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

( 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 , ⋯ ) , Use the symbol PS = [min Item] to represent it. If K i + 1 − K i = D > 1 then only I i + 1 − I i ≥ D is allowed.

The single ( I 1 , I 2 , ⋯ , I M ) is an item, I 1 I 2 ⋯ I M is the product. I_i 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 ]

|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

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

1.1) | S E T ( N , P S ) | = ( N − P H ( P S ) − 1 P B ( P S ) + 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 ] , B S = B A S E ( P S ) = [ 1 , G 1 , ⋯ , G M ]

Use the form ( G 1 + K 1 ) ( G 2 + K 2 ) ⋯ ( G M + K M ) = ∑ X 1 X 2 ⋯ X M , X_i = G_i or K_i.

The expansion has 2^M 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 ) I D X ( B S ) − q ) , q = count of X ∈ K

S U M ( N , P S ) = ∑ A q ( N − P H ( P S ) I D X ( B S ) − q ) , 2^M items in total.

A q = ∏ i = 1 M ( X i + D i ) , D i = { − m : X i = G i , m = countof   { X 1 , ⋯ , X i − 1 } ∈ K + m : X i = K i , m = countof   { X 1 , ⋯ , X i − 1 } ∈ G

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 ]

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 ) = N − { I D X ( P S ) − P B ( P S ) − 2 } = N − { K 3 + 1 − 3 − 2 } = N − K 3 + 4

I D X ( B S ) = 8

?

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 , F i = E i or F i = K i

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

Define

S U M _ K ( S E T ( N , P S ) , P F = F 1 F 2 ⋯ F M ) = Sum of one part of S U M ( N , P S )

PF indicates the part. P F = F 1 F 2 ⋯ F M , F i = E i or F i = K i

Rewrite 1.3) and add {braces}:

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

X i + D i = { { G i − D i } : X i = G 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 } ∈ G

Let S U M 1 ( N , P S ) = S U M ( N , P S ) expand by the {braces}:

1.4) S U M _ K ( S E T ( N , P S ) , P F ) = ∑ Expansionof S U M 1 ( N , P S ) with same { K i } ∈ P F = ∑ ∏ i = 1 M Y i ( A M q ) , Y i = { 0 : F i = K i , X i = G i K i : F i = K i , X i = K i G i − D i : F i = E i , X i = G 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 } ∈ G

Example:

S U M ( N , [ 1 , K 1 ≥ 3 , K 2 ≥ K 1 + 2 ] ) , form = ( 3 + K 1 ) ( 5 + K 2 ) ?

= 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 )

?

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

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

S U M _ K ( S E T ( N , P S ) , K 1 K 2 ) = ∑ allitems begin ∗ K 1 K 2 = K 1 K 2 ( N − K 2 + 3 4 )

This can explain why 1.3) has that strange form:

We can calculate every part of 1.3) by some way without 1.3). There may be complex relationships between the parts, but their sum just match a simple form.

1.5) Use the symbol of 1.3), when G i = K i , N 1 = Count   of   X ∈ K , N 1 + N 2 = M

H ( N 1 , N 2 , K ) = ∑ A q = N 1 = ∑ ∏ i = 1 M ( X i + D i ) = K 1 K 2 ⋯ K M ( M N 1 , N 2 )

Sum traverses all (N₁, N₂)-Choice of K

This ? 1.3) is compatible with 1.2)

1.6) P is a prime number, {PS1, PS2, …} are all of the Basic Shapes,

P M ( P S 1 ) = P M ( P S 2 ) = ⋯ , P B ( P S 1 ) = P B ( P S 2 ) = ⋯ > 0 , I D X ( P S 1 ) = I D X ( P S 2 ) = ⋯ = P ,

That is, them are Basic shapes, have same count of factors and same count of discontinuities > 0, and max factor = P − 1, then M I N ( P S 1 ) + M I N ( P S 2 ) + ⋯ ≡ 0   M O D   P

Example:

1 × 2 × 4 × 6 + 1 × 3 × 4 × 6 + 1 × 3 × 5 × 6 ≡ 1 × 2 × 3 × 4 × 6 + 1 × 2 × 3 × 5 × 6 + 1 × 2 × 4 × 5 × 6 + 1 × 3 × 4 × 5 × 6 ≡ 0   M O D   7

P S = [ 1 , K 1 , K 2 , ⋯ , K M ] , P T = [ 1 , T 1 , T 2 , ⋯ , T M ] , Item = { I 0 , I 1 , I 2 , ⋯ , I M } ∈ S E T ( N , P S )

If PB(PS) = 0, items ∈ S E T ( N , P S ) is very simple.

If PB(PS) > 0, some changes appear in SET(N, PS).

We can fix some discontinuities of the Shape to get subsets.

