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 OALibJ  Vol.7 No.9 , September 2020
Subdivide the Shape of Numbers and a Theorem of Ring
Abstract: This article is based on the concept of Shape of numbers, introduce more shapes, obtain the calculation formulas and find an explanation of the formulas. By observing and associating, show a law about the symmetry of Ring.

1. Introduction

Peng, J. introduced the definition of Shape of numbers in [1]:

( K 1 , K 2 , , K M ) , K i N , K 1 < K 2 < < K M

there are M-1 intervals between adjacent numbers. Use A for continuity and B for discontinuity, record as a string of M-1 characters (e.g. AABB…) to represents a catalog.

Define collection of a catalog as Shape of numbers. Use the symbol PX to represent a catelog (if M = 1 then PX = 1).

The single ( K 1 , K 2 , , K M ) is an Item, K 1 K 2 K M is the product of an item.

For example:

( 1 , 2 , 4 ) , ( 1 , 2 , 8 ) , ( 2 , 3 , 6 ) P X = A B , ( 1 , 3 , 5 ) , ( 1 , 3 , 6 ) , ( 2 , 4 , 6 ) P X = B B ,

( 1 , 2 , 5 , 6 ) , ( 2 , 3 , 6 , 7 ) , ( 2 , 3 , 7 , 8 ) , ( 1002 , 1003 , 6789 , 6790 ) P X = A B A

P M ( P X ) = Count of numbers of PX, P A ( P X ) = Count of A, P B ( P X ) = Count of B

P M ( P X ) = P A ( P X ) + P B ( P X ) + 1

| P X | = Count of items belonging to PX

M I N ( P X ) = Minimum product of PX: M I N ( A A ) = 1 × 2 × 3 , M I N ( A B ) = 1 × 2 × 4

I D X ( P X ) = 2 + P A ( P X ) + 2 × P B ( P X ) = P M ( P X ) + P B ( P X ) + 1 : I D X ( A A ) = 4 , I D X ( A B ) = 5

S U M ( N , P X ) = Sum of all product of items belonging to PX in [ 1 , N 1 ]

For example: S U M ( 6 , A B ) = 1 × 2 × 4 + 1 × 2 × 5 + 2 × 3 × 5

E N D ( N , P X ) = Set of items belonging to PX with the maximum factor = N-1

For example: E N D ( 6 , B ) = { ( 1 , 5 ) , ( 2 , 5 ) , ( 3 , 5 ) }

[1] obtain the conclusion:

| Itemsin S U M ( N , P X ) | = ( N P M ( P X ) P B ( P X ) + 1 ) = ( N I D X ( P X ) + P B ( P X ) + 1 P B ( P X ) + 1 ) (1)

S U M ( N , P X ) = M I N ( P X ) ( N I D X ( P X ) ) (2)

Definition: Subdividethe ( K 1 , K 2 , , K M ) by interval of adjacent numbers. If the discontinuity interval is D > 1, the interval of adjacent numbers ≥ D is classified into a same catelog. Use the symbol PY represent a catelog and represented by [min item]

For example:

P Y = [ 1 , 3 ] , ( 1 , 3 ) , ( 1 , 4 ) , ( 2 , 4 ) , ( 1 , 5 ) , ( 2 , 5 ) , ( 3 , 5 ) P Y . Same as PX = B

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

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

Redefinition: PB(PY) = Count of discontinuity intervals in PY, compatible with PX

Redefinition: IDX(PY) = The maximum factor of MIN(PY) + 1, compatible with PX

Definition: BASE(PY) = PX, If PB(PX) = PB(PY), PM(PX) = PM(PY), PX has discontinuity intervals at the same positions of PY.

