Expansion of the Shape of Numbers

Ji Peng

doi:10.4236/oalib.1107120

Open Access Library Journal > Vol.8 No.1, January 2021

Expansion of the Shape of Numbers

Ji Peng
Department of Electronic Information, Nanjing University, Nanjing, China.
DOI: 10.4236/oalib.1107120 PDF HTML XML 483 Downloads 1,472 Views Citations

Abstract

This article extends the concept of the shape of numbers. Originally, a shape was defined as [1, K₁, K₂, ···], 1<K₁<K₂< ···, K_i∈N. In this paper, the domain of a shape is extended from N to Z, the low bound is extended from 1 to Z, and K_i<K_i+1, K_i=K_i+1, K_i>K_i+1 are allowed, which prove that they can be calculated with the similar form (T₀+K₀)(T₁+K₁)(T₂+K₂) ···. In this way, a lot of calculation formulas can be obtained. At the end, the form is obtained to calculate K₁x···xK_M+(L+K₁)x···x(L+K_M)+(2L+K₁)x···x(2L+K_M)+(3L+K₁)x···x(3L+K_M)+···.

Keywords

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

Share and Cite:

Peng, J. (2021) Expansion of the Shape of Numbers. Open Access Library Journal, 8, 1-18. doi: 10.4236/oalib.1107120.

1. Introduction

Peng, J. has introduced Shape of numbers in [1] [2] [3]:

$(I_{1}, I_{2}, \dots, I_{M})$ , $I_{i} \in N$ , $I_{1} < I_{2} < \dots < I_{M}$ . There are M − 1 intervals between adjacent numbers. $I_{i + 1} - I_{i} = 1$ means continuity, $I_{i + 1} - I_{i} > 1$ means discontinuity.

Shape of numbers: collect $(I_{1}, I_{2}, \dots, I_{M})$ with the same continuity and discontinuity at the same position into a catalog, call it a Shape.

A shape has a min Item: $(1, K_{1}, K_{2}, \dots)$ that 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 $K_{i + 1} - K_{i} = 1$ , only $I_{i + 1} - I_{i} = 1$ is allowed.

The single $(I_{1}, I_{2}, \dots, I_{M})$ is an item, $I_{1} \times I_{2} \times \dots \times I_{M}$ is the product. I_i is a factor.

Example:

$P S = [1, 2] \to (1, 2), (2, 3), (3, 4), (1000, 1001) \in P S$

$P S = [1, 3] \to (1, 3), (1, 4), (2, 4), (1, 5), (2, 5), (3, 5), (1000, 2001) \in P S$

$P S = [1, 4] \to (1, 4), (1, 5), (2, 5), (1, 6), (2, 6), (3, 6) \in P S$ , $(3, 5), (4, 6) \notin P S$

$P S = [1, 4, 6] \to (1, 4, 7), (1, 5, 7), (2, 5, 7) \in P S$ , $(3, 5, 7) \notin P S$

Define:

SET(N, PS) = set of items belonging to PS in [1, N − 1]

PM(PS) = count of factors

PB(PS) = count of discontinuities

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

IDX(PS) = (max factor) + 1

PH(PS) = IDX(PS) − PB(PS) − 2

Basic Shape: 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.

Example:

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

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

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

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

End(N, PS) = set of items belonging to PS with the max factor = N − 1;

|SET(N, PS)| = count of items in SET(N, PS);

SUM(N, PS) = sum of all products in SET(N, PS).

Example:

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

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

[3] introduced the subset:

If PB(PS) = 0, SET(N, PS) is simple.

If PB(PS) > 0, then can fix some interval of discontinuities to get subsets.

SET(N, PS, PT) = subset of SET(N, PS), a valid

$P T = [1, T_{1}, \dots, T_{M}] = {\begin{cases} T_{i + 1} - T_{i} = 1 : K_{i + 1} - K_{i} = 1, means I_{i + 1} - I_{i} = 1 \\ T_{i + 1} - T_{i} = 1 : K_{i + 1} - K_{i} = D > 1, means I_{i + 1} - I_{i} = D \\ T_{i + 1} - T_{i} = 2 : K_{i + 1} - K_{i} = D > 1, means I_{i + 1} - I_{i} \geq D \end{cases}$ (*)

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:

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

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

SUM_SUBSET(N, PS, PT) is defined in [3] = sum of all products in SET(N, PS, PT)

Now, SUM() and SUM_SUBSET() are uniformly defined as SUM(N, PS, PT), SUM(N, PS, BASE(PS)) is abbreviated as SUM(N, PS)

Only valid PT is discussed below.

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

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

(1.2) $S U M (N, P S) = M I N (P S) (\begin{matrix} N \\ I D X (P S) \end{matrix})$ , PS is a Basic Shape

The following uses count of $X \in K$ for count of

${X_{1}, X_{2}, \dots, X_{M}} \in {K_{1}, K_{2}, \dots, K_{M}}$

(1.3) $P S = [1, K_{1}, \dots, K_{M}]$ , $P T = [1, T_{1}, T_{2}, \dots, T_{M}]$

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.

The expansion has 2^M items, don’t swap the factors of $X_{1} X_{2} \dots X_{M}$ , then each $X_{1} X_{2} \dots X_{M}$ corresponds to one expression =

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

$q = countof X \in K$ .

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

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

Example:

$P S = [1, K_{1} \geq 3, K_{2} \geq K_{1} + 2, K_{3} \geq K_{2} + 2]$ ,

$B S = B A S E (P S) = [1, 3, 5, 7]$ ,

$I D X (B S) = 8$

$\begin{matrix} Theform = (3 + K_{1}) (5 + K_{2}) (7 + K_{3}) \\ = 3 \times 5 \times 7 + 3 \times 5 \times K_{3} + 3 \times K_{2} \times 7 + 3 \times K_{2} \times K_{3} \\ + K_{1} \times 5 \times 7 + K_{1} \times 5 \times K_{3} + K_{1} \times K_{2} \times 7 + K_{1} \times K_{2} \times K_{3} \end{matrix}$

$\begin{matrix} P = N - P H (P S) - P C H G (P S, P T) \\ = N - {I D X (P S) - P B (P S) - 2} - 0 \\ = N - {K_{3} + 1 - 3 - 2} = N - K_{3} + 4 \end{matrix}$

$\begin{array}{l} S U M (N, P S) = 3 \times 5 \times 7 (\begin{matrix} P \\ 8 \end{matrix}) + 3 \times 5 \times (K_{3} + 2) (\begin{matrix} P \\ 7 \end{matrix}) \\ + 3 \times (K_{2} + 1) \times (7 - 1) (\begin{matrix} P \\ 7 \end{matrix}) + 3 \times (K_{2} + 1) \times (K_{3} + 1) (\begin{matrix} P \\ 6 \end{matrix}) \\ + K_{1} \times (5 - 1) \times (7 - 1) (\begin{matrix} P \\ 7 \end{matrix}) + K_{1} \times (5 - 1) \times (K_{3} + 1) (\begin{matrix} P \\ 6 \end{matrix}) \\ + K_{1} \times K_{2} \times (7 - 2) (\begin{matrix} P \\ 6 \end{matrix}) + K_{1} \times K_{2} \times K_{3} ( P 5 ) \end{array}$

$Anitem \in P S = {begin, K_{1} + E_{1}, \dots, K_{M} + E_{M}}$ , K is fixed, E is variable.

