Peng, J. has introduced Shape of numbers and three forms of calculation in [1] [2] [3] [4] [5]:

K i , D i ∈ CommutativeRing .

M series: S e r i e i = { K i , K i + D i , K i + 2 D i , ⋯ , K i + ( N − 1 ) D i } , i ∈ [ 1 , M ] .

Use P S = [ K 1 : D 1 , ⋯ , K M : D M ] to represent thems.

[ K 1 : 1 , ⋯ , K M : 1 ] is abbreviated as [ K 1 , ⋯ , K M ] .

Anitem = ( I 1 , I 2 , ⋯ , I M ) , I i comes from Serie_i. A product = ∏ i = 1 M I i .

Use P T = [ T 1 = 1 , T 2 , ⋯ , T M ] to indicate the item’s range:

I i = K i + a i D i , { T i + 1 − T i = 1 ,     means   a i = a i + 1 ,   continuity T i + 1 − T i = 2 ,     means   a i ≤ a i + 1 ,   discontinuity .

PS and PT are defined as Shape of numbers, they indicate some items.

S U M ( N , P S , P T ) = ∑ allitems product .

PB(PT) = count of discontinuity in PT, PM(PT) = count of factors in PT.

By default, the following uses:

P S = [ K 1 : D 1 , ⋯ , K M : D M ] , P T = [ T 1 , ⋯ , T M ] .

H(q) is short for H(PS, PT, q), SUM(N) is short for SUM(N, PS, PT).

The Form: ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M ) = ∑ ∏ i = 1 M X M , X i = T i or K i .

Don’t swap the factors. Each ∏ X i corresponds to one expression in SUM(…).

X ( T ) = countof { X 1 , ⋯ , X M } ∈ { T 1 , ⋯ , T M } ,

X K − 1 = countof { X 1 , ⋯ , X i − 1 } ∈ { K 1 , ⋯ , K M } .

1.1) q = X ( T ) , P M = P M ( P T ) ,

S U M ( N ) = → Form 1 ∑ q = 0 P M H 1 ( q ) ( N + T M − P M N − 1 − q ) , B i = { ( T i − X K − 1 ) D i ; X i = T i K i + X T − 1 D i ; X i = K i → Form 2 ∑ q = 0 P M H 2 ( q ) ( N + T M − P M + q N − 1 ) , B i = { ( T i − X K − 1 ) D i ; X i = T i K i + ( X K − 1 − T i ) D i ; X i = K i → Form 3 ∑ q = 0 P M H 3 ( q ) ( N + T M − q N − 1 − q ) , B i = { − K i + ( T i − X T − 1 ) D i ; X i = T i K i + X T − 1 D i ; X i = K i

H ( q ) = H ( P S , P T , q ) = ∑ all   of   the   ∏ X i   with   X ( T ) = q ∏ i = 1 M B i .

In particular:

1.2) S U M ( N , [ 1 , 2 , ⋯ , M ] , [ 1 , 3 , ⋯ , 2 M − 1 ] ) = S 1 ( N + M , N ) , unsigned Stirling number.

1.3) S U M ( N , [ 1 , 1 , ⋯ , 1 ] , [ 1 , 3 , ⋯ , 2 M − 1 ] ) = S 2 ( N + M , N ) , Stirling number of the second kind.

1.4) S U M ( N , [ 1 , 1 , ⋯ , 1 ] , [ 1 , 2 , ⋯ , M ] ) = 1 M + 2 M + ⋯ + N M .

The following change T_i’s domain to ℤ and T i + 1 − T i is not restricted.

P S 1 = [ P S , K M + 1 : D M + 1 ] , P T 1 = [ P T , T M + 1 ] , use H ( P S 1 , q ) = H ( P S 1 , P T 1 , q ) .

2.0) Recurrence relation

① H 1 ( P S 1 , q ) = H 1 ( q − 1 ) ( T M + 1 − [ M − ( q − 1 ) ] ) D M + 1 + H 1 ( q ) ( K M + 1 + q D M + 1 )

② H 2 ( P S 1 , q ) = H 2 ( q − 1 ) ( T M + 1 − [ M − ( q − 1 ) ] ) D M + 1                                     + H 2 ( q ) ( K M + 1 + [ − T M + 1 + M − q ] D M + 1 )

③ H 3 ( P S 1 , q ) = H 3 ( q − 1 ) ( − K M + 1 + [ T M + 1 − ( q − 1 ) ] D M + 1 )     + H 3 ( q ) ( K M + 1 + q D M + 1 )

2.1) ① H 1 ( q ) = ∑ x = q M H 2 ( x ) ( x q ) = ∑ x = 0 q H 3 ( x ) ( M − x M − q )

② H 2 ( q ) = ∑ x = q M ( − 1 ) x + q H 1 ( x ) ( x q ) , H 3 ( q ) = ∑ x = 0 q ( − 1 ) x + q H 1 ( x ) ( M − x M − q )

[Proof]

Suppose H 1 ( q ) = ∑ x = q M H 2 ( x ) ( x q ) = ∑ x = 0 M H 2 ( x ) ( x q ) , y = T M + 1 − M − 1 .

A : H 1 ( P S 1 , q ) = ( T M + 1 − M − 1 + q ) D M + 1 H 1 ( q − 1 ) + ( K M + 1 + q D M + 1 ) H 1 ( q ) = ( y + q ) D M + 1 ∑ x = 0 M H 2 ( x ) { ( x + 1 q ) − ( x q ) } + ( K M + 1 + q D M + 1 ) ∑ k = 0 M H 2 ( x ) ( x q ) = ( y + q ) D M + 1 ∑ x = 0 M H 2 ( x ) ( x + 1 q ) + ( K M + 1 − y D M + 1 ) ∑ k = 0 M H 2 ( x ) ( x q )

B : ∑ x = 0 M + 1 H 2 ( P S 1 , x ) ( x q ) = ∑ x = 0 M + 1 { H 2 ( x − 1 ) ( y + x ) D M + 1 ( x q ) + H 2 ( x ) ( K M + 1 + ( − y − 1 − x ) D M + 1 ) ( x q ) } = ∑ x = 0 M { H 2 ( x ) ( y + x + 1 ) D M + 1 ( x + 1 q ) + H 2 ( x ) ( K M + 1 + ( − y − 1 − x ) D M + 1 ) ( x q ) }

[ A − B ] = ∑ x = 0 M H 2 ( x ) ( q − 1 − x ) D M + 1 ( x + 1 q ) + ∑ x = 0 M H 2 ( x ) ( 1 + x ) D M + 1 ( x q ) = D M + 1 ∑ x = 0 M H 2 ( x ) q ( x + 1 q ) − D M + 1 ∑ x = 0 M H 2 ( x ) ( 1 + x ) ( x q − 1 ) = 0

→ H 1 ( q ) = ∑ x = q M H 2 ( x ) ( x q )

→ sameway H 1 ( q ) = ∑ x = 0 q H 3 ( x ) ( M − x M − q ) → Inversion ②

q.e.d.

In particular:

2.2) ① H 1 ( 0 ) = H 3 ( 0 ) = ∑ x = 0 M H 2 ( x ) = ∏ i = 1 M K i ;

② H 1 ( M ) = H 2 ( M ) = ∑ x = 0 M H 3 ( x ) = ∏ i = 1 M T i D i ;

③ H 2 ( 0 ) = ( − 1 ) M H 3 ( M ) = ∑ x = 0 M ( − 1 ) x H 1 ( x ) ;

④ H 1 ( 1 ) = ∑ x = 1 M H 2 ( x ) x = M H 3 ( 0 ) + H 3 ( 1 ) .

Calculation with 2.1):

2.3) ∑ q = 0 M H 1 ( q ) = ∑ q = 0 M H 2 ( q ) 2 q = ∑ q = 0 M H 3 ( q ) 2 M − q .

Use 2.1) → Form 1 = Form 2 = Form 3 .

2.4) ① ∑ q = 0 M H 1 ( q ) ( A B − q ) = ∑ q = 0 M H 2 ( q ) ( A + q B ) = ∑ q = 0 M H 3 ( q ) ( A + M − q B − q )

② ∑ q = 0 M H 1 ( q ) ( A q ) = ∑ q = 0 M H 2 ( q ) ( A + q q ) = ∑ q = 0 M H 3 ( q ) ( A + M − q M )

2.5) ∑ q = 0 M H 1 ( q ) q ( A + 1 B − q ) = ∑ q = 0 M H 2 ( q ) q ( A + q B − 1 ) = ∑ q = 0 M { H 3 ( q ) q ( A + M − q B − q ) + M H 3 ( q ) ( A + M − q B − 1 − q ) }

[Proof]

Suppose it’s holds when M, Let y = T M + 1 − M − 1 .

A : ∑ q = 0 M + 1 H 1 ( P S 1 , q ) ( A B − q ) = ∑ q = 0 M + 1 { ( y + q ) D M + 1 H 1 ( q − 1 ) + ( K M + 1 + q D M + 1 ) H 1 ( q ) } ( A B − q ) = ∑ q = 0 M { ( y + q + 1 ) D M + 1 H 1 ( q ) ( A B − 1 − q ) + ( K M + 1 + q D M + 1 ) H 1 ( q ) ( A B − q ) } = ( y + 1 ) D M + 1 ∑ q = 0 M H 1 ( q ) ( A B − 1 − q ) + K M + 1 ∑ q = 0 M H 1 ( q ) ( A B − q )     + D M + 1 ∑ q = 0 M H 1 ( q ) q ( A + 1 B − q )

B : ∑ q = 0 M + 1 H 2 ( P S 1 , q ) ( A + q B ) = ∑ q = 0 M + 1 { ( y + q ) D M + 1 H 2 ( q − 1 ) + ( K M + 1 + ( − y − 1 − q ) D M + 1 ) H 2 ( q ) } ( A + q B ) = ( y + 1 ) D M + 1 ∑ q = 0 M H 2 ( q ) ( A + q B − 1 ) + K M + 1 ∑ q = 0 M H 2 ( q ) ( A + q B )     + D M + 1 ∑ q = 0 M H 2 ( q ) q ( A + q B − 1 )

→ 2.4 ) A = B

→ 2.4 ) ∑ H 1 ( q ) ( A B − 1 − q ) = ∑ H 2 ( q ) ( A + q B − 1 ) ,                       ∑ H 1 ( q ) ( A B − q ) = ∑ H 2 ( q ) ( A + q B ) → left

q.e.d.

