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 OALibJ  Vol.8 No.3 , March 2021
Redefining the Shape of Numbers and Three Forms of Calculation
Abstract: This paper redefines the Shape of numbers, makes it more natural and concise, and the domain of definition is extended to ring. The inconvenient PCHG() and PH() are removed. The concept of subsets is also removed. The new definition can be used to calculate ∑n-0N-1Πi-1M (Ki+n×Di) ∑ni,j-0j-N-1Πi-1M (Ki+ni,j×Di), ni,j≤ni+1,j or ni,j=ni+1,j; Ki,Di∈ring. Three forms corresponding to three calculation methods are obtained. They can be used as a powerful tool for analysis. Some of the conclusions are: 1) Expressions and properties of two kinds of Stirling number, Lah number and Eulerian number; 2) Expression of power sum of natural numbers; 3) Vandermonde identity, Norlund identity; 4) New congruence and new proof of Wilson theorem; 5) ∑n-1P-1≡0 MOD P2, P>3; 6) ∑C-0C-M-1(-1)M-1-C∑PM(PS)-M,PB(PS)-CMIN(PS)=1.

1. Introduction

Peng, J. has introduced Shape of numbers in [1] [2] [3] [4]: ( I 1 , I 2 , , I M ) , I i Z . 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 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 , ) . Use the symbol PS = [min Item] to represent it.

If K i + 1 K i = D > 1 , only I i + 1 I i D is allowed.

If K i + 1 K i = D 1 , only I i + 1 I i = D is allowed.

The single ( I 1 , , I M ) is an item. I 1 × × I M is the product. Ii is a factor.

Example 1.1:

P S = [ 2 , 3 ] ( 2 , 3 ) , ( 3 , 4 ) , ( 1000 , 1001 ) P S

P S = [ 3 , 1 ] ( 3 , 1 ) , ( 3 , 0 ) , ( 2 , 0 ) , ( 3 , 1 ) , ( 2 , 1 ) , ( 1 , 1 ) , ( 1000 , 2007 ) P S

P S = [ 1 , 4 , 4 ] ( 1 , 4 , 4 ) , ( 1 , 5 , 5 ) , ( 2 , 5 , 5 ) , ( 1 , 6 , 6 ) , ( 2 , 6 , 6 ) , ( 3 , 6 , 6 ) P S , ( 3 , 5 , 5 ) P S

P S = [ 1 , 4 , 6 , 4 ] ( 1 , 4 , 7 , 5 ) , ( 1 , 5 , 7 , 5 ) , ( 2 , 5 , 7 , 5 ) P S , ( 1 , 4 , 7 , 6 ) , ( 3 , 5 , 7 , 5 ) P S

PM(PS) = Count of factors.

PB(PS) = Count of discontinuities.

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

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

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

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

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

P S = [ K 1 , , K M ] , B S = [ G 1 , , G M ] then K i + 1 K 1 G i + 1 G i = 1 ; K i + 1 K > 1 G i + 1 G i = 2

Example 1.2:

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

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

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

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

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

P S = [ 0 , 4 , 9 ] 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 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 , , G M ] , P T = [ T 1 , , T M ]

= { 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 (*)

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 × 2 × 4 + 1 × 2 × 5 + 2 × 3 × 5

S U M ( 9 , [ 1 , 4 , 7 ] ) = S U M ( 9 , [ 1 , 4 , 7 ] , [ 1 , 3 , 5 ] ) = 1 × 4 × 7 + 1 × 4 × 8 + 1 × 5 × 8 + 2 × 5 × 8

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

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

(1.1) | S E T ( N , P S , P T ) | = ( N P H ( P S ) P C H G ( P S , P T ) 1 P B ( P T ) + 1 )

The following uses count of X K for count of { X 1 , X 2 , , X M } { K 1 , K 2 , , K M }

(1.2) Use the form ( T 1 + K 1 ) ( T 2 + K 2 ) ( T M + K M ) = X 1 X 2 X M , X i = T i or K i .

Don’t swap the factors of X 1 X 2 X M , then each X 1 X 2 X M corresponds to a expression = A q ( N P H ( P S ) P C H G ( P S , P T ) 1 I D X ( P T ) q ) , q = count of X K .

S U M ( N , P S , P T ) = A q ( N P H ( P S ) P C H G ( P S , P T ) 1 I D X ( P T ) q )

A q = i = 1 M ( X i + D i ) , D i = { m : X i = T i , m = count of { X 1 , , X i 1 } K + m : X i = K i , m = count of { X 1 , , X i 1 } T

Example 1.5:

P S = [ K 1 , K 2 K 1 + 2 , K 3 K 2 + 2 ] , B S = B A S E ( P S ) = [ 1 , 3 , 5 ] ,

I D X ( B S ) = 6 ,

form = ( 1 + K 1 ) ( 3 + K 2 ) ( 5 + K 3 ) = 1 × 3 × 5 + 1 × 3 × K 3 + 1 × K 2 × 5 + 1 × K 2 × K 3 + K 1 × 3 × 5 + K 1 × 3 × K 3 + K 1 × K 2 × 5 + K 1 × K 2 × K 3

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

S U M ( N , P S ) = 1 × 3 × 5 ( P 6 ) + 1 × 3 × ( K 3 + 2 ) ( P 5 ) + 1 × ( K 2 + 1 ) × ( 5 1 ) ( P 5 ) + 1 × ( K 2 + 1 ) × ( K 3 + 1 ) ( P 4 ) + K 1 × ( 3 1 ) × ( 5 1 ) ( P 5 ) + K 1 × ( 3 1 ) × ( K 3 + 1 ) ( P 4 ) + K 1 × K 2 × ( 5 2 ) ( P 4 ) + K 1 × K 2 × K 3 ( P 3 )

An item = { K 1 + E 1 , , K M + E M } , K is fixed, E is variable.

A product = ( K 1 + E 1 ) ( K M + E M ) = F 1 F 2 F M , F i = E i or K i

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

Define S U M _ K ( N , P S , P T , P F = F 1 F M ) = Sum of one part, PF indicates the part.

