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(2021)
The golden ratio and Fibonacci sequence in music
(2010)
Phyllotaxis, 57
(2022)
A Fibonacci-like universe expansion on time-scale
Rasim Basak (2022)
GOLDEN RATIO AND FIBONACCI SEQUENCE: UNIVERSAL FOOTPRINTS OF THE GOLDEN FLOWTHE TURKISH ONLINE JOURNAL OF DESIGN ART AND COMMUNICATION
Guangmeng Guo (2021)
Fibonacci Sequence Found in Parkfield EarthquakeInternational Journal of Geosciences, 12
ARAB JOURNAL OF BASIC AND APPLIED SCIENCES University of Bahrain 2023, VOL. 30, NO. 1, 74–78 https://doi.org/10.1080/25765299.2022.2157951 ORIGINAL ARTICLE a b Saeree Wananiyakul and Janyarak Tongsomporn a b Department of Mathematics and Computer Science, Faculty of Science, Chulalongkorn University, Bangkok, Thailand; School of Science, Walailak University, Nakhon Si Thammarat, Thailand ABSTRACT ARTICLE HISTORY Received 2 February 2022 We consider linear recursions of length two and related gap recursions where the indices Revised 17 November 2022 may not be consecutive integers. Given a linear recursion of length two, we prove the exist- Accepted 9 December 2022 ence of an explicit linear recursion of length l with arbitrary distance between indices. Conversely, it is shown under some mild assumption that a linear recursion of length l can KEYWORDS be reduced to one of length two. Fibonacci NumberLinear RecursionRecurrence relation 1. Introduction the Fibonacci sequence to make predictions on the The sequence of Fibonacci numbers F is defined by n Parkfield earthquake sequence. Recently, Postavaru the recursion F ¼ F þ F for integer n 2 N, and Toma (2022) studied the time scale Fibonacci nþ1 n n1 starting with the initial values F ¼ 0 and F ¼ 1: sequences satisfying the Friedmann-Lemaıtre- 0 1 Livio (2002) provided an explicit formula for the nth Robertson-Walker (FLRW) dynamic equation on time scale, which is an exact solution of Einstein’s field Fibonacci number, i.e. equations of general relativity for expanding homo- F ¼ pffiffiffi G ðgÞ ðn 2 N Þ, (1) n 0 geneous and isotropic universe. They investigated the results based on the dynamic equations on time pffiffiffi where G :¼ 5 þ 1 is the golden ratio and g :¼ 2 scale describes the Fibonacci numbers. In the same pffiffiffi 1 1 G ¼ 5 1 : It is an easy exercise to prove this for- year, Basak (2022) proposed to construct a philo- mula by induction or using the associated generat- sophical theory based on unexplained natural phe- ing function. nomena, observing the golden ratio and Fibonacci Fibonacci numbers appear in various ways in dif- sequence in art and nature such as plants, animals, ferent areas. In botany, phyllotaxis is the study of storm and universe. He also discussed these phe- the arrangement of leaves on a plant stem. In 1830, nomena within his philosophical aspects based on Schimper (1830) observed that the ratio of the num- the literature evidences. Since Fibonacci sequence is in the form of a linear ber of circuits divided by the number of leaves—the recursion of length two, it inspires us to consider any so-called divergence—in the spiral configurations arbitrary recursions of length two and higher length. It often result as quotients of Fibonacci numbers may come as a surprise that any number of further F =F with gap; for example, the divergence is 1/3 k kþ2 recursions for Fibonacci numbers can be formed with for hazel, 2/5 for apricot, 3/8 for pear, and 5/13 for arbitrary distances between the indices. For example, almond (cf (Hellwig & Neukirchner, 2010)). The for every pair of positive integers k and l,there exist sequence of rational numbers F =F is called the k kþ2 coefficients c, d depending only on k, l such that Braun-Schimper sequence (named after Alexander Braun a colleague of Schimper). In 2015 Wang, Li, F ¼ cF þ dF (2) nþk n nl Kong, and He (2015) discovered that a fractal dimen- sion of polar bear hairs is close to the golden ratio for every n 2 N; we