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Formal Calculation of Q-Binomial

Peng Ji  *

Submitted:

07 October 2025

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08 October 2025

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Abstract
This article offers formulas for computing various q-binomial nested sums, give three forms of results, reveals the three forms of q-binomial and their interrelationships. It is a powerful tool for q-analysis, which can prove and generalize many classic conclusions in a simple way. This article also utilized it to obtain a large number of new results, including formulas for q-Eulerian numbers and polynomials. By taking the limit of q to 1, it can calculate general nested sums and analyze binomial coefficients.
Keywords: 
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1. Calculation Formula

q-binomial: M , N , K N , [ M N ] q = ( q N 1 ) ( q N 1 1 ) . . . ( q N M + 1 1 ) ( q M 1 ) ( q M 1 1 ) . . . ( q 1 1 ) , q 0 , 1 , abbreviated as G M N , [ N ] q = G 1 N .
( a ; q ) n = [ 1 , n = 0 ( 1 a ) ( 1 a q ) . . . ( 1 a q n 1 ) , n > 0 . G g 1 , g 2 . . . g p M = ( q ; q ) M i = 1 p ( q ; q ) g i , g i = M .
δ i j is Kronecker delta, i = j , δ i j = 1 ; i j , δ i j = 0 . The following relationship holds:
G 0 N = 1 , G M N = G N M N ; G M N = 0 , M > N , M < 0 .
G M N = q M G M N 1 + G M 1 N 1 = G M N 1 + q N M G M 1 N 1 .
n = 0 N 1 q n G M n + K = q M K G M + 1 N + K , M K 0 .
G K M = w Ω ( 0 M K , 1 K ) q i n v ( w ) , inv ( . ) denotes the inversion statistic . [1]
Lemma 1.
n = 0 N 1 q n [ n ] q G M n + K , M K 0 = q 2 ( M K ) + 1 G 1 M + 1 G M + 2 N + K + q M K G 1 M K G M + 1 N + K ( 1 ) = q M 2 K 1 G 1 M + 1 G M + 2 N + K + 1 + q M K ( G 1 M K q K 1 G 1 M + 1 ) G M + 1 N + K ( 2 ) = ( q 2 ( M K ) + 1 G 1 M + 1 q 2 M K + 2 G 1 M K ) G M + 2 N + K + q M K G 1 M K G M + 2 N + K + 1 ( 3 ) .
Proof. 
q n [ n ] q G M n + K = q n q n q n M + K q 1 G M n + K + q n q n M + K 1 q 1 G M n + K = q n ( q n M + K 1 ) q M K 1 q 1 G M n + K + q n q M K 1 q 1 G M n + K + q n q n M + K 1 q 1 G M n + K = q n ( q M + 1 1 ) G 1 M K G M + 1 n + K + q n G 1 M K G M n + K + q n G 1 M + 1 G M + 1 n + K = ( q M K 1 ) G 1 M + 1 G M + 2 N + K q M + 1 K + G 1 M K G M + 1 N + K q M K + G 1 M + 1 G M + 2 N + K q M + 1 K = q 2 ( M K ) + 1 G 1 M + 1 G M + 2 N + K + q M K G 1 M K G M + 1 N + K ( 1 ) = q 2 ( M K ) + 1 G 1 M + 1 q M 2 ( G M + 2 N + K + 1 G M + 1 N + K ) + q M K G 1 M K G M + 1 N + K = q M 2 K 1 G 1 M + 1 G M + 2 N + K + 1 + q M K ( G 1 M K q K 1 G 1 M + 1 ) G M + 1 N + K ( 2 ) = q 2 ( M K ) + 1 G 1 M + 1 G M + 2 N + K + q M K G 1 M K ( G M + 2 N + K + 1 q M + 2 G M + 2 N + K ) = ( q 2 ( M K ) + 1 G 1 M + 1 q 2 M K + 2 G 1 M K ) G M + 2 N + K + q M K G 1 M K G M + 2 N + K + 1 ( 3 ) .
Definition 1.
Recursively define q p , p Z ; S U M q ( N ) = S U M q ( N , P S , P T ) , K i , D i C , T i N .
q 0 f ( n ) = f ( n ) , n = 0 N 1 q n q 1 f ( n + 1 ) = f ( N ) , n = 0 N 1 q n f ( n + 1 ) = q 1 f ( N ) , q 1 = q . S U M q ( N , [ K 1 : D 1 ] , [ T 1 = 1 ] ) = n = 0 N 1 q n ( K 1 + [ n ] q D 1 ) . S U M q ( N , [ K 1 : D 1 , K 2 : D 2 ] , [ T 1 , T 2 = T 1 + 2 p ] ) = n = 0 N 1 q n ( K 2 + [ n ] q D 2 ) p S U M q ( n + 1 , [ K 1 : D 1 ] , [ T 1 ] ) .
Abbreviations : [ K 1 : D , K 2 : D . . . K M : D ] = [ K 1 , K 2 . . . K M ] : D , [ K 1 , K 2 . . . K M ] : 1 = [ K 1 , K 2 . . . K M ] . In this paper , the default P S = [ K 1 : D 1 , K 2 : D 2 . . . K M : D M ] , P T = [ T 1 , T 2 . . . T M ] , T i < T i + 1 . S U M q ( N , P S , [ 1 , 2 . . . M ] ) = n = 0 N 1 i = 1 M q n ( K i + [ n ] q D i ) . S U M q ( N , P S , [ 1 , 3 . . . 2 M 1 ] ) = n M = 0 N 1 q n M ( K M + [ n M ] q D M ) . . . n 2 = 0 n 3 q n 2 ( K 2 + [ n 2 ] q D 2 ) n 1 = 0 n 2 q n 1 ( K 1 + [ n 1 ] q D 1 ) . S U M q ( N , P S , [ 1 , 2 , 4 ] ) = n 3 = 0 N 1 q n 3 ( K 3 + [ n 3 ] q D 3 ) n = 0 n 3 q n ( K 1 + [ n ] q D 1 ) ( K 2 + [ n ] q D 2 ) . S U M q ( N , P S , [ 1 , 3 , 4 ] ) = n 3 = 0 N 1 q n 3 ( K 3 + [ n 3 ] q D 3 ) ( K 2 + [ n 3 ] q D 2 ) n = 0 n 3 q n ( K 1 + [ n ] q D 1 ) . S U M q ( N , P S , [ 1 , 4 ] ) = n 3 = 0 N 1 q n 3 ( K 2 + [ n 3 ] q D 2 ) n 2 = 0 n 3 q n 2 n 1 = 0 n 2 q n 1 ( K 1 + [ n 1 ] q D 1 ) . Use K , T to represent the set { K i } , { T i } . ( K 1 + T 1 ) ( K 2 + T 2 ) . . . ( K M + T M ) = i = 1 M X i , X i = T i o r K i
Definition 2.
X ( T ) =Number of { X 1 , X 2 . . . X M } T .
X T 1 =Number of { X 1 , X 2 . . . X i 1 } T , X K 1 =Number of { X 1 , X 2 . . . X i 1 } K .
X T =Number of { X 1 , X 2 . . . X i } T , X K =Number of { X 1 , X 2 . . . X i } K .
X T 1 + X K 1 = i 1 , X T + X K = i . Use the auxiliary form and each X i cannot be swapped:
Theorem 1.
H = T M M , S U M q ( N , P S , P T ) =
F o r m 1 g = 0 M H 1 q ( g ) G H + 1 + g N + H , B i = [ q ( T i T i 1 1 ) X T 1 ( K i + G 1 X T 1 D i ) , X i = K i q 1 + ( T i T i 1 ) X T 1 G 1 T i X K 1 D i , X i = T i , F o r m 2 g = 0 M H 2 q ( g ) G H + 1 + g N + H + g , B i = [ K i q ( T i X K 1 ) G 1 T i X K 1 D i , X i = K i q ( T i X K 1 ) G 1 T i X K 1 D i , X i = T i , F o r m 3 g = 0 M H 3 q ( g ) G T M + 1 N + T M g , B i = [ q ( T i T i 1 1 ) X T 1 ( K i + G 1 X T 1 D i ) , X i = K i q 1 + ( T i T i 1 1 ) X T 1 { ( q X T 1 G 1 T i q T i G 1 X T 1 ) D i K i q T i } , X i = T i . H i q ( g ) = H i q ( g , P S , P T ) = H i q ( g , M ) , is defined as X ( T ) = g i = 1 M B i .
Proof. 
S U M q ( 1 , [ K 1 : D 1 ] , [ 1 ] ) = n = 0 N 1 q n ( K 1 + [ n ] q D 1 ) = n = 0 N 1 q n G 1 n D 1 + n = 0 N 1 q n K 1 = q 1 D 1 G 2 N + K 1 G 1 N = q 1 D 1 G 2 N + 1 + ( K 1 q 1 D 1 ) G 1 N = ( q 1 D 1 K 1 q 2 ) G 2 N + K 1 G 2 N + 1 . It s holds when M = 1 , Assume that M holds . P S 1 = [ P S , K M + 1 : D M + 1 ] , P T 1 = [ P T , T M + 1 = T M + 2 p ] , X = T M M + 1 p = T M + 1 ( M + 1 ) . f ( n + 1 ) = A g G H + 1 + g n + H + 1 q p f ( n + 1 ) = A g G H + 1 + g p n + H + 1 p q p g . S U M q ( N , P S 1 , P T 1 ) = n = 0 N 1 q n ( K M + 1 + [ n ] q D M + 1 ) g = 0 M H 1 q ( g ) G X + g n + X q p g = g = 0 M ( K M + 1 q g + q g G 1 g D M + 1 ) H 1 q ( g ) G X + 1 + g N + X q p g + g = 0 M q 2 g + 1 G 1 X + g + 1 D M + 1 H 1 q ( g ) G X + 2 + g N + X q p g = g = 0 M ( K M + 1 + G 1 g D M + 1 ) H 1 q ( g ) G T M + 1 ( M + 1 ) + 1 + g N + T M + 1 ( M + 1 ) q ( 1 p ) g + g = 0 M q ( 2 p ) g + 1 G 1 T M + 1 ( M g ) D M + 1 H 1 q ( g ) G X + 2 + g N + X = g = 0 M + 1 H 1 q ( g , P S 1 , P T 1 ) G T M + 1 ( M + 1 ) + 1 + g N + T M + 1 ( M + 1 ) F o r m 1 . f ( n + 1 ) = A g G H + 1 + g n + H + 1 + g q p f ( n + 1 ) = A g G H + 1 + g p n + H + 1 + g p . S U M q ( N , P S 1 , P T 1 ) = n = 0 N 1 q n ( K M + 1 + [ n ] q D M + 1 ) q p S U M q ( n + 1 ) = n = 0 N 1 q n ( K M + 1 + [ n ] q D M + 1 ) g = 0 M H 2 q ( g ) G X + g n + X + g = g = 0 M ( K M + 1 q X g 1 G 1 X + g + 1 D M + 1 ) H 2 q ( g ) G X + 1 + g N + X + g + g = 0 M q X g 1 G 1 X + g + 1 D M + 1 H 2 q ( g ) G X + 2 + g N + X + g + 1 = g = 0 M ( K M + 1 q ( T M + 1 ( M g ) ) G 1 T M + 1 ( M g ) D M + 1 ) H 2 q ( g ) G T M + 1 ( M + 1 ) + 1 + g N + T M + 1 ( M + 1 ) + g + g = 0 M q ( T M + 1 ( M g ) ) G 1 T M + 1 ( M g ) D M + 1 H 2 q ( g ) G T M + 1 ( M + 1 ) + 1 + ( g + 1 ) N + T M + 1 ( M + 1 ) + ( g + 1 ) = g = 0 M + 1 H 2 q ( g , P S 1 , P T 1 ) G T M + 1 ( M + 1 ) + 1 + g N + T M + 1 ( M + 1 ) + g F o r m 2 . f ( n + 1 ) = A g G T M + 1 n + 1 + T M g q p f ( n + 1 ) = A g G T M + 1 p n + 1 + T M g p q p g . S U M q ( N , P S 1 , P T 1 ) = n = 0 N 1 q n ( K M + 1 + [ n ] q D M + 1 ) g = 0 M H 3 q ( g ) G T M + 1 p n + 1 + T M p g q p g = g = 0 M K M + 1 q g H 3 q ( g ) G T M + 2 p N + 1 + T M p g q p g + g = 0 M q g G 1 g D M + 1 H 3 q ( g ) G T M + 3 p N + T M p + 2 g q p g + g = 0 M ( q 2 g + 1 G 1 T M + 2 p q T M + 3 p + g G 1 g ) D M + 1 H 3 q ( g ) G T M + 3 p N + T M p + 1 g q p g = g = 0 M ( K M + 1 q g + q g G 1 g D M + 1 ) H 3 q ( g ) G T M + 3 p N + T M p + 2 g q p g + g = 0 M ( ( q 2 g + 1 G 1 T M + 2 p q T M + 3 p + g G 1 g ) D M + 1 q T M + 3 p K M + 1 q g ) H 3 q ( g ) G T M + 3 p N + T M p + 1 g q p g = g = 0 M ( K M + 1 + G 1 g D M + 1 ) H 3 q ( g ) G T M + 1 + 1 N + T M + 1 g q ( 1 p ) g + g = 0 M ( ( q g G 1 T M + 1 q T M + 1 G 1 g ) D M + 1 q T M + 1 K M + 1 ) H 3 q ( g ) G T M + 1 + 1 N + T M + 1 ( g + 1 ) q 1 + ( 1 p ) g = g = 0 M + 1 H 3 q ( g , P S 1 , P T 1 ) G T M + 1 + 1 N + T M + 1 g F o r m 3 .
n = 0 N 1 q n i = 1 M ( K i + D i q n ) = S U M ( N , [ K i + D i : D i ( q 1 ) ] , [ 1 , 2 . . . M ] ) = g = 0 M H 1 q ( g ) G 1 + g N .
B i = [ K i + D i + G 1 X T 1 D i ( q 1 ) = K i + D i + G 1 X T D i ( q 1 ) = K i + q X T D i , X i = K i q 1 + X T 1 G 1 i X K 1 D i ( q 1 ) = q X T G 1 1 + X T 1 D i ( q 1 ) = q X T ( q X T 1 ) D i , X i = T i . Following a similar form, induction proves:
Theorem 2.
n = 0 N 1 i = 1 M ( K i + D i q n ) = g = 1 M ( B i ) G g N + N K i , B i = [ K i , X T 1 = 0 ; K i + q X T 1 1 D i , X T 1 > 0 , X i = K i D i , X T 1 = 0 ; q X T 1 ( q X T 1 1 ) D i , X T 1 > 0 , X i = T i .
( K + D ) M = D g = 0 M 1 K g ( K + D ) M 1 g + K M . ( K + q D ) M = n = 0 1 ( K + q n D ) n = 0 0 ( K + q n D ) = = q ( q 1 ) D 2 a + b + c = M 2 , a , b , c 0 K a ( K + D ) b ( K + q D ) c + q D g = 0 M 1 K g ( K + D ) M 1 g + K M . P S = [ K + D , K + D . . . K + D ] : ( q 1 ) D , P T = [ 1 , 2 . . . M ] . B i = [ K + D q X T , X i = K i q X T ( q X T 1 ) D , X i = T i , H 1 q ( 0 ) = ( K + D ) M , H 1 q ( 1 ) = q 1 ( q 1 1 ) D a + b = M 1 , a , b 0 ( K + D ) a ( K + q D ) b . S U M q 2 S U M q 1 = H 1 q ( 1 ) G 2 2 + H 1 q ( 0 ) G 1 2 H 1 q ( 0 ) = H 1 q ( 1 ) + q H 1 q ( 0 ) = q ( K + q D ) M ( K + q D ) M = ( q 1 ) D g = 0 M 1 ( K + D ) g ( K + q D ) M 1 g + ( K + D ) M .
Definition 3.
lim q 1 H q ( g ) = H ( g ) , lim q 1 S U M q ( N ) = S U M ( N ) .
lim q 1 [ n ] q = n , lim q 1 G M N = [ M N , which yields the nested summation formula for K i + n D i .