Rewrite 1.2), add {braces}:

S U M ( N , P S , P T ) = i = 1 M ( X i + D i ) ( A M q ) ,

X i + D i = { { T i D i } : X i = T i , D i = count of { X 1 , , X i 1 } K { K i } + { D i } : X i = K i , D i = count of { X 1 , , X i 1 } T

Expand SUM() by {braces}:

(1.3) S U M _ K ( N , P S , P T , P F )

= Expansion of S U M ( ) with same { K i } P F = i = 1 M Y i ( A M q )

Y i = { 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 { , X i 1 } K D i : F i = E i , X i = K i , D i = count of { , X i 1 } T

Example 1.6:

Use the example above

S U M _ K ( E 1 E 2 E 3 ) = 1 × 3 × 5 ( P 6 ) + [ 1 × 3 × 2 + 1 × 1 × ( 5 1 ) ] ( P 5 ) + 1 3 ( P 4 )

S U M _ K ( E 1 E 2 K 3 ) = 1 × 3 × K 3 ( P 5 ) + 1 × 1 × K 3 ( P 4 )

S U M _ K ( E 1 K 2 E 3 ) = 1 × K 2 × ( 5 1 ) ( P 5 ) + 1 × K 2 × 1 ( P 4 )

S U M _ K ( K 1 E 2 E 3 ) = K 1 × ( 3 1 ) × ( 5 1 ) ( P 5 ) + K 1 × ( 3 1 ) × 1 ( P 4 )

S U M _ K ( E 1 K 2 K 3 ) = 1 × K 2 × K 3 ( P 4 )

S U M _ K ( K 1 E 2 K 3 ) = K 1 × ( 3 1 ) × K 3 ( P 4 )

S U M _ K ( K 1 K 2 E 3 ) = K 1 × K 2 × ( 5 2 ) ( P 4 )

S U M _ K ( K 1 K 2 K 3 ) = K 1 × K 2 × 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 Ti and Di all increase L times

(1.4) K 1 × × K M + ( L + K 1 ) × × ( L + K M ) +

+ ( ( N 1 ) L + K 1 ) × × ( ( N 1 ) L + K M )

P T = [ 1 × L , 2 × L , , M × L ] , can use the form ( T 1 + K 1 ) ( T M + K M ) = A q ( N M + 1 q ) , q = count of X K , 2M items in total.

A q = i = 1 M ( X i + D i ) , D i = { m L : X i = T i , m = count of { , X i 1 } K + m L : X i = K i , m = count of { , X i 1 } T

This paper starts from (1.4), tries to calculate

(1*) K 1 × × K M + ( D 1 + K 1 ) × × ( D M + K M ) +

+ ( ( N 1 ) D 1 + K 1 ) × × ( ( 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 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 , , ( N 1 ) D 1 + K 1 } = { K 1 + n × D 1 }

{ K 2 , D 2 + K 2 , 2 D 2 + K 2 , 3 D 2 + K 2 , , ( N 1 ) D 2 + K 2 } = { K 2 + n × D 2 }

{ K M , D M + K M , 2 D M + K M , 3 D M + K M , , ( N 1 ) D M + K M } = { K M + n × D M }

An item = ( I 1 , I 2 , , I M ) , I1 come from serie1, I2 come from serie2…

Use P S = [ K 1 : D 1 , K 2 : D 2 , , K M : D M ] to represent the Shape.

[ K 1 : 1 , K 2 : 1 , , K M : 1 ] is abbreviated as [ K 1 , K 2 , , K M ]

If K i Z , the new definition is similar to the old definition and allows D i 0 .

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

I i = K i + a × D i , I i + 1 = K i + 1 + b × D i + 1 , P T = [ T 1 = 1 , T 2 , , T M ] = { T i + 1 T i = 1 : means a = b T i + 1 T i = 2 , means a b

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 N , Di = 1

PT is always a Basic Shape.

MIN(PS) = Min product of a Basic Shape = K i

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

END(N, PS, PT) = Set of Items 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 , , M ] ) is abbreviated as SUM (N, PS)

S U M ( N , ) = Old Sum ( M a x ( K i ) + N , ) , 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 * ) = n = 0 N 1 i = 1 M ( K i + n × D i )

S U M ( N , P S , [ 1 , 3 , 5 , , 2 M 1 ] ) = i = 1 M ( K i + J i , j × D i ) , J 1. i J 2 , i J M , i , J i , j [ 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 ) | = ( N + P B ( P T ) P B ( P T ) + 1 )

3. Form1 of Calculation

Similar to [4], key points are:

Define f ( n ) = f ( n ) f ( n 1 ) : if f ( n ) = A i ( N i m i ) , then f ( n ) = A i ( N i 1 m i 1 )

1) n = 0 N 1 n ( n + K M ) = ( M + 1 ) ( N + K M + 2 ) + ( M K ) ( N + K M + 1 )

By definition:

2) E N D ( N , P S , P T ) = 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 ] )

= n = 0 N 1 ( K M + 1 + n × D M + 1 ) × 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 ] )

= n = 0 N 1 ( K M + 1 + n × D M + 1 ) × S U M ( n + 1 , P S , P T )

P S = [ K 1 : D 1 , , K M : D M ] , P T = [ T 1 , , T M ] , P S 1 = [ P S , K M + 1 : D M + 1 ]

3.1) Use the Form1 = ( T 1 + K 1 ) ( T M + K M ) = X 1 X M ,

S U M ( N , P S , P T ) = A q ( N + P B ( P T ) I D X ( P T ) q ) , q = count of X K .

A q = i = 1 M B i , B i = { ( T i m ) D i ; X i = T i , m = count of { X 1 , , X i 1 } K K i + m D i ; X i = K i , m = count of { X 1 , , X i 1 } T

[Proof]

Suppose S U M ( N , P S , P T ) = X 1 X M ( N + P B ( P T ) I D X ( P T ) q )

When P T 1 = [ P T , T M + 1 = 1 + T M ]

S U M ( N , P S 1 , P T 1 ) = n = 0 N 1 ( K M + 1 + n × D M + 1 ) × E N D ( n + 1 , P S , P T ) = n = 0 N 1 ( K M + 1 + n × D M + 1 ) × S U M ( n + 1 , P S , P T ) = n = 0 N 1 ( K M + 1 + n × D M + 1 ) × X 1 X M ( n + P B ( P T ) I D X ( P T ) q 1 ) ( 1 )

= X 1 X M D M + 1 ( I D X ( P T ) q ) ( N + P B ( P T ) I D X ( P T ) q + 1 ) + X 1 X M { K M + 1 + ( I D X ( P T ) q 1 P B ( P T ) ) D M + 1 } ( N + P B ( P T ) I D X ( P T ) q ) 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 = X 1 X M D M + 1 ( T M + 1 q ) ( N + P B ( P T 1 ) I D X ( P T 1 ) q ) + X 1 X M { K M + 1 + ( M q ) D M + 1 } ( N + P B ( P T 1 ) I D X ( P T 1 ) ( q + 1 ) )

The previous expression means X M + 1 = T M + 1

M ? q = Count of { X 1 X M } T . The following expression means X M + 1 = K M + 1

à Match the Form ( T 1 + K 1 ) ( T M + K M ) { T M + 1 + K M + 1 } .

