ffil~©©~ ~ffi)~ ~oo© OOoornru~~ - Alyoops! and Lucas Numbers.pdf1 Columbia University Library, D. E. Smith Collection 81 American Museum of Natural History, photo by Franklyn

~~~®ffil~©©~ ~ffi)~ ~oo©~~ OOoornru~~~~

Verner E. Hoggatt, Jr.

~~~@[ill~©©~ ~[ill~ ~00©~~ ~[IDffi]]~@~~

Verner E. Hoggatt, Jr.

Albert E. Meder, Jr. EDITORIAl ADVISER

A Publication of

THE FIBONACCI ASSOCIATION University of Santa Clara

Santa Clara, CA 95053

ABOUT THIS BOOK

This booklet offers an introduction to some of the interesting

properties of Fibonacci and Lucas numbers. In reading this

material, the student will have an opportunity to observe how

many mathematical generalizations can be derived from some

very simple notions.

ABOUT THE AUTHOR

The late Verner E. Hoggatt, Jr. was Professor of Mathematics at

San Jose State College in San Jose, California. Together with

Brother Alfred Brousseau, he founded T_he Fibonacci Quarterly

(the official journal of The Fibonacci Association) in 1963 and

was its General Editor until his death in August 1980.

© 1969 by Houghton Mifflin Company.

All rights reserved. Copyright reverted to the author in 1979, and The Fibonacci Association was granted permission to publish this book.

No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanica I, including photocopying and recording, or by an information storage or retrieval system without permission in writing from the publisher.

Contents

Section Page

1. Introduction 1

2. Rabbits, Fibonacci Numbers, and Lucas Numbers 2

3. The Golden Section and the Fibonacci Quadratic Equation 9

4. Some Geometry Related to the Golden Section 14

5. Some Fibonacci Algebra 26

6. Shortcuts to Large F11 and Ln 30

7. Divisibility Properties ofthe Fibonacci and Lucas Numbers 37

8. Periodicity of the Fibonacci and Lucas Numbers 42

9. Pascal's Triangle and the Fibonacci Numbers 48

10. Selected Identities Involving the Fibonacci and Lucas Numbers 52

11. Two-by-Two Matrices Related to the Fibonacci Numbers 62

12. Representation Theorems 69

13. Fibonacci Numbers in Nature 79

A List of the First 100 Fibonacci Numbers and the First 100 Lucas Numbers 83

Partial Solutions 85

Index 92

PICTURE CREDITS

Page

1 Columbia University Library, D. E. Smith Collection

81 American Museum of Natural History, photo by Franklyn K . Lauden

82 Courtesy of Dole Company

IV

1 • Introduction

Fibonacci

Who was Fibonacci? Leonardo Fibonacci, mathematical innovator of the thirteenth century,

was a solitary flame of mathematical genius during the Middle Ages. He was born in Pisa, Italy, and because of that circumstance, he was also known as Leonardo Pisano, or Leonardo of Pisa. While his father was a collector of customs at Bugia on the northern coast of Africa (now Bougie in Algeria), Fibonacci had a Moorish schoolmaster, who introduced him to the HinduArabic numeration system and computational methods.

After widespread travel and extensive study of computational systems, Fibonacci wrote, in 1202, the Liber Abaci, in which he explained the HinduArabic numerals and how they are used in computation. This famous book was instrumental in displacing the clumsy Roman numeration system and introducing methods of computation similar to those used today. 1t also included some geometry and algebra.

Although he wrote on a variety of mathematical topics, Fibonacci is remembered particularly for the sequence of numbers

l, 1, 2, 3, 5, 8, 13, 21, 34, 55, ... '

to which ·his name has been applied. This se·quence, even today, is the subject of continuing research, especially by ·the Fibonacci Association, which publishes The Fibonacci Quarterly.

We shall study some elementary and interesting aspects of the Fibonacci and related numbers in this booklet.

1

2 • Rabbits, Fibonacci Numbers,

and Lucas Nun1bers

Fibonacc;i introduced a problem in the Liber Abaci by a story that may be summarized as follows. Suppose that

(I) there is one pair of rabbits in an enclosure on the first day of January; (2) this pair will produce another pair of rabbits on February first and on

the first day of every month thereafter; and (3) each new pair will mature for one month and then produce a new pair

on the first day of the third month of its life and on the first day of every month thereafter.

The problem is to find the number of pairs of rabbits in the enclosure on the first day of the following January after the births have taken place on that day.

It will be helpful to make a chart to keep count of the pairs of rabbits. Let A denote an adult pair of rabbits and let B denote a "baby pair" of rabbits. Thus, on January first, we have on ly an A; on February first we have that A and a B; and on March first, we have the original A, a new B, and the former B, which has become an A:

Number of Number of Date Pairs A's B's

January I 0

February l

March l 2

To continue the chart conveniently, we condense our notation as follows. To get the next line of symbols, in any line we replace each A by A B and each B by A. Thus, we have the representation shown in the table at the top of the next page.

2

Rabbits, Fibonacci Numbers, and Lucas Numbers · 3

Number of Number of Date Pairs A's B's

March 1 ABA 2

April l ABAAB 3 2

May l ABAABABA 5 3

June 1 ABAABABAABAAB 8 5

We now see that the number of A's on July 1 will be the sum of the number of A's on June l and the number of B's born on that day (which become A's on July I). Tt~~ number of B's on _ _July 1 is the same as the number of A's on June l. We complete the table for the year:

I

2

3

4

5

6

7

8

9

10

ll

12

13

Month

January

After births on .first of

February

March

April

May

June

August

September

October

November

December

January

Number of A's

1

1

2

3

5

8

13

21

34

55

89

144

233

Number of B's

0

1

1

2

3

5

8

13

21

34

55

89

144

Total number of pairs

I

2

3

5

8

13

21

34

55

89

144

233

377

Thus, we see that under the conditions of the pro0lem, the number of pairs of rabbits in the enclosure one vear later would be 377.

We can draw some condusions by studying the table. It is clear that the number of A's on the following February l is 377. Of these, 376 were originally B's, descendants of the original A. Therefore, if we add all the numbers in the column headed "Number of B's," we have

S = 0 + 1 + I + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144 = 376.

From this, we observe that the sum of the first 12 entries in the column headed "Number of A's" is one less than 377, which would be the 14th entry in that column. This is a specific instance of a general result which we shall establish later in this section.

4 · Fibonacci and Lucas Numbers

Further examination of the table on page 3 reveals that each entry in the columns of numbers may be found in accordance with a pattern. For example, the entries in each line after the second may be found as the sum of the two preceding entries in that column. Those in line 3 are:

2 = I+ I 1=0+1 3=1+2

Those in line 4 are:

3=1+2 2 = 1 + 1 5 = 2 + 3

Can we describe this pattern by some kind of formula? Yes, as we shall now show.

In general, ordered sets of numbers such as those in the columns of the table on page 3 are called sequences. A sequence may be finite or infinite. An infinite sequence may be designated by symbols such as

where the subscripts indicate the order of the terms, with n a positive integer. An example of a sequence is the arithmetic progression

. . . '

2, 5, 8, ... ' 2 + (n - 1)3,

where a formula for the nth term is

Un = 2 + (n - 1)3.

Another way to specify this sequence would be to state the first term,

u 1 = 2,

and the formula

lin = Un-1 + 3, n > 1.

Such a definition is said to be a recursive definition, and the formula is called a recursion formula or a recurrence formula. (The words "recursive," "recursion," and "recurrence" all come from a Latin verb meaning "to run back.")

We can use an extension of this idea to specify the sequences in the columns of the table on page 3. For example, to specify the sequence in the column headed "Number of A's," we state the first two terms,

Ut = l,

and the recursive, or recurrence, formula

(R) 11 > 2.


This gives the sequence

1, I, 2, 3, 5, 8, 13, .. .

as we wished. For the column headed "Number of B's," we have u 1 0, u 2 = 1, and the same recurrence formula, yielding the sequence

0, I, 1, 2, 3, 5, 8, 13, .. ..

For the column headed "Total number of pairs," we have u 1 - 1, u2 = 2, and the sequence

1' 2, 3, 5, 8, 13, ....

Because of its source in Fibonacci's rabbit problem, the sequence

1' 1' 2, 3, 5, 8, 13, ..•

is called the Fibonacci sequence, and its terms are called Fibonacci numbers. We shall denote the nth Fibonacci number by Fn; thus,

F 1 = I, F2 = I, F3 = 2, F4 = 3, F5 = 5, F6 = 8, ....

Moreover, we may write these alternative forms:

or

or

F 1 - F 2 = I,

F 1 - F 2 = 1,

F 1 - F 2 = l ,

Fn+l - Fn + Fn-h

Fn+2 = Fn+ l + Fn ,

ll > 2,

n > 1,

n > 1.

We can now give a more formal d iscussion of the Fibonacci rabbit prob-lem. For all positive integraln, we define for the first day of the nth month:

A 11 = number of A's (adult pairs of rabbits) B,1 = number of B's (baby pairs of rabbits) T11 = total number of pairs of rabbits = An + B11

Only the A's on the first day of the nth month will produce B's on the first day of the (n + l)st month. Thus,

n > I.

In making up the table on page 3, we observed that the number of A's on the first day of the (n + 2)nd month is the sum of the number of A's on the first day of the (n + l)st month and the number of B's born on that day. Thus,

and since Bn+ 1 - A 11 , we have

An+ 2 = An+ l + A n, 11 > I.


We also observe from the table that A 1 = 1 and A 2 = 1. Thus, the sequence

is the Fibonacci sequence, and

11 > I.

Since Bn+t - An for n > I, we have

Bn = An-1 = Fn-1 for ll > 2.

If we now let n = I in this last formula, we have

B1 = Fo.

If we let n = I in the formula Fn+l = Fn + F,,_ 1, we have

F2 = F1 + F 0

or F 0 = F2- F 1 = I - I = 0,

which checks with B 1 = 0 in the table. Thus, we have now defined F,1 for 1l = 0.

Finally, the total number of pairs-on the first day of the nth month is

We can now establish the following result, already suggested by the specific instance shown at the bottom of page 3:

Symbolically:

The sum of the first n Fibonacci numbers is one less than the (n + 2)nd Fibonacci number.

1l > I.

We remember that Fn+ 2 = An+ 2 and that An+ 2 is the number of A's (adult pairs of rabbits) in the enclosure on the first day of the (n + 2)nd month.

Originally, we had only one A . Where did the extra A 's come from? Each of the extra A's was first a B.

How many more A's do we now have? The number of extra A's is

An+2 - l.

Now, one month after being born, each B became an A . If we add the number of B's from the first day of the first month to the first day of the (n + l)st month, the sum is the number of A 's other than the original pair that we have on the first day of the (n + 2)nd month. Thus,

BI + B2 + B3 + · · · + Bn+ l = An+2- 1.


But, remembering that B1 = 0, Bn = Fn-b and An+ 2 = F11 +2 , we have

Ft + Fz + · · · + F" = Fn+2- 1,

as we wished to show. This formula is an example of a Fibonacci number identity. We shall prove

this identity again later in three different ways (Section 10).

Many different sequences may be specified by using formula (R) on page 4 and choosing different numbers for the first two terms. For example, if we take llt = I and u2 = 3, we have

1, 3, 4, 7, 11, 18, 29, 47, ... '

which we shall call the Lucas sequence, in honor of the nineteenth-century French mathematician E. Lucas. Lucas did much work in recurrent sequences and gave the Fibonacci sequence its name. The terms of the Lucas sequence are called Lucas numbers, and we shall denote the nth Lucas number by Ln. The Lucas numbers are closely related to the Fibonacci numbers, as we shall show in this booklet.