2. Property

Definition 4.
[ n ] q = q n G 1 n , [ n ! ] q = [ n ] q . . . [ 2 ] q [ 1 ] q , [ 0 ! ] q = 1 , [ n ] q + = q n G 1 n , similarly defining [ n ! ] q + .
Theorem 3.
(1). q S U M q ( n + 1 , P S , [ 1 , 2 . . . M ] ) = i = 1 M ( K i + [ n ] q D i ) .
(2). At S U M q ( N , [ . . . P S . . . ] , [ . . . T , T + 1 . . . T + M 1 . . . ] ) , K i : D i can swap orders.
(3). q p S U M q ( N ) = g H 1 q ( g ) G X + 1 + g N + X q g p = g H 2 q ( g ) G X + 1 + g N + X + g = g H 3 q ( g ) G X + M + 1 N + X + M g q g p , X = T M M p .
(4). S U M q ( N , [ [ L 1 ] q , [ L 2 ] q . . . [ L Q ] q , P S 1 ] , [ L 1 , L 2 . . . L Q , P T 1 ] ) = [ L i ] q S U M q ( N , P S 1 , P T 1 ) . T 1 can great than 1.
(5). S U M q ( N , [ [ T 1 ] q , [ T 2 ] q . . . [ T M ] q ] , [ T 1 , T 2 . . . T M ] ) = [ T i ] q G T M + 1 N + T M .
(6) At H q ( g ) , X i K q X T = G M g M = G g M .
Proof. 
Definition of S U M q ( 1 ) ( 2 ) , n = 0 N 1 q n G M n + K = q M K G M + 1 N + K ( 3 ) , which has been used for the proof of [1].
At (4), H 2 q ( g < Q ) = 0 , H 2 q ( g Q ) = [ L i ] q H 2 q ( g Q , P S 1 , P T 1 ) ( 4 ) ( 5 ) . (6) is G K M = w Ω ( 0 M K , 1 K ) q i n v ( w ) . □
[ x ] q + [ n ] q = q x [ n + x ] q , P T = [ 2 i 1 ] , ( 5 ) n M = 0 N 1 . . . n 1 = 0 n 2 q n i [ n i + 2 i 1 ] q = [ 1 ] q [ 3 ] q . . . [ 2 M 1 ] q G 2 M N + 2 M 1 .
Theorem 4.
P T = [ 1 , 2 . . . M ] ,
(1). H 1 q ( g ) = q g ( g + 1 ) k H 2 q ( k ) G g k .
(2). H 1 q ( g ) = k H 3 q ( k ) G M g M k q ( g + 1 ) ( g k ) .
(3). H 2 q ( g ) = k ( 1 ) k + g G g k q g ( g + 1 ) k ( k + 3 ) 2 k g H 1 q ( k ) .
(4). H 3 q ( g ) = k ( 1 ) k + g G M g M k q g ( g + 3 ) k ( k + 3 ) 2 H 1 q ( k ) .
Proof. 
When M = 1 , verify directly . Assume that M holds . D i 0 , changing K i : D i to K i / D i : 1 does not affect the result . P S 1 = [ P S , K M + 1 ] , P T 1 = [ P T , T M + 1 ] . H 1 q ( g , M + 1 ) = H 1 q ( g 1 ) q 1 + X T G 1 1 + X T + H 1 q ( g ) ( K M + 1 + G 1 X T ) = q 1 + X T G 1 1 + X T q g ( g 1 ) k = 0 M H 2 q ( k ) G g 1 k + ( K M + 1 + G 1 X T ) q g ( g + 1 ) k = 0 M H 2 q ( k ) G g k = q g G 1 g q g ( g 1 ) k = 0 M H 2 q ( k ) ( G g k + 1 q g G g k ) + ( K M + 1 + G 1 g ) q g ( g + 1 ) k = 0 M H 2 q ( k ) G g k = q g ( g + 1 ) k = 0 M H 2 q ( k ) q g G 1 g G g k + 1 + K M + 1 q g ( g + 1 ) k = 0 M H 2 q ( k ) G g k . ( 1 * ) q g ( g + 1 ) k = 0 M + 1 H 2 q ( k , M + 1 ) G g k = q g ( g + 1 ) k = 0 M + 1 ( H 2 q ( k 1 ) q k G 1 k + H 2 q ( k ) ( K M + 1 q ( k + 1 ) G 1 k + 1 ) ) G g k = q g ( g + 1 ) k = 1 M + 1 ( H 2 q ( k 1 ) q k G 1 k G g k + q g ( g + 1 ) k = 0 M H 2 q ( k ) ( K M + 1 q ( k + 1 ) G 1 k + 1 ) ) G g k = q g ( g + 1 ) k = 0 M ( H 2 q ( k ) q ( k + 1 ) G 1 k + 1 G g k + 1 + q g ( g + 1 ) k = 0 M H 2 q ( k ) ( K M + 1 q ( k + 1 ) G 1 k + 1 ) ) G g k . ( 1 * * ) ( 1 * ) ( 1 * * ) = q g ( g + 1 ) k = 0 M H 2 q ( k ) { q g G 1 g G g k + 1 q ( k + 1 ) G 1 k + 1 G g k + 1 + q ( k + 1 ) G 1 k + 1 G g k } . { . . . } = 0 ( 1 ) . H 1 q ( g , M + 1 ) = q g G 1 g k = 0 M H 3 q ( k ) G M g + 1 M k q g ( g 1 k ) + ( K M + 1 + G 1 g ) k = 0 M H 3 q ( k ) G M g M k q ( g + 1 ) ( g k ) . ( 2 * ) k = 0 M + 1 H 3 q ( k , M + 1 ) G M + 1 g M + 1 k q ( g + 1 ) ( g k ) ( 2 * * ) = k = 0 M q ( q k G 1 M + 1 q M + 1 G 1 k K M + 1 q M + 1 ) H 3 q ( k ) G M g + 1 M k q ( g + 1 ) ( g k 1 ) + k = 0 M ( K M + 1 + G 1 k ) H 3 q ( k ) G M + 1 g M + 1 k q ( g + 1 ) ( g k ) . Items containing K M + 1 : K M + 1 G M + 1 g M + 1 k q ( g + 1 ) ( g k ) K M + 1 q M + 2 G M + 1 g M k q ( g + 1 ) ( g k 1 ) = K M + 1 G M g M k q ( g + 1 ) ( g k ) . Items does not contain K M + 1 in ( 2 * ) = q g G 1 g G M + 1 g M k q g ( g 1 k ) + G 1 g G M g M k q ( g + 1 ) ( g k ) . Divide by q g q g ( g 1 k ) ( q 1 ) 1 = ( q g 1 ) G M + 1 g M k + ( q g 1 ) G M g M k q g k = ( q g 1 ) G M + 1 g M + 1 k . Items does not contain K M + 1 in ( 2 * * ) = q M + 2 q k + 1 q 1 G M g + 1 M k q ( g + 1 ) ( g k 1 ) + G 1 k G M + 1 g M + 1 k q ( g + 1 ) ( g k ) . Divide by q g q g ( g 1 k ) ( q 1 ) 1 = ( q M + 1 k 1 ) G M + 1 g M k + ( q g q g k ) G M + 1 g M + 1 k = ( q g 1 ) G M + 1 g M + 1 k . ( 2 * ) = ( 2 * * ) ( 2 ) . H 2 q ( g , M + 1 ) = H 2 q ( g 1 ) q g G 1 g + H 2 q ( g ) ( K M + 1 q ( g + 1 ) G 1 g + 1 ) ( 3 * ) = q g G 1 g k = 0 M ( 1 ) k + g 1 H 1 q ( k ) G g 1 k q g ( g 1 ) k ( k + 3 ) 2 k ( g 1 ) + ( K M + 1 q ( 1 + g ) G 1 g + 1 ) k = 0 M ( 1 ) k + g H 1 q ( k ) G g k q g ( g + 1 ) k ( k + 3 ) 2 k g . k = 0 M + 1 ( 1 ) k + g H 1 q ( k , M + 1 ) G g k q g ( g + 1 ) k ( k + 3 ) 2 k g ( 3 * * ) = k = 0 M + 1 ( 1 ) k + g ( q k G 1 k H 1 q ( k 1 ) + ( K M + 1 + G 1 k ) H 1 q ( k ) ) G g k q g ( g + 1 ) k ( k + 3 ) 2 k g = k = 0 M ( 1 ) k + g 1 q 1 + k G 1 1 + k H 1 q ( k ) G g 1 + k q g ( g + 1 ) ( k + 1 ) ( k + 4 ) 2 ( k + 1 ) g + k = 0 M ( 1 ) k + g ( K M + 1 + G 1 k ) H 1 q ( k ) G g k q g ( g + 1 ) k ( k + 3 ) 2 k g . Items containing K M + 1 in ( 3 * ) and ( 3 * * ) = ( 1 ) k + g K M + 1 G g k q g ( g + 1 ) k ( k + 3 ) 2 k g . Items does not contain K M + 1 in ( 3 * ) : = q g G 1 g ( 1 ) k + g 1 G g 1 k q g ( g 1 ) k ( k + 3 ) 2 k ( g 1 ) q ( 1 + g ) G 1 1 + g ( 1 ) k + g G g k q g ( g + 1 ) k ( k + 3 ) 2 k g . Divide by q ( 1 + g ) ( 1 ) k + g 1 q g ( g + 1 ) k ( k + 3 ) 2 k g ( q 1 ) 1 = ( q g 1 ) G g 1 k q k g + 1 + ( q 1 + g 1 ) G g k = q k + 1 G g 1 k q k g + 1 G g 1 k + q 1 + g G g k G g k . ( A * ) Items does not contain K M + 1 in ( 3 * * ) : = ( 1 ) k + g 1 q 1 + k G 1 1 + k G g 1 + k q g ( g + 1 ) ( k + 1 ) ( k + 4 ) 2 ( k + 1 ) g + ( 1 ) k + g G 1 k G g k q g ( g + 1 ) k ( k + 3 ) 2 k g .
Divide by q ( 1 + g ) ( 1 ) k + g 1 q g ( g + 1 ) k ( k + 3 ) 2 k g ( q 1 ) 1 = ( q 1 + k 1 ) ( q g G g k + G g 1 k ) q 1 + g ( q k 1 ) G g k = q 1 + k G g 1 k q g G g k G g 1 k + q 1 + g G g k . ( A * * ) ( A * ) ( A * * ) = q k g + 1 G g 1 k G g k ( q g G g k G g 1 k ) = G g k + 1 + G g k + 1 = 0 . ( 3 * ) = ( 3 * * ) ( 3 ) . H 3 q ( g , M + 1 ) = ( q g G 1 M + 1 q M + 2 G 1 g 1 K M + 1 q M + 2 ) H 3 q ( g 1 ) + ( K M + 1 + G 1 g ) H 3 q ( g ) = ( q g G 1 M + 1 q M + 2 G 1 g 1 K M + 1 q M + 2 ) k = 0 M ( 1 ) k + g 1 G M g + 1 M k q ( g 1 ) ( g + 2 ) k ( k + 3 ) 2 H 1 q ( k ) + ( K M + 1 + G 1 g ) k = 0 M ( 1 ) k + g G M g M k q g ( g + 3 ) k ( k + 3 ) 2 H 1 q ( k ) . ( 4 * ) k = 0 M + 1 ( 1 ) k + g H 1 q ( k , M + 1 ) G M + 1 g M + 1 k q g ( g + 3 ) k ( k + 3 ) 2 ( 4 * * ) = k = 0 M + 1 ( 1 ) k + g ( q k G 1 k H 1 q ( k 1 ) + ( K M + 1 + G 1 k ) H 1 q ( k ) ) G M + 1 g M + 1 k q g ( g + 3 ) k ( k + 3 ) 2 = k = 0 M ( 1 ) k + g 1 q 1 + k G 1 1 + k H 1 q ( k ) G M + 1 g M k q g ( g + 3 ) ( k + 1 ) ( k + 4 ) 2 + k = 0 M ( 1 ) k + g ( K M + 1 + G 1 k ) H 1 q ( k ) G M + 1 g M + 1 k q g ( g + 3 ) k ( k + 3 ) 2 . Items containing K M + 1 in ( 4 * ) : = K M + 1 q M + 2 ( 1 ) k + g 1 G M g + 1 M k q ( g 1 ) ( g + 2 ) k ( k + 3 ) 2 + ( 1 ) k + g K M + 1 G M g M k q g ( g + 3 ) k ( k + 3 ) 2 . Divide by K M + 1 ( 1 ) k + g q g ( g + 3 ) k ( k + 3 ) 2 = q M + 1 g G M g + 1 M k + G M g M k = G M g + 1 M k + 1 = that in ( 4 * * ) divided by ( . . . ) . Items does not contain K M + 1 in ( 4 * ) : = ( q g G 1 M + 1 q M + 2 G 1 g 1 ) ( 1 ) k + g 1 G M g + 1 M k q ( g 1 ) ( g + 2 ) k ( k + 3 ) 2 + G 1 g ( 1 ) k + g G M g M k q g ( g + 3 ) k ( k + 3 ) 2 . Divide by ( 1 ) k + g 1 q g ( g + 3 ) k ( k + 3 ) 2 ( q 1 ) 1 = [ ( q g G 1 M + 1 q M + 2 G 1 g 1 ) G M g + 1 M k q g 1 G 1 g G M g M k ] ( q 1 ) 1 = ( q M g 1 q 1 ) G M g + 1 M k ( q g 1 ) G M g M k . ( B * ) Items does not contain K M + 1 in ( 4 * * ) : = ( 1 ) k + g 1 q 1 + k G 1 1 + k G M + 1 g M k q g ( g + 3 ) ( k + 1 ) ( k + 4 ) 2 + ( 1 ) k + g G 1 k G M + 1 g M + 1 k q g ( g + 3 ) k ( k + 3 ) 2 . Divide by ( 1 ) k + g 1 q g ( g + 3 ) k ( k + 3 ) 2 ( q 1 ) 1 = ( q 1 + k G 1 1 + k G M + 1 g M k q k 2 G 1 k G M + 1 g M + 1 k ) ( q 1 ) 1 = ( q k q 1 ) G M + 1 g M k ( q k 1 ) G M + 1 g M + 1 k . ( B * * ) ( B * ) ( B * * ) = ( q M g 1 1 ) G M g + 1 M k ( q g k 1 ) G M g M k = 0 . ( 4 * ) = ( 4 * * ) ( 4 ) .
Definition 5.
H q ( g , T ) = H q ( g , T , P S , P T ) = H q ( g , T , M ) = X i T B i , H q ( g , K ) = X i K B i
Definition 6.
F 0 K = E 0 K = E 0 N , q = E 0 N , q = 1 ,
F g K = 1 λ 1 < . . . < λ g M i = 1 g K λ i , F g N = F g { 1 , 2 . . . N } . E g K = 1 λ 1 . . . λ g M i = 1 g K λ i , E g N = E g { 1 , 2 . . . N } . E g N , q = 1 λ 1 . . . λ g N i = 1 g [ λ i ] q = S 2 q ( N + g , N ) . E g N , q = 1 λ 1 . . . λ g N i = 1 g [ λ i ] q = S 2 q ( N + g , N ) .
Theorem 5.
P T = [ T , T + 1 . . . T + M 1 ] , D i = 1 , H 1 q ( g , K ) = F M g K E 0 g , q + F M g 1 K E 1 g , q + . . . + F 0 K E M g g , q .
Proof. 
P S 1 = [ P S , K M + 1 ] , P T 1 = [ P T , T M + 1 ] , L = { K 1 , K 2 . . . K M , K M + 1 } , H 1 q ( g , K , M + 1 ) = H 1 q ( g 1 , K ) + H 1 q ( g , K ) ( K M + 1 + G 1 g ) . F A L has three sources . = F A K E M ( g 1 ) A g 1 , q + F A 1 K E M g ( A 1 ) g , q K M + 1 + F A K E M g A g , q G 1 g = F A K ( E M + 1 g A g 1 , q + E M g A g , q G 1 g ) + F A 1 K E M + 1 g A g , q K M + 1 = F A K E M + 1 g A g , q + F A 1 K E M + 1 g A g , q K M + 1 = F A L E M + 1 g A g , q .
In this article, g = 0 M a g G 1 + p + g X = g = 0 M b g G 1 + p + g X + g = g = 0 M c g G 1 + p + M X + M g , 1 + p 0 , a g * = a g q p g , c g * = c g q p g .
D i = 1 , P T = [ 1 , 2 . . . M ] , q p S U M q ( X p ) = g = 0 M H 1 q ( g ) q p g G 1 + p + g X , H 1 q ( M ) = [ M ! ] q + .
We can choose K such H 1 q ( g < M ) p p g can take any value, g = 0 M a g G 1 + p + g X can be converted to a M q p M [ M ! ] q + q p S U M q ( X p ) .
then a g * = c × H 1 q ( g ) , b g = c × H 2 q ( g ) , c g * = c × H 3 q ( g ) . c is a constant. Similarly, for any PT, S U M q ( N ) can be converted into constant × S U M q ( N , P S 1 , [ T M M + 1 . . . T M 1 , T M ] ) . From [4], [3(3)]:
Theorem 6.
a g * = q g ( g + 1 ) k = g M b k G g k = k = 0 g G M g M k q ( g + 1 ) ( g k ) c k * .
b g = ( 1 ) g q g ( g + 1 ) 2 k = g M ( 1 ) k G g k q k ( k + 3 ) 2 k g a k * . c g * = ( 1 ) g q g ( g + 3 ) 2 k = 0 g ( 1 ) k G M g M k q k ( k + 3 ) 2 a k * .
If g = 0 M a g G 1 + p + g X can be converted into g = 0 M R ( . . . ) G 1 + p + R + g X + R , 0 < R M , then
k ( 1 ) k G g k q k ( k + 3 ) 2 k g a k * = 0 , 0 g < R , k ( 1 ) k G M g M k q k ( k + 3 ) 2 a k * = 0 , 0 M g < R .
The latter part refers to the necessary and sufficient conditions for merging, which correspond to b g = c M g * = 0 .
S U M q ( N , [ [ T ] q , P S ] , [ T , P T ] ) = [ T ] q S U M q ( N ) . By utilizing this, we can extend S U M q ( N ) . As long as the Y of G Y X is greater than 1 , T M M 1 , then T i T i + 1 is also allowed. For Example:
S U M q ( N , [ K ] , [ 2 ] ) = q 1 G 1 2 G 3 N + 1 + K G 2 N + 1 = q 4 G 1 2 G 3 N + ( q 1 G 1 2 + K q 2 ) G 2 N + . . . = q 7 G 1 2 G 3 N 1 + { q 4 G 1 2 + q 3 G 1 2 + K q 4 } G 2 N 1 + . . .
= S U M q ( N , [ [ 1 ] q , [ 1 ] q , K ] , [ 1 , 1 , 2 ] ) [ 1 ] q [ 1 ] q = S U M q ( N , [ [ 1 ] q , [ 2 ] q , K ] , [ 1 , 2 , 2 ] ) [ 1 ] q [ 2 ] q = S U M q ( N , [ [ 2 ] q , [ 1 ] q , K ] , [ 2 , 1 , 2 ] ) [ 2 ] q [ 1 ] q .
H 1 q ( 2 , [ . . . ] , [ 2 , 1 , 2 ] ) = q 1 G 1 2 q 1 + 1 G 1 1 [ K + G 1 2 ] + q 1 G 1 2 [ q 2 ( q 1 + 1 ) ] q ( 2 1 ) + 1 G 1 2 1 + [ q 2 G 1 2 ] q 1 G 1 0 q 2 G 1 1 = q 1 G 1 1 q 2 G 1 2 { . . . } .
H 1 q ( g , [ . . . ] , [ 2 , 1 , 2 ] ) = H 1 q ( g , [ . . . ] , [ 1 , 2 , 2 ] ) . This way we can expand F o r m 1 .

