INTERNATIONAL JOURNAL OF MATHEMATICS AND COMPUTERS IN SIMULATION

About Fibonacci numbers and functions

Alina Bărbulescu

 be found in [6] [14]. Abstract— Fibonacci numbers and functions are topics of major Due to their importance in the applied sciences, the interest in mathematics, due to the importance of their applications in Fibonacci and Lucas numbers and their properties have been many sciences. In the first part of this article we present some extensively studied in the last years. Generalizations, as congruences involving Fibonacci and Lucas numbers. In the second Hyperfibonacci and Hyperlucas numbers have been done one we discuss the dimensions of the Fibonacci numbers, defined on different closed intervals, starting with the evaluation of the Box [2][7][8] and congruences modulo different numbers have dimension of this function defined on [0, 1] . been proved or conjectured. [11] [12][13][15] The present article is organized as follows. In the next Keywords—Fibonacci numbers, Fibonacci functions, chapter we give some basic definitions and results, which will quadratic residues, fractal dimension. be used in the chapters III – V. In Chapter III we prove some

congruences satisfied by the Fibonacci numbers Fp and Lucas I. INTRODUCTION numbers Lp , where p is a prime odd integer. The techniques he Fibonacci sequence was the outcome of a mathematical Tproblem about rabbit breeding that was posed in the Liber employed are combinatorial or of elementary number theory. Abaci (published in 1202) by Leonardo Fibonacci. [16] In Chapter IV we determine an upper bound of the Box- The Fibonacci numbers are Nature's numbering system. dimension of the Fibonacci function defined on the closed They appear everywhere in Nature, from the leaf arrangement interval [0, 1]. In the last chapter we present the results of the in plants, to the pattern of the florets of a flower, the bracts of evaluation of different types of dimensions of the same a pinecone, or the scales of a pineapple. function, using the software Benoit 1.3.1. Johann Kepler studied the Fibonacci sequence. Émile Léger II. MATHEMATICAL BACKGOUND appears to have been the first (or second, if the work of de Lagny is counted) to recognise that the worst case of the Let F )( nn 0 be the sequence defined by: Euclidean algorithm occurs when the inputs are consecutive Fibonacci numbers. [10]  10  ,1,0  12 nnn nFFFFF  ,0, In 1843, Jacques Philippe Marie Binet discovered a formula for finding the nth term of the Fibonacci series. [17] called the Fibonacci sequence and L )( , In 1870, the French mathematician Edouard Lucas gave nn 0 different results on Fibonacci numbers and proved that that   ,1,2 nLLLLL  ,0, number 127 12 is a prime one. The related Lucas sequences 10  12 nnn and Lucas numbers are named after him. The golden ratio is closely connected to the Fibonacci be the Lucas sequence and series, being the limit of the sequence of the ratios of two  x x  successive Fibonacci numbers. There are no extant records of    1   51   2  the Greek architects' plans for their most famous temples and xf )(       ,)cos( xx R 5  2   51 buildings. So we do not know if they deliberately used the      golden section in their architectural plans. The American mathematician Mark Barr used the Greek letter phi (φ) to be the Fibonacci function (Fig.1). represent the golden ratio, using the initial letter of the Greek It is known that Phidias who used the golden ratio in his sculptures. [17] More extended information on the history of Fibonacci n n      numbers, the golden ration and the relation between them can 1   51    51   Fn    , 5  2   2        Manuscript received January 29, 2012. Revised version received: A. Bărbulescu is with Ovidius University of Constanța, Faculty of and Mathematics and Computers Sciences, Department of Mathematics, 124, n n   51    51  Mamaia Bd., 900527, Constanța, Romania (phone: 0040 744 284444; e-mail: L    +   . alinadumitriu@ yahoo.com). n  2   2     

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and the Lucas number, Lp , satisfies:

Lp  1 (mod p).