For example:

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

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

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

Definition: if f ( n ) = K i ( N n i m i ) , then D 1 f ( n ) = K i ( N n i 1 m i 1 )

Definition: P H ( P Y ) = I D X ( P Y ) P B ( P Y ) 2 = Maximum factor of M I N ( P Y ) P B ( P Y ) 1

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

P H ( P Y ) = K M I D X ( B S ) + P M ( B S ) (1.0)

[Proof]

P H ( P Y ) = K M P B ( P Y ) 1 = K M P B ( B S ) 1 = K M G M + M = K M ( G M + 1 ) + ( M + 1 ) = Right

n = K N 1 n ( n K M ) = ( M + 1 ) ( N K M + 2 ) + ( M + K ) ( N K M + 1 ) (1.1)

[Proof]

Left = n = K N 1 ( n K + 1 ) ( n K M ) + ( K 1 ) n = K N 1 ( n K M ) = n = K N 1 ( M + 1 ) ( n K + 1 M + 1 ) + ( K 1 ) n = K N 1 ( n K M ) = ( M + 1 ) ( N K + 1 M + 2 ) + ( K 1 ) ( N K M + 1 ) = ( M + 1 ) ( N K M + 2 ) + ( M + 1 ) ( N K M + 1 ) + ( K 1 ) ( N K M + 1 ) = Right

By definition:

S U M ( N , P Y ) = n = 0 N E N D ( n , P Y ) (1.2)

Derived from (1.2)

E N D ( N , P Y ) = D 1 S U M ( N , P Y ) (1.3)

By definition:

S U M ( N , [ 1 , , K , K + 1 ] ) = n = 0 N 1 n × E N D ( n , [ 1 , , K ] ) (1.4)

S U M ( N , [ 1 , , K , X > K + 1 ] ) = n = 0 N 1 n × S U M ( n X + K + 1 , [ 1 , , K ] ) (1.5)

According to the method in [1]:

| Itemsin S U M ( N , P Y ) | = ( N I D X ( P Y ) + P B ( P Y ) + 1 P B ( P Y ) + 1 ) = ( N P H ( P Y ) 1 P B ( P Y ) + 1 ) , (1.6)

compatible with (1).

2. Calculation Formula

If P Y = B A S E ( P Y ) , the calculation formula has been given by (2).

Otherwise, it can be deduced from (1.1)-(1.5).

S U M ( N , [ 1 , K 3 ] ) = 3 ( N K + 2 4 ) + K ( N K + 2 3 ) (2.1)

[Proof]

Left 1.2 ) n = 0 N E N D ( n , [ 1 , K ] ) = n = K + 1 N ( n 1 ) ( 1 + 2 + + ( n K ) ) = n = K + 1 N ( n 1 ) ( n K + 1 2 ) = n = K N 1 n ( n K + 2 2 ) 1.1 ) Right

S U M ( N , [ 1 , 2 , K > 3 ] ) = 2 × 4 ( N K + 3 5 ) + 2 × K ( N K + 3 4 ) (2.2)

[Proof]

1.5 ) n = 0 N 1 n × S U M ( n K + 3 , [ 1 , 2 ] ) ( 2 ) n = 0 N 1 n × 2 ( n K + 3 3 ) 1.1 ) Right

S U M ( N , [ 1 , K 3 , X = K + 1 ] ) = 12 ( N X + 2 5 ) + 3 ( X + 1 ) ( N X + 2 4 ) + 3 K ( N X + 2 4 ) + K X ( N X + 2 3 ) (2.3)

[Proof]

1.2 ) 1.4 ) 2.1 ) n = 0 N 1 n × D 1 [ 3 ( n K + 2 4 ) + K ( n K + 2 3 ) ] = n = 0 N 1 n × [ 3 ( n K + 1 3 ) + K ( n K + 1 2 ) ] 1.1 ) = 12 ( N K + 1 5 ) + 3 ( K + 2 ) ( N K + 1 4 ) + 3 K ( N K + 1 4 ) + K ( K + 1 ) ( N K + 1 3 ) = Right

S U M ( N , [ 1 , K 3 , X > K + 1 ] ) = 15 ( N X + 3 6 ) + 3 ( X + 1 ) ( N X + 3 5 ) + 4 K ( N X + 3 5 ) + K X ( N X + 3 4 ) (2.4)

[Proof]

1.5 ) 2.1 ) n = 0 N 1 n × [ 3 ( n X + 3 4 ) + K ( n X + 3 3 ) ] 1.1 ) Right

2.1. S U M ( N , [ 1 , K 1 3 , , K M ] )

P Y = [ 1 , K 1 3 , , K M ] , B A S E ( P Y ) = B S = [ 1 , G 1 , , G M ]

1*) S U M ( N , P Y ) = A i ( P M i ) , 2 M items in total.

