Generalizations of Fibonacci and Lucas sequences

Transcription

1 Note di Matematica 1, n 1, 00, Generalizations of Fibonacci Lucas sequences Nihal Yilmaz Özgür Balikesir Universitesi, Fen-Edebiyat Fakultesi Matematik Bolumu, Balikesir/Turkey nihal@balikesiredutr Received: 9 September 001; accepted: 0 March 00 Abstract In this paper, we consider the Hecke groups H q, q primenumber, we find an interesting number sequence which is denoted by d nforq,wegetd n L n+1 d n+1 F n+ where L n+1 is n + 1th Lucas number F n+ is n + th Fibonacci number From this sequence, we obtain two new sequences which are, in a sense, generalizations of Fibonacci Lucas sequences Keywords: Hecke groups, Fibonacci numbers, Lucas numbers MSC 000 classification: 11B39, 0H10 Introduction Hecke groups Hλ are, in some sense, the generalizations of the well-known modular group { } az + b PSL, Z a, b, c, d Z, ad bc 1 cz + d They are the discrete subgroups of PSL, R the group of orientation preserving isometries of the upper half plane U generated by two linear fractional transformations Rz 1 T z z + λ z where λ is a fixed positive real number Hecke groups Hλ have been introduced by Erich Hecke, [4] Hecke asked the question that for what values of λ these groups are discrete In answering this question, he proved that Hλ has a fundamental region iff λ orλ λ q cos π q, q N, q 3 Therefore Hλ is discrete only for these values of λ The most interesting worked Hecke group is the modular group Hλ 3 PSL, Z obtained for q 3 In this case all coefficients of the elements of Hλ 3 are integers The next two important Hecke groups are H H 3 obtained for q 4 6, respectively For these groups, the underlying fields are the quadratic extensions of Q by the algebraic numbers 3, that is,

2 114 N Yilmaz Özgür Q Q 3 It is known that Hλ q is isomorphic to the free product of two finite cyclic groups of orders q, ie Hλ q C C q, [1] Here we are going to be interested in the case λ In this case, the element S RT is parabolic when λ, or hyperbolic when λ> It is known that when λ, Hλ is a free product of two cyclic groups of orders infinity, [8], so all such Hλ have the same algebraic structure, ie Hλ C Z so that the signature of Hλ is 0;, ; 1 In particular, we deal with the case where λ q, q prime number, denote the group by H q On the other h, Edouard Lucas made a deep study of sequences which is called generalized Fibonacci sequences, that begin with any two positive integers, each number thereafter being the sum of the preceding two The simplest such series, 0, 1, 1,, 3,, 8, 13, 1,, is called the Fibonacci sequence by Lucas The next simplest series,, 1, 3, 4, 7, 11, 18,, is then called the Lucas numbers in his honor The Fibonacci rule of adding the latest two to get the next is kept, but here we begin with 1 in this order The position of each number in this sequence is traditionally indicated by a subscript, so that F 0 0,F 1 1,F 1,F 3, so on F n refers to nth Fibonacci number Explicitly the Fibonacci sequence can be defined as follows: F n F n 1 + F n,n 1 with two boundary conditions: F 0 0,F 1 1 The Lucas sequence is defined as follows where we write its members as L n, for Lucas: L n L n 1 + L n, n > 1 L 0 L 1 1 The nth Fibonacci number is in the following formula: F n 1 [ 1+ n 1 n ] 3 The formula that gives the nth number of the Lucas sequence exactly is 1+ n 1 n L n + 4

3 Generalizations of Fibonacci Lucas sequences 11 Lucas numbers have lots of properties similar to those of Fibonacci numbers the Lucas numbers often occur in various formulas for the Fibonacci numbers For example: The nth Lucas number is equal to F n 1 + F n+1, ie L n F n 1 + F n+1 for all integers n The product of F n L n is equal to F n Another recurrence relations for the Fibonacci Lucas numbers are F n+4 3F n+ F n, 6 L n+4 3L n+ L n 7 For more details about the Fibonacci Lucas sequences, see [], [3], [] [7] We came across an interesting number sequence which is denoted by d n, when we were studying the principal congruence subgroups of the Hecke groups H q, q prime number The sequence d n is defined as follows: For q, we get the sequence d n qd n 1 d n, n 8 d 0 1, d 1 q d n d n 1 d n, n d 0 1,d 1 It is surprising that d n L n+1 d n+1 F n+,eachd n for all q has similar properties L n F n Insomesensed ns are generalizations of L n F n including both of them for q In this paper we will introduce this sequence get two new sequences from this Firstly we will give a brief information about the Hecke group H q 1 The Hecke Group H q In the case λ q, q prime, the underlying field is a quadratic extension of Q by q, ie Q q A presentation of H qis H q R, S; R S RS 1 where S RT By identifying the transformation az+b cz+d with the matrix ab cd, H q may be regarded as a multiplicative group of matrices in which a matrix is identified with its negative R S have matrix representations , q

