Journal of Computer Science & Computational Mathematics, Volume 9, Issue 4, December 2019 DOI: 10.20967/jcscm.2019.04.003

Hyperbolic k-Fibonacci Quaternions

Fügen Torunbalci Aydin

Department of Mathematical Engineering, Faculty of Chemical and Metallurgical Engineering, Yildiz Technical University, Istanbul, TURKEY [email protected]

Abstract: In this paper, hyperbolic k-Fibonacci quaternions are de- where fined. Also, some algebraic properties of hyperbolic k-Fibonacci e 2 = e 2 = e 2 = e e e = 1, quaternions which are connected with hyperbolic numbers and k- 1 2 3 1 2 3 − Fibonacci numbers are investigated. Furthermore, d’Ocagne’s iden- e1e2 = e2e1 = e3,e2e3 = e3e2 = e1, tity, the Honsberger identity, Binet’s formula, Cassini’s identity and − − e3e1 = e1e3 = e2, n 1. Catalan’s identity for these quaternions are given. − ≥ In 2015, Ramirez [11] defined the k-Fibonacci and the k- Keywords: Fibonacci number, k-Fibonacci number, k-Fibonacci Lucas quaternions as follows: quaternion, k-Fibonacci dual quaternion, hyperbolic quaternion. D = F + iF + jF + kF k,n { k,n k,n+1 k,n+2 k,n+3 | F , n th k-Fib. number , 1. Introduction k,n − } The quaternions constitute an extension of complex numbers P = L + iL + jL + kL into a four-dimensional space and can be considered as four- k,n { k,n k,n+1 k,n+2 k,n+3 | dimensional vectors, in the same way that complex numbers Lk,n, n th k-Lucas number are considered as two-dimensional vectors. − } Quaternions were first described by Irish mathematician where i, j, k satisfy the multiplication rules (2). Hamilton in 1843. Hamilton [1] introduced a set of quater- In 2015, Polatlı Kızılate¸sand Kesim [12] defined split k- nions which can be represented as Fibonacci and split k-Lucas quaternions (Mk,n) and (Nk,n) respectively as follows: H = q = q0 + iq1 + jq2 + kq3 qi R, i = 0,1,2,3 { | ∈ } M = F + iF + jF + kF (1) k,n { k,n k,n+1 k,n+2 k,n+3 | F , n th k-Fibonacci number k,n − } where where i, j, k are split quaternionic units which satisy the mul- i2 = j2 = k2 = 1, ij = ji = k, jk = kj = i, tiplication rules − − − ki = ik = j. 2 2 2 − i = 1, j = k = ijk = 1, (2) − ij = ji = k, jk = kj = i, ki = ik = j. − − − − Several authors worked on different quaternions and their In 2018, Aydın Torunbalcı [13] defined k-Fibonacci dual generalizations ([2], [3], [4], [5], [6], [7]). quaternions as follows: Horadam [8] defined complex Fibonacci and Lucas quater- nions as follows DFk,n = DFk,n = Fk,n + iFk,n+1 + jFk,n+2 + kFk,n+3 { | Fk,n, n th k-Fibonacci number Qn = Fn + iFn+1 + jFn+2 + kFn+3 − } where i, j, k are dual quaternionic units which satisy the mul- and tiplication rules

Kn = Ln + iLn+1 + jLn+2 + kLn+3 i2 = j2 = k2 = 0, ij = ji = jk = kj = ki = ik = 0. where − − − i2 = j2 = k2 = 1, ij = ji = k, jk = kj = i, In 2018, Kösal [14] defined hyperbolic quaternions (K) as − − − follows: ki = ik = j. − K = q =a0 + ia1 + ja2 + ka3 a0, a1, a2, a3 R, In 2012, Halıcı [9] gave generating functions and Binet’s { | ∈ i,j,k / R formulas for Fibonacci and Lucas quaternions. In 2013, ∈ } Halıcı [10] defined complex Fibonacci quaternions as fol- where i,j,k are hyperbolic quaternionic units which satisy lows: the multiplication rules

2 2 2 HF C = Rn = Cn + e1Cn+1 + e2Cn+2 + e3Cn+3 i = j = k = 1, { | C = F + iF ,i2 = 1 ij = k = ji, jk = i = kj, ki = j = ik. n n n+1 − } − − − 64 Hyperbolic k-Fibonacci Quaternions

