SOME SUBSEQUENCES OF THE GENERALIZED FIBONACCI AND LUCAS SEQUENCES EMRAH KILIC Abstract. We derive …rst-order nonlinear homogeneous recurrence relations for certain subsequences of generalized Fibonacci and Lucas sequences. We also present a polynomial representation for the terms of Lucas subsequence.
1. INTRODUCTION Let p and q be nonzero integers such that = p2 4q 6= 0: The generalized Fibonacci sequence fUn (p; q)g ; or brie‡y fUn g ; and Lucas sequence fVn (p; q)g ; or brie‡y fVn g ; are de…ned by for n > 1 Un = pUn
1
qUn
and Vn = pVn
2
qVn
1
2;
where U0 = 0; U1 = 1 and V0 = 2; V1 = p; respectively. When p = q = 1; Un = Fn (nth Fibonacci number) and Vn = Ln (nth Lucas number). The Binet forms of fUn g and fVn g are n
n
Un =
and Vn =
n
+
n
;
where and are the roots of x2 px + q = 0. In [5], the solution of the following …rst order cubic recursion was asked an+1 = 5a3n
3an ; a0 = 1:
(1.1)
Then the solution was given as an = F3n in [7]. After this, similarly the solution of recurrence Pn+1 = 25Pn5
25Pn3 + 5Pn ; P0 = 1
(1.2)
was also asked. Then the solution was given as Pn = F5n : 2000 Mathematics Subject Classi…cation. 11B37, 11B39. Key words and phrases. Generalized Fibonacci and Lucas sequences, subsequences, recurrence relation. 1
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EM RAH KILIC
As an addendum to the solution of the problem given in [7], Klamkin asked the solutions of recurrences: An+1
= A2n
Bn+1
=
Cn+1
=
Bn4 Cn6
2; A1 = 3; 4Bn2 + 2; B1 = 7; 6Cn4 + 9Cn2
2; C1 = 18:
Then the solutions of them were given as An = L2n ; Bn = L4n and Cn = L6n : In [1], the author give a recurrence relation for the Fibonacci subsequence fFkn g for positive odd k; which generalize (1.1) and (1.2). In [2], some generalizations of the results of [1] were obtained for the sequences fUn (p; 1)g and fVn (p; 1)g. Meanwhile Prodinger [3] proved a general expansion formula for a sum of powers of Fibonacci numbers, as considered by Melham, as well as some extensions. In this paper, we …nd …rst-order nonlinear recurrence relation for the subsequence fUkn g of generalized Fibonacci sequence fUn g for odd k; and …rst-order nonlinear recurrence relation for the subsequence fVkn g of generalized Lucas sequence fVn g for both odd and even k: We also give a polynomial representation for the generalized Lucas number Vkn in terms of generalized Fibonacci numbers Ukn of degree k for even k:
2. Recurrence Relations We …nd …rst-order nonlinear recursions for the sequences fUkn g and fVkn g for certain k 0 s. We need the following result for further use. Lemma 1. For n; t
0;
t X 2t + 1 t + k + 1 k n(t k) 2k q Un ; i) U(2t+1)n = Un t + k + 1 2k + 1 k=0 t X 2t t + k k n(t k) 2k ii) V2tn = q Un ; t + k 2k k=0 t X t + k + 1 n(t k) 2k t+k 2t + 1 Vn iii) V(2t+1)n = Vn ( 1) q t + k + 1 2k + 1 k=0 t X t + i (t i)n 2i t i 2t iv) V2tn = ( 1) q Vn : t + i 2i i=0
SOM E GENERALIZED FIBONACCI AND LUCAS SUBSEQUENCES
3
Proof. In order to prove the claimed identities, it is su¢ cient to use the following well-known formulas (for more details, see [6]): Xm + Y m =
bm=2c P
m
k
( 1)
k=0
m
m k
k
k
m 2k
(XY ) (X + Y )
k
; m
1; (2.1)
and Xm X
b(m P 1)=2c Ym k m = ( 1) Y k=0
k k
1
k
m 2k 1
(XY ) (X + Y )
For example, the claim (iii) follows from by taking X = m = 2t in (2.1). The other claims are similarly obtained.
