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\end{array}$

Define ${H}_{2}\left(PS,PT,C\right)={A}_{M-q}$ of 8.1), C = Count of $X\in T$

8.2) ${H}_{2}\left(PT,PT,q ; ${H}_{2}\left(PT,PT,PM\left(PT\right)\right)=\prod {T}_{i}$, This à 4.5)

8.3) ${H}_{2}\left(\left[1,\cdots ,M\right],\left[1,\cdots ,2M-1\right],C\right)$

$={\left(-1\right)}^{M-C-1}{H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],C-1\right)$

[Proof]

${H}_{2}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right],C\right)={\prod }_{i=1}^{M}{B}_{i}$,

$\begin{array}{l}{B}_{i}=\left\{\begin{array}{l}\left({T}_{i}-m\right);{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\\ 1+\left(m-2i+1\right);{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\end{array}\stackrel{m=i-1-q}{\to }\\ =\left\{\begin{array}{l}\left({T}_{i}-m\right);{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\\ -\left(\left(i-1\right)+q\right);{X}_{i}={K}_{i},q=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\end{array}\end{array}$

$\begin{array}{l}\to {\left(-1\right)}^{M-C}{H}_{2}\left(\cdots ,C\right)+{\left(-1\right)}^{M-C-1}{H}_{2}\left(\cdots ,C+1\right)\\ ={H}_{1}\left(\left[1,\cdots ,M\right],\left[1,\cdots ,2M-1\right],C\right)\stackrel{4.2\right)}{\to }\\ ={H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],C\right)+{H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],C-1\right)\end{array}$

$\begin{array}{l}\to {H}_{2}\left(\left[1,\cdots ,M\right],\left[1,\cdots ,2M-1\right],C\right)\\ ={\left(-1\right)}^{M-C-1}{H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],C-1\right)\end{array}$

q.e.d.

 obtains the unified expression of ${S}_{1}\left(N,N-M\right)$ and ${S}_{2}\left(N,N-M\right)$ by induction:

1) ${S}_{1}\left(N,N-M\right)={\sum }_{k=1}^{M}A\left(M,k\right)\left(\begin{array}{c}N\\ 2M-K+1\end{array}\right)$

2) ${S}_{2}\left(N,N-M\right)={\sum }_{k=1}^{M}{\left(-1\right)}^{k-1}A\left(M,k\right)\left(\begin{array}{c}N+M-K\\ 2M-K+1\end{array}\right)$

[Proof]

${S}_{1}\left(N,N-M\right)=SUM\left(N-M,\left[1,\cdots ,M\right],\left[1,\cdots ,2M-1\right]\right)$, ${\text{Form}}_{1}=\left({T}_{2}+{K}_{2}\right)\cdots \left({T}_{M}+{K}_{M}\right)$, 4.3) à

$={\sum }_{q=0}^{M-1}{H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],q\right)\left(\begin{array}{c}N-M+\left(M-1\right)+1\\ 2M-\left(M-q\right)\end{array}\right)\stackrel{k=M-q}{\to }$

$={\sum }_{k=1}^{M}{H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],M-k\right)\left(\begin{array}{c}N\\ 2M-k\end{array}\right)$ à 1)

${H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],M-K\right)=A\left(M,K\right)$

${S}_{2}\left(N,N-M\right)=SUM\left(N-M,\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right]\right)$, use the Form2

$={\sum }_{q=0}^{M}{H}_{2}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right],q\right)\left(\begin{array}{c}N-M+\left(2M-1\right)-\left(M-q\right)\\ 2M-\left(M-q\right)\end{array}\right)$

$={\sum }_{q=0}^{M}{H}_{2}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right],q\right)\left(\begin{array}{c}N-1+q\\ M+q\end{array}\right)\stackrel{{H}_{2}\left(\cdots ,0\right)=0}{\to }$

$={\sum }_{q=1}^{M}{H}_{2}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right],q\right)\left(\begin{array}{c}N-1+q\\ M+q\end{array}\right)\stackrel{K=M-q+1}{\to }$

$={\sum }_{K=1}^{M}{H}_{2}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right],M-k+1\right)\left(\begin{array}{c}N+M-K\\ 2M-K+1\end{array}\right)\stackrel{8.3\right)}{\to }$

$={\sum }_{k=1}^{M}{\left(-1\right)}^{k-1}{H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],M-k\right)\left(\begin{array}{c}N+M-k\\ 2M-k+1\end{array}\right)$ à 2)

q.e.d.

