A NOTE on BICOMPLEX FIBONACCI and LUCAS NUMBERS Semra

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International Journal of Pure and Applied Mathematics Volume 120 No. 3 2018, 365-377 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu AP doi: 10.12732/ijpam.v120i3.7 ijpam.eu

A NOTE ON BICOMPLEX FIBONACCI AND LUCAS NUMBERS

Semra Kaya Nurkan1 §, Ilkay˙ Arslan G¨uven2 1Department of Mathematics U¸sak University 64200, U¸sak, TURKEY 2Department of Mathematics Gaziantep University 27310, Gaziantep, TURKEY

Abstract: In this study, we deﬁne a new type of Fibonacci and Lucas numbers which are called bicomplex Fibonacci and 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 and negabicomplex Lucas numbers, Binet formulas and 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 Fn is deﬁned as the

Received: February 21, 2017 c 2018 Academic Publications, Ltd. Revised: March 28, 2018 url: www.acadpubl.eu Published: December 4, 2018

§Correspondence author 366 S.K. Nurkan, I.A.˙ G¨uven linear reccurence relation Fn = Fn−1 + Fn−2 with F = F = 1 and n Z. Also as a result of this relation we can deﬁne F = 1 2 ∈ 0 0. Fibonacci numbers are connected with the golden ratio as; the ratio of two consecutive Fibonacci numbers approximates the golden ratio 1, 61803399.... Fibonacci numbers are closely related to Lucas numbers which are named after the mathematician Francois Edouard Anatole Lucas who worked on both Fibonacci and Lucas numbers. The integer sequence of Lucas numbers denoted by Ln is given by 2, 1, 3, 4, 7, 11, 18, 29, 47, ... with the same reccurence relation

Ln = Ln−1 + Ln−2. where L = 2, L = 1, L = 3 and n Z. 0 1 2 ∈ There are many works on Fibonacci and Lucas numbers in literature. The properties, relations, results between Fibonacci and Lucas numbers can be found in, Dunlap[3], Güven and Nurkan[5], Koshy[14], Vajda[21], Verner and Hoggatt[22]. Also Fibonacci and Lucas quaternions were described by Horadam in [10], then many studies related with these quaternions were done in Akyi˘git [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 gener- alizations of the ordinary complex numbers. The complex numbers denoted as C are defined by

C = a + bi a, b R and i2 = 1 . | ∈ − The bicomplex numbers are deﬁned by

C = z + z j z , z C and j2 = 1 . 2 1 2 | 1 2 ∈ − Since z1 and z2 are complex numbers, writing z1 = a + bi and z2 = c+ di gives us another way to represent the bicomplex numbers as;

z1 + z2j = 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 and C = | ∈ . 2 i2 = 1, j2 = 1, ij = ji = k − − A NOTE ON BICOMPLEX FIBONACCI... 367

For any bicomplex numbers x = a1 + b1i + c1j + d1ij and y = a2 + b2i + c2j + d2ij, the addition and multiplication of these bicomplex numbers are given respectively by

x + y = (a1 + a2) + (b1 + b2)i + (c1 + c2)j + (d1 + d2)ij (1.1) and

x y = (a a b b c c d d ) + (a b + b a c d d c )i (1.2) × 1 2 − 1 2 − 1 2 − 1 2 1 2 1 2 − 1 2 − 1 2 + (a c + c a b d d b )j + (a d + d a + b c + c b )ij. 1 2 1 2 − 1 2 − 1 2 1 2 1 2 1 2 1 2 The multiplication of a bicomplex number by a real scalar λ is given by

λx = λa1 + λb1i + λc1j + λd1ij.

