Title: Some observations on the properties of Lucas sequence compared to Fibonacci sequence.

Driven by my fascination for different number series, I chose to explore Lucas numbers, a series closely connected with Fibonacci series of numbers. Edouardo Lucas (1842-1891) gave name to the number sequence developed by Leonardo of Pisa as Fibonacci Numbers. Later he derived the new number sequence having similar property as Fibonacci numbers known as Lucas numbers. Given the familiarly of Fibonacci numbers among the mathematicians’ community, I was enthusiastic about analyzing the Lucas number in relation with the Fibonacci series and particularly about ‘the golden ratio’, a concept so much heard but little known to me. The exploration moves through the connections between both the number series as I establish Binet’s formula, through the application of golden ratio, for a Fibonacci series and also prove a hypothetical connection between both the series. As the exploration reaches its conclusive part, I also create a few patterns employing Fibonacci and Lucas number. I also used Microsoft excel to generate the numbers and calculated the value of Golden Ratio to analyse the convergence of the values of Golden ratio for different number of terms.

The series 0 ,1 ,1 ,2 ,3 ,5 ,8 ,13. . . originally developed by Leonardo of Pisa was termed as Fibonacci series of numbers by Edward Lucas, a 19th century French mathematician. The Fibonacci series was based on the recursive formula given by Leonardo Bonacci (1170-1250 AD) known as Fibonacci. Lucas later gave another number series that followed the same formula but the starting values were different: 2 ,1,3 ,4 ,7 ,11 ,18.. . . it was named Lucas series. The series works as shown below:

Ln=Ln−1+Ln−2 ,n>1, n∈Z+¿¿

L0=2L1=1

The table below depicts some numbers of the series in comparison to Fibonacci number:-

n Ln Fn0 2 01 1 12 3 13 4 24 7 35 11 5

Page 1 of 12

6 18 87 29 138 47 21

It is apparent that the numbers occurring in Lucas series and those in the Fibonacci series have many similarities. Most importantly many of these numbers often feature in different formula based on Fibonacci series.

In a series of Fibonacci numbers if p and q are interrelated as factors, then their Fibonacci versions (Fp and Fq) will also share the same relation. For instance, we can take 3 and 6. Just like 3 and 6, their Fibonacci counterparts (F 3 =2 and F6 =8) also have the same relation, 2 being the factor of 8.

Another way of examining this relation would be to explore the evenly arranged numbers in Fibonacci series, which would be termed as F2n. Given the even nature of these numbers, all the numbers in the series can be divided by F2. F2 remains inconspicuous by being a universal factor in the series. Now I commence the investigation of Fn factors (with n being a factor for 2n)

n Fn 2n F2n m=F2nFn

1 1 2 1 12 1 4 3 33 2 6 8 44 3 8 21 75 5 10 55 116 8 12 144 187 13 14 377 29

We observe from the table that m is nothing but representing Lucas numbers for the corresponding n ,n∈Z+¿¿. Hence my conjecture is

F2n=Fn×LnNow, to prove the above conjecture we have to first find the formula for Fn using Golden ratio method. Let α be the golden ratio, which is defined ABAM

= AMMB

for the given below line segment. Let us assume that AB is a

line segment and M divides the line segment AB internally in the ratio x :1

such that ABAM

= AMMB

we get,

Page 2 of 12

x+1x

= x1

x+1=x2

x2−x−1=0Using the quadratic formula we are going to find x.

x=−b±√b2−4ac2a

a=1 , b=−1∧c=−1

x=−(−1)±√(−1)2−4×1×(−1)

2×1

x=1±√52

We got two values of x ,

One is negative and one is positive. Length can’t be negative, so I have to

choose positive one. Therefore, x=1+√52

, this is the value of the Golden

Ratio φ

φ=1+√52

squaring both the sides, we get

φ2=( 1+√52 )

2

¿ 1+2√5+54

¿ 6+2√54

¿ 3+√52

But φ+1=1+√52

+1

φ+1=3+√52

Now, we got the relationship between the Golden ratio and Fibonacci number

φ2=φ F2+F1

Now, cubing the value of φ,

φ3=( 1+√52 )

3

¿13+3×12×√5+3×1×√52+(√5 )3

8

¿ 1+3√5+15+5√58

Page 3 of 12

¿ 16+8√58

¿2+√5

¿2× 1+√52

+1=2φ+1=φ F3+F2

Therefore, we got the relationshipφ3=φF3+F2

After examining the above relationship, now I conclude the conjecture φn=φF n+Fn−1

Proof :- Proof for the conjecture using mathematical induction

Let P (n ) :φn=φF n+Fn−1

Case 1:- Put n=1φ1=φF 1+F1−1

1+√52

=1+√52

F1

+F0

which is true as F1=1∧F0=0.

