Recurrence relations for powers of q-Fibonacci polynomials Johann Cigler Fakultät für Mathematik Universität Wien A-1090 Wien, Nordbergstraße 15
[email protected]
Abstract We derive some q − analogs of Euler-Cassini-type identities and of recurrence formulas for powers of Fibonacci polynomials.
1. Introduction The Fibonacci numbers Fn are defined by the recurrence relation Fn = Fn −1 + Fn −2 with initial values F0 = 0 and F1 = 1.
(1.1)
The powers Fnk , k = 1, 2,3,", satisfy the recurrence relation k +1
∑ ( −1)
⎛ j +1⎞ ⎜ ⎟ ⎝ 2 ⎠
k +1 j
j =0
Fnk− j = 0,
(1.2)
k −1
n where = k
∏F i =0 k
n −i
∏F i =1
is a so called fibonomial coefficient.
i
E.g. the squares of the Fibonacci numbers satisfy the recurrence Fn2 − 2 Fn2−1 − 2 Fn2−2 + Fn2−3 = 0.
The triangle of Fibonomial coefficients ( see A010048 or A055870 in the On-Line Encyclopedia of Integer Sequences [7] ) begins with 1 1 1 1 1 1
1 2
3 5
1 1 2 6
15
1 3
15
1 5
1
The Fibonacci polynomials Fn ( x, s ) are defined by the recurrence relation Fn ( x, s ) = xFn −1 ( x, s ) + sFn −2 ( x, s )
(1.3)
1
with initial values F0 ( x, s ) = 0 and F1 ( x, s ) = 1. The first terms of this sequence are 0,1, x, x 2 + s, x 3 + 2 sx, x 4 + 3sx 2 + s 2 ," . The powers Fnk ( x, s ) , k = 1, 2,3,", satisfy the recurrence relation k +1
∑ ( −1)
⎛ j +1⎞ ⎛ j ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠
s
j =0
k +1 ( x, s ) Fnk− j ( x, s ) = 0, j
(1.4)
where the (polynomial-) fibonomial coefficients are defined by k −1
∏ n ( x, s ) = i =0k k
Fn −i ( x, s )
∏ Fi ( x, s)
.
(1.5)
i =1
E.g. for k = 2 we get the recurrence relation Fn ( x, s )2 − ( x 2 + s ) Fn −1 ( x, s ) 2 − s( x 2 + s ) Fn −2 ( x, s )2 + s 3 Fn −3 ( x, s )2 = 0.
The simplest proof of these facts depends on the Binet formula Fn ( x, s ) =
αn − β n , α −β
(1.6)
where
α=
x + x 2 + 4s x − x2 + 4s ,β = . 2 2
(1.7)
From (1.6) it is clear that Fn ( x, s ) k is a linear combination of α ( k − j ) n β jn , 0 ≤ j ≤ k . Let U
be the shift operator Uh (n ) = h (n − 1). The sequences (α ( k − j ) n β jn )
relation (1 − α k − j β jU )(α ( k − j ) n β jn ) = 0.
n ≥0
satisfy the recurrence
Since the operators 1 − α k − j β jU commute we get ⎛ n ⎞ k− j j k ⎜ ∏ (1 − α β U ) ⎟ Fn ( x, s ) = 0. ⎝ j =0 ⎠
(1.8)
As has been observed by L. Carlitz [2] we can now apply the q − binomial theorem (cf. e.g. [4]) n −1
n
∏ (1 − q x) = ∑ (−1) q j
j =0
k
k =0
⎛k ⎞ ⎜ ⎟ ⎝2⎠
⎡n ⎤ k ⎢k ⎥ x . ⎣ ⎦
(1.9)
⎡n ⎤ ⎡n ⎤ (1 − q n )(1 − q n −1 )" (1 − q n −k −1 ) denotes a q − binomial coefficient. Here ⎢ ⎥ = ⎢ ⎥ ( q) = (1 − q)(1 − q 2 )" (1 − q k ) ⎣k ⎦ ⎣k ⎦
2
For q =