The set C2 forms a commutative ring with addition and multiplication and it is a real vector space with addition and scalar multiplication [17]. In C, the complex conjugate of z = a + bi is z = a bi. In C , for a − 2 bicomplex number x = (a1 + b1i) + (c1 + d1i)j, there are diﬀerent conjugations which are;

⋆ x⋆ = [(a + b i) + (c + d i)j] = (a b i) + (c d i)j 1 1 1 1 1 − 1 1 − 1 x◦ = [(a + b i) + (c + d i)j]◦ = (a + b i) (c + d i)j (1.3) 1 1 1 1 1 1 − 1 1 † x† = [(a + b i) + (c + d i)j] = (a b i) (c d i)j 1 1 1 1 1 − 1 − 1 − 1 [18]. Then the following equation are written in [12];

x x⋆ = (a2 + b2 c2 d2) + 2(a c + b d )j × 1 1 − 1 − 1 1 1 1 1 x x◦ = (a2 b2 + c2 d2) + 2(a b + c d )i (1.4) × 1 − 1 1 − 1 1 1 1 1 x x† = (a2 + b2 + c2 + d2) + 2(a d b c )ij. × 1 1 1 1 1 1 − 1 1 Also in [18], the modulus of a bicomplex number x is deﬁned by

Rochon and Shapiro gave the detailed algebraic properties of bicomplex and hyperbolic numbers in [18]. The generalized bicomplex numbers were defined by Karaku¸sand Aksoyak in [12], they gave some algebratic properties of them and R4 R4 used generalized bicomplex number product to show that and 2 were Lie groups. Also Luna-Elizarraras et al [15], introduced the algebra of bicomplex numbers and described how to define elementary functions and their inverse functions in that algebra. In this paper, we define bicomplex Fibonacci and bicomplex Lucas numbers by combining bicomplex numbers and Fibonacci, Lucas numbers. We give some identities and Binet formulas of these new numbers.

2. Bicomplex Fibonacci Numbers

Deﬁnition 1: The bicomplex Fibonacci and bicomplex Lucas numbers are deﬁned respectively by

BFn = Fn + Fn+1i + Fn+2j + Fn+3k (2.1) and BLn = Ln + Ln+1i + Ln+2j + Ln+3k (2.2) th th where Fn is the n Fibonacci number, Ln is the n Lucas number and i, j, k are bicomplex units which satisfy the commutative multiplication rules:

i2 = 1, j2 = 1, k2 = 1 − − ij = ji = k, jk = kj = i, ik = ki = j. − − Starting from n = 0, the bicomplex Fibonacci and bicomplex Lucas numbers can be written respectively as;

BF0 = 1i + 1j + 2k, BF1 = 1 + 1i + 2j + 3k , BF2 = 1 + 2i + 3j + 5k,...

BL0 = 2 + 1i + 3j + 4k , BL1 = 1 + 3i + 4j + 7k , BL2 = 3 + 4i + 7j + 11k,...

Let BFn = Fn +Fn+1i+Fn+2j +Fn+3k and BFm = Fm +Fm+1i+Fm+2j + Fm+3k be two bicomplex Fibonacci numbers. By taking into account the equations (1.1) and (1.2), the addition, subtraction and multiplication of these numbers are given by

BFn BFm = (Fn Fm) + (Fn Fm ) i + (Fn Fm ) j ± ± +1 ± +1 +2 ± +2 A NOTE ON BICOMPLEX FIBONACCI... 369

+ (Fn Fm ) k +3 ± +3 and

BFn BFm = (FnFm Fn Fm Fn Fm + Fn Fm ) × − +1 +1 − +2 +2 +3 +3 + (FnFm + Fn Fm Fn Fm + Fn Fm ) i +1 +1 − +2 +3 +3 +2 + (FnFm + Fn Fm Fn Fm Fn Fm ) j +2 +2 − +1 +3 − +3 +1 + (FnFm+3 + Fn+3Fm + Fn+1Fm+2 + Fn+2Fm+1) k.