Define 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 (*)

others are invalid.

Example:

S E T ( N , [ 1 , 3 , 5 ] , [ 1 , 3 , 5 ] ) = S E T ( N , [ 1 , 3 , 5 ] )

S E T ( N , [ 1 , 3 , 5 ] , [ 1 , 4 , 5 ] ) , S E T ( N , [ 1 , 3 , 5 ] , [ 1 , 3 , 6 ] ) , S E T ( N , [ 1 , 2 , 9 ] , [ 1 , 3 , 4 ] ) is invalid.

S E T ( N , [ 1 , 3 , 5 ] , [ 1 , 2 , 4 ] ) = { ( 1 , 3 , 5 ) , ( 1 , 3 , 6 ) , ( 2 , 4 , 6 ) , ( 1 , 3 , 7 ) , ( 2 , 4 , 7 ) , ( 3 , 5 , 7 ) , ⋯ }

I 2 − I 1 = ( 3 − 1 ) = 2 , I 3 − I 2 ≥ ( 5 − 3 ) = 2

S E T ( N , [ 1 , 3 , 5 ] , [ 1 , 3 , 4 ] ) = { ( 1 , 3 , 5 ) , ( 1 , 4 , 6 ) , ( 2 , 4 , 6 ) , ( 1 , 5 , 7 ) , ( 2 , 5 , 7 ) , ( 3 , 5 , 7 ) , ⋯ }

I 3 − I 2 = ( 5 − 3 ) = 2 , I 2 − I 1 ≥ ( 3 − 1 ) = 2

S E T ( N , [ 1 , 3 , 5 ] , [ 1 , 2 , 3 ] ) = { ( 1 , 3 , 5 ) , ( 2 , 4 , 6 ) , ( 3 , 5 , 7 ) , ⋯ }

S E T ( N , [ 1 , 4 , 8 ] , [ 1 , 2 , 4 ] ) = { ( 1 , 4 , 8 ) , ( 1 , 4 , 9 ) , ( 2 , 5 , 9 ) , ( 1 , 4 , 10 ) , ( 2 , 5 , 10 ) , ( 3 , 6 , 10 ) , ⋯ }

I 2 − I 1 = ( 4 − 1 ) = 3 , I 3 − I 2 ≥ ( 8 − 4 ) = 4

PT only has the change at (*). When a change happens, make the interval fixed.

The more changes, the fewer items:

Define PCHG(PS, PT) = count of change from BASE(PS) to PT

Example:

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

2.1) S E T ( N , P S ) = S E T ( N , P S , B A S E ( P S ) )

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

If PT1 only change T_i of PT, Obvious: P C H G ( P S , P T 1 ) = P C H G ( P S , P T ) + 1

2.3) If PT1 only change T_i of PT, P T 1 = [ 1 , T 1 , ⋯ , T i − 1 , T i − 1 , T i + 1 − 1 , ⋯ , T M − 1 ] .

Let P S 1 = [ 1 , K 1 , ⋯ , K i − 1 , K i + 1 , K i + 1 + 1 , ⋯ , K M + 1 ] , then

S E T ( N , P S , P T ) = S E T ( N , P S , P T 1 ) ∪ S E T ( N , P S 1 , P T )

S E T ( N , P S , P T 1 ) = S E T ( N , P S , P T ) − S E T ( N , P S 1 , P T )

In particular: P C H G ( P S , P T ) = 1 ? S E T ( N , P S , P T ) = S E T ( N , P S ) − S E T ( N , P S 1 )

[Proof]

PT1 change T_i of PT ? T i − T i − 1 = 2 ? K i − K i − 1 > 1 ? P C H G ( P S , P T ) = P C H G ( P S 1 , P T )

| S E T ( N , P S , P T 1 ) | + | S E T ( N , P S 1 , P T ) | = ( N − P H ( P S ) − 1 − P C H G ( P S , P T 1 ) P B ( P T 1 ) + 1 )       + ( N − P H ( P S 1 ) − 1 − P C H G ( P S 1 , P T ) P B ( P T ) + 1 ) = ( N − P H ( P S ) − 2 − P C H G ( P S , P T ) P B ( P T ) )       + ( N − P H ( P S ) − 2 − P C H G ( P S 1 , P T ) P B ( P T ) + 1 )

= ( N − P H ( P S ) − 2 − P C H G ( P S , P T ) P B ( P T ) )       + ( N − P H ( P S ) − 2 − P C H G ( P S , P T ) P B ( P T ) + 1 ) = ( N − P H ( P S ) − 1 − P C H G ( P S , P T ) P B ( P T ) + 1 ) = | S E T ( N , P S , P T ) |

Count of the Items is equal.