Example 2.1

M = 1:

Form 1 = T 1 D 1 ( A B − 1 ) + K 1 ( A B ) ;

Form 2 = T 1 D 1 ( A + 1 B ) + ( K 1 − T 1 D 1 ) ( A B ) ;

Form 3 = ( − K 1 + T 1 D 1 ) ( A B − 1 ) + K 1 ( A + 1 B ) ;

M = 2:

Form 1 = T 1 T 2 D 1 D 2 ( A B − 2 ) + [ K 1 ( T 2 − 1 ) D 2 + T 1 D 1 ( K 2 + D 2 ) ] ( A B − 1 )                         + K 1 K 2 ( A B ) ;

Form 2 = T 1 T 2 D 1 D 2 ( A + 2 B ) + [ ( K 1 − T 1 D 1 ) ( T 2 − 1 ) D 2                         + T 1 D 1 ( K 2 − T 2 D 2 ) ] ( A + 1 B ) + ( K 1 − T 1 D 1 ) ( K 2 − T 2 D 2 + D 2 ) ( A B ) ;

Form 3 = ( − K 1 + T 1 D 1 ) ( − K 2 + T 2 D 2 − D 2 ) ( A B − 2 ) + [ K 1 ( − K 2 + T 2 D 2 )     + ( − K 1 + T 1 D 1 ) ( K 2 + D 2 ) ] ( A + 1 B − 1 ) + K 1 K 2 ( A + 2 B ) .

If f ( n ) = ∑ A i ( N i m i ) , m i is not changed with n, then define

∇ p f ( n ) = ∑ A i ( N i − p m i − p ) , P ∈ ℤ

∇ f ( n ) = f ( n ) − f ( n − 1 ) , this is a little different from the difference

∇ 0 f ( n ) = f ( n ) , ∇ − 1 f ( N ) = ∑ n = 0 N − 1 f ( n )

Eg: ∇ ( n + 1 n − 1 ) = ∇ ( n + 1 2 ) = ( n 1 ) ≠ ( n n − 2 ) .

In [1], 1.1) is proved by

(*) ∑ n = 0 N − 1 n ( n + K M ) = ( M + 1 ) ( N + K M + 2 ) + ( M − K ) ( N + K M + 1 )

S U M ( N , P S 1 , [ P T , T M + 1 ] ) = ∑ n = 0 N − 1 ( K M + 1 + n × D M + 1 ) × ∇ S U M ( n + 1 )

S U M ( N , P S 1 , [ P T , T M + 2 ] ) = ∑ n = 0 N − 1 ( K M + 1 + n × D M + 1 ) × ∇ 0 S U M ( n + 1 )

P S 1 = [ P S , K M + 1 : D M + 1 ] , P T 1 = [ P T , T M + 1 = T M + 2 − p ] .

Define: S U M ( N , P S 1 , P T 1 ) = ∑ n = 0 N − 1 ( K M + 1 + n × D M + 1 ) × ∇ p S U M ( n + 1 )

S U M ( N , P S 1 , P T 1 ) can be calculated using the same method of 1.1).

The Form: ( T 1 + K 1 ) ( T 2 + K 2 ) ⋯ ( T M + K M ) = ∑ ∏ i = 1 M X M , X i = T i or K i .

3.1) q = X ( T ) , P M = P M ( P T ) ,

S U M ( N ) = → Form 1 ∑ q = 0 P M H 1 ( q ) ( N + T M − P M N − 1 − q ) , B i = { ( T i − X K − 1 ) D i ; X i = T i K i + X T − 1 D i ; X i = K i → Form 2 ∑ q = 0 P M H 2 ( q ) ( N + T M − P M + q N − 1 ) , B i = { ( T i − X K − 1 ) D i ; X i = T i K i + ( X K − 1 − T i ) D i ; X i = K i → Form 3 ∑ q = 0 P M H 3 ( q ) ( N + T M − q N − 1 − q ) , B i = { − K i + ( T i − X T − 1 ) D i ; X i = T i K i + X T − 1 D i ; X i = K i

H ( q ) = H ( P S , P T , q ) = ∑ all   of   the   ∏ X i   with   X ( T ) = q ∏ i = 1 M B i .

[Proof]

S U M ( N ) → Form 1 ∑ q = 0 M H 1 ( q ) ( N + T M − M N − 1 − q ) = ∑ q = 0 M H 1 ( q ) ( N + T M − M T M − M + 1 + q )

∑ n = 0 N − 1 ( K M + 1 + n × D M + 1 ) × ∇ p S U M ( n + 1 ) = ∑ n = 0 N − 1 ( K M + 1 + n × D M + 1 ) × ∑ q = 0 M H 1 ( q ) ( n + 1 + T M − M − p T M − M + 1 + q − p ) → ( ∗ ) = ∑ q = 0 M ( T M − M + 2 + q − p ) D M + 1 × H 1 ( q ) ( N + 1 + T M − M − p T M − M + 3 + q − p )     + ∑ q = 0 M ( K M + 1 + q D M + 1 ) × H 1 ( q ) ( N + 1 + T M − M − p T M − M + 2 + q − p )

= ∑ q = 0 M ( T M + 1 − [ M − q ] ) D M + 1 × H 1 ( q ) ( N + T M + 1 − ( M + 1 ) N − 1 − ( q + 1 ) )     + ∑ q = 0 M ( K M + 1 + q D M + 1 ) × H 1 ( q ) ( N + T M + 1 − ( M + 1 ) N − 1 − q ) = ∑ q = 0 M + 1 H 1 ( P S 1 , P T 1 , q ) ( N + T M + 1 − ( M + 1 ) N − 1 − q ) → 2.4 ) threeforms

q.e.d.

3.2) P S 1 = [ D × A : D , P S ] , P T 1 = [ A , P T ]

① H 1 ( P S 1 , q ) = D × A × [ H 1 ( q ) + H 1 ( q − 1 ) ] ;

② H 2 ( P S 1 , 0 ) = 0 , H 2 ( P S 1 , q ) = D × A × H 2 ( q − 1 ) ;

③ H 3 ( P S 1 , M + 1 ) = 0 , H 3 ( P S 1 , q ) = D × A × H 3 ( q ) ;

④ H 1 ( [ D × T 1 : D , ⋯ , D × T M : D ] , [ T 1 , ⋯ , T M ] , q ) = D M ( M q ) ∏ i = 1 M T i ;

⑤ S U M ( N , P T , P T ) = ∏ i = 1 M T i ( N + T M T M + 1 ) ;

⑥ S U M ( N , [ L 1 , ⋯ , L P , P S ] , [ L 1 , ⋯ , L P , P T ] ) = ∏ i = 1 P L i S U M ( N ) .

These are conclusions of [1] and can be extended to the new PT.

3.3) P S 1 = [ 1 , 1 , ⋯ , 1 , P S ] , P T 1 = [ 1 , 1 , ⋯ , 1 , P T ] , S U M ( N , P S 1 , P T 1 ) = S U M ( N ) .

P = Count of 1 added

→ expands ∑ q = 0 M H 1 ( q ) ( A + P B − q ) to ∑ q = 0 M + P H 1 ( P S 1 , q ) ( A B − q ) .

Now PT’s domain is extended to ℕ and T i + 1 − T i is not restricted.

If T M ≤ 0 , S U M ( N ) = ∑ q = 0 M H 1 ( q ) ( A B ) , B < 0 , the formula has no meaning

when regardless of the actual meaning, Form 1 = Form 2 = Form 3 still established.

PT’s domain can be extended to ℂ .

Define

F M N + M − 1 = ∑ 1 ≤ I i ≤ I i + 1 ≤ N + M − 1 ∏ i = 1 M I i = S 1 ( N + M , N )

E M N = ∑ λ 1 + λ 2 + ⋯ + λ N = M 1 λ 1 2 λ 2 ⋯ N λ N = S 2 ( N + M , N )

E p q + 1 ⊙ ( P T , K ) = ∑ λ 1 + λ 2 + ⋯ + λ q + 1 = p 1 λ 1 2 λ 2 ⋯ ( q + 1 ) λ q + 1   × ( T 1 + λ 1 K ) ( T 2 + λ 1 K + λ 2 K ) ⋯ ( T q + λ 1 K + λ 2 K + ⋯ + λ q K ) , λ i ≥ 0

PT in section 1: T 1 = 1 , { T i + 1 − T i = 1 : meanscontinuity T i + 1 − T i = 2 , meansdiscontinuity .

[1] call them Basic Shapes and define:

P B ( P T ) = countofdiscontinuity ,   M I N ( P T ) = ∏ i = 1 M T i D i .

Expand the definition:

P S = [ K 1 : D 1 , K 2 : D 2 , ⋯ ] , Item = { K 1 + λ 1 D 1 , K 2 + λ 2 D 2 , ⋯ }

Specify λ 1 = 0 , { λ i = λ i + 1 : meanscontinuity λ i + 1 = λ i + 1 , meansdiscontinuity

PB (item of PS) = count of discontinuity in an item, value ∈ [ 0 , M − 1 ] .

M I N ( item   of   P S ) = ∏ i = 1 M ( K i + λ i D i )

M I N q ( P S ) = ∑ P B (   ) = q M I N ( item   of   P S ) ,   M I N q   is   short   for   M I N q ( [ 1 , ⋯ , M ] ) .