Deﬁnition 2: A bicomplex Fibonacci number can also be expressed as BFn = (Fn + Fn+1i) + (Fn+2 + Fn+3i) j. In that case,there is three diﬀerent conjugations with respect to i, j and k; for bicomplex Fibonacci numbers as follows:

i i BFn = [(Fn + Fn+1i) + (Fn+2 + Fn+3i) j]

= (Fn Fn i) + (Fn Fn i) j − +1 +2 − +3 j j BFn = [(Fn + Fn+1i) + (Fn+2 + Fn+3i) j]

= (Fn + Fn i) (Fn + Fn i) j +1 − +2 +3 k k BFn = [(Fn + Fn+1i) + (Fn+2 + Fn+3i) j]

= (Fn Fn i) (Fn Fn i) j. − +1 − +2 − +3 By the Deﬁnition 2 and the equations (1.4) and (2.1), we can write

i BFn BFn = (F n F n ) + 2 (F n ) j × 2 +1 − 2 +7 2 +3 j 2 2 2 2 BFn BFn = F F + F F + 2 (FnFn + Fn Fn ) i × n − n+1 n+3 − n+4 +1 +2 +3 k n+1 BFn BFn = (F n + F n ) + 2 ( 1) k. × 2 +1 2 +7 −

Deﬁnition 3: Let a bicomplex Fibonacci number be BFn = Fn + Fn+1i + Fn j + Fn k. The i modulus, j modulus, k modulus and real modulus of +2 +3 − − − BFn are given respectively as;

In [16], we proved some summation equations by supposing, for i 0 ≥ M M αmFm+i + βmLm+i = 0 (2.3) m=0 m=0 X X where αm and βm are ﬁxed numbers. By using the equations (2.1), (2.2), (2.3) and taking i = 0, 1, 2, 3, we can write the following equation clearly;

M M αmBFm + βmBLm = 0 (2.4) m=0 m=0 X X Now let us give theorems which give some properties.

Theorem 1. Let BFn and BLn be a bicomplex Fibonacci and a bicomplex Lucas number, respectively. For n 0 , the following relations hold: ≥

1) BFn + BFn+1 = BFn+2

2) BLn + BLn+1 = BLn+2 3) BLn = BFn−1 + BFn+1 4) BLn = BFn+2 BFn−2 2 2 − 5) BF + BF = BF n + F n + F n 3iF n n n+1 2 +1 2 +2 2 +5 − 2 +5 jF2n+6 + 3kF2n+4 2 2 − 6) BF BF = 2BF n + F n + F n− n+1 − n−1 2 2 +4 2 1 + 2 ( F n i F n j + F n k) − 2 +5 − 2 +4 2 +3 7) BFnBFm + BFn BFm = 2BFn m + 2Fn m Fn m +1 +1 + +1 + +4 − + +1 2Fn m i 2Fn m j + 2Fn m k − + +6 − + +5 + +4 8) BFn BFn i + BFn j BFn k = 5Fn − +1 +2 − +3 − +3 Proof. By using the equations of (2.1), (2.2), (2.4) and 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 2 2 2 2 in (2.1) , the equations F + F = F n , F F = F n and FnFm + n n+1 2 +1 n+1 − n−1 2 Fn+1Fm+1 = Fn+m+1 (see Vajda [21]), we get

2 2 2 BFn + BFn+1 = (Fn + Fn+1i + Fn+2j + Fn+3k) 2 + (Fn+1 + Fn+2i + Fn+3j + Fn+4k)

F n + 2 (FnFn Fn Fn ) i 2 +3 +1 − +2 +3 = +2 (FnFn Fn Fn ) j  +2 − +1 +3  +2 (FnFn+3 + Fn+1Fn+2) k   A NOTE ON BICOMPLEX FIBONACCI... 371

F n + 2 (Fn Fn Fn Fn ) i 2 +5 +1 +2 − +3 +4 + +2 (Fn Fn Fn Fn ) j  +1 +3 − +2 +4  +2 (Fn+1Fn+4 + Fn+2Fn+3) k   = F2n+1 + F2n+2i + F2n+3j + F2n+4k + F2n+2

+ F n 3F n i F n j + 3F n k 2 +5 − 2 +5 − 2 +6 2 +4 = BF n + F n + F n 3iF n 2 +1 2 +2 2 +5 − 2 +5 jF n + 3kF n . − 2 +6 2 +4

