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.
Peng, J. introduced the definition of Shape of numbers in [
( 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 ) }
[
| 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 [
| 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).
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
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 )
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 ) }
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 [
The author declares no conflicts of interest regarding the publication of this paper.
The author is very grateful to Mr. HanSan Z. (Department of Electronic Information, Nanjing University) for his suggestions on this paper.
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