2*) M i = I D X ( B S ) , I D X ( B S ) 1 , , I D X ( B S ) M

3*) P = N P H ( P Y )

Use the form ( G 1 + K 1 ) ( G 2 + K 2 ) ( G M + K M ) = X 1 X 2 X M . The expansion function has 2M items in total.

4*) M i = I D X ( B S ) ( Count of X K )

5*) A i = i = 1 M ( X i + D i ) ,

D i = { m , X i G , m = countof { X i X i 1 } K 5.1 * ) + m , X i K , m = countof { X i X i 1 } G 5.2 * )

[Proof]

P Y = [ 1 , K 1 3 , , K M ] , B S = B A S E ( P Y )

Suppose S U M ( N , P Y ) = X 1 X M ( P M i ) , P = N P H ( P Y ) .

According to inductive hypothesis:

( G 1 + K 1 ) ( G 2 + K 2 ) ( G M + K M ) { ( G M + 1 ) + ( K M + 1 ) } = X 1 X 2 X M { ( G M + 1 ) + ( K M + 1 ) }

C = Count of X K , M C = Count of X G , M i = I D X ( B S ) C

P Y 1 = [ P Y , K M + 1 ] , B S 1 = B A S E ( P Y 1 ) (2.1.1)

S U M ( N , P Y 1 ) 1 , 2 ) 1.3 ) n = 0 N 1 n × D 1 S U M ( n , P Y ) = n = 0 N 1 n × X 1 X M ( P 1 M i 1 ) 1.1 ) = ( X 1 X M M i ( P 1 M i + 1 ) + X 1 X M ( P H ( P Y ) + M i ) ( P 1 M i ) ) = X 1 X M ( I D X ( B S ) C ) ( P 1 M i + 1 ) + X 1 X M ( P H ( P Y ) + I D X ( B S ) C ) ( P 1 M i )

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

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

1*) is obvious.

Mi change form I D X ( B S ) , , I D X ( B S ) M to I D X ( B S ) + 1 , , I D X ( B S ) M = I D X ( B S 1 ) , , I D X ( B S 1 ) ( M + 1 )

2*) proved

P 1 = N P H ( P Y ) 1 = N ( K M I D X ( B S ) + P M ( B S ) ) 1 = N { ( K M + 1 ) ( I D X ( B S ) + 1 ) + ( P M ( B S ) + 1 ) } = N ( K M + 1 I D X ( B S 1 ) + P M ( B S 1 ) ) = N P H ( P Y 1 )

3*) proved

X 1 X M ( G M + 1 C ) ( P 1 I D X ( B S 1 ) C ) 4*) 5.1*)

X 1 X M ( G M + 1 C ) ( P 1 I D X ( B S 1 ) ( C + 1 ) ) 4*) 5.2*)

4*) 5*) proved

P Y 1 = [ P Y , K M + 1 > K M + 1 ] , B S 1 = B A S E ( P Y 1 ) (2.1.2)

S U M ( N , P Y 1 ) 1.5 ) n = 0 N 1 n × S U M ( n K M + 1 + K M + 1 , P Y ) = n = 0 N 1 n × X 1 X M ( n K M + 1 + K M + 1 P H ( P Y ) M i ) = n = 0 N 1 n × X 1 X M ( n K M + 1 + K M + 1 ( K M I D X ( B S ) + P M ( B S ) ) M i ) = n = 0 N 1 n × X 1 X M ( n ( K M + 1 I D X ( B S ) + P M ( B S ) 1 ) M i )

= X 1 X M ( M i + 1 ) ( P 1 M i + 2 ) + X 1 X M ( K M + 1 I D X ( B S ) + M + M i ) ( P 1 M i + 1 ) = X 1 X M ( I D X ( B S ) C + 1 ) ( P 1 M i + 2 ) + X 1 X M ( K M + 1 + M C ) ( P 1 M i + 1 )

= X 1 X M ( G M + 1 C ) ( P 1 I D X ( B S ) C + 2 ) + X 1 X M ( K M + 1 + M C ) ( P 1 I D X ( B S ) C + 1 ) = X 1 X M ( G M + 1 C ) ( P 1 I D X ( B S 1 ) C ) + X 1 X M ( K M + 1 + M C ) ( P 1 I D X ( B S 1 ) ( C + 1 ) )

1*) is obvious.

