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 OALibJ  Vol.8 No.1 , January 2021
Expansion of the Shape of Numbers
Abstract: This article extends the concept of the shape of numbers. Originally, a shape was defined as [1, K1, K2, ···], 1Ki+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···xKM+(L+K1)x···x(L+KM)+(2L+K1)x···x(2L+KM)+(3L+K1)x···x(3L+KM)+···.

1. Introduction

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

( I 1 , I 2 , , I M ) , I i N , I 1 < I 2 < < 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 , , 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 , ) that 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 = 1 , only I i + 1 I i = 1 is allowed.

The single ( I 1 , I 2 , , I M ) is an item, I 1 × I 2 × × I M is the product. Ii is a factor.

Example:

P S = [ 1 , 2 ] ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) , ( 1000 , 1001 ) P S

P S = [ 1 , 3 ] ( 1 , 3 ) , ( 1 , 4 ) , ( 2 , 4 ) , ( 1 , 5 ) , ( 2 , 5 ) , ( 3 , 5 ) , ( 1000 , 2001 ) P S

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

P S = [ 1 , 4 , 6 ] ( 1 , 4 , 7 ) , ( 1 , 5 , 7 ) , ( 2 , 5 , 7 ) P S , ( 3 , 5 , 7 ) 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 × 2 × 3 , M I N ( [ 1 , 2 , 4 ] ) = 1 × 2 × 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 ] B A S E ( P S ) = [ 1 , 2 ]

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

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

P S = [ 1 , 3 , 5 ] , [ 1 , 4 , 9 ] , [ 1 , K > 2 , X > K + 1 ] 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 × 2 × 4 + 1 × 2 × 5 + 2 × 3 × 5

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

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

(1.2) S U M ( N , P S ) = M I N ( P S ) ( N I D X ( P S ) ) , PS is a Basic Shape

The following uses count of X K for count of

{ X 1 , X 2 , , X M } { K 1 , K 2 , , K M }

(1.3) P S = [ 1 , K 1 , , K M ] , P T = [ 1 , T 1 , T 2 , , T M ]

Use the form ( T 1 + K 1 ) ( T 2 + K 2 ) ( T M + K M ) = X 1 X 2 X M , Xi = Ti or Ki.

The expansion has 2M items, don’t swap the factors of X 1 X 2 X M , then each X 1 X 2 X M corresponds to one expression =

A q ( N P H ( P S ) P C H G ( P S , P T ) I D X ( P T ) q )

q = countof 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 ) I D X ( P T ) q ) .

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

Example:

P S = [ 1 , K 1 3 , K 2 K 1 + 2 , K 3 K 2 + 2 ] ,

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

I D X ( B S ) = 8

Theform = ( 3 + K 1 ) ( 5 + K 2 ) ( 7 + K 3 ) = 3 × 5 × 7 + 3 × 5 × K 3 + 3 × K 2 × 7 + 3 × K 2 × K 3 + K 1 × 5 × 7 + K 1 × 5 × K 3 + K 1 × K 2 × 7 + K 1 × K 2 × K 3

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

à

S U M ( N , P S ) = 3 × 5 × 7 ( P 8 ) + 3 × 5 × ( K 3 + 2 ) ( P 7 ) + 3 × ( K 2 + 1 ) × ( 7 1 ) ( P 7 ) + 3 × ( K 2 + 1 ) × ( K 3 + 1 ) ( P 6 ) + K 1 × ( 5 1 ) × ( 7 1 ) ( P 7 ) + K 1 × ( 5 1 ) × ( K 3 + 1 ) ( P 6 ) + K 1 × K 2 × ( 7 2 ) ( P 6 ) + K 1 × K 2 × K 3 ( P 5 )

Anitem P S = { begin , K 1 + E 1 , , K M + E M } , K is fixed, E is variable.

Aproduct = begin × ( K 1 + E 1 ) ( K M + E M ) = begin × F 1 F 2 F M , Fi = Ei or Ki

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 2 F M ) = Sum of one part in SUM(N, PS).