⎡n ⎤ β we get ⎢ ⎥ = α k α ⎣k ⎦
2
− nk
n ( x, s ). k
This implies ⎛ j⎞
k
∏ (1 − α
k− j
j =0
j ⎜ ⎟ k +1 2 k +1 2 ⎛β ⎞ k j ⎛ β ⎞⎝ ⎠ ( x, s )α kjU j β U ) = ∏ (1 − ⎜ ⎟ (α U )) = ∑ ( −1) ⎜ ⎟ α j −( k +1) j j ⎝α ⎠ ⎝α ⎠ j =0 j =0
k +1
k
j
= ∑ ( −1) (αβ ) j
⎛ j⎞ ⎜ ⎟ ⎝ 2⎠
j =0
⎛ j +1⎞ ⎛ j ⎞ ⎟ ⎜ ⎟ 2 ⎠ ⎝ 2⎠
k +1 ⎜ k +1 ( x, s )U j = ∑ ( −1)⎝ j j =0
s
k +1 ( x, s )U j . j
By applying this operator to Fn ( x, s )k we get (1.4). 2. Recurrence relations for powers of q-Fibonacci polynomials
The (Carlitz-) q − Fibonacci polynomials f ( n, x, s ) are defined by f ( n, x, s ) = xf ( n − 1, x, s ) + q n −2 sf ( n − 2, x, s )
(2.1)
with initial values f (0, x, s ) = 0, f (1, x, s ) = 1 (cf. [3],[5]). The first values are 0,1, x, x 2 + qs, x 3 + ( qs + q 2 s ) x, x 4 + ( qs + q 2 s + q3s ) x 2 + q 4 s 2 ,". An explicit expression is f ( n, x , s ) =
⎡ n − 1 − k ⎤ k 2 n −1−2 k k q x s . k ⎥⎦ k ≤ n −1 ⎣
∑⎢
(2.2)
1 and then s → q n −1s we get q f ( n − k , x, q − k s ) → f ( n − k , x, q1−n s ) → f (n, x, s ).
If we change q →
This implies Remark 1 Each identity g ( x, s, q, f ( n, x, s ), f ( n − 1, x, s ), f ( n − 2, x, s ),") = 0
(2.3)
g ( x, q n −1s, q −1 , f ( n, x, s ), f ( n − 1, x, qs ), f ( n − 2, x, q 2 s ),") = 0.
(2.4)
is equivalent with
3
A special case is the well-known fact that (2.1) is equivalent with f ( n, x, s ) = xf ( n − 1, x, qs ) + qsf ( n − 2, x, q 2 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]) f ( − n, x, s ) = ( −1)
n −1
q
⎛ n +1⎞ ⎜ ⎟ ⎝ 2 ⎠
f ( n, x , q − n s ) . sn
(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 fibonomial coefficients by k
k ( x , s, q ) = j
∏ f (i , x , s ) i =1
j
∏ f (i , x , q
j −i
i =1
k− j
s )∏ f (i , x , q s )
.
(2.7)
j
i =1
Then the following recurrence relation holds for all n ∈ ] : k +1
∑ ( −1)
⎛ j +1⎞ ⎛ j ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠
s q
j ( j −1)(2 j −1) 6
j =0
k +1 j
( x, s, q) f ( n − j, x, q j s ) k = 0.
(2.8)
fibo( k + 1, x, s ) f ( n − j, x, s ) k = 0
(2.9)
By Remark 1 this theorem is equivalent to Corollary 1 k +1
∑ (−1)
⎛ j +1⎞ ⎛ j ⎞ ⎛ j ⎞ j ( j −1)(2 j −1) ⎜ ⎟ ⎜ ⎟ ( n −1)⎜ ⎟ − 6 ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 2⎠
s q
j =0
with k
k +1 fibo( k + 1, x, s ) = ( x, q n −1s, q −1 ) = j
∏ f (i , x , q i =1
j
∏ f (i , x , q i =1
n− j
k− j
n −i
s)
s )∏ f (i , x , q
. n −i − j
(2.10)
s)
i =1
4
Since there is no q − analogue of the Binet formula we use a q − analog of an extension of the Cassini - Euler identity for the proof of Theorem 1. Let k
fac( k , x, s ) = ∏ f (i, x, s )
(2.11)
i =1
and k
fac(k , x, s, m) = ∏ f (im, x, s ).
