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Journal of Integer Sequences, Vol. 12 (2009),
Article 09.1.6
23 11
On the Expansion of Fibonacci and Lucas
Polynomials
Helmut Prodinger
Department of Mathematics
University of Stellenbosch
7602 Stellenbosch
South Africa
hproding@sun.ac.za
Abstract
Recently, Belbachir and Bencherif have expanded Fibonacci and Lucas polynomials
using bases of Fibonacci- and Lucas-like polynomials. Here, we provide simplified
proofs for the expansion formulæthat in essence a computer can do. Furthermore, for
2 of the 5 instances, we find q-analogues.
1
Introduction
In [2], Belbachir and Bencherif studied the Fibonacci and Lucas polynomials:
U0 = 0, U1 = 1, Un = xUn−1 + yUn−2 ,
V0 = 2, V1 = x, Vn = xVn−1 + yVn−2 .
We prefer the modified polynomials
u0 = 0, u1 = 1, un = un−1 + zun−2 ,
v0 = 2, v1 = 1, vn = vn−1 + zvn−2 ,
so that
Un (x, y) = x
Then, with
n−1
y
un 2 ,
x
λ1,2 =
1±
1
y
Vn (x, y) = x vn 2 .
x
n
√
1 + 4z
,
2
1
(λn1 − λn2 ),
vn = λn1 + λn2 .
1 + 4z
2
Substituting z = t/(1 − t) , these formulæ become particularly nice:
un = √
un =
1 − (−t)n
,
(1 + t)(1 − t)n−1
The main result of [2] are the following 5 formulæ:
n
X
2u2n+1 =
an,k v2n−k ,
an,k
k=0
vn =
1 + (−t)n
.
(1 − t)n
n
X
n+k n
j+k j
.
− (−1)
=2
(−1)
k
k
j=0
n
X
n
.
u2n =
bn,k u2n−k ,
bn,k = (−1)
k
k=1
n
X
k+1 n
v2n−1 =
cn,k u2n−k ,
cn,k = 2(−1)
− [k = 1].
k
k=1
n
X
k+1 2n − k n
2v2n−1 =
dn,k v2n−1−k ,
dn,k = (−1)
.
n
k
k=1
2u2n =
n
X
k+1
en,k v2n−1−k ,
(1)
(2)
(3)
(4)
(5)
k=1
X
n−1
1
−k n
j
n+k n − 1
j+k−1
.
+
− (−1)
(−1)
k−1
2n
k
k−1
2
j=0
k+1 2n
en,k = (−1)
But the proofs of all these, using the simple forms for un and vn , can be done by a computer!
To give the reader an idea, let us do the last one, which seems to be the most complicated:
n
n
X
X
k+1 2n − k n
en,k v2n−1−k =
v2n−1−k
(−1)
2n
k
k=1
k=1
j+1
n
n−1 X
X
X
j
1
n+k n − 1
j+k−1
(−1)
v2n−1−k
(−1)
v2n−1−k −
+
k
−
1
2
k
−
1
j=0 k=1
k=1
1 − t2n−1
(−1)n tn−1
1 + t2n−1
(−1)n tn−1
−
+
+
(1 − t)2n−1 (1 − t)2n−2 (1 + t) (1 − t)2n−2 (1 − t)2n−2
2(1 − t2n )
= 2u2n .
=
(1 − t)2n−1 (1 + t)
=
The other proofs are similar/easier:
n
X
an,k v2n−k =
k=0
=
2[(−t)n (1 + t) + 1 + t2n+1 ]
2(−t)n
−
(1 − t)2n (1 + t)
(1 − t)2n
2[1 + t2n+1 ]
= 2u2n+1 .
(1 − t)2n (1 + t)
2
n
X
cn,k u2n−k =
k=1
1 + t2n−1
2(1 − t2n )
−
(1 − t)2n−1 (1 + t) (1 + t)(1 − t)2n−2
1 − t2n−1
=
= v2n−1 .
(1 − t)2n−1
Remark 1.
The polynomials un (x) and vn (x) are essentially Chebyshev polynomials.
The authors of [2] have also published a companion paper [3]; according to a remark in [2],
the results in [2] are more general than the ones in [3].
2
q-analogues
Now we are interested in q-analogues. For this, we replace un by
X
k+1
n
−
k
−
1
(
)
q 2
zk
Fibn =
k
q
n−1
0≤k≤
and vn by
Lucn =
X
2
k
[n]q
n
−
k
(
)
q 2
zk ,
k q [n − k]q
n
0≤k≤ 2
as suggested by Cigler [4]. We use standard q-notation here:
[n]q := 1 + q + · · · + q
n−1
,
[n]q ! := [1]q [2]q . . . [n]q ,
n
[n]q !