PF indicates the part. Fi = Ei or Ki

Rewrite 1.3), add {braces}:

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

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

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

(1.4) SUM_K(N, PS, PT, PF) = ∑Expansion of SUM() 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 = countof { X 1 , , X i 1 } K D i : F i = E i , X i = K i , D i = countof { X 1 , , X i 1 } T

Example:

S U M ( N , [ 1 , K 1 3 , K 2 K 1 + 2 ] ) ,

form = ( 3 + K 1 ) ( 5 + K 2 ) à

= 15 ( N K 2 + 3 6 ) + 3 ( { K 2 } + { 1 } ) ( N K 2 + 3 5 ) + K 1 ( { 5 1 } ) ( N K 2 + 3 5 ) + K 1 K 2 ( N K 2 + 3 4 )

Expand by the {braces}:

= { 15 ( N K 2 + 3 6 ) + 3 ( N K 2 + 3 5 ) } + 3 K 2 ( N K 2 + 3 5 ) + 4 K 1 ( N K 2 + 3 5 ) + K 1 K 2 ( N K 2 + 3 4 ) = begin = 1 N K 2 begin × ( K 1 + E 1 , begin ) ( K 2 + E 2 , begin )

à

S U M _ K ( N , P S , B S , E 1 E 2 ) = allitems begin E 1 , i E 2 , i = 15 ( N K 2 + 3 6 ) + 3 ( N K 2 + 3 5 )

S U M _ K ( N , P S , B S , E 1 K 2 ) = allitems begin E 1 , i K 2 = 3 K 2 ( N K 2 + 3 5 )

S U M _ K ( N , P S , B S , K 1 E 2 ) = allitems begin K 1 E 2 , i = 4 K 1 ( N K 2 + 3 5 )

S U M _ K ( N , P S , B S , K 1 K 2 ) = allitems begin K 1 K 2 = K 1 K 2 ( N K 2 + 3 4 )

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 , , K M , ] , Item = ( I 0 , , I M , ) ,

B A S E ( P S ) = B S = [ G 0 = 1 , G 1 , , G M , ]

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

2) allow K 0 K 1 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 ] B A S E ( P S ) = [ 1 , 3 ]

S E T ( 8 , [ 3 , 5 ] ) = { ( 3 , 5 ) , ( 3 , 6 ) , ( 4 , 6 ) , ( 3 , 7 ) , ( 4 , 7 ) , ( 5 , 7 ) } S E T ( 8 , [ 1 , 3 ] ) S E T ( 5 , [ 1 , 3 ] )

P S = [ 3 , 5 , 4 , 6 ] 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 ) = A i ( N n i m i ) , then D 1 f ( n ) = A i ( N n i 1 m i 1 )

2.1) | S E T ( N , P S , P T ) | = ( N K 0 P H ( P S ) P C H G ( P S , P T ) P B ( P T ) + 1 )

2.2) Specify ( N < M M ) = 0 , n = 0 N 1 n ( n K M ) = ( M + 1 ) ( N K M + 2 ) + ( M + K ) ( N K M + 1 )

2.3) P S = [ K 0 , K 1 , , K M ] , P T = [ 1 , T 1 , T 2 , , T M ] , can use the form ( T 0 + K 0 ) ( T M + K M )

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 = 0 M ( X i + D i ) ,

D i = { m : X i = T i , m = countof { X 0 , , X i 1 } K + m : X i = K i , m = countof { X 0 , , X i 1 } T

q = countof X 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 , , G M ]

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

(1*) S U M ( N , P S ) = n = N E N D ( n , P S )

(2*) 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 ] ) = n = N 1 n × E N D ( n , P S )

(4*) S U M ( N , [ P S , K M + 1 = K M ] ) = n = N 1 n × E N D ( n + 1 , P S )

(5*) S U M ( N , [ P S , K M + 1 > 1 + K M ] ) = n = N 1 n × S U M ( n ( K M + 1 K M ) + 1 , P S )

Suppose S U M ( N , P S ) = X 0 X 1 X M ( N P H ( P S ) 1 M i ) , Max factor of PS = KM

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 , , X M } 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 ]

S U M ( N , P S 1 ) = n = N 1 n × E N D ( n , P S ) = n = N 1 n × D 1 S U M ( n , P S ) = n = N 1 n × X 0 X M ( P 1 M i 1 ) = n = N 1 n × X 0 X M ( n [ K M P B ( B S ) + 1 ] M i 1 ) ( 2.2 ) = ( X 0 X M M i ( Q 1 M i + 1 ) + X 0 X M ( M i 1 + K M P B ( B S ) + 1 ) ( Q 1 M i ) )