$Aproduct = begin \times (K_{1} + E_{1}) \dots (K_{M} + E_{M}) = begin \times \sum F_{1} F_{2} \dots F_{M}$ , F_i = E_i or K_i

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

Define $S U M_K (N, P S, P T, P F = F_{1} F_{2} \dots F_{M})$ = Sum of one part in SUM(N, PS).

PF indicates the part. F_i = E_i or K_i

Rewrite 1.3), add {braces}:

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

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

Expand SUM(N, PS, PT) by {braces}:

(1.4) SUM_K(N, PS, PT, PF) = ∑Expansion of SUM() with same

${K_{i}} \in P F = \sum \prod_{i = 1}^{M} Y_{i} (\begin{matrix} A \\ M_{q} \end{matrix})$ ,

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

Example:

$S U M (N, [1, K_{1} \geq 3, K_{2} \geq K_{1} + 2])$ ,

$form = (3 + K_{1}) (5 + K_{2})$ à

$\begin{array}{l} = 15 (\begin{matrix} N - K_{2} + 3 \\ 6 \end{matrix}) + 3 ({K_{2}} + {1}) (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix}) \\ + K_{1} ({5 - 1}) (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix}) + K_{1} K_{2} (\begin{matrix} N - K_{2} + 3 \\ 4 \end{matrix}) \end{array}$

Expand by the {braces}:

$\begin{array}{l} = {15 (\begin{matrix} N - K_{2} + 3 \\ 6 \end{matrix}) + 3 (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix})} + 3 K_{2} (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix}) \\ + 4 K_{1} (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix}) + K_{1} K_{2} (\begin{matrix} N - K_{2} + 3 \\ 4 \end{matrix}) \\ = \sum_{begin = 1}^{N - K_{2}} \sum begin \times (K_{1} + E_{1, begin}) (K_{2} + E_{2, begin}) \end{array}$

$\begin{array}{l} S U M_K (N, P S, B S, E_{1} E_{2}) \\ = \sum_{allitems} begin * E_{1, i} E_{2, i} = 15 (\begin{matrix} N - K_{2} + 3 \\ 6 \end{matrix}) + 3 (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix}) \end{array}$

$S U M_K (N, P S, B S, E_{1} K_{2}) = \sum_{allitems} begin * E_{1, i} K_{2} = 3 K_{2} (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix})$

$S U M_K (N, P S, B S, K_{1} E_{2}) = \sum_{allitems} begin * K_{1} E_{2, i} = 4 K_{1} (\begin{matrix} N - K_{2} + 3 \\ 5 \end{matrix})$

$S U M_K (N, P S, B S, K_{1} K_{2}) = \sum_{allitems} begin * K_{1} K_{2} = K_{1} K_{2} (\begin{matrix} N - K_{2} + 3 \\ 4 \end{matrix})$

In this paper, we extend the definition of Shape of Numbers and generalize the corresponding results.

2. The Extension of Shape

Redefine:

$P S = [minItem] = [K_{0}, \dots, K_{M}, \dots]$ , $Item = (I_{0}, \dots, I_{M}, \dots)$ ,

$B A S E (P S) = B S = [G_{0} = 1, G_{1}, \dots, G_{M}, \dots]$

1) change factor’s domain of definition from N to Z, change K₀ from 1 to Z.

2) allow $K_{0} \leq K_{1} \leq \dots \leq K_{M}$ , If $K_{i + 1} = K_{i}$ , only $I_{i + 1} = I_{i}$ is allowed. $G_{i + 1} - G_{i} = 1$

3) allow $K_{i} > K_{i + 1}$ , only $I_{i + 1} = I_{i}$ is allowed. $G_{i + 1} - G_{i} = 1$ .

Example:

$P S = [3, 5] \to B A S E (P S) = [1, 3]$

$\begin{array}{l} S E T (8, [3, 5]) = {(3, 5), (3, 6), (4, 6), (3, 7), (4, 7), (5, 7)} \\ \neq S E T (8, [1, 3]) - S E T (5, [1, 3]) \end{array}$

$P S = [3, 5, 4, 6] \to B A S E (P S) = [1, 3, 4, 6]$

$S E T (8, P S) = {(3, 5, 4, 6), (3, 6, 5, 7), (4, 6, 5, 7), (3, 5, 4, 7)}$

Redefine:

Basic Shape: K₀ = 1 and intervals = 1 or 2

SET(N, PS) = set of items belonging to PS in [K₀, N − 1], Max Factor of item ≤ N − 1

PB(PS) = Count of discontinuities in BS

PH(PS) = (Max Factor) − 1 − PB(BS)

IDX(PS) = IDX of $B S = {\max factor of B S} + 1 = P M (B S) + P B (B S) + 1$

D¹f(n): if $f (n) = \sum A_{i} (\begin{matrix} N - n_{i} \\ m_{i} \end{matrix})$ , then $D^{1} f (n) = \sum A_{i} (\begin{matrix} N - n_{i} - 1 \\ m_{i} - 1 \end{matrix})$

2.1) $| S E T (N, P S, P T) | = (\begin{matrix} N - K_{0} - P H (P S) - P C H G (P S, P T) \\ P B (P T) + 1 \end{matrix})$

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

2.3) $P S = [K_{0}, K_{1}, \dots, K_{M}]$ , $P T = [1, T_{1}, T_{2}, \dots, T_{M}]$ , can use the form $(T_{0} + K_{0}) \dots (T_{M} + K_{M})$

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

$A_{q} = \prod_{i = 0}^{M} (X_{i} + D_{i})$ ,

$D_{i} = {\begin{cases} - m : X_{i} = T_{i}, m = countof {X_{0}, \dots, X_{i - 1}} \in K \\ + m : X_{i} = K_{i}, m = countof {X_{0}, \dots, X_{i - 1}} \in T \end{cases}$

$q = countof X \in K$

[Proof]

Here only prove SUM(N, PS), SUM(N, PS, PT) can use the same method.

$B S = B A S E (P S) = [1, G_{1}, G_{2}, \dots, G_{M}]$

Use the similar way of [2], by definition:

(1*) $S U M (N, P S) = \sum_{n = - \infty}^{N} \sum E N D (n, P S)$

(2*) $\sum E N D (N, P S) = D^{1} S U M (N, P S)$

(3*) $S U M (N, [P S, K_{M + 1} = 1 + K_{M}]) = \sum_{n = - \infty}^{N - 1} n \times \sum E N D (n, P S)$

(4*) $S U M (N, [P S, K_{M + 1} = K_{M}]) = \sum_{n = - \infty}^{N - 1} n \times \sum E N D (n + 1, P S)$

(5*) $S U M (N, [P S, K_{M + 1} > 1 + K_{M}]) = \sum_{n = - \infty}^{N - 1} n \times S U M (n - (K_{M + 1} - K_{M}) + 1, P S)$

Suppose $S U M (N, P S) = \sum X_{0} X_{1} \dots X_{M} (\begin{matrix} N - P H (P S) - 1 \\ M_{i} \end{matrix})$ , Max factor of PS = K_M