When P T 1 = [ P T , T M + 1 = 2 + T M ]

S U M ( N , P S 1 , P T 1 ) = n = 0 N 1 ( K M + 1 + n × D M + 1 ) × S U M ( n + 1 , P S , P T ) = n = 0 N 1 ( K M + 1 + n × D M + 1 ) × X 1 X M ( n + 1 + P B ( P T ) I D X ( P T ) q ) ( 1 ) = X 1 X M D M + 1 ( I D X ( P T ) q + 1 ) ( N + 1 + P B ( P T ) I D X ( P T ) q + 2 )

+ X 1 X M { K M + 1 + ( I D X ( P T ) q 1 P B ( P T ) ) D M + 1 } × ( N + 1 + P B ( P T ) I D X ( P T ) q + 1 ) P B ( P T 1 ) = P B ( P T ) + 1 , I D X ( P T 1 ) = I D X ( P T ) + 2

= X 1 X M D M + 1 ( T M + 1 q ) ( N + P B ( P T 1 ) I D X ( P T 1 ) q ) + X 1 X M { K M + 1 + ( M q ) D M + 1 } ( N + P B ( P T 1 ) I D X ( P T 1 ) ( q + 1 ) )

à Match the Form ( T 1 + K 1 ) ( 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 )

S U M ( N , P S , P T ) = 9867.0 ( N + 2 6 ) + 1084.6 ( N + 2 5 ) + 4044.7 ( N + 2 4 ) 808.5 ( N + 2 3 )

808.5 = 7 × ( 7.7 ) × 15

4044.7 = [ ( 1 0 ) × 13 ] × ( 7.7 + 2.2 × 1 ) × ( 15 23 × 1 ) + 7 × [ ( 3 1 ) × 2.2 ] × ( 15 23 × 1 ) + 7 × ( 7.7 ) × [ ( 5 2 ) × ( 23 ) ]

1084.6 = [ ( 1 0 ) × 13 ] × [ ( 3 0 ) × 2.2 ] × ( 15 23 × 2 ) + [ ( 1 0 ) × 13 ] × ( 7.7 + 2.2 × 1 ) × [ ( 5 1 ) × ( 23 ) ] + 7 × [ ( 3 1 ) × 2.2 ] × [ ( 5 1 ) × ( 23 ) ]

9867.0 = [ ( 1 0 ) × 13 ] × [ ( 3 0 ) × 2.2 ] × [ ( 5 0 ) × ( 23 ) ]

S U M ( 2 , P S , P T ) = 7 × ( 7.7 ) × 15 + 7 × ( 7.7 5.5 ) × ( 8 ) + 20 × ( 5.5 ) × ( 8 ) = 4044.7 ( 4 4 ) 808.5 × ( 4 3 ) = 810.7

S U M ( 3 , P S , P T ) = S U M ( 2 , P S , P T ) + 7 × ( 7.7 5.5 3.3 ) × ( 31 ) + 20 × ( 5.5 3.3 ) × ( 31 ) + 33 × ( 3.3 ) × ( 31 ) = 1084.6 ( 5 5 ) + 4044.7 ( 5 4 ) 808.5 ( 5 3 ) = 13223.1

S U M ( 4 , P S , P T ) = S U M ( 3 , P S , P T ) + 7 × ( 7.7 5.5 3.3 1.1 ) × ( 54 ) + 20 × ( 5.5 3.3 1.1 ) × ( 54 ) + 33 × ( 3.3 1.1 ) × ( 54 ) + 46 × ( 1.1 ) × ( 54 ) = 9867.0 ( 6 6 ) + 1084.6 ( 6 5 ) + 4044.7 ( 6 4 ) 808.5 ( 6 3 ) = 41141.1

Example 3.2:

P S = [ ( 7 3 1 2 ) : ( 1 3 4 2 ) , ( 2 1 1 2 ) : ( 2 3 1 2 ) ] , P T = [ 1 , 3 ]

Form 1 = ( 1 + ( 7 3 1 2 ) ) ( 3 + ( 2 1 1 2 ) )

( 17 13 4 5 ) = ( 7 3 1 2 ) ( 2 1 1 2 )

( 44 70 28 38 ) = 1 × [ ( 2 1 1 2 ) + ( 2 3 1 2 ) ] ( 1 3 4 2 ) + ( 7 3 1 2 ) × ( 3 1 ) ( 2 3 1 2 )

( 15 27 30 48 ) = 1 × ( 1 3 4 2 ) × 3 × ( 2 3 1 2 )

S U M ( N , P S , P T ) = ( 15 27 30 48 ) ( N + 1 4 ) + ( 44 70 28 38 ) ( N + 1 3 ) + ( 17 13 4 5 ) ( N + 1 2 )

S U M ( 2 , P S , P T ) = ( 7 3 1 2 ) ( 2 1 1 2 ) + [ ( 7 3 1 2 ) + ( 8 6 5 4 ) ] ( 4 4 2 4 ) = ( 2 + 1 3 ) ( 44 70 28 38 ) + ( 2 + 1 2 ) ( 17 13 4 5 ) = ( 95 109 40 53 )

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

A product = ( K 1 + E 1 ) ( K M + E M ) = F 1 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 , , 2 M parts.

S U M _ K ( N , P S , P T , P F = F 1 F 2 F M ) = Sum of one part, PF indicates the part.

Rewrite 3.1), add {braces}:

S U M ( N , P S , P T ) = i = 1 M A q ( A M q ) , A q = i = 1 M B i

B i = { { ( T i m ) D i } ; X i = T i , m = count of { , X i 1 } K { K i } + { m D i } ; F i R i , X i = K i , m = count of { , X i 1 } T { K i + m D i } ; F i = R i , X i = K i , m = count of { , X i 1 } T

Expand SUM() by {braces}:

3.2) S U M _ K ( N , P S , P T , P F ) = Expansion with same { K i , R i } 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 T , T Ring .

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

F M K = M productwithdifferentfactors K , the sum traverse all combinations.

E M K = M productwithfactors K , the sum traverse all combinations.