Every item in SET(N, PS1, PT) is in SET(N, PS, PT), and not in SET(N, PS, PT1).

q.e.d.

if P C H G ( P S , P T ) = 2 , PT changes at T_i and T_j, then

P T = [ 1 , G 1 , ⋯ , G i − 1 , G i − 1 , ⋯ , G j − 1 − 1 , G j − 2 , G j + 1 − 2 , ⋯ , G M − 2 ]

Let

P T A = [ 1 , G 1 , ⋯ , G i − 1 , G i − 1 , ⋯ , G M − 1 ] ,

P T B = [ 1 , G 1 , ⋯ , G j − 1 , G j − 1 , ⋯ , G M − 1 ]

P S A = [ 1 , K 1 , ⋯ , K i − 1 , K i + 1 , ⋯ , K M + 1 ] ,

P S B = [ 1 , K 1 , ⋯ , K j − 1 , K j + 1 , ⋯ , K M + 1 ]

P S 2 = [ 1 , K 1 , ⋯ , K i − 1 , K i + 1 , ⋯ , K j − 1 + 1 , K j + 2 , K j + 1 + 2 , ⋯ , K M + 2 ]

?

S E T ( N , P S , P T ) = S E T ( N , P S , P T A ) − S E T ( N , P S A , P T A ) = { S E T ( N , P S , B S ) − S E T ( N , P S B , B S ) }       – { S E T ( N , P S A , B S ) − S E T ( N , P S 2 , B S ) } = S E T ( N , P S , B S ) − [ S E T ( N , P S A , B S ) ∪ S E T ( N , P S B , B S ) ]       + S E T ( N , P S 2 , B S ) = S E T ( N , P S ) − [ S E T ( N , P S A ) ∪ S E T ( N , P S B ) ] + S E T ( N , P S 2 )

General:

2.4) The relationship between SET(N, PS, PT) and SET(N, PSX) is similar to the Inclusion Exclusion Principle.

Define:

SUM_SUBSET(N, PS, PT) = Sum of all products in SET(N, PS, PT)

When PT is invalid, SUM_SUBSET(N, PS, PT) = 0

Only valid PT is discussed below.

P S = [ 1 , K 1 , K 2 , ⋯ , K M ] , B S = B A S E ( P S ) = [ 1 , G 1 , G 2 , ⋯ , G M ] , P T = [ 1 , T 1 , T 2 , ⋯ , T M ]

3.1) Use the form ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M ) = ∑ X 1 X 2 ⋯ X M , then

S U M _ S U B S E T ( 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

q = count   of   X ∈ K

[Proof]

1) If PT = BS, then S U M _ S U B S E T ( N , P S , B S ) = S U M ( N , P S ) ? the formula holds.

2) If M = 1 and PT has 1 change, then P S = [ 1 , K 1 > 2 ] , P T = [ 1 , T 1 ] = [ 1 , 2 ] , B S = [ 1 , G 1 ] = [ 1 , 3 ] ,

Let P S 1 = [ 1 , K 1 + 1 ] , 2.3) →

S U M _ S U B S E T ( N , P S , P T ) = S U M ( N , P S ) − S U M ( N , P S 1 ) = { G 1 ( N − P H ( P S ) I D X ( B S ) ) + K 1 ( N − P H ( P S ) I D X ( B S ) − 1 ) }         − { G 1 ( N − P H ( P S 1 ) I D X ( B S ) ) + ( K 1 + 1 ) ( N − P H ( P S 1 ) I D X ( B S ) − 1 ) }

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

= T 1 ( N − P H ( P S 1 ) I D X ( B S ) − 1 ) + K 1 ( N − P H ( P S 1 ) I D X ( B S ) − 2 ) = T 1 ( N − P H ( P S ) − 1 I D X ( P T ) ) + K 1 ( N − P H ( P S ) − 1 I D X ( P T ) − 1 ) = T 1 ( N − P H ( P S ) − P C H G ( P S , P T ) I D X ( P T ) )         + K 1 ( N − P H ( P S ) − P C H G ( P S , P T ) I D X ( P T ) − 1 )

The form = (T₁ + K₁) ? The formula holds.