shall call this a gap recursion.In 1.618… . Sinha (2017) showed amazing applications fact, the equations for n ¼ l and n ¼ l þ 1 deter- of Fibonacci numbers in nature including construct- mine c and d and the verification of the gap recur- sion simply follows from the classical recursion ing security coding. Blankenship (2021) linked the formula by induction. This yields Fibonacci sequence to music and Guo (2021) used CONTACT Janyarak Tongsomporn tjanyarak@gmail.com School of Science, Walailak University, Nakhon Si Thammarat, 80160, Thailand 2023 The Author(s). Published by Informa UK Limited, trading as Taylor & Francis Group on behalf of the University of Bahrain. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. ARAB JOURNAL OF BASIC AND APPLIED SCIENCES 75 F F F any positive integers k, l and non-negative integers kþl kþl lþ1 F ¼ F þ F F (3) nþk n kþlþ1 nl m , m , :::, m where m m for all j, B 6¼ 0 for F F l l 1 2 l l j m m l i i ¼ 1, :::, l and B ¼ 0 for i ¼ l þ 1, :::, l 1, m m l i or we obtain F F ¼ F F þðF F F F ÞF (4) l nþk kþl n kþlþ1 l kþl lþ1 nl X ¼ A X þ B X , (10) nþk m þk nm m þk nm 1 l l l l for arbitrary positive integers k, l: X ¼ A X þ B X for 1 j < l: nm m m nm m m nm 1 j l j l l j l In this short note, we study the more general situ- (11) ation of a sequence of real numbers X satisfying a If l ¼ 0, it is easy to see that recursion of the form l1 l1 X ¼ aX þ bX , (5) X X n n1 n2 X ¼ X þ A þ A X nþk nm m þk m m nm i l l i l where a, b are again fixed real numbers. Our aims i¼1 i¼1 are to answer the following questions: þ B X : (12) m þk nm 1 l l 1. For every positive integers k, l and distinct 1 If B 6¼ 0 and l 6¼ 0, we derive m þk m , ... , m n, do there exist coefficients 1 l l l1 c , ... , c such that X X 1 l X ¼ X þ X þ nm 1 nm nm l i i l B X m m l i i¼1 i¼l þ1 0 1 X ¼ c X ? (6) (13) nþk nm i i l l1 X X m m i¼1 l i @ A A X : m m nm l i l l B m m l i i¼1 i¼l þ1 And, in case of their existence, is there a corre- sponding explicit formula for X in terms of the This implies gap recursion? 2. Conversely, let X ¼ c X for n l with a n ni X i¼1 i X ¼ B P X , (14) nþk m þk i nm positive integer l 2 and real numbers c . Are l i i¼1 there two real numbers a, b such that where X ¼ aX þ bX (7) n n1 n2 If B ¼ 0 and l 6¼ 0, we obtain similarly m þk for every integer n 2? if 1 i l , > l B m m < l i 1 if l þ 1 i < l, P ¼ (15) l l1 > X X > A A m þk m m > l l i þ A if i ¼ l, > m m l i B l B m þk m m l l i i¼1 i¼l þ1 We give the answer to the first and the second question in Sections 2 and 3, respectively by apply- l l1 X X ing fundamental techniques from linear algebra and X ¼ X þ X þ nm 1 nm nm l i i ðl 1ÞB elementary calculations with complex numbers. m m l i i¼2 i¼l þ1 0 1 l l1 X X m m l i @ A A X : m m nm l i l 2. Extension of the recursion of length two ðl 1ÞB m m l i i¼2 i¼l þ1 Given any sequence of real numbers X satisfying n (16) the following recursion This implies X ¼ aX þ bX for n 2, (8) n n1 n2 it follows that, for any positive integer l, we have X ¼ Q X , (17) nþk i nm X ¼ A X þ B X for n 1, (9) nþl l n l n1 i¼1 where A :¼ aA þ bA with A ¼ 1, A ¼ a, and n n1 n2 0 1 where B :¼ aB þ bB with B ¼ 0, B ¼ b: Hence, for n n1 n2 0 1 76 S. WANANIYAKUL AND J. TONGSOMPORN 1 if i ¼ 1, m m > l 1 if 2 i l , ðl 1ÞB m m l i Q ¼ : (18) i B if l þ 1 i < l, m m l 1 l l1 > X X > A m m > l i A þ A þ A if i ¼ l m þk m m m m > l l 1 l i ðl 1ÞB m m l i i¼2 i¼l þ1 This leads to 3. Reduction of the recursion of length l Theorem 1. Given real numbers a, b, let We consider the recursion X ¼ aX þ bX . Define n n1 n2 A :¼ aA þ bA with A ¼ 1, A ¼ a, and n n1 n2 0 1 X ¼ c X , (23) n ni B :¼ aB þ bB with B ¼ 0, B ¼ b: i¼1 n n1 n2 0 1 Then, for every pair of positive integers k, l and non- where l 2 is a positive integer and the c ’s are real negative integers m , m , :::, m ,where m m for all 1 2 l l j numbers such that c 6¼ 0: Recalling the ques- i¼1 i j l,let l l 1 such that B 6¼ 0 for i ¼ m m l i tion from the introduction, is there a relation X ¼ 1, :::, l and B ¼ 0 for i ¼ l þ 1, :::, l 1.Then m m l i aX þ bX for some real numbers a, b? n1 n2 B P X if B 6¼ 0, l 6¼ 0, > m þk i nm m þk l i l i¼1 X ¼ (19) Q X if B ¼ 0, l 6¼ 0, nþk i nm m þk i l i¼1 > X B X þ R X if l ¼ 0, > m þk nm 1 i nm l l i i¼1 if 1 i l , > l B m m l i 1 if l þ 1 i < l, P ¼ (20) l l1 X X A A m þk m m > l l i þ A if i ¼ l, > m m l i B l B m þk m m l l i i¼1 i¼l þ1 1 if i ¼ 1, > B m m > l 1 if 2 i l , ðl 1ÞB m m l i Q ¼ (21) i B if l þ 1 i < l, m m l 1 l l1 > X X > A m m > l i A þ A þ A if i ¼ l, > m þk m m m m l l 1 l i ðl 1ÞB m m l i i¼2 i¼l þ1 where In 2017, Wiboonton (2017) collected several meth- ods for research on the Fibonacci sequence. One of and them relies on matrix calculus. We shall follow this approach and consider: > 1 if 1 i < l, l1 X ab X nþ1 n R ¼ (22) ¼ : (24) A þ A if i ¼ l: > m þk m m l l i X 10 X n n1 i¼1 ARAB JOURNAL OF BASIC AND APPLIED SCIENCES 77 This implies, for any positive integer k, Summing up, we arrive Theorem 2. Let l 2 be a positive integer and X ¼ X ab X n nþk n ¼ : (25) c X be a linear homogeneous recursion with X 10 X nþk1 n1 ni i¼1 i initial conditions X for i ¼ 0, 1, :::, l 1 and suppose In the sequel we assume that a and b satisfy a þ that the sequence of X satisfies at least one of the fol- 4b 6¼ 0 (in order to have a non-vanishing discrimin- lowing statements: ab ant). We can diagonalize by an invertible 1. the corresponding characteristic equation has two k k real roots r and r such that r þ 4r 6¼ 0 and the 1 2 2 matrix P :¼ and a diagonal matrix D :¼ quadratic recursion k 0 Y ¼ðr þ r ÞY r r Y (32) n 1 2 n1 1 2 n2 such that 0 k with the initial conditions of Y ¼ X and Y ¼ 0 0 1 X leads to Y ¼ X for all i ¼ 2, 3:::, l 1, ab 1 i i ¼PDP , (26) 2. its characteristic equation has a complex root z such that z is not a real number and the quad- pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi 2 2 a a þ4b aþ a þ4b where k :¼ and k :¼ : ratic recursion 2 2 Y ¼ 2RðzÞY jzj Y (33) n n1 n2 nþ1 If we set X ¼ , then we can rewrite rela- tion (23) as with the initial conditions of Y ¼ X and Y ¼ 0 0 1 X leads to Y ¼ X for all i ¼ 2, 3:::, l 1: 1 i i X ¼ c X : (27) n ni i¼1 Then, Y ¼ X for all n. n n Applying the above matrices, we obtain Applying this to the second question from the beginning, it follows that there always exist two real l 1 li 1 numbers a, b satisfying X ¼ aX þ bX when- PD P X ¼ c PD P X , (28) n n1 n2 nl nl i¼1 ever the associated characteristic equation satisfies at least one of the two conditions in Theorem 2. or l li 1 PD c D P X ¼ 0: (29) nl 4. Applications and examples i¼1 To illustrate the theorems from above, we shall consider l li 1 Now, we set PðD c D ÞP ¼ 0: This i¼1 the well-known recursion for Fibonacci number F . l li implies D ¼ c D : Hence, we have i¼1 Example 3. It is easy to see that if the indices are 2 3 restricted to the set of non-negative odd or even li "# 6 7 c k 0 6 7 l integer, there are also recursions, namely k 0 6 7 i¼1 ¼ 6 7: (30) 0 k 6 X 7 ðÞ F ¼ 3F F n 2 : (34) nþ2 n n2 li 4 5 0 c k i¼1 Indeed, we may apply Theorem 1 with A ¼ F n nþ1 and B ¼ F for any positive integer n. We deduce n n Then, l l 1 F F kþlþ1 lþ1 X X l li l li F ¼ F F þ F (35) nþk kþl n nl k ¼ c k and k ¼ c k : (31) i i F F F þ þ l kþl l i¼1 i¼1 or To solve this system, we assume that there are F F ¼ F F þðÞ F F F F F (36) l nþk kþl n kþlþ1 l kþl lþ1 nl two complex numbers z and z satisfying x ¼ 1 2 pffiffiffiffiffiffiffiffiffiffi a a þ4b l for any positive integers k, l (that are formulae (Guo, li c x : If we set z :¼ k ¼ and z :¼ i 1 2 i¼1 pffiffiffiffiffiffiffiffiffiffi 2021) and (Hellwig & Neukirchner, 2010) from the aþ a þ4b k ¼ , then we get a ¼ z þ z and b ¼ introduction). For the special case k ¼ 2, l ¼ 2, we get þ 1 2 z z : In case that z and z are real numbers, both 1 2 1 2 F F ¼ F F þðÞ F F F F F (37) 2 nþ2 4 n 5 2 4 3 n2 a and b are also