A simple approach to some Hankel determinants

1 A simple approach to so me Hankel determinants Johann Cigler Fakultät für Mathematik Universität Wien A-1090 Wien, Nordbergstraße 15 johann.cigler@univie.ac.at Abstract I give simple elementary proofs for som e well-known Hankel dete rm inants and their q − analogues. My starting point is the following Lemma For given sequences () sn and () tn define (, ) an k by (0 , ) [ 0] ( , 0 ) (0 ) ( 1 , 0 ) (0) ( 1 ,1 ) (, ) ( 1 , 1 ) ()( 1 , ) () ( 1 , 1 ) . ak k an s an t an an k an k sk an k tk an k == =− + − =− − + − + − + (1) Then the Hankel determinant () 1 ,0 det ( , 0) n ij ai j − = + is given by () 11 1 ,0 10 det ( , 0) ( ). ni n ij ik ai j t k −− − = == += ∏∏ (2) Proof In order to prove this we show first that for all , mn ∈  1 0 (, )( , ) () ( , 0 ) . k kj an k am k t j am n − = =+ ∑∏ (3) The proof is obvious by induction. For 0 n = and arbitrary m it is trivially true. If it is true for 1 n − and arbitrary , m then we get 2 () () 1 1 0 0 1 0 1 0 (, )( , ) () ( 1 , 1 ) ()( 1 , ) ( )( 1 , 1 ) ( , ) () ( 1 , ) ( , 1 ) () ()( , ) ( , 1 ) () (1 , ) ( 1 , ) ( ) ( , 0 ) . k k kj k j k k j k kj a n ka m k t j a n k s ka n k t ka n k a m k t j a n k a mk t k s k a mk a mk t j an k am k t j an m − − = = − = − = =− − + − + − + =− + + + − =− + = + ∑∏ ∑ ∏ ∑∏ ∑∏ Consider now the matrices () 1 ,0 (, n n ij Aa i j − = = , 1 1 0 ,0 [] ( ) , n i n k ij Di j t j − − = = ⎛⎞ == ⎜⎟ ⎝⎠ ∏ and () 1 ,0 (( , 0 ) n n ij Ha i j − = =+ . Then (3) implies . t nn n n AD A H = Therefore we get () 11 10 1 ,0 det ( de ( . ,0 ) ) t ni n i i k n j Ht k ai j −− = = = − += = ∏∏ (4) Remark This Lemma is well- known and intim ately connected with the approach to Hankel determinants via orthogonal polynomials (see [6]) . It is especially useful if for a given sequence () 0 (, 0 ) n an ≥ explicit expressions fo r () , () sn t n and (, ) an k exist. To find such expressions it is often convenient to compute th e first values of the orthogonal polynomials (, ) p nx (cf. e.g. [4] (1.10)) and their Favard resolution [4] (1.11) and try to guess ( ) sn and () tn and the explicit form of ( , ) an k . Other approaches to this lemma can be found in [1] and [8]. If all ( ) 0 sn = identity (1) reduces to (0 , ) [ 0] (, 0 ) ( 0 ) ( 1 , 1 ) (, ) ( 1 , 1 ) ( ) ( 1 , 1 ) . Ak k An T An An k An k T k An k == =− =− − + − + (5) In this case (2 , 2 1 ) (2 1 , 2 ) 0 An k An k += + = for all , . nk 3 If we define (, ) ( 2 , 2) , an k A n k = (6) then it is easily verified that (1) ho lds with (0 ) (0) , () ( 2 1 ) ( 2) , () ( 2) ( 2 1 ) . sT sn T n T n t n Tn Tn = =− + =+ (7) Therefore it is convenient to cons ider first sequences of the form ( ) (0), 0, ( 1 ), 0 , ( 2), 0, . cc c  Let us first look at the situation for the famous example of the Catalan numbers 2 1 () . 