In general, if we take the first two terms of a sequence defined by (R) as arbitrary integers p and q, that is, lit = p and u 2 = q, then we have

p, q, p + q, p + 2q, 2p + 3q, 3p + 5q, . .. ' which is called a generalized Fibonacci sequence. We shall denote the nth term of this sequence by Hn. It may be shown by mathematical induction (see Exercise 17, Section 10) that this generalized Fibonacci sequence Is related to the Fibonacci sequence by the formula

n > 0, F0 = 0,

or, expressed in terms of the starting values, p and q,

Hn+2 = qFn+t + pFn.

EXERCISES

1. Compute the first 20 Fibonacci numbers.

2. Compute the first 20 Lucas numbers.

3. Study the results of Exercises 1 and 2, looking for any possible relationships or number patterns.

4. If Ht = 1, H2 = 4, and H,.+2 = Hn+t + Hn, n > 1, compute the first 20 terms of this generalized Fibonacci sequence.


5. Verify that:

a. Ls = FG + F4 c. L1 + Lg = 5F s

b. F9 = F~ + F~ d. H2o = (4)Ft9 + (l)Fts in Exercise 4 .

. 6. Verify that:

a. Fs = L4F4

FIO. . b. Fs IS an mteger.

F12. . c. F4 IS an mteger.

7. Verify that Ft + Fz + F3 + F4 + Fs + FG = Fs - 1.

8. Verify that Ft + Fz +Fa+ F4 + Fs + FG + F1 + Fs + Fg + F1o = 11F, . . · . 9. Show thai:

a. When Ft:~ is divided by Fs, the remainder is F3.

b. When F 15 is divided by F s, the remainder is F 1·

3 • The Golden Section and the

Fibonacci Quadratic Equation

Suppose that we are given a line segment AB, and that we are to find a point C on it (between A and B) such that the length of the greater part is the mean proportional between the length of the whole segment and the length of the lesser part; that is, in Figure I,

AB AC -=- · AC CB

A c B

where AB ¥= 0, AC ¥= 0, and CB ¥= 0. Figure I

We first find a positive numerical value for the ratio AB. For convenience, let AC

Then

AB X= AC (x > 0).

AB AC + CB CB I 1 I x = AC = AC - I + AC = I + AC = 1 + AB = 1 +;.

From I

x=1+x

CB AC

we obtain, by multiplying both members of the equation by x,

x 2 = x + I, or

(F) x 2 - x- 1 = 0.

The roots of this quadratic equation are (as you can verify, see Exercise 1, page 13)

a = 1 + -vs

2 and

I - VS {J = •

2

(a is the Greek letter alpha, and {J is the Greek letter beta.) You can verify 9


by computation that a > 0 and {3 < 0; a ..:.. 1.618 and {3 ..:.. -.618 (Exercise 2). Thus, we take the positive root, a, as the value of the desir.ed ratio:

AB AC=

I+ VS 2

D We can now use this numerical value to devise

a method for locating C on AB. Draw BD perpendicular to AB at B, but half its length. Draw AD. Make DE the same length as BD, and AC the same length as AE. Then ~

AB = 2BD, ED= BD A c B

and, by the Pythagorean theorem, Figure 2

AD= VS BD;

hence:

AC = AE = AD- ED = (VS- I)BD

AB 2BD 2(VS + I) VS + 1 - - -

AC (VS _ t)BD 5 - l 2

This computation verifies that the construction does indeed locate C on A B such that

AB AC

- 1 + V5 2

Since a is a root of equation (F) on page 9, we have

a 2 = a+ 1.

Multiplying both members of this equation by an (11 can be any integer) yields

(A)

If we let lin = an, n > I, then u1 = a and u2 = a 2 , and we have the sequence

... '

which satisfies the recursive formula (R) on page 4. Similarly, we have

(B)

and the sequence

2 I 3 z {3 , {3 = {3 + , {3 = {3 + {3 , •..

also satisfies (R).

The Golden Section and the Fibonacci Quadratic Equation · 11

You can easily verify (Exercise 3) that

a + 13 = 1 and a - 13 = VS. If we now subtract the members of equation (B) from the members of equation (A) and divide each member of the resulting equation by a- 13 (=VS ~ 0), we find

n+2 Rn+2 n+ 1 ,qn+ 1 n Rn a -p a -p a -p ------~--- +------a-/3 a-/3 a-/3

an - 13n If we now let Un = , n > 1, then we have

a-/3

and

a-/3 ZII = a - 13 = I,

2 ,q2 a - P

u2 = a- 13 - a-/3 (v5)(1) vs - 1.

(a - /3)(a + /3) -

Thus, this sequence u,~ is precisely the Fibonacci sequence defined in Section 2, and so

n {3n a -Fn = R ' a-p

n = 1, 2, 3, ....

This is called the Binet form for the Fibonacci numbers after the French mathematician Jacques-Phillipe-Marie Binet (1786-1856).

Because of the relationship of the roots, a and /3, of the equation (F),

x 2 - x- 1 = 0,

to the Fibonacci numbers, we shall call equation (F) the Fibonacci quadratic equation.

We shall call the positive root of (F),

a= 1 + vs

2

the Golden Section. [This is often represented by cJ> (Greek letter phi) or by some other symbol, but we shall continue to use a in this booklet.]

The point C in Figures 1 and 2, dividing AB such that

AB -=a= AC

1 + vs 2

is said to divide AB in the Golden Section.


Suppose that the rectangle ABCD in Figure 3 is such that if the square AEF D is removed from the rectangle, th~ lengths of the sides of the _remaining rectangle, BCFE, have the same ratio as the lengths of the sides of the rectangle ABCD. A x E Y B That is,

BC AB

EB DA

Then if DA = AE = BC = x and EB = y, we have

x x+y or

y X

X

y

y I+-.

X

Multiplying both members of

X by-, we find

.y

or

0r =~+I. Gr- ~-I= o.

X X

D F c Figure 3

which is in the form of equation (F), the variable now being 0) . Since

x andy are positive, we seek the positive value of ~. Thus, y

x 1 + ,rs -=a = --- · y 2

That is, the ratio of the length to the width for rectangle BCFE (and also for rectangle A BCD) is the number a, the Golden Section. Such a rectangle is called a Golden Rectangle.

The proportions of the Golden Rectangle appear often throughout dassical Greek art and architecture. As the German psychologists Gusta \ Theodor Fechner (1801-1887) and Wilhelm Max Wundt (1832-1920) ha\c shown in a series of psychological experiments, most people do unconsciously favor "golden dimensions" when selecting pictures, cards, mirrors, wrapped parcels, and other rectangular objects. For some reason not fully kno\vn by either artists or psychologists, the Golden Rectangle holds great aesthetic appeal.

The Golden Section and the Fibonacci Quadratic Equation · 13

EXERCISES

1. Solve the Fibonacci quadrat ic equation,

x 2 - x- 1 = 0,

I+Vs 1 -VS and verify that the roots are a = and {3 = as stated m

2 2

the text.

2. Verify that a > 0 and {3 < 0, using v5 ...:.... 2.236. (...:.... means "is approximately equal to.")

3. Verify that:

a. a+ {3 = 1

1 4. Verify that a = 1 + - ·

a

b. a- {3 = VS

5. Using a 2 = a + 1, verify that:

a. a 3 = 2a + 1 b. a 4 = 3a + 2

3 1 6. Verify that L3 = a - a 3

4 -4 a -a 7. Verify that F4 = VS

c. a{3 = -1

c. a 5 = Sa + 3

4 • Some Geometry Related

to the Golden Section

We shall now consider several geometric problems and their solutions.

PROBlEM 1 *

Suppose that we wish to remove from a rectangle, ABCD, three right triangles of equal area, 6.PAQ, 6.QBC, and 6.CDP, as shown in Figure 4. How shall we locate points P and Q?

X+Y

Figure 4

Solution. Let

A Q = x, QB = y, AP = w, and PD = z.

Then, since the areas of triangles PA Q, QBC, and CDP are to be equal, we have

!xw = !y(w + z) = !z(x + y),

or xw = yw + yz = xz + yz.

* J. A. H. Hunter, "Triangle Inscribed in a Rectangle," Tlze Fibo11acci Quarterly, Vol. 1, No. 3 (October, 1963), page 66.

14

Some Geometry Related to the Golden Section · I 5

From

yw + yz = xz + yz

we have

yw = xz,

or W X - = - · z y

Also, from

xw = y(w + z)

we have

X w+z_l+~ = 1+..!_· w w w y

z S. W X h mce - = - ,we ave

z y X I

1 +-' X - = y

y or

(x)2 x y - y- I = 0.

Again choosing the positive root, we have

But also

(A)

X - = a = 1 + -v:s

2 y

W X -=- =a z y '

and so the points P and Q must divide sides AD and AB, respectively, m the Golden Section. Thus:

AP AQ - = - =a PD QB

D c


PROBLEM 2*

If the rectangle ABCD in Problem I had been a Golden Rectangle, then the additional condition (B)

x+y w+z=a

would be imposed. Show that in this case D.PQC would be an isosceles right triangle with the right angle at vertex Q.

Solution. From (A) on page 15 we have

X= ay

and so, from (B) above, we have

and w = az,

X + y ay + y (a + I )y a= = = '

w + z az + z (a + I )z

But we had w = az, and so

w = y; that is,

Since we were given originally in Problem 1 that

!xw = !y(w + z),

the fact that w = y implies that

X= W + z, that is,

or y = az.

Therefore, right triangles P A Q and QBC are congruent. Thus,

and

Moreover,

Since

and

m0 LA QP = m0 LBCQ,

we have

m0 LPQC = 90.

Therefore, since in D.PQC we now have PQ r--J QC and 111° LPQC = 90, D.PQC is an isosceles right triangle, as pictured in Figure 5.

LA QP r-v LBCQ.

A X

z

D x+y

Figure 5

B

* H. E. Huntley, "Fibonacci Geometry," The Fibonacci Quarterly, Vol. 2, No. 2 (April, 1964), page 104.

Some Geometry Related to the Golden Section · 17

PROBLEM 3

Do two triangles exist which have measures of five of their six parts (three angles and three sides) equal and yet are not congruent? Your first impulsive answer may be a resounding No! However, we propose to show that this is indeed possible.

Solution. Clearly, if among the five parts are three sides, then the triangles must be congruent. The only possibility then is to have the three angles of one triangle congruent to the three angles of the other triangle (thus, the triangles are similar) and two sides of one triangle congruent to two sides of the other. (Notice that it is not specified that these sides be corresponding sides.) One example of such a pair of triangles (in this case with integral sides) is shown in Figure 6, where

27 18 12 ~ ~ 18 12 8

18 27 -=-=-·

Figure 6

Let us see how to find pairs of triangles having just five parts congruent. First, we know that the triangles must be similar ; that is, the measures of the sides must be related as shown in Figure 7, where r is the ratio of similarity:

~~ c rc

The additional conditions are

b = ra

Figure 7

and

Thus, the measures of the sides of the two triangles will be

and ra,

asshowninFi~ ~

Figure 8

If a = 8 and r = ~' you will find the measures of the sides of the triangles shown In Figure 6. Try other values to find other triangles having this property.


PROBLEM 4

If a = 4 and r = 2 in Problem 3, then the. measures of the sides of the triangles would be

4, 8, 16 and 8, 16, 32.