$P = n - P H (P S) - 1 = n - [K_{M} - 1 - P B (B S)] - 1 = n - [K_{M} - P B (B S)]$

$Q = N - P H (P S) - 1$

$C = Countof {X_{0}, \dots, X_{M}} \in K$ ,

$M_{i} = I D X (B S) - C$

1) $P S 1 = [P S, K_{M + 1} = 1 + K_{M}]$ , $B S 1 = B A S E (P S 1) = [B S, G_{M + 1} = 1 + G_{M}]$

$\begin{array}{l} S U M (N, P S 1) = \sum_{n = - \infty}^{N - 1} n \times \sum E N D (n, P S) = \sum_{n = - \infty}^{N - 1} n \times D^{1} S U M (n, P S) \\ = \sum_{n = - \infty}^{N - 1} n \times \sum X_{0} \dots X_{M} (\begin{matrix} P - 1 \\ M_{i} - 1 \end{matrix}) \\ = \sum_{n = - \infty}^{N - 1} n \times \sum X_{0} \dots X_{M} (\begin{matrix} n - [K_{M} - P B (B S) + 1] \\ M_{i} - 1 \end{matrix}) \overset{(2.2)}{\to} \\ = \sum^{} (X_{0} \dots X_{M} M_{i} (\begin{matrix} Q - 1 \\ M_{i} + 1 \end{matrix}) + X_{0} \dots X_{M} (M_{i} - 1 + K_{M} - P B (B S) + 1) (\begin{matrix} Q - 1 \\ M_{i} \end{matrix})) \end{array}$

$M_{i} = I D X (B S) - C = 1 + G_{M} - C = G_{M + 1} - C$

$\begin{array}{l} M_{i} - 1 + K_{M} - P B (B S) + 1 \\ = M_{i} + K_{M} - P B (B S) = I D X (B S) - C + K_{M} - P B (B S) \\ = (P M (B S) + P B (B S) + 1) - C + K_{M} - P B (B S) \\ = K_{M} + 1 + P M (B S) - C = K_{M + 1} + (M + 1) - C \end{array}$

$\begin{array}{l} S U M (N, P S 1) \\ = \sum X_{0} \dots X_{M} (G_{M + 1} - C) (\begin{matrix} Q - 1 \\ I D X (B S) - C + 1 \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M + 1} + M + 1 - C) (\begin{matrix} Q - 1 \\ I D X (B S) - C \end{matrix}) \\ = \sum X_{0} \dots X_{M} (G_{M + 1} - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - C \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M + 1} + M + 1 - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - (C + 1) \end{matrix}) \end{array}$

à Match the form $(G_{0} + K_{0}) (G_{1} + K_{1}) \dots (G_{M} + K_{M}) {G_{M + 1} + K_{M + 1}}$ .

2) $P S 1 = [P S, K_{M + 1} = K_{M}]$ , $B S 1 = B A S E (P S 1) = [B S, G_{M + 1} = 1 + G_{M}]$

$\begin{array}{l} S U M (N, P S 1) = \sum_{n = - \infty}^{N - 1} n \times \sum^{} E N D (n + 1, P S) \\ = \sum_{n = - \infty}^{N - 1} n \times D^{1} S U M (n + 1, P S) \\ = \sum_{n = - \infty}^{N - 1} n \times \sum X_{0} \dots X_{M} (\begin{matrix} P \\ M_{i} - 1 \end{matrix}) \\ = \sum_{n = - \infty}^{N - 1} n \times \sum X_{0} \dots X_{M} (\begin{matrix} n - [K_{M} - P B (B S)] \\ M_{i} - 1 \end{matrix}) \overset{( 2.2 )}{\to} \end{array}$

$\begin{array}{l} = \sum^{} (X_{0} \dots X_{M} M_{i} (\begin{matrix} Q \\ M_{i} + 1 \end{matrix}) + X_{0} \dots X_{M} (M_{i} - 1 + K_{M} - P B (B S)) (\begin{matrix} Q \\ M_{i} \end{matrix})) \\ = \sum X_{0} \dots X_{M} (G_{M + 1} - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - C \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M} + M + 1 - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - (C + 1) \end{matrix}) \end{array}$

à Match the form $(G_{0} + K_{0}) (G_{1} + K_{1}) \dots (G_{M} + K_{M}) {G_{M + 1} + K_{M + 1}}$ .

3) $P S 1 = [P S, K_{M + 1} > K_{M} + 1]$ , $B S 1 = B A S E (P S 1) = [B S, 2 + G_{M}]$

$\begin{array}{l} S U M (N, P S 1) = \sum_{n = - \infty}^{N - 1} n \times S U M (n - (K_{M + 1} - K_{M} - 1), P S) \\ = \sum_{n = - \infty}^{N - 1} n \times \sum X_{0} \dots X_{M} (\begin{matrix} n - (K_{M + 1} - K_{M} - 1) - P H (P S) - 1 \\ M_{i} \end{matrix}) \\ = \sum_{n = - \infty}^{N - 1} n \times \sum X_{0} \dots X_{M} (\begin{matrix} n - [K_{M + 1} - K_{M} + P H (P S)] \\ M_{i} \end{matrix}) \end{array}$

$\begin{matrix} Q 1 = N - [K_{M + 1} - K_{M} + P H (P S)] \\ = N - [K_{M + 1} - K_{M} + K_{M} - 1 - P B (B S)] \\ = N - [K_{M + 1} - 1 - P B (B S)] \\ = N - [K_{M + 1} - 1 - P B (B S 1)] - 1 \\ = N - P H (P S 1) - 1 \end{matrix}$

$\begin{array}{l} S U M (N, P S 1) \overset{(2.2)}{\to} \\ = \sum X_{0} \dots X_{M} (M_{i} + 1) (\begin{matrix} Q 1 \\ M_{i} + 2 \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M + 1} - K_{M} + P H (P S) + M_{i}) (\begin{matrix} Q 1 \\ M_{i} + 1 \end{matrix}) \end{array}$

$M_{i} + 1 = I D X (B S) - C + 1 = 1 + G_{M} - C + 1 = G_{M + 1} - C$

$\begin{array}{l} K_{M + 1} - K_{M} + P H (P S) + M_{i} = K_{M + 1} - K_{M} + P H (P S) + I D X (B S) - C \\ = K_{M + 1} - K_{M} + (K_{M} - 1 - P B (B S)) + (P M (B S) + P B (B S) + 1) - C \\ = K_{M + 1} + P M (B S) - C = K_{M + 1} + M + 1 - C \end{array}$

$\begin{array}{l} S U M (N, P S 1) \\ = \sum X_{0} \dots X_{M} (G_{M + 1} - C) (\begin{matrix} Q 1 \\ M_{i} + 2 \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M + 1} + M + 1 - C) (\begin{matrix} Q 1 \\ M_{i} + 1 \end{matrix}) \\ = \sum X_{0} \dots X_{M} (G_{M + 1} - C) (\begin{matrix} Q 1 \\ I D X (B S) - C + 2 \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M + 1} + M + 1 - C) (\begin{matrix} Q 1 \\ I D X (B S) - C + 1 \end{matrix}) \end{array}$

$\begin{array}{l} = \sum X_{0} \dots X_{M} (G_{M + 1} - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - C \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M + 1} + M + 1 - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - (C + 1) \end{matrix}) \end{array}$

à Match the form $(G_{0} + K_{0}) (G_{1} + K_{1}) \dots (G_{M} + K_{M}) {G_{M + 1} + K_{M + 1}}$ .

