PAijpam.eu A NOTE ON BICOMPLEX FIBONACCI AND LUCAS NUMBERS Semra Kaya Nurkan 1, İlkay Arslan Güven2

Transcription

1 International Journal of Pure Applied Mathematics Volume 120 No , ISSN: (printed version); ISSN: (on-line version) url: doi: /ijpam.v120i3.7 PAijpam.eu A NOTE ON BICOMPLEX FIBONACCI AND LUCAS NUMBERS Semra Kaya Nurkan 1, İlkay Arslan Güven2 1 Department of Mathematics Uşak University 64200, Uşak, TURKEY 2 Department of Mathematics Gaziantep University 27310, Gaziantep, TURKEY Abstract: In this study, we define a new type of Fibonacci Lucas numbers which are called bicomplex Fibonacci bicomplex Lucas numbers. We obtain the well-known properties e.g. D ocagnes, Cassini, Catalan for these new types. We also give the identities of negabicomplex Fibonacci negabicomplex Lucas numbers, Binet formulas relations of them. AMS Subject Classification: 11B39, 11E88, 11R52 Key Words: Fibonacci numbers, Lucas numbers, bicomplex numbers 1. Introduction Fibonacci numbers are invented by Italian mathematician Leonardo Fibonacci when he wrote his first book Liber Abaci in 1202 contains many elementary problems, including famous rabbit problem. The Fibonacci numbers are the numbers of the following integer sequence; 1,1,2,3,5,8,13,21,34,55,89,... The sequence of Fibonacci numbers which is denoted by F n is defined as the Received: February 21, 2017 Revised: March 28, 2018 Published: December 4, 2018 c 2018 Academic Publications, Ltd. url: Correspondence author

2 366 S.K. Nurkan, İ.A. Güven linear reccurence relation F n = F n 1 +F n 2 withf 1 = F 2 = 1n Z. Alsoasaresultofthisrelation wecefine F 0 = 0. Fibonacci numbers are connected with the golden ratio as; the ratio of two consecutive Fibonacci numbers approximates the golden ratio 1, Fibonacci numbers are closely related to Lucas numbers which are named after the mathematician Francois Edouard Anatole Lucas who worked on both Fibonacci Lucas numbers. The integer sequence of Lucas numbers denoted by L n is given by 2,1,3,4,7,11,18,29,47,... with the same reccurence relation L n = L n 1 +L n 2. where L 0 = 2, L 1 = 1, L 2 = 3 n Z. There are many works on Fibonacci Lucas numbers in literature. The properties, relations, results between Fibonacci Lucas numbers can be found in, Dunlap[3], Güven Nurkan[5], Koshy[14], Vajda[21], Verner Hoggatt[22]. Also Fibonacci Lucas quaternions were described by Horadam in [10], then many studies related with these quaternions were done in Akyiğit [1], Halici [6], Iyer [11], Nurkan[16], Swamy [20]. Corrado Segre introduced bicomplex numbers in 1892 [19]. The bicomplex numbers are a type of complex Clifford algebra, one of several possible generalizations of the ordinary complex numbers. The complex numbers denoted as C are defined by The bicomplex numbers are defined by C = { a+bi a,b R i 2 = 1 }. C 2 = { z 1 +z 2 j z 1,z 2 C j 2 = 1 }. Since z 1 z 2 are complex numbers, writing z 1 = a+bi z 2 = c+di gives us another way to represent the bicomplex numbers as; z 1 +z 2 j = a+bi+cj +dij. where a,b,c,d R. Thus the set of bicomplex numbers can be expressed by { } a+bi+cj +dij a,b,c,d R C 2 = i 2 = 1,j 2. = 1, ij = ji = k