3) If M > 1 and PT only has 1 change at T_M, then P T = [ 1 , G 1 , ⋯ , G M − 1 , G M − 1 ]

Let P S 1 = [ 1 , K 1 , ⋯ , K M − 1 , K M + 1 ] , 2.3) → P H ( P S 1 ) = P H ( P S ) + 1 :

S U M ( N , P S , P T ) = S U M ( N , P S ) − S U M ( N , P S 1 ) = ∑ i = I D X ( B S ) − M I D X ( B S ) ∑ A i ( N − P H ( P S ) i ) − ∑ i = I D X ( B S ) − M I D X ( B S ) ∑ B i ( N − P H ( P S 1 ) i ) = ∑ i = I D X ( B S ) − M I D X ( B S ) ∑ A i { ( N − P H ( P S 1 ) i ) + ( N − P H ( P S 1 ) i − 1 ) }         − ∑ i = I D X ( B S ) − M I D X ( B S ) ∑ B i ( N − P H ( P S 1 ) i )

= ∑ i = I D X ( B S ) − M I D X ( B S ) ( ∑ A i − ∑ B i ) ( N − P H ( P S 1 ) i ) + ∑ A i ( N − P H ( P S 1 ) i − 1 ) = ∑ i = I D X ( B S ) − M I D X ( B S ) ( ∑ A i − ∑ B i ) ( N − P H ( P S ) − 1 i ) + ∑ A i ( N − P H ( P S ) − 1 i − 1 ) = ∑ i = I D X ( B S ) − M − 1 I D X ( B S ) ∑ C J ( N − P H ( P S ) − 1 J ) , J = i − 1 , ∑ C J − 1 = ∑ A i + 1 + ∑ A i − ∑ B i = ∑ J = I D X ( P T ) − M I D X ( P T ) + 1 ∑ C J ( N − P H ( P S ) − 1 J )

When i = I D S ( B S ) , ∑ C i = M I N ( B S ) − M I N ( B S ) = 0

= ∑ J = I D X ( P T ) − M I D X ( P T ) ∑ C J ( N − P H ( P S ) − 1 J )

Use the symbol of (1.3)

i = I D X ( B S ) − C N T , C N T = Count   of   X ∈ K

Let R ( E ) = ∑ ∏ L = 1 M − 1 ( X L + D L ) , E = Count   of   { X 1 , ⋯ , X M − 1 } ∈ K

∑ A i + 1 = R ( C N T − 1 ) ( G M − ( C N T − 1 ) )                         + R ( C N T − 2 ) ( K M + ( M − 1 ) − ( C N T − 2 ) )

∑ A i = R ( C N T ) ( G M − C N T ) + R ( C N T − 1 ) ( K M + ( M − 1 ) − ( C N T − 1 ) )

∑ B i = R ( C N T ) ( G M − C N T ) + R ( C N T − 1 ) ( ( K M + 1 ) + ( M − 1 ) − ( C N T − 1 ) )

∑ C J − 1 = ∑ A i + 1 + ∑ A i − ∑ B i = R ( C N T − 1 ) ( G M − C N T ) + R ( C N T − 2 ) ( K M + M − C N T + 1 ) = R ( C N T − 1 ) ( T M − { C N T − 1 } ) + R ( C N T − 2 ) ( K M + { ( M − 1 ) − ( C N T − 2 ) } )

? Match of the form ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M )

4) If M > 1 and PT only has 1 change at T i < M , Let R ( E ) = ∑ ∏ L = 1 , l ≠ i M ( X L + D L ) , use the same method of (3).

5) if P C H G ( P S , P T ) > 1 , Use 2.3) → divide the Items into subset ? deducing by induction.

q.e.d.

Example:

N − P H ( [ 1 , 3 , 5 , 7 ] ) − P C H G ( [ 1 , 3 , 5 , 7 ] , [ 1 , 2 , 3 , 4 ] ) = N − ( 8 − 3 − 2 ) − 3 = N − 6

S U M _ S U B S E T ( N , [ 1 , 3 , 5 , 7 ] , [ 1 , 2 , 3 , 4 ] )

?

form = ( 2 + 3 ) ( 3 + 5 ) ( 4 + 7 ) = 24 ( N − 6 5 ) + 108 ( N − 6 4 ) + 174 ( N − 6 3 ) + 105 ( N − 6 2 ) = 1 × 3 × 5 × 7 + 2 × 4 × 6 × 8 + 3 × 5 × 7 × 9 + ⋯

Among:

24 = 2 × 3 × 4 ; 105 = 3 × 5 × 7

108 = 2 × 3 × ( 7 + 2 ) + 2 × ( 5 + 1 ) × ( 4 − 1 ) + 3 × ( 3 − 1 ) × ( 4 − 1 )

174 = 2 × ( 5 + 1 ) × ( 7 + 1 ) + 3 × ( 3 − 1 ) × ( 7 + 1 ) + 3 × 5 × ( 4 − 2 )

Use the same method of 3.1)

3.2) Calculation formula of SUM_K(SET(N, PS, PT), PF) is similar to 1.4).