1 n n cn C n n ⎛⎞ == ⎜⎟ + ⎝⎠ (8) The table ( ) (, ) An k can be found in [5], A053121. The first terms are 1 01 10 1 0201 2 0301 05040 1 5 0 9 0 501 0 1 40 1 40601 It is well known that () 1 Tn = and 1 1 (, ) , 1 2 n k An k nk n + ⎛⎞ + ⎜⎟ = − ⎜⎟ + ⎝⎠ if mod 2 nk ≡ and 0 else. This can also be written in the form 22 2 21 (2 , 2 ) 1 1 nn n k An k nk nk nk nk ⎛⎞ ⎛ ⎞ ⎛⎞ + =− = ⎜⎟ ⎜ ⎟ ⎜⎟ − −− − ++ ⎝⎠ ⎝ ⎠ ⎝⎠ and 21 21 21 22 (2 1 , 2 1 ) . 1 2 nn n k An k nk nk nk nk ++ + ⎛⎞ ⎛ ⎞ ⎛⎞ + ++ = − = ⎜⎟ ⎜ ⎟ ⎜⎟ −− − − ++ ⎝⎠ ⎝ ⎠ ⎝⎠ To prove this fact it suffices to verify (5). This is almost trivial: (, 0 ) ( 0 ) ( 1 , 1 ) An T An =− is equivalent with 2 2 21 21 (2 , 0 ) (2 1 , 1 ) 11 2 nn n n An An nn n n − − ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ =− = − = − ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ −− − ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ or 4 22 1 2 2 1 2 1 . 11 2 nn n n n nn n n n −− − ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ −= −= ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ −− − ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ (, ) ( 1 , 1 ) ( ) ( 1 , 1 ) An k An k T k An k =− − + − + reduces to ( 2 ,2 ) ( 2 1 ,2 1 ) ( 2 1 ,2 1 ) An k An k An k =− − + − + and (2 1 , 2 1 ) (2 , 2 ) (2 , 2 2) . An k An k An k ++ = + + The first identity is 2 2 21 21 21 21 11 1 2 nn n n n n nk nk nk nk nk nk −− − − ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ −= − + − ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ − −− − −− −− −− ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ and the second one 21 21 2 2 2 2 . 11 1 2 nn n n n n nk nk nk nk nk nk ++ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ −= − + − ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ − −− − −− −− −− ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ Both follow from the recurrences of the binomial coefficients. For the sequence ( , 0) n an C = we get 22 2 21 (, ) , 1 1 nn n k an k nk nk nk nk ⎛⎞ ⎛ ⎞ ⎛⎞ + =− = ⎜⎟ ⎜ ⎟ ⎜⎟ −− − − ++ ⎝⎠ ⎝ ⎠ ⎝⎠ (0 ) 1 , ( ) 2 ss n == for 0 n > and ( ) 1. tn = This gives the table [5], A039599 1 11 231 59 5 1 14 28 20 7 1 An immediate consequence of (4) is the well-kno wn result ( ) 1 ,0 det 1 n ij ij C − + = = for all . n ∈  It is also easy to verify that 0 (1 ) ( , ) [ 0 ] n k k an k n = − == ∑ (9) and 0 2 (, ) . n k n an k n = ⎛⎞ = ⎜⎟ ⎝⎠ ∑ (10) But we cannot immediately deduce that ( ) 1 1 ,0 det 1 n ij ij C − ++ = = or obtain the higher Hankel determinants () 1 ,0 det . n ij m ij C − ++ = 5 But observe that 1 0 12 4 42 n n n j j C j − = + = + ∏ and therefore 11 00 12 12 2 4. 