3 A NOTE ON BICOMPLEX FIBONACCI For any bicomplex numbers x = a 1 +b 1 i+c 1 j+d 1 ij y = a 2 +b 2 i+c 2 j+ d 2 ij, the addition multiplication of these bicomplex numbers are given respectively by x+y = (a 1 +a 2 )+(b 1 +b 2 )i+(c 1 +c 2 )j +(d 1 +d 2 )ij (1.1) x y = (a 1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 )+(a 1 b 2 +b 1 a 2 c 1 d 2 d 1 c 2 )i (1.2) +(a 1 c 2 +c 1 a 2 b 1 d 2 d 1 b 2 )j +(a 1 d 2 +d 1 a 2 +b 1 c 2 +c 1 b 2 )ij. The multiplication of a bicomplex number by a real scalar λ is given by λx = λa 1 +λb 1 i+λc 1 j +λd 1 ij. The set C 2 forms a commutative ring with addition multiplication it is a real vector space with addition scalar multiplication [17]. In C, the complex conjugate of z = a + bi is z = a bi. In C 2, for a bicomplex number x = (a 1 +b 1 i)+(c 1 +d 1 i)j, there are different conjugations which are; x = [(a 1 +b 1 i)+(c 1 +d 1 i)j] = (a 1 b 1 i)+(c 1 d 1 i)j x = [(a 1 +b 1 i)+(c 1 +d 1 i)j] = (a 1 +b 1 i) (c 1 +d 1 i)j (1.3) x = [(a 1 +b 1 i)+(c 1 +d 1 i)j] = (a 1 b 1 i) (c 1 d 1 i)j [18]. Then the following equation are written in [12]; x x = (a 2 1 +b2 1 c2 1 d2 1 )+2(a 1c 1 +b 1 d 1 )j x x = (a 2 1 b 2 1 +c 2 1 d 2 1)+2(a 1 b 1 +c 1 d 1 )i (1.4) x x = (a 2 1 +b 2 1 +c 2 1 +d 2 1)+2(a 1 d 1 b 1 c 1 )ij. Also in [18], the modulus of a bicomplex number x is defined by x i = x x x j = x x (1.5) x k = x x x = Re(x x ) where the names are i modulus, j modulus, k modulus real modulus, respectively.

4 368 S.K. Nurkan, İ.A. Güven Rochon Shapiro gave the detailed algebraic properties of bicomplex hyperbolic numbers in[18]. The generalized bicomplex numbers were defined by Karakuş Aksoyak in [12], they gave some algebratic properties of them used generalized bicomplex number product to show that R 4 R 4 2 were Lie groups. Also Luna-Elizarraras et al [15], introduced the algebra of bicomplex numbers described how to define elementary functions their inverse functions in that algebra. In this paper, we define bicomplex Fibonacci bicomplex Lucas numbers by combining bicomplex numbers Fibonacci, Lucas numbers. We give some identities Binet formulas of these new numbers. 2. Bicomplex Fibonacci Numbers Definition 1: The bicomplex Fibonacci bicomplex Lucas numbers are defined respectively by BF n = F n +F n+1 i+f n+2 j +F n+3 k (2.1) BL n = L n +L n+1 i+l n+2 j +L n+3 k (2.2) where F n is the n th Fibonacci number, L n is the n th Lucas number i,j,k are bicomplex units which satisfy the commutative multiplication rules: i 2 = 1, j 2 = 1, k 2 = 1 ij = ji = k, jk = kj = i, ik = ki = j. Starting from n = 0, the bicomplex Fibonacci bicomplex Lucas numbers can be written respectively as; BF 0 = 1i+1j +2k, BF 1 = 1+1i+2j +3k, BF 2 = 1+2i+3j +5k,... BL 0 = 2+1i+3j +4k, BL 1 = 1+3i+4j +7k, BL 2 = 3+4i+7j +11k,... Let BF n = F n +F n+1 i+f n+2 j+f n+3 k BF m = F m +F m+1 i+f m+2 j+ F m+3 k be two bicomplex Fibonacci numbers. By taking into account the equations (1.1) (1.2), the addition, subtraction multiplication of these numbers are given by BF n ±BF m = (F n ±F m )+(F n+1 ±F m+1 )i+(f n+2 ±F m+2 )j

5 A NOTE ON BICOMPLEX FIBONACCI (F n+3 ±F m+3 )k BF n BF m = (F n F m F n+1 F m+1 F n+2 F m+2 +F n+3 F m+3 ) +(F n F m+1 +F n+1 F m F n+2 F m+3 +F n+3 F m+2 )i +(F n F m+2 +F n+2 F m F n+1 F m+3 F n+3 F m+1 )j +(F n F m+3 +F n+3 F m +F n+1 F m+2 +F n+2 F m+1 )k. Definition 2: A bicomplex Fibonacci number can also be expressed as BF n = (F n +F n+1 i) + (F n+2 +F n+3 i)j. In that case,there is three different conjugations with respect to i, j k; for bicomplex Fibonacci numbers as follows: BF n i = [(Fn +F n+1 i)+(f n+2 +F n+3 i)j] i = (F n F n+1 i)+(f n+2 F n+3 i)j BF n j = [(Fn +F n+1 i)+(f n+2 +F n+3 i)j] j = (F n +F n+1 i) (F n+2 +F n+3 i)j BF n k = [(Fn +F n+1 i)+(f n+2 +F n+3 i)j] k = (F n F n+1 i) (F n+2 F n+3 i)j. By the Definition 2 the equations (1.4) (2.1), we can write BF n BF n i = (F2n+1 F 2n+7 )+2(F 2n+3 )j BF n BF n j = ( F 2 n F 2 n+1 +F 2 n+3 F 2 n+4) +2(Fn F n+1 +F n+2 F n+3 )i BF n BF n k = (F2n+1 +F 2n+7 )+2( 1) n+1 k. Definition 3: Let a bicomplex Fibonacci number be BF n = F n +F n+1 i+ F n+2 j + F n+3 k. The i modulus, j modulus, k modulus real modulus of BF n are given respectively as; i BF n i = BF n BF n j BF n j = BF n BF n k BF n k = BF n BF n BF n = F 2n+1 +F 2n+7

