Selçuk J. Appl. Math. Selçuk Journal of Vol. 13. No. 1. pp. 83-88, 2012 Applied Mathematics
On the k-Generalized Fibonacci Numbers Nazmiye Yılmaz, Yasin Yazlık, Necati Ta¸skara
Department of Mathematics, Science Faculty, Selcuk University, Campus, 42075, Konya, Turkiye
e-mail: nzyilm az@ selcuk.edu.tr,yyazlik@ selcuk.edu.tr,ntaskara@ selcuk.edu.tr
Received Date: December 15, 2011 Accepted Date: March 19, 2012
Abstract. In this paper, we define a new family of k-generalized Fibonacci numbers. Furthermore, we give sums and recurrence relations of this numbers and obtain generating functions of this numbers for k = 2.
Key words: k-generalized Fibonacci numbers;Sum; Generating function. 2000 Mathematics Subject Classification: 11B39, 11B65.
1. Introduction
Recently, Fibonacci, Lucas numbers and their applications have investigated very largely and authors tried to developed and give some directions to mathe-matical calculations using these type of special numbers. For rich applications of these numbers in science and nature, one can see the citations in [7-9]. For instance, the ratio of two consecutive of these numbers converges to the Golden section α = 1 +
√ 5
2 . (The applications of Golden ratio appears in many re-search areas, particularly in Physics, Engineering, Architecture, Nature and Art. Physicists Naschie and Marek-Crnjac gave some examples of the Golden ratio in Theoretical Physics and Physics of High Energy Particles [10-13]). The Fibonacci and Lucas sequences are defined by the following recurrence relations, for n ≥ 0,
(1) Fn+1= Fn+ Fn−1 and Ln+1= Ln+ Ln−1,
with initial conditions F−1 = 0, F1 = 1 and L−1 = 2 , L0 = 1 respectively.
[1-5]. For example, generalized Fibonacci numbers which were generalization of Fibonacci and Lucas numbers are defined by
Gn+1= Gn+ Gn−1, G−1= a, G0= b.
Thus, the Binet formula is obtained as
(2) Gn= Xαn− Y βn √ 5 , where X = a + bα, Y = a + bβ, α = 1+√5 2 and β = 1−√5 2 . Also, Fibonacci
numbers and generalized Fibonacci numbers are satisfied following property,
(3) Gn= bFn+ aFn−1.
Another of the generalizations, in [6], is generalized k-Fibonacci numbers Fn(k)
, which defined by (4) Fn(k)= ¡√1
5¢k ¡
αm+2− βm+2¢r¡αm+1− βm+1¢k−r,
where n = mk + r (0 ≤ r < k) and n, k (k 6= 0) ∈ N. These numbers satisfy following property,
(5) Fn(k)= Fmk−rFm+1r .
In the light of the above paragraph, the main goal of this paper is to improve the Fibonacci numbers with a different viewpoint . In order to do that we defined a new family of k-generalized Fibonacci numbers which was been generalization of new family of k-Fibonacci numbers presented in [6]. Then, sums and recurrence relations of this numbers have been obtained and generating functions of this numbers for k = 2 given.
2. Main Results
In this section, we define a new family k-generalized Fibonacci numbers. Also, we investigate recurrence relations, generating functions and sums of the new family.
Definition 1. For n, k (k 6= 0) ∈ N, the new family k-generalized Fibonacci numbers are defined by
(6) G(k)n = ¡√1 5¢k
¡
Xαm+1− Y βm+1¢r(Xαm− Y βm)k−r,
where n = mk + r (0 ≤ r < k) , X = a + bα, Y = a + bβ and m ∈ N. By considering Definition 1, we obtain the following equalities:
• For a = 0 and b = 1, G(k)n = Fn(k).
• For a = 2 and b = 1, G(k)n = L(k)n .
• For k = 1, r = 0 and n = m, it is clear that the k-generalized Fi-bonacci sequence is turn to the well-known generalized FiFi-bonacci sequence, that is G(1)n = Gn.
Also, for the special case k = 1, 2, 3, 4, n G(1)n o5 n=0= {b, a + b, a + 2b, 2a + 3b, 3a + 5b, 5a + 8b} = {Gn} 5 n=0, n G(2)n o5 n=0= n b2, b (a + b) , (a + b)2 , (a + b) (a + 2b) , (a + 2b)2, (a + 2b) (2a + 3b)o, n G(3)n o5 n=0= n b3, b2(a + b) , b (a + b)2 , (a + b)3, (a + b)2(a + 2b) , (a + b) (a + 2b)2o, n G(4)n o5 n=0= n b4, b3(a + b) , b2(a + b)2 , b (a + b)3, (a + b)4, (a + b)3(a + 2b)o. From (2) and (6), the relationship of between generalized Fibonacci numbers and k-generalized Fibonacci numbers is obtained by
G(k)n = Gkm−rGrm+1, n = mk + r. k-generalized Fibonacci numbersnG(k)n
o n∈N are hold: • G(k)0 = bk, G (k) 1 = bk−1(a + b) and G (k) k = (a + b) k , • G(2)2n = (Gn)2, G(3)3n = (Gn)3,. . . , G(k)kn = (Gn)k, • G(2)2n+1= GnGn+1, G(3)3n+1= Gn2Gn+1,. . . , G(k)kn+1= Gkn−1Gn+1.
