Recurrence relations for powers of q-Fibonacci polynomials

1 Recurrence relations for powers of q-Fibonacci polynomials Johann Cigler Fakultät für Mathematik Universität Wien A-1090 Wien, Nordbergstraße 15 johann.cigler@univie.ac.at Abstract We derive some q − analogs of Euler-Cassini-type identi ties and of recurrenc e formulas for powers of Fibonacci polynomials. 1. Introduction The Fibonacci numbers n F are defined by the recurrence relation 12 nn n FF F − − = + (1.1) with initial values 0 0 F = and 1 1. F = The powers k n F , 1 , 2, 3 , , k = " satisfy the recurrence relation 1 1 2 0 1 (1 ) 0 , j k k nj j k F j + ⎛⎞ + ⎜⎟ ⎝⎠ − = + − = ∑ (1.2) where 1 0 1 k ni i k i i F n k F − − = = = ∏ ∏ is a so called fibonomial coefficient. E.g. the squares of the Fibonacc i numbers satisfy the recurrence 22 2 2 12 3 22 0 . nn n n FF F F −− − − −+ = The triangle of Fibonomial co efficients ( see A010048 or A055870 in the On-Line Encyclopedia of Integer Sequences [7] ) begins with 1 1 1 1 1 1 1 2 2 1 1 3 6 3 1 1 5 15 15 5 1 The Fibonacci polynomials ( , ) n Fx s are defined by the recurrence relation 12 (,) (,) (,) nn n Fx s x F x s s F x s −− = + (1.3) 2 with initial values 0 (,) 0 Fx s = and 1 (,) 1 . Fx s = The first terms of this sequence are 23 4 2 2 0, 1 , , , 2 , 3 , xx sx s xx s x s ++ + + " . The powers (,) k n Fx s , 1 , 2, 3 , , k = " satisfy the recurrence relation 1 1 22 0 1 (1 ) ( , ) ( , ) 0 , jj k k nj j k sx s F x s j + ⎛⎞ ⎛ ⎞ + ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ − = + −= ∑ (1.4) where the (polynomial-) fibonomial coefficients are defined by 1 0 1 (,) (,) . (,) k ni i k i i Fx s n xs k Fx s − − = = = ∏ ∏ (1.5) E.g. for 2 k = we get the recurrence relation 22 2 2 2 3 2 12 3 (,) ( ) (,) ( ) (,) (,) 0 . nn n n F xs x s F xs s x sF xs sF xs −− − −+ − + + = The simplest proof of these facts depends on the Binet formula (,) , nn n Fx s α β α β − = − (1.6) where 22 44 ,. 22 x xs x xs αβ ++ −+ == (1.7) From (1.6) it is clear tha t ( , ) k n Fx s is a linear combination of () kj n j n α β − , 0 j k ≤≤ . Let U be the shift operator ( ) ( 1 ). Uh n h n =− The sequences ( ) () 0 kj n j n n αβ − ≥ satisfy the recurrence relation ( ) ( ) () 10 . kj j kj n j n U αβ α β −− −= Since the operators 1 kj j U α β − − commute we get 0 (1 ) ( , ) 0 . n kj j k n j UF x s αβ − = ⎛⎞ − = ⎜⎟ ⎝⎠ ∏ (1.8) As has been observed by L. Car litz [2] we can now apply the q − binomial theorem (cf. e.g. [4]) 1 2 00 (1 ) ( 1) . k nn jk k jk n qx q x k ⎛⎞ − ⎜⎟ ⎝⎠ == ⎡ ⎤ −= − ⎢ ⎥ ⎣ ⎦ ∏∑ (1.9) Here 11 2 (1 ) (1 ) (1 ) () (1 ) (1 ) (1 ) nn n k k nn qq q q kk qq q −− − ⎡⎤ ⎡⎤ −− − == ⎢⎥ ⎢⎥ −− − ⎣⎦ ⎣⎦ " " denotes a q − binomial coefficient. 