,
:=
[k]q ![n − k]q !
k q
compare [1]; the notions of the Introduction are the special instance q = 1.
Theorem 2.
Luc2n−1 =
n
X
dn,k Luc2n−1−k ,
k=1
with
q (2)
1 + q n−1
k
k−1
dn,k = (−1)
n−1
n−1 n
.
+q
k q
k q
Proof. We must prove that
X
k
2n
−
1
−
k
[2n − 1]q
zk
q (2)
k
q [2n − 1 − k]q
0≤k≤n−1
j
q (2)
n−1
n−1 n
=
(−1)
+q
n−1
j q
j
1
+
q
q
j=1
X
k
2n
−
j
−
1
−
k
[2n − j − 1]q
(
)
zk .
×
q 2
k
q [2n − j − 1 − k]q
2n−j−1
n
X
j−1
0≤k≤
2
3
Comparing coefficients, we have to prove that
k
2n
−
1
−
k
[2n − 1]q
(
)
q 2
k
q [2n − 1 − k]q
j
n
X
k
n−1
2n − j − 1 − k
[2n − j − 1]q
q (2)
n−1 n
j−1
(
)
2
.
+q
q
=
(−1)
n−1
j
1+q
j q
k
q
q [2n − j − 1 − k]q
j=1
Simplifying, we must prove that
n
X
j
2n − j − 2 − k
n−1
j (2)
n−1 n
[2n − j − 1]q = 0.
(−1) q
+q
j
k
−
1
j
q
q
q
j=0
Another form of this is
n
n 2n − j − 2 − k X
j (2j )
2n−1
n−j
n−1
2n−j−1
= 0.
(−1) q
1−q
−q
+q
1−q
k
−
1
j
q
q
j=0
Notice that
n
X
j
n −aj
(
)
(−1) q 2
q
=0
j
q
j=0
j
for 0 ≤ a ≤ n − 1. This follows from Rothe’s formula [1, p. 490]
n
X
j (2j ) n
(−1) q
xj = (1 − x)(1 − xq) . . . (1 − q n−1 ).
j
q
j=0
We write the desired identity as
n
X
n j
−j
−2j
(
)
(−1) q 2 A + Bq + Cq
D0 q −0 + · · · + Dk−1 q −j(k−1) = 0.
j q
j=0
j
Therefore, for k ≤ n − 2, the identity holds. For k = n − 1,
1
X
n n − j − 1
j
2n−j−1
2n−1
n−j
n−1
(
)
1−q
−q
+q
(−1) q 2 1 − q
=0
j
n
−
2
q
q
j=0
j
can be shown by inspection, and for k = n, the identity holds, since the sum is empty.
Theorem 3.
Fib2n =
n
X
bn,k Fib2n−k
k=1
with
k−1
bn,k = (−1)
4
k
n
(
)
q 2
.
k q
Proof. We must prove that
X
k+1
2n
−
k
−
1
(
)
q 2
zk
k
q
0≤k≤n−1
=
n
X
j=1
j−1
(−1)
j
n
(
)
q 2
j q
k+1
2n
−
j
−
k
−
1
(
)
q 2
zk .
k
q
2n−j−1
X
0≤k≤
2
Comparing coefficients, this means
n
X
j
n 2n − j − k − 1
(
)
2
(−1) q
= 0,
j q
k
q
j=0
j
which follows by a similar but simpler argument than before.
3
Conclusion
We found 2 q-analogues; for the remaining 3 instances we were not successful and leave this
as a challenge for anybody who is interested.
References
[1] G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press,
2000.
[2] H. Belbachir and F. Bencherif, On some properties of bivariate Fibonacci and Lucas
polynomials, J. Integer Sequences 11 (2008), Article 08.2.6.
[3] H. Belbachir and F. Bencherif, On some properties of Chebyshev polynomials, Discuss.
Math. Gen. Algebra Appl. 28 (2008), 121–133.
[4] J. Cigler, A new class of q-Fibonacci polynomials, Electronic J. Combinatorics 10
(2003), Article R19. Text available at
http://www.combinatorics.org/Volume 10/Abstracts/v10i1r19.html.
2000 Mathematics Subject Classification: Primary 11B39; Secondary 11B37.
Keywords: Fibonacci polynomials, Lucas polynomials, generating functions, q-analogues.
(Concerned with sequences A007318, A029653 and A112468.)
Received October 6 2008; revised version received December 15 2008. Published in Journal
of Integer Sequences, December 17 2008.
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