42 42 2 mn nm nm jj j mj C j mj −− + + == ++ + = ++ + ∏∏ If we can find explicit values of ( ) , ( ) sn t n and ( , ) an k for 1 0 12 2 () 42 2 n j mj cn mj − = ++ = + + ∏ , then we can compute these higher Hankel determinants. Th is is indeed the case, even for the more general case () () 1 0 1 0 () n j n j bj c cn aj c − = − = + = + ∏ ∏ . It also remains true for 1 0 1 () , 1 bj c n aj c j q cn q + − + = − = − ∏ which for 1 q → tends to () () 1 0 1 0 n j n j bj c aj c − = − = + + ∏ ∏ and more generally for 1 0 1 () 1 j n j j bq cn aq − = − = − ∏ for arbitrary , . ab We shall later use the trivial fact tha t (1) is equivalent with () () () () () () ( ) () () ( ) ( ) 11 12 2 (0 , ) [ 0] ( , 0) ( 0) ( 1 , 0) ( 0) ( 1 , 1 ) (, ) ( 1 , 1 ) () ( 1 , ) ( ) ( 1 , 1 ) . nn n nk nk n k nk ak k xan x s x an x t x a n x a n k x a nk x s k xa nk x t k x a nk −− −− − − − − == =− + − =− − + − + − + (11) I will not show here how to guess the following results (this the reader can do for him self using the above hint), but will only v erify the relevant identities. We shall use the usual q − notations, e.g. 1 0 (; ) ( 1 ) n j n j x qq x − = =− ∏ , 1 [] 1 n q n q − = − and () () () ; ;; n kn k qq n k qq qq − ⎡⎤ = ⎢⎥ ⎣⎦ for 0 kn ≤≤ and 0 n k ⎡⎤ = ⎢⎥ ⎣⎦ for 0 k < and . kn > Theorem 1 Let ( ) () ; (, , , ) . ; n n bq cna b q aq = (12) Condition (5) is satisfied if we set 1 21 2 1 21 2 (1 ) ( 1 ) (2 , , , ) , (1 ) (1 ) (1 ) ( ) (2 1 , , , ) (1 ) (1 ) nn n nn nn n nn qq b q a T nabq qa q a qq b q a Tn a b q qa q a − − + + −− = −− −− += −− (13) and 6 2 21 1 (2 , 2 , , , ) ( , , , ), (2 1 , 2 1 , , , ) ( , , , ) . kk kk n A n k abq cn k q a qbq k n An k a b q c n k q a q b q k ++ ⎡⎤ =− ⎢⎥ ⎣⎦ ⎡⎤ ++ = − ⎢⎥ ⎣⎦ (14) Proof We have to show that the identities ( 2 2 , 2 , ,, ) ( 2 1 , 2 1 , ,, ) ( 2 , ,, ) ( 2 1 , 2 1 , ,, ) 0 An k a b q An k a b q T k a b q An k a b q +− + − − + + = (15) and (2 1 , 2 1 , , , ) (2 , 2 , , , ) (2 1 , , , ) (2 , 2 2 , , , ) 0 A n k abq A n k abq T k abq A n k a bq ++ − − + + = (16) hold. The left-hand side of (15) is () () () () () () () () () () 22 1 21 1 11 1 22 1 11 1 11 1 1 (1 , , , ) (1 , , , ) 1 (2 , , , ) ( , , , ) ;; ;; ;; ; ; ;; (1 ) (1 kk k k kk kk nn k n n k kk kn k k n k nk nk kk k nn c n k q a qbq c n k q a qbq kk n Tk a b q c n k q a q b q k qbq qbq qq qq qq qq qq qq qa q q a q qq q − ++ +− + − + − +− − +− −+ −+ + + ⎡⎤ ⎡⎤ −+ − −+ ⎢⎥ ⎢⎥ − ⎣⎦ ⎣⎦ ⎡⎤ −− ⎢⎥ ⎣⎦ =− −− − () () () () () () () () () () () ( ) ( ) ( ) ( ) ( ) () 1 1 21 2 21 12 1 1 1 1 21 1 2 ; ; ) (1 ) (1 ) ; ; ; ; ; 11 1 1 1 1 ;; ; k nn k kk k kn k nk k nk k n k k k n k nn k k kn k nk qb q qq a q a q a qq qq qa q qbq qq qq a q q a q q a q qq qq qa q + − − − + − − +− + − + − −+ − +− −+ −− =− − − − − − − − Since () ( ) ( )( ) ( ) ( ) 12 1 1 1 11 1 1 1 1 0 , nk k n k k k n k qq a q q a q q a q +− + − + − −− − − − − − − = we see that (15) is true. In