M i = I D X ( B S ) C = 1 + G M C = G M + 1 C

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

à

S U M ( N , P S 1 ) = X 0 X M ( G M + 1 C ) ( Q 1 I D X ( B S ) C + 1 ) + X 0 X M ( K M + 1 + M + 1 C ) ( Q 1 I D X ( B S ) C ) = X 0 X M ( G M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) C ) + X 0 X M ( K M + 1 + M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) ( C + 1 ) )

à Match the form ( G 0 + K 0 ) ( G 1 + K 1 ) ( 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 ]

S U M ( N , P S 1 ) = n = N 1 n × E N D ( n + 1 , P S ) = n = N 1 n × D 1 S U M ( n + 1 , P S ) = n = N 1 n × X 0 X M ( P M i 1 ) = n = N 1 n × X 0 X M ( n [ K M P B ( B S ) ] M i 1 ) ( 2.2 )

= ( X 0 X M M i ( Q M i + 1 ) + X 0 X M ( M i 1 + K M P B ( B S ) ) ( Q M i ) ) = X 0 X M ( G M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) C ) + X 0 X M ( K M + M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) ( C + 1 ) )

à Match the form ( G 0 + K 0 ) ( G 1 + K 1 ) ( 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 ]

S U M ( N , P S 1 ) = n = N 1 n × S U M ( n ( K M + 1 K M 1 ) , P S ) = n = N 1 n × X 0 X M ( n ( K M + 1 K M 1 ) P H ( P S ) 1 M i ) = n = N 1 n × X 0 X M ( n [ K M + 1 K M + P H ( P S ) ] M i )

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

S U M ( N , P S 1 ) ( 2.2 ) = X 0 X M ( M i + 1 ) ( Q 1 M i + 2 ) + X 0 X M ( K M + 1 K M + P H ( P S ) + M i ) ( Q 1 M i + 1 )

M i + 1 = I D X ( B S ) C + 1 = 1 + G M C + 1 = G M + 1 C

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

à

S U M ( N , P S 1 ) = X 0 X M ( G M + 1 C ) ( Q 1 M i + 2 ) + X 0 X M ( K M + 1 + M + 1 C ) ( Q 1 M i + 1 ) = X 0 X M ( G M + 1 C ) ( Q 1 I D X ( B S ) C + 2 ) + X 0 X M ( K M + 1 + M + 1 C ) ( Q 1 I D X ( B S ) C + 1 )

= X 0 X M ( G M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) C ) + X 0 X M ( K M + 1 + M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) ( C + 1 ) )

à Match the form ( G 0 + K 0 ) ( G 1 + K 1 ) ( 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:

S U M ( N , [ P S , K M + 1 ] ) = n = N 1 ( n + K M + 1 K M ) E N D ( n + 1 , P S ) = n = N 1 ( n + K M + 1 K M ) × D 1 S U M ( n + 1 , P S ) = n = N 1 ( n + K M + 1 K M ) × X 0 X M ( p M i 1 ) = n = N 1 n × X 0 X M ( P M i 1 ) + n = N 1 ( K M + 1 K M ) × X 0 X M ( P M i 1 )

= X 0 X M { M i ( Q M i + 1 ) + ( M i 1 + K M P B ( B S ) ) ( Q M i ) + ( K M + 1 K M ) ( Q M i ) } = X 0 X M { ( I D X ( B S ) C ) ( Q M i + 1 ) + ( M i + K M + 1 P B ( B S ) 1 ) ( Q M i ) }

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

à

S U M ( N , P S 1 ) = X 0 X M ( G M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) C ) + X 0 X M ( K M + 1 + M + 1 C ) ( N P H ( P S 1 ) 1 I D X ( B S 1 ) ( C + 1 ) )

à Match the form ( G 0 + K 0 ) ( 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 ( N + 6 6 ) 118 ( N + 6 5 ) + 315 ( N + 6 4 ) 308 ( N + 6 3 )

15 = 1 × 3 × 5 ;