In this paper, the hyperbolic k-Fibonacci quaternions and Then, the addition and subtraction of two hyperbolic k- the hyperbolic k-Lucas quaternions will be defined respec- Fibonacci quaternions are defined in the obvious way, tively, as follows HFk,n HFk,m = (Fk,n + iFk,n+1 + jFk,n+2 + kFk,n+3) HFk,n = q = Fk,n + iFk,n+1 + jFk,n+2 + kFk,n+3 ± { | (Fk,m + iFk,m+1 + jFk,m+2 + kFk,m+3) Fk,n, nthk-Fib. num. ± } =(Fk,n Fk,m) + i(Fk,n Fk,m ) (3) ± +1 ± +1 + j(Fk,n+2 Fk,m+2) + k(Fk,n+3 Fk,m+3). (10) and ± ± Multiplication of two hyperbolic k-Fibonacci quaternions HLk,n = q = Lk,n + iLk,n+1 + jLk,n+2 + kLk,n+3 { | is defined by L , nthk-Lucas num. k,n } (4) HFk,nHFk,m = (Fk,n + iFk,n+1 + jFk,n+2 + kFk,n+3) where (Fk,m + iFk,m+1 + jFk,m+2 + kFk,m+3) =(Fk,nFk,m + Fk,n Fk,m + Fk,n Fk,m i2 = j2 = k2 = 1, +1 +1 +2 +2 (5) + Fk,n Fk,m ) ij = k = ji, jk = i = kj, ki = j = ik. +3 +3 − − − + i(Fk,nFk,m+1 + Fk,n+1Fk,m + Fk,n+2Fk,m+3 The aim of this work is to present in a unified manner Fk,n+3Fk,m+2) a variety of algebraic properties of both the hyperbolic k- − + j(F F + F F F F Fibonacci quaternions as well as the k-Fibonacci quater- k,n k,m+2 k,n+2 k,m − k,n+1 k,m+3 nions and hyperbolic quaternions. In accordance with these + Fk,n+3Fk,m+1) definitions, we given some algebraic properties and Binet’s + k(Fk,nFk,m+3 + Fk,n+3Fk,m + Fk,n+1Fk,m+2 formula for hyperbolic k-Fibonacci quaternions. More- Fk,n+2Fk,m+1) over, some sums formulas and some identities such as − =HFk,mHFk,n. (11) d’Ocagne’s, Honsberger, Cassini’s and Catalan’s identities 6 for these quaternions are given. The scaler and the vector parts of hyperbolic k-Fibonacci quaternion HFk,n are denoted by 2. Hyperbolic k-Fibonacci quaternions S F = Fk,n The k-Fibonacci sequence Fk,n n N [11] is defined as H k,n { } ∈ (12) VHFk,n = iFk,n+1 + jFk,n+2 + kFk,n+3. Fk,0 = 0,Fk,1 = 1 Fk,n+1 = kFk,n + Fk,n 1, n 1 Thus, hyperbolic k-Fibonacci quaternion HFk,n is given  − ≥ or by Fk,n = S F + V F . The conjugate of hyperbolic  H H k,n D k,n  2 3 4 2 k F F Fk,n n N = 0,1,k,k + 1,k + 2k,k + 3k + 1,... -Fibonacci quaternion H k,n is denoted by H k,n and it is { } ∈ { }  (6)  HF k,n = Fk,n iFk,n+1 jFk,n+2 kFk,n+3. (13) Here, k is a positive real number. In this section, firstly hy- − − − perbolic k-Fibonacci quaternions will be defined. Hyperbolic The norm of hyperbolic k-Fibonacci quaternion HFk,n is k-Fibonacci quaternions are defined by using the k-Fibonacci defined as follows numbers and hyperbolic quaternionic units as follows 2 HFk,n =HFk,nHF k,n k k (14) HFk,n = q = Fk,n + iFk,n+1 + jFk,n+2 + kFk,n+3 2 2 2 2 { | =Fk,n Fk,n+1 Fk,n+2 Fk,n+3. F , n th k-Fib. num. , − − − k,n − } (7) In the following theorem, some properties related to hy- perbolic k-Fibonacci quaternions are given. where Theorem 1: Let Fk,n and HFk,n be the n-th terms of 2 2 2 i = j = k = 1, k-Fibonacci sequence (Fk,n) and hyperbolic k-Fibonacci quaternion (HFk,n), respectively. In this case, for n 1 we ij = k = ji, jk = i = kj, ki = j = ik. ≥ − − − can give the following relations: Let HFk,n and HFk,m be two hyperbolic k-Fibonacci quaternions such that HFk,n+2 =kHFk,n+1 + HFk,n (15) 2 HFk,n =2Fk,nHFk,n HFk,nHF k,n (16) HFk,n = Fk,n + iFk,n+1 + jFn+2 + kFk,n+3 (8) − and HFk,n iHFk,n+1 jHFk,n+2 kHFk,n+3 F F F F F (9) − − − (17) H k,m = k,m + i k,m+1 + j k,m+2 + k k,m+3 = F F F F k,n − k,n+2 − k,n+4 − k,n+6 Fugen¨ Torunbalci Aydin 65