the same way the left-hand side of (16) reduces to 7 () () () () () () () () () () 21 1 2 1 22 1 21 2 1 21 2 (, , , ) (, , , ) (1 ) ( ) (1 , , , ) 1 (1 ) (1 ) ;; ;; ;; ;; ;; (1 kk k k kk k kk kk kk nn k n n k kk kn k kn k nk nk kk nn c n k q a q bq c n k q a qbq kk n qq b q a cn k q aq b q k qa q a qb q q b q qq qq qq qq qq qq qa q q a q qq ++ + ++ + + −− + −− −− ⎡⎤ ⎡⎤ −− − ⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦ ⎡⎤ −− −− − ⎢⎥ + −− ⎣⎦ =− − − () () () () () () () () () () () ( ) ( ) ( ) ( ) ( ) () 1 1 1 21 2 22 11 1 1 2 1 2 1 ; ; )( ) (1 ) (1 ) ; ; ; ; ; 11 11 1 0 . ;; ; k k nn k kk k kn k nk k nk k n k k k n k nn k k kn k nk qb q qq bq a q a q a qq qq qa q qb q qq qb q a qb q a q b qa q qq qq qa q + + −− − + +− − −− + +− −− − −+ − −− =− − − − − − − − = This implies Theorem 2 Let ( ) () ; (, , , ) . ; n n bq cna b q aq = (17) Then the Hankel determinants () 1 ,0 ( , ,, , ) d e t ( ,, ,) n ij dn m a b q c i j m a b q − = =+ + (18) are given by () ( ) () () () 1 1 2 3 0 1 1 2 ;; (, 0 , , , ) ;; k j n kk n j k k k k bq q q b qa dn a b q q qa q a q − ⎛⎞ − ⎜⎟ = ⎝⎠ − = − = ∏ ∏ (19) and ( ) () 1 2 1 0 ; (, , , , ) (, 0 , , , ) . ; n j m m n nj j n qbq dn m a b q dn a b q q qa q ⎛⎞ − ⎜⎟ ⎝⎠ −+ = = ∏ (20) Proof By (2) we have 11 11 10 10 ( , 0 ,, , ) ( ,, , ) ( 2 ,, ,) ( 2 1 ,, ,) . nk nk kj kj d n ab q t jabq T jabq T j abq −− −− == == == + ∏∏ ∏∏ From () () 1 2 1 0 ; (2 , , , ) ; k k k k j k bq Tj a b q q qa q ⎛⎞ − ⎜⎟ ⎝⎠ − = = ∏ and () () 11 2 00 2 ; (2 1 , , , ) ( ) ; k kk j k jj k qq Tj a b q q b q a aq ⎛⎞ −− ⎜⎟ ⎝⎠ == += − ∏∏ we deduce (19). 8 By (19) we get () () 2 1 ; (, 0 , , , ) 1 . (, 0 , , , ) 1 ; n n n n n bq d n qa qb q a q dn a b q b qa q ⎛⎞ ⎜⎟ ⎝⎠ − − ⎛⎞ = ⎜⎟ − ⎝⎠ This implies () () () () () () 1 1 1 ,0 2 1 ;) ;) (1 ) 1 (, 1 , , , ) d e t d e t (, 0 , , , ) (1 ) 1 ;) ;) ; (, 0 , , , ) . ; n n ij ij ij ij ij n n n n bq q bq bb dn a b q dn q a q b q aa aq q aq bq qd n a b q qa q − ++ + ++ + = ⎛⎞ ⎜⎟ ⎝⎠ − ⎛⎞ ⎛ ⎞ −− ⎛⎞ == = ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ −− ⎝⎠ ⎝⎠ ⎝ ⎠ = Iterating this argument we get () () () () 1 2 1 0 ; (, , , , ) (, 0 , , , ) ; ; (, 0 , , , ) . ; n mm m m n j m m n nj j n bq dn m a b q dn qa qb q aq qbq qd n a b q qa q ⎛⎞ − ⎜⎟ ⎝⎠ −+ = ⎛⎞ = ⎜⎟ ⎜⎟ ⎝⎠ = ∏ If we change , , abc aq b q qq →→→ and let 1 q → we get Corollary 1 Let () () 1 0 1 0 (, , , ) . n j n j bj c un ab c aj c − = − = + = + ∏ ∏ (21) Condition (5) is satisfied if we set (( 1 ) ) ( ) ( 2 , ,,) , (2 ) (( 2 1 ) ) (1 ) ( ) (2 1 , , , ) (2 ) (( 2 1 ) ) an c b n c T nabc an c a n c nc a b n c Tn a b c an c a n c + −+ = ++ − +− + += ++ + (22) and ( 2 , 2 , ,,) ( , 2 , ,) , (2 1 , 2 1 , , , ) ( , (2 1 ) , ( 1 ) , ). n A n ka b c un ka k c b k c c k n An k a b c u n k a k c b k c c k ⎛⎞ =− + + ⎜⎟ ⎝⎠ ⎛⎞ + + = −++ + + ⎜⎟ ⎝⎠ (23) 9 Corollary 2 Let () () 1 0 1 0 (, , , ) . n j n j bj c un ab c aj c − = − = + = + ∏ ∏ (24) Then the Hankel determinants () 1 ,0 ( , ,, , ) d e t ( ,, , ) n ij Dn m ab c u i j m ab c − = =+ + (25) are given by () ( ) 1 1 0 12 1 1 00 !