3 For q β α = we get 2 (,) . kn k nn x s kk α − ⎡⎤ = ⎢⎥ ⎣⎦ This implies () 2 2 (1 ) 00 0 1 22 2 00 1 (1 ) (1 ( ) ) ( 1) ( , ) 11 ( 1 ) (,) ( 1 ) (,) . j j kk k kj j k j j k j k j j jj j jj j kk jj j jj k UU x s U j kk xs U s xs U jj ββ αβ α α α αα αβ ⎛⎞ ⎜⎟ ⎝⎠ − −+ == = + ⎛⎞ ⎛ ⎞ ⎛⎞ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎝⎠ ⎝ ⎠ ⎝⎠ == + ⎛⎞ ⎛⎞ −= − = − ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ++ =− =− ∏∏ ∑ ∑∑ By applying this operator to ( , ) k n Fx s we get (1.4). 2. Recurrence relations for powe rs of q-Fibonacci polynomials The (Carlitz-) q − Fibonacci polynomials ( , , ) f nx s are defined by 2 (, ,) ( 1 , ,) ( 2 , ,) n f n x s x fn x s q s fn x s − =− + − (2.1) with initial values (0, , ) 0, ( 1 , , ) 1 fx s f x s == (cf. [3],[5]). The first values are 23 2 4 2 3 2 4 2 0, 1 , , , ( ) , ( ) , . x x q s x q sq s x x q sq sq s x q s ++ + + + + + " An explicit expression is 2 12 1 1 (, , ) . kn k k kn nk f nx s q x s k −− ≤− −− ⎡⎤ = ⎢⎥ ⎣⎦ ∑ (2.2) If we change 1 q q → and then 1 n sq s − → we get 1 (, , ) (, , ) ( , , ) . kn f nk x q s f nk x q s f n x s −− −→ − → This implies Remark 1 Each identity ( ) , ,, ( ,, ) , ( 1 ,, ) , ( 2 ,, ) , 0 g xsq f n xs f n xs f n xs − −= " (2.3) is equivalent with ( ) 11 2 , , , ( ,, ) , ( 1 ,, ) , ( 2 ,, ) , 0 . n gx q s q f n x s fn x q s fn x q s −− − −= " (2.4) 4 A special case is the well-known f act that (2.1) is equivalent with 2 (, ,) ( 1 , , ) ( 2 , , ) . f nx s x f n x q s q s f n x q s =− + − (2.5) The definition of the q − Fibonacci polynomials can be extended to all integers such that the recurrence (2.1) remains true. We then get (cf. [5]) 1 2 1 (, , ) (, , ) ( 1 ) . n n n n f nx q s fn x s q s + ⎛⎞ − ⎜⎟ − ⎝⎠ −= − (2.6) The main aim of this paper is the proof of the following q − analog of (1.4) which has been conjectured in [6]: Theorem 1 Define a q-analog of the fi bonomial coefficients by 1 11 (, , ) (,, ) . (, , ) (, , ) k i jk j ji j ii fi x s k xsq j f ix q s f ix q s = − − == = ∏ ∏∏ (2.7) Then the following recurr ence relation holds for all n ∈ ] : 1 (1 ) ( 21 ) 1 22 6 0 1 (1 ) ( , , ) ( , , ) 0 . jj jj j k jk j k sq x s q f n j x q s j + ⎛⎞ ⎛ ⎞ −− + ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ = + −− = ∑ (2.8) By Remark 1 this theorem is equivalent to Corollary 1 1 (1 ) ( 21 ) 1 (1 ) 22 2 6 0 ( 1 ) ( 1 , ,) ( , ,) 0 jj j jj j k n k j s q fibo k x s f n j x s + ⎛ ⎞ ⎛⎞ ⎛⎞ −− + −− ⎜ ⎟ ⎜⎟ ⎜⎟ ⎝ ⎠ ⎝⎠ ⎝⎠ = −+ − = ∑ (2.9) with 11 1 11 (, , ) 1 (1 , , ) ( , , ) . (, , ) (, , ) k ni n i jk j nj n ij ii fi x q s k fibo k x s x q s q j f ix q s f ix q s − −− = − −− − == + += = ∏ ∏∏ (2.10) 5 Since there is no q − analogue of the Binet formula we use a q − analog of an extension of the Cassini - Euler identity fo r the proof of Theorem 1. Let 1 (, ,) ( , ,) k i f ac k x s f i x s = = ∏ (2.11) and 1 (, ,, ) ( , ,) . k i f ac k x s m f im x s = = ∏ (2.12) Then the following theorem holds: Theorem 2 For all n ∈ ] and , m ∈ A` () 2 11 () ( ) 23 ,0 0 11 1 1 1 (2 ) ( 3 2 ) 22 3 3 2 2 3 