3. Application

Proposition 1.
(1). 0 λ 1 λ 2 . . . λ M N q λ i = G M N + M = g ( 1 ) g q M g + g ( g + 1 ) 2 G g M G 2 M N + 2 M g .
(2). A λ 1 < λ 2 < . . . < λ M B q λ i = q [ 2 M + A M G M B A + 1 , A , B Z , 1 λ 1 < . . . < λ g M q i = 1 M λ i = q [ 2 g + 1 G g M .
(3). i = 1 M ( a + q A + i z ) = g q [ 2 g + ( A + 1 ) g G g M z g a M g .
(4). g ( 1 ) g q g ( g + 1 ) 2 G g M G K N g = ( q M ; q 1 ) M K , N M K 0 .
(5). g ( 1 ) g q g ( g + 1 ) 2 G g M G M + 1 N g = i = 0 N M 1 q ( M + 1 ) i , N M + 1 , M > 0 .
(6). [ M N + M q 2 = g ( 1 ) g q g 2 [ g M q 2 G 2 M N + 2 M g .
Proof. 
P S = [ 1 , 1 . . . 1 ] : 0 , P T = [ 1 , 3 . . . 2 M 1 ] , H 1 q ( g > 0 ) = 0 , H 1 q ( 0 ) = 1 , S U M ( N + 1 ) = G M N + M pre - equation of ( 1 ) . A λ 1 < . . . < λ M B q λ i = q A + ( A + 1 ) + . . . + ( A + M 1 ) 0 λ 1 . . . λ M B A M + 1 q λ i ( 2 ) . At H 3 q ( g ) , B i = [ q X T = q i X K , X i = K i q X T ( q 2 i 1 ) , X i = T i , Extract q i form B i , after extraction , B i = [ q X K , X i = K i q X T ( q i 1 ) , X i = T i , H 3 q ( g ) = q [ 2 M + 1 [ 2 M g + 1 ( 1 ) g q [ 2 g + 1 g 1 λ 1 < . . . < λ g M q λ i = ( 1 ) g q [ 2 M + 1 [ 2 M g + 1 + [ 2 g + [ 2 g + 1 G g M . Using [ ( 4 ) ] is easier : H 3 q ( g ) = q g ( M 1 ) ( 1 ) g G M g M q g ( g + 3 ) 2 . F o r m 3 post - equation of ( 1 ) . Comparing the coefficients of z g on both sides of ( 3 ) , combined with ( 2 ) , proves that ( 3 ) . A = 1 , it s Rothe s q - Binomial Theorem i = 1 M ( a + q i 1 z ) = g q [ 2 g G g M z g a M g . ( 1 ) and q K G M K N K = g ( 1 ) g q ( M K ) g + g ( g + 1 ) 2 G g M G 2 M K N + M K g , N M K 0 . K = M , g ( 1 ) g q g ( g + 1 ) 2 G g M G M N g = 1 = g ( 1 ) g q g ( g + 1 ) 2 G g M ( q M G M N g 1 + G M 1 N g 1 ) = q M + g ( 1 ) g q g ( g + 1 ) 2 G g M G M 1 N g 1 g ( 1 ) g q g ( g + 1 ) 2 G g M G M 1 N g 1 = 1 q M . Continuing the same process ( 4 ) . Using the same recursive and inductive methods can obtain ( 5 ) . S U M q ( N + 1 , [ 1 , 1 . . . 1 ] : q 1 , [ 1 , 3 . . . 2 M 1 ] ) = 0 λ 1 . . . λ M N q 2 λ i = [ M N + M q 2 . A t H 1 q ( g ) , B i = [ q ( T i T i 1 1 ) X T 1 ( K i + G 1 X T 1 D i ) = q 2 X T 1 = q 2 X T , X i = K i q 1 + ( T i T i 1 ) X T 1 G 1 T i X K 1 D i = q 1 + 2 X T 1 ( q 2 i 1 X K 1 1 ) = q 2 X T 1 ( q i + X T 1 1 ) , X i = T i , Unable to derive a concise expression for H 1 q ( g ) , the same applies to H 2 q ( g ) . A t H 3 q ( g ) , B i = [ q X T 1 ( K i + G 1 X T 1 D i ) = q X T 1 ( 1 + G 1 X T 1 ( q 1 ) ) = q 2 X T 1 = q 2 X T , X i = K i q 1 + X T 1 { ( q X T 1 G 1 T i q T i G 1 X T 1 ) D i K i q T i } = q 1 + 2 X T 1 = q 1 + 2 X T , X i = T i . H 3 q ( g ) = ( 1 ) g q g + 2 ( i = 1 g i ) X i K q 2 X T = ( 1 ) g q g 2 [ g M q 2 .
(4) is unrelated to N; it is an effect of q K , just as the difference table of a polynomial series will have a row of constants.
This derivation : g ( 1 ) g q g ( g + 1 ) 2 p g G g M G K p N g p = 0 , N M > K 0 , 0 < p K . ( 5 ) q ( N M 1 ) q 1 ( g ( 1 ) g q g ( g + 1 ) 2 G g M G M + 1 N g ) = q ( M + 1 ) ( N M 1 ) g ( 1 ) g q g ( g 1 ) 2 G g M G M N g = q M ( N M ) , N M . A λ 1 < λ 2 < . . . < λ M B q λ i D = q [ 2 M D + A M D [ M B A + 1 q D . i = 1 A ( 1 + q C + D i z ) = g = 0 A q g C 1 λ 1 . . . λ g A q λ i D z g = g = 0 A q [ 2 g D + g D + g C [ g A q D z g . i = 1 A ( a + q ( 2 i 1 ) D z ) = g = 0 A q g 2 D [ g A q 2 D z g a A g , i = 1 B ( a 1 + q ( 2 i 1 ) D z 1 ) = g = 0 B q g 2 D [ g B q 2 D z g a ( B g ) . i = 1 A ( a + q ( 2 i 1 ) D z ) i = 1 B ( a 1 + q ( 2 i 1 ) D z 1 ) = k = B A z k f ( k ) . f ( k ) = i = 0 A + B [ i A q 2 D [ i k B q 2 D q i 2 D q ( i k ) 2 D a A i a ( B ( i k ) ) = q k 2 D a A B k i = 0 A + B [ i A q 2 D [ i k B q 2 D q 2 D ( i ( i k ) ) = q k 2 D a A B k [ A k A + B q 2 D . The last step used q - Vandermorde identity . a = D = 1 , it s MacMahon s q - binomial theorem  [2] p.74 | q | , | x | < 1 , 1 ( x ; q ) M + 1 = ( 1 + x + x 2 + . . . ) ( 1 + x q + x 2 q 2 + . . . ) ( 1 + x q 2 + + x 2 q 4 + . . . ) . . . ( 1 + x q M + + x 2 q 2 M + . . . ) . When we multiply this out , the coefficient of x 2 will be q a q b , 0 a b M , and so forth . ( 1 ) 1 ( x ; q ) M + 1 = k = 0 G k M + k x k , 1 ( x ; q ) = k = 0 x k ( q ; q ) k , it s Euler and Cauchy s identity  [2] pp121-123 ( 3 ) ( x , q ) M = g = 0 M q [ 2 g G g M x g ( x , q ) = k = 0 q [ 2 k x k ( q ; q ) k , it s Euler s identity  [2] p.129 C 2 , M > 1 , X = C ( M 1 ) M + 2 , A t H 3 q ( g , [ 1 , 1 . . . 1 ] : q 1 , [ 1 , C + 1 . . . C ( M 1 ) + 1 ] ) , B i = [ q C X T , X i = K i q C X T 1 + 1 , X i = T i , f i = [ 2 , i 1 ( mod C 1 ) 1 , i ¬ 1 ( mod C 1 ) , 0 λ 1 . . . λ X N 1 q f i λ i = g ( 1 ) g q g + C g ( g 1 ) 2 [ g M q C G C ( M 1 ) + 2 N + C ( M 1 ) + 1 g .
Proposition 2.
A Z ,
(1). G M + 1 N + M = g q ( g + 1 ) g G g M G 1 + g N ; G M + 1 + p N + M + p = g = 0 M q ( 1 + p + g ) g G g M G 1 + p + g N + p = g = 0 M + p q ( g + 1 ) g G g M + p G 1 + g N , 1 + p 0 .
(2). k ( 1 ) k + g G k M G g k q k ( k 1 ) 2 g k + A k z k = q g ( g + 1 ) 2 + A g G g M ( z q A ; q ) M g z g . Generalized Rothe’s q-Binomial Theorem.
(3). k ( 1 ) k G k M G M g M k q k ( k 1 ) 2 + A k z k = G g M ( z q A ; q ) g .
Proof. 
P S 1 = [ [ 1 ] q , [ 2 ] q . . . [ M ] q ] , P S = [ [ M ] q , [ M 1 ] q . . . [ 1 ] q ] , P T = [ 1 , 2 . . . M ] , S U M q ( N , P S 1 , P T ) = [ M ! ] q G M + 1 N + M = g H 1 q ( g ) G 1 + g N = S U M q ( N ) . I n H 1 q ( g ) , X i K , B i = [ M + 1 i ] q + G 1 X T 1 = q ( M i + 1 ) G 1 M + 1 i + X T 1 . H 1 q ( g , K ) = G 1 M G 1 M 1 . . . G 1 g + 1 q ( M + 1 ) ( M g ) 1 λ 1 < . . . < λ M g M q λ i = G 1 M . . . G 1 g + 1 q [ 2 M g + 1 ( M + 1 ) ( M g ) G M g M . H 1 q ( g ) [ M ! ] q = q i = 1 g i [ g ! ] q + H 1 q ( g , K ) [ M ! ] q = q [ 2 M g + 1 ( M + 1 ) ( M g ) + i = 1 M i + i = 1 g i G M g M = q g ( g + 1 ) G g M , this and [ ( 3 ) ] yields ( 1 ) . For ( 2 ) ( 3 ) , when g = 0 or M , verify directly [ ( 3 ) ] . Using induction to prove . Let f ( g ) = q g ( g + 1 ) 2 + A g , L ( k ) = q k ( k 1 ) 2 g k + A k . k ( 1 ) k + g ( G k M 1 + q M k G k 1 M 1 ) G g k q k ( k 1 ) 2 g k + A k z k = f ( g ) G g M 1 ( z q A ; q ) M 1 g z g + k ( 1 ) k + g G k 1 M 1 ( G g k 1 + q k g G g 1 k 1 ) L ( k ) q M k z k = f ( g ) G g M 1 ( z q A ; q ) M 1 g z g + k ( 1 ) k + 1 + g G k M 1 ( G g k + q k + 1 g G g 1 k ) L ( k + 1 ) q M ( k + 1 ) z k + 1 = f ( g ) G g M 1 ( z q A ; q ) M 1 g z g ( 1 z q A q M g 1 ) + q M + A 2 g z k ( 1 ) k + g 1 G k M 1 G g 1 k q k ( k 1 ) 2 ( g 1 ) k + A k z k = f ( g ) G g M 1 ( z q A ; q ) M g z g + q M + A 2 g z f ( g 1 ) G g 1 M 1 ( z q A ; q ) M g z g 1 = f ( g ) ( z q A ; q ) M g z g ( G g M 1 + q M g G g 1 M 1 ) ( 2 ) . k = 0 M ( 1 ) k ( q k G k M 1 + G k 1 M 1 ) G M g M k q k ( k 1 ) 2 + A k z k = k = 0 M ( 1 ) k G k M 1 G M g M k q k ( k 1 ) 2 + ( A + 1 ) k z k + k = 0 M ( 1 ) k G k 1 M 1 G M 1 ( g 1 ) M 1 ( k 1 ) q k ( k 1 ) 2 + A k z k . = k = 0 M ( 1 ) k G k M 1 ( G M 1 ( g 1 ) M 1 k + q g k G M 1 g M 1 k ) q k ( k 1 ) 2 + ( A + 1 ) k z k z k = 0 M ( 1 ) k G k M 1 G M 1 ( g 1 ) M 1 k q k ( k + 1 ) 2 + A ( k + 1 ) z k = G g 1 M 1 ( z q 1 + A ; q ) g 1 z q A G g 1 M 1 ( z q 1 + A ; q ) g 1 + q g G g M 1 ( z q A ; q ) g = G g 1 M 1 ( z q A ; q ) g + q g G g M 1 ( z q A ; q ) g ( 3 ) .