F M { 1 , 2 , , N } is abbreviated as F M N , E M { 1 , 2 , , N } is abbreviated as E M N ;

By definition:

E M N = N 1 + + N C = M 1 N 1 2 N 2 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 ) = K i , H 1 ( K , T , M ) = T i D i

[4] has proved:

4.2) H 1 ( [ D × A : D , K 1 : D 1 , , K M : D M ] , [ A , T 1 , , T M ] , C )

= D × A × H 1 ( [ K 1 : D 1 , , K M : D M ] , [ T 1 , , T M ] , C ) + D × A × H 1 ( [ K 1 : D 1 , , K M : D M ] , [ T 1 , , T M ] , C 1 )

this à

4.3) S U M ( N , [ 1 , 2 , , n , K 1 : D 1 , , K M : D M ] , [ 1 , 2 , , n , T 1 , , T M ] ) , P T = [ 1 , , n , T 1 , , T M ] can use the form:

( T 1 + K 1 ) ( T M + K M ) = n ! A q ( N + P B ( P T ) + n I D X ( P T ) q )

[2] has proved:

4.4) H 1 ( [ D × T 1 : D , , D × T M : D ] , [ T 1 , , T M ] , C ) = D M ( M C ) T i

D = 1, this à

4.5) S U M ( N , P T , P T ) = M I N ( P T ) ( N + P B ( P T ) + P M ( P T ) I D X ( P T ) )

This is a generalization of n = 0 N ( n M ) = ( N + 1 M + 1 )

Eg: S U M ( 2 , [ 1 , 2 , 4 ] ) = 1 × 2 × 4 + 1 × 2 × 5 + 2 × 3 × 5 = 1 × 2 × 4 ( 2 + 1 + 3 5 )

4.6) H 1 ( P S , P T , C ) = C S U M ( C , P S , P T ) , I D X ( P T ) S U M ( N > M , ) = T i 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

{ K i + 1 = K i + 1 , H 1 ( [ D × K 1 : D , ] , P T , C ) = D M ( M C ) T 1 T C K C + 1 K M K i 1 = K i + 1 , H 1 ( [ D × K 1 : D , ] , P T , C ) = D M ( M C ) T 1 T C K 1 K M C K i + 1 = K i + 1 , H 1 ( [ D × K 1 : D , ] , P T , C ) = ( 1 ) C D M ( M C ) T 1 T C K 1 K M C K i 1 = K i + 1 , H 1 ( [ D × K 1 : D , ] , P T , C ) = ( 1 ) C D M ( M C ) T 1 T C K C + 1 K M

[ x + y ] n = S U M ( y + 1 , [ x n + 1 , x n + 2 , , x ] ) = ( y n ) ( n n ) n ! [ x ] 0 + ( y n 1 ) ( n n 1 ) ( n 1 ) ! [ x ] 1 + + ( y 0 ) ( n 0 ) 0 ! [ x ] n = ( n n ) [ y ] n [ x ] 0 + ( n n 1 ) [ y ] n 1 [ x ] 1 + + ( n 0 ) [ y ] 0 [ x ] n

à Vandermonde identity [5]: [ x + y ] n = k = 0 n ( n k ) [ x ] k [ y ] n k

[ x + y ] n = S U M ( y + 1 , [ x : 1 , x + 1 : 1 , , x + n 1 : 1 ] ) = ( 1 ) n ( y n ) ( n n ) n ! [ x ] 0 + ( 1 ) n 1 ( y n 1 ) ( n n 1 ) ( n 1 ) ! [ x ] 1 + = ( n n ) [ y ] n [ x ] 0 + ( n n 1 ) [ y ] n 1 [ x ] 1 + + ( n 0 ) [ y ] 0 [ x ] n

à Norlund identity [5]: [ x + y ] n = k = 0 n ( n k ) [ x ] k [ y ] n k

( K 1 + N × D 1 ) ( K M + N × D M ) = S U M ( N + 1 , [ K 1 : D 1 .. K M : D M ] ) = q = 0 M A q ( N q )

4.8) ( K 1 + N × D 1 ) ( K M + N × D M ) can be decomposed to q = 0 M C q ( N q ) by 3.1)

N = x , K i = i 1 , P S = [ 0 , 1 , 2 , , M 1 ] à

[ x ] M = ( x M ) ( M M ) M ! [ M 1 ] 0 + ( x M 1 ) ( M M 1 ) ( M 1 ) ! [ M 1 ] 1 ( x 0 ) ( M 0 ) 0 ! [ M 1 ] M

4.9) [ x ] M = k = 0 M ( M k ) [ x ] k [ M 1 ] M k

N = x , K i = 1 i , P S = [ 0 : 1 , 1 : 1 , , M + 1 : 1 ] à

[ x ] M = ( 1 ) M + 0 ( X M ) ( M M ) M ! [ M 1 ] 0 + ( 1 ) ( M 1 ) + 1 ( X M 1 ) ( M M 1 ) ( M 1 ) ! [ M 1 ] 1

4.10) [ x ] M = k = 0 M ( 1 ) M ( M k ) [ x ] k [ M 1 ] M K

[4] has proved:

4.11) if T i + 1 = T i + 1 , then

H 1 ( [ D × K 1 : D , , D × K M : D ] , [ T 1 , , T M ] , C ) = D C T 1 T C [ F M C D × K E 0 D × { 1 , 2 , , C } + F M C 1 D × K E 1 D × { 1 , 2 , , C } + + F 0 D × K E M C D × { 1 , 2 , , C } ]

This à Ki can exchange order in S U M ( N , [ K 1 , , K M ] ) .

In fact, Ki can exchange order in S U M ( N , [ K 1 : D 1 , , K M : D M ] ) by definition.