n
;m n
,Y =
1: and
For odd k; we give a …rst-order nonlinear recurrence relation for the sequence fUkn g : Theorem 1. For odd k > 0 and n Ukn+1 = 4
k
1 2
Ukkn +
(k P 3)=2 i=0
0,
n k 1 2k (k + 1) =2 + i 4i q k ( 2 k + 2i + 1 2i + 1
i)
Uk2i+1 : n
Proof. From the Binet formula of fUn g and by the binomial theorem, we obtain !k kn kn k 1 P j jkn (k j)kn k k = k=2 Ukn = (2.2) j ( 1) 4 j=0 ! (k P 1)=2 j 1 k kn Ukn+1 + q U(k 2j)kn ; = j 4(k 1)=2 j=1 where
is de…ned as before. By (2.2), we obtain for odd k; Ukn+1 = 4(k
1)=2
(k P 1)=2
Ukkn
j=1
k j
qk
n
j
j
( 1) U(k
2j)kn :
(2.3)
By (i) in Lemma 1 and (2.3), we conclude Ukn+1
= 4(k k
1)=2 1 k
2 P
j=1
Ukkn
1
P
2
j
j
( 1)
i=0
k j
(k+1)=2+i j 2i+1
k 2j k+1 2 +i
j
n k 1 4i q k ( 2
i)
Uk2i+1 n
which, after reversing the summation order, can be rewritten as Ukn+1 = 4(k
1)=2
(k P 1)=2
Ukkn
i=0
where Ai;k =
(k P 1)=2 i j=1
j
( 1)
k j
n k 1 4i q k ( 2
i)
Ai;k Uk2i+1 ; n
k 2j (k+1)=2+i j 2i+1 (k+1)=2+i j :
(2.4)
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EM RAH KILIC
Since A(k
= 0, the equality (2.4) becomes
1)=2;k
Ukn+1 = 4(k
1)=2
(k P 3)=2
Ukkn
i=0
n k 1 4i Ai;k q k ( 2
From (pp. 58, [4]), it is known that for 1 m m P j ( 1) k k m2j j kj k mm j j =
(k
i)
3) =2
k m k k m m
j=1
Uk2i+1 : n
.
(2.5)
(k+1)=2+i 2k ; it is su¢ cient to replace m by In order to obtain Ai;k as k+2i+1 2i+1 (k 1) =2 i in (2.5). Thus we obtain the claimed result.
For example, when k = 7; we have that 3
U7n+1 =
n
n
2 7n
q U75n + 14 q 7 2 U73n + 7q 7 3 U7n :
U77n + 7
We now give a nonlinear …rst order recurrence relation for the sequence fVkn g for odd k: Theorem 2. For n > 0 and odd k > 1, k
Vkn+1 =
3
2 P
Vkkn
(k
n k 1 1) =2 + i 2k i+ k 1 ( 1) 2 q k ( 2 2i k + 1 2i + 1
i=0
i)
Vk2i+1 : n
Proof. It is easy to see that Vkkn =
k P
j=0
jkn
k j
(k j)kn
= Vkn+1 +
j=1
By (iii) in Lemma 1, we write k
Vkn+1
(k P 1)=2
q
1 k
2 P
= Vkkn
1
P
2
j=1
i=0
kn ( k 2 1
i)
k j
n
q jk V(k
j k j
(k+1)=2+i j 2i+1
( 1)
k
1 2
2j)kn :
j+i
k 2j 2i+1 (k+1)=2+i j Vkn
which, by reversing the summation order, becomes k
=
3 k
2 P
Vkkn
i=0
n k 1 qk ( 2
1
P
2
i
j=1 i)
k 2j (k+1)=2+i j 2i+1 (k+1)=2+i j
k j
( 1)
k
1 2
+i j
Vk2i+1 : n
For the sum k
1
P
2
i
j=1
if we take m = (k
j
( 1)
1) =2
k j
k 2j (k+1)=2+i j 2i+1 (k+1)=2+i j ;
i in (2.5), we obtain the claimed result.