8.4) 1) ${\sum }_{C=0}^{C=M-1}{\left(-1\right)}^{M-1-C}{\sum }_{PM\left(PS\right)=M,PB\left(PS\right)=C}MIN\left(PS\right)=1$

2) ${\sum }_{C=0}^{C=M-1}{\left(-1\right)}^{M-1-C}{\sum }_{PM\left(PS\right)=M,PB\left(PS\right)=C}\left(M+2+C\right)MIN\left(PS\right)={2}^{M+1}-1$

PS are Basic Shapes

[Proof]

${S}_{2}\left(M+1,1\right)=1$, this is a known property = $SUM\left(1,\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right]\right)$

$\begin{array}{l}=\sum {A}_{q}\left(\begin{array}{c}N+{T}_{M}-q\\ IDX\left(PT\right)-q\end{array}\right)=\sum {A}_{q}\left(\begin{array}{c}1+2M-1-q\\ 2M-q\end{array}\right)=\sum {A}_{q}\\ \stackrel{{A}_{0}=0}{\to }{\sum }_{q=1}^{q=M}{H}_{2}\left(\cdots ,q\right)\stackrel{8.3\right)}{\to }\end{array}$

$={\sum }_{C=0}^{C=M-1}{\left(-1\right)}^{M-1-C}{H}_{1}\left(\left[2,\cdots ,M\right],\left[3,\cdots ,2M-1\right],C\right)\stackrel{4.13\right)}{\to }$ 1)

${S}_{2}\left(M+2,2\right)={2}^{M+1}-1=SUM\left(2,\left[1,\cdots ,1\right],\left[1,\cdots ,2M-1\right]\right)$ à 2)

q.e.d.

Example 8.1:

$M=\text{1}:1=1$

$M=\text{2}:1×3-1×2=1$

$M=\text{3}:1×3×5-\left(1×3×4+1×2×4\right)+1×2×3=1$

$\begin{array}{l}M=\text{4}:1×3×5×7-\left(1×3×5+1×3×4+1×2×4\right)×6\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(1×2×3+1×3×4+1×2×4\right)×5-1×2×3×4=1\end{array}$

$\begin{array}{l}M=\text{5}:1×3×5×7×9-\left(1×3×5×7+1×3×4×6+1×2×4×6+1×3×5×6\right)×8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(1×2×3×5+1×3×4×5+1×2×4×5+1×3×5×6+1×2×4×6\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+1×3×4×6\right)×7-\left(1×2×3×4+1×2×3×5+1×2×4×5+1×3×4×5\right)×6\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+1×2×3×4×5=1\end{array}$

It can be concluded from the definition:

8.5) ${H}_{2}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,M\right],C\right)={\left(-1\right)}^{M-C}C!{E}_{M-C}^{C}={\left(-1\right)}^{M-C}C!{S}_{2}\left(M,C\right)$

${N}^{M}=\nabla SUM\left(N,\left[1,\cdots ,1\right],\left[1,\cdots ,M\right]\right)$

$\begin{array}{l}\stackrel{{\text{Form}}_{\text{2}}}{\to }\nabla \sum {A}_{q}\left(\begin{array}{c}N+{T}_{M}-q\\ IDX\left(PT\right)-q\end{array}\right)=\sum {A}_{c}\left(\begin{array}{c}N+M-\left(M-C\right)-1\\ M+1-\left(M-C\right)-1\end{array}\right)\\ =\sum {A}_{c}\left(\begin{array}{c}N+C-1\\ C\end{array}\right)\end{array}$