This à

4.12) if λ is a primitive unit root, λ M = 1 , then

S U M ( N , [ λ 1 , λ 2 , , λ M ] ) = M ! ( N M + 1 ) E 0 M + ( M 1 ) ! ( N M ) E 1 M 1 + + 1 ! ( N 2 ) E M 1 1 + 0 ! ( N 1 ) ( 1 ) M + 1

( λ 1 + 1 ) ( λ M + 1 ) = S U M ( 2 , [ ] ) S U M ( 1 , [ ] ) = 1 + ( 1 ) M + 1

It’s obvious when M is even; if M is odd then ( λ 1 + 1 ) ( λ M 1 + 1 ) = 1

( λ 1 + 2 ) ( λ M + 2 ) = S U M ( 3 , [ ] ) S U M ( 2 , [ ] ) = 2 M + ( 1 ) M + 1

It can be concluded from the definition:

4.13) 1) H 1 ( [ 1 , , M ] , [ 1 , , 2 M 1 ] , C )

= P M ( ) = M , P B ( ) = C M I N ( P S ) + P M ( ) = M , P B ( ) = C 1 M I N ( P S )

2) H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , C ) = 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 ] ) = n = 0 N 1 ( a + n d ) = d ( N 2 ) + a ( N 1 )

5.3) S U M ( N , [ 1 , 2 , , M ] ) = M ! ( N + M M + 1 )

5.4) S U M ( N , [ 1 , 2 , , M ] , [ 1 , 3 , , 2 M 1 ] ) = I 1 I 2 I M , 1 I 1 < I 2 < < I M N + M 1

5.5) S U M ( N , [ 1 , 1 , , 1 ] , [ 1 , 2 , , M ] ) = 1 M + 2 M + + N M

5.6) S U M ( N , [ 0 : D 1 , , 0 : D M ] , P T ) = S U M _ K ( N , P S , P T , E 1 E M )

5.7) S U M ( N , [ 1 , 1 , , 1 ] , [ 1 , 3 , , 2 M 1 ] ) = E M N

6. Stirling, Lah Number

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

[1] use 4.5) to calculate

S 1 ( N , N M ) = F M N 1 = M I N ( P S ) ( N + P B ( P T ) + P M ( P T ) I D X ( P T ) ) , 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 , , M ] , [ 1 , 3 , , 2 M 1 ] ) = S 1 ( N + M , N ) , this is 5.4)

E M N d e f N 1 + + N C = M 1 N 1 2 N 2 N N C , It’s a known property of S 2 ( N + M , N )

6.2) S U M ( N , [ 1 , , 1 ] , [ 1 , 3 , , 2 M 1 ] ) = E M N = S 2 ( N + M , N )

6.3) 1) H 1 ( [ 1 , , 1 ] , [ 1 , , M ] , K )

= K ! S 2 ( M + 1 , K + 1 ) = K ! i = 0 M K ( M i ) S 2 ( M i , K )

2) H 1 ( [ 1 , , 1 ] , [ 2 , , M ] , K ) = ( K + 1 ) ! S 2 ( M , K + 1 )

[Proof]

Definition of S2(M, K) is

N M = K = 0 M S 2 ( M , K ) [ N ] k = K = 0 M K ! S 2 ( M , K ) ( N K )

( N + 1 ) M = S U M ( N + 1 , [ 1 , , 1 ] , [ 1 , , M ] ) = q = 0 M A q ( N q ) = K = 0 M K ! S 2 ( M , K ) ( N + 1 K ) = K = 0 M K ! S 2 ( M , K ) ( N K + 1 ) + K = 0 M K ! S 2 ( M , K ) ( N K ) à

H 1 ( [ 1 , , 1 ] , [ 1 , , M ] , K ) = K ! S 2 ( M , K ) + ( K + 1 ) ! S 2 ( M , K + 1 ) à the first equation

4.11 ) H 1 ( , K ) = K ! [ F M K [ 1 , , 1 ] E 0 K + + F 0 [ 1 , , 1 ] E M K K ] F M K [ 1 , , 1 ] = ( M M K ) the second equation

4.2 ) H 1 ( [ 1 , , 1 ] , [ 1 , , M ] , K ) = H 1 ( [ 1 , , 1 ] , [ 2 , , M ] , K ) + H 1 ( [ 2 , , M ] , K 1 ) à 2)

q.e.d.

This à S 2 ( N + 1 , M ) = k = M 1 N ( N k ) S 2 ( K , M 1 ) , which is recorded in [5]

Example 6.1:

Directly according to the definition of H1()

H 1 ( , 1 ) = ( 1 × 2 × 2 × × 2 + 1 × 1 × 2 × × 2 + + 1 × 1 × × 1 × 1 ) S 2 ( M + 1 , 2 ) = 2 M 1

H 1 ( , M 1 ) = ( M 1 ) ! ( 1 + 2 + + M ) S 2 ( M + 1 , M ) = ( M + 1 2 )

6.4) ( M K ) M ! K ! = S 1 ( M + 1 , K + 1 ) S 2 ( K , K ) + + S 1 ( M + 1 , M + 1 ) S 2 ( M , K )

[Proof]

H 1 ( [ 1 , , M ] , [ 1 , , M ] , K ) 4.4 ) = ( M K ) M ! 4.11 ) = K ! [ F M K M E 0 K + + F 0 M E M K K ] = K ! [ S 1 ( M + 1 , K + 1 ) S 2 ( K , K ) + + S 1 ( M + 1 , M + 1 ) S 2 ( M , K ) ]

q.e.d.

Definition of Lah number [5] is [ X ] M = k = 0 M L M , K [ x ] k 4.10 )

6.5) L M , K = ( 1 ) M M ! K ! ( M 1 K 1 ) , 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 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 ) ( P I D X ( P T ) ) 0 M O D P , I D X ( P T ) < P

E.g.: 1 × 2 + 2 × 3 + 3 × 4 1 × 3 + 1 × 4 + 2 × 4 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 Z

1) M < P 1 , S U M ( P , [ K 1 : D , , K M : D ] ) 0 M O D P

2) M = P 1 , ( D , P ) = 1 , S U M ( P , [ K 1 : D , , K M : D ] ) 1 M O D P

S U M ( P , [ K 1 , , K M ] ) = K i + ( 1 + K i ) + + ( P 1 + K i ) . Exclude products≡0 MOD P:

Example 7.1:

1 Q + 2 Q + + ( P 1 ) Q 0 M O D P , Q < P 1 ; 1 M O D P , Q = P 1

1 2 × 2 × 3 + 2 2 × 3 × 4 + + ( P 3 ) 2 × ( P 2 ) × ( P 1 ) 1 M O D P , P = 5 ; 0 M O D P , P > 5

1 3 × 2 + 2 3 × 3 + + ( P 2 ) 3 × ( P 1 ) 1 M O D P , P = 5 ; 0 M O D P , P > 5

1 × 2 3 + 2 × 3 3 + + ( P 2 ) × ( P 1 ) 3 1 M O D P , P = 5 ; 0 M O D P , P > 5

1 2 × 2 2 + 2 2 × 3 2 + + ( P 2 ) 2 × ( P 1 ) 2 1 M O D P , P = 5 ; 0 M O D P , P > 5

In 1), let P S = [ K 1 = 1 , , K M ] is a Basic Shape and K M = P 1

Rewrite P S = L 1 L 2 L Q , L i = count of continuity. ( L i , L i + 1 ) means a discontinuity.