308 = ( 11 ) × ( 7 ) × ( − 4 )

118 = 1 × 3 × ( 4 + 2 ) + 1 × ( 7 + 1 ) × ( 5 1 ) + ( 11 ) × ( 3 1 ) × ( 5 1 )

315 = 1 × ( 7 + 1 ) × ( 4 + 1 ) + ( 11 ) × ( 3 1 ) × ( 4 + 1 ) + ( 11 ) × ( 7 ) × ( 5 2 )

S U M ( 2 , [ 11 , 7 , 4 ] ) = ( 11 ) × ( 7 ) × ( 4 ) + ( 11 ) × ( 7 ) × ( 3 ) + ( 11 ) × ( 6 ) × ( 3 ) + ( 10 ) × ( 6 ) × ( 3 ) = 315 308 × 4 = 917

S U M ( 1 , [ 11 , 7 , 4 ] ) = S U M ( 2 , [ 11 , 7 , 4 ] ) + ( 11 ) × ( 7 ) × ( 2 ) + ( 11 ) × ( 6 ) × ( 2 ) + ( 11 ) × ( 5 ) × ( 2 ) + ( 11 ) × ( 7 ) × ( 2 ) + ( 11 ) × ( 6 ) × ( 2 ) + ( 11 ) × ( 5 ) × ( 2 ) = 118 + 315 × 5 308 × 10 = 1623

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

= 8 ( N 10 5 ) + 62 ( N 10 4 ) + 200 ( N 10 3 ) + 308 ( N 10 2 )

62 = 1 × 2 × ( 11 + 2 ) + 1 × ( 7 + 1 ) × ( 4 1 ) + 4 × ( 2 1 ) × ( 4 1 )

200 = 1 × ( 7 + 1 ) × ( 11 + 1 ) + 4 × ( 2 1 ) × ( 11 + 1 ) + 4 × 7 × ( 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 11 7 = 4

S U M ( 15 , [ 4 , 7 , 11 ] , [ 1 , 2 , 4 ] ) = 4 × 7 × 11 + 4 × 7 × 12 + 5 × 8 × 12 + 4 × 7 × 13 + 5 × 8 × 13 + 6 × 9 × 13 + 4 × 7 × 14 + 5 × 8 × 14 + 6 × 9 × 14 + 7 × 10 × 14 = 8 + 62 × 5 + 200 × 10 + 308 × 10 = 5398

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 ( N 6 7 ) + 417 ( N 6 6 ) + 922 ( N 6 5 ) + 876 ( N 6 4 ) + 224 ( N 6 3 )

417 = 1 × 3 × 4 × ( 8 + 3 ) + 1 × 3 × ( 1 + 2 ) × ( 6 1 ) + 1 × ( 7 + 1 ) × ( 4 1 ) × ( 6 1 ) + 4 × ( 3 1 ) × ( 4 1 ) × ( 6 1 )

922 = 1 × 3 × ( 1 + 2 ) × ( 8 + 2 ) + 1 × ( 7 + 1 ) × ( 4 1 ) × ( 8 + 2 ) + 4 × ( 3 1 ) × ( 4 1 ) × ( 8 + 2 ) + 1 × ( 7 + 1 ) × ( 1 + 1 ) × ( 6 2 ) + 4 × ( 3 1 ) × ( 1 + 1 ) × ( 6 2 ) + 4 × 7 × ( 4 2 ) × ( 6 2 )

876 = 4 × 7 × 1 × ( 6 3 ) + 4 × 7 × ( 4 2 ) × ( 8 + 1 ) + 4 × ( 3 1 ) × ( 1 + 1 ) × ( 8 + 1 ) + 1 × ( 7 + 1 ) × ( 1 + 1 ) × ( 8 + 1 )

S U M ( 13 , [ 4 , 7 , 1 , 8 ] ) = 4 × 7 × 1 × 8 + 4 × 7 × 1 × 9 + ( 4 + 5 ) × 8 × 2 × 9 + 4 × 7 × 1 × 10 + ( 4 + 5 ) × 8 × 2 × 10 + ( 4 + 5 + 6 ) × 9 × 3 × 10 + 4 × 7 × 1 × 11 + ( 4 + 5 ) × 8 × 2 × 11 + ( 4 + 5 + 6 ) × 9 × 3 × 11 + ( 4 + 5 + 6 + 7 )