( ) ( ) (, 0 , , , ) (1 ) k k n j kk k jj k c bj c a bj c Dn ab c aj k c a j c − − = −− = == +− + = ++ − + ∏ ∏ ∏∏ (26) and ( ) () 11 00 () (, , , , ) (, 0 , , , ) . (1 ) mn ji bj i c Dn m ab c Dn ab c ai n j c −− == ++ = ++ + − ∏∏ (27) These Corollaries can of course also be proved directly. Hankel determinants of Catalan numbers Now let us consider again the Catalan numbers. Instead of n C we study 1 0 12 (, 4 , 1 , 2 ) . 42 n j j un j − = + = + ∏ (28) Here we get 1 () 4 Tn = and find again that ( 1 2) ( 3 2) ( 2 1 ) ( 2 ,2 ) ( ,4 4 , 1 2 ,2 ) ( 4 4) ( 6 4) ( 2 2 2 ) 2 1 ! (2 )! (2 1 )! 1 2 1 4 ! ( ) ! ( 2 ) ! ( 22 ) ( 42 ) ( 2 ) ( 1 ) ! 4 1 nk nk nn kk n An k u n k k k kk kk n k n nn k k nk kn k k k k n n k n k − − ⎛⎞ ⎛⎞ ++ − =− + + = ⎜⎟ ⎜⎟ ++ + + ⎝⎠ ⎝⎠ ⎛⎞ ++ == ⎜⎟ − −+ + + + + + ⎝⎠    and 21 12 2 (2 1 , 2 1 ) . 42 nk n k An k nk nk − + ⎛⎞ + ++ = ⎜⎟ − ++ ⎝⎠ It is evident that () 2 1 d e t ( ,4 , 1 ,2 ) . 4 n ui j ⎛⎞ ⎜⎟ ⎝⎠ += 10 In order to compute the higher Hankel determinants consid er the factor () () 1 0 () (1 ) n i bj i c ai n j c − = ++ ++ + − ∏ which occurs in (27). This reduces in our case to 11 00 1 1 12 2 1 12 2 1 ( 2 2 ) ! ( ) ! 42 2 2 2 2 1 2 ( 2 ) ! (2 )! (2 2 ) 1( 2 2 ) ! ! 1 2 . 4( 2 ) ! ( 2 ) ! 4 nn n nn ii j nn i j ij i j n n j in j n j i n j jj jn j n j i jn j j i −− == = = ++ ++ + + == ++ + − + + + + + ++ + == ++ ∏∏ ∏ ∏   Thus we get 1 01 2 12 (, , 4 , 1 , 2 ) . 4 j m n mn ji nj i Dn m j i − ⎛⎞ + ⎜⎟ == ⎝⎠ ++ = + ∏∏ Therefore we get for the Catalan numbers themse lves the well-known result (cf. [7], Theorem 33 and the historical comments given there) () 1 1 ,0 11 2 det . j m n ij m ij ji nj i C j i − − ++ = == + + = + ∏∏ (29) Since () 1 1 ,0 det 1 , n ij ij C − ++ = = we get also the well-known result, that the Catalan numbers are the uniquely determined numbers satisfying ( ) 1 ,0 det 1 n ij ij C − + = = and ( ) 1 1 ,0 det 1. n ij ij C − ++ = = Some applications of Theorem 2 1) Let () () ( , 0 , , ) ; . n cn cn xq xq == Then we get () () () () 1 2 1 32 2 ,0 0 det ; ; ; . nn n n m n ij m k m k ij k x qq x x q q q ⎛⎞ ⎛⎞ ⎛⎞ − + − ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ++ + = = = ∏ (30) As special case we note () 1 2( 1 ) 1 32 ,0 0 de t [ ]! [ ]! [ ]!. nn n m n ij k ij m q k m k ⎛⎞ ⎛⎞ − ++ ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ = = ++ = + ∏ (31) If we define [] () 1 0 :, n n j x xj − = =− ∏ (32) 11 then we get (1 ( 1) ) 1 ;. (1 ) 1 (1 ) n n n n qx x q qq x ⎛⎞ +− = ⎜⎟ −+ − ⎝⎠ Therefore we have () 1 2 1 23 