Can there be triangles with sides of these measures? No, because

4 + 8 < 16 and 8 + 16 < 32.

What restrictions must be placed on the values of r, assuming for the present that r > l?

Solution. If a, ra, and r 2a are to be the measures of the sides of a triangle, then the sum of each two must be greater than the third. Thus, we have three inequalities:

(i) a + ra > r 2a

(ii) ra + r 2a > a

(iii) r 2a + a > ra

or, since a > 0,

or, since a > 0,

or, since a > 0,

I+ r > r 2

r + r 2 > I

r 2 + 1 > r

We are looking for the solution set of these three inequalities. On the assumption that r > I, we have

r 2 > r > I,

and inequalities (ii) and (iii) hold. Thus, we need to consider inequality (i), which we shall write as

or r 2 - r- 1 < 0.

Recall that

x 2 - x- 1 = 0

is the Fibonacci quadratic equation with roots

a= l + V5

2

Therefore, we can write

For the second factor, we have

r-

and this is positive for r > I.

and l- V5

{3= . 2


Therefore, to have r 2 - r - 1 < 0

we must have

I+ VS 0 r- 2 < ' or

I+ v:S r < 2

that is, r < a, and so if r > I, we must have

l<r<a

in order to have a pair of triangles with just five parts congruent.

PROBLEM 5

Can a pair of right triangles have just five parts congruent?

Solution. Suppose that r > 1. Then r 2a is the measure of the longest side. If the triangles are to be right triangles, we have by the Pythagorean theorem that

Thus,

or, since a ~ 0,

Therefore,

,.4 - ,.z - 1 = 0.

,.z =a '

and so the positive value of r in this case is y~. How can we construct such a pair of right triangles? Recall that in a

right triangle, the altitude from the vertex of the right angle to the hypotenuse separates the given triangle into two triangles that are similar to each other and also to the given triangle. That is, in Figure 9, where angle C is a right angle,

D.ACD"'"' D.CBD"'"' D.ABC.

Notice that D.ACD and D.CBD are similar and have one side, CD, in common. If we had AD congruent to BC, we would have two right triangles that have just five parts congruent. These are shown as

c

~ A D B

Figure 9

D.A'C' D'

in Figure IO on the next page.

and D.C'B' D'


If we use the letters shown in Figure 10 to represent the measures of the sides, we have

x+y z x =- = -·

X W y

From

x x+y - = --"--y X

we have

Figure 10

l = 0, and so X -=a. y

Thus, the point D' must divide A' B' in the Golden Section. Let us find the ratio of similarity for 6A'C' D' and 6C' B' D', that is, the

value of

Since x = ay, we have

Therefore, since r > 0,

W X Z r= - =-= - ·

y W X

w ay - = _, or y w

r = w = v~, y

as predicted by our computation on page 19.

Some of the possible shapes for triangles having just five parts congruent are shown in Figure 11. Those sketched are right and oblique triangles. In order for such triangles to have only acute angles, we must have

1 < r < ~' ,ra .:_ 1.27.

r = Va = 1.27

aa2 =a( a:+ I) = 2.618a

Figure II

Triangles with sides of measures a, ar, ar! for 1 < r < a , a = 1.618

8 r = 5 = 1.6


r = v~ = L27 Triangles with sides of measures ar, ar2, ar, for ( < r <a, a = 1.618

aa3 = a(2a + 1) = 4.236a

Figure II (continued)

Up to now we have restricted ourselves to r > l. If r > I is the ratio of similarity of the larger triangle to the smaller triangle, then

I I r =-r

is the ratio of similarity of the smaller triangle to the larger triangle. Thus, in general, we may have

I -<r< a

that is, approximately

.6I8 < r < l

as well as I < r <a,

or 1 < r < 1.618.

The triangles pictured in Figure 12 have ratios that are the reciprocals of those for the triangles in Figure ll. Notice that each pair of triangles in Fi'gure II is similar to the pair in Figure I2 that has the reciprocal ratio (compare Exercise 6 on page 25). (If r were 1, then the triangles would be congruent equilateral triangles.)

\

5

"-----v--J -v= 8 = .625

a -:;= a(2- a)= .38a a-

Triangles with sides of measures 1 1

a, ar, ar2 for- < r < 1,- = .618 a a

Figure 12

'---v-J a - = a(2a - 3) = .24a al

Triangles with sides of measures 1 . 1

ar, ar2, ar3 for- < r < I, -= .618 a a

22 · ' Fibonacci and Lucas Numbers

In Section 3 we defined a Golden Rectangle. We shall now define a Golden Triangle. In Figure 10, the ratio of the area of L:lA' B'C' to the area of L:lA'C' D' can be found as follows:

Area L:lA' B'C' = !w(x + y)

Area L:lA'C' D~ = !wx

Area D.A'B'C' x + y ay + y a+ l - - -Area D.A'C' D' X ay

2 a =-=a

a

A triangle that has this property is called a Golden Triangle; that is, a Golden Triangle is one such that when a triangle similar to it is removed from it, the ratio of the area of the Golden Triangle to the area of the remaining triangle is a. That is, in Figure lO when D.C' B' D' is removed from D. A' B'C' '

Area L:lA'B'C' =a

Area 6.A'C' D' .

We also note that

Area L:lA'C' D' lwx x ----=~-=-=a

Area6.C'B'D' !wy y '

and

Area L:lA'B'C' Area L:lA' B'C' Area L:lA'C' D' 2 - · =a·a=a

Area L:lC'B' D' Area L:lA'C' D' Area L:lC'B' D' .

PROBlEM 6

Show that an isosceles triangle with vertex angle measunng 36° IS a Golden Triangle.

Solution. The base angles measure 72° each. If one of these base angles is bisected (see Figure 13), two isosceles triangles are formed. One triangle, L:lCDB, is similar to the given triangle, L:lABC, while the other, L:lACD, is not.

In L:lABC and L:lCDB,

x+y x ---=-'

X y

and again we have

(~)' X ---

y

and the positive result is X -=a. y

c

A

Figure 13

1=0


To find the areas of b.ABC and b.ADC, draw altitude CE. Then:

CE = ~x2 ~2 (CE r£ 0)

Area b.ABC = !(x + y)(C£)

Area b.ADC = !x(C£)

Area b.ABC x+y -

Area b.ADC =a.

X

Thus, b.ABC is a Golden Triangle. We also note that

Area b.ADC x =-=a

Area b.CDB y

and Area b.ABC x+y x+y x 2

-Area b.CDB

- ·-=a. y X y

Since the central angle of a regular decagon is 36° (see Figure 14), we know from Problem 6 that the ratio of the radius r to the measure s of the side of an inscribed decagon is a .

Figure 14 Figure 15

Also, in a regular inscribed pentagon, the angle between two adjacent diagonals at one vertex is 36° (see Figure 15), and so the ratio of the measure d of a diagonal to the measures of a side is also a .

AB A' B' Notice that in Figure 13, - = a and that in Figure 10, -- = a.

BC B'C' Thus, in each case the ratio of the measure of the longest side to the measure of the shortest side is a.

The Golden Triangle appears on pages 61-62 of Tobias Dantzig's The Bequest of the Greeks (New York: Charles Scribner's Sons, 1955) and also on page 42 ofN. N. Vorobyov's The Fibonacci Numbers (Boston: D. C. Heath and Company, 1963). Also, see the article, "Golden Triangles, Rectangles,


and Cuboids," by Marjprie Bicknell and Verner E. Hoggatt, Jr., m The Fibonacci Quarterly, Vol. 7, No. I (February, 1969), pages 73-9 I.

PROBLEM 7

Inscribe a square in a semicircle.

Solution. Figure 16 shows the completed construction where AM N B is a square. The construction makes use of the fact that right triangles

OAM, ODC, OEF, and OBN

are similar.

Now consider Figure 17, in which AF and FB have been drawn, forming similar right triangles ABF, AFE, and FBE. Thus,

t + s s --= _, s t

from which we obtain

l = 0,

11 and so the positive result is ' s

-=a. t

Thus, pointE divides DB in the Golden Section.

A D 0 E B

Figure 16

Figure 17

Two articles by Marvin Holt, which give further excellent material on the Golden Section and geometry, are "Mystery Puzzler and Phi .. in The Fibonacci Quarterly, Vol. 3, No. 2 (April, 1965), pages 135-138, and "The Golden Section" in the Pentagon, Spring, 1964, pages 80-104.

The Golden Cuboid is discussed in an article of that title by H. Huntley in The Fibonacci Quarterly, Vol. 2, No.3 (October, 1964), page 184.

Another interesting reference is Patterns in Space by Colonel R. S. Beard (available from Brother Alfred Brousseau. St. Mary's College, California 94575). This book describes many appearances of the Golden Section m variations of the regular solids.


EXERCISES

1. Show that there can be no triangle having three distinct Fibonacci numbers as measures of its sides.

2. Show that a pair of triangles which have the measures of five parts equal, but which are not congruent, cannot be isosceles.

3. Show that

a. if a < b, then y = (x - a)(x - b) is negative for a < x < b and nonnegative otherwise ;

b. if a = b, then y = (x - a) 2 ~ 0 for all x.

4. Using the results of Exercise 3a, show that, in general,

a. inequality (i) on page 18 holds when

r 2 - r - 1 < 0, that is, for {3 < r < a.

b. inequality (ii) on page 18 holds when

r2 + r - 1 > 0, that is, for r < -a or r > -{3.

c. inequalities (i) and (ii) both hold when

-{3 < r <a, or

since a/3 = - 1 (Exercise 3c, page 13).

1 - < r <a, a

5. Using the result of Exercise 3b, show that inequality (iii) on page 18 holds for r > 0, that-is, that

r 2 + 1 ~ 2r > r for r > 0.

6. The measures of the sides of one triangle are a, ar, and ar2 and those of a a a

second triangle are a, -, and ---:) . By suitably pairing the sides, show that the r r-

triangles are similar and find the ratio of similarity.

7. In Figun· 16 extend CF to meet BN in point G. Show that rectangle DCGB is a Golden Rectangle.

I . h AF 8. n FJgure 17 s ow that FB = a.

9. In Figure 17 show that: ?

s- + 2st a. s2 + t2 = a

?

r- - 2st R b . = tJ

s2 + t2 c.

2 ? s + 4sr- r- vs

Hint: ·Recall that a 2 = a+ 1, a+ {3 = 1, a - {3 = VS.

5 • Some Fibonacci Algebra

I+VS 1-y':S Recall from Section 3 that a = 2 and {1 = 2 are the roots of

(F) x 2 - x- I = 0 '

and so a 2 = a + 1 and {1 2 = {J + l. Also, a + {J = 1 and a - {J = y'S. Moreover,

(A)

and

(B)

and by using these equations, we found that the Fibonacci numbers can be expressed in the so-called Binet form:

(C) n Rn n Rn a-tJ a-tJ

Fn = = ' a-{J y'5

1l = 1,2,3, ...

Now suppose that we add the members of equation (B) to the members of equation (A), giving

(an+2 + {jn+2) = (an+ l + 13n+ I) + (an+ {Jn).

If we let Un = an + {jn, then we have

and

u1 = a+ {1 = I , u 2 = a 2 + {3 2 = a + I + {3 + l = (a + {1) + 2 = I + 2 = 3.

Thus, this sequence un is the sequence of Lucas numbers defined in Section 2, and so we have a Binet fo rm for the Lucas numbers:

(D) n = I, 2, 3, ...