A = 0 , z = 1 ( 2 ) ( 3 ) = 0 . ( 1 ) and [ ] arrived at this conclusion , and ( 2 ) ( 3 ) was inspired by them . ( 2 ) g + A < M , k ( 1 ) k G g k G k M q [ 2 k ( g + A ) k = 0 ; g < M , k ( 1 ) k G g k G k M q [ 2 k ( M 1 ) k = 0 . ( 3 ) g > A , k ( 1 ) k G M g M k G k M q [ 2 k A k = 0 ; g > 0 , k ( 1 ) k G M g M k G k M q [ 2 k ( g 1 ) k = 0 . ( 2 ) k G k M G g k q k ( k 1 ) 2 + A k z k = q g ( g 1 ) 2 + A g G g M ( z q g + A ; q ) M g z g . A = 1 , z = 1 , ( 3 ) is a special case of [ ( 5 ) ] : N = M , K = M g .
Proposition 3.
(1). k ( 1 ) k + g G k M G g k = [ 0 , M + g is odd G g M ( q ; q 2 ) M g 2 , M + g is even ; k ( 1 ) k + g G k M G M g M k = [ 0 , g is odd G g M ( q ; q 2 ) g 2 , g is even .
(2). k ( 1 ) k + g G k M G g k q k = [ q g G g M ( q ; q 2 ) M g + 1 2 , M + g is odd q g G g M ( q ; q 2 ) M g 2 , M + g is even .
(3). k ( 1 ) k + g G k M G g k q k = [ ( 1 ) M g q M G g M ( q ; q 2 ) M g + 1 2 , M + g is odd ( 1 ) M g q M G g M ( q ; q 2 ) M g 2 , M + g is even .
Proof. 
M + g is odd , ( 1 ) k + g G k M G g k = ( 1 ) k + g ( q ; q ) M ( q ; q ) g ( q ; q ) k g ( q ; q ) M k = ( 1 ) g + ( M k + g ) G M k + g M G g M k + g S u m = 0 . M + g is even , using induction . Sum = k ( 1 ) k + g G k M 1 G g k + k ( 1 ) k + g G k 1 M 1 G g k q M k = 0 + k ( 1 ) k + g G k 1 M 1 G g k q M k = k ( 1 ) k + g G k 1 M 1 G g k 1 q M k + k ( 1 ) k + g G k 1 M 1 G g 1 k 1 q M g = k ( 1 ) k + g 1 G k 1 M 1 G g k 1 ( 1 q M k ) + k ( 1 ) k + g G k 1 M 1 G g 1 k 1 q M g = k ( 1 ) k + ( g + 1 ) G k M 1 G g + 1 k ( 1 q g + 1 ) + k ( 1 ) ( k 1 ) + ( g 1 ) G k 1 M 1 G g 1 k 1 q M g = G g + 1 M 1 ( q ; q 2 ) M 1 ( g + 1 ) 2 ( 1 q g + 1 ) + q M g G g 1 M 1 ( q ; q 2 ) M 1 ( g 1 ) 2 = G g M 1 ( q ; q 2 ) M g 2 2 ( 1 q M g 1 ) + q M g G g 1 M 1 ( q ; q 2 ) M g 2 = ( G g M 1 + q M g G g 1 M 1 ) ( q ; q 2 ) M g 2 . g = 0 , it s Gauss s q - Binomial theorems [2] p.61. Replace k , g by M k , M k to obtain the second equation of (1). k ( 1 ) k + g G k M G g k ( 1 q k ) = k ( 1 ) k + g G k 1 M 1 G g k ( 1 q M ) = k ( 1 ) k + g G k M 1 G g k + 1 ( 1 q M ) = q g ( 1 q M ) k ( 1 ) k + g G k M 1 G g k + ( 1 q M ) k ( 1 ) k + g 1 G k M 1 G g 1 k . M + g is odd = q g ( 1 q M ) G g M 1 ( q ; q 2 ) M 1 g 2 = q g ( 1 q M g ) G g M ( q ; q 2 ) M 1 g 2 = q g G g M ( q ; q 2 ) M g + 1 2 . M + g is even = ( 1 q M ) G g 1 M 1 ( q ; q 2 ) M g 2 = ( 1 q g ) G g M ( q ; q 2 ) M g 2 . ( 1 ) ( 2 ) . Similarly , ( 3 ) can be obtained .
Another Gauss’s identity: k [ k M q 2 q k = ( q ; q ) M [2] p.65. Inspired by the above form:
Proposition 4.
k [ k M q 2 [ g k q 2 q k = q g [ g M q 2 ( q ; q ) M g . k [ k M q 2 [ g k q 2 q k = q M [ g M q 2 ( q ; q ) M g .
Proof. 
k [ k M q 2 [ g k q 2 q k = k [ k M 1 q 2 [ g k q 2 q k + k [ k 1 M 1 q 2 [ g k q 2 q 2 M 2 k + k = q g [ g M 1 q 2 ( q ; q ) M 1 g + { k [ k 1 M 1 q 2 [ g k 1 q 2 q 2 M k } + q 2 M 2 k + ( 2 k 2 g ) k [ k 1 M 1 q 2 [ g 1 k 1 q 2 q k = q g [ g M 1 q 2 ( q ; q ) M 1 g + q 2 M 2 g + g [ g 1 M 1 q 2 ( q ; q ) M g + { k [ k M 1 q 2 [ g k q 2 q 2 M k 1 } . { . . . } = k [ M 1 k , g , k g M 1 q 2 q 2 M k 1 , k : = M 1 ( k g ) k [ M 1 k , g , k g M 1 q 2 q M g + k = q M [ g M 1 q 2 ( q ; q ) M 1 g . k [ k M q 2 [ g k q 2 q k = q g ( q ; q ) M g ( [ g M 1 q 2 + q 2 ( M g ) [ g 1 M 1 q 2 ) . Similarly , the latter equation can be obtained .
Proposition 5.
(1). H 1 q ( g , [ [ T 1 ] q , [ T 2 ] q . . . [ T M ] q ] , [ T 1 , T 2 . . . T M ] ) = i = 1 M [ T i ] q q ( g + 1 + p ) g G g M , p = T M M .
(2). H 1 q ( g , [ [ K ] q , [ K + 1 ] q . . . [ K + M 1 ] q ] , [ T , T + 1 . . . T + M 1 ] ) = q [ 2 g + 1 i = 1 g [ T 1 + i ] q i = 1 M g [ K + M i ] q G g M .
(3). K = { [ K ] q , [ K + 1 ] q . . . [ K + M 1 ] q } , F M g K E 0 g , q + F M g 1 K E 1 g , q + . . . + F 0 K E M g g , q = i = 1 M g [ K + M i ] q G g M .
(4). H = T M M , H 1 q ( g , [ [ T ] q , P S ] , [ T , P T ] ) = [ T ] q ( H 1 q ( g ) + q H + g H 1 q ( g 1 ) ) .
Proof. 
g H 1 q ( g ) G 1 + p + g N + p = [ T i ] q G T M + 1 N + T M = [ T i ] q G M + p + 1 N + M + p , [ 2 ( 1 ) ] yields ( 1 ) , [ 6 ] can also reach the conclusion . At ( 2 ) , K i can swap order , let P S = [ [ K + M 1 ] q . . . [ K + 1 ] q , [ K ] q ] , P T = P T 1 . B i o f H 1 q ( g ) = [ q ( K + M ) q i G 1 K + M i + X T , X i = K i q X T G 1 T + X T 1 , X i = T i . H 1 q ( g ) = q [ 2 g + 1 i = 1 g [ T 1 + i ] q H 1 q ( g , K ) . H 1 q ( g , K ) = i = 1 M g [ K + M i ] q q ( K + M ) ( M g ) q [ 2 M g + 1 G g M [ 1 ( 2 ) ] , this and [ 5 ] yields ( 2 ) ( 3 ) . S U M q ( N , [ [ T ] q , P S ] , [ T , P T ] ) = g = 0 M + 1 H 1 q ( g , [ [ T ] q , P S ] , [ T , P T ] ) G H + g N + H 1 = [ T ] q g = 0 M H 1 q ( g ) G 1 + H + g N + H = [ T ] q g = 1 M + 1 H 1 q ( g 1 ) q H + g G H + g N + H 1 + [ T ] q g = 0 M H 1 q ( g ) G H + g N + H 1 ( 4 ) .
F o r H 1 ( g ) , ( 1 ) = T i [ g M , ( 2 ) = i = 1 g ( T 1 + i ) i = 1 M g ( K + M i ) [ g M . ( 1 q ) A + M ( q ; q ) A ( q ; q ) M q S U M ( N , [ [ 1 ] q , [ 2 ] q . . . [ A ] q , [ B + 1 ] q , [ B + 2 ] q . . . [ B + M ] q ] , [ 1 , 2 . . . A , A + 1 , A + 2 . . . A + M ] ) = G A n + A G M n + M + B q ( . . . ) = ( 1 q ) A + M ( q ; q ) A ( q ; q ) M [ A ! ] q q S U M ( N , [ [ B + 1 ] q . . . [ B + M ] q ] , [ A + 1 . . . A + M ] ) . So we can obtain the formula for G A n + A G M n + M + B . S U M q ( N , [ [ 1 ] q , [ 2 ] q . . . [ M 1 ] q , 1 : q 1 , 1 : q 1 . . . 1 : q 1 ] , [ 1 , 2 . . . M 1 , M + 1 , M + 2 . . . M + K 1 ] ) = [ ( M 1 ) ! ] q S U M q ( N , [ 1 , 1 . . . 1 ] : q 1 , [ M + 1 , M + 2 . . . M + K 1 ] ) = [ ( M 1 ) ! ] q n = 0 N 1 q K n G ( M 1 ) + 1 ( n + 1 ) + ( M 1 ) . So we can calculate n = 0 N 1 q K n G M n + M , K > 0 . P T 1 = [ 1 , 2 . . . M ] , P T 2 = [ T + 1 , T + 2 . . . T + M ] , q S U M ( n + 1 , P S , P T 1 ) = i ( K i + [ n ] q D i ) , q S U M ( n + 1 , P S , P T 2 ) = i ( K i + [ n ] q D i ) G T n + T , At H 1 q ( g ) , B i of P T 1 = [ K i + G 1 X T D i , X i = K i q X T G 1 X T D i , X i = T i , B i of P T 2 = [ K i + G 1 X T D i , X i = K i q X T G 1 T + X T D i , X i = T i . H 1 q ( g , P S , P T 2 ) = H 1 q ( g , P S , P T 1 ) G 1 T + 1 . . . G 1 T + g G 1 1 . . . G 1 g = H 1 q ( g , P S , P T 1 ) G g T + g If g = 0 M a g G g n = i = 1 M ( K i + [ n ] q D i ) then g = 0 M a g G g T + g G T + g n + T = i = 1 M ( K i + [ n ] q D i ) G T n + T . ( 1 ) is difficult to derive from its definition . Promote it , D i f f below is equivalent to X T 1 , X K 1 .
Definition 7.
Set T come from p Source: S 1 , S 2 . . . S p .
D i f f ( S x , S x ) = 0 , D i f f ( S x , S y ) = D i f f ( S y , S x ) = 1 , x > y . D i f f ( T i , T j ) = D i f f ( S x , S y ) , T i S x , T j S y .
Proposition 6.
g 1 + . . . + g p = M , g i = | S i | i = 1 M [ G 1 T i + j < i D i f f ( T j , T i ) q j < i , D i f f ( T j , T i ) = 1 1 = G g 1 , g 2 . . . g p M i = 1 M G 1 T i , T i i .
Proof. 