2 2 2 BF BF = (Fn + Fn i + Fn j + Fn k) n+1 − n−1 +1 +2 +3 +4 2 (Fn− + Fni + Fn j + Fn k) − 1 +1 +2 F n + 2 (Fn Fn Fn Fn ) i 2 +5 +1 +2 − +3 +4 = +2 (Fn Fn Fn Fn ) j  +1 +3 − +2 +4  +2 (Fn+1Fn+4 + Fn+2Fn+3) k   F n + 2 (Fn− Fn Fn Fn ) i 2 +1 1 − +1 +2 +2 (Fn− Fn FnFn ) j −  1 +1 − +2  +2 (Fn−1Fn+2 + FnFn+1) k   = 2BF2n + F2n+4 + F2n−1

+ 2 ( F n i F n j + F n k) − 2 +5 − 2 +4 2 +3

BFnBFm + BFn+1BFm+1 = (Fn + Fn+1i + Fn+2j + Fn+3k)

.(Fm + Fm+1i + Fm+2j + Fm+3k)

+ (Fn+1 + Fn+2i + Fn+3j + Fn+4k)

. (Fm+1 + Fm+2i + Fm+3j + Fm+4k)

= 2 (Fn+m+1 + Fn+m+2i)

+ 2 (Fn+m+3j + Fn+m+4k)

+ 2Fn m Fn m 2Fn m i + +4 − + +1 − + +6 2Fn m j + 2Fn m k − + +5 + +4 = 2BFn m + 2Fn m Fn m + +1 + +4 − + +1 2Fn m i 2Fn m j + 2Fn m k − + +6 − + +5 + +4 and

BFn BFn i + BFn j BFn k = (Fn + Fn i + Fn j + Fn k) − +1 +2 − +3 +1 +2 +3 372 S.K. Nurkan, I.A.˙ G¨uven

(Fn + Fn i + Fn j + Fn k) i − +1 +2 +3 +4 + (Fn+2 + Fn+3i + Fn+4j + Fn+5k) j

(Fn + Fn i + Fn j + Fn k) k − +3 +4 +5 +6 = Fn + Fn Fn Fn +2 − +4 − +6 = 5Fn . − +3

Theorem 2. For n, m 0 the D’ocagnes identity for bicomplex Fibonacci ≥ numbers BFn and BFm is given by n BFmBFn BFm BFn = ( 1) BFm−n +1 − +1 − Fm−n + Fm−n+1i n+1 + ( 1) (Fm−n− + 2Fm−n ) j . −  − 2 +2  +2Fm−n−1k   Proof. If we decide the equation (2.1) and the D’ocagnes identity for Fi- n bonacci numbers FmFn Fm Fn = ( 1) Fm−n (see Weisstein [23] ), we +1 − +1 − obtain the following calculations as;

BFmBFn BFm BFn = (Fm + Fm i + Fm j + Fm k) +1 − +1 +1 +2 +3 (Fn+1 + Fn+2i + Fn+3j + Fn+4k)

(Fm + Fm i + Fm j + Fm k) − +1 +2 +3 +4 (Fn + Fn+1i + Fn+2j + Fn+3k)

= [Fn+2 (Fm + Fm+4) Fm+2 (Fn + Fn+4)] i n − + [2 ( 1) (Fm−n−2 + Fm−n+2)] j − n + [( 1) (Fm−n−2 + Fm−n+2)] k − n n + ( 1) BFm−n ( 1) BFm−n − n − − = ( 1) BFm−n − Fm−n + Fm−n+1i n+1 + ( 1) (Fm−n− + 2Fm−n ) j . −  − 2 +2  +2Fm−n−1k  

Theorem 3. If BFn and BLn are bicomplex Fibonacci and bicomplex Lucas numbers respectively, then for n 0, the identities of negabicomplex Fibonacci and negabicomplex Lucas numbers≥ are n+1 n BF−n = ( 1) BFn + ( 1) Ln (i + j + 2k) − − A NOTE ON BICOMPLEX FIBONACCI... 373 and n n+1 BL−n = ( 1) BLn + ( 1) 5Fn (i + j + 2k) − − Proof. From the equations (2.1) and the identity of negaﬁbonacci numbers n+1 which is F−n = ( 1) Fn (see Knuth [13] , Dunlap [3]), we have − BF−n = F−n + F−n+1i + F−n+2j + F−n+3k