Proof: (15): By the equation (8) we get, Theorem 3: For n,m 0 the Honsberger identity for hyper- ≥ bolic k-Fibonacci quaternions HFk,n and HFk,m is given by HFk,n + kHFk,n+1 =(F + iF + jF + kF ) HFk,n+1HFk,m + HFk,nHFk,m 1 k,n k,n+1 k,n+2 k,n+3 − (19) + k(Fk,n+1 + iFk,n+2 + jFk,n+3 + kFk,n+4) = 2HFk,n+m + kFk,n+m+1 + Lk,n+m+5. Proof: (19) By using (11) =(Fk,n + kFk,n+1) + i(Fk,n+1 + kFk,n+2) + j(F + kF ) + k(F + kF ) k,n+2 k,n+3 k,n+3 k,n+4 HFk,n+1HFk,m =(Fk,n+1Fk,m + Fk,n+2Fk,m+1 =F + iF + jF + kF k,n+2 k,n+3 k,n+4 k,n+5 + Fk,n+3Fk,m+2 + Fk,n+4Fk,m+3) = F . H k,n+2 + i(Fk,n+1Fk,m+1 + Fk,n+2Fk,m + F F F F ) (16): By the equation (7) we get, k,n+3 k,m+3 − k,n+4 k,m+2 + j(Fk,n+1Fk,m+2 Fk,n+2Fk,m+3 2 2 2 2 2 − HFk,n =(Fk,n + Fk,n+1 + Fk,n+2 + Fk,n+3) + Fk,n+3Fk,m + Fk,n+4Fk,m+1)

+ 2Fk,n(iFk,n+1 + jFk,n+2 + kFk,n+3) + k(Fk,n+1Fk,m+3 + Fk,n+2Fk,m+2 2 =2Fk,nHFk,n 2F Fk,n+3Fk,m+1 + Fk,n+4Fk,m), − k,n − 2 2 2 2 HFk,nHFk,m 1 =(Fk,nFk,m 1 + Fk,n+1Fk,m + (Fk,n + Fk,n+1 + Fk,n+2 + Fk,n+3) − − + Fk,n+2Fk,m+1 + Fk,n+3Fk,m+2) =2Fk,nHFk,n HFk,nHF k,n. − + i(Fk,nFk,m + Fk,n+1Fk,m 1 − (17): By the equation (7) we get, + F F F F ) k,n+2 k,m+2 − k,n+3 k,m+1 + j(Fk,nFk,m+1 Fk,n+1Fk,m+2 HFk,n iHFk,n+1 jHFk,n+2 kHFk,n+3 − − − − + Fk,n+2Fk,m 1 + Fk,n+3Fk,m) = F F F F . − k,n − k,n+2 − k,n+4 − k,n+6 + k(Fk,nFk,m+2 + Fk,n+1Fk,m+1 Theorem 2: For m n + 1 the d’Ocagne’s identity for hy- Fk,n+2Fk,m + Fk,n+3Fk,m 1). ≥ − − perbolic k-Fibonacci quaternions HFk,m and HFk,n is given by Finally, adding by two sides to the side, we obtain

HFk,n+1HFk,m + HFk,nHFk,m 1 HFk,m HFk,n+1 HFk,m+1 HFk,n − n − = ( 1) [0 2iFk,m n 1 + 2jFk,m n 2 (18) =(Fk,n+m + Fk,n+m+2 − − 3 − − − − +k(Lk,m n + (k + 3k)Fk,m n)]. + Fk,n+m+4 + Fk,n+m+6) Proof: (18): By using− (7) − + 2iFk,n+m+1 + 2jFk,n+m+2

