Redefining the Shape of Numbers and Three Forms of Calculation

Ji Peng

doi:10.4236/oalib.1107277

Open Access Library Journal > Vol.8 No.3, March 2021

Redefining the Shape of Numbers and Three Forms of Calculation

Ji Peng
Department of Electronic Information, Nanjing University, Nanjing, China.
DOI: 10.4236/oalib.1107277 PDF HTML XML 198 Downloads 723 Views Citations

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-0^N-1Π_i-1^M (K_i+n×Di)
∑_{n_i,j-0}^j-N-1Π_i-1^M (K_i+n_i,j×Di), n_i,j≤n_i+1,j or n_i,j=n_i+1,j; K_i,D_i∈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-1^P-1≡0 MOD P², P>3; 6) ∑_C-0^C-M-1(-1)^M-1-C∑_{PM(PS)-M,PB(PS)-C}MIN(PS)=1.

Keywords

Shape of Numbers, Calculation Formula, Combinatorics, Congruence, Stirling Number, Eulerian Number

Share and Cite:

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.

1. Introduction

Peng, J. has introduced Shape of numbers in [1] [2] [3] [4]: $(I_{1}, I_{2}, \dots, I_{M})$ , $I_{i} \in Z$ . There are $M - 1$ intervals between adjacent numbers. $I_{i + 1} - I_{i} \leq 1$ means continuity; $I_{i + 1} - I_{i} > 1$ means discontinuity.

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

A Shape has a min Item: $(K_{1}, K_{2}, \dots)$ . Use the symbol PS = [min Item] to represent it.

If $K_{i + 1} - K_{i} = D > 1$ , only $I_{i + 1} - I_{i} \geq D$ is allowed.

If $K_{i + 1} - K_{i} = D \leq 1$ , only $I_{i + 1} - I_{i} = D$ is allowed.

The single $(I_{1}, \dots, I_{M})$ is an item. $I_{1} \times \dots \times I_{M}$ is the product. I_i is a factor.

Example 1.1:

$P S = [2, 3] \to (2, 3), (3, 4), (1000, 1001) \in P S$

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

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

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

PM(PS) = Count of factors.

PB(PS) = Count of discontinuities.

MIN(PS) = Min product: $M I N ([1, 2, 3]) = 1 \times 2 \times 3$ , $M I N ([1, 2, 4]) = 1 \times 2 \times 4$

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

$P S = [K_{1}, \dots, K_{M}]$ , $B S = [G_{1}, \dots, G_{M}]$ then $K_{i + 1} - K \leq 1 \to G_{i + 1} - G_{i} = 1$ ; $K_{i + 1} - K > 1 \to G_{i + 1} - G_{i} = 2$

Example 1.2:

$P S = [1, 2], [1001, 1002] \to B A S E (P S) = [1, 2]$

$P S = [1, 3], [- 9, 4], [1, K > 2] \to B A S E (P S) = [1, 3]$

$P S = [1, 3, 4], [- 8, 4, 5], [1, K > 2, X = K + 1] \to B A S E (P S) = [1, 3, 4]$

$P S = [1, 3, 5], [0, 4, 9], [1, K > 2, X > K + 1] \to B A S E (P S) = [1, 3, 5]$

$P S = [1001, 1002] \to P H (P S) = 1002 - 1 - 0 = 1001$ , $I D X (P S) = I D X ([1, 2]) = 3$

$P S = [0, 4, 9] \to P H (P S) = 9 - 1 - 2 = 6$ , $I D X ([0, 4, 9]) = I D X ([1, 3, 5]) = 6$

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

[3] introduced the subset: fix some interval of discontinuities.

SET(N,PS,PT) = Subset of SET(N, PS), $B A S E (P S) = [G_{1}, \dots, G_{M}]$ , $P T = [T_{1}, \dots, T_{M}]$

$= {\begin{cases} T_{i + 1} - T_{i} = 1 : G_{i + 1} - G_{i} = 1, means I_{i + 1} - I_{i} = K_{i + 1} - K_{i} \\ T_{i + 1} - T_{i} = 1 : G_{i + 1} - G_{i} = 2, means I_{i + 1} - I_{i} = K_{i + 1} - K_{i} \\ T_{i + 1} - T_{i} = 2 : G_{i + 1} - G_{i} = 2, means I_{i + 1} - I_{i} > = K_{i + 1} - K_{i} \end{cases}$ (*)

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:

$P C H G ([1, 3, 5], [1, 3, 5]) = 0$

$P C H G ([1, 3, 5], [1, 2, 4]) = P C H G ([1, 4, 7], [1, 2, 4]) = 1$ , changed at T₁

$P C H G ([1, 3, 5], [1, 3, 4]) = P C H G ([1, 4, 7], [1, 3, 4]) = 1$ , changed at T₂

$P C H G ([1, 3, 5], [1, 2, 3]) = P C H G ([1, 8, 10], [1, 2, 3]) = 2$ , 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:

$S U M (6, [1, 2, 4]) = 1 \times 2 \times 4 + 1 \times 2 \times 5 + 2 \times 3 \times 5$

$\begin{matrix} S U M (9, [1, 4, 7]) = S U M (9, [1, 4, 7], [1, 3, 5]) \\ = 1 \times 4 \times 7 + 1 \times 4 \times 8 + 1 \times 5 \times 8 + 2 \times 5 \times 8 \end{matrix}$

$S U M (9, [1, 4, 7], [1, 2, 4]) = 1 \times 4 \times 7 + 1 \times 4 \times 8 + 2 \times 5 \times 8$

[1] [2] [3] [4] came to the following conclusion:

(1.1) $| S E T (N, P S, P T) | = (\begin{matrix} N - P H (P S) - P C H G (P S, P T) - 1 \\ P B (P T) + 1 \end{matrix})$

The following uses count of $X \in K$ for count of ${X_{1}, X_{2}, \dots, X_{M}} \in {K_{1}, K_{2}, \dots, K_{M}}$

(1.2) Use the form $(T_{1} + K_{1}) (T_{2} + K_{2}) \dots (T_{M} + K_{M}) = \sum X_{1} X_{2} \dots X_{M}$ , $X_{i} = T_{i}$ or $K_{i}$ .

Don’t swap the factors of $X_{1} X_{2} \dots X_{M}$ , then each $X_{1} X_{2} \dots X_{M}$ corresponds to a expression = $A_{q} (\begin{matrix} N - P H (P S) - P C H G (P S, P T) - 1 \\ I D X (P T) - q \end{matrix})$ , q = count of $X \in K$ .

$S U M (N, P S, P T) = \sum A_{q} (\begin{matrix} N - P H (P S) - P C H G (P S, P T) - 1 \\ I D X (P T) - q \end{matrix})$

$A_{q} = \prod_{i = 1}^{M} (X_{i} + D_{i})$ , $D_{i} = {\begin{cases} - m : X_{i} = T_{i}, m = count of {X_{1}, \dots, X_{i - 1}} \in K \\ + m : X_{i} = K_{i}, m = count of {X_{1}, \dots, X_{i - 1}} \in T \end{cases}$

Example 1.5:

$P S = [K_{1}, K_{2} \geq K_{1} + 2, K_{3} \geq K_{2} + 2]$ , $B S = B A S E (P S) = [1, 3, 5]$ ,

$\to I D X (B S) = 6$ ,

$\begin{matrix} form = (1 + K_{1}) (3 + K_{2}) (5 + K_{3}) \\ = 1 \times 3 \times 5 + 1 \times 3 \times K_{3} + 1 \times K_{2} \times 5 + 1 \times K_{2} \times K_{3} + K_{1} \times 3 \times 5 \\ + K_{1} \times 3 \times K_{3} + K_{1} \times K_{2} \times 5 + K_{1} \times K_{2} \times K_{3} \end{matrix}$

$\begin{matrix} P = N - P H (P S) - P C H G (P S, B S) - 1 \\ = N - {K_{3} - 1 - 2} - 0 - 1 = N - K_{3} + 2 \end{matrix}$

$\begin{array}{l} \to S U M (N, P S) \\ = 1 \times 3 \times 5 (\begin{matrix} P \\ 6 \end{matrix}) + 1 \times 3 \times (K_{3} + 2) (\begin{matrix} P \\ 5 \end{matrix}) + 1 \times (K_{2} + 1) \times (5 - 1) (\begin{matrix} P \\ 5 \end{matrix}) \\ + 1 \times (K_{2} + 1) \times (K_{3} + 1) (\begin{matrix} P \\ 4 \end{matrix}) + K_{1} \times (3 - 1) \times (5 - 1) (\begin{matrix} P \\ 5 \end{matrix}) \\ + K_{1} \times (3 - 1) \times (K_{3} + 1) (\begin{matrix} P \\ 4 \end{matrix}) + K_{1} \times K_{2} \times (5 - 2) (\begin{matrix} P \\ 4 \end{matrix}) \\ + K_{1} \times K_{2} \times K_{3} ( P 3 ) \end{array}$

An item = ${K_{1} + E_{1}, \dots, K_{M} + E_{M}}$ , K is fixed, E is variable.

A product = $(K_{1} + E_{1}) \dots (K_{M} + E_{M}) = \sum F_{1} F_{2} \dots F_{M}$ , $F_{i} = E_{i}$ or $K_{i}$

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

Define $S U M_K (N, P S, P T, P F = F_{1} \dots F_{M})$ = Sum of one part, PF indicates the part.