By definition →

4.1) ① H 1 ( q ) = ( − 1 ) M − q H 2 ( [ − K i + T i D i − ( i − 1 ) D i ] , P T ) = ( − 1 ) q H 3 ( P S , [ K i D i − T i + ( i − 1 ) ] ) ;

② H 2 ( q ) = ( − 1 ) M − q H 1 ( [ − K i + T i D i − ( i − 1 ) D i ] , P T ) ;

③ H 3 ( q ) = ( − 1 ) q H 1 ( P S , [ K i D i − T i + ( i − 1 ) ] ) .

4.2) P S = [ 1 , 1 , ⋯ ] , P T = [ T i = T 1 + ( i − 1 ) ( K + 1 ) ]

① H 1 ( q ) = E M − q q + 1 ⊙ ( P T , K ) ;

② H 3 ( q ) = E M − q q + 1 ⊙ ( [ T i = ( T 1 − 1 ) + ( i − 1 ) K ] , K + 1 ) ;

③ E M − q q + 1 ⊙ ( P T , K ) = E M − 1 − q q + 1 ⊙ ( [ K + T i ] , K ) + T 1 E M − 1 − q q + 1 ⊙ ( [ K + T i ] , K ) .

[Proof]

H 1 ( q ) = ∑ ∏ ( X ∈ T ) ∏ ( X ∈ K ) → def ∑ ∏ ( X ∈ K ) = E M − q q + 1

∏ X = 1 λ 1 { ∏ x = 1 A − 1 X λ 1 + x } A λ A { ∏ x = 1 B − A X λ 1 + A − 1 + λ A + x } B λ B ⋯

∏ x = 1 A − 1 X λ 1 + x = ∏ x = 1 A − 1 ( T 1 + [ λ 1 + x − 1 ] [ K + 1 ] − λ 1 ) = ∏ x = 1 A − 1 ( T X + λ 1 K ) → λ 2 , λ 3 , ⋯ = 0 = ( T 1 + λ 1 K ) ⋯ ( T A − 1 + ⋯ + λ A − 1 K )

→ sameway ∏ x = 1 B − A X λ 1 + A − 1 + λ A + x = ( T A + λ 1 K + ⋯ ) ⋯ ( T B − 1 + ⋯ ) → ①

q.e.d.

4.3) { ①   P S 1 = [ K 2 : D 2 , ⋯ , K M : D M ] , P T 1 = [ K 2 D 2 + 1 , ⋯ , K M D M + M − 1 ] ②   P S 1 = [ D 2 : D 2 , ⋯ , D M : D M ] , P T 1 = [ K 2 D 2 + 1 , ⋯ , K M D M + M − 1 ] → ③   P S 1 = [ K 2 : D 2 , ⋯ , K M : D M ] , P T 1 = [ − 1 , ⋯ , − 1 ]

{ K 1 × H 1 ( P S 1 , q ) = M I N q ( P S ) , H 1 ( P S , [ K 1 D 1 , P T 1 ] , q ) = M I N q + M I N q − 1 K 1 × H 2 ( P S 1 , q ) = ( − 1 ) M − 1 − q M I N q ( P S ) K 1 × H 3 ( P S 1 , q ) = ( − 1 ) q M I N q ( P S )

Example 4.1: H 1 ( q ) , H 2 ( q ) , H 3 ( q ) are equal to:

① P S = [ 1 , 1 , ⋯ ] , P T = [ 1 , 1 , ⋯ ] { ( M q ) = E M − q q + 1 ⊙ ( P T , − 1 ) H 2 ( M ) = 1 , H 2 ( q < M ) = 0 H 3 ( 0 ) = 1 , H 3 ( q > 0 ) = 0

② P S = [ 1 , 2 , ⋯ , M ] , P T = [ 1 , 2 , ⋯ , M ] = { M ! ( M q ) H 2 ( M ) = M ! , H 2 ( q < M ) = 0 H 3 ( 0 ) = M ! , H 3 ( q > 0 ) = 0

③ P S = [ 1 , 1 , ⋯ ] , P T = [ 1 , 2 , ⋯ , M ] ,

〈 M q 〉 = Euleriannumber : N M = ∑ q = 0 M − 1 〈 M q 〉 ( N + q M )

{ q ! E M − q q + 1 = q ! S 2 ( M + 1 , q + 1 ) → def ( − 1 ) M − q q ! E M − q q = ( − 1 ) M − q q ! S 2 ( M , q ) 〈 M M − 1 − q 〉 = 〈 M q 〉 = E M − q q + 1 ⊙ ( [ 0 , 0 , ⋯ ] , 1 ) = E M − 1 − q q + 1 ⊙ ( [ 1 , 1 , ⋯ ] , 1 )

→ 2.2 ) H 1 ( M ) = ∑ q = 0 M H 3 ( q ) → ∑ q = 0 M 〈 M q 〉 = M !

→ 2.2 ) H 1 ( 1 ) = ∑ q = 0 M H 2 ( q ) q → ∑ q = 0 M ( − 1 ) M − q q × q ! S 2 ( M , q ) = 2 M − 1

④ P S = [ 1 , 1 , ⋯ ] , P T = [ 2 , 3 , ⋯ , M ]

{ ( q + 1 ) ! E M − 1 − q q + 1 = ( q + 1 ) ! S 2 ( M , q + 1 ) ( − 1 ) M − 1 − q ( q + 1 ) ! E M − 1 − q q + 1 = ( − 1 ) M − 1 − q ( q + 1 ) ! S 2 ( M , q + 1 ) 〈 M M − 1 − q 〉 = 〈 M q 〉 = E M − 1 − q q + 1 ⊙ ( [ 1 , 1 , ⋯ ] , 1 )

⑤ P S = [ 1 , 1 , ⋯ ] , P T = [ 1 , 3 , ⋯ , 2 M − 1 ] = { E M − q q + 1 ⊙ ( P T , 1 ) ( − 1 ) M − q M I N q − 1 E M − q q + 1 ⊙ ( [ 0 , 1 , 2 , ⋯ ] , 2 )

⑥ P S = [ 1 , 1 , ⋯ ] , P T = [ 3 , 5 , ⋯ , 2 M − 1 ] = { E M − 1 − q q + 1 ⊙ ( P T , 1 ) ( − 1 ) M − 1 − q M I N q E M − 1 − q q + 1 ⊙ ( [ 2 , 3 , 4 , ⋯ ] , 2 )

⑦ P S = [ 1 , 2 , ⋯ , M ] ,

P T = [ 1 , 3 , ⋯ , 2 M − 1 ] = { M I N q + M I N q − 1 → def 1 × ( − 1 ) M − q E M − q q ⊙ ( [ 3 , 5 , ⋯ ] , 1 ) → def 1 × E q M − q ⊙ ( [ 2 , 3 , 4 , ⋯ ] , 2 )

⑧ P S = [ 2 , 3 , ⋯ , M ] ,

P T = [ 3 , 5 , ⋯ , 2 M − 1 ] = { M I N q → def ( − 1 ) M − 1 − q E M − 1 − q q + 1 ⊙ ( [ 3 , 5 , ⋯ ] , 1 ) → def E q M − q ⊙ ( [ 2 , 3 , 4 , ⋯ ] , 2 )

③ ④, ⑤ ⑥, ⑦ ⑧ are in pairs, they can verify 3.2).

MIN_q appears in S U M ( N , [ 1 , 1 , ⋯ ] , [ 3 , 5 , ⋯ ] ) and S U M ( N , [ 2 , 3 , ⋯ ] , [ 3 , 5 , ⋯ ] ) . It’s easy to write out their items directly by continuity and discontinuity.

[1] has proved: ∑ q = 0 M − 1 ( − 1 ) M − 1 − q M I N q = 1 . Extand it:

→ 2.2 ) K 1 × H 1 ( 0 ) = K 1 ∑ q = 0 M − 1 H 2 ( q ) → 4.3 ) = ∑ q = 0 M − 1 ( − 1 ) M − 1 − q M I N q ( P S ) →

5.1) ∑ q = 0 M − 1 ( − 1 ) M − 1 − q M I N q ( P S ) = K 1 ∏ i = 2 M D i

Example 5.1:

Basic Shape, M = 3:

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

Basic Shape, M = 4:

1 × 3 × 5 × 7 − ( 1 × 3 × 5 + 1 × 3 × 4 + 1 × 2 × 4 ) × 6   + ( 1 × 2 × 3 + 1 × 3 × 4 + 1 × 2 × 4 ) × 5 − 1 × 2 × 3 × 4 = 1.

P S = [ 5 , 10 : 10 , 2 : 3 , 3 : 10 ] :

5 × 20 × 8 × 33 − ( 5 × 10 × 5 + 5 × 20 × 5 + 5 × 20 × 8 ) × 23 + ( 5 × 10 × 2 + 5 × 10 × 5 + 5 × 20 × 5 ) × 13 − 5 × 10 × 2 × 3 = 1500 = 5 × 10 × 3 × 10

P S = [ 1 , 1 , ⋯ ] , P T = [ 2 , ⋯ , M ] → ∑ q = 1 M ( − 1 ) M − q q ! S 2 ( M , q ) = 1 .

5.2) ① M I N M − 2 = 2 ( M − 1 ) 3 ( 2 M − 1 ) ! !

② 2 × M I N M − 1 + M I N M − 2 ≡ 0   M O D   ( M + 2 ) 2 , M > 3 , M is odd

[Proof]

P S = [ 2 , 3 , ⋯ , M ] , P T = [ 3 , 5 , ⋯ , 2 M − 1 ]

H 1 ( M − 2 ) = M I N M − 2 = ∑ i = 1 M − 1 ( ∏ X ∈ T ) × 2 i

H 2 ( M − 2 ) = − ∑ i = 1 M − 1 ( ∏ X ∈ T ) × i = − 1 2 H 1 ( M − 2 )

→ 2.1 ) H 2 ( M − 2 ) = − ( M − 1 ) H 1 ( M − 1 ) + H 1 ( M − 2 ) → ①

2 × M I N M − 1 + M I N M − 2 = 2 ( M + 2 ) 3 × ( 2 M − 1 ) ! ! → ②

q.e.d.