4) $P S 1 = [P S, K_{M + 1} < K_{M}]$ , $B S 1 = B A S E (P S 1) = [B S, 1 + G_{M}]$

By definition:

$\begin{array}{l} S U M (N, [P S, K_{M + 1}]) = \sum_{n = - \infty}^{N - 1} (n + K_{M + 1} - K_{M}) \sum E N D (n + 1, P S) \\ = \sum_{n = - \infty}^{N - 1} (n + K_{M + 1} - K_{M}) \times D^{1} S U M (n + 1, P S) \\ = \sum_{n = - \infty}^{N - 1} (n + K_{M + 1} - K_{M}) \times \sum X_{0} \dots X_{M} (\begin{matrix} p \\ M_{i} - 1 \end{matrix}) \\ = \sum_{n = - \infty}^{N - 1} n \times \sum X_{0} \dots X_{M} (\begin{matrix} P \\ M_{i} - 1 \end{matrix}) + \sum_{n = - \infty}^{N - 1} (K_{M + 1} - K_{M}) \\ \times \sum X_{0} \dots X_{M} (\begin{matrix} P \\ M_{i} - 1 \end{matrix}) \end{array}$

$\begin{array}{l} = \sum X_{0} \dots X_{M} {M_{i} (\begin{matrix} Q \\ M_{i} + 1 \end{matrix}) + (M_{i} - 1 + K_{M} - P B (B S)) (\begin{matrix} Q \\ M_{i} \end{matrix}) \\ + (K_{M + 1} - K_{M}) (\begin{matrix} Q \\ M_{i} \end{matrix})} \\ = \sum X_{0} \dots X_{M} {(I D X (B S) - C) (\begin{matrix} Q \\ M_{i} + 1 \end{matrix}) + (M_{i} + K_{M + 1} - P B (B S) - 1) (\begin{matrix} Q \\ M_{i} \end{matrix})} \end{array}$

$\begin{array}{l} M_{i} + K_{M + 1} - P B (P S) - 1 = I D X (B S) - C + K_{M + 1} - P B (B S) - 1 \\ = (P M (B S) + P B (B S) + 1) - C + K_{M + 1} - P B (B S) - 1 \\ = K_{M + 1} + (M + 1) - C \end{array}$

$\begin{array}{l} S U M (N, P S 1) \\ = \sum X_{0} \dots X_{M} (G_{M + 1} - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - C \end{matrix}) \\ + \sum X_{0} \dots X_{M} (K_{M + 1} + M + 1 - C) (\begin{matrix} N - P H (P S 1) - 1 \\ I D X (B S 1) - (C + 1) \end{matrix}) \end{array}$

à Match the form $(G_{0} + K_{0}) \dots (G_{M} + K_{M}) {G_{M + 1} + K_{M + 1}}$ .

q.e.d.

Example:

$N - P H ([- 11, - 7, - 4]) - 1 = N - (- 4 - 2 - 1) - 1 = N + 6$ ,

$B A S E ([- 11, - 7, - 4]) = [1, 3, 5]$

$S U M (N, [- 11, - 7, - 4])$ à $form = (1 - 11) (3 - 7) (5 - 4)$ à

$= 15 (\begin{matrix} N + 6 \\ 6 \end{matrix}) - 118 (\begin{matrix} N + 6 \\ 5 \end{matrix}) + 315 (\begin{matrix} N + 6 \\ 4 \end{matrix}) - 308 (\begin{matrix} N + 6 \\ 3 \end{matrix})$

$15 = 1 \times 3 \times 5$ ;

$- 308 = (- 11) \times (- 7) \times ( - 4 )$

$- 118 = 1 \times 3 \times (- 4 + 2) + 1 \times (- 7 + 1) \times (5 - 1) + (- 11) \times (3 - 1) \times (5 - 1)$

$315 = 1 \times (- 7 + 1) \times (- 4 + 1) + (- 11) \times (3 - 1) \times (- 4 + 1) + (- 11) \times (- 7) \times (5 - 2)$

$\begin{array}{l} S U M (- 2, [- 11, - 7, - 4]) \\ = (- 11) \times (- 7) \times (- 4) + (- 11) \times (- 7) \times (- 3) \\ + (- 11) \times (- 6) \times (- 3) + (- 10) \times (- 6) \times (- 3) \\ = 315 - 308 \times 4 = - 917 \end{array}$

$\begin{array}{l} S U M (- 1, [- 11, - 7, - 4]) \\ = S U M (- 2, [- 11, - 7, - 4]) + (- 11) \times (- 7) \times (- 2) + (- 11) \times (- 6) \times (- 2) \\ + (- 11) \times (- 5) \times (- 2) + (- 11) \times (- 7) \times (- 2) \\ + (- 11) \times (- 6) \times (- 2) + (- 11) \times (- 5) \times (- 2) \\ = - 118 + 315 \times 5 - 308 \times 10 = - 1623 \end{array}$

$S U M (N, [4, 7, 11], [1, 2, 4])$ à $form = (1 + 4) (2 + 7) (4 + 11)$ à

$= 8 (\begin{matrix} N - 10 \\ 5 \end{matrix}) + 62 (\begin{matrix} N - 10 \\ 4 \end{matrix}) + 200 (\begin{matrix} N - 10 \\ 3 \end{matrix}) + 308 (\begin{matrix} N - 10 \\ 2 \end{matrix})$

$62 = 1 \times 2 \times (11 + 2) + 1 \times (7 + 1) \times (4 - 1) + 4 \times (2 - 1) \times (4 - 1)$

$200 = 1 \times (7 + 1) \times (11 + 1) + 4 \times (2 - 1) \times (11 + 1) + 4 \times 7 \times (4 - 2)$

[1, 2, 4] means $I_{1} - I_{0} = K_{1} - K_{0} = 7 - 4 = 3$ , $I_{2} - I_{1} = K_{2} - K_{1} \geq 11 - 7 = 4$

$\begin{array}{l} S U M (15, [4, 7, 11], [1, 2, 4]) \\ = 4 \times 7 \times 11 + 4 \times 7 \times 12 + 5 \times 8 \times 12 + 4 \times 7 \times 13 + 5 \times 8 \times 13 \\ + 6 \times 9 \times 13 + 4 \times 7 \times 14 + 5 \times 8 \times 14 + 6 \times 9 \times 14 + 7 \times 10 \times 14 \\ = 8 + 62 \times 5 + 200 \times 10 + 308 \times 10 = 5398 \end{array}$

$N - P H ([4, 7, 1, 8]) - 1 = N - (8 - 1 - 2) - 1 = N - 6$ , $B A S E ([4, 7, 1, 8]) = [1, 3, 4, 6]$

$S U M (N, [4, 7, 1, 8])$ à $form = (1 + 4) (3 + 7) (4 + 1) (6 + 8)$ à

$= 72 (\begin{matrix} N - 6 \\ 7 \end{matrix}) + 417 (\begin{matrix} N - 6 \\ 6 \end{matrix}) + 922 (\begin{matrix} N - 6 \\ 5 \end{matrix}) + 876 (\begin{matrix} N - 6 \\ 4 \end{matrix}) + 224 (\begin{matrix} N - 6 \\ 3 \end{matrix})$