HFk,m HFk,n+1 HFk,m+1 HFk,n + 2kFk,n+m+3 − = (F F F F ) k,m k,n+1 − k,m+1 k,n =2(Fk,n+m + iFk,n+m+1 + jFk,n+m+2 +(Fk,m+1 Fk,n+2 Fk,m+2 Fk,n+1) − + kFk,n+m+3) Fk,n+m + Fk,n+m+2 +(F F F F ) − k,m+2 k,n+3 − k,m+3 k,n+2 +(F F F F ) + Lk,n+m+5) k,m+3 k,n+4 − k,m+4 k,n+3 +i[(F F F F ) =2HFk,n+m + kFk,n+m+1 + Lk,n+m+5. k,m k,n+2 − k,m+1 k,n+1 +(F F F F ) k,m+1 k,n+1 − k,m+2 k,n +(F F F F ) Here, the Honsberger identity of k-Fibonacci number k,m+2 k,n+4 − k,m+3 k,n+3 (Fk,m+3 Fk,n+3 Fk,m+4 Fk,n+2)] Fk,n+1Fk,m + Fk,nFk,m 1 = Fk,n+m in Falcon and Plaza − − − +j[(Fk,m Fk,n+3 Fk,m+1 Fk,n+2) [16] and Fk,n+1 + Fk,n 1 = Lk,n [11] was used. − − (Fk,m+1 Fk,n+4 Fk,m+2 Fk,n+3) Theorem 4: Let HFk,n and HLk,n be n-th terms of hy- − − +(Fk,m+2 Fk,n+1 Fk,m+3 Fk,n) perbolic k-Fibonacci quaternion (HFk,n) and hyperbolic k- − +(Fk,m+3 Fk,n+2 Fk,m+4 Fk,n+1)] Lucas quaternion (HLk,n), respectively. The following rela- − +k[(F F F F ) tion is satisfied k,m k,n+4 − k,m+1 k,n+3 +(F F F F ) k,m+1 k,n+3 − k,m+2 k,n+2 (Fk,m+2 Fk,n+2 Fk,m+3 Fk,n+1) HFk,n+1 + HFk,n 1 = HLk,n. (20) − − − +(Fk,m+3 Fk,n+1 Fk,m+4 Fk,n)] n − = ( 1) [0 2iFk,m n 1 + 2jFk,m n 2 and − − 3 − − − − +k(Lk,m n + (k + 3k)Fk,m n)]. − − HFk,n+2 HFk,n 2 = kHLk,n. (21) − − Here, d’Ocagne’s identity of k-Fibonacci number Proof: (20) From equation (8) n Fk,mFk,n+1 Fk,m+1Fk,n = ( 1) Fk,m n in Falcon and and identity Fk,n+1 + Fk,n 1 = Lk,n, n 1 Ramirez [11] − − − − ≥ Plaza [15] was used. between k-Fibonacci number and k-Lucas number, it follows 66 Hyperbolic k-Fibonacci Quaternions