× 10 × 4 × 11 + 4 × 7 × 1 × 12 + ( 4 + 5 ) × 8 × 2 × 12 + ( 4 + 5 + 6 ) × 9 × 3 × 12 + ( 4 + 5 + 6 + 7 ) × 10 × 4 × 12 + ( 4 + 5 + 6 + 7 + 8 ) × 11 × 5 × 12 = 72 + 417 × 7 + 922 × 21 + 876 × 35 + 224 × 35 = 60853

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

Anitem P S = { K 0 + E 0 , K 1 + E 1 , , K M + E M } , K is fixed, E is variable.

Aproduct = ( K 0 + E 0 ) × ( K 1 + E 1 ) ( K M + E M ) = F 0 F 1 F 2 F M , F i = E i or F i = K i

That is, a product can be broken down into 2M+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 )

Example:

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

S U M ( 13 , [ 4 , 7 , 11 ] ) = 4 × 7 × 11 + 4 × 7 × ( 11 + 1 ) + 4 × ( 7 + 1 ) × ( 11 + 1 ) + ( 4 + 1 ) × ( 7 + 1 ) × ( 11 + 1 ) = { 4 × 7 × 11 + 4 × 7 × 11 + 4 × 7 × 11 + 4 × 7 × 11 } + { 4 × 7 × 1 + 4 × 7 × 1 + 4 × 7 × 1 } + { 4 × 1 × 11 + 4 × 1 × 11 } + { 1 × 7 × 11 } + { 4 × 1 × 1 + 4 × 1 × 1 } + { 1 × 7 × 1 } + { 1 × 1 × 11 } + { 1 × 1 × 1 }

4 × 7 × 11 308

S U M _ K ( 13 , P S , B S , K 0 K 1 K 2 ) = { 4 × 7 × 11 + 4 × 7 × 11 + 4 × 7 × 11 + 4 × 7 × 11 } = 308 ( N 9 3 )

4 × 7 × ( 5 2 ) 4 × 7 × 3

S U M _ K ( 13 , P S , B S , K 0 K 1 E 2 ) = { 4 × 7 × 1 + 4 × 7 × 1 + 4 × 7 × 1 } = 4 × 7 × 3 ( N 9 4 )

4 × ( 3 1 ) × ( 11 + 1 ) 4 × 2 × 11

S U M _ K ( 13 , P S , B S , K 0 E 1 K 2 ) = { 4 × 1 × 11 + 4 × 1 × 11 } = 4 × 2 × 11 ( N 9 4 )

1 × ( 7 + 1 ) × ( 11 + 1 ) 1 × 7 × 11

S U M _ K ( 13 , P S , B S , E 0 K 1 K 2 ) = { 1 × 7 × 11 } = 1 × 7 × 11 ( N 9 4 )

4 × ( 3 1 ) × ( 5 1 ) + 4 × ( 3 1 ) × ( 11 + 1 ) 4 × 3 × 4 + 4 × 2 × 1

S U M _ K ( 13 , P S , B S , K 0 E 1 E 2 ) = { 4 × 1 × 1 + 4 × 1 × 1 } = 4 × 3 × 4 ( N 9 5 ) + 4 × 2 × 1 ( N 9 4 )

1 × ( 7 + 1 ) × ( 5 1 ) + 1 × ( 7 + 1 ) × ( 11 + 1 ) 1 × 7 × 4 + 1 × 7 × 1

S U M _ K ( 13 , P S , B S , E 0 K 1 E 2 ) = { 1 × 7 × 1 } = 1 × 7 × 4 ( N 9 5 ) + 1 × 7 × 1 ( N 9 4 )

1 × 3 × ( 11 + 2 ) + 1 × ( 7 + 1 ) × ( 11 + 1 ) 1 × 3 × 11 + 1 × 1 × 11

S U M _ K ( 13 , P S , B S , E 0 E 1 K 2 ) = { 1 × 1 × 11 } = 1 × 3 × 11 ( N 9 5 ) + 1 × 1 × 11 ( N 9 4 )