Record the sum as W q ( g 1 , g 2 , . . . g p , P T ) . W ( 1 , 1 , [ T 1 , T 2 ] ) = G 1 T 1 G 1 T 2 + 1 + G 1 T 1 G 1 T 2 1 q = G 1 T 1 G 1 T 2 G 1 2 , it s holds . Assume that W q ( g 1 , g 2 , P T ) holds , W q ( g 1 , g 2 + 1 , [ P T , T M + 1 ] ) = T M + 1 S o u r c e 1 + T M + 1 S o u r c e 2 = W q ( g 1 , g 2 , P T ) G 1 T M + 1 + g 1 + W q ( g 1 1 , g 2 + 1 , P T ) G 1 T M + 1 ( g 2 + 1 ) q g 2 + 1 . = ( i G 1 T i ) G g 1 , g 2 M G 1 T M + 1 + g 1 + ( i G 1 T i ) G g 1 1 , g 2 + 1 M G 1 T M + 1 ( g 2 + 1 ) q g 2 + 1 . Just need to prove : G g 1 M G 1 T M + 1 + g 1 + G g 1 1 M G 1 T M + 1 ( M g 1 + 1 ) q M g 1 + 1 = G 1 T M + 1 G g 1 M + 1 . ( R i g h t s i d e ) × ( q M g 1 + 1 1 ) / G g 1 M = ( q T M + 1 1 + . . . + q + 1 ) ( q M + 1 1 ) = ( 1 ) . ( L e f t s i d e ) × ( q M g 1 + 1 1 ) / G g 1 M = ( q M g 1 + 1 1 ) G 1 T M + 1 + g 1 + ( q g 1 1 ) G 1 T M + 1 ( M g 1 + 1 ) q M g 1 + 1 = ( q M g 1 + 1 1 ) ( q T M + 1 + g 1 1 + . . . + q + 1 ) + ( q g 1 1 ) ( q T M + 1 1 + . . . + q M g 1 + 2 + q M g 1 + 1 ) = ( 2 ) . ( 1 ) ( 2 ) = 0 It s holds when p = 2. W q ( g 1 , g 2 + g 3 , P T ) = G g 1 , g 2 + g 3 g 1 + g 2 + g 3 i G 1 T i . Every product has g 2 + g 3 factors come from S o u r c e 2 , divide them to g 2 × S o u r c e 2 + g 3 × S o u r c e 3 , g 1 - factors are invariant , ( g 2 + g 3 ) - factors are variant . ( variant factors ) = W q ( g 2 , g 3 , [ X 1 , X 2 . . . X g 2 + g 3 ] ) = G g 2 , g 3 g 2 + g 3 i = 1 g 2 + g 3 G 1 X i . W q ( g 1 , g 2 , g 3 , [ T 1 , T 2 . . . T M ] ) = G g 1 , g 2 + g 3 g 1 + g 2 + g 3 G g 2 , g 3 g 2 + g 3 i G 1 T i = G g 1 , g 2 , g 3 g 1 + g 2 + g 3 i G 1 T i .
Definition 8.
[ M M q = 0 , [ g M q = λ 1 + . . . + λ g + 1 = M g i = 1 g + 1 [ i ] q λ i [ 1 + λ 1 ] q [ 1 + λ 1 + λ 2 ] q . . . [ 1 + λ 1 + . . . + λ g ] q , λ i 0 .
Easy to obtain: [ g M q = [ M g ] q [ g 1 M 1 q + [ g + 1 ] q [ g M 1 q , [ g M q = [ M g 1 M q .
Proposition 7.
[ N ] q M = g = 1 M [ g ! ] q + S 2 q ( M , g ) q g G g N = g = 1 M ( 1 ) M g [ g ! ] q S 2 q ( M , g ) q g G g N + g 1 = g = 0 M 1 [ g M q q [ 2 M g G M N + g .
Proof. 
q 1 S U M ( N , [ [ 1 ] q , [ 1 ] q . . . [ 1 ] q ] , [ 1 , 2 . . . M ] ) = ( 1 q + q N 1 1 q 1 ) = q M [ N ] q M = q 1 q 1 S U M ( N , [ 1 ] q , [ 1 ] q . . . [ 1 ] q ] , [ 2 , 3 . . . M ] ) = q 1 g = 0 M 1 H 1 q ( g ) G 1 + g N q g . = q 1 g = 0 M 1 H 2 q ( g ) G 1 + g N + g . I n H 1 q ( g ) , B i = [ q 1 + G 1 X T 1 = q 1 G 1 1 + X T 1 = q 1 G 1 1 + X T , X i = K i q 1 + X T 1 G 1 ( i + 1 ) X K 1 = q 1 q 1 + X T G 1 1 + X T , X i = T i . H 1 q ( g ) = [ ( g + 1 ) ! ] q + [ 1 ] q + q g H 1 q ( g , K ) = [ ( g + 1 ) ! ] q + q g 1 q ( M 1 g ) E M 1 g g + 1 , q . q M [ N ] q M = q 1 g = 0 M 1 [ ( g + 1 ) ! ] q + q g 1 ( M g 1 ) S 2 q ( M , g + 1 ) G 1 + g N q g F o r m 1 . I n H 2 q ( g ) , B i = [ q 1 q ( i + 1 X K 1 ) G 1 i + 1 X K 1 = q 1 q ( 2 + X T ) G 1 2 + X T = q 1 q ( 1 + X T ) G 1 1 + X T , X i = K i q ( 1 + X T ) G 1 1 + X T , X i = T i . H 2 q ( g ) = [ ( g + 1 ) ! ] q [ 1 ] q H 2 q ( g , K ) = q [ ( g + 1 ) ! ] q ( 1 ) M 1 g q ( M 1 g ) E M 1 g g + 1 , q . q M [ N ] q M = q 1 g = 0 M 1 q [ ( g + 1 ) ! ] q ( 1 ) M 1 g q ( M 1 g ) E M 1 g g + 1 , q G 1 + g N + g F o r m 2 . q 1 S U M ( N , [ [ 1 ] q , [ 1 ] q . . . [ 1 ] q ] , [ 1 , 2 . . . M ] ) = g = 0 M H 3 q ( g ) G M N + M 1 g q g . B i = [ q 1 + G 1 X T 1 = q 1 G 1 1 + X T 1 = q 1 G 1 1 + X T , X i = K i q ( q X T 1 G 1 i q i G 1 X T 1 q i q 1 ) = q q i 1 q X T 1 q 1 = q X T G 1 i X T , X i = T i , H 3 q ( g ) = q ( M g ) q g ( g + 1 ) 2 [ g M q , H 3 q ( M ) = 0 . [ N ] q M = g = 0 M 1 [ g M q q g ( g + 1 ) 2 G M N + M 1 g = g = 0 M 1 [ M 1 g M q q ( M 1 g ) ( M g ) 2 G M N + g .
N M = g = 1 M g ! S 2 ( M , g ) [ g N = g = 1 M ( 1 ) M g g ! S 2 ( M , g ) [ g N + g 1 = g = 0 M 1 [ g M [ M N + g . S 2 ( M , g ) is Stirling number of the sec ond kind . S 2 ( M , g ) = E M g g , [ g M is Eulerian number , [ g M = λ 1 + λ 2 + . . + λ g + 1 = M g i = 1 g + 1 i λ i ( 1 + λ 1 ) ( 1 + λ 1 + λ 2 ) . . . ( 1 + λ 1 + . . + λ g )  [3] .
Proposition 8.
(1). 0 A < M , 0 T , g = 0 M G g M G A A + T + g G A + T + 1 + g N + A + T q g ( g + 1 + T ) = k = 0 A G k + T A + T G M + T M + T + k G M + T + 1 + k N + M + T q k ( k + 1 + T ) .
(2). 0 A , B , T , 0 A + B < M , g G g M G A A + T + g G B g q [ 2 g g ( A + B ) ( 1 ) g = 0 .
(3). 0 K , T , g ( 1 ) g G g M G M + K M + K + T + g q [ 2 M + 1 g + ( M g ) K = ( 1 ) M G K T + M + K .
Proof. 
P S = [ [ T + 1 ] q , [ T + 2 ] q . . . [ T + M ] q ] , P T = [ T + A + 1 , T + A + 2 . . . T + A + M ] ) , S U M q ( N ) = = g = 0 M G A + T + 1 + g N + A + T [ T + A + 1 ] q . . . [ T + A + g ] q × [ T + g + 1 ] q . . . [ T + M ] q q [ 2 g + 1 ( T + M + 1 ) ( M g ) + [ 2 M + 1 g G g M [ 5 ( 2 ) ] = ( q ; q ) A ( q T + A + 1 ; q ) M A ( 1 q ) M g = 0 M G A + T + 1 + g N + A + T q [ 2 g + 1 ( T + M + 1 ) ( M g ) + [ 2 M + 1 g G A A + T + g G g M ( * ) = i = 1 M A [ T + A + i ] q S U M q ( N , [ [ T + 1 ] q . . . [ T + A ] q ] , [ T + M + 1 . . . T + M + A ] ) . [ 3 ( 4 ) ] = q ( M A ) ( T + A + 1 + T + M ) 2 i = 1 M A [ T + A + i ] q k = 0 A G M + T + 1 + k N + M + T q [ 2 k + 1 ( T + A + 1 ) ( A k ) + [ 2 A + 1 k × [ T + M + 1 ] q . . . [ T + M + k ] q × [ T + 1 + k ] q . . . [ T + A ] q G k A = q ( M A ) ( T + A + 1 + T + M ) 2 k = 0 A G M + T + 1 + k N + M + T q [ 2 k + 1 ( T + A + 1 ) ( A k ) + [ 2 A + 1 k i = 1 M [ T + k + i ] q G k A . ( * * ) Compare ( * ) and ( * * ) : g = 0 M G A + T + 1 + g N + A + T q g ( 1 + g ) ( M g ) ( g + M + 1 + 2 T ) 2 G A A + T + g G g M = q ( M A ) ( T + A + 1 + T + M ) 2 k = 0 A G M + T + 1 + k N + M + T q k ( 1 + k ) ( A k ) ( k + A + 1 + 2 T ) 2 ( q T + k + 1 ; q ) M ( q T + A + 1 ; q ) M A ( q ; q ) k ( q ; q ) A k = q ( M A ) ( T + A + 1 + T + M ) 2 k = 0 A G M + T + 1 + k N + M + T q k ( 1 + k ) ( A k ) ( k + A + 1 + 2 T ) 2 G M + T M + T + k G k + T A + T ( 1 ) . [ 6 ] and ( 1 ) g = 0 M G g M G A A + T + g G B g q g ( 1 + g + T ) q g ( g + 3 ) 2 B g ( A + T ) g ( 1 ) g = 0 ( 2 ) . S U M q ( N , [ [ T + 1 ] q , [ T + 2 ] q . . . [ T + M ] q ] , [ T + K + M + 1 , T + K + M + 2 . . . T + K + 2 M ] ) . H 1 q ( g ) = q [ 2 g + 1 ( T + M + 1 ) ( M g ) + [ 2 M g + 1 i = 1 g [ T + K + M + i ] q i = 1 M g [ T + M + 1 i ] q G g M [ 5 ( 2 ) ] . H 2 q ( 0 ) = ( 1 ) M q M ( T + K + M + 1 ) i = 1 M [ K + i ] q = g ( 1 ) g H 1 q ( g ) q g ( g + 3 ) 2 ( T + K + M ) g [ 6 ( 3 ) ] . ( 1 ) M i = 1 M [ K + i ] q = g ( 1 ) g q [ 2 M g + 1 + ( M g ) K i = 1 g [ T + K + M + i ] q i = 1 M g [ T + M + 1 i ] q G g M . ( 1 ) M i = 1 M [ K + i ] q i = 1 K [ T + M + i ] q = g ( 1 ) g q [ 2 M g + 1 + ( M g ) K i = 1 K + M [ T + g + i ] q G g M . ( 1 ) M i = 1 K [ T + M + i ] q = g ( 1 ) g q [ 2 M g + 1 + ( M g ) K i = 1 K + M [ T + g + i ] q / i = 1 M [ K + i ] q × G g M . ( 1 ) M G K T + M + K = g ( 1 ) g q [ 2 M g + 1 + ( M g ) K × G M + K T + K + M + g G g M ( 3 ) , the case where A + B M of ( 2 ) .
Proposition 9.
n M = 0 N 1 . . . n 1 = 0 n 2 i = 1 M [ K 1 + i + 2 n i ] q q j = 1 M n j = q ( N 1 ) M i = 1 M [ K + N 2 + i ] q G M M + N 1 .
Proof. 
P S = [ [ K ] q , [ K + 1 ] q . . . [ K + M 1 ] q ] , P T = [ 1 , 2 . . . M ] . H 1 q ( g , K ) = q ( K + M ) ( M g ) + [ 2 M g + 1 i = 1 M g G 1 K + g 1 + i G g M . [ 5 ( 2 ) ] B i = K i o f H 1 q ( g ) = q ( K 1 + i ) G 1 K 1 + i + G X T = q ( K 1 ) i G 1 K 1 + i + X T . Expand by B i H 1 q ( g , K ) = n M g = 0 g . . . n 1 = 0 n 2 q ( K 1 ) ( M g ) j = 1 M g j j = 1 M g n j i = 1 M g G 1 K 1 + i + 2 n i . n M g = 0 g . . . n 1 = 0 n 2 i = 1 M g G 1 K 1 + i + 2 n i q g ( M g ) j = 1 M g n j = i = 1 M g G 1 K + g 1 + i G g M . n M = 0 g . . . n 1 = 0 n 2 i = 1 M G 1 K 1 + i + 2 n i q g M j = 1 M n j = i = 1 M G 1 K + g 1 + i G g M + g . n M = 0 N 1 . . . n 1 = 0 n 2 i = 1 M G 1 K 1 + i + 2 n i q ( N 1 ) M j = 1 M n j = i = 1 M G 1 K + N 2 + i G N 1 M + N 1 .
K = 1 , M = 1 n = 0 N 1 [ 1 + 2 n ] q q n = q ( N 1 ) [ N ] q 2 .