4 116 N Yilmaz Özgür respectively All elements of H q are of the following two types: a b q i c ; a, b, c, d Z, ad qbc 1 q d a q b ii c d ; a, b, c, d Z, qad bc 1 q But the converse is not true That is, all elements of type i orii need not be in H q see [6] Those of type i are called even while those of type ii are called odd The set of all even elements form a subgroup of index called the even subgroup It is denoted by H e q, [9] When we were determining the group structure of principal congruence subgroups of H q, we needed the powers of the transformation S which is one of the generators of H q We were looking for the answer of the question that for what values of n satisfy S n ±I mod p, p is an odd prime, we needed to compute nth power of S, for every integer n Notice that the transformation S is hyperbolic with fixed points q 4 q z 1 z q 4 q 9 therefore of infinite order It is hard to compute S n easily The next section we will try to compute nth power of S Note that S is an odd element of H q, so odd powers of S are odd even powers of S are even elements Powers of the Transformation S In this section we will prove that S n dn d n 1 d n 1 d n by induction Therefore the problem to compute S n is reduced to compute the explicit formula of d n Proposition 1 The n th power of S is S n dn d n 1 d n 1 where d 0 1, d 1 q d n qd n 1 d n, for n d n

5 Generalizations of Fibonacci Lucas sequences 117 Proof First we prove S n bn 1 d n 1 d n 1 qdn 1 + b n 1 Remember that S q a1 b S 1 c 1 d 1 Let us write S n an b n c n d n Then we have S q q 1 q b1 d 1 q q 1 d 1 qd1 + b 1 Hence assertion is true for n Let us assume that S n 1 bn d n d n qdn + b n Then we have S n bn d n d n qdn + b n bn 1 d n 1 d n 1 qdn 1 + b n q d n qd n + b n qdn + b n q qdn + b n d n Notice that b d 1, b n 1 d n b n d n 1 Ifwewrited 0 1,we get b 1 d 0 hence we proved that S n dn d n 1 d n 1 qdn 1 d n Thus we have a real number sequence which can be defined as d n qd n 1 d n, n 10 d 0 1, d 1 q Now we have to compute d n QED

6 118 N Yilmaz Özgür Proposition d n is in the following formula for all n: [ q n+1 n+1 ] 1 + q 4 q q 4 d n 11 q 4 Proof To solve the equation 10, let d n to be a characteristic polynomial r n, which appears to quickly reduce 10 to r n qr n 1 r n r qr +10 with the roots q ± q 4 r 1, The general solution of the equation d n r n will be all possible combinations of roots r 1 r Let us write n n q + q 4 q q 4 d n A + B Constants A B can be found from the boundary conditions d 0 1 d 1 qwehave d 0 1 A + B, d 1 q + q 4 q q 4 q A + B so q A q + q A q q 4 Hence we find q + q 4 q 4 q A B q 4 q 4 Then we get the formula of d n as follows: n q + q 4 q + q 4 d n + q 4 q 4 q q q 4 + q 4 [ q n+1 n ] 1 + q 4 q q 4 q 4 n QED

7 Generalizations of Fibonacci Lucas sequences 119 For q, we arrive at familiar expressions for Fibonacci Lucas numbers We have d n L n+1,n + 1th Lucas number, d n+1 F n+,n + th Fibonacci number Indeed, from 11 we get the sequence d n 1+ n+1 n+1 1 If n is even, then n + 1 is odd hence we can write d n 1+ n+1 1 n+1 + L n+1 If n is odd, then n + 1 is even we can write d n 1 1+ n+1 1 n+1 F n+1 So we get S n Ln 1 F n F n Ln+1 S n+1 Fn Ln+1 L n+1 F n+ in the Hecke group H q In general, the sequence d n has similar properties of F n L n for all q Proposition 3 We have d n q d n d n 4 1 d n+1 q d n 1 d n 3 13 Proof By10,wegetd n 3 qd n 4 d n so d n 3 + d n