Let us assume that P (k ) be true for some k∈ z+¿ ¿.This assumption gives us

φk=φF k+Fk−1

Now for P (k+1 ) , n=k+1We are to prove that

φk+1=φFk +1+Fk +1−1LHS:- φk+1 ¿φk×φ¿(φF¿¿k+Fk−1)φ ¿¿φ2Fk+φ(Fk+1−Fk ) as F k+1=Fk+Fk−1¿(φ2−φ)Fk+φF k+1¿ Fk (φ+1−φ )+φF k+1 as φ2−φ−1=0¿φF k+1+Fk

RHS¿φF k+1+Fk+1−1¿φF k+1+Fk¿ LHS

therefore P (k+1 )holds true.

Page 4 of 12

We have already shown P(1) was true and on assumption P(k) true ¿>¿ P(k+1) is also true. Hence by the principle of mathematical induction P(k) is true for all k∈Z+¿¿

Hence

φn=φF n+Fn−1

Now I am going to try negative reciprocal of golden ratio

−1φ

= −1

(1+√52 )

¿ −21+√5

×1−√51−√5 , rationalizing the denominator.

¿ −2+2√5(1−√5)(1+√5)

¿−2(1−√5)1−(√5)2

¿−2(1−√5)

−4

¿(1−√5)2

=φ

this value is equal to the conjugate value of φ found by solving the equation x2−x−1=0. But that time we discarded that value because length cannot be negative. But now i am going to make conjecture using the above value found.

F2(−1φ )+F1¿1×( 1−√5

2 )+1¿( 3−√5

2 )

(−1φ )2

=( 1−√52 )

2

¿ 1−2√5+54

¿ 6−2√54

¿ 3−√52

¿ F2(−1φ )+F1Page 5 of 12

we got the same pattern as aboveso we try for more values,

(−1φ )3

=F3(−1φ )+F2LHS=( 1−√5

2 )3

¿ 1−3√5+3×5−5√58

¿ 16−8√58

¿2−√5RHS=2× 1−√5

2+1

¿1−√5+1¿2−√5

after observing the above results, I conclude that the conjecture

(−1φ )n

=Fn(−1φ )+Fn−1∀n∈Z+¿ , n>1.¿

Now, I am going to prove the above conjecture using Principle of Mathematical Induction

P (n ) :(−1φ )n

=Fn(−1φ )+Fn−1P(2) is already proved to be true.Assuming that P (n )is true for n=k

Therefore P (k ) :(−1φ )k

=Fk(−1φ )+Fk−1For n=k+1, we are to prove that

(−1φ )k +1

=Fk+1(−1φ )+Fk+1−1LHS:-

¿(−1φ )k+1

¿(−1φ )k

(−1φ )¿(Fk (−1φ )+Fk−1)(−1φ )¿ Fk (−1φ )

2

+Fk−1(−1φ )

Page 6 of 12

¿ Fk (−1φ )2

+¿

¿ Fk ((−1φ )2

−−1φ )+Fk +1(−1φ )

¿ Fk ((−1φ )+1−(−1φ ))+Fk+1(−1φ )¿ Fk+1(−1φ )+Fk

RHS:-

¿ Fk+1(−1φ )+Fk+1−1¿ Fk+1(−1φ )+Fk

¿ LHStherefore P (k+1 )holds true.We already prove P(2) was true and assuming P(k) true ¿>P (k+1) is also true. Hence by the principle of mathematical induction P(n) is true for all n∈Z+¿ , n>1.¿

Now we have two equations,

Equation (1) φn=φF n+Fn−1

Equation (2) (−1φ )n

=Fn(−1φ )+Fn−1subtracting equation (2) from equation (1), we get

φn−(−1φ )n

=φF n+Fn−1−Fn(−1φ )−Fn−1

φn−(−1φ )n

=φF n+Fn( 1φ )φ2n−(−1 )n

φn=Fn(φ2+1φ )

Fn=(φ2n−(−1 )n

φn )×( φ

φ2+1 )Fn=(φ2nφn − (−1 )n

φn )×( 1+√52

3+√52

+1 )Page 7 of 12

Fn=¿

Fn=(φn−(−φ )−n )( (1+√5)(5−√5)(5+√5)(5−√5))

Fn=(φn−(−1φ )−n)( 4 √5

20 )Fn=(φn−(−1φ )

−n)( 4 √520 )

Fn=(( 1+√52 )

n

−(1−√52 )

−n)( 4 √520 )

Fn=( (1+√5 )n−(1−√5 )−n

2n )×( 1√5 )Fn=

(1+√5 )n−(1−√5 )−n

2n√5Fn=

φn−(−φ)−n

√5Here I have derived Binet’s formula by establishing the relationship between Fibonacci series of numbers and the golden ratio. This formula could be applied to establish a number’s value in Fibonacci series without knowing the previous two numbers.

The same could not be done with the Lucas series of numbers because that formula is closed in nature. Hence, I will be exploring the patterns that emerge through the use of golden ratio.

n φn (−1φ )n

φn+(−1φ )n

0 1.0000 1 21 1.6180 -0.6180 12 2.6180 0.3819 33 4.2360 -0.2360 44 6.8541 0.1459 75 11.090 -0.09017 116 17.94427 0.0557 18

We noticed that the sum of these numbers is the Lucas numbers.