26

Some Fibonacci Algebra · 27

Now look at the following comparison of the Fibonacci numbers and the Lucas numbers:

Notice that

F1 F2 Fa F4 Fs F6 F1 Fs Fg F1o 1 1 2 3 5 8 13 21 34 55

1 3 4 7 11 18 29 47 76 123

F1 + Fa = L2, F2 + F 4 = La, and so on.

It can be proved (Exercise 2, page 29) that in general

from which, since Fn+I = Fn + Fn-1> it follows that

(E)

You can verify this latter statement for specific examples; that is, you can show that L6 = F6 + 2F5 , and so on.

We now have Fn and Ln expressed in terms of an and {3n. We can also find an and {3n in terms of Fn and Ln ~ If we note that a - {3 = v's, then, from the Binet forms, we have

Adding, we find

or

subtracting, we find

v5 Fn = an - {3n,

Ln = an+ {3n.

n Ln + yS Fn a =

2

Recall that in Section 2 we had occasion to define F 0 as F 2 - F 1 (page 6). Similarly, we can define L 0 as L 2 - L 1 = 3 - 1 = 2. (Notice that this agrees with the definition by the Binet form, since a 0 + {3° = 1 + 1 = 2.) Since L 1 = a+ {3 = I, we can now write expression (E) for Ln (above) as

(E')

28 • Fibonacci and Lucas Numbers

The method of defining F 0 and L 0 suggests that we can also define F _ 1,

L_h and so on, by applying the formulas

Fn-1 = Fn+l - Fn and

Ln- 1 = Ln+l - Ln

repeatedly. Thus, we have :

F_ 4 F_ 3 F_ 2 F_ t Fo Ft F2 F3 F4 -3 2 -1 1 0 1 1 2 3

7 -4 3 -1 2 1 3 4 7 L_4 L _ 3 L _ 2 L _ l Lo Lt L2 L3 L4

We can derive a formula for F - rB n > 0, by assuming that the Binet form also holds for negative values of the exponents (compare derivation on pages 10- Il):

a-{3

Since a/3 = - I (Exercise 3c, page 13), we have

.!_ = - 13 and 1 -a 13

-a.

and so F - n = ( -ty+ 1 Fw

Similarly, you can show (Exercise 3, page 29) that for n > 0,

L_n = ( - IrL,!"

Now suppose that we compute the first 14 successive ratios Fn + 1 and Fn

Ln+ 1 • The values of the successive ratios as shown at the top of the next L n

page suggest that in both cases the value of the ratio becomes closer and closer to a as we take larger and larger values of n. However, we shall not undertake to prove this here. We can also observe that the first Fibonacc i ratio is less than a, the second is greater than a, and so on, while the first Lucas ratio is greater than a, the second is less than a, and so on. Moreover,

and so on.

Some Fibonacci Algebra ·

Fn+ 1 Ln+J Fn L"

1 1.0000 f = 3.0000 1 -

t= 2.0000 4 • 3= 1.3333 !l-2 - 1.5000 i= 1.7500 .§_~ 3- 1.6667 1 1 • 7= 1.5714 8 - 1.6000 18 1.6363 -s- IT -

13 -s- 1.6250 29 • n= 1.6111 21 • n= 1.6154 47 • 29 = 1.6207

!t ..:. 1.6190 76 • 47 = 1.6170

55 ~ 1.6176 123 ~ 1.6184 34 - 76-89 ss= 1.6182 199

123 = 1.6179 144 ~ 89- 1.6180 322 •

199 = 1.6181 233 ~ 1.6181 5 21 • 1.6180 144 - 322 = 377 ~ 1.6180 843 • 1.6180 233 - 521 = 610 ~ 1.6180 1364 ..:. 1.6180 377 - 843

a ..:. 1.61803398875 ...

EXERCISES

Using the Binet form:

1. Show that Fzn = F,L,., n > 1.

2. Show that Ln = F,_ I + F,+ 1, n > l. Hint: Use a{3 = -1.

3. Show that L_ 11 = ( - 1) 11L,. for n > 0.

4. Show that 5Fi = Lz,. - 2( -1)".

5. Show that L~ = L2,. + 2( - 1)".

6. Show that Fn+ JL, - Ln+ 1F, = 2( -1) 11 •

Using previously found results:

Sh h Fn+l Ln+l 2(-1)" 7. ow t at-----=---

F,. L" Fz"

8. Show that F _ ,. = ( -1)" + 1 F, holds also when n < 0.

9. Show that L_n = ( -l)"L,. holds also when n < 0.

10. Show that 5F~ = L~ - 4( -1)".

11. Show from L,. = Fu+l + Fn-1 that Ln = Fn+2 - Fn- 2·

29

6 • Shortcuts to Large F,, and L"

Since we shall be dealing extensively with inequalities in this section, we shall recall the following propositions.

For real numbers a, b, c, and d, we have:

a b (a) If a < band c > 0, then ac < be and- < - .

c c a b

(b) If a < band c < 0, then ac > be and- > -. c c

(c) If a < b, then a + c < b + c and a - c < b - c. a c b d

(d) If 0 < - < -d, then- > -,and conversely. b a c

a c c e a e (e) If- < - and - < - then - < -.

b d d J' b 1 (f) If lbl < I, then lbln < 1 for n = I, 2, 3, . . ..

(g) If a = b + c and c > 0, then b < a.

We shall also use an idea suggested by the following problem. Suppose that we have s pounds of sugar (s not necessarily an integer), and we ask how many one-pound sacks of sugar can be made from this quantity. Then we are interested in finding the greatest integer not greater than s. We denote this integer by [sJ. Thus,

[7.2] = 7 and [7.9] = 7.

Similarly, we have [-5.4] = -6 and (!] = 0.

We are now able to devise methods for finding values of Fn and L, (or at least estimates of them) without doing all the additions from the beginning. The first theorem that we shall prove is the following.

30

Shortcuts to Large F" and L" · 31

THEOREM I

for n = 1,2,3, . ...

Proof We have the Binet form n Qn a - JJ

F = ' n v5 n= 1,2,3, ... ,

where a= 1 +2 VS..:. 1 + ~-236 · 1.618 and {J = 1 - 2 VS · -.618.

We can write

Fn ~ ~ - ~ ~ (~; + n -G + ~). Since 0 < I.BI < I, we find by (f) on page 30 that

0 < j{Jjn < I.

Since 1 < V,:' we find by (e) that

0 < j{Jjn < VS. 2

Since V5 > 0, we find by (a) that

(A) 0 < I fJ I~ < ! . v~s 2

Then by (c) we find (by adding !) that

! < ! + j{Jjn < 1. 2 2 v'5

Although {:3 < 0, we have j{Jjn = fJn when n is even, and so

(B) ·r . I 1 ,sn I 1 n 1s even, 2 < 2 + V5 < .

On the other hand, we have j(JI" = -fJn when n is odd, and so from (A),

l {Jn -2 < V5 < 0.

Then by (c) we find (by adding !) that 1 {Jn 1

(C) if n is odd, 0 < - + - < - · 2 V5 2

Thus, from (B) and (C) we have, in general,

1 {Jn 0<2+ 0<1.


But we saw on page 31 that Fn can be expressed as

( n 1) (1 {3n) Fn = ~ + 2 - 2 + VS ·,

and so, since we have shown (page 31) that (~ + ~) is positive and less

than 1, we have

or

[ an 1] vs+2 for n = I, 2, 3, ... ,

and the theorem is proved.

Similarly, the following theorem can be proved, but we shall not give the proof here.

THEOREM II

for n = 2, 3, 4, ....

It is shown in The Fibonacci Numbers by N. N. Vorobyov* that Fn is the n

integer nearest to ~ , that is,

Using this result, Fn can be computed if logarithms to a sufficient number of places are used. A similar development can be given to show that

PROBlEM 1

16

Find F 16 ..:. a _ ·

vs Solution.

ILn- ani < !.

16 a .

log v5 = 16 log a - log 2.236

Since we are going to multiply log a by 16, we should find log a to more

* For one translation see the reference on page 23.

Shortcuts /o Large Fn and Ln · 33

decimal places than we plan to use for the remainder of the computation .

1 + -v:s a= -2--

_:_ 1 + 2223607 _:_ 1.6180

From a five-place table of common, or base 10, logarithms we find

log a -=- 0.20898.

Thus, we have:

16

log ~S · 16(0.20898) - 0.3494 _

_:_ 3.3437 - 0.3494 _:_ 2.9943 16 :rs _:_ 987.0

Therefore, F 16 = 987. You can check this answer in the list on page 83.

For larger indices we can find only the first three digits accurately if we are using four-place logarithms.

PROBLEM 2

Estimate F3 o.

Solution. 30

log~ ...:. 30(0.20898) - 0.3494 ...:. 5.9200 V5

30

~ _:_ 831800 V5

Therefore, F30 • 832,000. You can find the exact value in the list on page 83.

PROBLEM 3

Find L 14 ...:. a 14•

Solution. log a 14 -=- 14(0.20898) ...:. 2.9257

a 14 ...:. 842.8

Therefore: L 14 = 843. You can check this value in the list on page 83.


We shall now develop methods for finding Fn+ 1 from Fn and Ln+ 1 from Ln even ifwe do not know the value ofn. To achieve this, we shall prove the following new theorem.

THEOREM Ill '· Fn+ 1 - (aFn + !J, n = 2, 3, 4, .. . .

Proof Since

we have

n {3n a -F = '

n VS

an+I _ (a{3){3n-l _ {3n+1 + {3n+1 - v:s

Since a{3 = - I, we have

But

ll'+ I~ fl+2 ~ 1-1+4 ~ s-2vs ~ vs(vs2-1) ~ -·/S{J.

Therefore,

and

(A)

Now, since lf31 < .62, we have lf31 2 < !, and so

lf31n < ! for 11 > 2.

Also, lf3 111 - lf31 71 , and so

lf37ll < ! or _ 1. < {3 n < 1. 2 2•

Therefore, by (b) on page 30, we have

or

Shortcuts to Large F,. and L" · 35

and by (c) we have

0 < ~- {3 11 < I.

Now since ~ - {3 11 > 0, we find from equation (A) and (g) on page 30 that

Fn+l < aF11 + ~· Moreover, since ! - {3 11 < 1, we have

Fn+I + (!- f3 11 ) < Fn+I + 1.

Applying these to equation (A), we have

Fn+I < aF11 + ~ < Fn+I + I,

or

Fn+ 1 = [aFII + ~], and the theorem is proved.

11 = 2, 3, 4, ... '

COROllARY*

_ [Fn + 1 + V~] Fn+ 1 - 2 1 n > 2.

Proof From Theorem III, we have

This corollary shows that we car compute Fn+ 1 from Fn without using either n or a.

We could prove in a similar way the corresponding theorem and corollary for Lucas numbers.

THEOREM IV

n > 4.

COROllARY*

_ [Ln + 1 + V~] Ln+l - 2 1 n > 4.

* A slightly different form of these corollaries appears as Theorem 4 in Y. E. Hoggatt, Jr., and D. A. Lind, "The Heights of the Fibonacci Polynomials and an Associated Function," The Fibonacci Quarterly, Vol. 5, No. 2 (April, 1967), page 144.


PROBLEM 4

Given that 610 is a Fibonacci number, use the Corollary to Theorem III to find the next one.

Solution. F.+t = [610 +I i Vs(610)2 ]

= [611 + ~0500]

= [611 +21364.0]

= p9~5.0] = 987

You can check this value in the list on page 83. Alternatively, you may compute as follows:

Fn+I = [610 + I i V5 (610)]

= [611 + (2i236)(610)]

= [611 + ;363.96]

= p97;.96] = 987

For larger Fibonacci numbers you may need to find V5 to more decimal places.