that Proof: (24): Since n F = 1 (F + F 1) i=1 k,i k k,n+1 k,n − Falcon and Plaza [16], we get HFk,n+1 + HFk,n 1 P − n n n n =(Fk,n+1 + Fk,n 1) + i(Fk,n+2 + Fk,n) − HFk,s = Fk,s + i Fk,s+1 + j Fk,s+2 + j(Fk,n+3 + Fk,n+1) + k(Fk,n+4 + Fk,n+2) s=1 s=1 s=1 s=1 X Xn X X =(Lk,n + iLk,n+1 + jLk,n+2 + kLk,n+3) + k Fk,s L . +3 =H k,n s X=1 1 (21) From equation (7) and identity Fk,n+2 Fk,n 2 = = (Fk,n + Fk,n 1) − − k { +1 − kLk,n, n 1 between k-Fibonacci number and k-Lucas ≥ + i(F + F k 1) number, it follows that k,n+2 k,n+1 − − + j[F + F (k2 + 1) k] k,n+3 k,n+2 − − HFk,n+2 HFk,n 2 3 2 − − + k[Fk,n+4 + Fk,n+3 (k + 2k) (k + 1)] =(Fk,n+2 Fk,n 2) + i(Fk,n+3 Fk,n 1) − − } − − − − 1 + j(F F ) + k(F F ) = (Fk,n+1 + iFk,n+2 + jFk,n+3 + kFk,n+4) k,n+4 − k,n k,n+5 − k,n+1 k { =(kL + kL + kL + kL ) k,n i k,n+1 j k,n+2 k k,n+3 + (Fk,n + iFk,n+1 + jFk,n+2 + kFk,n+3) =kHLk,n. [1 + i(k + 1) + j(k2 + k + 1) + k(k3 + k2 + 2k + 1)] − } 1 = HFk,n+1 + HFk,n Theorem 5: Let HF k,n be conjugation of hyperbolic k- k { Fibonacci quaternion ( F ). In this case, we can give the [Fk,1 + i(Fk,2 + Fk,1) + j(Fk,3 + Fk,2) H k,n − following relations between these quaternions: + k(F + F )] k,4 k,3 } 1 HFk,n + HF k,n = 2Fk,n (22) = (HFk,n+1 + HFk,n HFk,1 HFk,0). k − − n F 1 F F F F F (25): Using k,2i 1 = k k,2n, Falcon and Plaza [16], H k,nH k,n + H k,n 1H k,n 1 i=1 − − − (23) we get = Fk,2n 1 Fk,2n+1 Fk,2n+3 Fk,2n+5 P − − − − Proof: (22): By using (14), we get n HFk,2s 1 − HFk,n + HF k,n s=1 X =(F + iF + jF + kF ) 1 k,n k,n+1 k,n+2 k,n+3 = (F ) + i(F 1) + j(F k) k { k,2n k,2n+1 − k,2n+2 − + (Fk,n iFk,n+1 jFk,n+2 kFk,n+3) − − − 2 + k[Fk,2n+3 (k + 1)] =2Fk,n. − } 1 = [F + iF + jF + kF ] (23): By using (15), we get k { k,2n k,2n+1 k,2n+2 k,2n+3 [i + kj + (k2 + 1)k] HFk,nHF k,n + HFk,n 1HF k,n 1 − } − − 1 2 2 2 2 = HFk,2n (Fk,0 + iFk,1 + jFk,2 + kFk,3) =(Fk,n + Fk,n 1) (Fk,n+1 + Fk,n) k { − } − − (F 2 + F 2 ) (F 2 + F 2 ) 1 k,n+2 k,n+1 k,n+3 k,n+2 = (HFk,2n HFk,0). − − k − =Fk,2n 1 Fk,2n+1 Fk,2n+3 Fk,2n+5. − n − − − 1 2 2 (26): Using Fk,2i = k (Fk,2n+1 1) Falcon and Plaza where the identity of k-Fibonacci number Fk,n + Fk,n+1 = i=1 − F , n 0 Ramirez [11] was used. [16], we obtainP k,2n+1 ≥ Theorem 6: Let HFk,n be hyperbolic k-Fibonacci quater- n nion. Then, we have the following identities HFk,2s s=1 n X 1 1 HFk,s = (HFk,n+1 + HFk,n HFk,1 HFk,0), = (Fk, n 1) + i[Fk, n k] k − − k { 2 +1 − 2 +2 − s=1 X + j[F (k2 + 1)] + k[F (k3 + 2k)] (24) 2n+3 − k,2n+4 − } n 1 1 = (Fk,2n+1 + iFk,2n+2 + jFk,2n+3 + kFk,2n+4) HFk,2s 1 = (HFk,2n HFk,0), (25) k { − k − s=1 [1 + ki + (k2 + 1)j + (k3 + 2k)k] X n − } 1 1 HFk,2s = (HFk,2n+1 HFk,1). (26) = ( F F ). k − H k,2n+1 H k,1 s k − X=1 Fugen¨ Torunbalci Aydin 67