6 370 S.K. Nurkan, İ.A. Güven In [16], we proved some summation equations by supposing, for i 0 M α m F m+i + m=0 M β m L m+i = 0 (2.3) m=0 where α m β m are fixed numbers. By using the equations (2.1), (2.2), (2.3) taking i = 0,1,2,3, we can write the following equation clearly; M α m BF m + m=0 M β m BL m = 0 (2.4) m=0 Now let us give theorems which give some properties. Theorem 1. LetBF n BL n beabicomplex Fibonacci abicomplex Lucas number, respectively. For n 0, the following relations hold: 1) BF n +BF n+1 = BF n+2 2) BL n +BL n+1 = BL n+2 3) BL n = BF n 1 +BF n+1 4) BL n = BF n+2 BF n 2 5) BF 2 n +BF 2 n+1 = BF 2n+1 +F 2n+2 +F 2n+5 3iF 2n+5 jf 2n+6 +3kF 2n+4 6) BF 2 n+1 BF2 n 1 = 2BF 2n +F 2n+4 +F 2n 1 +2( F 2n+5 i F 2n+4 j +F 2n+3 k) 7) BF n BF m +BF n+1 BF m+1 = 2BF n+m+1 +2F n+m+4 F n+m+1 2F n+m+6 i 2F n+m+5 j +2F n+m+4 k 8) BF n BF n+1 i+bf n+2 j BF n+3 k = 5F n+3 Proof. By using the equations of (2.1), (2.2), (2.4) taking appropriate fixed numbers in the equation (2.4), the proofs of 1), 2), 3), 4) are clear. From the definition of Fibonacci number, the bicomplex Fibonacci number in (2.1), the equations F 2 n +F2 n+1 = F 2n+1, F 2 n+1 F2 n 1 = F 2n F n F m + F n+1 F m+1 = F n+m+1 (see Vajda [21]), we get BF 2 n +BF 2 n+1 = (F n +F n+1 i+f n+2 j +F n+3 k) 2 +(F n+1 +F n+2 i+f n+3 j +F n+4 k) 2 F 2n+3 +2(F n F n+1 F n+2 F n+3 )i = +2(F n F n+2 F n+1 F n+3 )j +2(F n F n+3 +F n+1 F n+2 )k

7 A NOTE ON BICOMPLEX FIBONACCI F 2n+5 +2(F n+1 F n+2 F n+3 F n+4 )i +2(F n+1 F n+3 F n+2 F n+4 )j +2(F n+1 F n+4 +F n+2 F n+3 )k = F 2n+1 +F 2n+2 i+f 2n+3 j +F 2n+4 k +F 2n+2 +F 2n+5 3F 2n+5 i F 2n+6 j +3F 2n+4 k = BF 2n+1 +F 2n+2 +F 2n+5 3iF 2n+5 jf 2n+6 +3kF 2n+4. BF 2 n+1 BF 2 n 1 = (F n+1 +F n+2 i+f n+3 j +F n+4 k) 2 (F n 1 +F n i+f n+1 j +F n+2 k) 2 F 2n+5 +2(F n+1 F n+2 F n+3 F n+4 )i = +2(F n+1 F n+3 F n+2 F n+4 )j +2(F n+1 F n+4 +F n+2 F n+3 )k F 2n+1 +2(F n 1 F n F n+1 F n+2 )i +2(F n 1 F n+1 F n F n+2 )j +2(F n 1 F n+2 +F n F n+1 )k = 2BF 2n +F 2n+4 +F 2n 1 +2( F 2n+5 i F 2n+4 j +F 2n+3 k) BF n BF m +BF n+1 BF m+1 = (F n +F n+1 i+f n+2 j +F n+3 k).(f m +F m+1 i+f m+2 j +F m+3 k) +(F n+1 +F n+2 i+f n+3 j +F n+4 k).(f m+1 +F m+2 i+f m+3 j +F m+4 k) = 2(F n+m+1 +F n+m+2 i) +2(F n+m+3 j +F n+m+4 k) +2F n+m+4 F n+m+1 2F n+m+6 i 2F n+m+5 j +2F n+m+4 k = 2BF n+m+1 +2F n+m+4 F n+m+1 2F n+m+6 i 2F n+m+5 j +2F n+m+4 k BF n BF n+1 i+bf n+2 j BF n+3 k = (F n +F n+1 i+f n+2 j +F n+3 k)