Mi change form I D X ( B S ) , , I D X ( B S ) M to I D X ( B S ) + 2 , , I D X ( B S ) M + 1 = I D X ( B S 1 ) , , I D X ( B S 1 ) ( M + 1 )

2*) proved

P 1 = N { K M + 1 ( I D X ( B S ) + 2 ) + ( P M ( B S ) + 1 ) } = N P H ( P Y 1 )

3*) proved

X 1 X M ( G M + 1 C ) ( P 1 I D X ( B S 1 ) C ) 4*) 5.1*)

X 1 X M ( K M + 1 + M C ) ( P 1 I D X ( B S 1 ) ( C + 1 ) ) 4*) 5.2*)

4*) 5*) proved

q.e.d.

Example 2.1:

( 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 K 3 + I D X ( [ 1 , 3 , 5 , 7 ] ) P M ( [ 1 , 3 , 5 , 7 ] ) = N K 3 + 8 4 = N K 3 + 4

S U M ( N , [ 1 , K 1 3 , K 2 K 1 + 2 , K 3 K 2 + 2 ] ) = 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 7 ) + 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 )

2.2. S U M ( N , [ 1 , 2 , , n , K 1 n + 2 , , K M ] )

P Y = [ 1 , 2 , 3 , , n , K 1 n + 2 , , K M ] ,

B A S E ( P Y ) = B S = [ 1 , 2 , 3 , , n , G 1 , , G M ]

1*) S U M ( N , P Y ) = A i ( P M i ) , 2 M items in total.

2*) M i = I D X ( B S ) , I D X ( B S ) 1 , , I D X ( B S ) M

3*) P = N K M + I D X ( B S ) ( M + 1 ) = N ( K M P B ( P Y ) 1 ) = N P H ( P Y )

Use the form ( G 1 + K 1 ) ( G 2 + K 2 ) ( G M + K M ) = X 1 X 2 X M .

4*) M i = I D X ( B S ) ( Countof X K )

5*) A i = n ! × (Same as 2.1)

Example 2.2:

( 5 + K 1 ) ( 7 + K 2 ) = 5 × 7 + 5 × K 2 + K 1 × 7 + K 1 × K 2 , P = N K 2 + I D X ( [ 1 , 2 , 3 , 5 , 7 ] ) 3 = N K 2 + 5

S U M ( N , [ 1 , 2 , 3 , K 1 5 , K 2 K 1 + 2 ] ) = 3 ! { 5 × 7 ( P 8 ) + 5 × ( K 2 + 1 ) ( P 7 ) + K 1 × ( 7 1 ) ( P 7 ) + K 1 K 2 ( P 6 ) }

2.3. The Meaning of the Expansion of SUM (N, PY)

P Y = [ 1 , K 1 , , K M ] , anItem = { begin , K 1 + E 1 , , K M + E M }

( productofanitem ) = begin × ( K 1 + E 1 ) ( K M + E M ) = begin × F 1 F M

F i = E (means F i = E i ) or F i = K i ,

S U M ( N , P Y ) = product = begin × F 1 F M (2.3.1*)

Define 2.3. S U M _ K ( N , P Y , P F = F 1 F 2 F M ) = Items in (2.3.1*) with the same PF

Example 2.3:

S U M ( N , [ 1 , K 1 3 , K 2 K 1 + 2 ] ) = 15 ( N K 2 + 3 6 ) + 3 ( K 2 + 1 ) ( N K 2 + 3 5 ) + 4 K 1 ( N K 2 + 3 5 ) + K 1 K 2 ( N K 2 + 3 4 ) = 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 ) = n = 0 N K 2 n × ( K 1 + E 1 , i ) ( K 2 + E 2 , i )

then can prove

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

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

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

S U M ( N , P Y ) = i = 1 M ( X i + D i ) ( P M i ) ,

X i + D i = { { G i D i } , X i G , D i = countof { X 1 X i 1 } K { K i } + { D i } , X i K , D i = countof { X 1 X i 1 } G (2.3.2*)