Example 5.2:

2 × ( 1 × 3 × 5 ) + ( 1 × 2 × 4 + 1 × 3 × 4 ) = 50 ≡ 0   M O D   25

2 × ( 1 × 3 × 5 × 7 × 9 ) + ( 1 × 3 × 5 × 7 + 1 × 3 × 5 × 6 + 1 × 3 × 4 × 6 + 1 × 2 × 4 × 6 ) × 8 = 4410 ≡ 0   M O D   49

when PT is Basic Shape, items in SUM can be classified by continuity and discontinuity.

Eg: use A for continuity, B for discontinuity

[ 1 , 2 , 3 ] = AA ,   [ 1 , 2 , 4 ] = AB ,   [ 1 , 3 , 4 ] = BA ,   [ 1 , 3 , 5 ] = BB .

Products of F M N + M − 1 can be divided into 2 M − 1 categories.

It’s easy to write them intuitively. Eg: 1 × 2 × 4 , 2 × 3 × 5 , 1 × 2 × 5 , ⋯ , I 1 + 1 = I 2 , I 2 + 1 < I 3 ∈ [ 1 , 2 , 4 ] .

Each category has a simple formula → 3.2 ) − ⑤

S U M ( N − P B ( P T ) , P T , P T ) = M I N ( P T ) ( N + M T M + 1 ) .

This is the promotion of ∑ n = 0 N − 1 ( n M ) = ( N M + 1 )

→ Traverse F M N + M − 1 = S U M ( N , [ 2 , 3 , ⋯ ] , [ 3 , 5 , ⋯ ] ) = ∑ q = 0 M − 1 M I N q ( N + M M + 1 + q ) .

Similarly: for Basic PT, arbitrarily PS can use the classification.

Example 5.3:

S U M ( N , [ 1 , 1 , 1 ] , [ 1 , 3 , 5 ] ) = S U M ( N , [ 1 , 1 , 1 ] , [ 1 , 2 , 3 ] ) + S U M ( N − 1 , [ 1 , 1 , 2 ] , [ 1 , 2 , 4 ] )       + S U M ( N − 1 , [ 1 , 2 , 2 ] , [ 1 , 3 , 4 ] ) + S U M ( N − 2 , [ 1 , 2 , 3 ] , [ 1 , 3 , 5 ] )

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

The pairs of PSx and PTx compare with PS and [ 1 , 2 , ⋯ , M ] , P B ( P S x ) = P B ( P T x ) , and the discontinuity at the same position. They are called having the same shape.

5.3) For Basic PT, P S 1 = [ 1 , 1 , ⋯ , P S ] , P T 1 = [ 1 , 1 , ⋯ , P T ] , count of 1 added = PB(PT)

H 1 ( P S 1 , P T 1 , q ) = ∑ A + B = q , P B ( P S x ) = P B ( P T x ) = A , sameshape H 1 ( P S x , P T x , B ) .

P is Prime, P > 2, [1] has proved:

5.4) M I N q ≡ 0   M O D   P , q > 0 , q + M = P − 1

5.5) M I N q ≡ 0   M O D   P , { ①   M = P , q ≥ 0 ②   M = P − 1 , q > 0 ③   M < P − 1 , q + M ≥ P − 1

[Proof]

For ③: q + M = P − 1 → proved by 5.4)

q + M = P : P = Max   factor   of   M I N q → holds

q + M > P : M I N q = ( Items   has   P ) + ( Items   has   no   P )

( Items   has   no   P ) = ∑ ∏ ( factors ≥ P + 1 ) × { ∑ ∏ ( factors ≤ P − 1 ) }

{ ∑ ∏ ( factors ≤ P − 1 ) } is a MIN that match the conditions of 5.4)

q.e.d.

5.6) P = M + 2 , ① E M N ≡ ② F M M + N − 1 ≡ { 1 , N ≡ 1   M O D   P 0 , N ≡ 1   M O D   P   M O D   P

[Proof]

→ Example   4.1 − ⑥ E M N = ∑ q = 0 M − 1 ( − 1 ) M − 1 − q M I N q ( N + M + q N − 1 ) → 5.5 ) − ③

≡ M I N 0 ( N + M N − 1 ) ≡ ( P − 2 ) ! ( N + P − 2 N − 1 ) ≡ ( P + N − 2 N − 1 ) M O D   P → ①

→ Example   4.1 − ⑧ F M M + N − 1 = ∑ q = 0 M − 1 M I N q ( N + M N − 1 − q ) ≡ M I N 0 ( N + P − 2 P − 1 ) M O D   P → ②

q.e.d.

5.7) P = M + N , ① F M M + N − 1 ≡ ② E M N ≡ 0   M O D   P , 0 < M < P − 1

[Proof]

F M M + N − 1 = ∑ q = 0 M − 1 M I N q ( P M + 1 + q )

{ 2 M < P → ( P M + 1 + q ) ≡ 0   M O D   P , 0 ≤ q ≤ M − 1 2 M > P , M + q ≥ P − 1 → 5.5 ) M I N q ≡ 0   M O D   P 2 M > P , M + q < P − 1 → ( P M + 1 + q ) ≡ 0   M O D   P → ①

q.e.d.

N + M = P → def F M P − 1 = ( P − 1 ) F M − 1 P − 2 + F M P − 2 ; E M N = N E M − 1 N + E M N − 1 →

(1) S 1 ( P , N ) = ( P − 1 ) S 1 ( P − 1 , N ) + S 1 ( P − 1 , N − 1 )

(2) S 2 ( P , N ) = N S 2 ( P − 1 , N ) + S 2 ( P − 1 , N − 1 )

5.8) ① S 1 ( P − 1 , q ) ≡ 1   M O D   P ,   1 ≤ q ≤ P − 1

② q ! S 2 ( P − 1 , q ) ≡ ( − 1 ) q + 1   M O D   P ,   1 ≤ q ≤ P − 1

[Proof]

For ①: q = 1 , S 1 ( P − 1 , q ) = 1 , holds.

If q holds, S 1 ( P , q + 1 ) ≡ 0   M O D   P

S 1 ( P , q + 1 ) = ( P − 1 ) S 1 ( P − 1 , q + 1 ) + S 1 ( P − 1 , q ) ≡ − S 1 ( P − 1 , q + 1 ) + 1 ≡ 0   M O D   P → ①

For ②: q = 1 , q ! S 2 ( P − 1 , q ) = 1 , holds.

If q holds, S 2 ( P , q + 1 ) ≡ 0   M O D   P , q ! S 2 ( P , q + 1 ) ≡ 0   M O D   P

q ! S 2 ( P , q + 1 ) = ( q + 1 ) ! S 2 ( P − 1 , q + 1 ) + q ! S 2 ( P − 1 , q ) ≡ 0   M O D   P → ②

q.e.d.

Chart of 5.6), F M M + N − 1 = S 1 ( N + M , N ) , E M N = S 2 ( N + M , N )

5.9) ∑ n = 1 P − 1 n P − 2 ≡ 0   M O D   P 2 , P > 3

[Proof]

[1] has obtained this, but its proof is incorrect.

P P − 2 + ∑ n = 1 P − 1 n P − 2 = ∑ n = 1 P − 1 n P − 2 = S U M ( P , [ 1 , ⋯ , 1 ] , [ 1 , ⋯ , P − 2 ] ) = ∑ q = 0 P − 2 q ! S 2 ( P − 1 , q + 1 ) ( P q + 1 ) = ∑ q = 1 P − 1 ( q − 1 ) ! S 2 ( P − 1 , q ) ( P q ) = ∑ q = 1 P − 1 2 { ( q − 1 ) ! S 2 ( P − 1 , q ) ( P q ) + ( p − q − 1 ) ! S 2 ( P − 1 , p − q ) ( P p − q ) }

→ 5.8 ) − ② ≡ ∑ q = 1 P − 1 2 [ ( − 1 ) q + 1 + ( − 1 ) P − q + 1 ] ( P q ) , this step of [1] is wrong

∑ n = 1 P n P − 2 = ∑ q = 1 P − 1 ( q − 1 ) ! S 2 ( P − 1 , q ) ( P q ) = p ∑ q = 1 P − 1 q ! S 2 ( P − 1 , q ) q 2 ( P − 1 q − 1 ) → 5.8 ) − ②

( q − 1 ) ! S 2 ( P − 1 , q ) ( P q ) ∈ ℕ → q ! S 2 ( P − 1 , q ) q 2 ( P − 1 q − 1 ) ∈ ℕ

∑ n = 1 P n P − 2 p ≡ ∑ q = 1 P − 1 ( − 1 ) q + 1 q 2 ( P − 1 q − 1 ) → ( P − 1 q − 1 ) ≡ ( − 1 ) q − 1 ≡ ∑ q = 1 P − 1 1 q 2 ≡ ∑ q = 1 P − 1 q 2 ≡ 0   M O D   P

q.e.d.

S U M ( N , P S , P T ) = ∑ q = 0 M H 1 ( q ) ( N + T M − M N − 1 − q ) = ∑ q = 0 M H 1 ( q ) ( N + T M − M T M − M + 1 + q )

N starts from x to x + M, taking H₁(q) as variables, then get a linear equations.