We now give a nonlinear …rst order recurrence relation for the sequence fVkn g for even k:
SOM E GENERALIZED FIBONACCI AND LUCAS SUBSEQUENCES
5
Theorem 3. For n > 0 and even k > 1, k=2 1
X
Vkn+1 = Vkn+1 = Vkkn
i + k=2 2i
i+ k 2
( 1)
i=0
1
2k ( k2 q 2i k
i)kn
Vk2in :
Proof. By the binomial theorem, we have that for even k; k
Vkkn
Vkn+1 =
k k=2
+
q
2 X
kn+1 2
k j
n
q k j V(k
2j)kn :
j=1
From (iv) in Lemma 1, we write k
k
Vkn+1
Vkkn
=
k q( 2
+1 i 2 2X X i=0
j=1
i)kn
Vk2in :
(k
k j
2j) j+i
k 2
k 2
k j+i ( 1) 2 2i
j i
After reversing the summation order and by using (2.5), we get k
1 2 X
Vkkn
Vkn+1 =
i+
i+ k 2
( 1)
k 2
1
2i
i=0
2k ( k2 q 2i k
i)kn
Vk2in ;
as claimed. For example, when k = 6; we have that n
V6n+1 = V66n
n
n
6q 6 V64n + 9q 6 2 V62n
2q 6 3 :
(2.6)
3. A Polynomial Representation We give a polynomial representation for the Lucas number Vkn in terms of the generalized Fibonacci numbers Ukn for even k: Theorem 4. For even k > 0; n Vkn+1 = Proof. Consider
0 and
k=2 P
n i + k=2 2k 4i Uk2in q k (k=2 2i i=0 k + 2i
i)
:
Ukkn = =
k 1 P 4k=2 j=0
1 4k=2
k j
j
( 1)
Vkn+1
jkn
k 2
( 1) q
(k j)kn
kn+1 2
k k=2
+
k=2 P
j=1
j
( 1)
k j
V(k
2j)kn q
jk
n
!
:
6
EM RAH KILIC
By (ii) in Lemma 1 and reversing the summation order of the equation above, we write Ukkn
=
k n+1 1 (Vkn+1 + ( 1) 2 q k =2 k=2 4
j
k 2j k=2 j+i k=2 j+i 2i
( 1)
k
k k=2
+
2 k
2 2Pj P
j=1 i=0
i kn (k=2 i)
4q
k j
Uk2in ;
which becomes, k
1 = k=2 4
Vkn+1 +
j
( 1)
i=0 j=1
If we take m = form: Ukkn
2 k
2 2Pi P
k 2
1 = k=2 4
k 2j k k=2 j+i j
i in (2.5) for 1 k
Vkn+1
as claimed.
2
2 P
i=0
m
k=2 j+i 2i
q
kn (k=2 i)
i
4
Uk2in
!
:
k=2, the last equation takes the
n i + k=2 2k 4i Uk2in q k (k=2 2i k + 2i
i)
!
;
When k = 6; we get n
V6n+1 = 43 U66n + 642 U64n q 6 + 94U62n q 6
n
2
n
+ 2q 6 3 :
(3.1)
Notice that even the coe¢ cients of the formula in (3.1) and (2.6) appears to be the terms of the sequence A034807 in the OEIS. Conclusions Throughout the paper, we obtain recurrence relations for the sequences fUkn g and fVkn g for certain k’s (not all k’s) and obtain a polynomial representation for the generalized Lucas number Vkn in terms of generalized Fibonacci numbers Ukn of degree k for even k. In order to clear how the remaining cases could not be obtained, we note some facts here. Since we never reach at the statement Ukn+1 when we expand the k th powers of the statements Ukn and Vkn by the binomial theorem for even integer k; we can’t give a recurrence relation for Ukn+1 for even k: As a second remaining case, is there a polynomial representation of Ukn+1 in terms of Vkn for odd k ? Related with this question, we note that while doing required operations, there is a problem (in reversing the summation order) so that we couldn’t …nd a representation for the term Ukn+1 in terms of Vkn : Acknowledgement The author would thank to the anonymous referee for his/her valuable suggestions.
SOM E GENERALIZED FIBONACCI AND LUCAS SUBSEQUENCES
7
References [1] P. Filipponi, On the Fibonacci numbers whose subscript is a power, The Fibonacci Quart. 34(3) (1996), 271–276. [2] E. K¬l¬ç, Second order linear recursions whose subscripts are a power, Miskolc Math. Notes, 11(2) (2010), 151–161. [3] H. Prodinger, On a sum of Melham and its variants, The Fibonacci Quart. 46/47 (2008/2009), 207-215. [4] J. Riordan, Combinatorial identities. Wiley, New York, 1979. [5] A. Stinchcomb, Problem B-769: First order cubic recurrence relation, The Fibonacci Quart. 32(4) (1994) 373. [6] M. N. S. Swamy, On certain identities involving Fibonacci and Lucas numbers. The Fibonacci Quart. 35(3) (1997) 230–232. [7] D. C. Terr, Solution B-769: The recurrence of F3n , The Fibonacci Quart. 33(5) (1995) 468. TOBB Economics and Technology University Mathematics Department 06560 Ankara Turkey E-mail address :
[email protected] 