Rewrite 1.2), add {braces}:

$S U M (N, P S, P T) = \sum \prod_{i = 1}^{M} (X_{i} + D_{i}) (\begin{matrix} A \\ M_{q} \end{matrix})$ ,

$X_{i} + D_{i} = {\begin{cases} {T_{i} - D_{i}} : X_{i} = T_{i}, D_{i} = count of {X_{1}, \dots, X_{i - 1}} \in K \\ {K_{i}} + {D_{i}} : X_{i} = K_{i}, D_{i} = count of {X_{1}, \dots, X_{i - 1}} \in T \end{cases}$

Expand SUM() by {braces}:

(1.3) $S U M_K (N, P S, P T, P F)$

$= \sum Expansion of S U M () with same {K_{i}} \in P F = \sum \prod_{i = 1}^{M} Y_{i} (\begin{matrix} A \\ M_{q} \end{matrix})$

$Y_{i} = {\begin{cases} 0 : F_{i} = K_{i}, X_{i} = T_{i} \\ K_{i} : F_{i} = K_{i}, X_{i} = K_{i} \\ T_{i} - D_{i} : F_{i} = E_{i}, X_{i} = T_{i}, D_{i} = count of {\dots, X_{i - 1}} \in K \\ D_{i} : F_{i} = E_{i}, X_{i} = K_{i}, D_{i} = count of {\dots, X_{i - 1}} \in T \end{cases}$

Example 1.6:

Use the example above

$S U M_K (\dots E_{1} E_{2} E_{3}) = 1 \times 3 \times 5 (\begin{matrix} P \\ 6 \end{matrix}) + [1 \times 3 \times 2 + 1 \times 1 \times (5 - 1)] (\begin{matrix} P \\ 5 \end{matrix}) + 1^{3} ( P 4 )$

$S U M_K (\dots E_{1} E_{2} K_{3}) = 1 \times 3 \times K_{3} (\begin{matrix} P \\ 5 \end{matrix}) + 1 \times 1 \times K_{3} ( P 4 )$

$S U M_K (\dots E_{1} K_{2} E_{3}) = 1 \times K_{2} \times (5 - 1) (\begin{matrix} P \\ 5 \end{matrix}) + 1 \times K_{2} \times 1 ( P 4 )$

$S U M_K (\dots K_{1} E_{2} E_{3}) = K_{1} \times (3 - 1) \times (5 - 1) (\begin{matrix} P \\ 5 \end{matrix}) + K_{1} \times (3 - 1) \times 1 ( P 4 )$

$S U M_K (\dots E_{1} K_{2} K_{3}) = 1 \times K_{2} \times K_{3} ( P 4 )$

$S U M_K (\dots K_{1} E_{2} K_{3}) = K_{1} \times (3 - 1) \times K_{3} ( P 4 )$

$S U M_K (\dots K_{1} K_{2} E_{3}) = K_{1} \times K_{2} \times (5 - 2) ( P 4 )$

$S U M_K (\dots K_{1} K_{2} K_{3}) = K_{1} \times K_{2} \times K_{3} ( P 3 )$

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 T_i and D_i all increase L times

(1.4) $K_{1} \times \dots \times K_{M} + (L + K_{1}) \times \dots \times (L + K_{M}) + \dots$

$+ ((N - 1) L + K_{1}) \times \dots \times ((N - 1) L + K_{M})$

$P T = [1 \times L, 2 \times L, \dots, M \times L]$ , can use the form $(T_{1} + K_{1}) \dots (T_{M} + K_{M}) = \sum A_{q} (\begin{matrix} N \\ M + 1 - q \end{matrix})$ , q = count of $X \in K$ , 2^M items in total.

$A_{q} = \prod_{i = 1}^{M} (X_{i} + D_{i})$ , $D_{i} = {\begin{cases} - m L : X_{i} = T_{i}, m = count of {\dots, X_{i - 1}} \in K \\ + m L : X_{i} = K_{i}, m = count of {\dots, X_{i - 1}} \in T \end{cases}$

This paper starts from (1.4), tries to calculate

(1*) $K_{1} \times \dots \times K_{M} + (D_{1} + K_{1}) \times \dots \times (D_{M} + K_{M}) + \dots$

$+ ((N - 1) D_{1} + K_{1}) \times \dots \times ((N - 1) D_{M} + K_{M})$

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, $K_{i}, D_{i} \in Ring$ . $K_{i}$ no longer compares big or small.

M-series:

${K_{1}, D_{1} + K_{1}, 2 D_{1} + K_{1}, 3 D_{1} + K_{1}, \dots, (N - 1) D_{1} + K_{1}} = {K_{1} + n \times D_{1}}$

${K_{2}, D_{2} + K_{2}, 2 D_{2} + K_{2}, 3 D_{2} + K_{2}, \dots, (N - 1) D_{2} + K_{2}} = {K_{2} + n \times D_{2}}$

$⋮$

$\begin{array}{l} {K_{M}, D_{M} + K_{M}, 2 D_{M} + K_{M}, 3 D_{M} + K_{M}, \dots, (N - 1) D_{M} + K_{M}} \\ = {K_{M} + n \times D_{M}} \end{array}$

An item = $(I_{1}, I_{2}, \dots, I_{M})$ , I₁ come from serie1, I₂ come from serie2…

Use $P S = [K_{1} : D_{1}, K_{2} : D_{2}, \dots, K_{M} : D_{M}]$ to represent the Shape.

$[K_{1} : 1, K_{2} : 1, \dots, K_{M} : 1]$ is abbreviated as $[K_{1}, K_{2}, \dots, K_{M}]$

If $K_{i} \in Z$ , the new definition is similar to the old definition and allows $D_{i} \leq 0$ .

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

$I_{i} = K_{i} + a \times D_{i}$ , $I_{i + 1} = K_{i + 1} + b \times D_{i + 1}$ , $P T = [T_{1} = 1, T_{2}, \dots, T_{M}] = {\begin{cases} T_{i + 1} - T_{i} = 1 : means a = b \\ T_{i + 1} - T_{i} = 2, means a \leq b \end{cases}$

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, $K_{i} \in N$ , D_i = 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 $I_{M} = (N - 1) D_{M} + K_{M}$ .

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

$S U M (N, P S, [1, 2, \dots, M])$ is abbreviated as SUM (N, PS)

$S U M (N, \dots) = Old Sum (M a x (K_{i}) + N, \dots)$ , PH() and PCHG() are no longer needed.

The old does not allow $S U M ([2, 3], [1, 3])$ , $S U M ([3, 2], [1, 3])$ . The new it is allowed.

$S U M (N, P S) = (1 *) = \sum_{n = 0}^{N - 1} \prod_{i = 1}^{M} (K_{i} + n \times D_{i})$

$S U M (N, P S, [1, 3, 5, \dots, 2 M - 1]) = \sum \prod_{i = 1}^{M} (K_{i} + J_{i, j} \times D_{i})$ , $J_{1. i} \leq J_{2, i} \leq \dots \leq J_{M, i}$ , $J_{i, j} \in [0, N - 1]$

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

$I D X (P T) = T_{M} + 1 = P B (P T) + P M (P T) + 1$

2.1) $| S E T (N, P S, P T) | = (\begin{matrix} N + P B (P T) \\ P B (P T) + 1 \end{matrix})$

3. Form₁ of Calculation

Similar to [4], key points are:

Define $\nabla f (n) = f (n) - f (n - 1)$ : if $f (n) = \sum A_{i} (\begin{matrix} N_{i} \\ m_{i} \end{matrix})$ , then $\nabla f (n) = \sum A_{i} (\begin{matrix} N_{i} - 1 \\ m_{i} - 1 \end{matrix})$

1) $\sum_{n = 0}^{N - 1} n (\begin{matrix} n + K \\ M \end{matrix}) = (M + 1) (\begin{matrix} N + K \\ M + 2 \end{matrix}) + (M - K) (\begin{matrix} N + K \\ M + 1 \end{matrix})$

By definition:

2) $\sum E N D (N, P S, P T) = \nabla S U M (N, P S, P T)$

3) $S U M (N, [P S, K_{M + 1} : D_{M + 1}], [P T, T_{M} = T_{M} + 1])$

$= \sum_{n = 0}^{N - 1} (K_{M + 1} + n \times D_{M + 1}) \times \sum E N D (n + 1, P S, P T)$

4) $S U M (N, [P S, K_{M + 1} : D_{M + 1}], [P T, T_{M} = T_{M} + 2])$

$= \sum_{n = 0}^{N - 1} (K_{M + 1} + n \times D_{M + 1}) \times S U M (n + 1, P S, P T)$

$P S = [K_{1} : D_{1}, \dots, K_{M} : D_{M}]$ , $P T = [T_{1}, \dots, T_{M}]$ , $P S 1 = [P S, K_{M + 1} : D_{M + 1}]$

3.1) Use the Form₁ = $(T_{1} + K_{1}) \dots (T_{M} + K_{M}) = \sum X_{1} \dots X_{M}$ ,

$S U M (N, P S, P T) = \sum A_{q} (\begin{matrix} N + P B (P T) \\ I D X (P T) - q \end{matrix})$ , q = count of $X \in K$ .

$A_{q} = \prod_{i = 1}^{M} B_{i}$ , $B_{i} = {\begin{cases} (T_{i} - m) D_{i}; X_{i} = T_{i}, m = count of {X_{1}, \dots, X_{i - 1}} \in K \\ K_{i} + m D_{i}; X_{i} = K_{i}, m = count of {X_{1}, \dots, X_{i - 1}} \in T \end{cases}$

[Proof]

Suppose $S U M (N, P S, P T) = \sum X_{1} \dots X_{M} (\begin{matrix} N + P B (P T) \\ I D X (P T) - q \end{matrix})$

When $P T 1 = [P T, T_{M + 1} = 1 + T_{M}]$

$\begin{array}{l} S U M (N, P S 1, P T 1) \\ = \sum_{n = 0}^{N - 1} (K_{M + 1} + n \times D_{M + 1}) \times \sum E N D (n + 1, P S, P T) \\ = \sum_{n = 0}^{N - 1} (K_{M + 1} + n \times D_{M + 1}) \times \nabla S U M (n + 1, P S, P T) \\ = \sum_{n = 0}^{N - 1} (K_{M + 1} + n \times D_{M + 1}) \times \sum X_{1} \dots X_{M} (\begin{matrix} n + P B (P T) \\ I D X (P T) - q - 1 \end{matrix}) \overset{( 1 )}{\to} \end{array}$

$\begin{array}{l} = \sum X_{1} \dots X_{M} D_{M + 1} (I D X (P T) - q) (\begin{matrix} N + P B (P T) \\ I D X (P T) - q + 1 \end{matrix}) \\ + \sum X_{1} \dots X_{M} {K_{M + 1} + (I D X (P T) - q - 1 - P B (P T)) D_{M + 1}} (\begin{matrix} N + P B (P T) \\ I D X (P T) - q \end{matrix}) \\ \overset{P B (P T 1) = P B (P T), I D X (P T 1) = I D X (P T) + 1, I D X (P T) = 1 + T_{M} = T_{M + 1}, I D X (P T) - P B (P T) - 1 = M}{\to} \\ = \sum X_{1} \dots X_{M} D_{M + 1} (T_{M + 1} - q) (\begin{matrix} N + P B (P T 1) \\ I D X (P T 1) - q \end{matrix}) \\ + \sum X_{1} \dots X_{M} {K_{M + 1} + (M - q) D_{M + 1}} (\begin{matrix} N + P B (P T 1) \\ I D X (P T 1) - (q + 1) \end{matrix}) \end{array}$

The previous expression means $X_{M + 1} = T_{M + 1}$

M ? q = Count of ${X_{1} \dots X_{M}} \in T$ . The following expression means $X_{M + 1} = K_{M + 1}$

à Match the Form $(T_{1} + K_{1}) \dots (T_{M} + K_{M}) {T_{M + 1} + K_{M + 1}}$ .