EXERCISES

In Exercises 1-4, use four-place tables of logarithms except for log a = 0.20898.

1. Find F 12, using the method of Problem 1.

2. Find L 12, using the method of Problem 3.

3. Find the first three digits of F34, using the method of Problem 2.

4. Find the first three digits of L:{:3.

S. Given that 1597 is a Fibonacci number, find the next one.

6. Given that 2207 is a Lucas number, find the next one.

7 • Divisibility Properties of the

Fibonacci and Lucas Numbers

Let us look at the first few Fibonacci numbers:

F1 F2 Fa F4 F5 F6 F1 Fs Fg F1o F11 F12 F1a F14 F15 .. .

1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 .. .

We observe the following:

I. Every third Fn is even; that is, Fa = 2 divides F6 = 8, F9 = 34, FI2 = 144, Fl5 = 610, ....

2. F 4 = 3 divides F8 = 21, F 12 = 144, ....

3. F5 = 5 divides F 10 = 55, F 15 = 610, ....

4. F6 = 8dividesF12 = 144, ... .

5. F7 = 13 divides F 14 = 377, ... .

These examples suggest the following theorem.

THEOREM I

Every Fibonacci number Fk divides. every Fibonacci number F11 k for

n = 1, 2, 3, ... ; or if r is divisible by s, then Fr is divisible by f 5 •

First, consider, for example, aiO _ {310

FlO= a-{3

f.. The difference of equal powers of two numbers can be factored, often in various ways. Here we may divide a 10 - {3 1 ~ by a 2 - {3 2 or by a 5 - (3 5

as shown on the following page. 37


10 ,qlO a - fJ Fto = ----'-

a-{3 2 2

= aa = ~ (a8 + a6{32 + a4{34 + a2{36 + {38)

= F2[(a8 + {38) + a2{32(a4 + {34) + a4{34]

Since a{J = -I, we have

(A)

Also, we have

(B)

10 ,q lO a - fJ

F1o = ---a-{3

a 5 - f3 5 5 5 = ---;;-~{3 (a + {3 ), or

F1o = F5Ls.

Thus, since (L 8 + L 4 + I) in equation (A) is an integer, F 2 is a factor of F10 ; also F5 is a factor -of F 10 as shown by equation (B).

In general, to see why Fnk is divisible by Fk. consider the factorization

nk ,qnk a - fJ

Fnk = {3 a-(C)

k k = a - f3 (a<n-l)k + a<n-2>k{Jk

. a-{3

+ a<n -3)k{32k + ... + ak{J(n-2)k + f3(n-l)k)

In the right-hand parenthesis, the first and last terms may be paired to form a Lucas number:

a (n-Ok + ,q(n-Ok _ L fJ - (n-1 }k

The second and next-to-last terms may be paired to form a product of ( - l)k and a Lucas number:

a(n-2)k{3k + ak{J(n-2}k = ak(3''(a(n - 3)k + {J(n - 3)k)

= (a{3)kL(n-3}k = ( - l)kL(n-3>k

And so on. Notice that the number of terms in the right-hand parenthesis of (C) is n. If n is even, then the terms match up in symmetric pairs to make Lucas numbers, clea rly adding up to an integer. See, for examrle, equation (B), where

with k = 5 and 11 = 2, which is even. If 11 is odd, then the terms still match

Divisibility Properties of the Fibonacci and Lucas Numbers · 39

up in symmetric pairs except for the middle term, which is of the form (a{3)(n~l>k, clearly an integer. Again the right-hand parenthetical expression in (C) is an integer. See, for example, equation (A) where

with k = 2 and n = 5, which is odd.

Another interesting theorem is given in terms of the greatest common divisor of two positive integers. The greatest common divisor of two positive integers a and b is denoted by

(a, b) = d,

meaning that dis the greatest positive integer dividing both a and b. For example,

(14, 2) = 2, (24, 15) = 3, and (6765, 610) = 5.

This theorem can now be stated as:

THEOREM II

This means that the greatest common divisor of two Fibonacci numbers is a Fibonacci number whose subscript (index) is the greatest common divisor of the subscripts (indices) of the other two Fibonacci numbers. Thus,

(F1s, F14) = (610, 377) = Fos. 14> = F1 = l;

(F9 , F6 ) = (34, 8) = F< 9 • 6> = F3 = 2;

(F12, F 6 ) = (144, 8) = F02, 6> = F6 = 8. Here F6 divides F12·

Theorem II can be proved by using the Euclidean Algorithm* or as the solution to a Diophantine equation.t

Theorem I and Theorem II may be combined as:

THEOREM Ill

F11 is divisible by Fm if and only if n is divisible by m.

* N. N. V~robyov, Fibonacci Numbers (Boston: D. C. Heath and Co., 1963), pp. 22-24. t Glenn Michael, "A New Proof for an Old Property," The Fibonacci Quarterly, Vol. 2,

No. I (February, 1964), pp. 57-58.


Let us now look at the first few Lucas numbers:

L 1 L 2 L3 £ 4 L 5 L 6 L7 L8 L 9 Lto Lu L 12 L 13 L14 L 15 •••

l 3 4 7 II 18 29 47 76 123 199 322 521 843 1364 ...

We immediately notice that every third Ln is even, as was the case for the Fibonacci numbers.

We next state without proof two interesting theorems.

THEOREM IV (l. Carlitz)*

Ln divides Fm if and only if m = 2kn, n > 1.

For example, £ 3 = 4 divides F 6 = 8, F 12 = 144, and so on.

THEOREM V (l. Carlitz)*

Ln divides Lm if and only if m = (2k - l)n, n > 1.

For example, L3 = 4 divides L 9 = 76, £ 15 = 1364, and so on.

It is easy to see that two consecutive Fibonacci numbers have no common factor greater than one. If Fn+ 2 and Fn+ 1 had a common factor d, then Fn = Fn+ 2 - Fn+ 1 would also be divisible by d. Thus, we could progress down to F 2 = I, which d would have to divide. Since dis a positive integer, d must be 1. A similar argument applies to Lucas numbers. Thus, we have:

THEOREM VI THEOREM VII

EXERCISES

Refer to the list of Fibonacci and Lucas numbers on page 83.

1. Verify that F; divides F14, F21, and F23.

2. Verify that F1o divides F2o, F3o, and F4o.

* L. Carlitz, "A Note on Fibonacci Numbers," Tire Fibonacci Quarterly, Vol. 2, No. 1 (February, 1964), pages 15-28.

Divisibility Properties of the Fibonacci and Lucas Numbers · 41

3. Verify that F24 is divisible by FJ, F4, Fu, Fs, and F12.

4. Verify that F3o is divisible by FJ, Fs, Fu, Fto, and Fts.

5. Verify that L4 divides Fs and Fu5.

6. Verify that L1 divides F14 and F2s.

7. Verify that £4 divides L12 and L2o.

8. Verify that Ls divides L 15 and L25.

9. Verify that

10. Verify that

F12 = FJ(Ln - L3) = F3L3Lu

= F4(Ls + 1)

= FoL6.

F1s = F3(L1s - Ln + L3)

= Fu(Lt2 + 1)

= FuLg.

11. Find the greatest common divisor ofF 16 and F 24·

12. Find the greatest common divisor of F24 and F3G.

8 • Periodicity of the Fibonacci and

Lucas Numbers

What else can we discover from the list of Fibonacci numbers?

Ft F2 F3 F4 Fs F6 F1 Fs Fg Fto F11 F12 F13 Ft4 Fts .. .

l 1 2 3 5 8 13 21 34 55 89 144 233 377 610 .. .

Let us consider the sequence of primes. We observe that

2 divides F 3 , F6 , and so on,

3. divides F4 , F8 , and so on,

5 divides F 5, F 1 0 , and so on,

7 divides F 8 , and so divides F 16, and so on

11 divides F 10, and so divides F 20, and so on

13 divides F7 , F 14, and so on.

We can speculate as to whether every prime number divides some Fibonacci number (and hence divides infinitely many of them). Later in this section we shall prove that every integer divides some Fibonacci number.

To do this, we shall use a kind of periodicity property of the Fibonacci numbers. But first we consider the following.

You are, no doubt, familiar with periodic decimals, 1 4 2 8 5 7 such as the decimal representation of 7 ) 1 : o o o o o o o

1 7 7·

When 1 is divided by 7, the only possible remainders are

0, 1, 2, 3, 4, 5, 6.

If we had a zero remainder, our division would be exact. However, when we get any other of those seven digits, the division continues as shown, leading to the periodic decimal

t = .142857142857142857 .. . .

47

30 2 8

2 0 I 4 .6 0

5 6

4 0 3 5

5 0 4 9

1 0 7

-3

)

Periodicity of the Fibonacci and Lucas Numbers · 43

Every rational number has a periodic decimal expansion and every periodic decimal expansion represents a rational number. (Even ! = .5000 .. . = .4999 .... )

In our present discussion we shall use a similar argument, but this time we shall be discussing repeating ordered pairs of remainders.

When F0 through Fa 1 are each divided by 7, we obtain the following sequence displaying the quotients and remainders.

Fo = 0= 0·7 + 0 F16 = 987 = I4I·7+0

Ft = 1 = 0 · 7 + I F11 = 1597 = 228. 7 + 1

F2 = 1 = 0 · 7 +I FI B= 2584 = 369. 7 + 1

Fa = 2 = 0 · 7 + 2 F19 = 4181 = 597.7 + 2

F4 = 3 = 0· 7 + 3 F2o = 6765 = 966.7 + 3

Fs = 5 = 0·7 + 5 F21 = I0946 = 1563. 7 + 5

F6 = 8 = l · 7 + I F22 = 17711 = 2530 · 7 + I

F7 = 13 = I· 7 + 6 F2a = 28657 = 4093 ·1 + 6

Fs = 21 = 3 . 7 + 0 F24 = 46368 = 6624.7 + 0 Fg = 34 = 4·7 + 6 F2s = 75025 = 107I7 . 7 + 6

Flo = 55 = 7. 7 + 6 F26 = 12I393 = 1734I . 7 + 6

F11 = 89 = I2. 7 + 5 F21 = 1964I8 = 28059.7 + 5

F12 = I44 = 20.7 + 4 F2s = 3I781l = 4540I. 7 + 4 F I a = 233 = 33 . 7 + 2 F29 = 514229 = 73461 . 7 + 2

F t4 = 377 = 53 · 7 + 6 Fao = 832040 = 118862. 7 + 6

F I 5 = 610 = 87 . 7 + l Fat = 1346269 = 192324 · 7 + I

You can see by the zero remainders that 7 divides F0 , F8 , F16, and F24 in this list.

Notice the pattern of the remainders (since each F11 is the sum of the two preceding Fibonacci numbers):

0 + I = I, l + 1 = 2, . . . , 3 + 5 = 8 = 7 + I, I + 6 = 7 + 0, . . .

That is, each remainder after the second is the sum of the two preceding remainders, decreased as necessary by 7.

Consider the sequences of remainders

1' 6, 0, 6 (beginning with F 6 )

and 6, 1, 0, 1 (beginning with F 14) .