8 10 N Yilmaz Özgür qdn 4 Using this last equation 10, we obtain d n qd n 1 d n q qd n d n 3 d n q 1d n qd n 3 q 1d n q qd n 4 d n q 1d n qd n 4 + qd n q d n + qd n 3 d n 4 qd n 4 + qd n q d n d n 4 qd n 4 + qd n 3 + d n q d n d n 4 qd n 4 + q qd n 4 q d n d n 4 Similarly, from 10, we find d n + d n 4 qd n 3 then d n+1 qd n d n 1 q qd n 1 d n d n 1 q 1d n 1 qd n q 1d n 1 q qd n 3 d n 4 q 1d n 1 qd n 3 + qd n 4 q d n 1 + qd n d n 3 qd n 3 + qd n 4 q d n 1 d n 3 qd n 3 + qd n + d n 4 q d n 1 d n 3 qd n 3 + q qd n 3 q d n 1 d n 3 Note that for q,wehave QED d n 3d n d n 4 d n+1 3d n 1 d n 3, that is, we have L n+1 3L n 1 L n 3 F n+ 3 F n F n, so F n+ 3F n F n which are coincide 7 6, respectively From d n we have two subsequences Indeed, let us consider the polynomials u 0 1, u 1 q 14 u n q u n 1 u n,n

9 Generalizations of Fibonacci Lucas sequences 11 Then we have d n+1 u n q 1 Now we will prove this by induction As d 1 u 0 q, d3 u 1 q, assume that d n 1 d n 1+1 u n 1 qfrom13,wehave d n+1 q d n 1 d n 3 q d n 1+1 d n +1 q u n 1 q un q [q un 1 u n ] q u n q Similarly, if we consider the polynomials v 0 1,v 1 q 1 16 v n q v n 1 v n, n, we have d n v n 17 Indeed,aswehaved 0 v 0, d v 1, assume that d n d n 1 v n 1 From 1, we have d n q d n d n 4 q d n 1 d n q v n 1 v n v n Note that for q,v n s are Lucas numbers with odd index u n s are Fibonacci numbers with even index Now we want to generalize the sequences u n v n First consider the sequence u n generalize into a sequence of which elements of even index are u n s Let us define U n+ u n compute U n using the equalities U n U n 1 + U n U n+1 q U n 1 U n 3 Then we have U 0 0 U 1 1 U 1 U 3 q 3 U 4 q U q q + U 6 q 4q +3

10 1 N Yilmaz Özgür For even indexed elements of the sequence, the condition U n U n 1 + U n 18 is hold We will investigate whether this condition holds for odd indexed elements or not From U n+ U n+1 + U n, we have U n+1 U n+ U n q U n U n U n q 3U n U n q 3U n 1 + U n U n q 3U n 1 +q 4U n U n 1 +q 4U n 1 + U n U n 1 +q 4U n Therefore only for q, we have the Fibonacci sequence Let us now generalize v n into a sequence of which elements of odd index are v n s Similarly, let us define V n+1 v n compute V n using the equalities Then we have V n+1 V n + V n 1 V n q V n V n 4 V 0 V 1 1 V q V 3 q 1 V 4 q 4q + V q 3q +1 For odd indexed elements of the sequence, the condition V n+1 V n + V n 1 19 is hold Again similarly an easy calculation shows that V n q 4V n 1 + V n