L 1 L Q can been slided to [ L Q , L 1 , L 2 , ] , [ L Q 1 , L Q , L 1 , L 2 , ] by S U M ( P , P S ) M O D P

[3] has proved:

7.3) M I N ( P S ) 0 M O D P

{PS} = All of the Basic Shapes [ 1 , , K M = P 1 ] [ 1 , 2 , , P 1 ] can slide to.

Example 7.2:

1 × ( 3 × 4 × 5 ) × ( 7 × 8 ) × 10 + 1 × 3 × ( 5 × 6 × 7 ) × ( 9 × 10 ) + ( 1 × 2 ) × 4 × 6 × ( 8 × 9 × 10 ) + ( 1 × 2 × 3 ) × ( 5 × 6 ) × 8 × 10 = 139260 0 M O D 11

This à [1] has proved:

7.4) M I N ( P S ) 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 × 2 2 + 2 2 × 3 2 + 3 3 × 4 2 ) + ( 1 2 × 3 2 + 2 2 × 4 2 + 4 2 × 1 2 )

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 K M ] ) 0 M O D P , M < P 1 ;

7.5) F ( λ ( P S ) ) ( P 1 L E N ( P S ) ) ( μ 1 + μ 2 + ) ! μ 1 ! μ 2 ! P L E N ( P S ) M O D P , M = P 1

[Proof]

Use CNT(PS) = Count of P S F ( P S ) à F ( P S ) C N T ( P S ) M O D P

Use LEN(PS) =Count of different K i P S

Obviously:

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

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

Use λ ( P S ) to classify PS.

λ ( P S ) : = 1 λ 1 2 λ 2 ( P 1 ) λ p 1 , P M ( P S ) = 1 × λ 1 + 2 × λ 2 + + ( P 1 ) × λ p 1

λ ( P S ) : = 4 1 : 1 4 + 2 4 + 3 4 +

λ ( P S ) : = 1 4 : 1 × 2 × 3 × 4 +

λ ( P S ) : = 1 1 3 1 : 1 3 × 2 + 2 3 × 3 + ; 1 3 × 3 + 2 3 × 4 + ; 1 × 2 3 + 2 × 3 3 + ; 1 × 3 3 + 2 × 4 3 +

λ ( P S ) : = 2 2 : 1 2 × 2 2 + 2 2 × 3 2 + ; 1 2 × 3 2 + 2 2 × 4 2 + ; 1 2 × 4 2 + 2 2 × 5 2 +

λ ( P S ) : = 1 2 2 1 : 1 2 × 2 × 3 + ; 1 × 2 2 × 3 + ; 1 × 2 × 3 2 + ; 1 2 × 3 × 5 +

1) The combination number of LEN(PS) in P 1 = ( P 1 L E N ( P S ) )

2) Record λ ( P S ) as { μ 1 , μ 2 , }

There are μ 1 numbers in λ 1 , λ 2 , equal to x > 0

There are μ 2 numbers in λ 1 , λ 2 , equal to y > 0 , y x

The combination number of μ 1 , μ 2 , = ( μ 1 + μ 2 + ) ! μ 1 ! μ 2 !

Count of products F(PS) = ( P 1 L E N ( P S ) ) ( μ 1 + μ 2 + ) ! μ 1 ! μ 2 !

q.e.d.

Example 7.3:

( 1 3 × 2 + 2 3 × 3 + 3 3 × 4 ) + ( 1 3 × 3 + 2 3 × 4 + 4 3 × 1 ) + ( 1 3 × 4 + 3 3 × 1 + 4 3 × 2 ) + ( 1 × 2 3 + 2 × 3 3 + 3 × 4 3 ) 4 ! 2 ! 2 ! × 2 ! 1 ! 1 ! / ( 5 2 ) 4 M O D P

( 1 2 × 2 2 + 2 2 × 3 2 + 3 2 × 4 2 ) + ( 1 2 × 3 2 + 2 2 × 4 2 + 4 2 × 1 2 ) 4 ! 2 ! 2 ! × 2 ! 2 ! / ( 5 2 ) 2 M O D P

( 1 2 × 2 × 3 + 2 2 × 3 × 4 ) + ( 1 2 × 2 × 4 + 3 2 × 4 × 1 ) + ( 1 2 × 3 × 4 + 4 2 × 1 × 2 ) + ( 1 × 2 2 × 3 + 2 × 3 2 × 4 ) + ( 1 × 2 2 × 4 + 3 × 4 2 × 1 ) + ( 1 × 2 × 3 2 + 2 × 3 × 4 2 ) 4 ! 3 ! 1 ! × 3 ! 2 ! 1 ! / ( 5 3 ) 6 M O D P

7.6) 1) K ! S 2 ( P 1 , K ) = K ! E P 1 K K ( 1 ) K + 1 M O D P , 1 K P 1

2) S 2 ( P , K ) 0 M O D P , 1 < K < P

[Proof]

S U M ( N , [ 1 , 2 , , P 1 ] ) 4.11 ) ( 7 * ) ( P 1 ) ! ( N P ) E 0 P 1 + + 1 ! ( N 2 ) E P 2 1 + 0 ! ( N 1 ) ( P 1 ) ! M O D P

S U M ( N , [ 1 , , P 1 ] ) ( P 1 ) ! ( N 1 P 1 ) E 0 P 1 + + 1 ! ( N 1 1 ) E P 2 1 + 0 ! ( N 1 0 ) ( p 1 ) ! M O D P

[ P ] P 1 = S U M ( 2 , [ 1 , , P 1 ] ) 1 + ( P 1 ) ! 0 M O D P ( P 1 ) ! 1 M O D P

This is a new proof of Wilson theorem.

1 ! E P 2 1 = 1 1 M O D P

[ P + 1 ] P 1 = S U M ( 3 , [ 1 , , P 1 ] ) 0 M O D P 2 ! E P 3 2 1 M O D P

[ P + 2 ] P 1 = S U M ( 4 , [ 1 , , P 1 ] ) 0 M O D P 3 ! E P 4 3 1 M O D P

à 1)

S 2 ( P , K ) = S 2 ( P 1 , K 1 ) + K × S 2 ( P 1 , K ) à

( K 1 ) ! S 2 ( P , K ) = ( K 1 ) ! S 2 ( P 1 , K 1 ) + K ! S 2 ( P 1 , K ) ( 1 ) K + ( 1 ) K + 1 0 M O D P à 2)

q.e.d.