real, however, it remains to verify or whether a þ 4b is non-vanishing or not. F ¼ 3F F (38) nþ2 n n2 For otherwise, we choose z ¼ z 2 C n R which is l li a root of x ¼ c x and its conjugate z ¼ z is 2 (which is formula (34) above). i¼1 i also a root. In this case the values a and b are 2RðzÞ On the other hand, by Theorem 2, the correspond- and jzj , respectively. Furthermore, a þ 4b 6¼ 0 ing characteristic equation of the recursion 4 2 since z is not a real number. F ¼ 3F F is x 3x þ 1 which has four n n2 n4 78 S. WANANIYAKUL AND J. TONGSOMPORN distinct real roots, namely Disclosure statement pffiffiffi pffiffiffi pffiffiffi No potential conflict of interest was reported by 1 1 1 1 5 , 1 5 , 1 þ 5 , the authors. 2 2 2 (39) pffiffiffi 1 þ 5 : Funding This work was supported by the National Research Council If the initial values are F ¼ 0, F ¼ 1, F ¼ 1, 0 1 2 of Thailand (NRCT) and Walailak University under Grant F ¼ 2, we get that Y ¼ Y þ Y : 3 n n1 n2 number NRCT5-RSA63019-06 (2563NRCT322800). If the initial conditions are F ¼ 0, F ¼ 1, F ¼1, 0 1 2 F ¼ 2, we get that Y ¼Y þ Y : 3 n n1 n2 ORCID Janyarak Tongsomporn http://orcid.org/0000-0001- 5. Conclusions 8744-2298 From Sections 2 and 3, we conclude the main results as follows: References Remark 4 In Theorem 1 (Section 2), we extend the Basak, R. (2022). Golden ratio and Fibonacci sequence: length of a linear recursion of length two to one of Universal footprints of the golden flow. The Turkish Online Journal of Design Art and Communication, 12(4), length l. Conversely, we can reduce the length of a 1092–1107. doi:10.7456/11204100/013 linear recursion of length l to one of length two in Blankenship, R. A. (2021). The golden ratio and Fibonacci Theorem 2 (Section 3). sequence in music (Diss). Ohio Dominican University, Ohio. Guo, G. (2021). Fibonacci sequence found in Parkfield Remark 5 We observe that a linear recursion of earthquake. International Journal of Geosciences, 12(01), length l that is constructed from Theorem 1 (Section 1–5. doi:10.4236/ijg.2021.121001 2) can always be reduced to linear recursion of length Hellwig, H., & Neukirchner, T. (2010). Phyllotaxis. Die math- ematische Beschreibung und Modellierung von two by using the method in Theorem 2 (Section 3). Blattstellungsmustern. Math. Sember, 57,17–56. However, for any linear recursion of length l,ifit is Livio, M. (2002). The golden ratio: The story of Phi, the World’s written in any form of linear recursions of length two most astonishing number. New York: Broadway Books. obtained by Theorem 2 but there is no initial condi- Postavaru, O., & Toma, A. (2022). A Fibonacci-like universe tion of each recursion of length two which corre- expansion on time-scale. Chaos, Solitons and Fractals, 154, 111619. doi:10.1016/j.chaos.2021.111619 sponds to one of length l, we cannot conclude that Schimper, K. F. (1830). Beschreibung des Symphytum there is not a linear recursion of length two which Zeyheri und seiner zwei deutschen Verwandten der S. represents the same sequence. bulbosum Schimper und S. tuberosum Jacq, Geiger’s Mag. Pharm, 20,1–92. Sinha, S. (2017). The Fibonacci numbers and its amazing applications. International Journal of Engineering Science Acknowledgments Invention, 6(9), 7–14. The authors are grateful to the anonymous referee for her Wang, Q., Li, Z., Kong, H., & He, J. (2015). Fractal analysis of or his valuable comments and corrections. The authors polar bear hairs. Thermal Science, 19(Suppl. 1), 143–144. would like to thank National Research Council of Thailand doi:10.2298/TSCI15S1S43W (NRCT) and Walailak University for facilities and finan- Wiboonton, K. (2017). The Fibonacci sequence in mathem- cial supports. atics courses, MJ-Math, 60(691), 1–11.
Arab Journal of Basic and Applied Sciences – Taylor & Francis
Published: Dec 31, 2023
Keywords: Fibonacci Number; Linear Recursion; Recurrence relation
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