Example:

S U M _ S U B S E T ( N , [ 1 , 3 , 7 ] , [ 1 , 2 , 3 ] ) = 6 ( N − 6 4 ) + 2 × ( { 7 } + { 1 } ) ( N − 6 3 ) + 3 × ( { 3 − 1 } ) ( N − 6 3 ) + 3 × 7 ( N − 6 2 )

S U M _ S U B S E T ( 10 , [ 1 , 3 , 7 ] , [ 1 , 2 , 3 ] ) = 1 × 3 × 7 + 2 × 4 × 8 + 3 × 5 × 9 = 1 × 3 × 7 + 2 × ( 3 + 1 ) × ( 7 + 1 ) + 3 × ( 3 + 2 ) × ( 7 + 2 ) = { 1 × 3 × 7 + 2 × 3 × 7 + 3 × 3 × 7 } + { 2 × 3 × 1 + 3 × 3 × 2 }         + { 2 × 1 × 7 + 3 × 2 × 7 } + { 2 × 1 × 1 + 3 × 2 × 2 }

?

{ 1 × 3 × 7 + 2 × 3 × 7 + 3 × 3 × 7 } = 3 × 7 ( 10 − 6 2 )

{ 2 × 3 × 1 + 3 × 3 × 2 } = 3 × ( { 3 − 1 } ) ( N − 6 3 ) = 6 ( 10 − 6 3 )

{ 2 × 1 × 7 + 3 × 2 × 7 } = 2 × 7 ( N − 6 3 ) = 14 ( N − 6 3 )

{ 2 × 1 × 1 + 3 × 2 × 2 } = 6 ( 10 − 6 4 ) + 2 × { 1 } ( 10 − 6 3 )

3.3) Use the form ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M ) = ∑ X 1 X 2 ⋯ X M

S U M _ K ( S E T ( N , P S , P T ) , E 1 E 2 ⋯ E M ) = ∑ 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 D i , D i = { T i − m : X i = T i , m = countof   { X 2 , ⋯ , X i − 1 } ∈ K + m : X i = K i , m = countof   { X 2 , ⋯ , X i − 1 } ∈ T

q = count   of   X ∈ K , X 1 = T 1 , 2^M-1 Items in total.

In particular:

If P T 1 = [ 1 , T 1 , ⋯ , T M ] = [ 1 , 2 , ⋯ , M + 1 ] , then P B ( P S ) = P C H G ( P S , P T 1 )

N − P H ( P S ) − P C H G ( P S , P T 1 ) = N − [ I D X ( P S ) − 2 − P B ( P S ) ] − P B ( P S ) = N − ( K M − 1 )

I D X ( P T 1 ) = M + 2

S U M ( S E T ( N , P S , P T 1 ) , E 1 ⋯ E M ) = 2 × 1 M + 3 × 2 M + ⋯ + ( N − K M ) × ( N − K M − 1 ) M

?

S U M ( S E T ( N + K M + 1 , P S , P T 1 ) , E 1 ⋯ E M ) = 2 × 1 M + 3 × 2 M + ⋯ + ( N + 1 ) × N M

3.4) ∑ n = 1 N ( n + 1 ) n M = S U M _ K ( S E T ( N + K M + 1 , P S , P T 1 ) , E 1 ⋯ E M )

P S = [ 1 , K 1 , K 2 , ⋯ , K M ] , P T 1 = [ 1 , 2 , 3 , ⋯ , M + 1 ] . The simplest subset of PS is SET(N, PS, PT1).

S U M _ S U B S E T ( N , P S , P T 1 ) = ∑ q = 0 M C q ( N − ( K M − 1 ) M + 2 − q ) = ∑ n = 1 N − K M ∑ q = 0 M C q ( n M + 1 − q ) = 1 × K 1 × ⋯ × K M + 2 × ( 1 + K 1 ) × ⋯ × ( 1 + K M ) + ⋯         + ( N − K M ) × ( [ N − K M − 1 ] + K 1 ) × ⋯ × ( N − 1 ) = ∑ n = 1 N − K M n ( n + K 1 − 1 ) ( n + K 2 − 1 ) ⋯ ( n + K M − 1 ) (1*)

Solve (1*) in a normal way:

Decompose n ( n + K 1 − 1 ) ⋯ ( n + K M − 1 ) to ∑ j = 1 M + 1 D j ( n j ) ? D j = C q , q = M + 1 − j

4.1) 1 < K 1 < ⋯ < K M , 3.1) can decompose n ( n + K 1 − 1 ) ⋯ ( n + K M − 1 ) to ∑ j = 1 M + 1 D j ( n j )

In particular, 1.5) ? C q = ( M + 1 ) ! ( M M − q ) ? D j = ( M + 1 ) ! ( M j − 1 ) ?