When $P T 1 = [P T, T_{M + 1} = 2 + T_{M}]$

$\begin{array}{l} S U M (N, P S 1, P T 1) \\ = \sum_{n = 0}^{N - 1} (K_{M + 1} + n \times D_{M + 1}) \times S U M (n + 1, P S, P T) \\ = \sum_{n = 0}^{N - 1} (K_{M + 1} + n \times D_{M + 1}) \times \sum X_{1} \dots X_{M} (\begin{matrix} n + 1 + P B (P T) \\ I D X (P T) - q \end{matrix}) \overset{(1)}{\to} \\ = \sum X_{1} \dots X_{M} D_{M + 1} (I D X (P T) - q + 1) (\begin{matrix} N + 1 + P B (P T) \\ I D X (P T) - q + 2 \end{matrix}) \end{array}$

$\begin{array}{l} + \sum X_{1} \dots X_{M} {K_{M + 1} + (I D X (P T) - q - 1 - P B (P T)) D_{M + 1}} \\ \times (\begin{matrix} N + 1 + P B (P T) \\ I D X (P T) - q + 1 \end{matrix}) \overset{P B (P T 1) = P B (P T) + 1, I D X (P T 1) = I D X (P T) + 2}{\to} \end{array}$

$\begin{array}{l} = \sum X_{1} \dots X_{M} D_{M + 1} (T_{M + 1} - q) (\begin{matrix} N + P B (P T 1) \\ I D X (P T 1) - q \end{matrix}) \\ + \sum X_{1} \dots X_{M} {K_{M + 1} + (M - q) D_{M + 1}} (\begin{matrix} N + P B (P T 1) \\ I D X (P T 1) - (q + 1) \end{matrix}) \end{array}$

à Match the Form $(T_{1} + K_{1}) \dots (T_{M} + K_{M}) {T_{M + 1} + K_{M + 1}}$ .

q.e.d.

The proof process can be extended to ring.

Example 3.1:

$P S = [7 : 13, - 7.7 : 2.2, 15 : - 23]$ , $P T = [1, 3, 5]$ , ${Form}_{1} = (1 + 7) (3 - 7.7) (5 + 15)$

$\begin{array}{l} S U M (N, P S, P T) \\ = - 9867.0 (\begin{matrix} N + 2 \\ 6 \end{matrix}) + 1084.6 (\begin{matrix} N + 2 \\ 5 \end{matrix}) + 4044.7 (\begin{matrix} N + 2 \\ 4 \end{matrix}) - 808.5 (\begin{matrix} N + 2 \\ 3 \end{matrix}) \end{array}$

$- 808.5 = 7 \times (- 7.7) \times 15$

$\begin{matrix} 4044.7 = [(1 - 0) \times 13] \times (- 7.7 + 2.2 \times 1) \times (15 - 23 \times 1) \\ + 7 \times [(3 - 1) \times 2.2] \times (15 - 23 \times 1) \\ + 7 \times (- 7.7) \times [(5 - 2) \times (- 23)] \end{matrix}$

$\begin{matrix} 1084.6 = [(1 - 0) \times 13] \times [(3 - 0) \times 2.2] \times (15 - 23 \times 2) \\ + [(1 - 0) \times 13] \times (- 7.7 + 2.2 \times 1) \times [(5 - 1) \times (- 23)] \\ + 7 \times [(3 - 1) \times 2.2] \times [(5 - 1) \times (- 23)] \end{matrix}$

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

$\begin{array}{l} S U M (2, P S, P T) \\ = 7 \times (- 7.7) \times 15 + 7 \times (- 7.7 - 5.5) \times (- 8) + 20 \times (- 5.5) \times (- 8) \\ = 4044.7 (\begin{matrix} 4 \\ 4 \end{matrix}) - 808.5 \times (\begin{matrix} 4 \\ 3 \end{matrix}) = 810.7 \end{array}$

$\begin{array}{l} S U M (3, P S, P T) \\ = S U M (2, P S, P T) + 7 \times (- 7.7 - 5.5 - 3.3) \times (- 31) \\ + 20 \times (- 5.5 - 3.3) \times (- 31) + 33 \times (- 3.3) \times (- 31) \\ = 1084.6 (\begin{matrix} 5 \\ 5 \end{matrix}) + 4044.7 (\begin{matrix} 5 \\ 4 \end{matrix}) - 808.5 (\begin{matrix} 5 \\ 3 \end{matrix}) = 13223.1 \end{array}$

$\begin{array}{l} S U M (4, P S, P T) \\ = S U M (3, P S, P T) + 7 \times (- 7.7 - 5.5 - 3.3 - 1.1) \times {(- 54)}_{_{}}^{^{}} \\ + 20 \times (- 5.5 - 3.3 - 1.1) \times (- 54) + 33 \times (- 3.3 - 1.1) \times {(- 54)}_{_{}}^{^{}} \\ + 46 \times (- 1.1) \times {(- 54)}_{_{}} \\ = - 9867.0 (\begin{matrix} 6 \\ 6 \end{matrix}) + 1084.6 (\begin{matrix} 6 \\ 5 \end{matrix}) + 4044.7 (\begin{matrix} 6 \\ 4 \end{matrix}) - 808.5 (\begin{matrix} 6 \\ 3 \end{matrix}) = 41141.1 \end{array}$

Example 3.2:

$P S = [(\begin{matrix} 7 & 3 \\ 1 & 2 \end{matrix}) : (\begin{matrix} 1 & 3 \\ 4 & 2 \end{matrix}), (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}) : (\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix})]$ , $P T = [1, 3]$

${Form}_{1} = (1 + (\begin{matrix} 7 & 3 \\ 1 & 2 \end{matrix})) (3 + (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}))$

$(\begin{matrix} 17 & 13 \\ 4 & 5 \end{matrix}) = (\begin{matrix} 7 & 3 \\ 1 & 2 \end{matrix}) (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})$

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

$(\begin{matrix} 15 & 27 \\ 30 & 48 \end{matrix}) = 1 \times (\begin{matrix} 1 & 3 \\ 4 & 2 \end{matrix}) \times 3 \times (\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix})$

$S U M (N, P S, P T) = (\begin{matrix} 15 & 27 \\ 30 & 48 \end{matrix}) (\begin{matrix} N + 1 \\ 4 \end{matrix}) + (\begin{matrix} 44 & 70 \\ 28 & 38 \end{matrix}) (\begin{matrix} N + 1 \\ 3 \end{matrix}) + (\begin{matrix} 17 & 13 \\ 4 & 5 \end{matrix}) (\begin{matrix} N + 1 \\ 2 \end{matrix})$

$\begin{matrix} S U M (2, P S, P T) = (\begin{matrix} 7 & 3 \\ 1 & 2 \end{matrix}) (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}) + [(\begin{matrix} 7 & 3 \\ 1 & 2 \end{matrix}) + (\begin{matrix} 8 & 6 \\ 5 & 4 \end{matrix})] (\begin{matrix} 4 & 4 \\ 2 & 4 \end{matrix}) \\ = (\begin{matrix} 2 + 1 \\ 3 \end{matrix}) (\begin{matrix} 44 & 70 \\ 28 & 38 \end{matrix}) + (\begin{matrix} 2 + 1 \\ 2 \end{matrix}) (\begin{matrix} 17 & 13 \\ 4 & 5 \end{matrix}) \\ = (\begin{matrix} 95 & 109 \\ 40 & 53 \end{matrix}) \end{matrix}$

$\begin{matrix} S U M (3, P S, P T) = S U M (2, P S, P T) + [(\begin{matrix} 7 & 3 \\ 1 & 2 \end{matrix}) + (\begin{matrix} 8 & 6 \\ 5 & 4 \end{matrix}) + (\begin{matrix} 9 & 9 \\ 9 & 6 \end{matrix})] (\begin{matrix} 6 & 7 \\ 3 & 6 \end{matrix}) \\ = (\begin{matrix} 3 + 1 \\ 4 \end{matrix}) (\begin{matrix} 15 & 27 \\ 30 & 48 \end{matrix}) + (\begin{matrix} 3 + 1 \\ 3 \end{matrix}) (\begin{matrix} 44 & 70 \\ 28 & 38 \end{matrix}) + (\begin{matrix} 3 + 1 \\ 2 \end{matrix}) (\begin{matrix} 17 & 13 \\ 4 & 5 \end{matrix}) \\ = (\begin{matrix} 293 & 385 \\ 166 & 230 \end{matrix}) \end{matrix}$

A product = $(K_{1} + E_{1}) \dots (K_{M} + E_{M}) = \sum F_{1} \dots F_{M}$ , $F_{i} = E_{i}$ or $K_{i}$

Here’s another extension: Let $F_{i} = E_{i}$ or $K_{i}$ or $R_{i}$ , $R_{i}$ means $(K_{i} + E_{i})$

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

$S U M_K (N, P S, P T, P F = F_{1} F_{2} \dots F_{M})$ = Sum of one part, PF indicates the part.