Now, we can give some properties of te k-Generalized Fibonacci numbers. Lemma 1. Relations of between Fn(2)and G(2)n are given by
i) G(2)2n = a2F(2) 2n−2+ 2abF (2) 2n−1+ b2F (2) 2n, ii) G(2)2n+1 = a2F(2) 2n−1+ ab ³ F2n(2)+ F2n(2)−1+ F2n(2)−2´+ b2F(2) 2n+1. Proof.
i) From (3), (5) and above properties, we have G(2)2n = (Gn)2
= (aFn−1+ bFn)2
= a2(Fn−1)2+ 2abFn−1Fn+ b2(Fn)2
= a2F2n(2)−2+ 2abF2n(2)−1+ b2F2n(2). ii) As similar to (i), we can write
G(2)2n+1 = GnGn+1 = (aFn−1+ bFn) (aFn+ bFn+1) = a2Fn−1Fn+ abFn−1Fn+1+ ab (Fn)2+ b2FnFn+1 = a2Fn−1Fn+ abFn−1(Fn+ Fn−1) + ab (Fn)2+ b2FnFn+1 = a2F(2) 2n−1+ abF (2) 2n−1+ abF (2) 2n−2+ abF (2) 2n + b2F (2) 2n+1.
The following theorem, for k = 2, gives us recurrence relation and generating function of k-generalized Fibonacci numbers.
Theorem 1. For k = 2, we have the following recurrence relation and the generating function of the k-generalized Fibonacci numbers:
i) G(2)n = G(2)n−1+ G (2) n−3+ G (2) n−4, n = 4, 5, . . . ii) g(2)n (x) = b2+ abx +¡a2+ ab¢x2+ abx3 1 − x − x3− x4 . Proof.
i) There are two cases of subscript n. If n is even integer, then we have G(2)2m = (Gm)2
= Gm(Gm−1+ Gm−2)
= GmGm−1+ Gm−2(Gm−1+ Gm−2)
= GmGm−1+ Gm−1Gm−2+ Gm−2Gm−2
= G(2)2m−1+ G(2)2m−3+ G(2)2m−4. If n is odd integer, then we have
G(2)2m+1 = GmGm+1
= Gm(Gm+ Gm−1)
= GmGm+ Gm−1(Gm−1+ Gm−2)
= GmGm+ Gm−1Gm−1+ Gm−1Gm−2
= G(2)2m+ G(2)2m−2+ G(2)2m−3.
ii) Let g(2)n (x) be generating function for the k-generalized Fibonacci
numbers, where (7) g(2)n (x) = ∞ X n=0 G(2)n xn.
If g(2)n (x) given in (7) multiply with x, x3 and x4, respectively, then we get
(8) xgn(2)(x) =P∞n=1G (2) n−1xn x3g(2) n (x) =P∞n=3G (2) n−3xn x4g(2) n (x) =P∞n=4G (2) n−4xn. ⎫ ⎪ ⎬ ⎪ ⎭
Consequently, by subtracting (8) from (7), it is obtained the equation gn(2)(x) = b
2+ abx +¡a2+ ab¢x2+ abx3
1 − x − x3− x4 ,
Theorem 2. We have the following recurrence relation and the generating function of the k-generalized Fibonacci numbers:
i) G(k)n = G(k)n−1+ G(k)n−2, n = mk + 1,
ii) g(k)n (x) =
bk+ abk−1x
1 − x − x2 , n = mk + 1.
Proof. The proof is done similarly to Theorem 1.
The following theorem gives us the sum of the k-generalized Fibonacci numbers. Theorem 3. Sum of k-generalized Fibonacci numbers is
k−1 X i=0 G(k)mk+i= Gm Gm−1 ³ G(k)(m+1)k− G(k)mk´.
Proof. By considering the equality G(k)mk+i= Gk−i
m Gim+1, we can write kX−1 i=0 G(k)mk+i = kX−1 i=0 µ Gm+1 Gm ¶i Gkm = Gkm ⎛ ⎜ ⎝ ³G m+1 Gm ´k − 1 Gm+1 Gm − 1 ⎞ ⎟ ⎠ = Gm Gm−1 ¡ Gkm+1− Gkm¢ = Gm Gm−1 ³ G(k)(m+1)k− G(k)mk´. References
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