$\begin{array}{l} 417 = 1 \times 3 \times 4 \times (8 + 3) + 1 \times 3 \times (1 + 2) \times (6 - 1) \\ + 1 \times (7 + 1) \times (4 - 1) \times (6 - 1) + 4 \times (3 - 1) \times (4 - 1) \times (6 - 1) \end{array}$

$\begin{array}{l} 922 = 1 \times 3 \times (1 + 2) \times (8 + 2) + 1 \times (7 + 1) \times (4 - 1) \times (8 + 2) \\ + 4 \times (3 - 1) \times (4 - 1) \times (8 + 2) + 1 \times (7 + 1) \times (1 + 1) \times (6 - 2) \\ + 4 \times (3 - 1) \times (1 + 1) \times (6 - 2) + 4 \times 7 \times (4 - 2) \times (6 - 2) \end{array}$

$\begin{array}{l} 876 = 4 \times 7 \times 1 \times (6 - 3) + 4 \times 7 \times (4 - 2) \times (8 + 1) \\ + 4 \times (3 - 1) \times (1 + 1) \times (8 + 1) + 1 \times (7 + 1) \times (1 + 1) \times (8 + 1) \end{array}$

$\begin{array}{l} S U M (13, [4, 7, 1, 8]) \\ = 4 \times 7 \times 1 \times 8 + 4 \times 7 \times 1 \times 9 + (4 + 5) \times 8 \times 2 \times 9 + 4 \times 7 \times 1 \times 10 \\ + (4 + 5) \times 8 \times 2 \times 10 + (4 + 5 + 6) \times 9 \times 3 \times 10 + 4 \times 7 \times 1 \times 11 \\ + (4 + 5) \times 8 \times 2 \times 11 + (4 + 5 + 6) \times 9 \times 3 \times 11 + (4 + 5 + 6 + 7) \end{array}$

$\begin{array}{l} \times 10 \times 4 \times 11 + 4 \times 7 \times 1 \times 12 + (4 + 5) \times 8 \times 2 \times 12 + (4 + 5 + 6) \times 9 \times 3 \times 12 \\ + (4 + 5 + 6 + 7) \times 10 \times 4 \times 12 + (4 + 5 + 6 + 7 + 8) \times 11 \times 5 \times 12 \\ = 72 + 417 \times 7 + 922 \times 21 + 876 \times 35 + 224 \times 35 = 60853 \end{array}$

2.2. SUM_K(N, PS, PT, PF)

$Anitem \in P S = {K_{0} + E_{0}, K_{1} + E_{1}, \dots, K_{M} + E_{M}}$ , K is fixed, E is variable.

$Aproduct = (K_{0} + E_{0}) \times (K_{1} + E_{1}) \dots (K_{M} + E_{M}) = \sum F_{0} F_{1} F_{2} \dots F_{M}$ , $F_{i} = E_{i}$ or $F_{i} = K_{i}$

That is, a product can be broken down into 2^M⁺¹ parts.

Use the same method of [2]

2.4) SUM_K(N, PS, PT, PF) is similar to (1.4), except the form = $(T_{0} + K_{0}) \dots$

Example:

$B A S E ([4, 7, 11]) = B S = [1, 3, 5]$

$\begin{array}{l} S U M (13, [4, 7, 11]) \\ = 4 \times 7 \times 11 + 4 \times 7 \times (11 + 1) + 4 \times (7 + 1) \times (11 + 1) + (4 + 1) \times (7 + 1) \times (11 + 1) \\ = {4 \times 7 \times 11 + 4 \times 7 \times 11 + 4 \times 7 \times 11 + 4 \times 7 \times 11} \\ + {4 \times 7 \times 1 + 4 \times 7 \times 1 + 4 \times 7 \times 1} + {4 \times 1 \times 11 + 4 \times 1 \times 11} + {1 \times 7 \times 11} \\ + {4 \times 1 \times 1 + 4 \times 1 \times 1} + {1 \times 7 \times 1} + {1 \times 1 \times 11} + {1 \times 1 \times 1} \end{array}$

$4 \times 7 \times 11 \to 308$

$\begin{array}{l} S U M_K (13, P S, B S, K_{0} K_{1} K_{2}) \\ = {4 \times 7 \times 11 + 4 \times 7 \times 11 + 4 \times 7 \times 11 + 4 \times 7 \times 11} = 308 (\begin{matrix} N - 9 \\ 3 \end{matrix}) \end{array}$

$4 \times 7 \times (5 - 2) \to 4 \times 7 \times 3$

$\begin{array}{l} S U M_K (13, P S, B S, K_{0} K_{1} E_{2}) \\ = {4 \times 7 \times 1 + 4 \times 7 \times 1 + 4 \times 7 \times 1} = 4 \times 7 \times 3 (\begin{matrix} N - 9 \\ 4 \end{matrix}) \end{array}$

$4 \times (3 - 1) \times (11 + 1) \to 4 \times 2 \times 11$

$S U M_K (13, P S, B S, K_{0} E_{1} K_{2}) = {4 \times 1 \times 11 + 4 \times 1 \times 11} = 4 \times 2 \times 11 (\begin{matrix} N - 9 \\ 4 \end{matrix})$

$1 \times (7 + 1) \times (11 + 1) \to 1 \times 7 \times 11$

$S U M_K (13, P S, B S, E_{0} K_{1} K_{2}) = {1 \times 7 \times 11} = 1 \times 7 \times 11 (\begin{matrix} N - 9 \\ 4 \end{matrix})$

$4 \times (3 - 1) \times (5 - 1) + 4 \times (3 - 1) \times (11 + 1) \to 4 \times 3 \times 4 + 4 \times 2 \times 1$

$\begin{array}{l} S U M_K (13, P S, B S, K_{0} E_{1} E_{2}) \\ = {4 \times 1 \times 1 + 4 \times 1 \times 1} = 4 \times 3 \times 4 (\begin{matrix} N - 9 \\ 5 \end{matrix}) + 4 \times 2 \times 1 (\begin{matrix} N - 9 \\ 4 \end{matrix}) \end{array}$

$1 \times (7 + 1) \times (5 - 1) + 1 \times (7 + 1) \times (11 + 1) \to 1 \times 7 \times 4 + 1 \times 7 \times 1$

$\begin{array}{l} S U M_K (13, P S, B S, E_{0} K_{1} E_{2}) \\ = {1 \times 7 \times 1} = 1 \times 7 \times 4 (\begin{matrix} N - 9 \\ 5 \end{matrix}) + 1 \times 7 \times 1 (\begin{matrix} N - 9 \\ 4 \end{matrix}) \end{array}$

$1 \times 3 \times (11 + 2) + 1 \times (7 + 1) \times (11 + 1) \to 1 \times 3 \times 11 + 1 \times 1 \times 11$

$\begin{array}{l} S U M_K (13, P S, B S, E_{0} E_{1} K_{2}) \\ = {1 \times 1 \times 11} = 1 \times 3 \times 11 (\begin{matrix} N - 9 \\ 5 \end{matrix}) + 1 \times 1 \times 11 (\begin{matrix} N - 9 \\ 4 \end{matrix}) \end{array}$