Rewrite 3.1), add {braces}:

$S U M (N, P S, P T) = \sum \prod_{i = 1}^{M} A_{q} (\begin{matrix} A \\ M_{q} \end{matrix})$ , $A_{q} = \prod_{i = 1}^{M} B_{i}$

$B_{i} = {\begin{cases} {(T_{i} - m) D_{i}}; X_{i} = T_{i}, m = count of {\dots, X_{i - 1}} \in K \\ {K_{i}} + {m D_{i}}; F_{i} \neq R_{i}, X_{i} = K_{i}, m = count of {\dots, X_{i - 1}} \in T \\ {K_{i} + m D_{i}}; F_{i} = R_{i}, X_{i} = K_{i}, m = count of {\dots, X_{i - 1}} \in T \end{cases}$

Expand SUM() by {braces}:

3.2) $S U M_K (N, P S, P T, P F) = \sum Expansion with same {K_{i}, R_{i}} \in P F$

Use the similar method of [2] to prove.

4. Coefficient Analysis

Define $H_{1} (P S, P T, C) = A_{M - C}$ of 3.1), C = Count of $X \in T$ , $T \in Ring$ .

It’s the coefficient in the expression. Here $T \in R i n g$ is just for analysis.

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

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

$F_{M}^{{1, 2, \dots, N}}$ is abbreviated as $F_{M}^{N}$ , $E_{M}^{{1, 2, \dots, N}}$ is abbreviated as $E_{M}^{N}$ ;

By definition:

$E_{M}^{N} = \sum_{N_{1} + \dots + N_{C} = M} 1^{N_{1}} 2^{N_{2}} \dots N^{N_{C}}$

$F_{0}^{K} = E_{0}^{K} = 1; F_{M > | K |}^{K} = 0$

$E_{M}^{N + 1} = (N + 1) E_{M - 1}^{N + 1} + E_{M}^{N}; F_{M}^{[K, K_{M + 1}]} = K_{M + 1} F_{M - 1}^{K} + F_{N}^{K}$

4.1) $H_{1} (K, T, 0) = \prod K_{i}$ , $H_{1} (K, T, M) = \prod T_{i} \prod D_{i}$

[4] has proved:

4.2) $H_{1} ([D \times A : D, K_{1} : D_{1}, \dots, K_{M} : D_{M}], [A, T_{1}, \dots, T_{M}], C)$

$\begin{array}{l} = D \times A \times H_{1} ([K_{1} : D_{1}, \dots, K_{M} : D_{M}], [T_{1}, \dots, T_{M}], C) \\ + D \times A \times H_{1} ([K_{1} : D_{1}, \dots, K_{M} : D_{M}], [T_{1}, \dots, T_{M}], C - 1) \end{array}$

this à

4.3) $S U M (N, [1, 2, \dots, n, K_{1} : D_{1}, \dots, K_{M} : D_{M}], [1, 2, \dots, n, T_{1}, \dots, T_{M}])$ , $P T = [1, \dots, n, T_{1}, \dots, T_{M}]$ can use the form:

$(T_{1} + K_{1}) \dots (T_{M} + K_{M}) = n! \sum A_{q} (\begin{matrix} N + P B (P T) + n \\ I D X (P T) - q \end{matrix})$

[2] has proved:

4.4) $H_{1} ([D \times T_{1} : D, \dots, D \times T_{M} : D], [T_{1}, \dots, T_{M}], C) = D^{M} (\begin{matrix} M \\ C \end{matrix}) \prod T_{i}$

D = 1, this à

4.5) $S U M (N, P T, P T) = M I N (P T) (\begin{matrix} N + P B (P T) + P M (P T) \\ I D X (P T) \end{matrix})$

This is a generalization of $\sum_{n = 0}^{N} (\begin{matrix} n \\ M \end{matrix}) = (\begin{matrix} N + 1 \\ M + 1 \end{matrix})$

Eg: $S U M (2, [1, 2, 4]) = 1 \times 2 \times 4 + 1 \times 2 \times 5 + 2 \times 3 \times 5 = 1 \times 2 \times 4 (\begin{matrix} 2 + 1 + 3 \\ 5 \end{matrix})$

4.6) $H_{1} (P S, P T, C) = \nabla^{C} S U M (C, P S, P T)$ , $\nabla^{I D X (P T)} S U M (N > M, \dots) = \prod T_{i} \prod D_{i}$

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

Comparison with 4.4), [4] has proved:

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

${\begin{cases} K_{i} + 1 = K_{i + 1}, H_{1} ([D \times K_{1} : D, \dots], P T, C) = D^{M} (\begin{matrix} M \\ C \end{matrix}) T_{1} \dots T_{C} K_{C + 1} \dots K_{M} \\ K_{i} - 1 = K_{i + 1}, H_{1} ([D \times K_{1} : D, \dots], P T, C) = D^{M} (\begin{matrix} M \\ C \end{matrix}) T_{1} \dots T_{C} K_{1} \dots K_{M - C} \\ K_{i} + 1 = K_{i + 1}, H_{1} ([D \times K_{1} : - D, \dots], P T, C) = {(- 1)}^{C} D^{M} (\begin{matrix} M \\ C \end{matrix}) T_{1} \dots T_{C} K_{1} \dots K_{M - C} \\ K_{i} - 1 = K_{i + 1}, H_{1} ([D \times K_{1} : - D, \dots], P T, C) = {(- 1)}^{C} D^{M} (\begin{matrix} M \\ C \end{matrix}) T_{1} \dots T_{C} K_{C + 1} \dots K_{M} \end{cases}$

$\begin{matrix} {[x + y]}_{n} = \nabla S U M (y + 1, [x - n + 1, x - n + 2, \dots, x]) \\ = (\begin{matrix} y \\ n \end{matrix}) (\begin{matrix} n \\ n \end{matrix}) n! {[x]}_{0} + (\begin{matrix} y \\ n - 1 \end{matrix}) (\begin{matrix} n \\ n - 1 \end{matrix}) (n - 1)! {[x]}_{1} + \dots + (\begin{matrix} y \\ 0 \end{matrix}) (\begin{matrix} n \\ 0 \end{matrix}) 0! {[x]}_{n} \\ = (\begin{matrix} n \\ n \end{matrix}) {[y]}_{n} {[x]}_{0} + (\begin{matrix} n \\ n - 1 \end{matrix}) {[y]}_{n - 1} {[x]}_{1} + \dots + (\begin{matrix} n \\ 0 \end{matrix}) {[y]}_{0} {[x]}_{n} \end{matrix}$

à Vandermonde identity [5]: ${[x + y]}_{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) {[x]}_{k} {[y]}_{n - k}$

$\begin{matrix} {[x + y]}^{n} = \nabla S U M (- y + 1, [x : - 1, x + 1 : - 1, \dots, x + n - 1 : - 1]) \\ = {(- 1)}^{n} (\begin{matrix} - y \\ n \end{matrix}) (\begin{matrix} n \\ n \end{matrix}) n! {[x]}^{0} + {(- 1)}^{n - 1} (\begin{matrix} - y \\ n - 1 \end{matrix}) (\begin{matrix} n \\ n - 1 \end{matrix}) (n - 1)! {[x]}^{1} + \dots \\ = (\begin{matrix} n \\ n \end{matrix}) {[y]}^{n} {[x]}^{0} + (\begin{matrix} n \\ n - 1 \end{matrix}) {[y]}^{n - 1} {[x]}^{1} + \dots + (\begin{matrix} n \\ 0 \end{matrix}) {[y]}^{0} {[x]}^{n} \end{matrix}$

à Norlund identity [5]: ${[x + y]}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) {[x]}^{k} {[y]}^{n - k}$

$\begin{array}{l} (K_{1} + N \times D_{1}) \dots (K_{M} + N \times D_{M}) \\ = \nabla S U M (N + 1, [K_{1} : D_{1} .. K_{M} : D_{M}]) = \sum_{q = 0}^{M} A_{q} ( N q ) \end{array}$

4.8) $(K_{1} + N \times D_{1}) \dots (K_{M} + N \times D_{M})$ can be decomposed to $\sum_{q = 0}^{M} C_{q} (\begin{matrix} N \\ q \end{matrix})$ by 3.1)

$N = x$ , $K_{i} = i - 1$ , $P S = [0, 1, 2, \dots, M - 1]$ à

$\begin{array}{l} {[x]}^{M} = (\begin{matrix} x \\ M \end{matrix}) (\begin{matrix} M \\ M \end{matrix}) M! {[M - 1]}_{0} \\ + (\begin{matrix} x \\ M - 1 \end{matrix}) (\begin{matrix} M \\ M - 1 \end{matrix}) (M - 1)! {[M - 1]}_{1} \dots (\begin{matrix} x \\ 0 \end{matrix}) (\begin{matrix} M \\ 0 \end{matrix}) 0! {[M - 1]}_{M} \end{array}$

4.9) ${[x]}^{M} = \sum_{k = 0}^{M} (\begin{matrix} M \\ k \end{matrix}) {[x]}_{k} {[M - 1]}_{M - k}$

$N = x$ , $K_{i} = 1 - i$ , $P S = [0 : - 1, - 1 : - 1, \dots, - M + 1 : - 1]$ à

$\begin{array}{l} {[- x]}_{M} = {(- 1)}^{M + 0} (\begin{matrix} X \\ M \end{matrix}) (\begin{matrix} M \\ M \end{matrix}) M! {[M - 1]}_{0} \\ + {(- 1)}^{(M - 1) + 1} (\begin{matrix} X \\ M - 1 \end{matrix}) (\begin{matrix} M \\ M - 1 \end{matrix}) (M - 1)! {[M - 1]}_{1} \end{array}$

4.10) ${[- x]}_{M} = \sum_{k = 0}^{M} {(- 1)}^{M} (\begin{matrix} M \\ k \end{matrix}) {[x]}_{k} {[M - 1]}_{M - K}$

[4] has proved:

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

$\begin{array}{l} H_{1} ([D \times K_{1} : D, \dots, D \times K_{M} : D], [T_{1}, \dots, T_{M}], C) \\ = D^{C} T_{1} \dots T_{C} [F_{M - C}^{D \times K} E_{0}^{D \times {1, 2, \dots, C}} + F_{M - C - 1}^{D \times K} E_{1}^{D \times {1, 2, \dots, C}} + \dots + F_{0}^{D \times K} E_{M - C}^{D \times {1, 2, \dots, C}}] \end{array}$

This à K_i can exchange order in $S U M (N, [K_{1}, \dots, K_{M}])$ .