(2.12)
i =1
Then the following theorem holds: Theorem 2 For all n ∈ ] and m, A ∈ `
det ( f ( n + mi − Aj, x, q s ) Aj
)
⎛ k +1⎞ ⎛ k +1⎞ ⎟( n − k A ) + ⎜ ⎟( A + m ) 2 ⎠ ⎝ 3 ⎠
⎜ ⎛k ⎞ = ∏ ⎜ ⎟( − s ) ⎝ j =0 ⎝ j ⎠ k
k k i , j =0 2
q
⎛ k +1⎞⎛ n ⎞ ⎛ k +1⎞ ⎛ k +1⎞ m ( km − 2) ⎛ k +1⎞ ⎛ A ⎞ ⎛ k +1⎞ A (3k + 2) −⎜ ⎜ ⎟⎜ ⎟ + ⎜ ⎟mn + ⎜ ⎟ ⎟ ⎜ ⎟−⎜ ⎟ 4 4 ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 3 ⎠
k −1
∏ fac(k − j, x, q j =0
mj + n
(2.13)
k −1
s, m )∏ fac( k − j, x, q Aj s, A). j =0
First we prove the special case k = 1: Lemma 1 For all n ∈ ] and m, A ∈ ` ⎛ f ( n, x, s ) det ⎜ ⎝ f ( n + m, x , s )
⎛n⎞ ⎛ A⎞
⎜ ⎟−⎜ ⎟ f ( n − A, x, q A s ) ⎞ n −A ⎝ 2 ⎠ ⎝ 2 ⎠ = − ( s ) q f (A, x, s ) f (m, x, q n s ). (2.14) ⎟ f ( n + m − A, x, q A s ) ⎠
Various versions of this lemma are well known (cf. [1] and [5]). Since f ( n + m, x, s ) = xf ( n + m − 1, x, s ) + q n + m −2 sf ( n + m − 2, x, s ) and f ( n + m − A, x, q A s ) = xf ( n + m − 1 − A, x, q A s ) + q n +m −2 sf ( n + m − 2 − A, x, q A s ) we see that ⎛ f ( n, x , s ) f ( n − A, x, q A s ) ⎞ g ( m ) := det ⎜ ⎟ A ⎝ f ( n + m, x, s ) f ( n + m − A, x, q s ) ⎠ satisfies g ( m ) = xg ( m − 1) + q m + n −2 sg ( m − 2) and g (0) = 0. Therefore g ( m) = cf ( m, x, q n s ) for some constant c. To compute c we set m = −n. This gives g ( −n ) = f ( n, x, s ) f ( − A, x, q A s ) = cf ( −n, x, q n s ) or A −1
f ( n, x, s )( −1) q and therefore g (m) = ( − s)
n −A
q
⎛ A +1⎞ 2 ⎜ ⎟−A −A ⎝ 2 ⎠
⎛n⎞ ⎛A⎞ ⎜ ⎟ −⎜ ⎟ ⎝ 2⎠ ⎝2⎠
s f ( A, x, s ) = cf ( −n, x, q s ) = c( −1) n
f ( A, x, s ) f ( m, x, q n s ) = ( − s ) n −A q
n −1
q
( n − A )( A + n −1) 2
⎛ n +1⎞ 2 ⎜ ⎟−n −n ⎝ 2 ⎠
s f ( n, x , s )
f ( A, x, s ) f ( m, x, q n s ). 5
As a special case we get Corollary 2
For each k ∈ ` there is a representation of f ( n − k , x, q k s ) as a linear combination of f ( n, x, s ) and f ( n − 1, x, qs ) : 1 f ( n − k , x, q k s ) = (2.15) ( f (k − 1, x, qs ) f (n, x, s) − f (k , x, s ) f (n − 1, x, qs) ) v(k ) with ⎛k ⎞ ⎜ ⎟ ⎝ 2 ⎠ k −1
v ( k ) = ( −1) q s . k
(2.16)
Proof of Theorem 2
Using (2.15) we get det ( f ( n + mi − Aj, x, q Aj s ) k )
k i , j =0
(
= det ( v ( Aj ) −1 ( f ( Aj − 1, x, qs ) f ( n + mi, x, s ) − f ( Aj, x, s ) f ( n + mi − 1, x, qs ) ) ) = ( −1)
⎛ k +1⎞ k A⎜ ⎟ ⎝ 2 ⎠
1 s
= ( −1)
⎛ k +1⎞ k A⎜ ⎟ ⎝ 2 ⎠
s
⎛ kA ⎞ ⎞ ⎛ k +1⎞ ⎛ ⎛ A ⎞ ⎛ 2 A ⎞ ( k A − 2)⎜ ⎟ ⎜ ⎜ ⎟ + ⎜ ⎟ +"⎜ 2 ⎟ ⎟k ⎝ ⎠⎠ ⎝ 2 ⎠ ⎝ ⎝ 2⎠ ⎝ 2 ⎠