Theorem 2.3 S U M _ K ( N , P Y , P F ) = Items in (2.3.2*) expand by {}, factors has the same { K i } F

= i = 1 M Y i ( P M i ) ,

Y i = { 0 , F i K , X i G K i , F i K , X i K G i D i , F i = E , X i G , D i = countof { X i 1 } K D i , F i = E , X i K , D i = countof { X i 1 } G (2.3.3*)

P = N P H ( P Y ) , M i = I D X ( B S ) ( Count of X K )

[Proof]

Suppose S U M _ K ( N , P Y , P Z ) = i = 1 M Y i ( P M i ) , P = N P H ( P Y ) .

P Y 1 = [ P Y , K M + 1 ] , C = Count of X K ,

M C = Count of X G , M i = I D X ( B S ) C

When K M + 1 = K M + 1

S U M ( N , P Y 1 ) 1 , 2 ) 1.3 ) n = 0 N 1 n × D 1 S U M ( n , P Y ) n = 0 N 1 n × D 1 S U M _ K ( n , P Y , P F ) = n = 0 N 1 n × i = 1 M Y i ( P 1 M i 1 ) reference ( 2.1.1 )

= i = 1 M Y i ( G M + 1 C ) ( P 1 I D X ( B S 1 ) C ) + i = 1 M Y i ( M C ) ( P 1 I D X ( B S 1 ) ( C + 1 ) ) + i = 1 M Y i K M + 1 ( P 1 I D X ( B S 1 ) ( C + 1 ) ) = S U M _ K ( N , P Y 1 , [ P F , E ] ) + S U M _ K ( N , P Y 1 , [ P F , K M + 1 ] )

P 1 = N P H ( P Y 1 ) (2.3.3*)

S U M _ K ( N , P Y 1 , [ P F , K M + 1 ] ) = i = 1 M Y i K M + 1 ( P 1 I D X ( B S 1 ) ( C + 1 ) ) (2.3.3*)

S U M _ K ( N , P Y 1 , [ P F , E ] ) = i = 1 M Y i ( G M + 1 C ) ( P 1 I D X ( B S 1 ) C )

+ i = 1 M Y i ( M C ) ( P 1 I D X ( B S 1 ) ( C + 1 ) ) (2.3.3*)

When K M + 1 > K M + 1

S U M ( N , P Y 1 ) 1 , 5 ) n = 0 N 1 n × S U M ( n K M + 1 + K M + 1 , P Y ) n = 0 N 1 n × S U M _ K ( n K M + 1 + K M + 1 , P Y , P F ) = n = 0 N 1 n × i = 1 M Y i ( P 1 M i ) reference ( 2.1.2 )

= i = 1 M Y i ( G M + 1 C ) ( P 1 I D X ( B S 1 ) C ) + i = 1 M Y i ( M C ) ( P 1 I D X ( B S 1 ) ( C + 1 ) ) + i = 1 M Y i K M + 1 ( P 1 I D X ( B S 1 ) ( C + 1 ) ) = S U M _ K ( N , P Y 1 , [ P F , E ] ) + S U M _ K ( N , P Y 1 , [ P F , K M + 1 ] )

P 1 = N P H ( P Y 1 ) (2.3.3*)

S U M _ K ( N , P Y 1 , [ P F , K M + 1 ] ) = i = 1 M Y i K M + 1 ( P 1 I D X ( B S 1 ) ( C + 1 ) ) (2.3.3*)

S U M _ K ( N , P Y 1 , [ P F , E ] ) = i = 1 M Y i ( G M + 1 C ) ( P 1 I D X ( B S 1 ) C )

+ i = 1 M Y i ( M C ) ( P 1 I D X ( B S 1 ) ( C + 1 ) ) (2.3.3*)

q.e.d.