In fact, K_i can exchange order in $S U M (N, [K_{1} : D_{1}, \dots, K_{M} : D_{M}])$ by definition.

This à

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

$\begin{array}{l} S U M (N, [λ^{1}, λ^{2}, \dots, λ^{M}]) \\ = M! (\begin{matrix} N \\ M + 1 \end{matrix}) E_{0}^{M} + (M - 1)! (\begin{matrix} N \\ M \end{matrix}) E_{1}^{M - 1} + \dots \\ + 1! (\begin{matrix} N \\ 2 \end{matrix}) E_{M - 1}^{1} + 0! (\begin{matrix} N \\ 1 \end{matrix}) {(- 1)}^{M + 1} \end{array}$

$(λ^{1} + 1) \dots (λ^{M} + 1) = S U M (2, [\dots]) - S U M (1, [\dots]) = 1 + {(- 1)}^{M + 1}$

It’s obvious when M is even; if M is odd then $(λ^{1} + 1) \dots (λ^{M - 1} + 1) = 1$

$(λ^{1} + 2) \dots (λ^{M} + 2) = S U M (3, [\dots]) - S U M (2, [\dots]) = 2^{M} + {(- 1)}^{M + 1}$

It can be concluded from the definition:

4.13) 1) $H_{1} ([1, \dots, M], [1, \dots, 2 M - 1], C)$

$= \sum_{P M () = M, P B () = C} M I N (P S) + \sum_{P M () = M, P B () = C - 1} M I N ( P S )$

2) $H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], C) = \sum_{P M (P S) = M, P B (P S) = C} M I N ( P S )$

PS are Basic Shapes.

5. Special Functions

5.1) $S U M (N, [a : 0]) = a ( N 1 )$

5.2) $S U M (N, [a : d]) = \sum_{n = 0}^{N - 1} (a + n d) = d (\begin{matrix} N \\ 2 \end{matrix}) + a ( N 1 )$

5.3) $S U M (N, [1, 2, \dots, M]) = M! (\begin{matrix} N + M \\ M + 1 \end{matrix})$

5.4) $S U M (N, [1, 2, \dots, M], [1, 3, \dots, 2 M - 1]) = \sum I_{1} I_{2} \dots I_{M}$ , $1 \leq I_{1} < I_{2} < \dots < I_{M} \leq N + M - 1$

5.5) $S U M (N, [1, 1, \dots, 1], [1, 2, \dots, M]) = 1^{M} + 2^{M} + \dots + N^{M}$

5.6) $S U M (N, [0 : D_{1}, \dots, 0 : D_{M}], P T) = S U M_K (N, P S, P T, E_{1} \dots E_{M})$

5.7) $S U M (N, [1, 1, \dots, 1], [1, 3, \dots, 2 M - 1]) = E_{M}^{N}$

6. Stirling, Lah Number

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

[1] use 4.5) to calculate

$S_{1} (N, N - M) = F_{M}^{N - 1} = \sum M I N (P S) (\begin{matrix} N + P B (P T) + P M (P T) \\ I D X (P T) \end{matrix})$ , PS traverses all Basic Shapes, $P M (P S) = M$

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

6.1) $S U M (N, [1, 2, \dots, M], [1, 3, \dots, 2 M - 1]) = S_{1} (N + M, N)$ , this is 5.4)

$E_{M}^{N} \overset{d e f}{\to} \sum_{N_{1} + \dots + N_{C} = M} 1^{N_{1}} 2^{N_{2}} \dots N^{N_{C}}$ , It’s a known property of $S_{2} (N + M, N)$

6.2) $S U M (N, [1, \dots, 1], [1, 3, \dots, 2 M - 1]) = E_{M}^{N} = S_{2} (N + M, N)$

6.3) 1) $H_{1} ([1, \dots, 1], [1, \dots, M], K)$

$= K! S_{2} (M + 1, K + 1) = K! \sum_{i = 0}^{M - K} (\begin{matrix} M \\ i \end{matrix}) S_{2} (M - i, K)$

2) $H_{1} ([1, \dots, 1], [2, \dots, M], K) = (K + 1)! S_{2} (M, K + 1)$

[Proof]

Definition of S₂(M, K) is

$N^{M} = \sum_{K = 0}^{M} S_{2} (M, K) {[N]}_{k} = \sum_{K = 0}^{M} K! S_{2} (M, K) ( N K )$

$\begin{matrix} {(N + 1)}^{M} = \nabla S U M (N + 1, [1, \dots, 1], [1, \dots, M]) \\ = \sum_{q = 0}^{M} A_{q} (\begin{matrix} N \\ q \end{matrix}) = \sum_{K = 0}^{M} K! S_{2} (M, K) (\begin{matrix} N + 1 \\ K \end{matrix}) \\ = \sum_{K = 0}^{M} K! S_{2} (M, K) (\begin{matrix} N \\ K + 1 \end{matrix}) + \sum_{K = 0}^{M} K! S_{2} (M, K) (\begin{matrix} N \\ K \end{matrix}) \end{matrix}$ à

$H_{1} ([1, \dots, 1], [1, \dots, M], K) = K! S_{2} (M, K) + (K + 1)! S_{2} (M, K + 1)$ à the first equation

$\overset{4.11)}{\to} H_{1} (\dots, K) = K! [F_{M - K}^{[1, \dots, 1]} E_{0}^{K} + \dots + F_{0}^{[1, \dots, 1]} E_{M - K}^{K}]$ $\overset{F_{M - K}^{[1, \dots, 1]} = (\begin{matrix} M \\ M - K \end{matrix})}{\to}$ the second equation

$\begin{array}{l} \overset{4.2)}{\to} H_{1} ([1, \dots, 1], [1, \dots, M], K) \\ = H_{1} ([1, \dots, 1], [2, \dots, M], K) + H_{1} ([2, \dots, M], K - 1) \end{array}$ à 2)

q.e.d.

This à $S_{2} (N + 1, M) = \sum_{k = M - 1}^{N} (\begin{matrix} N \\ k \end{matrix}) S_{2} (K, M - 1)$ , which is recorded in [5]

Example 6.1:

Directly according to the definition of H₁()

$\begin{array}{l} H_{1} (\dots, 1) = (1 \times 2 \times 2 \times \dots \times 2 + 1 \times 1 \times 2 \times \dots \times 2 + \dots + 1 \times 1 \times \dots \times 1 \times 1) \\ \to S_{2} (M + 1, 2) = 2^{M} - 1 \end{array}$

$H_{1} (\dots, M - 1) = (M - 1)! (1 + 2 + \dots + M) \to S_{2} (M + 1, M) = (\begin{matrix} M + 1 \\ 2 \end{matrix})$

6.4) $(\begin{matrix} M \\ K \end{matrix}) \frac{M!}{K!} = S_{1} (M + 1, K + 1) S_{2} (K, K) + \dots + S_{1} (M + 1, M + 1) S_{2} (M, K)$

[Proof]

$\begin{array}{l} H_{1} ([1, \dots, M], [1, \dots, M], K) \overset{4.4)}{\to} = (\begin{matrix} M \\ K \end{matrix}) M! \\ \overset{4.11)}{\to} = K! [F_{M - K}^{M} E_{0}^{K} + \dots + F_{0}^{M} E_{M - K}^{K}] \\ = K! [S_{1} (M + 1, K + 1) S_{2} (K, K) + \dots + S_{1} (M + 1, M + 1) S_{2} (M, K)] \end{array}$

q.e.d.

Definition of Lah number [5] is ${[- X]}_{M} = \sum_{k = 0}^{M} L_{M, K} {[x]}_{k} \overset{4.10)}{\to}$

6.5) $L_{M, K} = {(- 1)}^{M} \frac{M!}{K!} (\begin{matrix} M - 1 \\ K - 1 \end{matrix})$ , this is recorded in [5].

7. Congruence Analysis

P is a prime number, we already know:

(7*) $S_{1} (P, P - K) = F_{K}^{P - 1} \equiv 0 M O D P, 0 < K < P - 1$

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

7.1) $S U M (P - P B (P T) - P M (P T), P T, P T)$

$= M I N (P T) (\begin{matrix} P \\ I D X (P T) \end{matrix}) \equiv 0 M O D P, I D X (P T) < P$

E.g.: $1 \times 2 + 2 \times 3 + 3 \times 4 \equiv 1 \times 3 + 1 \times 4 + 2 \times 4 \equiv 0 M O D 5$

This is the promotion of (7*).

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

7.2) For arbitrary $K_{i} \in Z$

1) $M < P - 1$ , $S U M (P, [K_{1} : D, \dots, K_{M} : D]) \equiv 0 M O D P$

2) $M = P - 1$ , $(D, P) = 1$ , $S U M (P, [K_{1} : D, \dots, K_{M} : D]) \equiv - 1 M O D P$

$S U M (P, [K_{1}, \dots, K_{M}]) = \prod K_{i} + \prod (1 + K_{i}) + \dots + \prod (P - 1 + K_{i})$ . Exclude products≡0 MOD P:

Example 7.1:

$1^{Q} + 2^{Q} + \dots + {(P - 1)}^{Q} \equiv 0 M O D P$ , $Q < P - 1$ ; $\equiv - 1 M O D P$ , $Q = P - 1$

$1^{2} \times 2 \times 3 + 2^{2} \times 3 \times 4 + \dots + {(P - 3)}^{2} \times (P - 2) \times (P - 1) \equiv - 1 M O D P$ , $P = 5$ ; $\equiv 0 M O D P$ , $P > 5$

$1^{3} \times 2 + 2^{3} \times 3 + \dots + {(P - 2)}^{3} \times (P - 1) \equiv - 1 M O D P$ , $P = 5$ ; $\equiv 0 M O D P$ , $P > 5$

$1 \times 2^{3} + 2 \times 3^{3} + \dots + (P - 2) \times {(P - 1)}^{3} \equiv - 1 M O D P$ , $P = 5$ ; $\equiv 0 M O D P$ , $P > 5$

$1^{2} \times 2^{2} + 2^{2} \times 3^{2} + \dots + {(P - 2)}^{2} \times {(P - 1)}^{2} \equiv - 1 M O D P$ , $P = 5$ ; $\equiv 0 M O D P$ , $P > 5$

In 1), let $P S = [K_{1} = 1, \dots, K_{M}]$ is a Basic Shape and $K_{M} = P - 1$

Rewrite $P S = 〚 L_{1} L_{2} \dots L_{Q} 〛$ , $L_{i}$ = count of continuity. $(L_{i}, L_{i + 1})$ means a discontinuity.