4.2) [ x ] M = M ! M ! ( M − 1 M − 1 ) [ x ] M + M ! ( M − 1 ) ! ( M − 1 M − 2 ) [ x ] M − 1 + ⋯ + M ! 1 ! ( M − 1 0 ) [ x ] 1

4.3) P is a prime number, N − ( K M − 1 ) = P

1) | S E T ( N , P S , P T 1 ) | = P − 1 ,

2) S E T ( N , P S , P T 1 ) = { ( 1 , K 1 , ⋯ , K M ) , ( 2 , 1 + K 1 , ⋯ , 1 + K M ) ⋯ ( P − 1 , ⋯ , N − 1 ) }

3) if K M ≤ P − 1 and P S ≠ [ 1 , 2 , ⋯ , P − 1 ] , then S U M _ S U B S E T ( N , P S , P T 1 ) ≡ 0   M O D   P

[Proof]

| S E T ( N , P S , P T 1 ) | = ( N − P H ( P S ) − 1 − P C H G ( P S , P T 1 ) P B ( P T 1 ) + 1 ) = ( ( P + K M − 1 ) − ( K M + 1 − P B ( P S ) − 2 ) − 1 − P C H G ( P S , P T 1 ) 1 ) → P B ( P S ) = P C H G ( P S , P T 1 ) P − 1 → ( 1 ) ( 2 )

S U M _ S U B S E T ( N , P S , P T 1 ) = ∑ q = 0 M C q ( P I D X ( P T 1 ) − q )

K M ≤ P − 1 and P S ≠ [ 1 , 2 , ⋯ , P − 1 ] → I D X ( P T 1 ) < P → ( 3 )

q.e.d.

When K M = P − 1 and P S ≠ [ 1 , 2 , ⋯ , P − 1 ]

S U M _ S U B S E T ( N , P S , P T 1 ) = 1 × K 1 × ⋯ × K M + 2 × ( 1 + K 1 ) × ⋯ × P     + 3 × ( 2 + K 1 ) ⋯ ( 2 + K M − 1 ) ( 2 + K M )     + 4 × ( 3 + K 1 ) ⋯ ( 3 + K M − 1 ) ( 3 + K M ) + ⋯     + ( P − 1 ) × ( P − 2 + K 1 ) × ⋯ × ( P − 2 + K M )

3 × ( 2 + K 1 ) ⋯ ( 2 + K M − 1 ) ( 2 + K M ) ≡ 1 × 3 × ( 2 + K 1 ) ⋯ ( 2 + K M − 1 )

4 × ( 3 + K 1 ) ⋯ ( 3 + K M − 1 ) ( 3 + K M ) ≡ 2 × ( 1 + [ 3 ] ) × ( 1 + [ 2 + K 1 ] ) ⋯ ( 1 + [ 2 + K M − 1 ] )

⋯

( P − 1 ) × ( P − 2 + K 1 ) ⋯ ( P − 2 + K M ) ≡ ( P − 1 ) × ( P − 2 + K 1 ) ⋯ ( P − 2 + K M − 1 ) ( 2 P − 3 ) ≡ ( P − 4 + [ 1 ] ) ( P − 4 + [ 3 ] ) ( P − 4 + [ 2 + K 1 ] ) ⋯ ( P − 4 + [ 2 + K M − 1 ] )

1 × K 1 × ⋯ × K M ≡ 1 × K 1 × ⋯ × ( P − 1 ) ≡ ( P + 1 ) ( P + K 1 ) ⋯ ( P + P − 1 ) ≡ ( P − 1 ) ( P − 2 + 3 ) ( [ P − 2 ] + [ 2 + K 1 ] ) ⋯ ( [ P − 2 ] + [ 2 + K M − 1 ] )

?

S U M _ S U B S E T ( N , P S , P T 1 ) ≡ S U M _ S U B S E T ( N , [ 1 , 3 , 2 + K 1 , ϵ , 2 + K M − 1 ] , P T 1 )   M O D   P

P S = [ 1 , K 1 ⋯ K M ] can be slided to [ 1 , 3 , 2 + K 1 , ⋯ , 2 + K M − 1 ] by MOD P

If K M = P − 1 and exists K i + 1 − K i > 2 , PS can be slided to [ 1 , ⋯ , X < P − 1 ]

If K M = P − 1 and not exists K i + 1 − K i > 2 , PS must be a Basic Shape, can only be slided to [ 1 , ⋯ , X = P − 1 ]

Define:

A Basic Shape P S = [ 1 , K 1 ⋯ K M ] = 〚 L 1 L 2 ⋯ L Q 〛 , among:

L_i = count of continuity. (L_i, L_i+1) means a discontinuity, there are Q-1 discontinuities.