Let P = x + T M − M , Q = x − 1 , each row from left to right, Q is from small to large

→ Form 1 A 1 ( P , Q , M ) = [ ( P Q ) ⋯ ( P Q − M ) ⋮ ⋱ ⋮ ( P + M Q + M ) ⋯ ( P + M Q ) ]

→ Form 2 A 2 ( P , Q , M ) = [ ( P Q ) ⋯ ( P + M Q ) ⋮ ⋱ ⋮ ( P + M Q + M ) ⋯ ( P + 2 M Q + M ) ]

→ Form 3 A 3 ( P , Q , M ) = [ ( P + M Q ) ⋯ ( P Q − M ) ⋮ ⋱ ⋮ ( P + 2 M Q + M ) ⋯ ( P + M Q ) ]

They are ( M + 1 ) × ( M + 1 ) matrices

6.1) ① ‖ A 2 ( P , Q , M ) ‖ = ( P Q ) ( P 0 ) ( P + 2 Q + 1 ) ( P + 2 1 ) ( P + 4 Q + 2 ) ( P + 4 2 ) ⋯ ( P + 2 M Q + M ) ( P + 2 M M ) = ‖ A 2 ( P , P − Q , M ) ‖

② Upper triangle: col_q of A 2 ( 1 , 0 , M ) = [ ( q 0 ) , ( q 1 ) , ( q 2 ) , ⋯ , ( q q ) , ⋯ ] T

[Proof]

Row M + 1 : = Row M + 1 − Row M P + M Q + M = [ 0     ( P + M + 1 Q + M ) P + M + 1 2 ( P + M + 2 Q + M ) P + M + 2 ⋯ M ( P + 2 M Q + M ) P + 2 M ]

Row M : = Row M − Row M − 1 P + M − 1 Q + M − 1 = [ 0     ( P + M Q + M − 1 ) P + M 2 ( P + M + 1 Q + M − 1 ) P + M + 1 ⋯ M ( P + 2 M − 1 Q + M − 1 ) P + 2 M − 1 ]

⋯

Row 2 : = Row 2 − Row 1 P + 1 Q + 1 = [ 0     ( P + 2 Q + 1 ) P + 2 2 ( P + 3 Q + 1 ) P + 3 ⋯ M ( P + M + 1 Q + 1 ) P + M + 1 ]

Repeat the above process and change it into upper triangle.

( P Q ) ( P 0 ) ( P + 1 Q ) ( P + 1 0 ) ( P + 2 Q ) ( P + 2 0 ) ( P + 3 Q ) ( P + 3 0 ) ⋯ ( P + M Q ) ( P + M 0 )

( P + 2 Q + 1 ) p + 2 1 ( P + 3 Q + 1 ) p + 3 2 ( P + 4 Q + 1 ) p + 4 3 ⋯ ( P + M + 1 Q + 1 ) p + M + 1 M

( P + 4 Q + 2 ) ( p + 4 ) ( p + 3 ) 2 ! ( P + 5 Q + 2 ) ( p + 5 ) ( p + 4 ) 3 × 2 ⋯ ( P + M + 2 Q + 2 ) ( p + M + 2 ) ( p + M + 1 ) M ( M − 1 )

⋯

when the original matrix is transformed into an upper triangular matrix,

Row K + 1 : = Row K + 1 − Row K P + K Q + K , repeat the operation K times.

q.e.d.

6.2) ‖ A 1 ( P , Q , M ) ‖ = ‖ A 2 ( P , Q , M ) ‖ = ‖ A 3 ( P , Q , M ) ‖

[Proof]

A 2 → Row i + 1 : = Row i + 1 − Row i   and   repeat change   ( row 1 , col 1 )   to   ( P Q )

[ ( P Q ) ⋯ ( P + M Q ) ⋮ ⋱ ⋮ ( P Q + M ) ⋯ ( P + M Q + M ) ] → transpose [ ( P Q ) ⋯ ( P Q + M ) ⋮ ⋱ ⋮ ( P + M Q ) ⋯ ( P + M Q + M ) ] = A 1 ( P , P − Q , M )

A 1 → col i : = col i + col i + 1   and   repeat → change   ( row 1 , col 1 )   to   ( P + M Q ) = A 3

q.e.d.

Use ‖ A ‖ for ‖ A 1 , 2 , 3 ‖ .

6.3) ‖ A ( P , 0 , M ) ‖ = 1 ,

‖ A ( P , 1 , M ) ‖ = ( P + M 1 + M ) , ‖ A ( P , Q > 0 , M ) ‖ = ∏ q = 0 Q − 1 ( P + M − q 1 + M ) ( 1 + M + q 1 + M )

[Proof]

→ 6.1 ) ‖ A ( P , 1 , M ) ‖ = P 1 P + 1 2 P + 2 3 ⋯ P + M M + 1 = ( P + M 1 + M )

q.e.d.

T M ≥ M , N ∈ [ 1 , M + 1 ] , then

Matrix of S U M ( N ) = A ( 1 + T M − M , 0 , M ) , Matrix of ∇ S U M ( N ) = A ( T M − M , 0 , M )

Use Cramer’s law, let y ( n ) = S U M ( N ) or ∇ S U M ( N )

H 1 ( q ) = ‖ A 1 ( P , 0 , M ) → replace   col q + 1 [ y ( 1 ) , ⋯ , y ( M + 1 ) ] T ‖

A 1 ( P , 0 , M ) = [ ( P 0 ) ⋯ 0 ⋮ ⋱ ⋮ ( P + M M ) ⋯ ( P + M 0 ) ] = [ 1 ⋯ 0 ⋮ ⋱ ⋮ ( P + M M ) ⋯ 1 ]

when col_q+1 replace with [ y ( 1 ) , ⋯ ] T , calculate from col_q+1, only { y ( 1 ) , ⋯ , y ( q + 1 ) } work.

From algebraic cofactor, y(k) corresponds to 1 K − 1 × ‖ A T m p ‖ × 1 M − q count of rows of A T m p = ( M + 1 ) − ( K − 1 ) − ( M − q ) − 1 = q − K + 1

( row k , col q + 1 ) = ( P + K − 1 K − 1 − q ) , ( q + 1 ) − ( q − k + 1 ) − 1 = k − 1

( row 0 , col 0 )   of   A T m p = ( row k + 1 , col k − 1 ) = ( P + K 1 )

‖ A T m p ‖ = ‖ A 1 ( P + k , 1 , q − k ) ‖ = ( P + q q + 1 − k ) →

6.4) T M ≥ M , H 1 ( q ) = { ∑ k = 1 q + 1 ( − 1 ) q + 1 + k ( 1 + T M − M + q q + 1 − k ) S U M ( k ) ∑ k = 1 q + 1 ( − 1 ) q + 1 + k ( T M − M + q q + 1 − k ) ∇ S U M ( k )

∇ S U M ( N , [ 1 , ⋯ , 1 ] , [ 2 , ⋯ , M ] ) = N M → A 1 ( 1 , 0 , M − 1 )

( q + 1 ) ! S 2 ( M , q + 1 ) = ∑ k = 1 q + 1 ( − 1 ) q + 1 + k ( q + 1 q + 1 − k ) k M

→ q ! S 2 ( M , q ) = ∑ k = 0 q ( − 1 ) q + k ( q q − k ) k M

x = q − k → q ! S 2 ( M , q ) = ∑ x = 0 q ( − 1 ) x ( q x ) ( q − x ) M ,

this is a known formula.

S U M ( N , [ 1 , ⋯ , M ] , [ 1 , ⋯ , M ] ) = ( M + N M + 1 ) → ( M q ) = ∑ k = 0 q ( − 1 ) k + q ( q k ) ( M + k M + 1 )

S U M ( N , [ 2 , 3 , ⋯ , M ] , [ 3 , 5 , ⋯ , 2 M − 1 ] ) → M I N q − 1 = ∑ k = 0 q ( − 1 ) k + q ( M + q q − k ) S 1 ( k + M , k )

Similarly,

A 3 ( P , 0 , M ) = [ ( P + M 0 ) ⋯ 0 ⋮ ⋱ ⋮ ( P + 2 M M ) ⋯ ( P + M 0 ) ] = [ 1 ⋯ 0 ⋮ ⋱ ⋮ ( P + 2 M M ) ⋯ 1 ] →

y(k) corresponds to 1 K − 1 × ‖ A T m p ‖ × 1 M − q , count of rows of A T m p = q − k + 1

( row k , col q + 1 ) = ( P + M + K − 1 − q K − 1 − q )

( row 0 , col 0 )   of   A T m p = ( row k + 1 , col k − 1 ) = ( P + M + 1 1 )

P + M + 1 − ( q − k ) = P + M + 1 − q + k

‖ A T m p ‖ = ‖ A 3 ( P + M + 1 − q + K , 1 , q − k ) ‖ = ( P + M + 1 q + 1 − k ) →

6.5) T M ≥ M , H 3 ( q ) = { ∑ k = 1 q + 1 ( − 1 ) q + 1 + k ( 2 + T M q + 1 − k ) S U M ( k ) ∑ k = 1 q + 1 ( − 1 ) q + 1 + k ( 1 + T M q + 1 − k ) ∇ S U M ( k )

∇ S U M ( N , [ 1 , ⋯ , 1 ] , [ 2 , ⋯ , M ] ) = N M → A 3 ( 1 , 0 , M − 1 ) →

〈 M q 〉 = ∑ k = 1 q + 1 ( − 1 ) q + k − 1 ( M + 1 q + 1 − k ) k M → x = q + 1 − k ∑ x = 0 q ( − 1 ) x ( M + 1 x ) ( q + 1 − x ) M

This is a known formula too.

→ 6.1 ) A 2 ( 1 , 0 , M ) = [ ( 0 0 ) ⋯ ( M 0 ) ⋮ ⋱ ⋮ 0 ⋯ ( M M ) ]

In the algebraic cofactor, ( row q + 1 , row q + 2 , ⋯ , row M + 1 ) will work.