8.6) ${N}^{M}={\sum }_{K=0}^{M}K!{S}_{2}\left(M,K\right)\left(\begin{array}{c}N\\ K\end{array}\right)={\sum }_{K=0}^{M}{\left(-1\right)}^{M-K}K!{S}_{2}\left(M,K\right)\left(\begin{array}{c}N+K-1\\ K\end{array}\right)$

It can be concluded from the definition:

8.7) ${H}_{2}\left(\left[1+y,2+y,\cdots ,M+y\right],\left[1,2,\cdots ,M\right],C\right)=C!{\left[y\right]}^{M-C}\left( M C \right)$

${\left[x+y\right]}^{M}=\nabla SUM\left(x,\left[1+y,2+y,\cdots ,M+y\right]\right)$

$\stackrel{{\text{Form}}_{2}}{\to }\nabla {\sum }_{c=0}^{M}\left(\begin{array}{c}M\\ c\end{array}\right)C!{\left[y\right]}^{M-c}\left(\begin{array}{c}x+M-\left(M-C\right)\\ M+1-\left(M-C\right)\end{array}\right)$

$=\nabla {\sum }_{c=0}^{M}\left(\begin{array}{c}M\\ c\end{array}\right)c!{\left[y\right]}^{M-c}\left(\begin{array}{c}x+C\\ C+1\end{array}\right)={\sum }_{c=0}^{M}\left(\begin{array}{c}M\\ c\end{array}\right)c!{\left[y\right]}^{M-c}\left(\begin{array}{c}x+C-1\\ C\end{array}\right)$

$={\sum }_{K=0}^{M}\left(\begin{array}{c}M\\ K\end{array}\right){\left[y\right]}^{M-K}{\left[x+k-1\right]}_{K}={\sum }_{K=0}^{M}\left(\begin{array}{c}M\\ K\end{array}\right){\left[y\right]}^{M-K}{\left[x\right]}^{K}$ à Norlund identity

9. Form3 and Eulerian Number

Rewrite (1) of section 3 as

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

$PS=\left[{K}_{1}:{D}_{1},\cdots ,{K}_{M}:{D}_{M}\right]$, $PT=\left[{T}_{1},\cdots ,{T}_{M}\right]$, use the same method of 3.1) to prove:

9.1) Use the ${\text{Form}}_{\text{3}}=\left({T}_{1}+{K}_{1}\right)\cdots \left({T}_{M}+{K}_{M}\right)=\sum {X}_{1}\cdots {X}_{M}$,

$SUM\left(N,PS,PT\right)=\sum {A}_{q}\left(\begin{array}{c}N+{T}_{M}-q\\ IDX\left(PT\right)\end{array}\right)$, q = count of $X\in T$. ${A}_{q}={\prod }_{i=1}^{M}{B}_{i}$,

${B}_{i}=\left\{\begin{array}{l}-{K}_{i}+\left[{T}_{i}-m\right]{D}_{i};{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\\ {K}_{i}+m{D}_{i};{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\end{array}$

Define ${H}_{3}\left(PS,PT,C\right)={A}_{q}$ of 9.1), C = Count of $X\in T$

$〈\begin{array}{c}M\\ k\end{array}〉$ is Eulerian number. Worpitzky identity: ${N}^{M}={\sum }_{k=0}^{M-1}〈\begin{array}{c}M\\ k\end{array}〉\left(\begin{array}{c}N+k\\ M\end{array}\right)$

Already known 1) $〈\begin{array}{c}M\\ M-q-1\end{array}〉=〈\begin{array}{c}M\\ k\end{array}〉$, 2) $〈\begin{array}{c}M\\ q\end{array}〉=\left(M-q\right)〈\begin{array}{c}M-1\\ q-1\end{array}〉+\left(q+1\right)〈\begin{array}{c}M-1\\ q\end{array}〉$

9.2) ${H}_{3}\left(PT,PT,q>0\right)=0$ ; ${H}_{3}\left(PT,PT,0\right)=\prod {T}_{i}$ This à 4.5)

9.3) ${H}_{3}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,M\right],q\right)=〈\begin{array}{c}M\\ M-q-1\end{array}〉=〈\begin{array}{c}M\\ q\end{array}〉$, ${H}_{3}\left(\left[0,\cdots ,0\right],\left[1,\cdots ,M\right],q\right)=〈\begin{array}{c}M\\ q-1\end{array}〉$