4. Extensions of q-Euler Polynomials and Relationships between Three Forms

In this section, a 0 , 1 , q 1 , q 2 . . . q M . . . .
Lemma 2.
n = 0 N 1 a n G M n + A = a N g = 0 M q ( N + A M ) g G M g N + A 1 g ( a ; q ) g + 1 + a M A ( a ; q ) M + 1 , 0 A M , N > M A .
Proof. 
A = M , M = 0 , n = 0 N 1 a n = a N 1 a + 1 1 a , holds . A = M , M = 1 , n = 0 N 1 a n G 1 n + 1 = n = 0 N 1 a n ( 1 + q + . . . + q n ) = ( 1 + q + . . . + q N 1 ) n = 0 N 1 a n q n = 0 0 a n q 2 n = 0 1 a n . . . q N 1 n = 0 N 2 a n = G 1 N 1 a N 1 a q ( 1 a ) 1 a q 2 ( 1 a 2 ) 1 a . . . q N 1 ( 1 a N 1 ) 1 a = G 1 N 1 a N 1 a + 1 + a q + . . . + a N 1 q N 1 1 a 1 + q + . . . + q N 1 1 a = G 1 N G 1 N a N 1 a + 1 a N q N ( 1 a ) ( 1 a q ) G 1 N 1 a = a N q N ( 1 a ) ( 1 a q ) + a N G 1 N 1 a + 1 ( 1 a ) ( 1 a q ) , holds . A = M , N = 1 , a 1 g = 0 M q g ( a ; q ) g + 1 + 1 ( a ; q ) M + 1 = 1 a q M ( a ; q ) M + 1 a 1 g = 0 M 1 q g ( a ; q ) g + 1 = ( 1 a q M ) ( 1 a q M 1 ) ( a ; q ) M + 1 a 1 g = 0 M 2 q g ( a ; q ) g + 1 = . . . = 1 , holds . Assume that N holds . n = 0 N a n G M n + M = n = 0 N a n ( q M G M n 1 + M + G M 1 n 1 + M ) = a q M n = 0 N 1 a n G M n + M + n = 0 N a n G M 1 n + M 1 = q M a N + 1 g = 0 M q N g G M g N + M 1 g ( a ; q ) g + 1 + a q M ( a ; q ) M + 1 a N + 1 g = 0 M 1 q ( N + 1 ) g G M 1 g N + M 1 g ( a ; q ) g + 1 + 1 ( a ; q ) M = a N + 1 g = 0 M q N g + M G M g N + M 1 g + q ( N + 1 ) g G M 1 g N + M 1 g ( a ; q ) g + 1 + 1 ( a ; q ) M + 1 = a N + 1 g = 0 M q ( N + 1 ) g q M g G M g N + M 1 g + q ( N + 1 ) g G M 1 g N + M 1 g ( a ; q ) g + 1 + 1 ( a ; q ) M + 1 = a N + 1 g = 0 M q ( N + 1 ) g G M g ( N + 1 ) + M 1 g ( a ; q ) g + 1 + 1 ( a ; q ) M + 1 . Proof of A = M completed . n = 0 N 1 a n G M n + A = a M A n = 0 N M + A 1 a n G M n + M , complete the remaining proofs .
Theorem 7.
X = T M M y 1 , 0 Y 1 , f ( g ) = ( a q 2 + X + g ; q ) M g = ( a q 2 + T M M y + g ; q ) M g ,
(1). g H 1 q ( g ) a g q y g f ( g ) = g H 2 q ( g ) f ( g ) = g H 3 q ( g ) a g q y g ,define as A a q ( P S , P T , y ) .
(2). n = 0 N 1 a n q y S U M q ( n + Y ) = a N k = 0 M q ( N + Y 1 ) k q y + k S U M q ( N + Y 1 ) ( a ; q ) k + 1 + a 1 Y A a q ( P S , P T , y ) ( a ; q ) T M + 2 y .
(3). | a | , | q | < 1 , n = 0 a n q y S U M q ( n + Y ) = a 1 Y A a q ( P S , P T , y ) ( a ; q ) T M + 2 y .
(4). q y S U M q ( ) = A q q ( P S , P T , 1 + y ) ( q ; q ) T M + 1 y , k = 0 R q ( N R ) ( k + 1 ) G R k N 1 k ( q ; q ) k + 1 + 1 ( q ; q ) 1 + R = G 1 + R N .
Proof. 
n = 0 N 1 a n q y S U M q ( n + Y ) = g = 0 M H 1 q ( g ) n = 0 N 1 a n G 1 + X + g n + Y + X q y g = g = 0 M H 1 q ( g ) ( a N k = 0 1 + X + g q ( N + Y 1 g ) k q y g G 1 + X + g k N + Y + X 1 k ( a ; q ) k + 1 + a 1 + g Y q y g ( a ; q ) 2 + X + g ) . = a N k = 0 M q ( N + Y 1 ) k g = 0 M H 1 q ( g ) q ( y + k ) g G 1 + X + g k N + Y + X 1 k ( a ; q ) k + 1 + g = 0 M H 1 q ( g ) a 1 + g Y q y g f ( g ) ( a ; q ) T M + 2 y = a N k = 0 M q ( N + Y 1 ) k q y + k S U M q ( N + Y 1 ) ( a ; q ) k + 1 + g = 0 M H 1 q ( g ) a 1 + g Y q y g f ( g ) ( a ; q ) T M + 2 y . n = 0 N 1 a n q y S U M q ( n + Y ) = g = 0 M H 2 q ( g ) n = 0 N 1 a n G 1 + X + g n + Y + X + g = g = 0 M H 2 q ( g ) ( a N k = 0 1 + X + g q ( N + Y 1 ) k G 1 + X + g k N + Y 1 + X + g k ( a ; q ) k + 1 + a 1 Y ( a ; q ) 2 + X + g ) . = a N k = 0 M q ( N + Y 1 ) k q y + k S U M q ( N + Y 1 ) ( a ; q ) k + 1 + a 1 Y g = 0 M H 2 q ( g ) f ( g ) ( a ; q ) T M + 2 y . n = 0 N 1 a n q y S U M q ( n + Y ) = g = 0 M H 3 q ( g ) n = 0 N 1 a n G T M + 1 y n + Y + T M y g q y g = g = 0 M H 3 q ( g ) ( a N k = 0 T M + 1 y q ( N + Y 1 ) k q ( y + k ) g G T M + 1 y k N + Y + T M y g 1 k ( a ; q ) k + 1 + a 1 + g Y q y g ( a ; q ) T M + 2 y ) . = a N k = 0 M q ( N + Y 1 ) k q y + k S U M q ( N + Y 1 ) ( a ; q ) k + 1 + g = 0 M H 3 q ( g ) a 1 + g Y q y g ( a ; q ) T M + 2 y . Three summations are identical ( 1 ) ( 2 ) . ( 3 ) is obvious . S U M q ( N ) = n = 0 N 1 q n q 1 S U M q ( n + 1 ) ( 4 ) . Calculations show that the equations derived by F o r m 2 and F o r m 3 are the same as the latter half of ( 4 ) .
n = 0 a n [ n ] q M = E M ( a , q ) ( a ; q ) M + 1 , E M ( a , q ) is q-Eularian polynomials[2] p.332. [7] → three expressions for E M ( a , q ) .
Eularian polynomials: n 1 a n n M = a A M ( a ) ( 1 a ) M + 1 . A M ( a ) = g H 1 ( g ) a g ( 1 a ) M g = g H 2 ( g ) ( 1 a ) M g = g H 3 ( g ) a g
= g g ! S 2 ( M , g ) a g ( 1 a ) M g = g ( 1 ) M g g ! S 2 ( M , g ) ( 1 a ) M g = g [ g M a g .
At [6], some relationships have been obtained, and now the remaining ones can be deduced:
Theorem 8.
(1). c g * = ( 1 ) g q g ( g + 3 ) 2 k b k G g M k q g k = ( 1 ) g q g ( g + 3 ) 2 + M g k b M k G g k q g k .
(2). b M g = ( 1 ) g q g ( g 1 ) 2 k c k * G g k q ( M + 1 ) k , b g = ( 1 ) M g q ( M g ) ( M g 1 ) 2 k c k * G M g k q ( M + 1 ) k .
(3). If g = 0 M b g G 1 + p + g X + g can be converted into g = 0 M R ( . . . ) G 1 + p + R + g X + g , 0 < R M , then
k G g k b k = 0 , 0 g < R , k G g M k q g k b k = 0 , 0 g < R .
(4). If g = 0 M c g G 1 + p + M X + M g can be converted into g = 0 M R ( . . . ) G 1 + p + M R X + M R g , 0 < R M , then
k G M g M k q k ( g + 1 ) c k * = 0 , 0 M g < R , k G g k q ( M + 1 ) k c k * = 0 , 0 g < R .
Proof. 
At [ 7 ] , P T = [ 1 , 2 . . . M ] , f ( g ) = ( 1 a q 2 + g ) ( 1 a q 3 + g ) . . . ( 1 a q M + 1 ) , g b g f ( g ) = g b g k G k M g ( a ) k q g ( g 1 ) 2 + ( 2 + g ) k = g c g q p g a g . Compare a g on both sides ( 1 ) . x c x = x ( 1 ) x q x ( x + 1 ) 2 + ( M + p + 1 ) x k b M k G x k q x k . x ( x + 1 ) 2 ( M + p + 1 ) x + ( x g ) ( x g 1 ) 2 + g x + x = g ( g + 1 ) 2 ( M + p + 1 ) x . x c x G g x q g ( g + 1 ) 2 ( M + p + 1 ) x = x ( 1 ) x G g x q ( x g ) ( x g 1 ) 2 + g x + x k b M k G x k q x k . = k b M k x ( 1 ) x G g x G x k q ( x g ) ( x g 1 ) 2 q ( g + 1 k ) x = k b M k q ( . . . ) x ( 1 ) x G g x G x k q x ( x 1 ) 2 ( k 1 ) x . ( * ) [ 2 ( 3 ) ] k > g , x = g k ( . . . ) = 0 ( * ) = ( 1 ) g b M g q g ( 2 ) . ( 3 ) is correspond to a g * = c g * = 0 , ( 4 ) is correspond to a M g * = b M g = 0 .
Theorem 9.
A Z ,
(1). g a g * q g ( g + 1 ) 2 + A g a M g z g = a M g b g ( q A + 1 z a ; q ) g = g c g * q g ( g + 1 ) 2 + A g a M g z g ( q A + 1 z a ; q ) M g .
(2). g b g q A g a M g z g = a M g ( 1 ) g a g * q g ( g + 3 ) 2 ( q A g + 1 z a ; q ) g = q A M z M g c g * q ( M + 1 ) g ( q A a z ; q ) g .
(3). g c g * q A g a M g z g = g a g * q A g a M g z g ( q A + 2 + g z a ; q ) M g = a M g b g ( q A + 2 + g z a ; q ) M g .
Proof. 
g a g * q g ( g + 1 ) q g ( g 1 ) 2 + ( A + 1 ) g a M g z g = g k b k G g k q g ( g 1 ) 2 + ( A + 1 ) g a M g z g = k b k a M k g G g k q g ( g 1 ) 2 + ( A + 1 ) g a k g z g = k b k a M k i = 1 k ( a + q A + i z ) . g a g * q g ( g + 3 ) 2 + ( A + 1 ) g a M g z g = k c k * q k ( k + 3 ) 2 + ( A + 1 ) k z k g G g k M k q [ 2 g k + ( A + 1 ) ( g k ) a ( M k ) ( g k ) z g k = k c k * q k ( k + 1 ) 2 + A k z k i = 1 M k ( a + q A + i z ) ( 1 ) . g b g q A g a M g z g = g q A g a M g z g k ( 1 ) k + g G g k q g ( g + 1 ) k ( k + 3 ) 2 k g a k * = k ( 1 ) k a k * q k ( k + 3 ) 2 a M k g G g k q [ 2 g + ( A k + 1 ) g a k g ( z ) g = k ( 1 ) k a k * q k ( k + 3 ) 2 a M k i = 1 k ( a q A k + i z ) . g b g q A g a M g z g = g q A g a M g z g ( 1 ) M g q ( M g ) ( M g 1 ) 2 k G M g k q ( M + 1 ) k c k * = k q ( M + 1 ) k c k * g a M g z g ( 1 ) M g G M g k q ( M g ) ( M g 1 ) 2 + A g , replace g with M - g = k q ( M + 1 ) k c k * g z M g ( a ) g G g k q g ( g 1 ) 2 + A ( M g ) = q A M k q ( M + 1 ) k c k * z M k i = 1 k ( z q A 1 + i a ) ( 2 ) . g c g * q A g a M g z g = g k ( 1 ) k + g a k * G g k M k q g ( g + 3 ) 2 k ( k + 3 ) 2 + A g a M g z g = k ( 1 ) k a k * q k ( k + 3 ) 2 g ( 1 ) g G g k M k q g ( g + 3 ) 2 + A g a M g z g , replace g with g + k = k ( 1 ) k a k * q k ( k + 3 ) 2 g ( 1 ) g + k G g M k q ( g + k ) ( g + k + 3 ) 2 + A ( g + k ) z g + k a M k g = k a k * q A k z k g G g M k q g ( g 1 ) 2 + ( A + k + 2 ) g ( z ) g a M k g = k a k * q A k z k i = 1 M k ( a q A + k + 1 + i z ) . g c g * q A g a M g z g = k b k g ( 1 ) g q g ( g + 3 ) 2 G g M k q g k + A g a M g z g = k b k a k g ( 1 ) g q g ( g 1 ) 2 + ( A + k + 2 ) g G g M k a M k g z g = k b k a k i = 1 M k ( a q A + k + 1 + i z ) ( 3 ) .