)
k k i , j =0
(
det ( a j f ( n + mi, x, s ) + b j f ( n + mi − 1, x, qs ) )
)
k k
(2.17)
i , j =0
q
⎛ k +1⎞ − ( k A − 2)⎜ ⎟ ⎝ 2 ⎠
q
−
k 2 ( k +1) A (2 k A + A −3) 12
(
det ( a j f ( n + mi, x, s ) + b j f ( n + mi − 1, x, qs ) )
)
k k i , j =0
with a j = f (Aj − 1, x, qs ), b j = − f (Aj, x, s ). Since the determinant is multilinear and alternating we get
(
det ( a j f ( n + mi, x, s ) + b j f ( n + mi − 1, x, qs ) )
)
k k i , j =0 k
⎛ k ⎛k ⎞ h k −h ⎞ = det ⎜ ∑ ⎜ ⎟ ( a j f ( n + mi, x, s ) ) ( b j f ( n + mi − 1, x, qs ) ) ⎟ ⎝ h =0 ⎝ h ⎠ ⎠i , j =0
( (
k k −π ( j ) π ( j) ⎛k ⎞ = ∏ ⎜ ⎟∑ det ( a j f ( n + mi, x, s ) ) ( b j f ( n + mi − 1, x, qs ) ) j =0 ⎝ j ⎠ π k k −π ( j ) π ( j) ⎛k ⎞ = ∏ ⎜ ⎟∑ det ( a j f ( n + mi, x, s ) ) ( b j f ( n + mi − 1, x, qs ) ) j =0 ⎝ j ⎠ π
) )
(
)
k k ⎛k ⎞ π ( j) k −π ( j ) = ∏ ⎜ ⎟∑∏ aπj ( j )bkj −π ( j ) det ( f ( n + mi, x, s ) ) ( f ( n + mi − 1, x, qs ) ) j =0 ⎝ j ⎠ π j =0 k k ⎛k ⎞ j k− j = ∏ ⎜ ⎟∑ sgn(π )∏ aπj ( j )bkj −π ( j ) det ( f ( n + mi, x, s ) ) ( f ( n + mi − 1, x, qs ) ) j =0 ⎝ j ⎠ π j =0
(
k ⎛k ⎞ = ∏ ⎜ ⎟ det ( aij bik − j ) det j =0 ⎝ j ⎠
)
(( f (n + mi, x, s)) ( f (n + mi − 1, x, qs)) ) . j
k− j
6
Now we need Lemma 2
For m ∈ `
D (n, m, s, k ) = det ( f ( n + mi, x, s ) j f ( n + mi − 1, x, qs ) k − j ) = ( −1)
⎛ k +1⎞ ⎛ k +1⎞ ⎛ k +1⎞ ⎛ k +1⎞ ⎜ ⎟n + ⎜ ⎟m ⎜ ⎟( n −1) + ⎜ ⎟m 2 3 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 3 ⎠
s
k −1
∏ fac(k − j, x, q
mj + n
q
⎞ ⎛ k +1⎞ 2 ⎛ k +1⎞⎛ n ⎞ ⎛ k +1⎞⎛ ⎛ m ⎞ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎜ ⎟ + mn ⎟ + ⎜ ⎟m ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 3 ⎠⎝ ⎝ 2 ⎠ ⎠ ⎝ 4 ⎠
(2.18)
s, m ).
j =0
Proof
Using formula f ( n + mi, x, s ) = xf ( n + mi − 1, x, qs ) + qsf ( n + mi − 2, x, q 2 s ) we get as above D ( n , m, s , k ) = ( − s )
⎛ k +1⎞ n⎜ ⎟ ⎝ 2 ⎠
q
⎛ k +1⎞⎛ n +1⎞ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠
D (0, m, q n s, k ).
(2.19)
For D ( n , m, s , k ) =
(
det f ( n + mi − 1, x, qs ) k − j ( xf ( n + mi − 1, x, qs ) + qsf ( n + mi − 2, x, q 2 s ) )
(
= det f ( n + mi − 1, x, qs ) k − j ( qsf ( n + mi − 2, x, q 2 s ) ) = ( qs )
⎛ k +1⎞ ⎜ ⎟ ⎝ 2 ⎠
= (− s)
j
)
det ( f ( n + mi − 2, x, q s ) f ( n + mi − 1, x, qs )
⎛ k +1⎞ n⎜ ⎟ ⎝ 2 ⎠
2
q
⎛ k +1⎞⎛ n +1⎞ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠
j
k− j
) = ( −qs)
⎛ k +1⎞ ⎜ ⎟ ⎝ 2 ⎠
j
)
D ( n − 1, m, qs, k )
D (0, m, q n s, k ).