Row_K corresponds to 1 q × ‖ A T m p ‖ × 1 M + 1 − K count of rows of A T m p = ( M + 1 ) − q − ( M + 1 − k ) − 1 = k − q − 1

( row k , col q + 1 ) = ( q k − 1 ) , k − ( k − q − 1 ) = q + 1

( row 0 , col 0 )     of   A T m p = ( row q + 1 , col q + 2 ) = ( q + 1 q )

‖ A T m p ‖ = ‖ A 2 ( q + 1 , q , k − q − 2 ) ‖ = ‖ A 2 ( q + 1 , 1 , k − q − 2 ) ‖ = ( k − 1 q )

It can be concluded by induction:

→ 6.1 ) y ( k )   willchangeto   z ( k ) = ∑ x = 1 k ( − 1 ) k − x y ( x ) ( K x )

H 2 ( q ) = ∑ k = q + 1 M + 1 ( − 1 ) q + k − 1 ( k − 1 q ) z ( k )

∇ S U M ( N , [ 1 , ⋯ , 1 ] , [ 2 , ⋯ , M ] ) = N M → Z ( k ) = k ! S 2 ( M , k ) →

q ! S 2 ( M , q ) = ∑ k = q M ( − 1 ) M + k ( k − 1 q − 1 ) k ! S 2 ( M , k ) , this matches 2.1)-②

We need an expression similar to ( N M ) , which is Gaussian coefficient [ N M ] q

[ N M ] q = ( q N − 1 ) ( q N − 1 − 1 ) ⋯ ( q N − ( M − 1 ) − 1 ) ( q M − 1 ) ( q M − 1 − 1 ) ⋯ ( q − 1 ) ,   q ≠ 1 ,   [ N 0 ] q = 1 ,   Abbreviated   as   [ N M ]

1) [ N M ] = [ N M − q ]

2) [ N M ] = [ N − 1 M − 1 ] + q M [ N − 1 M ]

7.1) ① ∑ n = 0 N − 1 q n [ n + M M ] = [ N + M M + 1 ]

② ∑ n = 0 N − 1 q n [ n + K M ] = q M − K [ N + K M + 1 ] ; ∑ n = 0 N − 1 q n [ n M ] = q M [ N M + 1 ]

[Proof]

When n = 0, ① is obviously true. Suppose it holds when N − 1,

∑ n = 0 N q n [ n + M M ] = [ N + M M + 1 ] + q N [ N + M M ] = ( q N + M − 1 ) ( q N + M − 1 − 1 ) ⋯ ( q N − 1 ) ( q M + 1 − 1 ) ( q M − 1 ) ( q M − 1 − 1 ) ⋯ ( q − 1 ) + q N ( q N + M − 1 ) ( q N + M − 1 − 1 ) ⋯ ( q N + 1 − 1 ) ( q M − 1 ) ( q M − 1 − 1 ) ⋯ ( q − 1 ) = [ N + M + 1 M + 1 ]

∑ n = 0 N − 1 q n [ n + K M ] → x = n + K − M q M − K ∑ x = 0 N − 1 + K − M q x [ x + M M ] = q M − K [ N + K M + 1 ]

q.e.d.

∑ n = 0 N − 1 q n = q N − 1 q − 1 = [ N 1 ]

∑ n = 0 N − 1 q n ( K 1 + D 1 q n ) = ∑ n = 0 N − 1 ( q n ( q n − 1 ) D 1 + ( K 1 + D 1 ) q n ) = ∑ n = 0 N − 1 ( q − 1 ) q n [ N 1 ] D 1 + ( K 1 + D 1 ) q n [ N 0 ] = q ( q − 1 ) D 1 [ N 2 ] + ( K 1 + D 1 ) [ N 1 ]

∑ n = 0 N − 1 q n ( K 1 + D 1 q n ) ( K 2 + D 2 q n ) = ∑ n = 0 N − 1 ( D 1 D 2 q 3 n + ( K 1 D 2 + K 2 D 1 ) q 2 n + K 1 K 2 q n ) = ∑ n = 0 N − 1 ( D 1 D 2 q 2 n ( q n − 1 ) + ( K 1 D 2 + K 2 D 1 + D 1 D 2 ) q 2 n + K 1 K 2 q n ) = ∑ n = 0 N − 1 [ D 1 D 2 q n + 1 ( q n − 1 ) ( q n − 1 − 1 ) + ( K 1 D 2 + K 2 D 1 + D 1 D 2     + q D 1 D 2 ) q n ( q n − 1 ) + ( K 1 K 2 + K 1 D 2 + K 2 D 1 + D 1 D 2 ) q n ]

= D 1 D 2 q 3 ( q 2 − 1 ) ( q − 1 ) [ N 3 ] + ( K 1 D 2 + K 2 D 1 + D 1 D 2     + q D 1 D 2 ) q ( q − 1 ) [ N 2 ] + ( K 1 K 2 + K 1 D 2 + K 2 D 1 + D 1 D 2 ) [ N 1 ] = q ( q − 1 ) D 1 × q 2 ( q 2 − 1 ) D 2 [ N 3 ] + q ( q − 1 ) { ( K 1 + D 1 ) D 2     + D 1 ( K 2 + q D 2 ) } [ N 2 ] + ( K 1 + D 1 ) ( K 2 + D 2 ) [ N 1 ]

T i is arbitrary, use the Form ∏ i = 1 M ( K i + T i ) , X T = count   of   { X 1 , ⋯ , X i } ∈ T

7.2) ∑ n = 0 N − 1 q n ∏ i = 1 M ( K i + D i q n ) = ∑ g = 0 M G ( M , g ) [ N g + 1 ]

G ( M , g ) = ∑ ∏ X i   with   X ( T ) = g ∏ i = 1 M f ( X i )

f ( X i ) = { q X T ( q X T − 1 ) D i ; X ∈ T K i + q X T − 1 D i ; X ∈ K

[Proof]

When M = 1, 2, it’s true. Let G ( q ) = G ( M , q )

Suppose F ( N ) = ∑ n = 0 N − 1 q n ∏ i = 1 M ( K i + D i q n ) = ∑ g = 0 M G ( g ) [ N g + 1 ] →

q n ∏ i = 1 M ( K i + D i q n ) = F ( n + 1 ) − F ( n ) = ∑ g = 0 M G ( g ) { [ n + 1 g + 1 ] − [ n g + 1 ] } = ∑ g = 0 M G ( g ) { ( q g + 1 − 1 ) [ n g + 1 ] + [ n g ] }

∑ n = 0 N − 1 q n ∏ i = 1 M + 1 ( K i + D i q n ) = K M + 1 ∑ n = 0 N − 1 q n ∏ i = 1 M ( K i + D i q n ) + D M + 1 ∑ n = 0 N − 1 q n ∏ i = 1 M + 1 ( K i + D i q n ) q n = K M + 1 ∑ g = 0 M G ( g ) [ N g + 1 ] + D M + 1 ∑ g = 0 M G ( g ) ∑ n = 0 N − 1 q n { ( q g + 1 − 1 ) [ n g + 1 ] + [ n g ] }

→ 7.1 ) ∑ g = 0 M G ( g ) ( K M + 1 + q g D M + 1 ) [ N g + 1 ]   + ∑ g = 0 M G ( g ) q g + 1 ( q g + 1 − 1 ) D M + 1 [ N g + 2 ] = ∑ g = 0 M + 1 G ( M + 1 , g ) [ N g + 1 ]

q.e.d.

In the same way, use the Form = ( T 1 + K 1 ) ⋯ ( T M + K M ) :

7.3) ∑ n = 0 N − 1 ∏ i = 1 M ( K i + D i q n ) = ∑ g = 1 M G ( M , g ) [ N g ] + N ∏ i = 1 M K i

G ( M , g ) = ∑ ∏ X i   with   X ( T ) = g ∏ i = 1 M f ( X i )

f ( X i ) = { 1 ; X ∈ T , X T − 1 = 0 q X T − 1 ( q X T − 1 − 1 ) D i ; X ∈ T , X T − 1 > 0 K i ; X ∈ K , X T − 1 = 0 K i + q X T − 1 − 1 D i ; X ∈ K , X T − 1 > 0

7.4) q M N − 1 q M − 1 = ∑ g = 1 M ( ∏ i = 1 g − 1 q i ( q i − 1 ) ) [ N g ] [ M − 1 M − g ]

[Proof]

∑ n = 0 N − 1 ∏ i = 1 M ( 0 + q n ) = ∑ n = 0 N − 1 q M n = q M N − 1 q M − 1 = ∑ g = 1 M G ( M , g ) [ N g ]

In G ( M , g ) , ∏ ( X ∈ T ) = ∏ i = 1 g − 1 q i ( q i − 1 ) , X 1 must be T 1 , count of ( X ∈ K ) = M − g , M − 1 positions can be placed.

In 1916 MacMahon [6] observed that [ N K ] = ∑ w ∈ Ω ( N , K ) q i n v ( w ) , Ω ( N , K ) denotes all permutations of the multiset { 0 N − K , 1 K } , that is, all words w = w 1 , ⋯ , w n

with n - k zeroes and k ones, and inv(・) denotes the inversion statistic defined by i n v ( w 1 , ⋯ , w n ) = | { ( i , j ) : 1 ≤ i < j ≤ n , w i > w j } | .

So in G ( M , g ) , ∑ ∏ ( X ∈ K ) = [ M − 1 M − g ]

q.e.d.

7.5) ① ( K + D ) M = K M + D ∑ g = 0 M − 1 k g ( K + D ) M − g

② ( K + q ) M = K M + q ∑ g = 0 M − 1 k g ( K + 1 ) M − 1 − g     + q ( q − 1 ) ∑ a + b + c = M − 2 , a , b , c ≥ 0 k a ( K + 1 ) b ( K + q ) c

[Proof]

∑ n = 0 0 ∏ i = 1 M ( K + D q n ) = K M + G ( M , 1 ) [ 1 1 ] → ①

∑ n = 0 1 ∏ i = 1 M ( K + q n ) − ∑ n = 0 0 ∏ i = 1 M ( K + q n ) = 2 K M + G ( M , 1 ) [ 2 1 ] + G ( M , 2 ) [ 2 2 ] − ( K M + G ( M , 1 ) [ 1 1 ] ) = K M + G ( M , 1 ) ( q 2 − 1 ) ( q − 1 ) + G ( M , 2 ) − ( K M + G ( M , 1 ) ) → ②

q.e.d.