[Proof]

Obviously: ${H}_{3}\left(\left[{K}_{1}=1,\cdots ,{K}_{M}\right],\left[{T}_{1}=1,\cdots ,{T}_{M}\right],M\right)=0$

$\begin{array}{c}{N}^{M}=\nabla SUM\left(N,\left[1,\cdots ,1\right]\right)={\sum }_{q=0}^{M}{A}_{q}\left(\begin{array}{c}N+M-q-1\\ M\end{array}\right)\\ ={\sum }_{q=0}^{M}{A}_{q}\left(\begin{array}{c}N+M-q-1\\ M\end{array}\right)={\sum }_{k=0}^{M-1}〈\begin{array}{c}M\\ k\end{array}〉\left(\begin{array}{c}N+k\\ M\end{array}\right)\end{array}$

q.e.d.

It’s easy to deduce:

(*) ${H}_{3}\left(q\right)={H}_{2}\left(\left[0,\cdots ,0\right],\left[1,\cdots ,M\right],q\right)=\sum \prod {B}_{i}$, ${B}_{i}=\left\{\begin{array}{l}m+1:{X}_{i}={T}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in K\\ m:{X}_{i}={K}_{i},m=\text{count of}\text{\hspace{0.17em}}\left\{\cdots ,{X}_{i-1}\right\}\in T\end{array}$

(*1) ${H}_{3}\left(q\right)={H}_{2}\left(M-q-1\right)$ à 1)

(*2) ${H}_{3}\left(q\right)=\left(M-q\right){H}_{3}\left(q-1\right)+\left(q+1\right){H}_{3}\left(\left[0,\cdots ,0\right],\left[1,\cdots ,M-1\right],q\right)$ à 2)

(*3) ${H}_{3}\left(q\right)=\sum \prod {B}_{i}$, then ${B}_{1}+{B}_{2}+\cdots +{B}_{M}=q\left(M-q+1\right)$

$\begin{array}{l}{H}_{3}\left(\left[1,\cdots ,1\right],\left[1,\cdots ,M\right],1\right)\\ ={1}^{M-1}×\left(M-1\right)+\text{}{1}^{M-2}×\left(M-2\right)×2+{1}^{M-3}×\left(M-3\right)×{2}^{2}+\cdots \end{array}$

$={2}^{0}×M+{2}^{1}×M+\cdots +{2}^{M-2}×M-\left({2}^{0}×1+{2}^{1}×2+\cdots +{2}^{M-2}×\left(M-1\right)\right)$

$=M\left({2}^{M-1}-1\right)-{\sum }_{i=0}^{M-2}{2}^{i}\left(i+1\right)=M\left({2}^{M-1}-1\right)-\left({2}^{M-1}-1\right)-{\sum }_{i=0}^{M-2}{2}^{i}i$

$=〈\begin{array}{c}M\\ 1\end{array}〉={2}^{M}-\left(M+1\right)$, the final equation is a known property of $〈\begin{array}{c}M\\ 1\end{array}〉$ à

9.4) ${\sum }_{i=0}^{M}{2}^{i}i=\left(M-1\right){2}^{M+1}+2$

9.5) $〈\begin{array}{c}M\\ q\end{array}〉={\sum }_{{t}_{1}+\cdots +{t}_{q+1}=M-q-1}{1}^{{t}_{1}}{2}^{{t}_{2}}\cdots {\left(q+1\right)}^{{t}_{q+1}}\left(1+{t}_{1}\right)\cdots \left(1+{t}_{1}+\cdots +{t}_{q}\right),{t}_{i}\ge 0$

[Proof]

${H}_{3}\left(\left[0,\cdots ,0\right],\left[1,\cdots ,M\right],q+1\right)=〈\begin{array}{c}M\\ q\end{array}〉=\sum \Pi \left(X\in T\right)\Pi \left(X\in K\right)$

If $X\in K$ is certain, then $X\in T$ is certain and ${X}_{1}\cdots {X}_{M}$ is certain.