Finally we expand D (0, m, s, k ) with respect to the first column and get D (0, m, s, k ) = det ( f (mi, x, s ) j f ( mi − 1, x, qs ) k − j ) 2 k −2 f ( m, x , s ) f ( m − 1, x , qs ) " ⎛ f ( m, x , s ) f ( m − 1, x, qs ) k −1 ⎜ k −1 2 k −2 f (2 m, x , s ) f (2m − 1, x , qs ) f (2m, x , s ) f (2m − 1, x , qs ) " = f ( −1, x, qs ) k det ⎜ ⎜ " " " ⎜ k −1 2 k −2 f ( km, x , s ) f ( km − 1, x , qs ) ⎝ f ( km, x, s ) f ( km − 1, x, qs ) k = f ( −1, x, qs ) f ( m, x, s ) f (2m, x, s )" f ( km, x, s ) D ( m, m, s, k − 1). Thus we have
⎞ ⎟ f (2m, x , s ) ⎟ ⎟ " k ⎟ f ( km, x , s ) ⎠ f ( m, x , s )
D (0, m, s, k ) = f ( −1, x, qs ) k f (m, x, s ) f (2m, x, s )" f (km, x, s ) D ( m, m, s, k − 1).
k
k
(2.20)
7
For k = 1 we get from (2.14) that ⎛n⎞ ⎜ ⎟ n −1 ⎝ 2 ⎠
D ( n, m, s,1) = ( −1) s q n
f ( m, x, q n s ).
Therefore Lemma 2 is true for k = 1. The general case follows by using (2.19) and (2.20) D ( n , m, s , k ) = ( − s ) = (−s) ( −1)
⎛ k +1⎞ n⎜ ⎟ ⎝ 2 ⎠
q
⎛k ⎞ ⎛k ⎞ ⎜ ⎟ m + ⎜ ⎟m ⎝ 2⎠ ⎝ 3⎠
⎛ k +1⎞ n⎜ ⎟ ⎝ 2 ⎠
⎛ k +1⎞⎛ n +1⎞ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠
n
(q s)
q
⎛ k +1⎞⎛ n +1⎞ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠
( q n s ) − k f ( m, x, q n s ) f (2m, x, q n s )" f ( km, x, q n s ) D ( m, m, q n s, k − 1)
( q n s ) − k f ( m, x, q n s ) f (2m, x, q n s )" f ( km, x, q n s )
⎛k ⎞ ⎛k ⎞ ⎜ ⎟( m −1) + ⎜ ⎟m ⎝ 2⎠ ⎝ 3⎠
q
⎛ k ⎞⎛ m ⎞ ⎛ k ⎞⎛ ⎛ m ⎞ 2 ⎞ ⎛ k ⎞ 2 ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎜ ⎟ + m ⎟ + ⎜ ⎟m k − 2 ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 3 ⎠⎝ ⎝ 2 ⎠ ⎠ ⎝ 4⎠
∏ fac(k − j − 1, x, q
mj + m + n
s, m )
j =0
= ( −1)
⎛ k +1⎞ ⎛ k +1⎞ ⎛ k +1⎞ ⎛ k +1⎞ n⎜ ⎟+ ⎜ ⎟ m ( n −1)⎜ ⎟+⎜ ⎟m ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠
s
q
⎛ k +1⎞⎛ n ⎞ ⎛ k +1⎞ ⎛ k +1⎞⎛ ⎛ m ⎞ ⎞ ⎛ k +1⎞ 2 ⎜ ⎟⎜ ⎟ + nm ⎜ ⎟+⎜ ⎟⎜ ⎜ ⎟ ⎟ + ⎜ ⎟m k −1 ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠⎝ ⎝ 2 ⎠ ⎠ ⎝ 4 ⎠
∏ fac(k − j, x, q
mj + n
s, m).
j =0
A special case is Lemma 3
(
D (0, A, s, k ) = det ( aij bik − j ) = det f ( Ai − 1, x, qs ) j ( − f ( Ai, x, s ) ) = ( −1)
⎛ k +1⎞ ⎜ ⎟A ⎝ 3 ⎠
s
⎛ k +1⎞ ⎛ k +1⎞ −⎜ ⎟+⎜ ⎟A ⎝ 2 ⎠ ⎝ 3 ⎠
k −1
∏ fac(k − j, x, q
Aj
q
)
k− j k i , j =0
⎛ k +1⎞⎛ A ⎞ ⎛ k +1⎞ 2 ⎜ ⎟⎜ ⎟ + ⎜ ⎟A ⎝ 3 ⎠⎝ 2 ⎠ ⎝ 4 ⎠
(2.21)
s, A).