7.2) can be understood as use the Form = ( T 1 + K 1 ) ⋯ ( T M + K M ) , P T = [ 1 , 2 , ⋯ , M ]

f ( X i ) = { q T i − X K − 1 ( q T i − X K − 1 − 1 ) D i ; X i = T i K i + q X T − 1 D i ; X i = K i

But it can not be simply extended to something like 3.1).

8.1) ∑ n = 0 N − 1 q n ( n + M M ) = q N ∑ g = 0 M ( − 1 ) g ( N + M − 1 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 ( q − 1 ) M + 1

[Proof]

M = 0 ,   ∑ n = 0 N − 1 q n ( n 0 ) = q N − 1 q − 1 = q N ( N − 1 0 ) q − 1 − 1 q − 1 ,   holds

M = 1 , ∑ n = 0 N − 1 q n ( n + 1 1 ) = ∑ n = 0 N − 1 q n ( n + 1 ) = ∑ n = 0 N − 1 q n + ∑ n = 1 N − 1 q n + ⋯ + ∑ n = N − 1 N − 1 q n = N ∑ n = 0 N − 1 q n − ( ∑ n = 0 0 q n + ∑ n = 0 1 q n + ⋯ + ∑ n = 0 N − 2 q n ) = N q N − 1 q − 1 − ( q − 1 q − 1 + q 2 − 1 q − 1 + ⋯ + q N − 1 − 1 q − 1 ) = N q N − 1 q − 1 − ( 1 + q + q 2 + ⋯ + q N − 1 ) − N q − 1 = q N N q − 1 − q N − 1 ( q − 1 ) 2 = q N ( N 1 ) q − 1 − q N ( N 0 ) ( q − 1 ) 2 + 1 ( q − 1 ) 2 ,   holds

Suppose it holds at M − 1; When M and N = 1, ∑ n = 0 N − 1 q n ( n + M M ) = 1

A = ∑ g = 0 M ( − 1 ) g q N ( N + M − 1 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 ( q − 1 ) M + 1 = q ( q − 1 ) 1 + ⋯ + ( − 1 ) M q ( q − 1 ) M + 1 + ( − 1 ) M + 1 ( q − 1 ) M + 1

( q − 1 ) M + 1 A = q { ( q − 1 ) M − ( q − 1 ) M − 1 + ⋯ + ( − 1 ) M − 1 } + ( − 1 ) M + 1

→ 7.5 ) − ② , K = − 1 = ( q − 1 ) M + 1 → A = 1 ; It holds when M and N = 1.

Suppose it holds at M and N

∑ n = 0 N q n ( n + M M ) = ∑ n = 0 N q n { ( n + M − 1 M ) + ( n + M − 1 M − 1 ) } = q ∑ n = 0 N − 1 q n ( n + M M ) + ∑ n = 0 N q n ( n + M − 1 M − 1 ) = ∑ g = 0 M ( − 1 ) g q N + 1 ( N + M − 1 − g M − g ) ( q − 1 ) g + 1 + q ( − 1 ) M + 1 ( q − 1 ) M + 1

    + ∑ g = 0 M − 1 ( − 1 ) g q N + 1 ( N + M − 1 − g M − 1 − g ) ( q − 1 ) g + 1 + ( − 1 ) M ( q − 1 ) M = ∑ g = 0 M ( − 1 ) g q N + 1 ( ( N + 1 ) + M − 1 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 ( q − 1 ) M + 1

It holds when M and N + 1.

q.e.d.

Example 8.1

∑ n = 0 N − 1 q n ( n 0 ) = q N q − 1 − 1 q − 1

∑ n = 0 N − 1 q n ( n + 1 1 ) = q N ( N 1 ) q − 1 − q N ( q − 1 ) 2 + 1 ( q − 1 ) 2

∑ n = 0 N − 1 q n ( n + 2 2 ) = q N ( N + 1 2 ) q − 1 − q N ( N 1 ) ( q − 1 ) 2 + q N ( q − 1 ) 3 − 1 ( q − 1 ) 3

8.2) ∑ n = 0 N − 1 q n ( n + M − K M ) = q N ∑ g = 0 M ( − 1 ) g ( N − K + M − 1 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 q K ( q − 1 ) M + 1 , N ≥ 1 + K

[Proof]

∑ n = 0 N − 1 q n ( n + M − K M ) = ∑ n = k N − 1 q n ( n + M − K M ) = q K ∑ n = 0 N − 1 − K q n ( n + M M ) = q K { q N − K ∑ g = 0 M ( − 1 ) g ( N − K + M − 1 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 ( q − 1 ) M + 1 }

q.e.d.

→ ∑ n = 0 N − 1 q n ( n M ) = q N ∑ g = 0 M ( − 1 ) g ( N − 1 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 q M ( q − 1 ) M + 1 , N ≥ 1 + M

→ ∑ n = 0 N − 1 q n ( n + M − 1 M ) = q N ∑ g = 0 M ( − 1 ) g ( N + M − 2 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 q ( q − 1 ) M + 1 , N ≥ 2

→ ∑ n = 0 N − 1 q n ( n − 1 M ) = q N ∑ g = 0 M ( − 1 ) g ( N − 2 − g M − g ) ( q − 1 ) g + 1 + ( − 1 ) M + 1 q M + 1 ( q − 1 ) M + 1 , N ≥ 2 + M

Define A q M = ∑ k = 0 M ( 1 − q ) M − k q K S 2 ( M , k ) k ! ,   q ≠ 0 , 1 ,   M ≥ 0

A 2 M = 1 , 2 , 6 , 26 , 150 , 1082 , 9366 , ⋯ http://oeis.org/A000629

A 3 M = 1 , 3 , 12 , 66 , 480 , 4368 , 47712 , ⋯ http://oeis.org/A123227

A 4 M = 1 , 4 , 20 , 132 , 1140 , 12324 , 160020 , ⋯ http://oeis.org/A201355

A q 0 = 1 , A q 1 = q

n M = ∇ S U M ( n , [ 1 , ⋯ , 1 ] [ 2 , ⋯ , M ] ) → Form 1 ∑ K = 0 M k ! S 2 ( M , k ) ( n K )

∇ g ( N − 1 ) M = ∑ K = g M k ! S 2 ( M , k ) ( N − 1 − g K − g ) .

9.1) ∑ n = 0 N − 1 q n n M = q N ( q − 1 ) M + 1 ∑ g = 0 M ( q − 1 ) M − g ( − 1 ) g ▽ g ( N − 1 ) M + ( − 1 ) M + 1 A q M ( q − 1 ) M + 1

[Proof]

∑ n = 0 N − 1 q n n 0 = q N q − 1 − 1 q − 1 ,   holds

∑ n = 0 N − 1 q n n 1 = ∑ n = 0 N − 1 q n ( n 1 ) = q N ∑ g = 0 1 ( − 1 ) g ( N − 1 − g 1 − g ) ( q − 1 ) g + 1 + q ( q − 1 ) 2 = q N ( N − 1 ) ( q − 1 ) 1 − q N ( q − 1 ) 2 + q ( q − 1 ) 2 = q N ( q − 1 ) 2 ( ( q − 1 ) ( N − 1 ) 1 − 1 ) + A q 1 ( q − 1 ) 2 ,   holds

∑ n = 0 N − 1 q n n M = ∑ n = 0 N − 1 q n ∑ K = 0 M k ! S 2 ( M , k ) ( n K ) = ∑ K = 0 M K ! S 2 ( M , K ) { q N ∑ g = 0 k ( − 1 ) g ( N − 1 − g k − g ) ( q − 1 ) g + 1 + ( − 1 ) k + 1 q k ( q − 1 ) k + 1 }

∑ K = 0 M K ! S 2 ( M , K ) q k ( − 1 ) K + 1 ( q − 1 ) K + 1 = ( − 1 ) M + 1 ( q − 1 ) M + 1 ∑ K = 0 M K ! S 2 ( M , K ) q k ( q − 1 ) M − K ( − 1 ) − ( M − K ) = ( − 1 ) M + 1 A q M ( q − 1 ) M + 1

∑ K = 0 M K ! S 2 ( M , K ) q N ∑ g = 0 k ( − 1 ) g ( N − 1 − g k − g ) ( q − 1 ) g + 1 = q N ( ... ) ( q − 1 ) M + 1 = q N ( q − 1 ) M + 1 ∑ K = 0 M k ! S 2 ( M , k ) ∑ g = 0 k ( − 1 ) g ( N − 1 − g K − g ) ( q − 1 ) M − g

( ... ) = K ! S 2 ( M , M ) { ( N − 1 M ) ( q − 1 ) M − ( N − 2 M − 1 ) ( q − 1 ) M − 1 + ⋯ + ( − 1 ) M ( N − 1 − M 0 ) }   + K ! S 2 ( M , M − 1 ) { ( N − 1 M − 1 ) ( q − 1 ) M − ( N − 2 M − 2 ) ( q − 1 ) M − 1 + ⋯ } + ⋯   + K ! S 2 ( M , 0 ) { ( N − 1 0 ) ( q − 1 ) M }

Verticalcalculation = ( q − 1 ) M ∇ 0 ( N − 1 ) M − ( q − 1 ) M − 1 ∇ 1 ( N − 1 ) M ⋯

q.e.d.