When ${X}_{1}\cdots {X}_{M}>0$,then ${X}_{1}={T}_{1}=1$, ${K}_{i}\in \left[1,q+1\right]$

Record $\Pi \left(X\in K\right)={1}^{{t}_{1}}{2}^{{t}_{2}}\cdots {\left(q+1\right)}^{{t}_{q+1}}$, ${t}_{1}+{t}_{2}+\cdots +{t}_{q+1}=M-q-1$

Take out factors > 0, record as $\left\{{P}_{1},{P}_{2},\cdots \right\}$,(*)à

$\Pi \left(X\in T\right)={1}^{{P}_{1}}{\left(1+{t}_{p1}\right)}^{{P}_{2}-{P}_{1}}{\left(1+{t}_{p1}+{t}_{p2}\right)}^{{P}_{3}-{P}_{2}}\cdots$, it can be rewritten as

$\Pi \left(X\in T\right)=\left(1+{t}_{1}\right)\left(1+{t}_{1}+{t}_{2}\right)\left(1+{t}_{1}+{t}_{2}+{t}_{3}\right)\cdots \left(1+{t}_{1}+\cdots +{t}_{q}\right)$

q.e.d.

This is the conclusion of , which is obtained by guess and proved by induction.

Example 9.1:

$\begin{array}{l}〈\begin{array}{c}6\\ 2\end{array}〉=\underset{{t}_{1}+{t}_{2}+{t}_{3}=3}{\sum }{1}^{{t}_{1}}{2}^{{t}_{2}}{3}^{{t}_{3}}\left(1+{t}_{1}\right)\left(1+{t}_{1}+{t}_{2}\right)\\ ={1}^{3}{2}^{0}{3}^{0}\left(1+3\right)\left(1+3+0\right)+{1}^{2}{2}^{1}{3}^{0}\left(1+2\right)\left(1+2+1\right)+{1}^{2}{2}^{0}{3}^{1}\left(1+2\right)\left(1+2+0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{1}^{1}{2}^{1}{3}^{1}\left(1+1\right)\left(1+1+1\right)+{1}^{1}{2}^{2}{3}^{0}\left(1+1\right)\left(1+1+2\right)+{1}^{1}{2}^{0}{3}^{2}\left(1+1\right)\left(1+1+0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{1}^{0}{2}^{1}{3}^{2}\left(1+0\right)\left(1+0+1\right)+{1}^{0}{2}^{2}{3}^{1}\left(1+0\right)\left(1+0+2\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{1}^{0}{2}^{3}{3}^{0}\left(1+0\right)\left(1+0+3\right)+{1}^{0}{2}^{0}{3}^{3}\left(1+0\right)\left(1+0+0\right)\\ =302\end{array}$

Cite this paper: Peng, J. (2021) Redefining the Shape of Numbers and Three Forms of Calculation. Open Access Library Journal, 8, 1-22. doi: 10.4236/oalib.1107277.
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

   Peng, J. (2020) Subdivide the Shape of Numbers and a Theorem of Ring. Open Access Library Journal, 7: e6719.
https://doi.org/10.4236/oalib.1106719

   Peng, J. (2020) Subset of the Shape of Numbers. Open Access Library Journal, 7: e7040.
https://doi.org/10.4236/oalib.1107040

   Peng, J. (2021) Expansion of the Shape of Numbers. Open Access Library Journal, 8: e7120.
https://doi.org/10.4236/oalib.1107120

   Editorial Board of Handbook of Modern Applied Mathematics (2002) Handbook of Modern Applied Mathematics Discrete Mathematics Volume. Tsinghua University Press, Beijing.

   Wang, Y.K. and Li, S.X. (1999) The Uniform Expression for Two Kinds of Stirling Number and Bernoulli Number. Journal of Guangxi Teachers College (Natural Science Edition), 1, 49-53.

   Qi, D.-J. (2012) A New Explicit Expression for the Eulerian Numbers. Journal of Qingdao University of Science and Technology (Natural Science Edition), 33, 439-440.

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