We remember the following theorems that will be used to prove the results in the next chapter:

Fermat’s little theorem. If p is a prime number, then for any integer a, a p  a is evenly divisible by p. Remark. A variant of this theorem is the following: If p is a prime and a is an integer coprime to p, then

a p1  1 (mod p).

Fig.1. Graph of the Fibonacci function defined on [-20, 20] Euler’s criterion. Let p be an odd prime and a an integer coprime to p. Then Let p be a prime odd integer, n - an odd integer and a a p1 natural number. The following notation will be used in the  a  following:    a 2 (mod p).  p   

 a  -   - the Legendre symbol; The quadratic reciprocity law [9]. Let p and q be two  p  distinct odd prime integers. Then  a  -   - the Jacobi quadratic symbol. 1 qp 1  n   p   q        2 2 .1 We recall that if p is an odd prime integer and a is an  q   p  integer number, the Legendre symbol is defined by: The fractal character [3] [4] of Fibonacci series and  pa  and1),(if,1 function may be studied by calculating different types of  dimensions (ruler, box, information, Hausdorff) or the Hurst  a is a quadratic residue mod p  a   coefficient [1].     if,1 pa  and1),( Let F be a nonempty and bounded subset of R2, (FN ) the  p      a is anot quadratic residue mod p  least number of sets whose union covers F with and diameters that do not exceed a given  .0 The upper bound box  if,0 ap dimension of F is defined by:

and if n is an odd natural number, ... pppn , where 21 r FN )(log dim F  suplim  , 21 ,...,, ppp r are prime integers, then the quadratic Jacobi B  0 log symbol is defined by: and the lower bound box dimension of F, by:  a   a   a   a    =       ....  n   p   p21   pr   FN )(log dim B F  inflim .  0 log The following result about the Fibonacci and Lucas numbers is known: If dim F  dim F , the common value is called the box- B B Theorem 2.1 (Legendre, Lagrange). Let p be a prime odd dimension of F and is denoted by dim B F or b (FD ). integer. Then the Fibonacci number, Fp , has the property: Let : If  R be a function defined on the interval I and  p   , tt 21  be a subinterval of I . We denote by: Fp    (mod p)  5 

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f ttR 21  )(sup),(  uftf .  Calculate the mean value of the rescaled range for all sub- 1 ,  tutt 2 series of length m: and by  f )( the graph of f. 1 d SR )/(  / SR . m d  nn n1 Lemma 2.1 [5] Let f be a continuous function defined on   ,10,1,0 and m be the least integer greater than or Hurst found that (R/S) scales by power - law as time H equal to ./1 If N is the number of the squares of the increases, which indicates )/( t  tcSR .  mesh that intersect  f ).( Then: In practice, in classical R/S analysis, H can be estimated as

the slope of log-log plot of (versusSR )/ t t. m1 m1 The fractal dimension of the trace can then be calculated 1 1   f  2)1(, mNjjR   f jjR  .)1(, from the relationship between the Hurst exponent H and the j 0 j 0 fractal dimension: rs   HD ,2 where Drs denotes the fractal dimension estimated from the rescaled range method. Remark. In practice, the box dimension is determined to be the exponent Db such that d  DdN d , where N d is the III. CONGRUENCES OF FIBONACCI NUMBERS number of boxes of linear size d necessary to cover a data set Lemma 3.1 Let n be a positive integer, n  2 and p a prime of points distributed in a two-dimensional plane.  p  To measure D one counts the number of boxes of linear odd integer. Then p divides    n. b np size d necessary to cover the set for a range of values of d and   plot the logarithm of N on the vertical axis versus the Proof. d The proof is simple, using the congruence: logarithm of d on the horizontal axis. If the set is indeed fractal, this plot will follow a straight line with a negative  pb b slope that equals D .      (mod p), b  pa a A choice to be made in this procedure is the range of values     of d. In Benoit 1.3.1 software, a conservative choice may be to use as the smallest d ten times the smallest distance between with a and b - positive integers, b  a and p - a prime integer, * points in the set, and as the largest d the maximum distance or using induction after nN and the identity between points in the set divided by ten. We shall discuss this aspect, in detail in the last chapter. m n t    tn        . k   km   m  The Hurst exponent can be calculated by rescaled range k 0 analysis (R/S analysis), by the following algorithm [1]. Lemma 3.2 Let p be a prime odd integer. Then p divides A time series X k )( is divided into d sub-series of  ,1 Nk 2 p length m. For each sub-series n = 1, … , d :   , for 1  k  2p - 1, k  p.  k   Find the mean, En and the standard deviation, Sn ; Proof.  Normalize the data ( X ) by subtracting the sub-series in 2 p p)!2(  2...... 21 pp mean:    = .  k   kpk )!2(!  kpk !2!