j =0
With the use of these lemmas we get det ( f ( n + mi − Aj, x, q Aj s ) k ) = ( −1) = ( −1) = ( −1)
⎛ k +1⎞ k A⎜ ⎟ ⎝ 2 ⎠
⎛ k +1⎞ k A⎜ ⎟ ⎝ 2 ⎠
⎛ k +1⎞ k A⎜ ⎟ ⎝ 2 ⎠
s s
s
⎛ k +1⎞ − ( k A − 2)⎜ ⎟ ⎝ 2 ⎠
⎛ k +1⎞ − ( k A − 2)⎜ ⎟ ⎝ 2 ⎠
⎛ k +1⎞ − ( k A − 2)⎜ ⎟ ⎝ 2 ⎠
k −1
∏ fac(k − j, x, q
mj + n
q q
q
k i , j =0
−
k 2 ( k +1) A (2 k A + A − 3) 12
−
k ( k +1) A (2 k A + A − 3) 12
−
2
k ( k +1) A (2 k A + A − 3) 12 2
s, m)( −1)
⎛ k +1⎞ ⎜ ⎟A ⎝ 3 ⎠
s
(
det ( a j f (n + mi, x, s ) + b j f ( n + mi − 1, x, qs ) ) k
⎛k ⎞
j =0
⎝ ⎠
)
k k i , j =0
∏ ⎜ j ⎟D(n, m, s, k ) D(0, A, s, k ) ⎛ k +1⎞ ⎛ k +1⎞ ⎛ k +1⎞ ⎛ k +1⎞ ⎟n + ⎜ ⎟m ⎜ ⎟( n −1) + ⎜ ⎟m 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠
⎜ ⎛k ⎞ ⎝ − ( 1) ∏ ⎜ j⎟ j =0 ⎝ ⎠ k
⎛ k +1⎞ ⎛ k +1⎞ −⎜ ⎟+⎜ ⎟A ⎝ 2 ⎠ ⎝ 3 ⎠
q
s
q
⎞ ⎛ k +1⎞ 2 ⎛ k +1⎞⎛ n ⎞ ⎛ k +1⎞⎛ ⎛ m ⎞ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎜ ⎟ + mn ⎟ + ⎜ ⎟m ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 3 ⎠⎝ ⎝ 2 ⎠ ⎠ ⎝ 4 ⎠
⎛ k +1⎞⎛ A ⎞ ⎛ k +1⎞ 2 ⎜ ⎟⎜ ⎟ + ⎜ ⎟A ⎝ 3 ⎠⎝ 2 ⎠ ⎝ 4 ⎠
j =0
k −1
∏ fac(k − j, x, q
Aj
s, A )
j =0
8
⎛ k +1⎞ ⎛ k +1⎞ ⎟( n − k A ) + ⎜ ⎟( A + m ) 2 ⎠ ⎝ 3 ⎠
⎜ ⎛k ⎞ = ∏ ⎜ ⎟( − s ) ⎝ j =0 ⎝ j ⎠ k
k −1
∏ j =0
2
q
⎛ k +1⎞⎛ n ⎞ ⎛ k +1⎞ ⎛ k +1⎞ m ( km − 2) ⎛ k +1⎞ ⎛ A ⎞ ⎛ k +1⎞ A (3k + 2) −⎜ ⎜ ⎟⎜ ⎟ + ⎜ ⎟mn + ⎜ ⎟ ⎟ ⎜ ⎟−⎜ ⎟ 4 4 ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 3 ⎠
k −1
fac( k − j, x, q mj + n s, m)∏ fac(k − j, x, q Aj s, A). j =0
Thus Theorem 2 is proved. We will also need some modifications of these results. Let d ( n, m, s, k , j ) = det ( f ( n + m(i + [i ≥ j ]), x, s ) j f ( n + m(i + [i ≥ j ]) − 1, x, qs ) k − j ) ,
(2.22)
where [ P ] denotes the Iverson symbol, i.e. [ P ] = 1 if property P is true and [ P ] = 0 else. Then d (n, m, s, k ,0) = D(n + m, m, s, k ). (2.23) From (2.18) we get d (0, m, s, k ,0) = ( −1)
⎛ k +1⎞ ⎛ k +1⎞ ⎜ ⎟m ⎜ ⎟m −k ⎝ 2 ⎠ ⎝ 2 ⎠
s
q
km (( km + m + k −3) 4
fac( k , x, q m s, m )d (0, m, q m s, k − 1,0). (2.24)
Furthermore we get in the same way as above for j > 0 ⎛k ⎞
d (0, m, s, k , j ) = s
−k
⎛ k ⎞⎛ m +1⎞ ⎟ 2 ⎠
m⎜ ⎟ ⎜ ⎟⎜ fac( k + 1, x, s, m ) ( − s ) ⎝ 2 ⎠ q ⎝ 2 ⎠⎝ f ( jm, x, s )
d (0, m, q m s, k − 1, j − 1).