Notice that the pair (1, 6) gives a different sequence from that following the pair (6, 1). Since there are seven possible remainders (0, 1, 2, 3, 4, 5, 6) when numbers are divided by 7, there can be at most 7 X 7, or 49, different ordered pairs of remainders. Therefore, in a set of 49 + 1, or 50, ordered

' • .


pairs, at least two of the pairs must be the same. In our example, not all the possible ordered pairs of remainders appear, and there .are many repetitions between the first ordered pair of remainders, (0, I) for F0 and F 1,

and the fiftieth, (l, I) for F 49 and F 50•

We shall now show that the first repeated pair of remainders is (0, 1) when we consider Fibonacci numbers with positive or zero subscripts, as we did in the table above for m = 7. In general, let the pairs of remainders (rk. rk+ 1) be obtained by dividing the Fibonacci numbers Fk and Fk+ 1 by some integer m. Consider the sequence of pairs

~'. We· shall say that pairs (a 11 b 1) and (a 2 , b2) are equal if and only if a 1 = a 2

and b 1 = ·b2. Clearly, in the first m 2 + I of these pairs obtained from the division of consecutive Fibonacci numbers by m, there must be at least two which are equal. (Conceivably, there could be more than two equal pairs as in the example above.)

Let us assume that the first pair to be repeated is (rk, ~"k+ 1) for k > 0. Then in our sequence of pairs there is a later pair (rn, ~"n+ I) equal to (rk, rk+ I) with m 2 + I > n + l > k + I. But smce the pairs are equal, rk = rn

and rk+I =r n+I· Since

Fk-l = Fk+l - Fk

upon dividing by m, we get

and

and

although it may be necessary to add m to ~"k-t and r n- I to make them positive. Thus, the pairs (rk-b rk) and (rn-h rn) are also equal. Therefore, every repeated pair has a predecessor that is also repeated except the first (0, 1), which has no predecessor. This first (0, l) is the first pair in our list ; see the table made for divisor m = 7 on page 43.

Let (r11 , ~"n+.) = (0, 1), n > 0, be the second appearance of (0, I) in the sequence of ordered pairs of remainders. Then rn = 0, so that Fn is divisible by m where I < n + 1 < m 2 + I, or 0 < n < m 2• Therefore, we have proved the following theorem.

THEOREM I

Every integer m divides some Fibonacci number ( > F0 ) whose subscript

does not exceed m 2•


As we progress up the sequence of remainders from the second (0, 1), we simply repeat the pairs in the same order as before so that F 2 n is also divisible by m. We define n as the period of m in the Fibonacci sequence and denote it by Km. For example,

as you can see by looking back at the remainder pairs obtained by dividing the Fibonacci numbers by 7 (page 43). The pair (0, I) occurred first with F 0 and F 1. The next occurrence of these remainders was for F 16 and F 17,

and F 1 6 was divisible by 7. The pairs of remainders now repeat in cycles of 16; hence, K7 = 16.

However, we noticed that F 8 also has a zero remainder and is divisible by 7. The subscript of the first positive Fibonacci number divisible by m is called the rank of apparition or entry point of the number m in the Fibonacci numbers. Thus, the entry point of 7 in the Fibonacci numbers is 8.

\Ve can assert that 7 divides a Fibonacci number if and only if its subscript is divisible by 8. We note that the only if follows from the periodicity and not from Theorem Ill in Section 7, since 7 is not a Fibonacci number. (We also note that 7 does not divide F7 . ) If a prime divides some Fibonacci number Fd, then it divides every Fibonacci number Fnd, since Fd divides F,,rf. (Note that, since the number 7, which we chose to illustrate periodicity, happens to be a Lucas number, for this particular divisor we could have applied Theorem IV in Section 7.)

Now let us turn to the Lucas numbers. It can easily be shown that 5 does not divide any Lucas number:

L0 = 2 = 0 · 5 + 2

L 1 =1=0·5+ 1 L 2 = 3 = 0 · 5 + 3

L3 = 4 = 0 · 5 + 4

L 4 = 7 = I· 5 + 2 L 5 = II = 2 · 5 + I

L 6 = 18 = 3 · 5 + 3 L 7 = 29 = 5 · 5 + 4

Thus, in considering the remainder pairs when the Lucas numbers are divided by 5, we come to a repeated remainder pair (2, I) again at L 4 and L 5 •

V-./e note that in the period of length 4 there were no zeros. (Zero is not a Lucas number, and so the remainder 0 may not occur at all.) Thus, we see that no Lucas number is divisible by 5.

Of course, some Lucas numbers are divisible by other integers, for example 6 (see Exercise 4, page 47).

In the booklet An Introduction to Fibonacci Discovery by Brother U. Alfred (available from the author, St. Mary's College, California 94575) there is a nice table (Table 3, page 55) of entry points and periods of primes 2 to 269 in both the Fibonacci numbers and the Lucas numbers.


The Fibonacci and Lucas numbers have interesting remainder properties. We state the following theorems without proof.

THEOREM II

If Fn is divided by Fm {n > m), then either the remainder R is a fibonacci number or F m - R is a fibonacci number.

For example,

89 = 2 . 34 + 21'

144 = 6 . 21 + 18,

or

or

F11 = 2 · F9 + Fs;

-F12 = 6 · F 8 + 18.

In the second case, although 18 is not a Fibonacci number, the difference

F8 - 18 = 21 - 18 = 3 = F4

is a Fibonacci number. We can write

F12 = 6 · Fs + (Fs - F4).

Of course, F 0 = 0 is a Fibonacci number, and thus when F 10 is divided by F 5 , the zero remainder is a Fibonacci number.

With one exception, the property given in Thf"orem II ·is shared by the Lucas numbers. We have this theorem.

THEOREM Ill

If Ln is divided by Lm, n > m, then the remainder R is zero, or R is a Lucas number, or Lm - R is a lucas number.

For example,

76 = 19.4 + 0, or

where R = 0, which is not a Lucas number but is the first possibility given in the theorem. Also,

18 = 1·11+7, 47 = 6. 7 + 5,

or

or

L6 = 1 · Ls + L4;

Ls = 6 · L4 + (L4 - Lo).

A proof of the "Remainder Property" occurs in a short article by John H. Halton in The Fibonacci Quarterly, Vol. 2, No.3 (October, 1964), pages 217-218, titled "Fibonacci Residues." The Lucas property is discussed in a paper by Laurence Taylor in The Fibonacci Quarterly, Vol. 5, No.3 (October, 1967), pages 298-304, titled "Residues of Fibonacci-Like Sequences"; in the same paper, we find that the Fibonacci and Lucas sequences are the only sequences which satisfy the recurrence relation Un+Z = Un+I + u11 and have the properties given in Theorems I I and I II.


EXE RC ISES

1. Show that K 13 = 28 for the Fibonacci sequence, and find the entry point of 13 in that sequence.

2. Show that 13 does not divide any Lucas number, but K 13 = 28 a lso for the Lucas sequence.

3. Show that K 1 o = 60 for the Fibonacci sequence, and find the entry point of 10 in that sequence.

4. Show that Kr. = 24 for the Lucas sequence, and state the entry point of 6 in that sequence.

5. Show that K10 = 12 for the Lucas sequence, and state why 10 has no entry point in that sequence.

9 • Pascal's Triangle and the

Fibonacci Numbers

If we consider the expansions of the binomial (x + y)n for n = 0, I, 2, 3, 4, 5, . . . , we can write them in the form :

(x + y)o = xoyo

(x + y)I = xtyo + xoy1

(x + y)2 = xzyo + 2x1y1 + xoyz

(x + y)a = xayo + 3x2y1 + 3x 1y2 + xoya

(x + y)4 = x4yo + 4xayt + 6xzyz + 4x•ya + xoy4

(x + y)5 = xsyo + 5x4yt + lOxayz + lOxzya + 5x•y4 + xoys

Recalling that x 0 = y 0 = l (x and y nonzero), we can write the array of coefficients, which is called Pascal's triangle, as follows:

1

l 1 2 1

1 3 3 1 4 6 4

5 10 10 5

Before we proceed further, let us introduce some special symbolism . If we write

1 + 2 + 3 + 4 + ... + 11,

we mean that starting with one, we add the consecutive integers until we come to n, which is the last number added . By using a notation based on

Pascal's Triangle and the Fibonacci Numbers · 49

:L. the Greek letter sigma for s, the first letter in summation, we can write

n

L i = 1 + 2 + 3 + · · · + n, i= l

where the index is i and the limits of summation are l and n. The index progresses from i = I to i = 11 in unit increases. A summand is a quantity obtained by putting a certain value of i into a formula that yields the quantities to be added. For example, in

6

L i = I + 2 + 3 + 4 + 5 + 6 = 21 i= l

the summands are I, 2, 3, 4, 5. and 6. You may recall from your study of arithmetic and geometric progressions

the generol formulas

t i = 1 + 2 + 3 + · · · + n = n(n + 1)' i =l 2

n - 1 1 n L / = I + r + r2 + ... + rn-I = - r ' i=O I - r

r ~ I.

Returning now to Pascal's triangle, let (;:) represent the mth term in

the nth row (n > m > 0). Thus, for n = 4, we have

(x + y)4 ~ (~) x' + G) x 3y + (~) x 2y 2 + (~) xy• + (:) y'

- (l)x4 + (4)x3y + (6)x 2y 2 + (4)xy3 + (l)y4•

The numbers represented by the symbols (n) are called the binomial coeffim

cients. In the expansion of (x + y) 4 the numbers are the binomial coefficients for the fourth-power expansion.

In general, from a study of Pascal's triangle, we can define

(0) _ (n) _ (n) _ 1 ( n) _ (n - 1) + ( n - 1) 0 - 0 - n - ' m - m m - 1 for n > m > 1.

\Ve see that these interesting numbers are thus defined by a recurrence formula for each positive integer n. For fixed n, the binomial coefficients are the entries across the nth level of Pascal's triangle. We can now write

(x + y)4 = t (~) x 4-iyi (since y 0 = x 0 = I) i=O 1

and, in general (as cap be proved by mathematical induction),

(A) n n n-i i n () (x + y) = ~ i x y .


Although we can generate any npmber of the binomial coefficient~ by writing down successive levels of Pascal's triangle, it is possible to calculate them directly. Let l · 2 · 3 · · · · · n = n! (called "n factorial") and 0! = L Then it can be shown that

(n) n! m = m! (n - m )! ·

For example,

( 5) l . 2 . 3 . 4 . 5 3 = ( 1 . 2 . 3 )(1 . 2) = 10.

Compare th~~ ~ith the coefficient for m = 3 in the expansion of (x + y) 5

How are the Fibonacci numbers related to Pascal's triangle? Let us write Pascal's triangle in the following form:

10

4

10 5 l

We can now see that the sums a long the rising diagonals are the Fibonacci numbers:

F 1 = 1, F 2 = I, F 3 = 1 + I, F 4 = 1 + 2,

F5 = 1 + 3 + 1, F6 = 1 + 4 + 3, F7 = l + 5 + 6 + I , etc.

It can be proved that in general,

[n/2] ( ·) Il- l Fn-i-1 = L · '

i=O I

where [x] denotes the greatest ·integer not greater than x (Section 6). For example,

Fs ~ t. e 7 i) ~ (~) + (D + (;) ~ I + 3 + 5,

Fs = t. C 7 i) = (~) + (7) + G) + G) 6! 5! 4!

- l + (1!)(5! ) + (2!)(3!) + (3!)(1!)

- I + 6 + 10 + 4 = 21.