2.3.1) S U M _ K ( N , P Y = [ 1 , 2 , , n , K 1 n + 2 , , K M ] , P F ) has the similar conclusions.

2.3.2) S U M _ K ( N , P Y = [ 1 , 2 , , n , K 1 n + 2 , , K M ] , K 1 K 2 K M ) ,

B S = B A S E ( P Y ) = M I N ( P Y ) ( N P H ( P Y ) I D X ( B S ) M ) = M I N ( P Y ) n = 0 N 1 ( n P H ( P Y ) 1 I D X ( B S ) M 1 ) = M I N ( P Y ) n = 0 N 1 ( n P H ( P Y ) 1 P B ( B S ) + 1 ) = M I N ( P Y ) n = 0 N 1 ( n P H ( P Y ) 1 P B ( P Y ) + 1 ) = M I N ( P Y ) n = 0 N 1 | Itemsin S U M ( n + 1 , P Y ) |

3. A Theorem of Ring

P Y = [ 1 , K 1 3 , , K M ] , B A S E ( P Y ) = B S = [ 1 , G 1 , , G M ]

S U M ( N , P Y ) = A i ( P M i ) ,

When PY = BS, S U M ( N , P Y ) = M I N ( B S ) ( N I D X ( B S ) )

This inspired us:

In the form ( G 1 + K 1 ) ( G 2 + K 2 ) ( G M + K M ) , K i = G i M I N ( B S ) | A i (Mi is same).

Definition 3.1:

S , { K i } Ring ; K i = K j , K i < K j , K i > K j are allowed.

Choice N from ( K 1 , K 2 , , K M ) , R = R 1 R M

R i = + 1 Indicates that Ki was selected, R i = 1 Indicates that Ki was unselected,

F ( N , M , R , K , S ) = i = 1 M ( K i + D i ) ,

D i = { m S , R i = 1 , m = count of { K 1 , , K i 1 } selected + m S , R i = + 1 , m = count of { K 1 , , K i 1 } unselected (*)

When S = 1, abbreviated as F ( N , M , R , K )

Definition 3.2:

H ( N 1 , N 2 , K , S ) = F ( N 1 , N 1 + N 2 , R , K , S ) , Sum traverses all N1-Choice of K

Theorem 3.1: H ( N 1 , N 2 , K , S ) = ( M N 1 , N 2 ) K 1 K M , M = N 1 + N 2

[Proof]

When N 1 = 0 or N 2 = 0 , it is obvious.

H ( 1 , 1 , K , S ) = select K 1 + select K 2 = unselect K 2 + select K 2 = K 1 ( K 2 S ) + K 1 ( K 2 + S ) = 2 K 1 K 2

Suppose H ( 1 , M 1 , K , S ) = ( M 1 ) K 1 K M

H ( 1 , M , K , S ) = select K M + 1 + unselect K M + 1 = H ( 0 , M , K , S ) ( K M + 1 + M S ) + H ( 1 , M 1 , K , S ) ( K M + 1 S ) = ( M 0 ) K 1 K M ( K M + 1 + M S ) + ( M 1 ) K 1 K M ( K M + 1 S ) = { ( M 0 ) + ( M 1 ) } K 1 K M K M + 1 + { M ( M 0 ) ( M 1 ) } S × K 1 K M K M + 1 = ( M + 1 1 ) K 1 K M K M + 1

H ( 1 , N 2 , K , S ) holds Symmetry H ( N 2 , 1 , K , S ) holds

Suppose H ( N 1 , M N 1 , K , S ) = ( M N 1 ) K 1 K M

H ( N 1 + 1 , M N 1 , K , S ) = select K M + 1 + unselect K M + 1 = H ( N 1 , M N 1 , K , S ) ( K M + 1 + ( M N 1 ) S ) + H ( N 1 + 1 , M N 1 1 , K , S ) ( K M + 1 ( N 1 + 1 ) S ) = ( M N 1 ) K 1 K M ( K M + 1 + ( M N 1 ) S ) + ( M N 1 + 1 ) K 1 K M ( K M + 1 ( N 1 + 1 ) S )

= { ( M N 1 ) + ( M N 1 + 1 ) } K 1 K M + 1 + { ( M N 1 ) ( M N 1 ) ( N 1 + 1 ) ( M N 1 + 1 ) } S × K 1 K M + 1 = ( M + 1 N 1 + 1 ) K 1 K M K M + 1

H ( N 1 , N 2 , K , S ) holds

q.e.d.