$〚 L_{1} \dots L_{Q} 〛$ can been slided to $[L_{Q}, L_{1}, L_{2}, \dots]$ , $[L_{Q - 1}, L_{Q}, L_{1}, L_{2}, \dots]$ by $S U M (P, P S) M O D P$

[3] has proved:

7.3) $\sum M I N (P S) \equiv 0 M O D P$

{PS} = All of the Basic Shapes $[1, \dots, K_{M} = P - 1] \neq [1, 2, \dots, P - 1]$ can slide to.

Example 7.2:

$\begin{array}{l} 1 \times (3 \times 4 \times 5) \times (7 \times 8) \times 10 + 1 \times 3 \times (5 \times 6 \times 7) \times (9 \times 10) \\ + (1 \times 2) \times 4 \times 6 \times (8 \times 9 \times 10) + (1 \times 2 \times 3) \times (5 \times 6) \times 8 \times 10 \\ = 139260 \equiv 0 M O D 11 \end{array}$

This à [1] has proved:

7.4) $\sum M I N (P S) \equiv 0 M O D P$ , {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.: $(1^{2} \times 2^{2} + 2^{2} \times 3^{2} + 3^{3} \times 4^{2}) + (1^{2} \times 3^{2} + 2^{2} \times 4^{2} + 4^{2} \times 1^{2})$

$\equiv S U M (5, [1, 1, 2, 2]) + S U M (5, [1, 1, 3, 3]) M O D 5$

Use F(PS) = The symmetrical express. Obviously: $F ([K_{1} \dots K_{M}]) \equiv 0 M O D P$ , $M < P - 1$ ;

7.5) $F (λ (P S)) \equiv - \frac{(\begin{matrix} P - 1 \\ L E N (P S) \end{matrix}) \frac{(μ_{1} + μ_{2} + \dots)!}{μ_{1}! μ_{2}! \dots}}{P - L E N (P S)} M O D P, M = P - 1$

[Proof]

Use CNT(PS) = Count of $P S \in F (P S)$ à $F (P S) \equiv - C N T (P S) M O D P$

Use LEN(PS) =Count of different $K_{i} \in P S$

Obviously:

$L E N (P S \in F (P S))$ is same, count of products $\in$ F(PS) = P-LEN(PS)

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

Use $λ (P S)$ to classify PS.

$λ (P S) : = 1^{λ_{1}} 2^{λ_{2}} \dots {(P - 1)}^{λ_{p - 1}}, P M (P S) = 1 \times λ_{1} + 2 \times λ_{2} + \dots + (P - 1) \times λ_{p - 1}$

$λ (P S) : = 4^{1} : 1^{4} + 2^{4} + 3^{4} + \dots$

$λ (P S) : = 1^{4} : 1 \times 2 \times 3 \times 4 + \dots$

$\begin{array}{l} λ (P S) : = 1^{1} 3^{1} : 1^{3} \times 2 + 2^{3} \times 3 + \dots; 1^{3} \times 3 + 2^{3} \times 4 + \dots; \\ 1 \times 2^{3} + 2 \times 3^{3} + \dots; 1 \times 3^{3} + 2 \times 4^{3} + \dots \end{array}$

$λ (P S) : = 2^{2} : 1^{2} \times 2^{2} + 2^{2} \times 3^{2} + \dots; 1^{2} \times 3^{2} + 2^{2} \times 4^{2} + \dots; 1^{2} \times 4^{2} + 2^{2} \times 5^{2} + \dots$

$λ (P S) : = 1^{2} 2^{1} : 1^{2} \times 2 \times 3 + \dots; 1 \times 2^{2} \times 3 + \dots; 1 \times 2 \times 3^{2} + \dots; 1^{2} \times 3 \times 5 + \dots$

1) The combination number of LEN(PS) in $P - 1 = (\begin{matrix} P - 1 \\ L E N (P S) \end{matrix})$

2) Record $λ (P S)$ as ${μ_{1}, μ_{2}, \dots}$

There are $μ_{1}$ numbers in $λ_{1}, λ_{2}, \dots$ equal to $x > 0$

There are $μ_{2}$ numbers in $λ_{1}, λ_{2}, \dots$ equal to $y > 0, y \neq x$

The combination number of $μ_{1}, μ_{2}, \dots$ = $\frac{(μ_{1} + μ_{2} + \dots)!}{μ_{1}! μ_{2}! \dots}$

Count of products $\in$ F(PS) = $(\begin{matrix} P - 1 \\ L E N (P S) \end{matrix}) \frac{(μ_{1} + μ_{2} + \dots)!}{μ_{1}! μ_{2}! \dots}$

q.e.d.

Example 7.3:

$\begin{array}{l} (1^{3} \times 2 + 2^{3} \times 3 + 3^{3} \times 4) + (1^{3} \times 3 + 2^{3} \times 4 + 4^{3} \times 1) + (1^{3} \times 4 + 3^{3} \times 1 + 4^{3} \times 2) \\ + (1 \times 2^{3} + 2 \times 3^{3} + 3 \times 4^{3}) \equiv - \frac{4!}{2! 2!} \times \frac{2!}{1! 1!} / (5 - 2) \equiv - 4 M O D P \end{array}$

$\begin{array}{l} (1^{2} \times 2^{2} + 2^{2} \times 3^{2} + 3^{2} \times 4^{2}) + (1^{2} \times 3^{2} + 2^{2} \times 4^{2} + 4^{2} \times 1^{2}) \\ \equiv - \frac{4!}{2! 2!} \times \frac{2!}{2!} / (5 - 2) \equiv - 2 M O D P \end{array}$

$\begin{array}{l} (1^{2} \times 2 \times 3 + 2^{2} \times 3 \times 4) + (1^{2} \times 2 \times 4 + 3^{2} \times 4 \times 1) + (1^{2} \times 3 \times 4 + 4^{2} \times 1 \times 2) \\ + (1 \times 2^{2} \times 3 + 2 \times 3^{2} \times 4) + (1 \times 2^{2} \times 4 + 3 \times 4^{2} \times 1) + (1 \times 2 \times 3^{2} + 2 \times 3 \times 4^{2}) \\ \equiv - \frac{4!}{3! 1!} \times \frac{3!}{2! 1!} / (5 - 3) \equiv - 6 M O D P \end{array}$

7.6) 1) $K! S_{2} (P - 1, K) = K! E_{P - 1 - K}^{K} \equiv {(- 1)}^{K + 1} M O D P, 1 \leq K \leq P - 1$

2) $S_{2} (P, K) \equiv 0 M O D P, 1 < K < P$

[Proof]

$\begin{array}{l} S U M (N, [1, 2, \dots, P - 1]) \overset{4.11) (7 *)}{\to} \\ \equiv (P - 1)! (\begin{matrix} N \\ P \end{matrix}) E_{0}^{P - 1} + \dots + 1! (\begin{matrix} N \\ 2 \end{matrix}) E_{P - 2}^{1} + 0! (\begin{matrix} N \\ 1 \end{matrix}) (P - 1)! M O D P \end{array}$

$\begin{array}{l} \nabla S U M (N, [1, \dots, P - 1]) \\ \equiv (P - 1)! (\begin{matrix} N - 1 \\ P - 1 \end{matrix}) E_{0}^{P - 1} + \dots + 1! (\begin{matrix} N - 1 \\ 1 \end{matrix}) E_{P - 2}^{1} + 0! (\begin{matrix} N - 1 \\ 0 \end{matrix}) (p - 1)! M O D P \end{array}$

$\begin{array}{l} {[P]}_{P - 1} = \nabla S U M (2, [1, \dots, P - 1]) \equiv 1 + (P - 1)! \equiv 0 M O D P \\ \to (P - 1)! \equiv - 1 M O D P \end{array}$

This is a new proof of Wilson theorem.

$1! E_{P - 2}^{1} = 1 \equiv 1 M O D P$

${[P + 1]}_{P - 1} = \nabla S U M (3, [1, \dots, P - 1]) \equiv 0 M O D P \to 2! E_{P - 3}^{2} \equiv - 1 M O D P$

${[P + 2]}_{P - 1} = \nabla S U M (4, [1, \dots, P - 1]) \equiv 0 M O D P \to 3! E_{P - 4}^{3} \equiv 1 M O D P$

$\dots$

à 1)

$S_{2} (P, K) = S_{2} (P - 1, K - 1) + K \times S_{2} (P - 1, K)$ à

$\begin{array}{l} (K - 1)! S_{2} (P, K) = (K - 1)! S_{2} (P - 1, K - 1) + K! S_{2} (P - 1, K) \\ \equiv {(- 1)}^{K} + {(- 1)}^{K + 1} \equiv 0 M O D P \end{array}$ à 2)

q.e.d.