1 × 3 × 5 + [ 1 × 3 × ( 11 + 2 ) + 1 × ( 7 + 1 ) × ( 5 1 ) + 4 × ( 3 1 ) × ( 5 1 ) ] + [ 1 × ( 7 + 1 ) × ( 11 + 1 ) + 4 × ( 3 1 ) × ( 11 + 1 ) + 4 × 7 × ( 5 2 ) ] 1 × 3 × 5 + [ 1 × 3 × 2 + 1 × 1 × 4 + 0 ] + [ 1 × 1 × 1 + 0 + 0 ] = 15 + [ 10 ] + [ 1 ]

S U M _ K ( 13 , P S , B S , E 0 E 1 E 2 ) = { 1 × 1 × 1 } = 15 ( N 9 6 ) + 10 ( N 9 5 ) + ( N 9 4 )

3. Coefficient Analysis

K = [ K 1 , , K M ] ,

T = [ T 1 , , T M ]

Use the form ( T 1 + K 1 ) ( T M + K M ) = X 1 X 2 X M , Xi = Ti or Ki

Define H ( K , T , N , S ) = B N , 0 N M , N = countof X T

B N = i = 1 M ( X i + D i ) , D i = { m S : X i = T i , m = countof { X 1 , , X i 1 } K + m S : X i = K i , m = countof { X 1 , , X i 1 } T

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

3.1) H ( K , T , M ) = T 1 × T 2 × × T M , H ( K , T , 0 ) = K 1 × K 2 × × K M

[3] has proved:

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

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

3.2) H ( [ 1 , 1 , , 1 ] , [ 2 , 3 , , M ] , N ) = ( M N ) ! × S 2 ( M , M N )

S U M ( N , [ K 0 = 1 , K 1 , , K M ] , [ T 0 = 1 , T 1 , , T M ] )

can use the form = ( T 1 + K 1 ) ( T M + K M ) or ( T 0 + K 0 ) ( T 1 + K 1 ) ( T M + K M )

For arbitrary K, T:

3.3) H ( [ P , K ] , [ P , T ] , N , S ) = P × H ( K , T , N , S ) + P × H ( K , T , N 1 , S )

[Proof]

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 ] × S ) + H ( [ P , K ] , [ P , T ] , N , S ) ( T M + 1 [ M + 1 N ] × S ) = { P × H ( K , T , N + 1 , S ) + P × H ( K , T , N , S ) } ( K M + 1 + [ N + 1 ] × S ) + { P × H ( K , T , N , S ) + P × H ( K , T , N 1 , S ) } ( T M + 1 [ M + 1 N ] × S )

= P × { H ( K , T , N + 1 , S ) ( K M + 1 + [ N + 1 ] × S ) + H ( K , T , N , S ) ( T M + 1 [ M N ] × S ) } + P × { H ( K , T , N , S ) ( K M + 1 + N × S ) + H ( K , T , N 1 , S ) ( T M + 1 [ M ( N 1 ) ] × S ) } = P × H ( [ K , K M + 1 ] , [ T , T M + 1 ] , N + 1 , S ) + P × H ( [ K , K M + 1 ] , [ T , T M + 1 ] , N , S )

q.e.d.

this à

3.4) S U M ( N , [ 1 , 2 , , n , K 1 , , K M ] , [ 1 , 2 , , n , T 1 , , T M ] )

can use the form: ( T 1 + K 1 ) ( T M + K M ) = n ! A q ( N P H ( P S ) P C H G ( P S , P T ) + n 1 I D X ( P T ) q )

[2] has proved:

3.5) H ( K , K , N , S ) = ( M N ) K 1 × K 2 × × 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 ) = ( M N ) T 1 T N × K N + 1 K M

[Proof]

Suppose H ( K , T , N , S ) = ( M N ) T 1 T N K N + 1 K N + 2 K M

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 ] × S ) + H ( K , T , N , S ) ( T M + 1 [ M N ] × S )

= ( M N + 1 ) T 1 T N + 1 K N + 2 K M ( K M + 1 + [ N + 1 ] × S ) + ( M N ) T 1 T N K N + 1 K M ( T M + 1 [ M N ] × S ) = T 1 T N K N + 2 K M ( M N + 1 ) T N + 1 ( K M + 1 + [ N + 1 ] × S ) + T 1 T N K N + 2 K M ( M N ) K N + 1 ( T M + 1 [ M N ] × S )