$\begin{array}{l} 1 \times 3 \times 5 + [1 \times 3 \times (11 + 2) + 1 \times (7 + 1) \times (5 - 1) + 4 \times (3 - 1) \times (5 - 1)] \\ + [1 \times (7 + 1) \times (11 + 1) + 4 \times (3 - 1) \times (11 + 1) + 4 \times 7 \times (5 - 2)] \\ \to 1 \times 3 \times 5 + [1 \times 3 \times 2 + 1 \times 1 \times 4 + 0] + [1 \times 1 \times 1 + 0 + 0] = 15 + [10] + [1] \end{array}$

$\begin{array}{l} S U M_K (13, P S, B S, E_{0} E_{1} E_{2}) \\ = {1 \times 1 \times 1} = 15 (\begin{matrix} N - 9 \\ 6 \end{matrix}) + 10 (\begin{matrix} N - 9 \\ 5 \end{matrix}) + (\begin{matrix} N - 9 \\ 4 \end{matrix}) \end{array}$

3. Coefficient Analysis

$K = [K_{1}, \dots, K_{M}]$ ,

$T = [T_{1}, \dots, T_{M}]$

Use the form $(T_{1} + K_{1}) \dots (T_{M} + K_{M}) = \sum X_{1} X_{2} \dots X_{M}$ , X_i = T_i or K_i

Define $H (K, T, N, S) = \sum B_{N}$ , $0 \leq N \leq M$ , $N = countof X \in T$

$B_{N} = \prod_{i = 1}^{M} (X_{i} + D_{i})$ , $D_{i} = {\begin{cases} - m S : X_{i} = T_{i}, m = countof {X_{1}, \dots, X_{i - 1}} \in K \\ + m S : X_{i} = K_{i}, m = countof {X_{1}, \dots, X_{i - 1}} \in T \end{cases}$

H(K, T, N, 1) is abbreviated as H(K, T, N)

3.1) $H (K, T, M) = T_{1} \times T_{2} \times \dots \times T_{M}$ , $H (K, T, 0) = K_{1} \times K_{2} \times \dots \times K_{M}$

[3] has proved:

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

S₂(M, K) is Stirling number of the second kind. à

3.2) $H ([1, 1, \dots, 1], [2, 3, \dots, M], N) = (M - N)! \times S_{2} (M, M - N)$

$S U M (N, [K_{0} = 1, K_{1}, \dots, K_{M}], [T_{0} = 1, T_{1}, \dots, T_{M}])$

can use the form = $(T_{1} + K_{1}) \dots (T_{M} + K_{M})$ or $(T_{0} + K_{0}) (T_{1} + K_{1}) \dots (T_{M} + K_{M})$

For arbitrary K, T:

3.3) $H ([P, K], [P, T], N, S) = P \times H (K, T, N, S) + P \times H (K, T, N - 1, S)$

[Proof]

$\begin{array}{l} H ([P, K, K_{M + 1}], [P, T, T_{M + 1}], N + 1, S) \\ = (X_{M + 1} = K_{M + 1}) + (X_{M + 1} = T_{M + 1}) \\ = H ([P, K], [P, T], N + 1, S) (K_{M + 1} + [N + 1] \times S) \\ + H ([P, K], [P, T], N, S) (T_{M + 1} - [M + 1 - N] \times S) \\ = {P \times H (K, T, N + 1, S) + P \times H (K, T, N, S)} (K_{M + 1} + [N + 1] \times S) \\ + {P \times H (K, T, N, S) + P \times H (K, T, N - 1, S)} (T_{M + 1} - [M + 1 - N] \times S) \end{array}$

$\begin{array}{l} = P \times {H (K, T, N + 1, S) (K_{M + 1} + [N + 1] \times S) \\ + H (K, T, N, S) (T_{M + 1} - [M - N] \times S)} \\ + P \times {H (K, T, N, S) {(K_{M + 1} + N \times S)}_{}^{} \\ + H (K, T, N - 1, S) (T_{M + 1} - [M - (N - 1)] \times S)} \\ = P \times H ([K, K_{M + 1}], [T, T_{M + 1}], N + 1, S) + P \times H ([K, K_{M + 1}], [T, T_{M + 1}], N, S) \end{array}$

q.e.d.

this à

3.4) $S U M (N, [1, 2, \dots, n, K_{1}, \dots, K_{M}], [1, 2, \dots, n, T_{1}, \dots, T_{M}])$

can use the form: $(T_{1} + K_{1}) \dots (T_{M} + K_{M}) = n! \sum A_{q} (\begin{matrix} N - P H (P S) - P C H G (P S, P T) + n - 1 \\ I D X (P T) - q \end{matrix})$

[2] has proved:

3.5) $H (K, K, N, S) = (\begin{matrix} M \\ N \end{matrix}) K_{1} \times K_{2} \times \dots \times K_{M}$

1.3) can derive 1.2) from this.

3.6) if $K_{i} + S = K_{i + 1}$ , $T_{i} + S = T_{i + 1}$ , then $H (K, T, N, S) = (\begin{matrix} M \\ N \end{matrix}) T_{1} \dots T_{N} \times K_{N + 1} \dots K_{M}$

[Proof]

Suppose $H (K, T, N, S) = (\begin{matrix} M \\ N \end{matrix}) T_{1} \dots T_{N} K_{N + 1} K_{N + 2} \dots K_{M}$

$\begin{array}{l} H ([K, K_{M + 1} = S + K_{M}], [T, T_{M + 1} = S + T_{M}], N + 1, S) \\ = (X_{M + 1} = K_{M + 1}) + (X_{M + 1} = T_{M + 1}) \\ = H (K, T, N + 1, S) (K_{M + 1} + [N + 1] \times S) \\ + H (K, T, N, S) (T_{M + 1} - [M - N] \times S) \end{array}$

$\begin{array}{l} = (\begin{matrix} M \\ N + 1 \end{matrix}) T_{1} \dots T_{N + 1} K_{N + 2} \dots K_{M} (K_{M + 1} + [N + 1] \times S) \\ + (\begin{matrix} M \\ N \end{matrix}) T_{1} \dots T_{N} K_{N + 1} \dots K_{M} (T_{M + 1} - [M - N] \times S) \\ = T_{1} \dots T_{N} K_{N + 2} \dots K_{M} (\begin{matrix} M \\ N + 1 \end{matrix}) T_{N + 1} (K_{M + 1} + [N + 1] \times S) \\ + T_{1} \dots T_{N} K_{N + 2} \dots K_{M} (\begin{matrix} M \\ N \end{matrix}) K_{N + 1} (T_{M + 1} - [M - N] \times S) \end{array}$

$\begin{array}{l} = T_{1} \dots T_{N} K_{N + 2} \dots K_{M} (\begin{matrix} M \\ N + 1 \end{matrix}) [T_{N + 1} K_{M + 1} + T_{N + 1} (N + 1) \times S] \\ + T_{1} \dots T_{N} K_{N + 2} \dots K_{M} (\begin{matrix} M \\ N \end{matrix}) [K_{M + 1} - [M - N] \times S] \\ \times ([T_{N + 1} + [M - N] \times S] - [M - N] \times S) \end{array}$