44 det ( , , ) ( ) (, , , ) (, , kk k nk m k jk ij j kn k k k k mk m k mn mj n j k fn m i j x q s s j q fac k j x q s m fac k j x q s ++ ⎛⎞ ⎛⎞ −+ + ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ = = ++ + + + ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ −+ ++ − − ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ + ⎛⎞ +− = − ⎜⎟ ⎝⎠ −− ∏ AA A A A A A 11 00 ,) . kk jj −− == ∏∏ A (2.13) First we prove the special case 1: k = Lemma 1 For all n ∈ ] and , m ∈ A` 22 (, ,) ( , , ) det ( ) , , ) ( , , ). (, , ) ( , , ) ( n nn fn x s fn x q s sq x s f m x q s fn m x s fn m x q s f ⎛⎞ ⎛⎞ − ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ ⎛⎞ − =− ⎜⎟ ++ − ⎝⎠ A A A A A A A (2.14) Various versions of this lemma are well known (cf. [1] and [5]). Since 2 ( , ,) ( 1 , ,) ( 2 , ,) nm f nm x s x f nm x s q s f nm x s +− += + − + + − and 2 ( , ,) ( 1 , , ) ( 2 , ,) nm f n m x q s x fn m x q s q s fn m x q s +− +− = +− − + +− − AA A AA A we see that (, ,) ( , , ) () : d e t (, , ) ( , , ) fn x s fn x q s gm f nm x s f nm x q s ⎛⎞ − = ⎜⎟ ++ − ⎝⎠ A A A A satisfies 2 () ( 1 ) ( 2 ) mn gm x gm q s gm +− =− + − and ( 0) 0. g = Therefore ( ) ( , , ) n gm c f mx qs = for some constant . c To compute c we set . mn = − This gives () ( , , ) ( , , ) (, , ) n gn f n x s f x q s c f n x q s −= − = − A A or 2 2 11 22 11 ( ,, ) (1 ) ( ,, ) ( ,, ) ( 1 ) ( ,, ) n n nn n f nx s q s f x s c f nx q s c q s f nxs ++ ⎛⎞ ⎛ ⎞ −− ⎜⎟ ⎜ ⎟ −− − − ⎝⎠ ⎝ ⎠ −= − = − A A AA A and therefore () ( 1 ) 22 2 ( ) () ( , , ) ( , , ) () ( , , ) ( , , ) . n nn nn n n g m sq fx s f m x q s sq fx s f m x q s ⎛⎞ ⎛⎞ −+ − − ⎜⎟ ⎜⎟ −− ⎝⎠ ⎝⎠ =− =− A AA AA AA 6 As a special case we get Corollary 2 For each k ∈ ` there is a representation o f (, , ) k f nk x q s − as a linear combination of (, ,) f n x s and (1 , , ) : f nx q s − () 1 ( ,, ) ( 1 ,, ) ( ,, ) ( ,, ) ( 1 ,, ) () k f n k x q s fk x q sfn x s fk x sfn x q s vk −= − − − (2.15) with 2 1 () ( 1 ) . k kk vk q s ⎛⎞ ⎜⎟ − ⎝⎠ =− (2.16) Proof of Theorem 2 Using (2.15) we get () () () ( ) () ( ) ,0 1 ,0 1 2 2 1 , (2 ) 22 2 2 det ( , , ) d e t ( ) ( 1 , , ) ( , ,) ( , ,) ( 1 , , ) 1 (1 ) d e t ( , ,) ( 1 , , ) k jk ij k k ij k k k jj k k i k k fn m i j x q s v j f j x q sfn m i x s f j x sf n m i x q s af n m i x s b f n m i x q s sq = − = + ⎛⎞ ⎜⎟ ⎝⎠ ⎛⎞ + ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ++ − ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ +− =− + − + − =− + + + − A A AA A " A A AA A () ( ) 2 0 11 (1 ) ( 2 3 ) (2 ) 22 12 ,0 (1 ) d e t ( , ,) ( 1 , , ) k j kk kk k k kk k jj ij sq a f n m i x s b f n m i x q s = ++ ⎛⎞ ⎛⎞ ++ − −− ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ = =− + + + − AA A AA (2.17) with ( 1 , , ), ( , , ). jj af j x q s b f j x s =− = − AA Since the determ inant is multilin ear and alternating we get () ( ) () ( ) () ( ) ( ) () ,0 0 ,0 () () 0 () 0 det ( , , ) ( 1 , , ) det ( , , ) ( 1 , , ) det ( , , ) ( 1 , , ) det ( , , ) ( k k jj ij k k hk h jj h ij k jk j jj j k j jj j af n m i x s b f n m i x q s k af n m i x s b f n m i x q s h k af n m i x s b f n m i x q s j k af n m i x s b f n j ππ π π = − = = − = = ++ + − ⎛⎞ ⎛⎞ =+ + − ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎛⎞ =+ + − ⎜⎟ ⎝⎠ ⎛⎞ =+ + ⎜⎟ ⎝⎠ ∑ ∏∑ ∏ () ( ) () 1, , ) kj mi x qs π π − − ∑ () ( ) () () ( ) () () () ( ) () () () () 00 () () 00 0 det ( , , ) ( 1 , , ) sgn( ) det ( , , ) ( 1 , , ) det det ( , , ) ( 1 , , ) kk jk j jk j jj jj kk jk j jk j jj jj k j jk j ii j k a b fn m i x s fn m i x q s j k a b fn m i x s fn m i x q s j k a b fn m i x s fn m i x q s j ππ ππ π ππ π π − − == − − == − = ⎛⎞ =+ + − ⎜⎟ ⎝⎠ ⎛⎞ =+ + − ⎜⎟ ⎝⎠ ⎛⎞ =+ + − ⎜⎟ ⎝⎠ ∏∑ ∏ ∏∑ ∏ ∏ () . kj − 7 Now we need Lemma 2 For m ∈ ` ( ) 2 11 1 11 1 1 (1 ) 22 3 2 4 23 2 3 1 0 (, ,, ) d e t ( , , ) ( 1 , , ) (1 ) (, , , ) . jk j kn k m k kk k k mn m nm n m k mj n j Dn ms k f n m i xs f n m i xq s sq fac k j x q s m − ++ + ⎛⎞ ++ + + ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞⎛⎞ ⎛⎞ ++ + +− + ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠⎝⎠ ⎝⎠ ⎝ ⎠ − + = =+ + − =− − ∏ (2.18) Proof Using formula 2 (, , ) ( 1 , , ) ( 2 , , ) f nm i x s x f nm i x q s q s f nm i x q s += + − + + − we get as above 11 1 22 2 (, ,, ) ( ) ( 0 , , , ) . kk n n n Dn ms k s q D m qs k ++ + ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ =− (2.19) For () ( ) () ( ) () 2 2 1 1 2 2 2 1 2 (, ,, ) det ( 1 , , ) ( 1 , , ) ( 2 , , ) det ( 1 , , ) ( 2, , ) () d e t ( 2 , , ) ( 1 , , ) ( ) ( 1 , , , ) () j kj j kj k k jk j k n Dn ms k fn m i x q s x fn m i x q s q s fn m i x q s fn m i x q s q s fn m i x q s qs f n mi x q s f n mi x qs qs D n m qs k s − − + + ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ + ⎛⎞ ⎜⎟ ⎝⎠ = +− +− + +− =+ − + − = + − + −= −− =− 11 22 (0 , , , ). kn n qD m q s k ++ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ Finally we expand ( 0, , , ) Dm s k with respect to the first column and get () 12 2 12 2 1 ( , ,) ( 1 , , ) ( , ,) ( 1 , , ) ( , ,) (2 , , ) (2 1 , , ) (2 , , ) (2 1 , , ) (2 , , ) det (, , ) ( 1 , , ) ( (0 , , , ) de t ( , , ) ( 1 , , ) (1 , , ) kk k kk k k jk j k fm x sfm x q s f m x s fm x q s fm x s fm x s fm x q s fm x s fm x q s fm x s fk m x sfk m x q s fk Dm s k f m i x s f m i x q s fx q s −− −− − − −− −− − =− =− " " "" " " 22 ,, ) ( 1 ,, ) ( ,, ) ( 1 ,, ) ( ,, ) ( 2 ,, ) ( ,, ) ( , , , 1 ) . kk k m x s f km x qs f km x s fx q s f m x s f m x s f k m x s D m m s k − − ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ =− − " Thus we have ( 0 , , ,) ( 1 ,, ) ( ,, ) ( 2 ,, ) ( ,, ) ( , , , 1 ) . k D ms k f xq s f m xs f mxs f k mxs Dm ms k =− − " (2.20) 8 For 1 k = we get from (2.14) that 2 1 (, ,, 1 ) ( 1 ) ( , , ) . n nn n Dn ms s q f m xqs ⎛⎞ ⎜⎟ − ⎝⎠ =− Therefore Lemma 2 is true for 1. k = The general case follows by using (2.19) and (2.20) 11 1 22 2 11 1 22 2 23 ( , , , ) ( ) ( ) ( , ,) ( 2 , ,) ( , ,) ( , ,,1 ) ( ) ( ) ( ,, ) ( 2 ,, ) ( ,, ) (1 ) kk n n nk n n n n kk n n nk n n n kk m Dnms k s q qs f m xqs f m xqs f k mxqs Dm mqsk s q qs f m xqs f m xqs f k m xqs ++ + ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ ⎝⎠ ++ + ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ ⎝⎠ ⎛⎞ ⎛⎞ + ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ =− − =− − " " 22 2 (1 ) 22 3 2 4 23 0 11 11 11 (1 ) 22 3 23 23 () ( 1 , , , ) (1 ) km k m k kk k mm mm m n mj m n j kn k kk kk nm nm n m qs q f a ck j xq sm sq ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞ − ++ + −+ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟ ++ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ = ++ ++ + + ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ + +− + ⎜⎟ ⎜ ⎟ ⎜ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ −− =− ∏ 2 11 1 32 4 0 (, , , ) . km k k m mj n j fac k j x q s m ++ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ − ++ ⎜⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ = − ∏ A special case is Lemma 3 () () ( ) 2 ,0 11 1 1 1 32 3 3 2 4 1 0 ( 0 ,,, ) d e t d e t ( 1 , , ) ( , , ) (1 ) (, , , ) . k kj jk j j ii ij kk k k k k j j Ds k a b f i x q s f i x s sq fac k j x q s − − = ++ + + + ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ −+ + ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ − = == − − =− − ∏ A AA A A AA A A (2.21) With the use of these lemmas we get () () ( ) 2 2 ,0 11 (1 ) ( 2 3 ) (2 ) 22 12 ,0 11 (1 ) ( 2 3 ) (2 ) 22 12 0 det ( , , ) (1 ) d e t ( , ,) ( 1 , , ) (1 ) ( , , , k jk ij kk kk k k kk k jj ij kk kk k k kk j fn m i j x q s sq a f n m i x s b f n m i x q s k sq D n m s j = ++ ⎛⎞ ⎛⎞ ++ − −− ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ = ++ ⎛⎞ ⎛⎞ ++ − −− ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ = +− =− + + + − ⎛⎞ =− ⎜⎟ ⎝⎠ ∏ A AA A AA AA A AA A )( 0 , , , ) kD s k A 2 2 11 1 11 1 1 1 1 (1 ) ( 2 3 ) (2 ) ( 1 ) 22 3 2 4 22 2 3 2 3 12 0 (1 ) (1 ) (, , kn k m k kk k k k k kk k k mn m kk n m n m j mj n k sq s q j fac k j x q s ++ + ⎛⎞ ++ + + + + ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞⎛⎞ ⎛⎞ ++ − ++ + −− + − + ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ − ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎝⎠ ⎝⎠ ⎝ ⎠ = + ⎛⎞ =− − ⎜⎟ ⎝⎠ − ∏ AA A AA 2 11 1 1 1 1 32 3 3 2 4 0 1 0 ,) ( 1 ) (, , , ) kk k k k k j k j j msq fac k j x q s ++ + + + ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ − −+ + ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ = − = − − ∏ ∏ A AA A A A 9 2 11 1 1 1 1 1 (2 ) ( 3 2 ) () ( ) 23 2 2 3 3 2 2 3 44 0 11 00 () (, , , ) (, , , ) . kk k n k k k k mk m k k nk m m n j kk mj n j jj k sq j fac k j x q s m fac k j