$N^{P - 1} \equiv 1 M O D P = \sum_{K = 0}^{P - 1} K! S_{2} (P - 1, K) (\begin{matrix} N \\ K \end{matrix}) \overset{7.6), S_{2} (P - 1, 0) = 0}{\to}$

7.7) $- (\begin{matrix} N \\ P - 1 \end{matrix}) + (\begin{matrix} N \\ P - 2 \end{matrix}) + \dots - (\begin{matrix} N \\ 2 \end{matrix}) + (\begin{matrix} N \\ 1 \end{matrix}) \equiv 1 M O D P, (N, P) = 1$

7.8) $\sum_{n = 1}^{P - 1} n^{P - 2} \equiv 0 M O D P^{2}, P > 3$

[Proof]

$\begin{array}{l} P^{P - 2} + \sum_{n = 1}^{P - 1} n^{P - 2} = \sum_{n = 1}^{P} n^{P - 2} = S U M (P, [1, \dots, 1], [1, \dots, P - 2]) \\ \overset{6.3)}{\to} = \sum_{k = 0}^{P - 2} k! S_{2} (P - 1, k + 1) (\begin{matrix} P \\ K + 1 \end{matrix}) \\ = \sum_{k = 1}^{\frac{P - 1}{2}} [(k - 1)! S_{2} (P - 1, k) (\begin{matrix} P \\ k \end{matrix}) + (P - k - 1)! S_{2} (P - 1, P - k) (\begin{matrix} P \\ P - k \end{matrix})] \\ \overset{7.6)}{\to} \equiv \sum_{k = 1}^{\frac{P - 1}{2}} [{(- 1)}^{K + 1} + {(- 1)}^{P - K + 1}] (\begin{matrix} P \\ K \end{matrix}) \\ \equiv \sum_{k = 1}^{\frac{P - 1}{2}} P \times λ_{k} (\begin{matrix} P \\ K \end{matrix}) M O D P \equiv 0 M O D P^{2} \end{array}$

q.e.d.

8. Form₂ and Analysis

Rewrite (1) of section 3 as

$\sum_{n = 0}^{N - 1} n (\begin{matrix} n + K \\ M \end{matrix}) = (M + 1) (\begin{matrix} N + K + 1 \\ M + 2 \end{matrix}) - (1 + K) (\begin{matrix} N + K \\ M + 1 \end{matrix})$

$P S = [K_{1} : D_{1}, \dots, K_{M} : D_{M}]$ , $P T = [T_{1}, \dots, T_{M}]$ , use the same method of 3.1) to prove:

8.1) Use the ${Form}_{2} = (T_{1} + K_{1}) \dots (T_{M} + K_{M}) = \sum X_{1} \dots X_{M}$ ,

$S U M (N, P S, P T) = \sum A_{q} (\begin{matrix} N + T_{M} - q \\ I D X (P T) - q \end{matrix})$ , q = count of $X \in K$ . $A_{q} = \prod_{i = 1}^{M} B_{i}$

$B_{i} = {\begin{cases} (T_{i} - m) D_{i}; X_{i} = T_{i}, m = count of {\dots, X_{i - 1}} \in K \\ K_{i} + (m - T_{i}) D_{i}; X_{i} = K_{i}, m = count of {\dots, X_{i - 1}} \in K \end{cases}$

Define $H_{2} (P S, P T, C) = A_{M - q}$ of 8.1), C = Count of $X \in T$

8.2) $H_{2} (P T, P T, q < P M (P T)) = 0$ ; $H_{2} (P T, P T, P M (P T)) = \prod T_{i}$ , This à 4.5)

8.3) $H_{2} ([1, \dots, M], [1, \dots, 2 M - 1], C)$

$= {(- 1)}^{M - C - 1} H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], C - 1)$

[Proof]

$H_{2} ([1, \dots, 1], [1, \dots, 2 M - 1], C) = \prod_{i = 1}^{M} B_{i}$ ,

$\begin{array}{l} B_{i} = {\begin{cases} (T_{i} - m); X_{i} = T_{i}, m = count of {\dots, X_{i - 1}} \in K \\ 1 + (m - 2 i + 1); X_{i} = K_{i}, m = count of {\dots, X_{i - 1}} \in K \end{cases} \overset{m = i - 1 - q}{\to} \\ = {\begin{cases} (T_{i} - m); X_{i} = T_{i}, m = count of {\dots, X_{i - 1}} \in K \\ - ((i - 1) + q); X_{i} = K_{i}, q = count of {\dots, X_{i - 1}} \in T \end{cases} \end{array}$

$\begin{array}{l} \to {(- 1)}^{M - C} H_{2} (\dots, C) + {(- 1)}^{M - C - 1} H_{2} (\dots, C + 1) \\ = H_{1} ([1, \dots, M], [1, \dots, 2 M - 1], C) \overset{4.2)}{\to} \\ = H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], C) + H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], C - 1) \end{array}$

$\begin{array}{l} \to H_{2} ([1, \dots, M], [1, \dots, 2 M - 1], C) \\ = {(- 1)}^{M - C - 1} H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], C - 1) \end{array}$

q.e.d.

[6] obtains the unified expression of $S_{1} (N, N - M)$ and $S_{2} (N, N - M)$ by induction:

1) $S_{1} (N, N - M) = \sum_{k = 1}^{M} A (M, k) (\begin{matrix} N \\ 2 M - K + 1 \end{matrix})$

2) $S_{2} (N, N - M) = \sum_{k = 1}^{M} {(- 1)}^{k - 1} A (M, k) (\begin{matrix} N + M - K \\ 2 M - K + 1 \end{matrix})$

[Proof]

$S_{1} (N, N - M) = S U M (N - M, [1, \dots, M], [1, \dots, 2 M - 1])$ , ${Form}_{1} = (T_{2} + K_{2}) \dots (T_{M} + K_{M})$ , 4.3) à

$= \sum_{q = 0}^{M - 1} H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], q) (\begin{matrix} N - M + (M - 1) + 1 \\ 2 M - (M - q) \end{matrix}) \overset{k = M - q}{\to}$

$= \sum_{k = 1}^{M} H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], M - k) (\begin{matrix} N \\ 2 M - k \end{matrix})$ à 1)

$H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], M - K) = A (M, K)$

$S_{2} (N, N - M) = S U M (N - M, [1, \dots, 1], [1, \dots, 2 M - 1])$ , use the Form₂

$= \sum_{q = 0}^{M} H_{2} ([1, \dots, 1], [1, \dots, 2 M - 1], q) (\begin{matrix} N - M + (2 M - 1) - (M - q) \\ 2 M - (M - q) \end{matrix})$

$= \sum_{q = 0}^{M} H_{2} ([1, \dots, 1], [1, \dots, 2 M - 1], q) (\begin{matrix} N - 1 + q \\ M + q \end{matrix}) \overset{H_{2} (\dots, 0) = 0}{\to}$

$= \sum_{q = 1}^{M} H_{2} ([1, \dots, 1], [1, \dots, 2 M - 1], q) (\begin{matrix} N - 1 + q \\ M + q \end{matrix}) \overset{K = M - q + 1}{\to}$

$= \sum_{K = 1}^{M} H_{2} ([1, \dots, 1], [1, \dots, 2 M - 1], M - k + 1) (\begin{matrix} N + M - K \\ 2 M - K + 1 \end{matrix}) \overset{8.3)}{\to}$

$= \sum_{k = 1}^{M} {(- 1)}^{k - 1} H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], M - k) (\begin{matrix} N + M - k \\ 2 M - k + 1 \end{matrix})$ à 2)

q.e.d.

8.4) 1) $\sum_{C = 0}^{C = M - 1} {(- 1)}^{M - 1 - C} \sum_{P M (P S) = M, P B (P S) = C} M I N (P S) = 1$

2) $\sum_{C = 0}^{C = M - 1} {(- 1)}^{M - 1 - C} \sum_{P M (P S) = M, P B (P S) = C} (M + 2 + C) M I N (P S) = 2^{M + 1} - 1$

PS are Basic Shapes

[Proof]

$S_{2} (M + 1, 1) = 1$ , this is a known property = $S U M (1, [1, \dots, 1], [1, \dots, 2 M - 1])$ $\overset{{Use Form}_{3}}{\to}$

$\begin{array}{l} = \sum A_{q} (\begin{matrix} N + T_{M} - q \\ I D X (P T) - q \end{matrix}) = \sum A_{q} (\begin{matrix} 1 + 2 M - 1 - q \\ 2 M - q \end{matrix}) = \sum A_{q} \\ \overset{A_{0} = 0}{\to} \sum_{q = 1}^{q = M} H_{2} (\dots, q) \overset{8.3)}{\to} \end{array}$

$= \sum_{C = 0}^{C = M - 1} {(- 1)}^{M - 1 - C} H_{1} ([2, \dots, M], [3, \dots, 2 M - 1], C) \overset{4.13)}{\to}$ 1)

$S_{2} (M + 2, 2) = 2^{M + 1} - 1 = S U M (2, [1, \dots, 1], [1, \dots, 2 M - 1])$ à 2)

q.e.d.

Example 8.1:

$M = 1 : 1 = 1$

$M = 2 : 1 \times 3 - 1 \times 2 = 1$

$M = 3 : 1 \times 3 \times 5 - (1 \times 3 \times 4 + 1 \times 2 \times 4) + 1 \times 2 \times 3 = 1$

$\begin{array}{l} M = 4 : 1 \times 3 \times 5 \times 7 - (1 \times 3 \times 5 + 1 \times 3 \times 4 + 1 \times 2 \times 4) \times 6 \\ + (1 \times 2 \times 3 + 1 \times 3 \times 4 + 1 \times 2 \times 4) \times 5 - 1 \times 2 \times 3 \times 4 = 1 \end{array}$

$\begin{array}{l} M = 5 : 1 \times 3 \times 5 \times 7 \times 9 - (1 \times 3 \times 5 \times 7 + 1 \times 3 \times 4 \times 6 + 1 \times 2 \times 4 \times 6 + 1 \times 3 \times 5 \times 6) \times 8 \\ + (1 \times 2 \times 3 \times 5 + 1 \times 3 \times 4 \times 5 + 1 \times 2 \times 4 \times 5 + 1 \times 3 \times 5 \times 6 + 1 \times 2 \times 4 \times 6 \\ + 1 \times 3 \times 4 \times 6) \times 7 - (1 \times 2 \times 3 \times 4 + 1 \times 2 \times 3 \times 5 + 1 \times 2 \times 4 \times 5 + 1 \times 3 \times 4 \times 5) \times 6 \\ + 1 \times 2 \times 3 \times 4 \times 5 = 1 \end{array}$

It can be concluded from the definition:

8.5) $H_{2} ([1, \dots, 1], [1, \dots, M], C) = {(- 1)}^{M - C} C! E_{M - C}^{C} = {(- 1)}^{M - C} C! S_{2} (M, C)$

