Expansion of the Shape of Numbers

This article extends the concept of the shape of numbers. Originally, a shape was defined as [1, K ₁, K ₂, ···], 1< K ₁< K ₂< ···, K _i∈ N. In this paper, the domain of a shape is extended from N to Z, the low bound is extended from 1 to Z, and K _i< K _i+1, K _i= K _i+1, K _i> K _i+1 are allowed, which prove that they can be calculated with the similar form ( T ₀+ K ₀)( T ₁+ K ₁)( T ₂+ K ₂) ···. In this way, a lot of calculation formulas can be obtained. At the end, the form is obtained to calculate K ₁x···x K _M+( L+ K ₁)x···x( L+ K _M)+( 2L+ K ₁)x···x( 2L+ K _M)+( 3L+ K ₁)x···x( 3L+ K _M)+···.

• Shape of Numbers
• Calculation Formula
• Combinatorics
• Congruence
• Stirling Number

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. 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 ]

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 , X_i = T_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 ) − 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 , F_i = E_i or K_i

That is, a product can be broken down into 2^M 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. F_i = E_i or K_i

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 = K_M

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 2^M+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 , X_i = T_i or K_i

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 ] ) , K_i 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 = K_M

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 2^M 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 T_i and D_i 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 , 2^M 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 2^K−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 ) .

Conflicts of Interest

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

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. <br />https://doi.org/10.4236/oalib.1107040