x q s ++ + + + + + ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ − + −+ + + + − − ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ = −− + == ⎛⎞ =− ⎜⎟ ⎝⎠ −− ∏ ∏∏ A A AA A A Thus Theorem 2 is proved. We will also need some m odifications of these results. Let () ( , , ,,) d e t ( ( [ ] ) ,, ) ( ( [ ] ) 1 ,, ) , jk j d n m s k j fn m i i j x s fn m i i j x q s − =+ + ≥ + + ≥ − (2.22) where [ ] P denotes the Iverson symbol, i.e. [ ] 1 P = if property P is true and [ ] 0 P = else. Then (, ,, , 0 ) ( , ,, ) . d n msk D n mms k = + (2.23) From (2.18) we get 11 (( 3 ) 22 4 ( 0 ,, , , 0 ) ( 1 ) ( , , , ) ( 0 ,, , 1 , 0 ) . kk km km m k mm k mm d m sk s q f a ck xq smd m q sk ++ ⎛⎞⎛⎞ ++ − − ⎜⎟⎜⎟ ⎝⎠⎝⎠ =− − (2.24) Furthermore we get in the same way as above for 0 j > 1 22 2 (1 , , , ) (0 , , , , ) ( ) ( 0 , , , 1 , 1 ). (, , ) kk m m km fac k x s m dm s k j s s q dm q s k j fj m x s + ⎛⎞ ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜⎟ ⎜ ⎟ − ⎝⎠ ⎝⎠ ⎝ ⎠ + =− − − (2.25) Proof of Theorem 1 The above argument implies that ( ) 1 ,0 det ( , , ) 0. k jk ij fn i j x q s + = + −= (2.26) If we denote by j A the matrix obtained by crossi ng out the first row and the j − th column of () 1 ,0 (, , ) , k jk ij fn i j x q s + = +− we get () 1 0 (, , ) ( 1 ) d e t 0 k jk j j j fn j x q s A + = −− = ∑ or ( ) () 1 0 0 det (, , ) ( 1 ) 0 . det k j jk j j A fn j x q s A + = −− = ∑ (2.27) To compute () det j A we use the same m ethod as in Theorem 1. We get () () () ,0 () d e t d e t ( , ) ( ,, ) ( , ) ( 1 ,, ) , (0) ( 1 ) ( 1 ) k k k jh h ih vj A a js f n i xs b js f n i xq s vv v k = ⎛⎞ =+ + + − ⎜⎟ + ⎝⎠ " where 10 ( ,) ( 1 , , ) , (,) (, ,) hh aj s f h x q s b j s f h x s =− = − for hj < and (, ) (, , ) , (, ) ( 1 , ,) hh aj s f h x q s b j s f h x s == − + for . hj ≥ Therefore ( ) det ( , ) ( , ) ( 0 , 1 , , , ). ik i hh aj s b j s d s k j − = (2.28) By (2.24) we have 1 22 2 (0 ,1 , , , 0) ( 1 ) ( , , ) (0 ,1 , , 1 , 0). kk k d s k s q fac k x qs d qs k + ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ ⎝ ⎠ =− − For 0 j > we get from (2.25) 22 (1 , , ) (0 ,1 , , , ) ( ) (0 ,1 , , 1 , 1 ) . (, ,) kk k fac k x s ds k j s s q d q s k j fj x s ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ − ⎝⎠ ⎝⎠ + =− − − (2.29) Therefore ( 0 , 1 , ,,) ( 1 ,, ) ( 0 , 1 , , 1 , 1 ) () . ( 0 , 1 , ,, 0 ) ( ,, ) ( ,, ) ( 0 , 1 , , 1 , 0 ) k ds k j f a c k x s dq s k j s d s k f j x s fac k x qs d qs k − + −− =− − (2.30) This implies 11 00 1 2 0 () () 2 2 1 (0 ,1 , , , ) (0 ,1 , , , 0) () ( , , ) (0 ,1 , , , 0) (0 ,1 , , , 0 ) 1 () ( , , ) . jj ii j i j ki i ki j j j ki kj k ds k j dq s k j sq x s q j ds k dq s k j k sq x s q j −− == − = −− − − ⎛⎞ ⎛⎞ −+ −+ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ∑∑ + − =− − ∑ + =− Therefore we get () () 1 2 0 1 2 0 2 22 0 1 22 det 1 () ( 0 , 1 ,, , ) (1 ) (1 ) (1 ) ( ) ( , , ) det ( 0) (0,1 , , , 0) 1 (1 ) ( , , ) . j i j i j jj k ki kk j j jj j k j k j jj i A k vj d s kj sq s q x s q j Av d s k k sq x s q j − = − = ⎛⎞ ⎛⎞ ⎛⎞ −+ −+ ⎜⎟ ⎜⎟ ⎜⎟ + ⎝⎠ ⎝⎠ ⎝⎠ + ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ ∑ + ⎛⎞ −= − = − − ⎜⎟ ⎝⎠ ∑ + =− Thus Theorem 1 is proved. 