$\begin{array}{l} = T_{1} \dots T_{N} K_{N + 2} \dots K_{M} [(\begin{matrix} M \\ N + 1 \end{matrix}) T_{N + 1} K_{M + 1} + (\begin{matrix} M \\ N \end{matrix}) T_{N + 1} K_{M + 1}] \\ + T_{1} \dots T_{N} K_{N + 2} \dots K_{M} [(\begin{matrix} M \\ N + 1 \end{matrix}) T_{N + 1} (N + 1) - (\begin{matrix} M \\ N \end{matrix}) T_{N + 1} (M - N)] \times S \\ = T_{1} \dots T_{N} K_{N + 2} \dots K_{M} [(\begin{matrix} M \\ N + 1 \end{matrix}) T_{N + 1} K_{M + 1} + (\begin{matrix} M \\ N \end{matrix}) T_{N + 1} K_{M + 1}] \\ = (\begin{matrix} M + 1 \\ N + 1 \end{matrix}) T_{1} \dots T_{N + 1} K_{N + 2} \dots K_{M + 1} \end{array}$

à $H ([K, K_{M + 1}], [T, T_{M + 1}], N + 1, S)$ holds

q.e.d.

Define

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

$E_{Q}^{K} = \sum Q - product with factors \in K$ , the sum traverse all combinations.

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

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

By definition:

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

3.7) if $T_{i} + 1 = T_{i + 1}$ , then $H (K, T, N) = T_{1} \dots T_{N} [F_{M - N}^{K} E_{0}^{N} + F_{M - N - 1}^{K} E_{1}^{N} + \dots + + F_{0}^{K} E_{M - N}^{N}]$

[Proof]

Suppose H(K, T, N) holds

$\begin{array}{l} H ([K, K_{M + 1}], [T, T_{M + 1} = 1 + T_{M}], N + 1) \\ = H (K, T, N + 1) (K_{M + 1} + N + 1) + H (K, T, N) (T_{M + 1} - [M - N]) \\ = T_{1} \dots T_{N + 1} [F_{M - N - 1}^{K} E_{0}^{N + 1} + F_{M - N - 2}^{K} E_{1}^{N + 1} + \dots + F_{0}^{K} E_{M - N - 1}^{N + 1}] (K_{M + 1} + N + 1) \\ + T_{1} \dots T_{N} [F_{M - N}^{K} E_{0}^{N} + F_{M - N - 1}^{K} E_{1}^{N} + \dots + F_{0}^{K} E_{M - N}^{N}] (T_{M + 1} - [M - N]) \\ = T_{1} \dots T_{N + 1} [F_{M - N - 1}^{K} E_{0}^{N + 1} + F_{M - N - 2}^{K} E_{1}^{N + 1} + \dots + F_{0}^{K} E_{M - N - 1}^{N + 1}] (K_{M + 1} + N + 1) \\ + T_{1} \dots T_{N + 1} [F_{M - N}^{K} E_{0}^{N} + F_{M - N - 1}^{K} E_{1}^{N} + \dots + F_{0}^{K} E_{M - N}^{N}] \end{array}$

$\begin{array}{l} H ([K, K_{M + 1}], [T, T_{M + 1} = 1 + T_{M}], N + 1) / T_{1} \dots T_{N + 1} \\ = [F_{M - N - 1}^{K} E_{0}^{N + 1} + F_{M - N - 2}^{K} E_{1}^{N + 1} + \dots + F_{0}^{K} E_{M - N - 1}^{N + 1}] (K_{M + 1} + N + 1) \\ + [F_{M - N}^{K} E_{0}^{N} + F_{M - N - 1}^{K} E_{1}^{N} + \dots + F_{0}^{K} E_{M - N}^{N}] \end{array}$

$\begin{array}{l} Sumofallitemswithcountoffactors \in K = M - N - Q \\ = F_{M - [N + 1] - Q}^{K} E_{Q}^{N + 1} K_{M + 1} + F_{M - [N + 1] - (Q - 1)}^{K} E_{Q - 1}^{N + 1} (N + 1) + F_{M - N - Q}^{K} E_{Q}^{N} \\ = F_{M - [N + 1] - Q}^{K} E_{Q}^{N + 1} K_{M + 1} + F_{M - N - Q}^{K} [(N + 1) E_{Q - 1}^{N + 1} + E_{Q}^{N}] \\ = F_{M - [N + 1] - Q}^{K} E_{Q}^{N + 1} K_{M + 1} + F_{M - N - Q}^{K} E_{Q}^{N + 1} \\ = F_{M - N - Q}^{[K, K_{M + 1}]} E_{Q}^{N + 1} = F_{[M + 1] - [N + 1] - Q}^{[K, K_{M + 1}]} E_{Q}^{N + 1} \end{array}$

à $H ([K, K_{M + 1}], [T, T_{M + 1}], N + 1)$ holds

q.e.d.

Example:

$\begin{array}{l} H ([A, B, C, D], [1, 2, 3, 4], 2) \\ = 1 \times 2 \times (C + 2) (D + 2) + 1 \times (B + 1) \times (3 - 1) (D + 2) \\ + A \times (2 - 1) \times (3 - 1) (D + 2) + 1 \times (B + 1) (C + 1) (4 - 2) \\ + A \times (2 - 1) \times (C + 1) (4 - 2) + A B (3 - 1) (4 - 2) \end{array}$

$\begin{array}{l} = 1 \times 2 [(C + 2) (D + 2) + (B + 1) (D + 2) + A (D + 2) \\ + (B + 1) (C + 1) + A (C + 1) + A B] \\ = 1 \times 2 [(C D + B D + A D + B C + A C + A B) \\ + (C + D + B + A) (1 + 2) + (1 \times 2 + 1 \times 1 + 2 \times 2)] \end{array}$

3.8) In $S U M (N, [K_{1}, \dots, K_{M}], [1, 2, \dots, M])$ , K_i can switch the order.

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

$\begin{array}{l} H (K, T, N, S) \\ = T_{1} \dots T_{N} [F_{M - N}^{K} E_{0}^{{S, 2 S, \dots, N S}} + F_{M - N - 1}^{K} E_{1}^{{S, 2 S, \dots, N S}} + \dots + F_{0}^{K} E_{M - N}^{N {S, 2 S, \dots, N S}}] \end{array}$

$F_{M - N}^{M} = S_{1} (M + 1, N + 1)$ , S₁ is the first kind of unsigned Stirling number.