N P 1 1 M O D P = K = 0 P 1 K ! S 2 ( P 1 , K ) ( N K ) 7.6 ) , S 2 ( P 1 , 0 ) = 0

7.7) ( N P 1 ) + ( N P 2 ) + ( N 2 ) + ( N 1 ) 1 M O D P , ( N , P ) = 1

7.8) n = 1 P 1 n P 2 0 M O D P 2 , P > 3

[Proof]

P P 2 + n = 1 P 1 n P 2 = n = 1 P n P 2 = S U M ( P , [ 1 , , 1 ] , [ 1 , , P 2 ] ) 6.3 ) = k = 0 P 2 k ! S 2 ( P 1 , k + 1 ) ( P K + 1 ) = k = 1 P 1 2 [ ( k 1 ) ! S 2 ( P 1 , k ) ( P k ) + ( P k 1 ) ! S 2 ( P 1 , P k ) ( P P k ) ] 7.6 ) k = 1 P 1 2 [ ( 1 ) K + 1 + ( 1 ) P K + 1 ] ( P K ) k = 1 P 1 2 P × λ k ( P K ) M O D P 0 M O D P 2

q.e.d.

8. Form2 and Analysis

Rewrite (1) of section 3 as

n = 0 N 1 n ( n + K M ) = ( M + 1 ) ( N + K + 1 M + 2 ) ( 1 + K ) ( N + K M + 1 )

P S = [ K 1 : D 1 , , K M : D M ] , P T = [ T 1 , , T M ] , use the same method of 3.1) to prove:

8.1) Use the Form 2 = ( T 1 + K 1 ) ( T M + K M ) = X 1 X M ,

S U M ( N , P S , P T ) = A q ( N + T M q I D X ( P T ) q ) , q = count of X K . A q = i = 1 M B i

B i = { ( T i m ) D i ; X i = T i , m = count of { , X i 1 } K K i + ( m T i ) D i ; X i = K i , m = count of { , X i 1 } K

Define H 2 ( P S , P T , C ) = A M q of 8.1), C = Count of X 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 ) ) = T i , This à 4.5)

8.3) H 2 ( [ 1 , , M ] , [ 1 , , 2 M 1 ] , C )

= ( 1 ) M C 1 H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , C 1 )

[Proof]

H 2 ( [ 1 , , 1 ] , [ 1 , , 2 M 1 ] , C ) = i = 1 M B i ,

B i = { ( T i m ) ; X i = T i , m = count of { , X i 1 } K 1 + ( m 2 i + 1 ) ; X i = K i , m = count of { , X i 1 } K m = i 1 q = { ( T i m ) ; X i = T i , m = count of { , X i 1 } K ( ( i 1 ) + q ) ; X i = K i , q = count of { , X i 1 } T

( 1 ) M C H 2 ( , C ) + ( 1 ) M C 1 H 2 ( , C + 1 ) = H 1 ( [ 1 , , M ] , [ 1 , , 2 M 1 ] , C ) 4.2 ) = H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , C ) + H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , C 1 )

H 2 ( [ 1 , , M ] , [ 1 , , 2 M 1 ] , C ) = ( 1 ) M C 1 H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , C 1 )

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 ) = k = 1 M A ( M , k ) ( N 2 M K + 1 )

2) S 2 ( N , N M ) = k = 1 M ( 1 ) k 1 A ( M , k ) ( N + M K 2 M K + 1 )

[Proof]

S 1 ( N , N M ) = S U M ( N M , [ 1 , , M ] , [ 1 , , 2 M 1 ] ) , Form 1 = ( T 2 + K 2 ) ( T M + K M ) , 4.3) à

= q = 0 M 1 H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , q ) ( N M + ( M 1 ) + 1 2 M ( M q ) ) k = M q

= k = 1 M H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , M k ) ( N 2 M k ) à 1)

H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , M K ) = A ( M , K )

S 2 ( N , N M ) = S U M ( N M , [ 1 , , 1 ] , [ 1 , , 2 M 1 ] ) , use the Form2

= q = 0 M H 2 ( [ 1 , , 1 ] , [ 1 , , 2 M 1 ] , q ) ( N M + ( 2 M 1 ) ( M q ) 2 M ( M q ) )

= q = 0 M H 2 ( [ 1 , , 1 ] , [ 1 , , 2 M 1 ] , q ) ( N 1 + q M + q ) H 2 ( , 0 ) = 0

= q = 1 M H 2 ( [ 1 , , 1 ] , [ 1 , , 2 M 1 ] , q ) ( N 1 + q M + q ) K = M q + 1

= K = 1 M H 2 ( [ 1 , , 1 ] , [ 1 , , 2 M 1 ] , M k + 1 ) ( N + M K 2 M K + 1 ) 8.3 )

= k = 1 M ( 1 ) k 1 H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , M k ) ( N + M k 2 M k + 1 ) à 2)

q.e.d.

8.4) 1) C = 0 C = M 1 ( 1 ) M 1 C P M ( P S ) = M , P B ( P S ) = C M I N ( P S ) = 1

2) C = 0 C = M 1 ( 1 ) M 1 C 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 , , 1 ] , [ 1 , , 2 M 1 ] ) Use Form 3

= A q ( N + T M q I D X ( P T ) q ) = A q ( 1 + 2 M 1 q 2 M q ) = A q A 0 = 0 q = 1 q = M H 2 ( , q ) 8.3 )

= C = 0 C = M 1 ( 1 ) M 1 C H 1 ( [ 2 , , M ] , [ 3 , , 2 M 1 ] , C ) 4.13 ) 1)

S 2 ( M + 2 , 2 ) = 2 M + 1 1 = S U M ( 2 , [ 1 , , 1 ] , [ 1 , , 2 M 1 ] ) à 2)

q.e.d.