8 372 S.K. Nurkan, İ.A. Güven (F n+1 +F n+2 i+f n+3 j +F n+4 k)i +(F n+2 +F n+3 i+f n+4 j +F n+5 k)j (F n+3 +F n+4 i+f n+5 j +F n+6 k)k = F n +F n+2 F n+4 F n+6 = 5F n+3. Theorem 2. For n,m 0 the D ocagnes identity for bicomplex Fibonacci numbers BF n BF m is given by BF m BF n+1 BF m+1 BF n = ( 1) n BF m n F m n +F m n+1 i +( 1) n+1 (F m n 2 +2F m n+2 )j. +2F m n 1 k Proof. If we decide the equation (2.1) the D ocagnes identity for Fibonacci numbers F m F n+1 F m+1 F n = ( 1) n F m n (see Weisstein [23] ), we obtain the following calculations as; BF m BF n+1 BF m+1 BF n = (F m +F m+1 i+f m+2 j +F m+3 k) (F n+1 +F n+2 i+f n+3 j +F n+4 k) (F m+1 +F m+2 i+f m+3 j +F m+4 k) (F n +F n+1 i+f n+2 j +F n+3 k) = [F n+2 (F m +F m+4 ) F m+2 (F n +F n+4 )]i +[2( 1) n (F m n 2 +F m n+2 )]j +[( 1) n (F m n 2 +F m n+2 )]k +( 1) n BF m n ( 1) n BF m n = ( 1) n BF m n F m n +F m n+1 i +( 1) n+1 (F m n 2 +2F m n+2 )j. +2F m n 1 k Theorem 3. If BF n BL n are bicomplex Fibonacci bicomplex Lucas numbers respectively, then for n 0, the identities of negabicomplex Fibonacci negabicomplex Lucas numbers are BF n = ( 1) n+1 BF n +( 1) n L n (i+j +2k)

9 A NOTE ON BICOMPLEX FIBONACCI BL n = ( 1) n BL n +( 1) n+1 5F n (i+j +2k) Proof. From the equations (2.1) the identity of negafibonacci numbers which is F n = ( 1) n+1 F n (see Knuth [13], Dunlap [3]), we have BF n = F n +F n+1 i+f n+2 j +F n+3 k = F n +F (n 1) i+f (n 2) j +F (n 3) k = ( 1) n+1 F n +( 1) n F n 1 i+( 1) n+1 F n 2 j +( 1) n F n 3 k = ( 1) n+1 (F n +F n+1 i+f n+2 j +F n+3 k) ( 1) n+1 F n+1 i ( 1) n+1 F n+2 j ( 1) n+1 F n+3 k+( 1) n F n 1 i +( 1) n+1 F n 2 j +( 1) n F n 3 k = ( 1) n+1 BF n +( 1) n (F n+1 +F n 1 )i +( 1) n (F n+2 F n 2 )j +( 1) n (F n+3 +F n 3 )k In this equation, we take into account that F n 1 +F n+1 = L n, F n+2 F n 2 = L n, F n+3 +F n 3 = 2L n (see Vajda [21]) the identity of negalucas numbers L n = ( 1) n L n (see Knuth [13],Dunlap [3]), thus we obtain; BF n = ( 1) n+1 BF n +( 1) n L n i+( 1) n L n j +( 1) n 2L n = ( 1) n+1 BF n +( 1) n L n (i+j +2k) Now by using the equation (2.2) the identity of negalucas numbers, we get BL n = L n +L n+1 i+l n+2 j +L n+3 k = L n +L (n 1) i+l (n 2) j +L (n 3) k = ( 1) n L n +( 1) n 1 L n 1 i+( 1) n L n 2 j +( 1) n 1 L n 3 k = ( 1) n (L n +L n+1 i+l n+2 j +L n+3 k) ( 1) n L n+1 i ( 1) n L n+2 j ( 1) n L n+3 k+( 1) n 1 L n 1 i+( 1) n L n 2 j +( 1) n 1 L n 3 k = ( 1) n BL n ( ) +( 1) n+1 (L n+1 +L n 1 )i +(L n+2 L n 2 )j +(L n+3 +L n 3 )k { 5F m F n, if n is odd, Here if we use the identity L m+n +L m n = (see Koshy, L m L n, otherwise [14]), namely L n 1 +L n+1 = 5F n, the definition of dual Lucas number the