Combining [ 6 ] , [ 8 ] , [ 9 ] and a g * q p g G 1 + p + g X = b g G 1 + p + g X + g = c g * q p g G 1 + p + M X + M g , we can arbitrarily specify one set of values , calculate the other two sets , and treat A , a , z as an independent variable or part of a g , b g , c g to derive the corresponding relationship . a g * , 1 = q g ( g + 1 ) + A g G g M , a g * , 2 = a g * , 1 z g [ 2 ( 2 ) ( 3 ) ] b g * , 1 = q A g G g M ( q A ; q ) M g , b g * , 2 = q A g G g M ( z q A ; q ) M g z g , c g * , 1 = ( 1 ) g q g ( g + 3 ) 2 G g M ( q A ; q ) g , c g * , 2 = ( 1 ) g q g ( g + 3 ) 2 G g M ( z q A ; q ) g g q g ( g + 1 + A + p ) G g M z g G 1 + p + g N = g q A g ( z q A ; q ) M g G g M G 1 + p + g N + g = g ( 1 ) g q g ( g + 3 ) 2 + p g G g M ( z q A ; q ) g G 1 + p + M N + M g . g a g * , 1 q g ( g + 1 ) 2 z g = g a g * , 2 q g ( g + 1 ) 2 = ( z q 1 + A ; q ) M = g q A g G g M ( q A ; q ) M g ( z q ; q ) g ( * * ) = g q A g z g G g M ( z q A ; q ) M g ( q ; q ) g = g ( 1 ) g q g z g G g M ( q A ; q ) g ( z q ; q ) M g = g ( 1 ) g q g G g M ( z q A ; q ) g ( q ; q ) M g . A = 1 , replace z q by z , ( * * ) ( z q ; q ) ( q ; q ) = g q g ( z ; q ) g ( q ; q ) g . A = 0 , a g * , 2 = q g ( g + 1 ) G g M z g , b g * , 2 = G g M ( z ; q ) M g z g , calculate a g * , 2 using b g * , 2 g ( z ; q ) M g z g G g M G k g = z k G k M . It s equivalent to g q g ( q ; q ) g k = q k ( q ; q ) M k g q g ( q ; q ) g = 1 ( q ; q ) M ( * ) , a known formula  [2] [ 9 ( 1 ) ] g a g * , 2 q g ( g + 1 ) 2 g z g x g ( 1 ) g = ( 1 x ) ( 1 x q ) . . . ( 1 x q M 1 ) = g b g * , 2 ( x z ; q ) g = g G g M ( 1 z ) ( 1 z q ) . . . ( 1 z q M g 1 ) ( z x ) ( z x q ) . . . ( z x q g 1 ) . Replace x by a b and z by c b , then multiply through by b M to get Jacobi s q - binomial theorem [2] p.71 ( b a ) ( b a q ) . . . ( b a q M 1 ) = g G g M ( b c ) ( b c q ) . . . ( b c q M g 1 ) ( c a ) ( c a q ) . . . ( c a q g 1 ) . A = 0 , c g * , 1 = δ g 0 g a g * , 1 q ( r 1 ) g a g ( a q 1 + r + g ; q ) M g = 1 g G g M q g 2 + g r a g ( a q 1 + r + g ; q ) M g ( a q ; q ) M = 1 ( a q ; q ) M . It s a finite form of Jacobi s Durfee square identity [2] pp.158-159 g q g 2 + g r a g ( q ; q ) g ( a q ; q ) g + r = 1 ( a q ; q ) . b g = q g , k = 0 M G g k q k g = G g + 1 M + 1 a g * = q g ( g + 2 ) G g + 1 M + 1 ; k G g M k q ( g + 1 ) k = G g + 1 M + 1 [ 2 ] p . 22 c g * = ( 1 ) g q g ( g + 3 ) 2 G g + 1 M + 1 . S o g q g ( g + 2 + p ) G g + 1 M + 1 G 1 + p + g N = g q g G 1 + p + g N + g = g ( 1 ) g q g ( g + 3 ) 2 + p g G g + 1 M + 1 G 1 + p + M N + M g . calculate b g by a g * , c g * k ( 1 ) k + g G g k q g ( g + 1 ) k ( k + 1 ) 2 g k G k + 1 M + 1 = q [ 2 M g k ( 1 ) M + k + g G M g k q k ( k + 3 ) 2 ( M + 1 ) k G k + 1 M + 1 = q g . calculate a g * , c g * by b g G g + 1 M + 1 = k ( 1 ) k G M g M k G k + 1 M + 1 q k ( k + 1 ) 2 = k ( 1 ) k G M g M k G k + 1 M + 1 q k ( k + 1 ) 2 g ( k + 1 ) . [ ( 3 ) ] g q g ( q 1 + g ; q ) M g = g q g ( g + 1 ) G g + 1 M + 1 ( q 1 + g ; q ) M g = g ( 1 ) g q g ( g + 1 ) 2 G g + 1 M + 1 = 1 ( 1 ; q ) M + 1 = 1 . g c g * q g z g = z 1 z 1 ( z ; q ) M + 1 = g q g ( z q 1 + g ; q ) M g 1 + g = 0 M z q g ( z ; q ) g + 1 = 1 ( z ; q ) M + 1 , promotion of ( * ) [2] p.113 g b g q g = M + 1 = q M g c g * q ( M + 1 ) g ( q ; q ) g = g ( 1 ) g G g + 1 M + 1 q ( g + 1 ) g 2 M ( g + 1 ) ( q ; q ) g .
If q 1 replaces q , the result will be M + 1 , which is Euler s identity : g = 1 M G g M ( q ; q ) g 1 = M [2] p.83 c g * , 1 = ( 1 ) g q g ( g + 1 ) 2 + B g G g M , c g * , 2 = c g * , 1 z g [ ( 3 ) ] a g * , 1 = q g ( g + 1 ) G g M ( q B g ; q ) g , a g * , 2 = q g ( g + 1 ) G g M ( q B g z ; q ) g . [ 9 ( 1 ) ] g ( 1 ) g q A g + B g G g M z g ( q A + 1 z ; q ) M g = g q g ( g + 1 ) 2 + A g G g M ( q B g ; q ) g z g . [ 9 ( 1 ) ] g ( 1 ) g q A g + B g G g M z g ( q A + 1 ; q ) M g = g q g ( g + 1 ) 2 + A g G g M ( q B g z ; q ) g . [ 9 ( 2 ) ] q A M z M g ( 1 ) g q g ( g + 1 ) 2 + B g ( M + 1 ) g G g M ( q A z 1 ; q ) g = g ( 1 ) g q g ( g 1 ) 2 G g M ( q B g ; q ) g ( q A + 1 g z ; q ) g . [ 9 ( 2 ) ] q A M g ( 1 ) g q g ( g + 1 ) 2 + B g ( M + 1 ) g G g M z g ( q A ; q ) g = g ( 1 ) g q g ( g 1 ) 2 G g M ( q B g z ; q ) g ( q A + 1 g ; q ) g . B = 1 g ( 1 ) g q A g + g G g M z g ( q A + 1 z ; q ) M g = q A M z M g ( 1 ) g q g ( g + 1 ) 2 M g G g M ( q A z 1 ; q ) g = 1 . A = 1 ; A = 0 g ( 1 ) g G g M z g ( z ; q ) M g = z M g ( 1 ) g q g ( g + 1 ) 2 M g G g M ( z 1 ; q ) g = 1 . z = q ; z = q 1 , g : = M g g q g ( q ; q ) g = 1 ( q ; q ) M ; g = 0 M ( 1 ) g q g ( g 1 ) 2 ( q ; q ) g = ( 1 ) M q M ( M + 1 ) 2 ( q ; q ) M ( 4.1 * ) . In the previous text , T i N , but excluding the actual meaning of S U M q ( N ) , T i can be any number . a = b q T , P S = [ a , a . . . a ] : a ( q 1 ) , P T = [ T , T + 1 . . . . T + M 1 ] , due to T , a and b are independent . At H 1 q ( g ) , B i = [ q ( T i T i 1 1 ) X T 1 ( K i + G 1 X T 1 D i ) = a q X T , X i = K i q X T G 1 T + i 1 X K 1 a ( q 1 ) = q X T ( b q X T 1 a ) , X i = T i , H 1 q ( g ) = G g M q g ( g + 1 ) 2 ( b a ) ( b q a ) . . . ( b q g 1 a ) a M g . At H 3 q ( g ) , B i = [ ( K i + G 1 X T 1 D i ) = a q X T , X i = K i q 1 { ( q X T 1 G 1 T i q T i G 1 X T 1 ) D i K i q T i } = a q X T , X i = T i , H 3 q ( g ) = G g M ( 1 ) g q g ( g + 1 ) 2 a M . a g * = H 1 q ( g ) q ( T 1 ) g , c g * = H 3 q ( g ) q ( T 1 ) g . g c g * ( b x ) g q ( T 1 ) g ( T + 1 ) g = g H 3 q ( g ) ( b x ) g q ( T + 1 ) g = a M g G g M ( 1 ) g q g ( g 1 ) 2 ( b q T x ) g = a M ( a x ; q ) M = g a g * ( b x ) g q ( T 1 ) g ( T + 1 ) g ( b x q g ; q ) M g = g H 1 q ( g ) ( b x ) g q ( T + 1 ) g ( b x q g ; q ) M g = g G g M q g ( g + 1 ) 2 ( b a ) ( b q a ) . . . ( b q g 1 a ) ( b q T ) M g ( b x ) g q ( T + 1 ) g ( b x q g ; q ) M g . = a M g G g M q g ( g 1 ) 2 ( b a ) ( b q a ) . . . ( b q g 1 a ) x g ( b x q g ; q ) M g . That is to say : ( a x ; q ) M ( b x ; q ) M = g G g M q g ( g 1 ) 2 ( b a ) ( b q a ) . . . ( b q g 1 a ) x g ( b x × q g ; q ) M g ( b x ; q ) M . This proves Cauchy s identity [2] p.260 : ( a x ; q ) ( b x ; q ) = g = 0 q g ( g 1 ) 2 x g ( b a ) ( b q a ) . . . ( b q g 1 a ) ( q ; q ) g ( b x ; q ) g . M = , | x | < 1 , b g = x g , [ 2 ] a g * = q g ( g + 1 ) x g ( x ; q ) g + 1 , [ 9 ( 1 ) ( 2 ) ] g = 0 q g ( g + 1 ) 2 + A g x g z g ( x ; q ) g + 1 = g = 0 x g ( q A + 1 z ; q ) g A 0 , g = 0 ( 1 ) g q g ( g 1 ) 2 A g x g ( x ; q ) g + 1 = g = 0 A x g ( q A ; q ) g . g = 0 x g z g q A g = 1 1 x z q A = g = 0 ( 1 ) g q g ( g 1 ) 2 x g ( x ; q ) g + 1 ( q A + 1 g z ; q ) g A 0 , 1 1 x q A = g = 0 A ( 1 ) g q g ( g 1 ) 2 x g ( x ; q ) g + 1 ( q A ; q 1 ) g . a 0 * = K i = b g , we can arbitrarily specify q , T i and D i to compute H 2 q ( g ) , or specify q and a g * , g > 0 to compute b g , thereby obtaining an arbitrary number of expansions of K i . Other forms also have similar situations .