Example 8.1:

M = 1 : 1 = 1

M = 2 : 1 × 3 1 × 2 = 1

M = 3 : 1 × 3 × 5 ( 1 × 3 × 4 + 1 × 2 × 4 ) + 1 × 2 × 3 = 1

M = 4 : 1 × 3 × 5 × 7 ( 1 × 3 × 5 + 1 × 3 × 4 + 1 × 2 × 4 ) × 6 + ( 1 × 2 × 3 + 1 × 3 × 4 + 1 × 2 × 4 ) × 5 1 × 2 × 3 × 4 = 1

M = 5 : 1 × 3 × 5 × 7 × 9 ( 1 × 3 × 5 × 7 + 1 × 3 × 4 × 6 + 1 × 2 × 4 × 6 + 1 × 3 × 5 × 6 ) × 8 + ( 1 × 2 × 3 × 5 + 1 × 3 × 4 × 5 + 1 × 2 × 4 × 5 + 1 × 3 × 5 × 6 + 1 × 2 × 4 × 6 + 1 × 3 × 4 × 6 ) × 7 ( 1 × 2 × 3 × 4 + 1 × 2 × 3 × 5 + 1 × 2 × 4 × 5 + 1 × 3 × 4 × 5 ) × 6 + 1 × 2 × 3 × 4 × 5 = 1

It can be concluded from the definition:

8.5) H 2 ( [ 1 , , 1 ] , [ 1 , , M ] , C ) = ( 1 ) M C C ! E M C C = ( 1 ) M C C ! S 2 ( M , C )

N M = S U M ( N , [ 1 , , 1 ] , [ 1 , , M ] )

Form 2 A q ( N + T M q I D X ( P T ) q ) = A c ( N + M ( M C ) 1 M + 1 ( M C ) 1 ) = A c ( N + C 1 C )

8.6) N M = K = 0 M K ! S 2 ( M , K ) ( N K ) = K = 0 M ( 1 ) M K K ! S 2 ( M , K ) ( N + K 1 K )

It can be concluded from the definition:

8.7) H 2 ( [ 1 + y , 2 + y , , M + y ] , [ 1 , 2 , , M ] , C ) = C ! [ y ] M C ( M C )

[ x + y ] M = S U M ( x , [ 1 + y , 2 + y , , M + y ] )

Form 2 c = 0 M ( M c ) C ! [ y ] M c ( x + M ( M C ) M + 1 ( M C ) )

= c = 0 M ( M c ) c ! [ y ] M c ( x + C C + 1 ) = c = 0 M ( M c ) c ! [ y ] M c ( x + C 1 C )

= K = 0 M ( M K ) [ y ] M K [ x + k 1 ] K = K = 0 M ( M K ) [ y ] M K [ x ] K à Norlund identity

9. Form3 and Eulerian Number

Rewrite (1) of section 3 as

n = 0 N 1 n ( n + K M ) = ( M K ) ( N + K + 1 M + 2 ) + ( 1 + K ) ( N + K M + 2 )

P S = [ K 1 : D 1 , , K M : D M ] , P T = [ T 1 , , T M ] , use the same method of 3.1) to prove:

9.1) Use the Form 3 = ( T 1 + K 1 ) ( T M + K M ) = X 1 X M ,

S U M ( N , P S , P T ) = A q ( N + T M q I D X ( P T ) ) , q = count of X T . A q = i = 1 M B i ,

B i = { K i + [ T i m ] D i ; X i = T i , m = count of { , X i 1 } T K i + m D i ; X i = K i , m = count of { , X i 1 } T

Define H 3 ( P S , P T , C ) = A q of 9.1), C = Count of X T

M k is Eulerian number. Worpitzky identity: N M = k = 0 M 1 M k ( N + k M )

Already known 1) M M q 1 = M k , 2) M q = ( M q ) M 1 q 1 + ( q + 1 ) M 1 q

9.2) H 3 ( P T , P T , q > 0 ) = 0 ; H 3 ( P T , P T , 0 ) = T i This à 4.5)

9.3) H 3 ( [ 1 , , 1 ] , [ 1 , , M ] , q ) = M M q 1 = M q , H 3 ( [ 0 , , 0 ] , [ 1 , , M ] , q ) = M q 1

[Proof]

Obviously: H 3 ( [ K 1 = 1 , , K M ] , [ T 1 = 1 , , T M ] , M ) = 0

N M = S U M ( N , [ 1 , , 1 ] ) = q = 0 M A q ( N + M q 1 M ) = q = 0 M A q ( N + M q 1 M ) = k = 0 M 1 M k ( N + k M )

q.e.d.

It’s easy to deduce:

(*) H 3 ( q ) = H 2 ( [ 0 , , 0 ] , [ 1 , , M ] , q ) = B i , B i = { m + 1 : X i = T i , m = count of { , X i 1 } K m : X i = K i , m = count of { , X i 1 } T

(*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 , , 0 ] , [ 1 , , M 1 ] , q ) à 2)

(*3) H 3 ( q ) = B i , then B 1 + B 2 + + B M = q ( M q + 1 )

H 3 ( [ 1 , , 1 ] , [ 1 , , M ] , 1 ) = 1 M 1 × ( M 1 ) + 1 M 2 × ( M 2 ) × 2 + 1 M 3 × ( M 3 ) × 2 2 +

= 2 0 × M + 2 1 × M + + 2 M 2 × M ( 2 0 × 1 + 2 1 × 2 + + 2 M 2 × ( M 1 ) )

= M ( 2 M 1 1 ) i = 0 M 2 2 i ( i + 1 ) = M ( 2 M 1 1 ) ( 2 M 1 1 ) i = 0 M 2 2 i i

= M 1 = 2 M ( M + 1 ) , the final equation is a known property of M 1 à

9.4) i = 0 M 2 i i = ( M 1 ) 2 M + 1 + 2

9.5) M q = t 1 + + t q + 1 = M q 1 1 t 1 2 t 2 ( q + 1 ) t q + 1 ( 1 + t 1 ) ( 1 + t 1 + + t q ) , t i 0

[Proof]

H 3 ( [ 0 , , 0 ] , [ 1 , , M ] , q + 1 ) = M q = Π ( X T ) Π ( X K )

If X K is certain, then X T is certain and X 1 X M is certain.

When X 1 X M > 0 ,then X 1 = T 1 = 1 , K i [ 1 , q + 1 ]

Record Π ( X K ) = 1 t 1 2 t 2 ( q + 1 ) t q + 1 , t 1 + t 2 + + t q + 1 = M q 1

Take out factors > 0, record as { P 1 , P 2 , } ,(*)à

Π ( X T ) = 1 P 1 ( 1 + t p 1 ) P 2 P 1 ( 1 + t p 1 + t p 2 ) P 3 P 2 , it can be rewritten as

Π ( X T ) = ( 1 + t 1 ) ( 1 + t 1 + t 2 ) ( 1 + t 1 + t 2 + t 3 ) ( 1 + t 1 + + t q )

q.e.d.

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

Example 9.1:

6 2 = 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

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

[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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