(2.25)
Proof of Theorem 1 The above argument implies that det ( f ( n + i − j, x, q j s ) k )
k +1 i , j =0
= 0.
(2.26)
If we denote by Aj the matrix obtained by crossing out the first row and the j − th column of
( f ( n + i − j, x, q s ) )
k k +1
j
i , j =0
k +1
∑ f (n − j, x, q s) ( −1) j
j =0
k
j
, we get det ( Aj ) = 0
or k +1
∑ f (n − j, x, q s) ( −1) j
j =0
k
j
det ( Aj ) det ( A0 )
= 0.
(2.27)
To compute det ( Aj ) we use the same method as in Theorem 1. We get k
(
⎛ ⎞ v( j ) k det ( Aj ) = ⎜ det ( ah ( j, s ) f ( n + i, x, s ) + bh ( j, s ) f (n + i − 1, x, qs ) ) ⎟ ⎝ v (0)v (1)" v ( k + 1) ⎠ where
)
k i ,h =0
,
9
ah ( j, s ) = f ( h − 1, x, qs ), bh ( j, s ) = − f ( h, x, s ) for h < j and ah ( j, s ) = f ( h, x, qs ), bh ( j, s ) = − f (h + 1, x, s ) for h ≥ j. Therefore det ( ah ( j, s )i bh ( j, s ) k −i ) = d (0,1, s, k , j ).
(2.28)
By (2.24) we have d (0,1, s, k ,0) = ( −1)
⎛ k +1⎞ ⎛ k ⎞ ⎛ k ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠ ⎝2⎠
s q
fac(k , x, qs )d (0,1, qs, k − 1,0).
For j > 0 we get from (2.25) ⎛k ⎞
⎛k ⎞
⎜ ⎟ ⎜ ⎟ fac( k + 1, x, s ) ( − s )⎝ 2 ⎠ q⎝ 2 ⎠ d (0,1, qs, k − 1, j − 1). f ( j , x, s )
(2.29)
d (0,1, s, k , j ) fac( k + 1, x, s ) d (0,1, qs, k − 1, j − 1) = ( − s)− k . d (0,1, s, k ,0) f ( j, x, s ) fac(k , x, qs ) d (0,1, qs, k − 1,0)
(2.30)
d (0,1, s, k , j ) = s
−k
Therefore
This implies j −1
j −1
− ∑ ( k − i ) − ∑ i ( k −i ) k + 1 d (0,1, s, k , j ) d (0,1, q j s, k − j,0) = ( − s ) i =0 q i =0 ( x , s, q ) j d (0,1, s, k ,0) d (0,1, q j s, k − j, 0)
= (− s)
⎛ j⎞ − kj + ⎜ ⎟ ⎝ 2⎠
q
j −1 ⎛ j⎞ − k ⎜ ⎟+ i2 2 ⎝ ⎠ i =0
∑
k +1 j
( x, s, q).
Therefore we get det ( Aj ) j
⎛ j⎞
⎛ j⎞
⎛ j⎞
j −1
k⎜ ⎟ − kj + ⎜ ⎟ − k ⎜ ⎟ + ∑ i ⎛ v ( j ) ⎞ d (0,1, s, k , j ) 2 j + kj kj ⎝ 2⎠ ⎝ 2⎠ = ( −1) j ⎜ = − − ( −1) ( 1) s q ( s ) q ⎝ ⎠ i =0 ⎟ det ( A0 ) ⎝ v (0) ⎠ d (0,1, s, k ,0)
= ( −1)
⎛ j +1⎞ ⎛ j ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠
k
2
k +1 ( x , s, q ) j
j −1
∑i2 k + 1 s q i =0 ( x, s, q). j
Thus Theorem 1 is proved.
10
If we use the fact that det ( f (n + A(i − j ), x, q jA s ) k ) k +1
∑ f (n − jA, x, q
jA
s ) k ( −1) j
j =0
det ( B j ) det ( B0 )
k +1 i , j =0
= 0, we get in the same way that
= 0,
where we denote by B j the matrix obtained by crossing out the first row and the j − th column of ( f ( n + A(i − j ), x, q jA s ) k )
k +1 i , j =0
.