2 ,0 0 det ( 1 ) [ ]! . nn n n m n ij m j m ij j xq j x ⎛⎞ ⎛⎞ ⎛⎞ − + − ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ++ + = = =− ∏ (33) 2) Let ( ) () 1 1 ; () ( , , , ) . ; d d n n qq cn cnq q q qq + + == Here we get () () () () () () ( ) () ( ) 1 11 1 1 2 3 0 1 1 2 1 1 1 2 32 0 1 1 ;; (, 0 , , , ) ;; ; . ;; k dd j n k n k j d k k k k k dd j nn n k j kk k kk qq q q q q dn q q q q qq q q qq qq q qq q q − ++ + ⎛⎞ − ⎜⎟ = + ⎝⎠ = − + ⎛⎞ ⎛⎞ − + ⎜⎟ ⎜⎟ = ⎝⎠ ⎝⎠ + = − = − = ∏ ∏ ∏ ∏ For dm =∈  we get ( ) . nm cn m + ⎡⎤ = ⎢⎥ ⎣⎦ The Hankel determinant reduces to 2 1 (1 ) 1 2 2 0 ,0 2 det ( 1 ) . 21 n n nn n k ij mk ij m k q k m k − ⎛⎞ − − ⎜⎟ ⎝⎠ = = + ⎡ ⎤ ⎢ ⎥ ++ ⎛⎞ ⎡⎤ ⎣ ⎦ =− ⎜⎟ ⎢⎥ − ⎡ ⎤ ⎣⎦ ⎝⎠ ⎢ ⎥ ⎣ ⎦ ∏ (34) This result is due to Carlitz [3] . For some ge neralizations see Krattenthaler [6], especially Theorem 26. 3) Choose [] 1 11 () ( , 2 , 1 , 1 ) . 11 n q cn cn qn + − == = −+ Then (20) gives a q − analogue of the determinant of the Hilbert matrix [] () () 1 23 (1 ) ( 21 ) 11 2 6 00 ,0 [] ! [ ] ! 1 det . 1 [ ]! [ 2 ]! [ ]! n n nn n mn m jj ij jn j q ij m j nj nj − ⎛⎞ −− −− + ⎜⎟ ⎝⎠ == = ⎛⎞ + = ⎜⎟ ++ + + + ⎝⎠ ∏∏ (35) 12 4) Let () (1 ) 1 () ( 1 ) (, 1 , 0 , 1 ) . ;[ ] ! n n n q cn q cn qq n − =− = = Then 2 1 (1 ) 2 2 2 0 ,0 1[ ] ! det ( 1 ) . [] ! [ ] ! n n nn nm j ij j q ij m nj − ⎛⎞ − +− ⎜⎟ ⎝⎠ = = ⎛⎞ =− ⎜⎟ ++ + ⎝⎠ ∏ (36) For 2 1 (1 ) 2 2 2 0 ,0 1[ ] ! det ( 1 ) [] ! [ ] ! n n nn n j ij j q ij nj − ⎛⎞ − − ⎜⎟ ⎝⎠ = = ⎛⎞ =− ⎜⎟ ++ ⎝⎠ ∏ and 2 11 1 2 0 ,0 ,0 (1 ) 2 2 2 0 1[ 1 ] ! 1 det det [] ! [ 2 1 ] ! [ ] ! [] ! (1 ) . [] ! n nn m m j ij ij n nn nm j nj q ij m n j ij j q nj −− ⎛⎞ − ⎜⎟ ⎝⎠ = == ⎛⎞ − +− ⎜⎟ ⎝⎠ = ⎛⎞ ⎛ ⎞ +− = ⎜⎟ ⎜ ⎟ ++ +− + ⎝⎠ ⎝ ⎠ =− + ∏ ∏ 5) Now we choose 2 2 21 22 2 2 1 1 1 2 11 () ( , , , ) (1 ) . 2 1 (1 ) j n nn n j j j q j n q cn cnq qq q n q q n − = = ⎡ ⎤ − ⎡⎤ − ⎢ ⎥ == = = − ⎢⎥ ⎢ ⎥ − ⎣⎦ + ⎣ ⎦ ∏ ∏ Here we get (1 ) ( 4 5 ) 22 6 22 21 1 1 (, 0 , , , ) (1 ) nn n n jn j j dn q q q q q −− − − − = = + ∏ (37) and ( ) () 21 2 1 2 2 22 22 22 2 0 ; (, , , , ) (, 0 , , , ) . ; n j m m n nj j n qq dn m q q q q dn q q q qq + ⎛⎞ − ⎜⎟ ⎝⎠ + = = ∏ (38) The right-hand side can be simplified: () () [] [] [] [] [] [ ] [ ] [ ] [ ] [] [] [ ] [ ] [ ] [] [ ] [] [ ] [ ] 21 2 11 1 1 22 2 00 0 0 1 22 1 0 1 22 1 0 ; 22 1 22 1 ! 22 2 2 4 422 1 3 2 1 ; 22 1 ! 21 ! 1 3 2 1 ( 1 ) ( 1 ) ( 1 ) 22 2 1 . 13 2 1 ( 1 ) ( 1 ) ( 1 ) j mm n m n nj jj i j n m nj j m nj j qq ji jn jin n j j qq nj nj j q q q nj n j jq q q + −− − − + == = = − +− = − +− = ++ + − == ++ + − − +− = +− − + + + ++ − = −+ + + ∏∏ ∏ ∏ ∏ ∏     13 Observe that [ ] [ ] [ ] [] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [] 12 2 2 4 2 12 2 (1 ) (1 ) . 