From 3.5) and 3.7)à

$\begin{array}{l} H ([1, \dots, M - 1], [1, \dots, M - 1], N) = (M - 1)! (\begin{matrix} M - 1 \\ N \end{matrix}) \\ = N! [F_{M - 1 - N}^{M - 1} E_{0}^{N} + F_{M - 1 - N - 1}^{M - 1} E_{1}^{N} + \dots + F_{0}^{M - 1} E_{M - 1 - N}^{N}] \\ = N! [S_{1} (M, N + 1) E_{0}^{N} + S_{1} (M, N + 2) E_{1}^{N} + \dots + S_{1} (M, M) E_{M - 1 - N}^{N}] \end{array}$ à

3.10) $\begin{array}{l} S_{1} (M, N + 1) E_{0}^{N} + S_{1} (M, N + 2) E_{1}^{N} + \dots + S_{1} (M, M) E_{M - 1 - N}^{N} \\ = (\begin{matrix} M - 1 \\ N \end{matrix}) \frac{(M - 1)!}{N!} \end{array}$

4. $\begin{array}{l} K_{1} \times \dots \times K_{M} + (L + K_{1}) \times \dots \times (L + K_{M}) \\ + (2 L + K_{1}) \times \dots \times (2 L + K_{M}) + \dots \end{array}$

$K = [K_{1}, \dots, K_{M}]$ , $T = [T_{1}, \dots, T_{M}]$ , Max Factor = K_M

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

When $N = K_{M} + 1$

$\begin{array}{l} N - P H (P S) - P C H G (P S, P T) - 1 \\ = N - (K_{M} - 1 - P B (B S)) - P C H G (P S, P T) - 1 \\ = P B (B S) - P C H G (P S, P T) + 1 \\ = P B (P T) + 1 = I D X (P T) - M \end{array}$

à $S U M (N, P S, P T) = A_{M}$

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

SUM(N, PS, PT) can be broken down into 2^M parts.

SUM_K(N, PS, PT, PF) can explain why SUM(N, PS, PT) 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 just match a simple form.

SUM_K(N, PS, PT, PF) use the form =

$(T_{1} + K_{1}) \dots (T_{M} + K_{M}) = \sum \prod_{i = 1}^{M} Y_{i} (\begin{matrix} A \\ M_{q} \end{matrix})$

When T_i and D_i all increase L times. If $P T = [1, 2, \dots, M]$ , when N increase,

$| E n d (N, P S, P T) | = 1$ , match the corresponding SUM_K().

Define

$S U M L (N, P S, 1) = S U M (N + K_{M}, P S, [1, 2, \dots, M])$

$\begin{matrix} S U M L (N, P S, L) = K_{1} \times \dots \times K_{M} + (L + K_{1}) \times \dots \times (L + K_{M}) + \dots \\ + ((N - 1) L + K_{1}) \times \dots \times ((N - 1) L + K_{M}) \end{matrix}$

SUML_K(N, PS, PF, L) = corresponding part of SUML(N, PS, L)

Above à

4.1) SUML(N, PS, L), SUML_K(N, PS, PF, L),

$P T = [T_{1}, \dots, T_{M}] = [1 \times L, 2 \times L, \dots, M \times L]$ , can use the from $(T_{1} + K_{1}) \dots (T_{M} + K_{M})$

$S U M L (N, P S, L) = \sum A_{q} (\begin{matrix} N \\ M + 1 - q \end{matrix})$ , $q = countof X \in K$ , 2^M items in total.

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

Example:

$S U M L (N, [3, 5, 8], 4)$ à

$form = (1 \times 4 + 3) (2 \times 4 + 5) (3 \times 4 + 8) = (4 + 3) (8 + 5) (12 + 8)$ à

$= 384 (\begin{matrix} N \\ 4 \end{matrix}) + 896 (\begin{matrix} N \\ 3 \end{matrix}) + 636 (\begin{matrix} N \\ 2 \end{matrix}) + 120 ( N 1 )$

$384 = 4 \times 8 \times 12$ ; $120 = 3 \times 5 \times 8$

$\begin{array}{l} 896 = 4 \times 8 \times (8 + 2 \times 4) + 4 \times (5 + 1 \times 4) \times (12 - 1 \times 4) \\ + 3 \times (8 - 1 \times 4) \times (12 - 1 \times 4) \end{array}$

$\begin{array}{l} 636 = 3 \times 5 \times (12 - 2 \times 4) + 3 \times (8 - 1 \times 4) \times (8 + 1 \times 4) \\ + 4 \times (5 + 1 \times 4) \times (8 + 1 \times 4) \end{array}$

$\begin{array}{l} S U M L (6, [3, 5, 8], 4) \\ = 3 \times 5 \times 8 + 7 \times 9 \times 12 + 11 \times 13 \times 16 + 15 \times 17 \times 20 \\ + 19 \times 21 \times 24 + 23 \times 25 \times 28 \\ = 384 \times 15 + 896 \times 20 + 636 \times 15 + 120 \times 6 = 33940 \end{array}$

$S U M L (N, [1], 2) = 1 + 3 + \dots + (2 N - 1) = 2 (\begin{matrix} N \\ 2 \end{matrix}) + (\begin{matrix} N \\ 1 \end{matrix}) = N^{2}$

4.2) P is a prime number, For arbitrary $K_{1}, K_{2}, \dots, K_{M}$ :

If $M < P - 1$ , then

$\begin{array}{l} K_{1} \times K_{2} \times \dots \times K_{M} + (L + K_{1}) \times \dots \times (L + K_{M}) + \dots \\ + ((P - 1) L + K_{1}) \times \dots \times ((P - 1) L + K_{M}) \equiv 0 M O D P \end{array}$

If $M = P - 1$ and $(L, P) = 1$ then

[Proof]

$Theexpression = S U M L (P, P S, L) = \sum A_{q} (\begin{matrix} P \\ M + 1 - q \end{matrix})$

If $M < P - 1$ , then $M + 1 < P$

If $M = P - 1$ and $(L, P) = 1$ , then

$\begin{matrix} S U M L (N, P S, L) = L^{P - 1} (P - 1)! (\begin{matrix} P \\ P \end{matrix}) + \sum A_{q} (\begin{matrix} P \\ M + 1 - q \end{matrix}), q > 0 \\ \equiv L^{P - 1} (P - 1)! \equiv - 1 M O D P \end{matrix}$

q.e.d.

5. Conclusions

The whole process of [1-3] is reviewed and this paper:

[1] tries to calculate all products of k distinct integers in [1, N − 1], introduces the concept of Shape of numbers. The idea divides all products of k distinct integers in [1, N − 1] into 2^K⁻¹ catalogs and derives the calculation formula of every catalog, that is 1.2).

[1] only introduces the basic shape. The conclusion is obtained through the derivation process.

[2] introduces the shape $P S = [1, K_{1}, \dots, K_{M}]$ , tries to calculate SUM(N, PS), the form $(G_{1} + K_{1}) (G_{2} + K_{2}) \dots (G_{M} + K_{M})$ is guessed by observation and proved by induction.

At the same time, SUM_K() is introduced.

[3] introduces the subset, and shows the way to calculate $1^{M} + 2^{M} + 3^{M} + \dots + N^{M}$ .

In this paper, the Shape and the form are further extended. So a lot of numbers’s series can be calculated.

Some new congruences are also obtained in [1] [2] [3] and this article.

The whole foundation is just $(\begin{matrix} N \\ M \end{matrix}) + (\begin{matrix} N \\ M + 1 \end{matrix}) = (\begin{matrix} N + 1 \\ M + 1 \end{matrix})$ .

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, 1-11.[CrossRef]
[2]	Peng, J. (2020) Subdivide the Shape of Numbers and a Theorem of Ring. Open Access Library Journal, 7, 1-14.[CrossRef]
[3]	Peng, J. (2020) Subset of the Shape of Numbers. Open Access Library Journal, 7, 1-15. [CrossRef]

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