ninin  miEXZ ;...,,1, If 1  k  p - 1, it results that p does not divide k! and  Create a cumulative time series: p +1  2p - k  2p - 1. i in   jn  miZY ;,...,1, This implies that: j1 p|(2p - k)! and p2 doesn’t divide (2p - k)!  Find the range:

2 p n max jn  YYR jn ;min Thus: p   .  ,1 mj  ,1 mj  k  If p + 1  k  2p - 1, it results that  Rescale the range / SR nn ;

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p|k! and p2 doesn’t divide k!. p1 1 F  5 2 (mod p). 2 p p1 We have 1  2p - k  p - 1, so 4

p doesn’t divide (2p - k)!. Applying Fermat’s little theorem and Euler’s criterion, it results that:

2 p Therefore, p   .  5   k  F2 p    (mod p).  p 

Proposition 3.3 Let p be a prime odd integer. Then the Applying the quadratic reciprocity law, we have: Lucas number L2 p  3 (mod p). Proof.  p   5  p1  5   p        11      , 5  p   p  5 2 p 2 p           51    51  L        2 p      2   2  so  p  F2 p    (mod p). 2 p k  5  1 2 p L     5 2 . 2 p p12   k  2 k 0   evenk Corollary 3.5 Let p be a prime odd integer. Then:

  Using Lemma 3.2, we obtain that   Fp  p   .1  2 p    2  1 p   L2 p    51  (mod p)  2 p12 Proof. 2 Since between Fibonacci and Lucas numbers there is the  51 p L2 p  p    (mod p). relationship 4   FFL , Applying Fermat’s small theorem, we obtain: nnn  11

it results that L2 p  3 (mod p).

 pp   FFL p12122 . Proposition 3.4 Let p be a prime odd positive integer. Then, the Fibonacci number Applying Proposition 3.3, Proposition 3.4 and the properties of Fibonacci and Lucas numbers, we obtain:  p  F2 p    (mod p).  5   p  3     FL 22 pp 5 F     (mod p), Proof. p12 2 2  2 p 2 p  1   51    51  F          2 p      p  5  2   2   3       FL 22 pp  5  F p12   (mod p). 2 2 2 p k 1 1 2 p F    5 2 . Using the properties of Legendre’ symbol it results that: 2 p p12   k  2 k 0   oddk     Fp  p   .1 Using Lemma 3.2 and Lemma 3.1 we obtain that  2 p     5  

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IV. ON THE BOX DIMENSION OF FIBONACCI FUNCTIONS m1 1 DEFINED ON [0, 1]  2mN   f jjR )1(,  In the following we shall use Lemma 2.1, to estimate the j0 upper bound of the Fibonacci function f, defined on [0, 1]. m1 1  j     2mN   aa 1    Theorem 4.1. The upper Box dimension of the graph of the  5   a j  Fibonacci function defined on [0, 1] is less or equal than 1. j0 Proof. In order to simplify the calculation we denote by  m1 m1  1  j 1  2  1 aamN     51  5   a j  a  .  j 0 j0  2 If  ,10 then: 1   m )1(   2    ...11 aaamN    5 11  1 m 1 m    1 1 .      m )1(    ...1  aa  5 We evaluate the difference  1 jfjf  , in order to apply Lemma 1. m 1 m 1 a 1 jfjf   2  amN 1    5 5 1 a 