Ln=φn+(−1φ )

n

this is the formula for the Lucas numbers.Ln=φ

n+(−φ )−n

Page 8 of 12

We observe that, F2k=Fk×LkLet us prove the conjecture using mathematical induction.

Put n=1 ,F2× 1=F1×L1=1×1=1

assume that P(k) be true,

P (k ) :F2k=F k×Lk for somekϵ Z+¿ ¿

P (k+1 ) :F2 (k+1)=F (k+1)×L(k +1) must be true.

RHS:-

(φn+1−(−φ)−n−1

√5 )×(φn+1+ (−φ )−n−1)

¿( φn+1−(−φ)−n−1

√5 )×(φn+1+(−φ )−n−1)

¿ 1√5

[φn+1φn+1+(−φ )−n−1φn+1−(−φ )−n−1φn+1− (−φ )−n−1φ−n+1 ]

¿ 1√5

(φ2(n+1)−(−φ )2 (−n−1))

¿φ2 (n+1)−(−φ)2(−n−1)

√5

LHS ¿φ2 (n+1)−(−φ)−2 (n+1)

√5Hence P(k+1) is also true.

We already prove P(1) was true and assuming P(k) true ¿>P (k+1) is also true. Hence by the principle of mathematical induction P(n) is true for all n∈Z+¿ , n≥1.¿

Hence Proved.

I also investigate new patterns that is listed below,

n Ln Fn0 2 01 1 12 3 13 4 24 7 35 11 5

Page 9 of 12

6 18 87 29 13

We find from the table the formula connecting Lucas numbers and Fibonacci number.

Ln=Fn−1+Fn+1 , n≥1For example,

L6=18=5+13=F5−1+F5+1Finally I have graphed the value of the Golden Ratio using both Lucas sequence and Fibonacci sequence to see their relevance. An excel sheet is added in the appendix with the values of the Golden Ratio calculated using n up to 27. It is observed from the table as well as from the graphs that even though there were more fluctuations of the values of φ initially with Lucas numbers but the convergence is a bit faster using Lucas numbers than using Fibonacci numbers. It is interesting to note that even though both the series consisting of different terms but both gives the Golden ratio as ratio of the two consecutive terms. However, Lucas series gives faster convergence compared to Fibonacci sequence.

1 3 5 7 9 11 13 15 17 19 21 23 25 270

0.5

1

1.5

2

2.5

3

3.5

Convergence of phi using Ln

Phi

Conclusion:-

While I have always been in awe of those great minds, it was self-fulfilling experience to take up the subject and pursue it to such a length. I used the basic premises, formed by own conjectures or hypothesis and substantiated or validated them through my exploration. Deriving Binet’s formula or understanding golden ratio in a better way were the highlights

Page 10 of 12

of this experience, an achievement that gave me great self-confidence. There are many relations exists between the terms of Fibonacci sequence and Lucas sequence. As observed from the table one such relation could be Ln=Fn−1+Fn+1 . As I was more concerned about comparing the possible relation with reference to the Golden Ratio hence further such observation isn’t relevant to my research. However, these intuitions and searching more such relations could prove to be of great interest to mathematicians even though the terms are apparently different, there must be a good number of various relations may exist as I see basic structure of creating such sequences were essentially the same, namely a member of the sequence is sum of its two previous terms. It was an opportunity to use my mathematical ability for a planned investigative task, and it was both an assessment and enhancement of my knowledge and skills. Bibliography:-

1. http://www.homeschoolmath.net/teaching/ fibonacci_golden_section.php

2. http://www.homeschoolmath.net/teaching/fibonacci_golden_section.php

AppendixCalculation of Golden Ratio using Lucas number versus Fibonacci

numbersn Ln Phi Fn Phi

0 2   0  

1 10.50000000

0 1  

2 33.00000000

0 11.00000000

0

3 41.33333333

3 22.00000000

0

4 71.75000000

0 31.50000000

0

5 111.57142857

1 51.66666666

7

6 181.63636363

6 81.60000000

0

7 291.61111111

1 131.62500000

0

8 471.62068965

5 211.61538461

5

9 761.61702127

7 341.61904761

9

10 1231.61842105

3 551.61764705

9

Page 11 of 12

11 1991.61788617

9 891.61818181

8

12 3221.61809045

2 1441.61797752

8

13 5211.61801242

2 2331.61805555

6

14 8431.61804222

6 3771.61802575

1

15 13641.61803084

2 6101.61803713

5

16 22071.61803519

1 9871.61803278

7

17 35711.61803353

0 15971.61803444

8

18 57781.61803416

4 25841.61803381

3

19 93491.61803392

2 41811.61803405

6

20 151271.61803401

4 67651.61803396

3

21 244761.61803397

9 109461.61803399

9

22 396031.61803399

2 177111.61803398

5

23 640791.61803398

7 286571.61803399

0

24 1036821.61803398

9 463681.61803398

8

25 1677611.61803398

9 750251.61803398

9

26 2714431.61803398

9 1213931.61803398

9

27 4392041.61803398

9 1964181.61803398

9

Page 12 of 12