Proof: (28): By using (7) and (11), we get

Theorem 7 (Binet’s Formula): Let HFk,n be hyperbolic k- 2 HFk,n 1HFk,n+1 (HFk,n) Fibonacci quaternion. For n 1, Binet’s formula for these − − ≥ 2 quaternions is as follows Falcon and Plaza [16]: =[(Fk,n 1Fk,n+1 Fk,n) − − + (F F F 2 ) k,n k,n+2 − k,n+1 1 2 n ˆ n + (Fk,n+1Fk,n+3 Fk,n+2) HFk,n = ααˆ ββ (27) − √ 2 − k + 4 + (F F F 2 )]   k,n+2 k,n+4 − k,n+3 + i[(Fk,n 1Fk,n+2 Fk,nFk,n+1) − − where (F F F F )] − k,n+1 k,n+4 − k,n+2 k,n+3 + j[(Fk,n 1Fk,n+3 Fk,nFk,n+2) 2 − − αˆ = 1 + i[(k β)] + j[(k + 1) kβ] (F F F F ) − − − k,n k,n+4 − k,n+1 k,n+3 + k[(k3 + 2k) (k2 + 1)β], + (F F F F ) − k,n+1 k,n+1 − k,n+2 k,n + (F F F F )] k,n+2 k,n+2 − k,n+3 k,n+1 and + k[(Fk,n 1Fk,n+4 Fk,nFk,n+3) − − + (F F F F ) k,n k,n+3 − k,n+1 k,n+2 ˆ 2 + (Fk,n+2Fk,n+1 Fk,n+3Fk,n)] β = 1 + i(α k) + j[kα (k + 1)] − − − − n 2 3 + k[(k2 + 1)α (k3 + 2k)]. =( 1) [2ki + 2(k + 1)j + (k + 2k)k]. − − Proof: The characteristic equation of recurrence relation Here, the identity of the k-Fibonacci number HFk,n+2 = kHFk,n+1 + HFk,n is 2 n Fk,n 1Fk,n+1 Fk,n = ( 1) , Falcon and Plaza [16] − − − was used. (Catalan’s Identity): Let F be hyperbolic t2 kt 1 = 0. Theorem 9 H k,n − − k-Fibonacci quaternion. For n 1, Catalan’s identity for ≥ HFk,n is as follows:

The roots of this equation are 2 HFk,n+r 1HFk,n+r+1 HFk,n+r − − (29) = ( 1)n+r[0 + 2ki + 2(k2 + 1)j + (k3 + 2k)k]. − k + √k2 + 4 k √k2 + 4 Proof: (29): By using (7) and (11), we get α = and β = − 2 2 2 HFk,n+r 1HFk,n+r+1 HFk,n+r − − 2 2 =[(Fk,n+r 1Fk,n+r+1 Fk,n+r) where α + β = k , α β = √k + 4, αβ = 1. − − − − 2 + (Fk,n+rFk,n+r+2 Fk,n+r+1) Using recurrence relation and initial values HFk,0 = − 2 2 (0,1,k,k + 1), + (Fk,n+r+1Fk,n+r+3 Fk,n+r+2) 2 3 − HFk,1 = (1,k,k +1,k +2k), the Binet formula for HFk,n 2 + (Fk,n+r+2Fk,n+r+4 F )] is − k,n+r+3 + i[(Fk,n+r 1Fk,n+r+2 Fk,n+rFk,n+r+1) − − + (Fk,n+r+1Fk,n+r+4 Fk,n+r+2Fk,n+r+3)] n n 1 n n − Fk,n = Aα + Bβ = ααˆ ββˆ , H 2 + j[(Fk,n+r 1Fk,n+r+3 Fk,n+rFk,n+r+2) √k + 4 − − − h i + (F F F F ) k,n+r+1 k,n+r+1 − k,n+r+2 k,n+r (Fk,n+rFk,n+r+4 Fk,n+r+1Fk,n+r+3) HFk,1 βHFk,0 αHFk,0 HFk,1 − − where A = − , B = − and + (F F F F )] α β α β k,n+r+2 k,n+r+2 − k,n+r+3 k,n+r+1 2 −3 ˆ 2 −3 + k[(Fk,n+r 1Fk,n+r+4 Fk,n+rFk,n+r+3) αˆ = 1 + iα + jα + kα , β = 1 + iβ + jβ + kβ . − − + (Fk,n+r+2Fk,n+r+1 Fk,n+r+3Fk,n+r) Theorem 8 (Cassini’s Identity): Let HFk,n be hyperbolic k- − + (F F F F )] Fibonacci quaternion. For n 1, Cassini’s identity for k,n+r k,n+r+3 − k,n+r+1 k,n+r+2 ≥ n+r 2 3 HFk,n is as follows: =( 1) [0 + 2ki + 2(k + 1)j + (k + 2k)k]. − Here, the identity of the k-Fibonacci number 2 HFk,n 1HFk,n+1 (HFk,n) 2 n+r Fk,n+r 1Fk,n+r+1 Fk,n r = ( 1) , Falcon and − − (28) − − + − = ( 1)n[2ki + 2(k2 + 1)j + (k3 + 2k)k]. Plaza [16] was used. − 68 Hyperbolic k-Fibonacci Quaternions

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