11 If we use the fact that () 1 ,0 det ( ( ), , ) 0 , k jk ij fn i j x q s + = +− = A A we get in the same way that () () 1 0 0 det (, , ) ( 1 ) 0 , det k j jk j j B fn j x q s B + = −− = ∑ A A where we denote by j B the matrix obtained by crossi ng out the first row and the j − th column of () 1 ,0 (( ) , , ) . k jk ij fn i j x q s + = +− A A Here we have () () 2 0 det () ( 0 , , , , ) ( 0 , , , , ) (1 ) (1 ) (1 ) . d e t ( 0 ) ( 0 ,,, , 0 ) ( 0 ,,, , 0 ) j k k j jj j k j k j B v j ds k j ds k j sq Bv d s k d s k ⎛⎞ ⎜⎟ + ⎝⎠ ⎛⎞ −= − = − ⎜⎟ ⎝⎠ A AA AA A AA If we define 1 () 11 (, , ) (, , ,) (, , ) (, , ) k i jk j ji j ii fi x s k xsq j f ix q s f ix q s = − − == = ∏ ∏∏ AA A A AA (2.31) we get from (2.24) and (2.25) 1 22 2 11 (( 3) 22 4 2 (1 , , , ) () ( 0 , , , 1 , 1 ) ( 0 ,,, , ) (, , ) ( 0 ,,, , 0 ) (1 ) ( , , ,)( 0 , , , 1 , 0 ) (1 ) ( kk k kk kk k k k k k fac k x s ss q d q s k j ds k j fj x s ds k s q f a ck xq s d q sk fac q s + ⎛⎞ ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜⎟ ⎜ ⎟ − ⎝⎠ ⎝⎠ ⎝ ⎠ ++ ⎛⎞ ⎛⎞ ++ − − ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎛⎞ − ⎜⎟ ⎝⎠ + −− − = −− − = A A A AA A AA AA A A A A A A A A AA 1 2 0 () 22 22 1, , , ) ( 0 , , , 1, 1 ) ( , , ) ( , ,, ) ( 0 , ,, 1 , 0 ) 1 (1 ) ( , , ,) . j i j jj kj i k i kj kj kx s d q s k j fj x s f a c k x q s d q s k k sq x s q j − = ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ −− − − −− ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ +− − − ∑ + =− A AA A A AA A A AA AA A A Therefore () () 1 2 0 () 22 22 2 0 (4 1 ) 3 22 6 det 1 (1 ) (1 ) (1 ) ( , , , ) det 1 (1 ) ( ) ( , , , ) . j i j jj j kj i k i kk j k j j j j kj kj jj j j B k sq s q x s q j B k qs x s q j − = ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ −− − − −− ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ + ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎛⎞ ⎛⎞ +− + ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ∑ + −= − − + =− A A A AA A A AA A AA A A Thus we get 12 Theorem 3 For , k ∈ A` the following recurrence relation holds: (4 1 ) 3 1 22 6 0 1 (1 ) ( ) ( , ,,) ( , , ) 0 . jj j k j jk j k qs x s q f n j x q s j ⎛⎞ ⎛⎞ +− + + ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ = + −− = ∑ A AA A AA (2.32) For the special case 1 k = this reduces to (3 1 ) 2 2 (2 , , ) ( , , ) (, ,) ( , , ) ( 1 ) ( 2 , , ) 0 , (, , ) (, , ) fx s f x s fn x s fn x q s q s fn x q s fx q s fx q s − −− + − − = AA AA A A A A AA AA AA (2.33) which has already been proved in [6]. 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