11 Generalizations of Fibonacci Lucas sequences 13 So only for q, we get the Lucas sequence Therefore we get two new sequences, U n V n, which are not generalized Fibonacci sequences except q Explicitly U n can be defined as follows: U n q U n U n 4,n 4 0 with the boundary conditions: U 0 0, U 1 1, U 1 U 3 q 3 V n can be defined as follows: V n q V n V n 4,n 4 1 with the boundary conditions: V 0, V 1 1, V q V 3 q 1 For q, we have polynomial representations of Fibonacci Lucas numbers In a sense, U n is a generalization of F n V n is a generalization of L n Furthermore, the sequence d n defined in 10, contains both of even indexed elements of U n of odd indexed elements of V n The sequences U n V n have similar properties of Fibonacci Lucas numbers F n L n For example we have the following propositions: Proposition 4 V n U n+1 + U n 1 Proof Since V 1 U 0 +U 0+11V U 1 +U 3 1+q 3 q, assume that the identity is true for n 1,, 3,,k 1 show that it holds for n k Wehave V k q V k V k 4 q U k 1 + U k 3 U k + U k 3 q U k 1 U k 3 +q U k 3 U k U k+1 + U k 1 Proposition The product of U n V n is equal to U n, that is, QED U n U n V n 3 Proof As we have U 1U 1 V 1 U 4 q U V, assume that U n U n 1 V n 1 Now we can compute U n V n From,wehaveU n V n U n U n+1 + U n 1 using definitions we find U n V n U n q U n 1 U n 3 +U n 1 q U n U n 4 q U n U n 1 +q U n 1 U n U n U n 3 U n 1 U n 4 q U n 1 U n + U n U n U n 3 U n 1 U n 4 q U n 1 V n 1 U n U n 3 U n 1 U n 4

12 14 N Yilmaz Özgür If n is even, we know that U n U n 1 + U n Using this n iseven,we have U n V n q U n 1 V n 1 U n 3 U n 1 + U n U n 1 U n U n 3 q U n 1 V n 1 U n 3 U n 1 U n U n 3 + U n 1 +U n 3 U n 1 q U n 1 V n 1 U n V n By the assumption, we get that U n V n q U n U n 4 U n If n is odd, we know that U n q 4U n 1 + U n Sincen is odd, we have U n q 4U n 3 + U n 4 Hence, we obtain U n V n q U n 1 V n 1 U n 3 q 4U n 1 + U n U n 1 U n q 4U n 3 q U n 1 V n 1 U n U n 3 + U n 1 U n 3 U n 1 q 4 + U n 3 U n 1 q 4 q U n 1 V n 1 U n V n Then by assumption, we have Therefore assertion is true for all n U n V n q U n U n 4 U n Finally we have the following theorem: Theorem 1 In the Hecke group H q, q prime, the nth powers of S are obtained as the following forms: S n Vn 1 U n q U n q Vn+1 S n+1 Un q Vn+1 V n+1 U n+ q Proof From Proposition 1, we know that S n dn d n 1 d n 1 By the definitions U n V n the equalities 1 17, we have d n d n+1 u n q,sodn+1 U n+ q QED

13 Generalizations of Fibonacci Lucas sequences 1 d n v n,sod n V n+1 Thus we have d n v n 1 V n 1 d n 1 u n 1 q Un q Therefore, as we have S n dn d n 1, we find Similarly, we have d n 1 d n S n Vn 1 U n q U n q Vn+1 S n+1 dn 1 d n d n d n+1 so, S n+1 Un q Vn+1 U n+ q V n+1 QED References [1] I N Cangul, D Singerman: Normal Subgroups of Hecke Groups Regular Maps, Math Proc Camb Phil Soc, 13, 1998, 9 74 [] R A Dunlap: The Golden Ratio Fibonacci Numbers, World Scientific Press, 1997 [3] M Gardner: Mathematical Circus: More Puzzles, Games, Paradoxes Other Mathematical Entertainments from Scientific American, New York, Knopf, 1979 [4] E Hecke: Über die Bestimmung Dirichletscher Reihen durch ihre Funktionalgleichungen, Math Ann, 11, 1936, [] V E Hogatt Jr: The Fibonacci Lucas Number, Boston, MA, Houghton Mifflin, 1969 [6] D Rosen: A Class of Continued Fractions Associated with Certain Properly Discontinuous Groups, Duke Math J, 1 194, [7] S Vajda: Fibonacci Lucas Numbers, the Golden Section: Theory Applications, Halsted Press, 1989 [8] N Yilmaz Özgür, I N Cangul: On the Group Structure Parabolic Points of Hecke Group Hλ, Proc Estonian Acad Sci Phys Math, 1 1, [9] N Yilmaz Özgür: Principal Congruence Subgroups of Hecke Groups H q, to be submitted