$N^{M} = \nabla S U M (N, [1, \dots, 1], [1, \dots, M])$

$\begin{array}{l} \overset{{Form}_{2}}{\to} \nabla \sum A_{q} (\begin{matrix} N + T_{M} - q \\ I D X (P T) - q \end{matrix}) = \sum A_{c} (\begin{matrix} N + M - (M - C) - 1 \\ M + 1 - (M - C) - 1 \end{matrix}) \\ = \sum A_{c} (\begin{matrix} N + C - 1 \\ C \end{matrix}) \end{array}$

8.6) $N^{M} = \sum_{K = 0}^{M} K! S_{2} (M, K) (\begin{matrix} N \\ K \end{matrix}) = \sum_{K = 0}^{M} {(- 1)}^{M - K} K! S_{2} (M, K) (\begin{matrix} N + K - 1 \\ K \end{matrix})$

It can be concluded from the definition:

8.7) $H_{2} ([1 + y, 2 + y, \dots, M + y], [1, 2, \dots, M], C) = C! {[y]}^{M - C} ( M C )$

${[x + y]}^{M} = \nabla S U M (x, [1 + y, 2 + y, \dots, M + y])$

$\overset{{Form}_{2}}{\to} \nabla \sum_{c = 0}^{M} (\begin{matrix} M \\ c \end{matrix}) C! {[y]}^{M - c} (\begin{matrix} x + M - (M - C) \\ M + 1 - (M - C) \end{matrix})$

$= \nabla \sum_{c = 0}^{M} (\begin{matrix} M \\ c \end{matrix}) c! {[y]}^{M - c} (\begin{matrix} x + C \\ C + 1 \end{matrix}) = \sum_{c = 0}^{M} (\begin{matrix} M \\ c \end{matrix}) c! {[y]}^{M - c} (\begin{matrix} x + C - 1 \\ C \end{matrix})$

$= \sum_{K = 0}^{M} (\begin{matrix} M \\ K \end{matrix}) {[y]}^{M - K} {[x + k - 1]}_{K} = \sum_{K = 0}^{M} (\begin{matrix} M \\ K \end{matrix}) {[y]}^{M - K} {[x]}^{K}$ à Norlund identity

9. Form₃ and Eulerian Number

Rewrite (1) of section 3 as

$\sum_{n = 0}^{N - 1} n (\begin{matrix} n + K \\ M \end{matrix}) = (M - K) (\begin{matrix} N + K + 1 \\ M + 2 \end{matrix}) + (1 + K) (\begin{matrix} N + K \\ M + 2 \end{matrix})$

$P S = [K_{1} : D_{1}, \dots, K_{M} : D_{M}]$ , $P T = [T_{1}, \dots, T_{M}]$ , use the same method of 3.1) to prove:

9.1) Use the ${Form}_{3} = (T_{1} + K_{1}) \dots (T_{M} + K_{M}) = \sum X_{1} \dots X_{M}$ ,

$S U M (N, P S, P T) = \sum A_{q} (\begin{matrix} N + T_{M} - q \\ I D X (P T) \end{matrix})$ , q = count of $X \in T$ . $A_{q} = \prod_{i = 1}^{M} B_{i}$ ,

$B_{i} = {\begin{cases} - K_{i} + [T_{i} - m] D_{i}; X_{i} = T_{i}, m = count of {\dots, X_{i - 1}} \in T \\ K_{i} + m D_{i}; X_{i} = K_{i}, m = count of {\dots, X_{i - 1}} \in T \end{cases}$

Define $H_{3} (P S, P T, C) = A_{q}$ of 9.1), C = Count of $X \in T$

$〈 \begin{matrix} M \\ k \end{matrix} 〉$ is Eulerian number. Worpitzky identity: $N^{M} = \sum_{k = 0}^{M - 1} 〈 \begin{matrix} M \\ k \end{matrix} 〉 (\begin{matrix} N + k \\ M \end{matrix})$

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

9.2) $H_{3} (P T, P T, q > 0) = 0$ ; $H_{3} (P T, P T, 0) = \prod T_{i}$ This à 4.5)

9.3) $H_{3} ([1, \dots, 1], [1, \dots, M], q) = 〈 \begin{matrix} M \\ M - q - 1 \end{matrix} 〉 = 〈 \begin{matrix} M \\ q \end{matrix} 〉$ , $H_{3} ([0, \dots, 0], [1, \dots, M], q) = 〈 \begin{matrix} M \\ q - 1 \end{matrix} 〉$

[Proof]

Obviously: $H_{3} ([K_{1} = 1, \dots, K_{M}], [T_{1} = 1, \dots, T_{M}], M) = 0$

$\begin{matrix} N^{M} = \nabla S U M (N, [1, \dots, 1]) = \sum_{q = 0}^{M} A_{q} (\begin{matrix} N + M - q - 1 \\ M \end{matrix}) \\ = \sum_{q = 0}^{M} A_{q} (\begin{matrix} N + M - q - 1 \\ M \end{matrix}) = \sum_{k = 0}^{M - 1} 〈 \begin{matrix} M \\ k \end{matrix} 〉 (\begin{matrix} N + k \\ M \end{matrix}) \end{matrix}$

q.e.d.

It’s easy to deduce:

(*) $H_{3} (q) = H_{2} ([0, \dots, 0], [1, \dots, M], q) = \sum \prod B_{i}$ , $B_{i} = {\begin{cases} m + 1 : X_{i} = T_{i}, m = count of {\dots, X_{i - 1}} \in K \\ m : X_{i} = K_{i}, m = count of {\dots, X_{i - 1}} \in T \end{cases}$

(*1) $H_{3} (q) = H_{2} (M - q - 1)$ à 1)

(*2) $H_{3} (q) = (M - q) H_{3} (q - 1) + (q + 1) H_{3} ([0, \dots, 0], [1, \dots, M - 1], q)$ à 2)

(*3) $H_{3} (q) = \sum \prod B_{i}$ , then $B_{1} + B_{2} + \dots + B_{M} = q (M - q + 1)$

$\begin{array}{l} H_{3} ([1, \dots, 1], [1, \dots, M], 1) \\ = 1^{M - 1} \times (M - 1) + 1^{M - 2} \times (M - 2) \times 2 + 1^{M - 3} \times (M - 3) \times 2^{2} + \dots \end{array}$

$= 2^{0} \times M + 2^{1} \times M + \dots + 2^{M - 2} \times M - (2^{0} \times 1 + 2^{1} \times 2 + \dots + 2^{M - 2} \times (M - 1))$

$= M (2^{M - 1} - 1) - \sum_{i = 0}^{M - 2} 2^{i} (i + 1) = M (2^{M - 1} - 1) - (2^{M - 1} - 1) - \sum_{i = 0}^{M - 2} 2^{i} i$

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

9.4) $\sum_{i = 0}^{M} 2^{i} i = (M - 1) 2^{M + 1} + 2$

9.5) $〈 \begin{matrix} M \\ q \end{matrix} 〉 = \sum_{t_{1} + \dots + t_{q + 1} = M - q - 1} 1^{t_{1}} 2^{t_{2}} \dots {(q + 1)}^{t_{q + 1}} (1 + t_{1}) \dots (1 + t_{1} + \dots + t_{q}), t_{i} \geq 0$

[Proof]

$H_{3} ([0, \dots, 0], [1, \dots, M], q + 1) = 〈 \begin{matrix} M \\ q \end{matrix} 〉 = \sum Π (X \in T) Π (X \in K)$

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

When $X_{1} \dots X_{M} > 0$ ,then $X_{1} = T_{1} = 1$ , $K_{i} \in [1, q + 1]$

Record $Π (X \in K) = 1^{t_{1}} 2^{t_{2}} \dots {(q + 1)}^{t_{q + 1}}$ , $t_{1} + t_{2} + \dots + t_{q + 1} = M - q - 1$

Take out factors > 0, record as ${P_{1}, P_{2}, \dots}$ ,(*)à

$Π (X \in T) = 1^{P_{1}} {(1 + t_{p 1})}^{P_{2} - P_{1}} {(1 + t_{p 1} + t_{p 2})}^{P_{3} - P_{2}} \dots$ , it can be rewritten as

$Π (X \in T) = (1 + t_{1}) (1 + t_{1} + t_{2}) (1 + t_{1} + t_{2} + t_{3}) \dots (1 + t_{1} + \dots + t_{q})$

q.e.d.

This is the conclusion of [7], which is obtained by guess and proved by induction.

Example 9.1:

$\begin{array}{l} 〈 \begin{matrix} 6 \\ 2 \end{matrix} 〉 = \sum_{t_{1} + t_{2} + t_{3} = 3} 1^{t_{1}} 2^{t_{2}} 3^{t_{3}} (1 + t_{1}) (1 + t_{1} + t_{2}) \\ = 1^{3} 2^{0} 3^{0} (1 + 3) (1 + 3 + 0) + 1^{2} 2^{1} 3^{0} (1 + 2) (1 + 2 + 1) + 1^{2} 2^{0} 3^{1} (1 + 2) (1 + 2 + 0) \\ + 1^{1} 2^{1} 3^{1} (1 + 1) (1 + 1 + 1) + 1^{1} 2^{2} 3^{0} (1 + 1) (1 + 1 + 2) + 1^{1} 2^{0} 3^{2} (1 + 1) (1 + 1 + 0) \\ + 1^{0} 2^{1} 3^{2} (1 + 0) (1 + 0 + 1) + 1^{0} 2^{2} 3^{1} (1 + 0) (1 + 0 + 2) \\ + 1^{0} 2^{3} 3^{0} (1 + 0) (1 + 0 + 3) + 1^{0} 2^{0} 3^{3} (1 + 0) (1 + 0 + 0) \\ = 302 \end{array}$

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

References

[1]	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
[2]	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
[3]	Peng, J. (2020) Subset of the Shape of Numbers. Open Access Library Journal, 7: e7040. https://doi.org/10.4236/oalib.1107040
[4]	Peng, J. (2021) Expansion of the Shape of Numbers. Open Access Library Journal, 8: e7120. https://doi.org/10.4236/oalib.1107120
[5]	Editorial Board of Handbook of Modern Applied Mathematics (2002) Handbook of Modern Applied Mathematics Discrete Mathematics Volume. Tsinghua University Press, Beijing.
[6]	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.
[7]	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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