13 2 1 2 ! j jj j j jj j qq jj ++ ++ == + + − … …   This implies [ ] [ ] [] [ ] [ ] [ ] [ ] [] [ ] () 2 1 1 22 1 22 1 0 0 1 1 12 2 1 1 0 0 21 22 2 1 22 2 1 ( 1 ) ( 1 ) ( 1 ) 1 3 2 1 (1 ) (1 ) (1 ) (1 ) (1 ) (1 ) [ 1] [ 2 ] 22 2 1 2 1 (1 ) (1 ) (1 ) [ 1] [ 2 ] ;) j m m nj nj j j m m jj n j j j j n nj n j nj n j q q q jq q q q q q j j nj n j qq q j j qq − − +− +− = = − − ++ + − + = = − ++ − ++ − + + + = −+ + + + + + + ++ − == ++ + + − ∏∏ ∏∏        [] 1 1 . [] j i nj i ji = ++ − + ∏ Finally we get () [] 1 2 2 22 22 1 01 21 21 1 (, , , , ) (, 0 , , , ) . [] ;) n j m m j ji n nj i d nmq qq q d n q qq ji qq ⎛⎞ − ⎜⎟ ⎝⎠ + == − ++ − = + − ∏∏ (39) We note also that 22 1 (0 , , , ) 1 Tq q q q = + (40) and 22 1 (, , , ) (1 ) (1 ) n nn q Tn q q q qq + = ++ (41) for 0 n > . If we let 1 q → we obtain 2 1 () . 4 n n cn n ⎛⎞ → ⎜⎟ ⎝⎠ This implies that the Hankel dete rminants of the central binomial coefficients are given by 1 1 1 01 ,0 22 2 21 det 2 . n j m nm ji ij ij m nj i ij m ji − − −+ == = ++ ⎛⎞ ⎛⎞ + +− = ⎜⎟ ⎜⎟ ++ + ⎝⎠ ⎝⎠ ∏∏ (42) Similar determinants have also been obtained with other m ethods by Krattenthaler in [6], especially Theorem 26. From (40) and (41) and (7) we get for () cn the values 1 () 2 sn = , 1 (0) 8 t = and 2 1 () 4 tn = for 0. n > Therefore we get for 2 (, 0 ) 4 ( ) n n an cn n ⎛⎞ == ⎜⎟ ⎝⎠ using (11) that 1 () 4 2 2 sn = ⋅= , 2 1 (0) 4 2 8 t =⋅ = and 2 2 1 () 4 1 4 tn =⋅ = for 0. n > 14 This leads to the table [5], A094527, 1 21 641 20 15 6 1 70 56 28 8 1 Here it is easily verified that 2 (, ) . n an k nk ⎛⎞ = ⎜⎟ − ⎝⎠ Observe that 32 22 4 3 2 42 1 2 1 21 (; ) 1 1 (1 ) ( 1, , , ) ( , , , ) . (; ) ( 1 ) (1 ) n n n j n j n qq q qcn q q q cn q q q n qq q q + = + ⎡⎤ + ++ = = = ⎢⎥ + ⎣⎦ + ∏ Therefore () ( ) 2 43 2 2 2 d e t ( ,, ,) ( 1 ) d e t ( 1 ,, ,) . n ci j mq q q q ci j m q qq ⎛⎞ ⎜⎟ ⎝⎠ ++ = + ++ + This implies that the Ha nkel determinants for the binom ial coefficients 21 n n + ⎛⎞ ⎜⎟ ⎝⎠ are 11 ,0 ,0 22 2 1 22 2 2 1 det det . 1 2 nn n ij ij ij m ij m ij m ij m −− == ++ + ++ + ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ = ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ++ ++ + ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ (43) 6) Another interesting case is the f ollowing q − analogue of the Catalan numbers which has first been studied by George Andrews [2]: [] 2 2 2 2 1 1 4 1 2 11 (1 ) ( 1 ) . 2 () ( , 1 (1 ) ( 1 , ) ) 1 , nn n nj q j c n q qq n nc n n qq n qq q + = ⎡⎤ ⎡⎤ + ⎢⎥ −+ = ⎢⎥ ⎢⎥ + ⎣⎦ ++ + ⎣⎦ == ∏ Here we get (1 ) ( 4 5 ) 6 42 23 12 2 2 0 (, 0 , , , ) (1 ) (1 ) nn n n nj n j j q dn q q q qq −− − − +− − = = ++ ∏ (44) and with a similar argument as abov e () [] [] 1 2 2 42 42 1 01 2 2 1 (, , , , ) (, 0 , , , ) . ;) n j m m j ji n nj i d nmq qq q d n q qq ji qq ⎛⎞ − ⎜⎟ ⎝⎠ + == ++ = + − ∏∏ (45) Letting 1 q → and simplifying we obtain again (29). 