= F−n + F−(n−1)i + F−(n−2)j + F−(n−3)k n+1 n n+1 n = ( 1) Fn + ( 1) Fn− i + ( 1) Fn− j + ( 1) Fn− k − − 1 − 2 − 3 n+1 n+1 = ( 1) (Fn + Fn i + Fn j + Fn k) ( 1) Fn i − +1 +2 +3 − − +1 n+1 n+1 n ( 1) Fn j ( 1) Fn k + ( 1) Fn− i − − +2 − − +3 − 1 n+1 n + ( 1) Fn− j + ( 1) Fn− k − 2 − 3 n+1 n = ( 1) BFn + ( 1) (Fn+1 + Fn−1) i − n − n + ( 1) (Fn Fn− ) j + ( 1) (Fn + Fn− ) k − +2 − 2 − +3 3 In this equation , we take into account that Fn− + Fn = Ln, Fn Fn− = 1 +1 +2 − 2 Ln,Fn+3 + Fn−3 = 2Ln(see Vajda [21]) and the identity of negalucas numbers n L−n = ( 1) Ln (see Knuth [13] ,Dunlap [3]), thus we obtain; − n+1 n n n BF−n = ( 1) BFn + ( 1) Lni + ( 1) Lnj + ( 1) 2Ln − − − − n+1 n = ( 1) BFn + ( 1) Ln (i + j + 2k) − − Now by using the equation (2.2) and the identity of negalucas numbers, we get

BL−n = L−n + L−n+1i + L−n+2j + L−n+3k

= L−n + L−(n−1)i + L−(n−2)j + L−(n−3)k n n−1 n n−1 = ( 1) Ln + ( 1) Ln−1i + ( 1) Ln−2j + ( 1) Ln−3k − n − − n− = ( 1) (Ln + Ln i + Ln j + Ln k) ( 1) Ln i − +1 +2 +3 − − +1 n n n−1 n ( 1) Ln j ( 1) Ln k + ( 1) Ln− i + ( 1) Ln− j − − +2 − − +3 − 1 − 2 n−1 + ( 1) Ln−3k − n = ( 1) BLn − (L + L ) i + ( 1)n+1 n+1 n−1 − + (Ln Ln− ) j + (Ln + Ln− ) k +2 − 2 +3 3

5FmFn, if n is odd, Here if we use the identity Lm+n + Lm−n = (see Koshy, (LmLn, otherwise [14]), namely Ln−1 + Ln+1 = 5Fn , the definition of dual Lucas number and the 374 S.K. Nurkan, I.A.˙ Güven identity of negafibonacci number in last equation, we complete the proof as;

n n+1 n+1 n+1 BL−n = ( 1) BLn + ( 1) 5Fni + ( 1) 5Fnj + ( 1) 10Fnk − − − − n n+1 = ( 1) BLn + ( 1) 5Fn (i + j + 2k) − −

The Binet formulas for Fibonacci and Lucas numbers are given by (see Koshy, [14]) αn βn F = − and L = αn + βn n α β n − where 1 + √5 1 √5 α = and β = − . 2 2 Theorem 4. Let BFn and BLn be bicomplex Fibonacci and bicomplex Lucas numbers, respectively. For n 0 , the Binet formulas for these numbers are given as; ≥ ααn ββn BF = − n α β − and n n BLn = αα + ββ where α = 1 + iα + jα2 + kα3 and β = 1 + iβ + jβ2 + kβ3.

Proof. By using the Binet formulas for Fibonacci and Lucas numbers, taking α = 1 + iα + jα2 + kα3 and β = 1 + iβ + jβ2 + kβ3, we ﬁnd the result clearly as;

BFn = Fn + Fn+1i + Fn+2j + Fn+3k αn βn αn+1 βn+1 αn+2 βn+2 αn+3 βn+3 = − + − i + − j + − α β α β α β α β − − − − αn 1 + iα + jα2 + kα3 βn 1 + iβ + jβ2 + kβ3 = − α β − ααn ββn = − α β − and

BLn = Ln + Ln+1i + Ln+2j + Ln+3k A NOTE ON BICOMPLEX FIBONACCI... 375

= αn + βn + (αn+1 + βn+1)i + (αn+2 + βn+2)j + (αn+3 + βn+3)k = αn(1 + iα + jα2 + kα3) + βn(1 + iβ + jβ2 + kβ3) = ααn + ββn.