9.2) A q M = q ∑ k = 0 M ( q − 1 ) M − k S 2 ( M , k ) k ! , M > 0

[Proof]

n M = ∇ S U M ( n , [ 1 , ⋯ , 1 ] [ 1 , ⋯ , M ] ) → Form 2 ∑ K = 0 M ( − 1 ) M − K K ! S 2 ( M , K ) ( n + K − 1 K )

∑ n = 0 N − 1 q n n M = ∑ n = 0 N − 1 q n ∑ K = 0 M ( − 1 ) M − K K ! S 2 ( M , K ) ( n + K − 1 K ) = ∑ K = 0 M ( − 1 ) M − K K ! S 2 ( M , K ) { q N ∑ g = 0 K ( − 1 ) g ( N + K − 2 − g K − g ) ( q − 1 ) g + 1 + ( − 1 ) K + 1 q ( q − 1 ) K + 1 }

∑ K = 0 M ( − 1 ) M − K K ! S 2 ( M , K ) ∑ g = 0 K ( − 1 ) g q N ( N + K − 2 − g K − g ) ( q − 1 ) g + 1 → Samewayas   9.1 ) = q N ( q − 1 ) M + 1 ∑ g = 0 M ( q − 1 ) M − g ( − 1 ) g ∇ g ( N − 1 ) M

Compare with 9.1) → ∑ K = 0 M ( − 1 ) M − K K ! S 2 ( M , K ) ( − 1 ) K + 1 q ( q − 1 ) K + 1 = ( − 1 ) M + 1 A q M ( q − 1 ) M + 1

q.e.d.

n M = ∇ S U M ( n , [ 1 , ⋯ , 1 ] [ 1 , ⋯ , M ] ) → Form 1 ∑ K = 0 M k ! S 2 ( M + 1 , k + 1 ) ( n − 1 K ) →

9.3) A q M = ∑ k = 0 M ( 1 − q ) M − k q K + 1 S 2 ( M + 1 , k + 1 ) k ! ,   M > 0

9.4) ∇ g ( N − 1 ) M = ∑ k = 0 M − g ( − 1 ) M − g − k ( M k ) N K S 2 ( M − K + 1 , g + 1 ) g !

[Proof]

S 2 ( M , g ) = 1 g ! ∑ k = 0 g ( − 1 ) k ( g k ) ( g − k ) M → j = g − k = 1 g ! ∑ j = 0 g ( − 1 ) g − j ( g j ) j M = 1 ( g − 1 ) ! ∑ j = 1 g ( − 1 ) g − j ( g − 1 j − 1 ) j M − 1 = 1 ( g − 1 ) ! ∑ j = 0 g − 1 ( − 1 ) g − 1 − j ( g − 1 j ) ( j + 1 ) M − 1

∇ g ( N − 1 ) M → Bydefinition ∑ j = 0 g ( − 1 ) j ( g j ) ( N − j − 1 ) M = ∑ j = 0 g ( − 1 ) j ( g j ) ∑ K = 0 M ( M k ) N k ( j + 1 ) M − k ( − 1 ) M − K = ∑ k = 0 M ( − 1 ) M − k ( M k ) N k ( ∑ j = 0 g ( − 1 ) j ( g j ) ( j + 1 ) M − K ) = ∑ k = 0 M ( − 1 ) M − k − g ( M k ) N k ( ∑ j = 0 g ( − 1 ) g + j ( g j ) ( j + 1 ) M − K ) = ∑ k = 0 M ( − 1 ) M − k − g ( M k ) N K S 2 ( M − K + 1 , g + 1 ) g !

q.e.d.

9.5) ① A q M = ∑ k = 0 M ( q − 1 ) M − k S 2 ( M + 1 , k + 1 ) k !

② ∑ n = 0 N − 1 q n n M = q N ( q − 1 ) M + 1 ∑ g = 0 M ( − 1 ) M − g A q M − g ( M g ) ( q − 1 ) g N g + ( − 1 ) M + 1 A q M ( q − 1 ) M + 1

[Proof]

f ( N ) = ∑ n = 0 N − 1 q n n M = q N ( q − 1 ) M + 1 { … } + ( − 1 ) M + 1 A q M ( q − 1 ) M + 1 = q N ( q − 1 ) M + 1 { ∑ g = 0 M ( q − 1 ) M − g ( − 1 ) g ∇ g ( N − 1 ) M } + ( − 1 ) M + 1 A q M ( q − 1 ) M + 1

{ … } → 9.4 ) = ∑ g = 0 M ( q − 1 ) M − g ( − 1 ) g ∑ k = 0 M − g ( − 1 ) M − g − k ( M k ) N K S 2 ( M − K + 1 , g + 1 ) g ! = ∑ g = 0 M ( q − 1 ) M − g ∑ k = 0 M − g ( − 1 ) M − k ( M k ) N K S 2 ( M − K + 1 , g + 1 ) g ! = ( q − 1 ) M { ( − 1 ) M ( M 0 ) N 0 S 2 ( M + 1 , 1 ) 0 !   + ( − 1 ) M − 1 ( M 1 ) N 1 S 2 ( M , 1 ) 0 !   + ⋯ }     + ( q − 1 ) M − 1 { ( − 1 ) M ( M 0 ) N 0 S 2 ( M + 1 , 2 ) 1 !   + ( − 1 ) M − 1 ( M 1 ) N 1 S 2 ( M , 2 ) 1 !   + ⋯ }     + ⋯ + ( q − 1 ) 0 { ( − 1 ) M ( M 0 ) N 0 S 2 ( M + 1 , M + 1 ) M ! }

Arrange by N^g

= ∑ g = 0 M ( q − 1 ) g N g ( M g ) ( − 1 ) M − g ∑ k = 0 M − g ( q − 1 ) M − g − k S 2 ( M − g + 1 , k + 1 ) k !

take g = 0 , { … } = ( − 1 ) M ∑ k = 0 M ( q − 1 ) M − k S 2 ( M + 1 , k + 1 ) k !

f ( 0 ) = 0 → ①

① → ∑ k = 0 M − g ( q − 1 ) M − g − k S 2 ( M − g + 1 , k + 1 ) k ! = A q M − g → ②

q.e.d.

Example 9.1

∑ n = 0 N − 1 q n i = q N ( ( q − 1 ) N − q ) + q ( q − 1 ) 2

∑ i = 0 N − 1 2 i = 2 N − 1

∑ i = 0 N − 1 2 i i = 2 N ( N − 2 ) + 2

∑ i = 0 N − 1 2 i i 2 = 2 N ( N 2 − 4 N + 6 ) − 6

∑ i = 0 N − 1 2 i i 3 = 2 N ( N 3 − 6 N 2 + 18 N − 26 ) + 26

∑ i = 0 N − 1 2 i i 4 = 2 N ( N 4 − 8 N 3 + 36 N 2 − 104 N + 150 ) − 150

∑ i = 0 N − 1 2 i i 5 = 2 N ( N 5 − 10 N 4 + 60 N 3 − 260 N 2 + 750 N − 1082 ) + 1082

∑ i = 0 N − 1 3 i = 3 N − 1 2

∑ i = 0 N − 1 3 i i = 3 N ( 2 N − 3 ) + 3 4

∑ i = 0 N − 1 3 i i 2 = 3 N ( 4 N 2 − 2 × 3 × 2 N + 12 ) − 12 8 ,   2 × 3 × 2 = ( 3 − 1 ) A 3 1 ( 2 1 )

∑ i = 0 N − 1 3 i i 3 = 3 N ( 8 N 3 − 4 × 3 × 3 N 2 + 2 × 12 × 3 N − 66 ) + 66 16 , 4 × 3 × 3 = ( 3 − 1 ) 2 A 3 1 ( 3 2 )

∑ i = 0 N − 1 3 i i 4 = 3 N ( 16 N 4 − 8 × 3 × 4 N 3 + 4 × 12 × 6 N 2 − 2 × 66 × 4 N + 480 ) − 480 32

∑ i = 0 N − 1 4 i = 4 N − 1 3

∑ i = 0 N − 1 4 i i = 4 N ( 3 N − 4 ) + 4 9

∑ i = 0 N − 1 4 i i 2 = 4 N ( 9 N 2 − 3 × 4 × 2 N + 20 ) − 20 27

∑ i = 0 N − 1 4 i i 3 = 4 N ( 27 N 3 − 9 × 4 × 3 N 2 + 3 × 20 × 3 N − 132 ) + 132 81

It is hard to imagine that the four expressions of A q M are equal, but they can stand verification.

A q M = ∑ k = 0 M ( 1 − q ) M − k q K S 2 ( M , k ) k ! = ∑ k = 0 M ( 1 − q ) M − k q K + 1 S 2 ( M + 1 , k + 1 ) k ! , M > 0 = ∑ k = 0 M ( q − 1 ) M − k S 2 ( M + 1 , k + 1 ) k ! = q ∑ k = 0 M ( q − 1 ) M − k S 2 ( M , k ) k ! , M > 0

9.6) M > 0 , A 2 M = 2 ∑ k = 0 M S 2 ( M , k ) k ! = ∑ k = 0 M S 2 ( M + 1 , k + 1 ) k !

= ∑ k = 0 M ( − 1 ) M − k S 2 ( M , k ) k ! 2 K = ∑ k = 0 M ( − 1 ) M − k S 2 ( M + 1 , k + 1 ) k ! 2 K + 1

→ 2.3 ) P S = [ 1 , 1 , ⋯ ] , P T = [ 1 , 2 , ⋯ , M ] ∑ q = 0 M H 1 ( q ) = ∑ q = 0 M H 2 ( q ) 2 q →

∑ k = 0 M k ! S 2 ( M + 1 , k + 1 ) = ∑ k = 0 M ( − 1 ) M − k k ! S 2 ( M , k ) 2 K → match 9.6).

The author declares no conflicts of interest.

Peng, J. (2021) Further Study of the Shape of the Numbers and More Calculation Formulas. Open Access Library Journal, 8: e7969. https://doi.org/10.4236/oalib.1107969