5. Inferences of Relationships Among the Three Forms

Simplifying the mutual expressions yields the inversion formulas.
Theorem 10.
Sum from 0 to M,
(1). a g = k b k G g k , b g = k ( 1 ) k + g G g k q k ( k 1 ) + g ( g + 1 ) 2 g k a k .
(2). a g = k c k G M g M k q g k , c g = k ( 1 ) k + g G M g M k q k ( k 1 ) + g ( g + 1 ) 2 a k
(3). c g = k b k G g M k q g k , b g = k ( 1 ) M g k c k G M g k q M 2 + k 2 + x 2 M + k + x 2 M g M k .
Arbitrariness of a g , b g , c g can derive the formulas of δ x g .
Theorem 11.
Sum from 0 to M, 0 x , g M ,
(1). k ( 1 ) k + x G g k G k x q x ( x 1 ) + k ( k + 1 ) 2 x k = k ( 1 ) k + x G M g M k G M k M x q x ( x 1 ) + k ( k + 1 ) 2 g k = δ x g .
(2). k ( 1 ) k + g G g k G k x q k ( k 1 ) + g ( g + 1 ) 2 g k = k ( 1 ) M g k G M g k G k M x q M 2 + g 2 + k 2 M + g + k 2 M g M k + x k = δ x g .
(3). k ( 1 ) k + g G M g M k G M k M x q k ( k 1 ) + g ( g + 1 ) 2 x k = k ( 1 ) M x k G g M k G M k x q M 2 + g 2 + k 2 M + g + k 2 M k M x + g k = δ x g .
Combining [6] and [8(3)(4)] , b g = 0 c M g = 0 ; a g = 0 c g = 0 ; a M g = 0 b M g = 0 . That is to say:
Theorem 12.
Sum from 0 to M,
(1). k ( 1 ) k G g k q k ( k + 3 ) 2 g k x k = 0 , 0 g < R k ( 1 ) k G M g M k q k ( k + 3 ) 2 x k = 0 , 0 M g < R .
(2). k G g k x k = 0 , 0 g < R k G M g M k q g k x k = 0 , 0 M g < R .
(3). k G M g M k q k ( g + 1 ) x k = 0 k G M g k q ( M + 1 ) k x k = 0 , 0 M g < R .
From [ 1 ] , 0 g < R , 1 i R , D i = 1 , K i = [ T i ] q b g = c M g = 0 D i = 1 , K i = [ i 1 ] q a g = c g = 0 D i = 0 a M g = b M g = 0 . This is not a necessary condition . P T = [ T , T + 1 . . . T + M 1 ] , it s a necessary and sufficient condition by adjusting K i : D i and the order of P S . a g * q p g G 1 + p + g X = b g G 1 + p + g X + g = c g * q p g G 1 + p + M X + M g can yields q - Vandermorde identity and its generalizations .
Theorem 13.
Sum from 0 to M, p 1 , 0 k M ,
(1). g q g ( g + 1 + p ) G g k G 1 + p + g N = G 1 + p + k N + k , g q g ( g + 1 + p ) k ( g + 1 + p ) G M g M k G 1 + p + g N = G 1 + p + M N + M k .
(2). g ( 1 ) k + g q g ( g + 1 ) k ( k + 3 ) 2 g k p k G g k G 1 + p + g N + g = G 1 + p + k N , g ( 1 ) M g q [ 2 M g ( M + 1 + p ) k G M g k G 1 + p + g N + g = G 1 + p + M N + M k .
(3). g ( 1 ) k + g q g ( g + 3 ) k ( k + 3 ) 2 p k + p g G M g M k G 1 + p + M N + M g = G 1 + p + k N , g ( 1 ) g q g ( g + 3 ) 2 + g k + p g G g M k G 1 + p + M N + M g = G 1 + p + k N + k .
This yields : if g a g [ Y + g X = g b g [ Y + g X + g = g c g [ Y + M X + M g , then ( 1 ) g a g [ Y + g X + g = ( 1 ) g b g [ Y + g X = ( 1 ) M c M g [ Y + M X + M g . ( 1 ) g a g [ Y + M X + M g = b M g [ Y + g X + g = ( 1 ) g c g [ Y + g X . b g [ Y + M X + M g = ( 1 ) M g a M g [ Y + g X + g = ( 1 ) M g c M g [ Y + g X . c g [ Y + g X + g = ( 1 ) g a M g [ Y + M X + M g = ( 1 ) g b M g [ Y + g X . [ 9 ] a g a M g z g = b g a M g ( a + z ) g = c g ( a + z ) M g z g , a 0 . From above , there are : ( 1 ) g a g a M g ( a + z ) g = ( 1 ) g b g a M g z g = ( 1 ) M c M g ( a + z ) M g z g . ( 1 ) g a g ( a + z ) M g z g = b M g a M g ( a + z ) g = ( 1 ) g c g a M g z g . ( 1 ) M g a M g a M g ( a + z ) g = b g ( a + z ) M g z g = ( 1 ) M g c M g a M g z g . ( 1 ) g a M g ( a + z ) M g z g = ( 1 ) g b M g a M g z g = c g a M g ( a + z ) g .
Theorem 14.
g b g q ( 1 + p + M ) g G 1 + p + M N + p + M g = g ( 1 ) M g c M g * q ( 1 + p + M ) g + g ( g + 3 ) M ( M + 3 ) 2 G 1 + p + g N + p .
Proof. 
g b g G 1 + p + M N + p + M g = g G 1 + p + M N + p + M g ( 1 ) M g q ( M g ) ( M g 1 ) 2 k c k * G M g k q ( M + 1 ) k = k c k q * g G M g k q ( M + 1 ) k G 1 + p + M N + p + M g ( 1 ) M g q ( M g ) ( M g 1 ) 2 = k c M k q * g G M g M k q ( M + 1 ) ( M k ) G 1 + p + M N + p + M g ( 1 ) M g q ( M g ) ( M g 1 ) 2 , [ 13 ( 3 ) ] g b g q g ( g + 3 ) 2 + p g ( M g ) ( M g 1 ) 2 G 1 + p + M N + p + M g = g ( 1 ) M g c M g * q ( M + 1 ) ( M g ) + p g + g ( g + 3 ) 2 G 1 + p + g N + p . The reason why other forms could not be obtained is because q g k appeared .
Theorem 15.
Sum from 0 to M, 0 y M ,
(1). ( x ; q ) y = g ( x ; q ) M g x g G g M y q g y , x y ( x ; q ) M y = g ( x ; q ) g ( 1 ) M g y G M g y q ( M g ) ( M g 1 ) 2 ( M + 1 ) y + y ( y + 3 ) 2 .
(2). ( x ; q ) y = x M g ( x 1 q 1 y ; q ) g ( 1 ) g G M g M y q g ( g + 3 ) 2 ( M + 1 ) g + ( y 1 ) M , ( x ; q ) y = x M g ( x 1 q 1 g ; q ) g ( 1 ) g G M g M y q g ( g 1 ) 2 + ( M g ) y .
(3). x y ( x ; q ) M y = g ( q g y x ; q ) M g ( 1 ) g + y G g y q g ( g + 1 ) + y ( y + 1 ) 2 g y , ( x ; q ) M y = g ( q g y x ; q ) M g x g G g y q g ( g y 1 ) .
(4). z y = g ( 1 ) y + g q g ( g + 1 ) 2 y ( g + 1 ) G g y ( z q ; q ) g = g ( 1 ) y + g q g y z g G M g M y ( z q ; q ) M g .
(5). z y = g ( 1 ) g q g ( g 1 ) 2 G g y ( z q 1 g ; q ) g = z M g ( 1 ) g q g ( g + 1 ) 2 M g + g y G g M y ( z 1 ; q ) g .
(6). z y = g z g q g ( g + 1 ) y ( g + 1 ) G M g M y ( z q 2 + g ; q ) M g = g ( 1 ) M g q ( M g ) ( M g 1 ) 2 ( M + 1 ) y G M g y ( z q 2 + g ; q ) M g .
Proof. 
[ 9 ] g b g ( q z ; q ) g = g c g * q g ( g + 1 ) 2 z g ( q z ; q ) M g = g q g ( g + 1 ) 2 z g ( q z ; q ) M g ( 1 ) g q g ( g + 3 ) 2 k b k G g M k q g k ( 1 * ) . g c g * q g ( g + 1 ) 2 z g ( q z ; q ) M g = g ( q z ; q ) g ( 1 ) M g q ( M g ) ( M g 1 ) 2 k c k * G M g k q ( M + 1 ) k ( 2 * ) . x = q z , b g = δ y g and ( 1 * ) ; c g * = δ y g and ( 2 * ) ( 1 ) . Similarly , ( 2 ) and ( 3 ) can be proven . [ 9 ] g a g * q g ( g + 1 ) 2 z g = g b g ( z q ; q ) g = g ( z q ; q ) g q g ( g + 1 ) 2 k ( 1 ) k + g G g k q k ( k + 3 ) 2 g k a g * = g c g * q g ( g + 1 ) 2 z g ( z q ; q ) M g = g q g ( g + 1 ) 2 z g ( z q ; q ) M g q g ( g + 3 ) 2 k ( 1 ) k + g G M g M k q k ( k + 3 ) 2 a k * a g * = δ y g ( 4 ) . Similarly , ( 5 ) and ( 6 ) can be proven .
(1) or (4) x M = g ( 1 ) g ( x ; q ) g G g M q g ( g + 1 ) 2 g M , x = q (4.1*).

6. An Example

P S = [ A , B , C ] , P T = [ 1 , 3 , 5 ] , S U M q ( N ) = g H 1 q ( g ) G 3 + g N + 2 = g H 2 q ( g ) G 3 + g N + 2 + g = g H 3 q ( g ) G 6 N + 5 g . H 1 q ( 0 ) = A B C . H 1 q ( 1 ) = A B q 1 G 1 3 + A q 1 G 1 2 ( C + 1 ) + q 1 G 1 1 q 1 ( B + 1 ) q 1 ( C + 1 ) . H 1 q ( 2 ) = q 1 G 1 1 q 3 G 1 3 q 2 ( C + G 1 2 ) + q 1 G 1 1 q 1 ( B + 1 ) q 3 G 1 4 + A q 1 G 1 2 q 3 G 1 4 . H 1 q ( 3 ) = q 1 G 1 1 q 3 G 1 3 q 5 G 1 5 . H 2 q ( 0 ) = ( A q 1 G 1 1 ) ( B q 2 G 1 2 ) ( C q 3 G 1 3 ) . H 2 q ( 1 ) = ( A q 1 G 1 1 ) ( B q 2 G 1 2 ) q 3 G 1 3 + ( A q 1 G 1 1 ) q 2 G 1 2 ( C q 4 G 1 4 ) + q 1 G 1 1 ( B q 3 G 1 3 ) ( C q 4 G 1 4 ) . H 2 q ( 2 ) = q 1 G 1 1 q 3 G 1 3 ( C q 5 G 1 5 ) + q 1 G 1 1 ( B q 3 G 1 3 ) q 4 G 1 4 + ( A q 1 G 1 1 ) q 2 G 1 2 q 4 G 1 4 . H 2 q ( 3 ) = q 1 G 1 1 q 3 G 1 3 q 5 G 1 5 . H 3 q ( 0 ) = A B C . H 3 q ( 1 ) = A B q 1 ( G 1 5 q 5 C ) + A q 2 ( G 1 3 q 3 B ) ( C + 1 ) + q 3 ( 1 q 1 A ) ( B + 1 ) ( C + 1 ) . H 3 q ( 2 ) = q 5 ( 1 q 1 A ) ( q 1 G 1 3 q 3 q 3 B ) ( C + G 1 2 ) + q 4 ( 1 q 1 A ) ( B + 1 ) ( q 1 G 1 5 q 5 q 5 C ) + A q 3 ( G 1 3 q 3 B ) ( q 1 G 1 5 q 5 q 5 C ) . H 3 q ( 3 ) = q 6 ( 1 q 1 A ) ( q 1 G 1 3 q 3 q 3 B ) ( q 2 G 1 5 q 5 G 1 2 q 5 C ) . P S = [ A , B ] , P T = [ 1 , 3 ] . S U M ( N ) = g H 1 ( g ) [ 2 + g N + 1 = g H 2 ( g ) [ 2 + g N + 1 + g = g H 3 ( g ) [ 4 N + 3 g . H 1 ( 0 ) = A B , H 1 ( 1 ) = A × 2 + 1 × ( B + 1 ) , H 1 ( 2 ) = 1 × 3 . H 2 ( 0 ) = ( A 1 ) ( B 2 ) , H 2 ( 1 ) = ( A 1 ) × 2 + 1 × ( B 3 ) , H 2 ( 2 ) = 1 × 3 . H 3 ( 0 ) = A B , H 3 ( 1 ) = A ( 3 B ) + ( 1 A ) ( B + 1 ) , H 3 ( 2 ) = ( 1 A ) ( 2 B ) .

Conflicts of Interest

The authors declare that they have no conflict of interest.

References

  1. P.A. MacMahon. The Indices of Permutations and the Derivation Therefrom of Functions of a Single Variable Associated with the Permutations of Any Assemblage of Objects, American Journal of Mathematics. 35 (1913) 281-322.
  2. Warren P. Johnson , An Introduction to q-analysis. American Mathematical Society. (2020).
  3. QI Deng-Ji. A New Explicit Expression for the Eulerian Numbers, Journal of Qingdao University of Science and Technology: Natural Science Edition. 04 (2012) 33.
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