Theorem 5. Let BFn and BLn be bicomplex Fibonacci and bicomplex Lucas numbers. For n 1 ,the Cassini identities for BFn and BLn are given by; ≥ 2 n BFn BFn− BF = 3 ( 1) (2j + k) +1 1 − n − and 2 n−1 BLn BLn− BL = 5( 1) (2j + k) +1 1 − n − Proof. From the equation (2.1), we have

2 BFn BFn− BF = (Fn + Fn i + Fn j + Fn k) +1 1 − n +1 +2 +3 +4 (Fn−1 + Fni + Fn+1j + Fn+2k) 2 (Fn + Fn i + Fn j + Fn k) . − +1 +2 +3 n If we use the identity FmFn+1 Fm+1Fn = ( 1) Fm−n (see Weisstein [23] ) n+1 − − and F−n = ( 1) Fn (see Knuth [13] ) in the above equation , we get − 2 n+2 n+2 BFn BFn− BF = 2 ( 1) 3j + 3 ( 1) k +1 1 − n − − = 3h ( 1)n (2ji+ k) h i − Similarly for bicomplex Lucas numbers we can simply get;

2 BLn BLn− BL = (Ln + Ln i + Ln j + Ln k)(Ln− +1 1 − n +1 +2 +3 +4 1 + Lni + Ln+1j + Ln+2k) 2 (Ln + Ln i + Ln j + Ln k) . − +1 +2 +3 2 n−1 By using the identity of Lucas numbers which are Ln− Ln L = 5( 1) 1 +1 − n − (see Koshy [14]) and Ln+2 = Ln+1 +Ln (see Dunlap [3]) in the above equation, we compute the following expression;

2 n−1 BLn BLn− BL == 5( 1) (1 + ε) +1 1 − n − = 5( 1)n−1 (2j + k) . − 376 S.K. Nurkan, I.A.˙ G¨uven

Theorem 6. Let BFn be a bicomplex Fibonacci number. The Catalan’s identity for BFn is given by 2 2 2 n−r 2 Fr−2 + Fr j BF BFn rBFn−r = ( 1) . n − + − + F 2 + F 2 F 2 F 2 k r+1 r−2 − r − r−3 Proof. From the equation (1.4), we have

2 2 2 2 Fn Fn+1 Fn+2 + Fn+3 2 − − BF BFn rBFn−r = Fn rFn−r + Fn r Fn−r− n − +  − + + +1 1  +Fn r Fn−r Fn r Fn−r + +2 +2 − + +3 +3   2FnFn 2Fn Fn +1 − +2 +3 + Fn rFn−r Fn r Fn−r i  − + +1 − + +1  +Fn+r+2Fn−r+3 + Fn+r+3Fn−r+2   2FnFn 2Fn Fn +2 − +1 +3 + Fn rFn−r Fn r Fn−r j  − + +2 − + +2  +Fn+r+1Fn−r+3 + Fn+r+3Fn−r+1   2FnFn+3 + 2Fn+1Fn+2 + Fn rFn−r Fn r Fn−r k  − + +3 − + +3  Fn r Fn−r Fn r Fn−r − + +1 +2 − + +2 +1   2 Here using the Catalan’s identity for Fibonacci numbers Fn Fn−rFn+r = n−r 2 2 2 − ( 1) F (see Weisstein [23]) , F F = F n (see Vajda [21]) and doing − r n+1 − n−1 2 necessary calculations, we obtain that;

2 2 2 n−r 2 Fr−2 + Fr j BF BFn rBFn−r = ( 1) n − + − + F 2 + F 2 F 2 F 2 k r+1 r−2 − r − r−3

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