Here we have ( −1)
j
det ( B j ) det ( B0 )
⎛ jA ⎞
k
⎜ ⎟k ⎛ v ( jA) ⎞ d (0, A, s, k , j ) j + kjA kjA ⎝ 2 ⎠ d (0, A, s, k , j ) = ( −1) ⎜ = − ( 1) s q . ⎟ d (0, A, s, k ,0) ⎝ v (0) ⎠ d (0, A, s, k ,0) j
If we define k
k ( x, s, q, A) = j
∏ f (iA, x, s) i =1
j
∏ f (iA, x, q
( j −i ) A
i =1
k− j
s )∏ f (iA, x, q s )
(2.31)
jA
i =1
we get from (2.24) and (2.25) ⎛k ⎞
d (0, A, s, k , j ) = d (0, A, s, k ,0)
s
−k
( −1)
⎛ k ⎞⎛ A +1⎞ ⎟ 2 ⎠
A⎜ ⎟ ⎜ ⎟⎜ fac( k + 1, x, s, A) ( − s ) ⎝ 2 ⎠ q ⎝ 2 ⎠⎝ f ( jA, x, s )
⎛ k +1⎞ ⎛ k +1⎞ ⎜ ⎟A ⎜ ⎟A −k ⎝ 2 ⎠ ⎝ 2 ⎠
s
q
k A (( k A + A + k − 3) 4
d (0, A, q A s, k − 1, j − 1)
fac( k , x, q A s, A)d (0, A, q A s, k − 1, 0)
⎛A⎞
( −1) k A − k ⎜ ⎟ fac( k + 1, x, s, A) d (0, A, q A s, k − 1, j − 1) = kA q ⎝ 2 ⎠ s f ( jA, x, s ) fac( k , x, q A s, A) d (0, A, q A s, k − 1,0) = ( −1)
⎛ j⎞ ⎛ j⎞ kjA − A⎜ ⎟ A⎜ ⎟ − kjA ⎝ 2⎠ ⎝ 2⎠
s
j −1
q
⎛ A ⎞⎛ ⎛ j ⎞⎞ −⎜ ⎟⎜ kj −⎜ ⎟ ⎟ − A2 i ( k −i ) ⎝ 2 ⎠⎝ ⎝ 2⎠⎠ i =0
∑
k +1 j
( x, s, q, A).
Therefore ( −1) j
det ( B j ) det ( B0 )
= ( −1)
⎛ j⎞ j + A⎜ ⎟ ⎝ 2⎠
(q
= ( −1)
(4 j +1) A − 3 6
⎛ jA ⎞ ⎜ ⎟k j + kjA kjA ⎝ 2 ⎠
s)
s q
⎛ j⎞ A⎜ ⎟ ⎝ 2⎠
( −1)
⎛ j⎞ kjA − A⎜ ⎟ ⎝ 2⎠
s
⎛ j⎞ A⎜ ⎟ − kjA ⎝ 2⎠
j −1
q
⎛ A ⎞⎛ ⎛ j ⎞⎞ − ⎜ ⎟⎜ kj − ⎜ ⎟ ⎟ − A2 i ( k −i ) ⎝ 2 ⎠⎝ ⎝ 2⎠⎠ i =0
∑
k +1 ( x , s , q, A ) j
k +1 ( x, s, q, A). j
Thus we get
11
Theorem 3 For k , A ∈ ` the following recurrence relation holds: k +1
∑ ( −1)
⎛ j⎞ j + A⎜ ⎟ ⎝ 2⎠
(q
(4 j +1) A − 3 6
s)
⎛ j⎞ A⎜ ⎟ ⎝ 2⎠
k +1 j
j =0
( x, s, q, A) f (n − jA, x, q jA s ) k = 0.
(2.32)
For the special case k = 1 this reduces to f ( n, x, s ) −
f (2A, x , s ) A
f ( A, x , q s )
A ( 3 A −1)
f ( n − A, x , q s ) + ( −1) q A
A
2
s
A
f ( A, x , s ) A
f ( A, x , q s )
f ( n − 2A, x , q s ) = 0, 2A
(2.33)
which has already been proved in [6].
References
[1] G. Andrews, A. Knopfmacher, P. Paule, An infinite family of Engel expansions of Rogers-Ramanujan type, Adv. Appl. Math. 25 (2000), 2-11 [2] L. Carlitz, The characteristic polynomial of a certain matrix of binomial coefficients, Fibonacci Quarterly 3 (1965), 81-89 [3] L. Carlitz, Fibonacci notes 4: q-Fibonacci polynomials, Fibonacci Quarterly 13 (1975), 97-102 [4] J. Cigler, Elementare q-Identitäten, Séminaire Lotharingien de Combinatoire, B05a (1981) [5] J. Cigler, q-Fibonacci polynomials, Fibonacci Quarterly 41 (2003), 31-40 [6] J. Cigler, Some conjectures about q-Fibonacci polynomials, arXiv:0805.0415 [7] N.J.A. Sloane, The On-Line Encyclopedia of Integer Sequences
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