Example 3.1: { K 1 , K 2 , K 3 } = { A , B , C }

H ( 1 , 2 , K ) = select A + select B + select C = A ( B 1 ) ( C 1 ) + A ( B + 1 ) ( C 1 ) + A B ( C + 2 ) = 3 A B C

H ( 2 , 1 , K ) = select A B + select B C + select A C = A B ( C 2 ) + A ( B + 1 ) ( C + 1 ) + A ( B 1 ) ( C + 1 ) = 3 A B C

Definition 3.3:

Ki come from q sources: S 1 , S 2 , , S q , K i S j indicates Ki come from Sj d i f f ( K i , K j ) = d i f f ( K i S a , K j S b ) = d i f f ( a , b ) , d i f f ( a , b ) = S , ( a b ); d i f f ( a , b ) = d i f f ( b , a ) ; d i f f ( a , a ) = 0 then can change related parts of definition 3.1 to

i = 1 M ( K i + D i ) , D i = j < i d i f f ( K i , K j )

and define H ( N 1 , N 2 , , N q , K , S )

Theorem 3.2 H ( N 1 , N 2 , , N q , K , S ) = ( M N 1 , N 2 , , N q ) K 1 K M , M = N 1 + N 2 + + N q

[Proof]

Only need to prove S=1, and specify d i f f ( a , b ) = 1 , ( a < b ) H ( N 1 , N 2 , K ) = Item , K i { S 1 , S 2 } . An item has M factors,

Choice N1 factors, Ki in these factors, K i S 1 , It is called invariant factor{F}.

Others are called variable factors {V}

L 2 + L 3 = N 2 , choice L2 factors in {V} to S2, L3 to S3

By definition, H ( L 2 , L 3 , V ) = { ( L 2 , L 3 ) Choice of V }

( { F } × { ( L 2 , L 3 ) Choice of V } ) = ( { F } × H ( L 2 , L 3 , V ) ) = ( { F } × ( N 2 L 1 , L 2 ) × { V } ) = ( N 2 L 2 , L 3 ) Item = ( N 2 L 2 , L 3 ) ( M N 1 , N 2 ) K 1 K M = ( M N 1 , L 2 , L 3 ) K 1 K M = H ( N 1 , L 2 , L 3 , K )

and items gives all (N1, L2, L3)-Choice of K.

Prove ( { F } × { ( L 2 , L 3 ) Choice of V } ) satisfies the definition of H ( N 1 , , K )

Let Item = i = 1 M ( K i + D i ) , ( anitemchanged ) = i = 1 M ( H i + E i ) , K i = H i

A 1 = Count of { K 1 , , K i 1 } S 1 , A 2 = Count of { K 1 , , K i 1 } S 2

B 1 = Count of { , H i 1 } S 1 , B 2 = Count of { , H i 1 } S 2 ,

B 3 = Count of { , H i 1 } S 3

A 1 = B 1 , A 2 = B 2 + B 3

K i + 1 S 1 H i + 1 S 1 D i + 1 = A 2 = i A 1 = i B 1 = B 2 + B 3 = E i + 1

The invariant factors match the definition

K i + 1 S 2 , H i + 1 S 2 definition of  H ( L 2 , L 3 , V ) E i = D i + B 3 = A 1 + B 3 = B 1 + B 3

definitionof H ( N , L 2 , L 3 , V ) Match the definition

K i + 1 S 2 , H i + 1 S 3 definition of  H ( L 2 , L 3 , V ) E i = D i B 2 = A 1 B 2 = B 1 B 2

definitionof H ( N , L 2 , L 3 , V ) Match the definition

Match the definition of H ( N 1 , N 2 , N 3 , K ) recursion

q.e.d.