= T 1 T N K N + 2 K M ( M N + 1 ) [ T N + 1 K M + 1 + T N + 1 ( N + 1 ) × S ] + T 1 T N K N + 2 K M ( M N ) [ K M + 1 [ M N ] × S ] × ( [ T N + 1 + [ M N ] × S ] [ M N ] × S )

= T 1 T N K N + 2 K M [ ( M N + 1 ) T N + 1 K M + 1 + ( M N ) T N + 1 K M + 1 ] + T 1 T N K N + 2 K M [ ( M N + 1 ) T N + 1 ( N + 1 ) ( M N ) T N + 1 ( M N ) ] × S = T 1 T N K N + 2 K M [ ( M N + 1 ) T N + 1 K M + 1 + ( M N ) T N + 1 K M + 1 ] = ( M + 1 N + 1 ) T 1 T N + 1 K N + 2 K M + 1

à H ( [ K , K M + 1 ] , [ T , T M + 1 ] , N + 1 , S ) holds

q.e.d.

Define

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

E Q K = Q product with factors K , the sum traverse all combinations.

F Q { 1 , 2 , , N } is abbreviated as F Q N , E Q { 1 , 2 , , 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 T N [ F M N K E 0 N + F M N 1 K E 1 N + + + F 0 K E M N N ]

[Proof]

Suppose H(K, T, N) holds

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 T N + 1 [ F M N 1 K E 0 N + 1 + F M N 2 K E 1 N + 1 + + F 0 K E M N 1 N + 1 ] ( K M + 1 + N + 1 ) + T 1 T N [ F M N K E 0 N + F M N 1 K E 1 N + + F 0 K E M N N ] ( T M + 1 [ M N ] ) = T 1 T N + 1 [ F M N 1 K E 0 N + 1 + F M N 2 K E 1 N + 1 + + F 0 K E M N 1 N + 1 ] ( K M + 1 + N + 1 ) + T 1 T N + 1 [ F M N K E 0 N + F M N 1 K E 1 N + + F 0 K E M N N ]

H ( [ K , K M + 1 ] , [ T , T M + 1 = 1 + T M ] , N + 1 ) / T 1 T N + 1 = [ F M N 1 K E 0 N + 1 + F M N 2 K E 1 N + 1 + + 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 + + F 0 K E M N N ]

Sumofallitemswithcountoffactors 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

à H ( [ K , K M + 1 ] , [ T , T M + 1 ] , N + 1 ) holds

q.e.d.

Example:

H ( [ A , B , C , D ] , [ 1 , 2 , 3 , 4 ] , 2 ) = 1 × 2 × ( C + 2 ) ( D + 2 ) + 1 × ( B + 1 ) × ( 3 1 ) ( D + 2 ) + A × ( 2 1 ) × ( 3 1 ) ( D + 2 ) + 1 × ( B + 1 ) ( C + 1 ) ( 4 2 ) + A × ( 2 1 ) × ( C + 1 ) ( 4 2 ) + A B ( 3 1 ) ( 4 2 )

= 1 × 2 [ ( C + 2 ) ( D + 2 ) + ( B + 1 ) ( D + 2 ) + A ( D + 2 ) + ( B + 1 ) ( C + 1 ) + A ( C + 1 ) + A B ] = 1 × 2 [ ( C D + B D + A D + B C + A C + A B ) + ( C + D + B + A ) ( 1 + 2 ) + ( 1 × 2 + 1 × 1 + 2 × 2 ) ]

3.8) In S U M ( N , [ K 1 , , K M ] , [ 1 , 2 , , M ] ) , Ki can switch the order.

3.9) if T i + S = T i + 1 , then

H ( K , T , N , S ) = T 1 T N [ F M N K E 0 { S , 2 S , , N S } + F M N 1 K E 1 { S , 2 S , , N S } + + F 0 K E M N N { S , 2 S , , N S } ]

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

H ( [ 1 , , M 1 ] , [ 1 , , M 1 ] , N ) = ( M 1 ) ! ( M 1 N ) = N ! [ F M 1 N M 1 E 0 N + F M 1 N 1 M 1 E 1 N + + 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 + + S 1 ( M , M ) E M 1 N N ] à

3.10) S 1 ( M , N + 1 ) E 0 N + S 1 ( M , N + 2 ) E 1 N + + S 1 ( M , M ) E M 1 N N = ( M 1 N ) ( M 1 ) ! N !