10 374 S.K. Nurkan, İ.A. Güven identity of negafibonacci number in last equation, we complete the proof as; BL n = ( 1) n BL n +( 1) n+1 5F n i+( 1) n+1 5F n j +( 1) n+1 10F n k = ( 1) n BL n +( 1) n+1 5F n (i+j +2k) The Binet formulas for Fibonacci Lucas numbers are given by (see Koshy, [14]) F n = αn β n L n = α n +β n α β where α = 1+ 5 β = Theorem 4. Let BF n BL n be bicomplex Fibonacci bicomplex Lucas numbers, respectively. For n 0, the Binet formulas for these numbers are given as; BF n = ααn ββ n α β BL n = αα n +ββ n where α = 1+iα+jα 2 +kα 3 β = 1+iβ +jβ 2 +kβ 3. Proof. By using the Binet formulas for Fibonacci Lucas numbers, taking α = 1 + iα + jα 2 + kα 3 β = 1 + iβ + jβ 2 + kβ 3, we find the result clearly as; BF n = F n +F n+1 i+f n+2 j +F n+3 k = αn β n α β + αn+1 β n+1 α β i+ αn+2 β n+2 j + αn+3 β n+3 α β α β = αn( 1+iα+jα 2 +kα 3) β n( 1+iβ +jβ 2 +kβ 3) α β = ααn ββ n α β BL n = L n +L n+1 i+l n+2 j +L n+3 k

11 A NOTE ON BICOMPLEX FIBONACCI = α n +β n +(α n+1 +β n+1 )i+(α n+2 +β n+2 )j +(α n+3 +β n+3 )k = α n (1+iα+jα 2 +kα 3 )+β n (1+iβ +jβ 2 +kβ 3 ) = αα n +ββ n. Theorem 5. Let BF n BL n be bicomplex Fibonacci bicomplex Lucas numbers. For n 1,the Cassini identities for BF n BL n are given by; BF n+1 BF n 1 BF 2 n = 3( 1) n (2j +k) BL n+1 BL n 1 BL 2 n = 5( 1)n 1 (2j +k) Proof. From the equation (2.1), we have BF n+1 BF n 1 BF 2 n = (F n+1 +F n+2 i+f n+3 j +F n+4 k) (F n 1 +F n i+f n+1 j +F n+2 k) (F n +F n+1 i+f n+2 j +F n+3 k) 2. If we use the identity F m F n+1 F m+1 F n = ( 1) n F m n (see Weisstein [23] ) F n = ( 1) n+1 F n (see Knuth [13] ) in the above equation, we get BF n+1 BF n 1 BFn [2( 1) 2 = n+2] [ 3j + 3( 1) n+2] k = 3( 1) n (2j +k) Similarly for bicomplex Lucas numbers we can simply get; BL n+1 BL n 1 BL 2 n = (L n+1 +L n+2 i+l n+3 j +L n+4 k)(l n 1 +L n i+l n+1 j +L n+2 k) (L n +L n+1 i+l n+2 j +L n+3 k) 2. By using the identity of Lucas numbers which are L n 1 L n+1 L 2 n = 5( 1)n 1 (see Koshy [14]) L n+2 = L n+1 +L n (see Dunlap [3]) in the above equation, we compute the following expression; BL n+1 BL n 1 BL 2 n == 5( 1) n 1 (1+ε) = 5( 1) n 1 (2j +k).