Example: [ 1 , 3 , 5 ] = 〚 1 , 1 , 1 〛 , [ 1 , 2 , 3 , 5 ] = 〚 3 , 1 〛 , [ 1 , 2 , 4 , 5 ] = 〚 2 , 2 〛

Obvious: 〚 L 1 L 2 ⋯ L Q 〛 can been slided to [ L Q , L 1 , L 2 , ⋯ ] , [ L Q − 1 , L Q , L 1 , L 2 , ⋯ ] , ⋯

4.4) PS is a Basic Shape, I D X ( P S )   =   P , P S   ≠   [ 1 , 2 , ⋯ , P − 1 ] , { P S 1 , P S 2 , ⋯ } are all shapes that PS can scroll to, then M I N ( P S 1 ) + M I N ( P S 2 ) + ⋯ ≡ 0   M O D   P . This is a promotion of 1.6)

[Proof]

3 × ( 2 + K 1 ) ⋯ ( 2 + K M − 1 ) ( 2 + K M ) ≡ 1 × 3 × ( 2 + K 1 ) ⋯ ( 2 + K M − 1 ) M O D   P

If 2 + K M − 1 ≠ P , [ 1 , 3 , ( 2 + K 1 ) , ⋯ , ( 2 + K M − 1 ) ] ∈ { P S 1 , P S 2 , ⋯ }

…

S U M _ S U B S E T ( N , P S 1 , P T 1 ) ≡ M I N ( P S 1 ) + M I N ( P S 2 ) + ⋯ ≡ 0   M O D   P

q.e.d.

Example:

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

Use the form ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M ) .

In general, K_i and K_j cannot be exchanged, but when P T = P T 1 = [ 1 , 2 , ⋯ , M + 1 ] ,

S U M _ S U B S E T ( N , P S , P T 1 ) = ∑ q = 0 M C q ( N − P H ( P S ) − P C H G ( P S , P T 1 ) I D X ( P T 1 ) − q )

C q = ∑ ∏ i = 1 M ( X i + D i ) = ∑ ∏ choice   M − q   from   T ( T − D ) ∏ choice   q   from   K ( K + D )

Easy to see: ∏ choice   M − q   from   T ( T − D ) = ( M + 1 − q ) !

Can prove: ( K 1 , K 2 , ⋯ , K M ) is permutable in ∑ ∏ choice   q   from   K ( K + D )

? ( K 1 , K 2 , ⋯ , K M ) is permutable in the form.

S U M _ S U B S E T ( N , P S , P T 1 ) = ∑ q = 0 M A q ( N − K M + 1 M + 2 − q ) = ∑ n = K M + 1 N ∑ q = 0 M A q ( n − K M M + 1 − q ) = 1 × K 1 × ⋯ × K M + 2 × ( 1 + K 1 ) × ⋯ × ( 2 + K M ) + ⋯

Add one more factor K_i to the end:

1 × K 1 × ⋯ × K M × K i + 2 × ( 1 + K 1 ) × ⋯ × ( 1 + K M ) × ( 1 + K i ) + ⋯ = ∑ n = K M + 1 N ∑ q = 0 M A q ( n − K M M + 1 − q ) ( n − 1 − K M + K i ) = ∑ n = K M + 1 N ∑ q = 0 M A q ( n − K M M + 1 − q ) ( [ n − K M + 1 ] + [ K i − 2 ] ) = ∑ n = K M + 1 N ∑ q = 0 M { A q ( M + 2 − q ) ( n − K M + 1 M + 2 − q ) + A q ( K i − 2 ) ( n − K M M + 1 − q ) } = ∑ q = 0 M A q ( M + 2 − q ) ( N − K M + 2 M + 3 − q ) + ∑ q = 0 M A q ( K i − 2 ) ( N − K M + 1 M + 2 − q )

= ∑ q = 0 M A q ( M + 2 − q ) ( N − K M + 1 M + 3 − q )         + ∑ q = 0 M { A q ( M + 2 − q ) + A q ( K i − 2 ) } ( N − K M + 1 M + 2 − q ) = ∑ q = 0 M A q ( M + 2 − q ) ( N − K M + 1 M + 3 − q )         + ∑ q = 0 M A q ( K i + ( M − q ) ) ( N − K M + 1 M + 3 − ( q + 1 ) )

Let P S 2 = [ 1 , K 1 , ⋯ , K M , K M + 1 = K i ] ,

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

I D X ( P T 2 ) = M + 3 , N − P H ( P S 2 ) − P C H G ( P S 2 , P T 2 ) = N − K M + 1

q = count   of   { X 1 , ⋯ , X M } ∈ K

A q ( M + 2 − q ) = A q ( T M + 1 − q ) means X M + 1 = T M + 1 , A q ( K i + ( M − q ) ) means X M + 1 = K M + 1

It’s match the form ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M ) ( T M + 1 + K i )

Recursion ?

5.1) K i < K j , K i = K j , K i > K j are allowed, S U M _ S U B S E T ( N , P S , P T 1 ) can use the form ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M )

Example:

The form = ( 2 + 3 ) ( 3 + 5 ) ( 4 + 3 ) ?