4. K 1 × × K M + ( L + K 1 ) × × ( L + K M ) + ( 2 L + K 1 ) × × ( 2 L + K M ) +

K = [ K 1 , , K M ] , T = [ T 1 , , T M ] , Max Factor = KM

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

When N = K M + 1

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

à S U M ( N , P S , P T ) = A M

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

SUM(N, PS, PT) can be broken down into 2M 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 ) ( T M + K M ) = 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 = countof { X 1 , , X i 1 } K D i : F i = E i , X i = K i , D i = countof { X 1 , , X i 1 } T

When Ti and Di all increase L times. If P T = [ 1 , 2 , , 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 , , M ] )

S U M L ( N , P S , L ) = K 1 × × K M + ( L + K 1 ) × × ( L + K M ) + + ( ( N 1 ) L + K 1 ) × × ( ( N 1 ) L + K M )

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 , , T M ] = [ 1 × L , 2 × L , , M × L ] , can use the from ( T 1 + K 1 ) ( T M + K M )

S U M L ( N , P S , L ) = A q ( N M + 1 q ) , q = countof X K , 2M items in total.

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

Example:

S U M L ( N , [ 3 , 5 , 8 ] , 4 ) à

form = ( 1 × 4 + 3 ) ( 2 × 4 + 5 ) ( 3 × 4 + 8 ) = ( 4 + 3 ) ( 8 + 5 ) ( 12 + 8 ) à

= 384 ( N 4 ) + 896 ( N 3 ) + 636 ( N 2 ) + 120 ( N 1 )

384 = 4 × 8 × 12 ; 120 = 3 × 5 × 8

896 = 4 × 8 × ( 8 + 2 × 4 ) + 4 × ( 5 + 1 × 4 ) × ( 12 1 × 4 ) + 3 × ( 8 1 × 4 ) × ( 12 1 × 4 )

636 = 3 × 5 × ( 12 2 × 4 ) + 3 × ( 8 1 × 4 ) × ( 8 + 1 × 4 ) + 4 × ( 5 + 1 × 4 ) × ( 8 + 1 × 4 )

S U M L ( 6 , [ 3 , 5 , 8 ] , 4 ) = 3 × 5 × 8 + 7 × 9 × 12 + 11 × 13 × 16 + 15 × 17 × 20 + 19 × 21 × 24 + 23 × 25 × 28 = 384 × 15 + 896 × 20 + 636 × 15 + 120 × 6 = 33940

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

4.2) P is a prime number, For arbitrary K 1 , K 2 , , K M :

If M < P 1 , then

K 1 × K 2 × × K M + ( L + K 1 ) × × ( L + K M ) + + ( ( P 1 ) L + K 1 ) × × ( ( P 1 ) L + K M ) 0 M O D P

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

K 1 × K 2 × × K M + ( L + K 1 ) × × ( L + K M ) + + ( ( P 1 ) L + K 1 ) × × ( ( P 1 ) L + K M ) 1 M O D P

[Proof]

Theexpression = S U M L ( P , P S , L ) = A q ( P M + 1 q )

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

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

S U M L ( N , P S , L ) = L P 1 ( P 1 ) ! ( P P ) + A q ( P M + 1 q ) , q > 0 L P 1 ( P 1 ) ! 1 M O D P

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 2K−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 , , K M ] , tries to calculate SUM(N, PS), the form ( G 1 + K 1 ) ( G 2 + K 2 ) ( 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 + + 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 ( N M ) + ( N M + 1 ) = ( N + 1 M + 1 ) .

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

[1]   Peng, J. (2020) Shape of Numbers and Calculation Formula of Stirling Numbers. Open Access Library Journal, 7, 1-11. 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, 1-14. https://doi.org/10.4236/oalib.1106719

[3]   Peng, J. (2020) Subset of the Shape of Numbers. Open Access Library Journal, 7, 1-15.
https://doi.org/10.4236/oalib.1107040

 
 
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