1 1 1  a j )1(   1cos aj j cosj  1 j )1(  j 1 5 a a 1  m 2  amN m 1  a    1  5 5 1 1 1  j 1cos     j aa  1   j  .cos a 5 a j  a   1 1 m Since 1 m  a  2  amN 1    5 5 1 1  a  01 a   a11,10 aa   1  1 m   a 1 a 1  m and 2  amN m 1  a      5 5 a 1 coscos ,, yxyxyx  R, a 

m then 1 m a 1   2  amN 1    5   )1( aa m )1(  5 1 jfjf 

1 m    1  1   2  amN 11     j aa  1 j cos1cos j   m )1(   j    5   )1( aa  5  a 

  1  1   1  1 m   j aa  1 1 jj    2 1  a 11     j    m )1(  5  a      5   )1( aa 

1  j  1   11    1  aajfjf 1  .  2  a m 11    j    m )1(   5  a    5   )1( aa 

Applying Lemma 2.1 we obtain:

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1  1     1     )1(2 a m 11  . 22lim  a  1 11      m )1(     1     5   )1( aa   0 5   )1( aa 

log N  1     2 a 11    ln aa  1    5   22log a m 11   .log    m )1(    5   )1( aa  N )log( lim   0 log log N   1    log   22log  a  1 1)1(     1    5   )1( aa   1 m     lim1   22log a 11   m )1(   log   0 log  5   )1( aa    log   1    1 2log1 a 11  lim 1.  5  ln aa   0 log  1 m     22log a 11   m )1(   log N  5   )1( aa   1   . Thus, log  log  N )log( lim   1  0 1 m    log  22 a  11     m )1(  5   )1( aa  and B f  .1)(dim  1    1    1       V. ESTIMATIONS OF DIMENSIONS OF FIBONACCI FUNCTION 22  a   11   5      )1( aa m )1(   DEFINED ON A SYMMETRIC INTERVAL     It is well known that the Fibonacci sequence have fractal

properties. 1    22 a  1 11  , The software used was Benoit 1.3.1.  1   5   )1( aa  In the following we determine the box - dimension and the fractal dimension Drs of the Fibonacci function, defined on a because: domain  aa ],[ , where a > 0. For exemplification, a has be chosen to be 20. 1 11   1 m 11   m 1   In theory, to determine the box – dimension, for each box     size, the grid should be overlaid in such a way that the minimum number of boxes is occupied. This is accomplished m   1)1(1  in Benoit by rotating the grid for each box size through 90 degrees and plotting the minimum value of N d . 111 Benoit permits the user to select the angular increments of m )1(1  aaa  . a m aa 1)1(  rotation. For our calculation, the increment of the rotation

angle has been set to 15 degrees, the coefficient of box- But, decrease, 1.1, and the size of the largest box 70 pixels (Fig.3). The value calculated for the box dimension was 1.91362.  1 Looking to the left hand side of Fig. 3, we see that the chart of lim  lim, 1   aa , the number of boxes versus the boxes - side length is linear,  0  ln a  0 a 1 proving that fractal character of the Fibonacci number defined on [-20, 20] . So, For the same function the value of Hurst coefficient has

been determined to be H = 1.536 and Drs = 0.464 (Fig.4).  1    22lim a m 11   Therefore, the function has the long range dependence    m )1(    0 5   )1( aa  property on the studied interval.

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Fig.3. Determination of the box-dimension of the Fibonacci function defined on [-20, 20]

Fig.4. Determination of the Hurst coefficient for the Fibonacci function defined on [-20, 20]

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