S U M _ S U B S E T ( N , [ 1 , 3 , 5 , 3 ] , [ 1 , 2 , 3 , 4 ] ) = 24 ( N − 4 5 ) + 84 ( N − 4 4 ) + 102 ( N − 4 3 ) + 45 ( N − 4 2 )

Among:

45 = 3 × 5 × 3

102 = 3 × 5 × ( 4 − 2 ) + 3 × ( 3 − 1 ) × ( 3 + 1 ) + 2 × ( 5 + 1 ) × ( 3 + 1 )

84 = 2 × 3 × ( 3 + 2 ) + 2 × ( 5 + 1 ) × ( 4 − 1 ) + 3 × ( 3 − 1 ) × ( 4 − 1 )

S U M _ S U B S E T ( 9 , [ 1 , 3 , 5 , 3 ] , [ 1 , 2 , 3 , 4 ] ) = 1 × 3 × 3 × 5 + 2 × 4 × 4 × 6 + 3 × 5 × 5 × 7 + 4 × 6 × 6 × 8 = 1914 = 24 ( 5 5 ) + 84 ( 5 4 ) + 102 ( 5 3 ) + 45 ( 5 2 )

The form = ( 2 + 4 ) ( 3 + 7 ) ( 4 + 7 ) ?

S U M _ S U B S E T ( N , [ 1 , 4 , 7 , 7 ] , [ 1 , 2 , 3 , 4 ] ) = 24 ( N − 6 5 ) + 126 ( N − 6 4 ) + 248 ( N − 6 3 ) + 196 ( N − 6 2 )

Among:

196 = 4 × 7 × 7

248 = 4 × 7 × ( 4 − 2 ) + 4 × ( 3 − 1 ) × ( 7 + 1 ) + 2 × ( 7 + 1 ) × ( 7 + 1 )

126 = 2 × 3 × ( 7 + 2 ) + 2 × ( 7 + 1 ) × ( 4 − 1 ) + 4 × ( 3 − 1 ) × ( 4 − 1 )

S U M _ S U B S E T ( 12 , [ 1 , 4 , 7 , 7 ] , [ 1 , 2 , 3 , 4 ] ) = 1 × 4 × 7 × 7 + 2 × 5 × 8 × 8 + 3 × 6 × 9 × 9 + 4 × 7 × 10 × 10 + 5 × 8 × 11 × 11 = 9934 = 24 ( 6 5 ) + 126 ( 6 4 ) + 248 ( 6 3 ) + 196 ( 6 2 )

In particular:

5.2) ∑ n = 1 M n M = S U M _ S U B S E T ( N + 1 , [ 1 , 1 , ⋯ , 1 ] , [ 1 , 2 , ⋯ , M ] )

Example:

( N + 1 ) − P H ( P S ) − P C H G ( P S , P T 1 ) = ( N + 1 ) − 0 − 0 = N + 1

The form = ( 2 + 1 )

? 1 2 + 2 2 + ⋯ + N 2 = 2 ( N + 1 3 ) + ( N + 1 2 ) = N ( N + 1 ) ( 2 N + 1 ) 6

The form = ( 2 + 1 ) ( 3 + 1 )

? 1 3 + 2 3 + ⋯ + N 3 = 6 ( N + 1 4 ) + 6 ( N + 1 3 ) + ( N + 1 2 ) = N 2 ( N + 1 ) 2 4

The form = ( 2 + 1 ) ( 3 + 1 ) ( 4 + 1 )

? 1 4 + 2 4 + ⋯ + N 4 = 24 ( N + 1 5 ) + 36 ( N + 1 4 ) + 14 ( N + 1 3 ) + ( N + 1 2 )

Among:

14 = 2 × ( 1 + 1 ) × ( 1 + 1 ) + 1 × 1 × ( 4 − 2 ) + 1 × ( 3 − 1 ) × ( 1 + 1 )

36 = 2 × 3 × ( 1 + 2 ) + 2 × ( 1 + 1 ) × ( 4 − 1 ) + 1 × ( 3 − 1 ) × ( 4 − 1 )

S(M, K) is Stirling number of the second kind,

Definition of S(M, K) ? ∑ n = 1 N n M = ∑ K = 1 M K ! S ( M , K ) ( N + 1 K + 1 )

It’s equal to 5.2), so we have a way to calculate S(M, K).

3.4) can be seen as ∑ n = 1 N ( n + 1 ) n M = S U M _ S U B S E T ( N + 1 , [ 1 , 0 , ⋯ , 0 ] , [ 1 , 2 , ⋯ , M + 1 ] )

The author declares no conflicts of interest regarding the publication of this paper.

Peng, J. (2020) Subset of the Shape of Numbers. Open Access Library Journal, 7: e7040. https://doi.org/10.4236/oalib.1107040