Pascal's Triangle and the Fibonacci Numbers · 51

11 a -- ~ n ( ) i r.~i Now suppose that we look at ,8) i Fi, where Fi = a __ {3 · Thus,

L n F· = L n i - L n {3i n () 1 [n () n () ] i=O i 1 a - {3 1=0 i a i =O i

Since from formula (A) on page 49 we have ~(~)xi = (l + xf (see

Exercise I below), we can write

t (~) Fi = _ I _ [(1 + af - (1 + f3f] i=O I a - {3

But. 1 + a = a 2 and 1 + {3 = {3 2 : therefore,

t ('.1) Fi = a2n - ~211 = F2n· i=O 1 a -

For example,

t., (;) F, = (~) Fo + m F 1 + (D F2 + (;) F3 + G) F4 + (;) F 5

= (I )(0) + (5)(1) + (IO)(l) + (10)(2) + (5)(3) + (l )(5)

= 0 + 5 + lO + 20 + 15 + 5 =55 = F10.

EXERCISES

Demonstrate the following:

" () 11 i n 1. ~ i x = (I + x) 2. t (~) = 2"

i=O 1 3. t (~) (-l)i .= 0,

i = O 1 n>l

4. Verify that the two formulas for binomial coefficients agree by using the factorial notation to demonstrate that

and

5. Verify that ± (~) Fi = Fs. i=O I

Using the Binet form show that

(n) (" - 1) (n - l) m = m + m-1.

n () n ; i+l

8. L . ( - 1)Fi+j = ( - 1) Fn-j i=O I

10 • Selected Identities Involving the

Fibonacci and Lucas Nun1bers

A great many identities have been discovered that reveal interesting patterns of relationship between the Fibonacci and Lucas numbers.

In Section 2 we developed our first Fibonacci identity:

We shall now give several other proofs of this.

Derivation of (l 1) from the definition. Since for all positive integers 11 the Fibonacci recurrence formula

holds, we may write the sequence of equations:

F 1 = F 3 - F 2

F2 = F.! - F3

F3 = F 5 - F4

Fu-1 = Fn+ 1 - Fn

Fn = Fn+2- F,l+1

By adding these term by term and noting the cancellations on the ri!!.ht, we find

F 1 + F 2 + Fa + · · · + Fu = Fu + 2 - F 2

= Fn +'2 - I. 52

Selected Identities lm:oluing the Fibonacci and Lucas Numbers · 53

Derication of (I 1) using the Binet form. Since

n {Jn a -F" = ----'----

a-{3

we have

{3 2 {3? n {Jn _ a -_ + ~-=-------= + ... + a -

a-{3 a-{3 a-{3

(a + a2 + ... + a") - ({3 + {32 + ... + {Jn) _________________________ , a-{3

or

F1 + F? + · · · + Fn

But

I ? 2 - [(a+ a- + · · · + a 11 ) - ({3 + {3 + · · · + {3n)]. a-{3

a+ a 2 + · · · + a 71 = (I +a+ a 2 + ···+ an)- J, {3 + {32 + ... + {3'1 = (I + {3 + {32 + ... + {3n) _ 1.

From the formula for the sum of a geometric progression (recalled on page 49), we see that

I n+l 2 -a

1 + a + a + · · · + an = ' l-a

l _ {Jn+l 1 + {3 + {32 + ... + {Jn = ____ .

1-{3

Therefore,

Ft + F2 + · · · + Fn

= --~- [(!_~+I _ I) _ (1 _ ~n+l I)] ' a-{3 1 -a l -{3

and since I - a = {3 and I - {3 = a, we have

a(l _ an+l) _ {J(I _ {Jn+l) F1 + Fz + · · · + Fn = a{J(a _ {3)

Since a{3 = - J, we have

n+2 {3 + {3n+2 a- a -F 1 + F 2 + · · · + Fn = ------.,-( - 1-)(,_.a - -/3-,--)

n+2 {3n+2 a --

a-{3

=Fn+2-l.


\ Proof of(l 1) by mathematical induction. We wish to prove

P(n): F 1 + F 2 + Fa + · · · + Fn = Fu+ 2 - I.

There are two parts to the proof:

I. The statement P(l) is found true by trial. Here P(l) is FI = Fa - I, which is true since I = 2 - I.

2. One proves: If P(k) is true for some integer k (k > 1), then P(k + I) must also be true. In this case, we assume

and must prove

By adding Fk+ I to both sides of

Fz + F2 + · · · + Fk = F~.-+z - I,

we have

since Fk+a = Fk+2 + Fk+ I· Therefore P(k + I) is true, and the proof is complete by mathematical induction.

The first part of the proof, where P(l) is verified by direct trial, is often called the _basis for induction. Clearly, if no statements can be found to be true by trial, we have no basis for our attempted proof. The second part is often called the inductive transition or inductive step. The assumption that P(k) is true for some integer k (k > I) is the inductive hypothesis. If we can show that the inductive hypothesis is suflicient to prove that P(k + I) is true, then we have completed the inductive transition.

It is important that both part I and part 2 be satisfied. The declaration, "The proof is complete by mathematical induction,"

following such a proof is a simple statement telling the reader what method has been used.

There is a corresponding identity for Luca:; numbers:

This is to be proved in Exercise 2, page 60.

Selected Identities Involving the Fibonacci and Lucas Numbers · 55

As our next identity, let us consider:

n > 1

\Vc shall prove this by mathematical induction. Here we have

P(n): Fi + F~ + F~ + · · · + F~ = FnFn+ t· Then

P(l): Fi = F IF 2

is easily seen to be true, since (1) 2 = (I)( I). Thus, we have completed the basis for induction.

Now we suppose that

P(k): Fi + F~ + · · · + F'fc = FkFk+I

is true (the inductive hypothesis), and we undertake to prove:

P(k + 1): Fi + F~ + · · · + F;+I = Fk+IFk-t-2

To do so. add Fr+ I to both sides of equality P(k), obtaining

(Ff + F~ + · · · + F~) + F'f+l- FkFk+I + Ft+I = Fk+I(Fk + Fk+I)

- Fk+IFk+2·

Therefore, we have shown that if P(k) is true, then P(k + I) is true, and we .have completed the inductive transition. The proof .is complete by mathematical induction.

Formula (1 3 ) may be pictured geometrically as follows:

Fi + Fi = F 2 X F 3 B Fi + F~ + F5 = F 3 X F 4 ITTl (1):? + (1) 2 = I X 2 (1) 2 + (1) 2 + (2) 2 = 2 X 3 D..lJ

Fi + Fi + F5 + F~ = F 4 X F 5

(1) 2 + (1) 2 + (2) 2 + (3) 2 = 3 X 5

Fi + F~ + F~ + F~ + F~ = F 5 X F 6

(1):? + (1) 2 + (2) 2 + (3) 2 + (5) 2 = 5 X 8 ~ · I

II There is a corresponding identity for Lucas numbers:

I I I

I I I I

I

L i + L~ + L~ + · · · + L~ = LnLIL+ 1 - 2, n > 1

This is to be proved in Exercise 5, page 60.

I I I

I I

I

I J

I


Now consider the following pair of identities (to be proved in Exercises 6 and 8, page 60):

(15) F1 + £3 + £5 + · · · + F2n-l = F21u II > 1

(16) F2 + £4 + F6 + · · · + F2u = F2u+l - 1, II > 1

Verify by adding term by term that the sum of (1 5 ) and (1 6 ) is the same as (I I) with n replaced by 2n.

In Section 5 we found two interesting identities connecting the Fibonacci and Lucas numbers:

These were to be proved in Exercises 1 and 2, page 29. We also have:

(I g)

This can be proved by applying (1 8 ) (see Exercise 12, page 61 ). From (1 7) and (1 8) we can derive:

(II o)

See Exercise 13, page 61. By writing

and applying (I 10), we can derive:

F2n+l = F~+l + F~, II > 1

See Exercise 14, page 61. Also in Section 5 we found this identity:

SF~ . L~ - 4( - 1)11

This was to be proved in Exercise I 0 9n page 29. Let us now look again at the Fibonacci sequence:

Fl F2 F3 F4 F5 Fu F7 F'd FfJ FlO ...

1 1 2 3 5 8 13 21 34 55 ...

Notice that:

FIF3 - Fi = F2F4 - Fa = - I

F3F5 - F~ = I

F4Fu - F~ = - I

and so on.

Selected Identities /m;o!t.:ing the Fibonacci and Lucas Numbers · 57

In general, we may·write:

\Vc shall prove this by mathematical induction, starting with n = I.

P( I): F 0 F 2 - Fi = 0 · I - I 2 = - I

\Vc have established the inducth:e basis. Assume for some integer k > I that

P(k): F"+ tFk-t - F'f = ( - 1/. Let us look for P(k + I):

Thus,

Fk+2Fk- FI+t = (F!.+I + Fk)Fk- Ff+t

- F'f + Fk+ I (Fk - Fk+ I)

Fi - Fk+IFk-I

- ( -I)(Fk+IFk-I - Fi) - (- I)( - I )k = (- I )k+ t

P(k + 1): Fk+2Fk - F'f+ 1 = (- I)k+ 1

is true. The proof is complete by mathematical induction. The mathematician Charles Lutwidge Dodgson, whose pen name was

Lewis Carroll, liked to entertain his friends with puzzles. According. to his nephew,* one of his favorites was this geometrical paradox. ·Let us cut the square on the left in Figure 18 as marked and reassemble the pieces to form the rectangle on the right.

Figure 18

The area of the square is 8 X 8 = 64, while the area of the rectangle is 5 X 13 = 65. What happe·ned?

* Dicersions and Digressions of Lewis Carroll, edited by Stuart Dodgson Collingwood (New York: Dover Publications, Inc., 1961), pp. 316-317.


If you draw the diagrams pictured in Figure 18 making the square units rather large, you can discover what happened. The points A, B, C, and D, which in the small.figure appear to lie on a straight line, actually are the vertices of a parallelogram, the area of which is the unexpectedly added unit. If you make the unit squares I inch on a side, the height of the added parallelogram will be nearly k inch (see Exercise 15, page 61).

How is such a puzzle discovered? Have you noticed that the numbers 5, 8, and 13 are the consecutive Fibonacci numbers F 5 , F6 , and F7 ? Therefore from identity (1 13) we have

5 · 13 - 8 2 = ( - I ) (\

or 65 - 64 = I.

Any triple of Fibonacci numbers such that the middle one has an even subscript (index) will produce this difference of I. However, the larger the numbers used, the less noticeable is the added parallelogram.

In Figure 19 the situation is shown schematically with the parallelogram exaggerated:

,...1·---F--F'!.,.+l -------t•l '!.n F2n-1

F2n-2 F2n-l

Area: F~n

Figure 19

Since

F 2 11- 1 F 211 + I - Fi II = (- I ) :2 II = I'

the area of the parallelogram is one square unit. The greatest width, the height h, of the parallelogram is easily found . Since the area equals the product of the height and the length of the base, we have

(by the Pythagorean theorem), or

lz =

Selected Identities Involving the Fibonacci and Lucas Numbers · 59

A grea t many more identities have been discovered. The following additional list will i!ldicate the various types.

(I I.,)

Ot ,.)

Otd

(I I i)

(ltl')

(I I o)

(I 2 o)

(12 I)

(I 2 2)

(In)

(12-t)

(12;.)