12 376 S.K. Nurkan, İ.A. Güven Theorem 6. Let BF n be a bicomplex Fibonacci number. The Catalan s identity for BF n is given by [ BFn 2 BF n+r BF n r = ( 1) n r 2 ( Fr 2 2 ] r) +F2 j + ( Fr+1 2 ). +F2 r 2 F2 r F2 r 3 k Proof. From the equation (1.4), we have F 2 BFn 2 n F2 n+1 F2 n+2 +F2 n+3 BF n+r BF n r = F n+r F n r +F n+r+1 F n r 1 +F n+r+2 F n r+2 F n+r+3 F n r+3 2F n F n+1 2F n+2 F n+3 + F n+r F n r+1 F n+r+1 F n r i +F n+r+2 F n r+3 +F n+r+3 F n r+2 2F n F n+2 2F n+1 F n+3 + F n+r F n r+2 F n+r+2 F n r j +F n+r+1 F n r+3 +F n+r+3 F n r+1 2F n F n+3 +2F n+1 F n+2 + F n+r F n r+3 F n+r+3 F n r F n+r+1 F n r+2 F n+r+2 F n r+1 k Here using the Catalan s identity for Fibonacci numbers Fn 2 F n r F n+r = ( 1) n r Fr 2 (see Weisstein [23]), Fn+1 2 F2 n 1 = F 2n (see Vajda [21]) doing necessary calculations, we obtain that; [ BFn 2 BF n+rbf n r = ( 1) n r 2 ( Fr 2 2 ] r) +F2 j + ( Fr+1 2 +F2 r 2 F2 r Fr 3 2 ) k References [1] M. Akyiğit, H.H. Kösal M. Tosun, Split Fibonacci quaternions, Adv. in Appl. Clifford Algebras, 23 (2013), , doi /s [2] W. K. Clifford, Preliminary sketch of bi-quaternions, Proc. London Math. Soc., 4 (1873), [3] R. A. Dunlap, The Golden Ratio Fibonacci Numbers, World Scientific Pub. Co. Pte. Ltd., (1997). [4] H. W. Guggenheimer, Differential Geometry, McGraw-Hill Comp., New York (1963). [5] İ. A. Güven S. K. Nurkan, A new approach to Fibonacci, Lucas numbers dual vectors, Adv. in Appl. Clifford Algebras, 25 (2015), , doi /s

13 A NOTE ON BICOMPLEX FIBONACCI [6] S. Halıcı, On Fibonacci quaternions, Adv. in Appl. Clifford Algebras, 22 (2012), , doi /s [7] S. Halıcı, On complex Fibonacci quaternions, Adv. in Appl. Clifford Algebras, 23 (2013), , doi /s [8] W. R. Hamilton, Elements of Quaternions, Longmans, Green Co., London, (1866). [9] A. F. Horadam, A generalized Fibonacci sequence, American Math. Monthly, 68 (1961), [10] A. F. Horadam, Complex Fibonacci numbers Fibonacci quaternions, American Math. Monthly, 70 (1963), [11] M. R. Iyer, Some results on Fibonacci quaternions, The Fibonacci Quarterly, 7(2)(1969), [12] S. Ö. Karakuş F. K. Aksoyak, Generalized bicomplex numbers Lie groups, Adv. in Appl. Clifford Algebras, 25 (2015), , doi /s x. [13] D. Knuth, Negafibonacci Numbers Hyperbolic Plane, Annual Meeting of the Math. Association of America, (2013). [14] T. Koshy, Fibonacci Lucas Numbers With Applications, A Wiley-Intersience Publication, USA (2001). [15] M.E. Luna-Elizarraras, M. Shapiro, D.C. Struppa A. Vajiac, Bicomplex numbers their elementary functions, CUBO A Math. Jour., 14(2) (2012), 61-80, doi.org/ /s [16] S. K. Nurkan İ. A. Güven, Dual Fibonacci quaternions, Adv. in Appl. Clifford Algebras, 25 (2015), , doi /s [17] G. B. Price, An Intoduction to Multicomplex Spaces Functions, Marcel Dekker Inc., New York (1990). [18] D. Rochon M. Shapiro, On algebraic properties of bicomplex hyperbolic numbers, Anal. Univ. Oradea Fascicula Matematica, 11 (2004), [19] C. Segre, Le rappresentazioni reali delle forme complesse e gli enti iperalgebrici, Mathematische Annalen, 40 (1892), [20] M. N. Swamy, On generalized Fibonacci quaternions, The Fibonacci Quarterly, 5 (1973), [21] S. Vajda, Fibonacci Lucas Numbers And The Golden Section, Ellis Horwood Limited Publ., Engl (1989). [22] E. Verner Jr. Hoggatt, Fibonacci Lucas Numbers, The Fibonacci Association, (1969). [23] E. W. Weisstein, Fibonacci Number, MathWorld (

14 378