15 Some identities connected with this method As a generalization of (9) we show, that with ( , ) An k defined by (5) the following identity holds: [] 1 00 (1 ) ( 2 , 2 ) ( 2 ) 0 . nk k kj An k T j n − == −= = ∑∏ (46) For the proof let (1 ) 0 T −= and observe that (2 , 2) (2 , 2 2) 0 An An n − =+ = . Then we get () ( ) () 11 00 1 1 0 0 1 00 (1 ) ( 2 2 , 2 ) ( 2 ) ( 1 ) (2 , 2 2 (2 1 ) (2 ) (2 , 2 ) (2 ) (2 1 ) (2 , 2 2 ) (2 ) ( 1 ) ( 2 , 2) ( 2) ( 2 1 ) ( 2) ( 2) ( 2 1 ) 0 . nk k kj n k k k j nk k kj An k T j An k T k T k An k T k T k a n k T j A n k T j T k T kT kT k +− == + − = = − == −+ =− − + − + + + + =− − + − − − = ∑∏ ∑∏ ∑∏ If we choose () () ; (, , , ) ; n n bq cna b q aq = then formula (46) is equivalent to () () [] 2 2 0 1 1 (1 ) 1 0 . ; k n kk k k n n qq a n k qaq ⎛⎞ ⎜⎟ ⎝⎠ = + ⎡⎤ −− = = ⎢⎥ ⎣⎦ ∑ (47) For the left-hand side of (46) is () () () () () () 2 21 0 2 21 1 0 1 ;) ;) (1 ) ;) ;) ;) (1 ) ( 1 ) ;) k k n nk k k kk k nk k k n kk n k k n qbq n bq q k qa q q a q n bq qq a k qa q ⎛⎞ ⎜⎟ − ⎝⎠ − = − ⎛⎞ ⎜⎟ − ⎝⎠ − = + ⎡⎤ − ⎢⎥ ⎣⎦ ⎡⎤ =− − ⎢⎥ ⎣⎦ ∑ ∑ Now replace aq a → and factor through ( ; ) . n bq A companion to (47) is () () () 2 2 2 0 1 1; (1 ) . ;; k k n n k k nn q n qa q k qaq q aq ⎛⎞ ⎜⎟ ⎝⎠ = + − ⎡⎤ − = ⎢⎥ ⎣⎦ ∑ (48) 16 This is easily proved using Zeilber ger’s algorithm. E.g. qZeil gives qZeil @ q ^ Binomial @ k, 2 D qBinomial @ n, k , q DH 1 − aq ^ H 2 k LL qPochhammer @ qa , q ^ 2 , n Dê qPochhammer @ q ^ k a ,q ,n + 1 D , 8 k, 0 , n < ,n ,1 D SUM @ n D  H 1 + q − 1 + n L SUM @ − 1 + n D The proof depends on the following certificate Cert @D q − k + n H − 1 + aq k LH − 1 + aq 1 + 2 k LH − q k + q n L       H − 1 + aq 2k LH − 1 + q n LH q − a q 2n L Identity (48) is equivalen t with ( ) ( ) () 1 2 00 ;1 ; (2 , 2 , , , ) (2 , , , ) . ; nk nn kj n bq q An k a b q T j a b q aq − == − = ∑∏ (49) For the left-hand side is () () ( ) () () () () () () ( ) () 2 2 00 21 1 0 1 22 11 2 2 ; (, , , ) ; ;( 1 ) ; ;; ;; ;1 ; . ; kk kk kk kk k nn kk nk k k kk nk k n n n nn k n k qbq nn cn k q aqb q kk qa q bq q a n k bq bq qq qa q qa q bq q q qa q a q ⎛⎞ ⎛⎞ ⎜ − ⎟⎜ ⎟ ⎝⎠ ⎝⎠ −− ⎛⎞ ⎜⎟ ⎝⎠ == − − − = + ⎡⎤ ⎡⎤ − ⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦ − ⎡⎤ = ⎢⎥ ⎣ − ⎦ = = ∑∑ ∑ An interesting special case is the follow ing q − analogue of (10) [] [] () () 2 1 42 4 2 00 21 1 0 22 22 2 1 (2 , 2 , , , ) (2 , , , ) 2 21 1 11 ; ; 2 21 2 (2 , 0 , , , ). 11 (1 ) nk kj kk k n nk k nk nk n nn j j An k q q q T j q q q n k qq nk nk q q q q q n An q q q n qq q − == +− ++ = −+ = + ⎡⎤ + = ⎢⎥ − ++ + − − ⎣⎦ ⎡⎤ == ⎢⎥ ++ ⎣⎦ + ∑∏ ∑ ∏ Another special case of (49) is ( ) () 2 2 11 22 22 21 2 00 0 1; 1 (2 , 2 , , , ) (2 , , , ) . 1 ; j nk n n j kj j n q q An k q q q T j q q q q qq −− + == = − + == + − ∑∏ ∏ 17 References [1] M. 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