Example 3.2 { K 1 , K 2 , K 3 } = { A , B , C }

H ( 1 , 2 , K ) = A ( B 1 ) ( C 1 ) + A ( B + 1 ) ( C 1 ) + A B ( C + 2 ) = 3 A B C

A ( B 1 ) ( C 1 ) S 1 S 2 S 2 { F } = { A } A { ( B 1 ) ( C 2 ) + ( B 1 ) C }

A ( B 1 ) ( C 2 ) + A ( B 1 ) C S 1 S 2 S 3 + S 1 S 3 S 2

A ( B + 1 ) ( C 1 ) S 2 S 1 S 2 { F } = { B + 1 } ( B + 1 ) { A ( C 2 ) + A C }

A ( B + 1 ) ( C 2 ) + A ( B + 1 ) C S 2 S 1 S 3 + S 3 S 1 S 2

A B ( C + 2 ) S 2 S 2 S 1 { F } = { C + 2 } ( C + 2 ) { A ( B 1 ) + A ( B + 1 ) }

A ( B 1 ) ( C + 2 ) + A ( B + 1 ) ( C + 2 ) S 2 S 3 S 1 + S 3 S 2 S 1

H ( 2 , 1 , K ) = A B ( C 2 ) + A ( B + 1 ) ( C + 1 ) + A ( B 1 ) ( C + 1 ) = 3 A B C

A B ( C 2 ) S 1 S 1 S 2 { F } = { C 2 } ( C 2 ) { A ( B 1 ) + A ( B + 1 ) }

A ( B 1 ) ( C 2 ) + A ( B + 1 ) ( C 2 ) S 1 S 2 S 3 + S 2 S 1 S 3

A ( B + 1 ) ( C + 1 ) S 2 S 1 S 1 { F } = { A } A { ( B + 1 ) C + ( B + 1 ) ( C + 2 ) }

A ( B + 1 ) C + A ( B + 1 ) ( C + 2 ) S 3 S 1 S 2 + S 3 S 2 S 1

A ( B 1 ) ( C + 1 ) S 1 S 2 S 1 { F } = { B 1 } ( B 1 ) { A C + A ( C + 2 ) }

A ( B 1 ) C + A ( B 1 ) ( C + 2 ) S 1 S 3 S 2 + S 2 S 3 S 1

H ( 1 , 1 , 1 , K ) = A ( B 1 ) ( C 2 ) + A ( B 1 ) C + A ( B + 1 ) ( C 2 ) + A ( B + 1 ) C + A ( B 1 ) ( C + 2 ) + A ( B + 1 ) ( C + 2 ) = 6 A B C

3.1) ( K 1 , K 2 , , K M ) is permutable in H ( N 1 , N 2 , , N q , K , S )

3.2) H ( N 1 , N 2 , , N q , K , S ) = i = 1 M ( K i + D i )

H ( N q , N q 1 , , N 1 , K , S ) = i = 1 M ( K i D i )

3.3) In the section 2, S U M ( N , P Y ) = A i ( P M i )

A i is generated by the form ( G 1 + K 1 ) ( G 2 + K 2 ) ( G M + K M ) . It can be understood asgenerated by 2-SET: { G i } , { K i } .

If K i = J 1 , i + J 2 , i + + J x , i , ( J x , i > 0 , J x , i = 0 , J x , i < 0 are allowed),

S U M ( N , P Y ) = B i ( P M i )

B i can be understood asgenerated by (X + 1)-SET: { G i } , { J 1 } , { J 2 } , , { J X }

3.4) K i = J 1 , i + J 2 , i + + J x , i , Theorem 2.3 has the similar promotion.

4. Conclusion

Review the whole process, ( N M ) + ( N M + 1 ) = ( N + 1 M + 1 ) Basic Shapes in [1] More Shapes in this article.

Acknowledgements

The author is very grateful to Mr. HanSan Z. (Department of Electronic Information, Nanjing University) for his suggestions on this paper.

Cite this paper: Peng, J. (2020) Subdivide the Shape of Numbers and a Theorem of Ring. Open Access Library Journal, 7, 1-14. doi: 10.4236/oalib.1106719.
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

 
 
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