(I 2 r,)

(127)

(12s)

(12o)

(I3o)

(f3 I)

(I 3 2)

(133)

(134)

(13 ;})

(I 3 r,)

(13 7)

(I 3s)

(13o)

Ln = Fn+2 - Fn -2

L4n + 2 = Lin

L4n- 2 = SF~n

L-tn+z + 2 = SF~n+ 1

L4n+2 - 2 = Lin+ l

F F F 2 _ ( l)n+k+IF2 n-k n+k - 11 - - k

Fn+p + Fn-p = LnFp, p odd

F,+Jl - Fn-p = . FnLp, p odd

Fn +11 - Fn-p = LnFp, p even

F;;+P- F~-P = F2nF2p

Fm+n+ I - Fm+ tFn+ l + FmFn

Lm+n+1 - Fm+ILn+l + FmLn

FnLn+k- Fn+kLn . 2(-l)n+lFk

F~,;Fk+IFk+3Fk+4 = Fk+2 - 1

L2nL2n+2 - l = SF~n+t

LnLn+ 1 - L2n+ 1 = ( -l)n

SFn = Ln+2 - Ln-2

L~ + 4Ln- ILn+ 1 = 25F,~

L,_ ILn+ l + Fn-tFn+I = 6F?,

F,~ + 4Fn - tFn+I = L~

F?,+I - 4FnFn-l = F~-2

Fm - lFm+l- Fm+2Fm-2 = 2(-i)'11

LnFm - n + FnLm- n = 2Fm

11

L kFk = (1Z + 1)Fn+2 - Fn+4 + 2 k=I


(145)

2~2 (2'z + 2) L2 _ sn+IL ~ k k - 2n+2

k=O

2~ 1 (2" + 1) p2 - snF ~ k k - 211+1

k=O

EXERCISES

1. Write (I 1) using summation notation for the left-hand member.

2. Prove (h) by mathematical induction.

3. Write (13) using summation notation for the left-hand member.

4. The diagram at the right illustrates the statement

F~ = F~ + 3F~ + 2(F! + Fi + F~ + Fi).

Use (b) to prove that, in general,

Show that the number of squares is 2n.

5. Prove (14) by mathematical induction.

6. Derive (1.<>) by using the sequence of equations F 1 = F :2, F:1

F5 = FG - F.t, and so on .

7. Write (15) using summation notation for the left-hand member.

a. Prove (I G) by mathematical induction.

9. Write (1 0 ) using summation notation for the left-hand member.

F.t - !':!_,

Selected Identities /m;o/ving the Fibonacci and Lucas Numbers · 61

10. Find and prove an identity for Lucas numbers corresponding to (1;;).

11. Find and prove an identity for Lucas numbers corresponding to (I r.).

12. Use (I~) to prove (In).

13. Use (I,) and (I ~) to prove (I 1 o).

14. Use (I to) to prove (Itt ) .

15. Verify that if the small squares in F igure 18 are drawn 1 inch square on a side, the height It of the parallelogram is less than -A- inch.

16. On graph paper, carefully draw a square with each side Fs = 21 units, cut it as indicated in Figure 19, and assemble the pieces as a rectangle as shown there. (You will be able to surprise your friends with the result.)

17. Recall the definit ion of a generalized Fibonacci sequence on page 7. Show by mathematical induction that if H 1 = p, H2 = q, and Hn+2 = Hn+l + Hn, 11 ~ I, !hen Hn+2 = qFn+ 1 + pF,.

18. Using 02u) on page 59, show by mathematical induction that Fn divides Fnk. k > 0. Hint: Let 111 + 1 = nk. Then 02u) becomes F<k+lht = FknFn+l + Fkn-lFn.

11 • Two-by-Two Matrices Related

to the Fibonacci N u1nbers

A two-by-two matrix is represented by a symbol such as

where a, b, c, and d represent any real numbers. These numbers are called elements. In this section, all matrices mentioned are to be considered to be two-by-two matrices.

Two matrices,

and

are equal if and only if their corresponding elements are equal. That is,

A=B if and only if

a = e, b = f, c = g, and d = h.

The matrix which is the sum of two matrices A and B is defined to be

A + B = (a b) + (e !) = (a + e b + !) . c d g h c+g d+h

The zero matrix Z is defined as

z = (~ ~) It can be shown (Exercise I, page 68) that

A+ Z = Z +A= A,

and so Z is the additive identity element for the system of two-by-two matrices. 62

Two-by-Two Matrices Related to the Fibonacci Numbers · 63

Also, it can be shown (Exercise 2, page 68) that, in general,

A+ B = B +A.

Thus, matrix addition is commutative. It can also be shown (Exercise 3, page 68) that matrix addition is associative.

\Ve define the negative, or additive inverse, of A to be

-A= ( -a -b) - c -d

and the difference of A and Bas

A - B = A + (-B).

You can verify (Exercise 4, page 68) that

A+ (-A)= Z.

\Vhen we are dealing with matrices, we refer to the real numbers as scalars. The product of a scalars and a matrix A is defined to be

sA = s (; ~) = (~; ~!) · It can be shown (Exercise 5, page 68) that

(sr)A = s(rA).

The matrix which is the product of two matrices A and B is defined to be

A 8 = (~ ~) (; {.) = (~; ! ~~ ~! ~~) · The identity matrix I is defined as

I= (b ~). It can be shown (Exercise 7, page 68) that

AI= /A = A,

and so I is the multiplicative identity element for this system. It can also be shown that matrix multiplication is associative.

Let A = (! ~) and B = (~ !) : then:

A B = (I I) (3 I) = (I · 3 + I · 0 I 0 0 I I · 3+0·0

I · 1 + I · 1) = (3 2) I·I+0·1 3 I

BA = (3 I) (I I) = (3 · I + I · 1 0 I I 0 · O·J+I·I

3 . 1 + 1 . 0) = (4 3) 0 · 1+1·0 I 0

Therefore, in this case AB ~ BA. This demonstrates that matrix multiplication is not always commutative.


If A = ( _: -:)and B = (: :) , then we have

·AB = (I · I + ( -1)1 I · I + ( -1)1) = (0 0) z (-1)1+1·1 (-1)1+1·1 0 0 = .

Here we note that the product AB is the zero matrix although neither A nor B is the zero matrix. In elementary mathematics, if the product of two numbers is zero, then at least one of the numbers has to be zero. For matrices (which are not numbers) this rule does not apply.

If A, B, and C are matrices, it can be shown (Exercise 8, page 68) that matrix multiplication is distributive over matrix addition. That is,

A(B+C)=AB+AC and (B + C)A = BA +CA.

Since matrix multiplication is not always commutative, these two distributive laws do not necessarily say the same thing.

Associated with each matrix A is a number, called the determinant of matrix A, and denoted by det A, which is defined as

a b det A = c d = ad- be.

The matrix A is said to be nonsingular if det A ~ 0; otherwise, matrix A is singular.

We now prove a simple theorem.

THEOREM I

The determinant, det AB, of the product of two matrices, A and B, is the product of the determinants, det A and det B. That is,

det AB = (det A)(det B).

Proof Using the matrices A and B shown at the beginning of this section, we have

det A = I ~ ; I = ad - be, det B = I; {I = eh - fg,

det A B = I ae + bg af + biz I ce + dg cJ+ dh

= (ae + bg)(cf + dh) - (af + bh)(ce + dg)

= acef + adeh + bcfg + bdgh - (acef + adfg + bcelz + bdgh)

= adelz + bcjg - adfg - bcelz

= (ad- bc)(eh - fg)

= (det A)(det B).


Before looking at powers of a two-by-two matrix, let us recall how powers were defined for real numbers. If a is a nonzero real number, then one can define all integral powers of a as follows:

and -11 a for positive integraln

The definition a"+ 1 = a" a 1 is called an inductive definition. We shall also use an inductive definition for powers of matrices. For any nonzero matrix A,

and

for positive integraln. We shall now consider a particular matrix

Q = (! ~), withdetQ =-I,

and we shall prove the following theorem.

THEOREM II

For n > 1, the nth power of Q is given by

Proof We shall use mathematical induction. Clearly, for 11 = I,

Fk ) (I I) Fk-t 1 0

The proof is complete by mathematical induction.


THEOREM Ill n > 1.

Proof The proof is left to you (Exercise 9, page 68).

From Theorem II and Theorem III we have

Thus, we have still another derivation of (1 13) of Section 10. Of much technical interest in the algebra of matrices is the characteristic

polynomial. For a matrix A, the characteristic polynomial is defined to be

P(x) = det (A - xl) = ( ac db) - x (ol 01) - (~ !) + (-z -~)

a-x b -

C d- X

= x 2 - (a + d)x + (ad- be).

The equation P(x) = 0, or

x 2 - (a+ d)x + (ad- be) = 0,

is called the characteristic equation of matrix A. Note that the constant term is the determinant of matrix A and the linear term has coefficient -(a + d), which is the negative of (a + d). The sum (a + d) of the elements on the diagonal of matrix ~ is cal lee! the trace of matrix A. The roots of the characteristic equation are called the characteristic roots of matrix A.

For Q, the characteristic polynomial is

det(Q- xl) ~ IG ~)- (~ ~)I -x

and the characteristic equation is

II = X:? - X- I, -x

x 2 - x- I = 0,

which we have called the Fibonacci quadratic equation (page ll ). The characteristic roots of Q are then, of course,

a= l + ~./5

2 and •- vs (3=---- ·

2


Now what are the characteristic equation and the characteristic roots of Qn?

det(Qn - xl) = IFn+l- X • Fn I Fn Fn - 1 - X

= x 2 - (Fn+I + Fn_J)x + (Fn+IFn-I - F1~)

But Fn+I + Fn-I = Ln [from (Ig) of Section 10] and Fn+lFn-I- F?; (-If' [(113)]. Thus, ·

det (Q 11 - xl) = x 2 - L 11 X + ( -l)'Z,

and the characteristic equation is

X 2 - L 11 X + ( -l)n = 0.

The characteristic roots are given by (using the quadratic formula)

Ln ± V L~ - 4(- 1 f X= .

2

\Ve now recall from (I 12) in Section 10 that L~ - 4( - l)n = 5F~; thus, the roots are

Ln + v5 Fn - - - ---

2 and

However~ we recall that these are the expressions that we found for a 11 and {3 11 on page 27, and so we have the following theorem.

THEOREM IV

or

The characteristic roots of matrix on are the nth powers of the charac

teristic roots of Q, and the trace of on is L11 •

You can easily verify (Exercise 10, page 68) that

Qz = Q +I,

Q 2 - Q- I= Z.

Thus, Q may be said to satisfy its characteristic equation

x 2 - x - I = 0.

This is an instance of the following theorem, which we state without proof.

THEOREM V (Hamilton-Cayley* Theorem)

Every square matrix satisfies its own characteristic equation.

*Sir W. R. Hamilton (1805-1865), Irish mathematician; Arthur Cayley (1821-1895), English mathematician.

68 · Fibonacci and Lucas. Numbers

EXERCISES

1. Using

and

and the definition of addition, show that

A+ Z = Z+ A= A.

2. Using

A=(: !) . and B = (e I) g h


A+ B = B +A.

3. Using

A = (: !) , B = (; ~) , and c = (i j) k I


(A + B) + C = A + (B + C).

4. Show that A + (-A) = Z.

5. Using the definition of scalar multiplication, show that (sr)A = s(rA).

6. Show that -A = ( -l)(A).

7. Using

A=(; !) and I = (~ ~) and the definition of multiplication, show that

AI= lA = A.

8. Using A, B, and Cas given in Exercise 3, show that

A (B + C) = A B + A C and (B + C)A = BA + CA.

9. Prove that det (Q") = (-1)" for Q = ( 1 ~), n ~ I. I '-'1

10. Show that Q2 = Q + I, or Q'2 - Q - I = Z.

12 • Representation Theorems

A sequence of positive integers, at. a 2 , ••• , an, ... , is complete with respect to the positive integers if and only if every positive integer m is the sum of a finite number of the members of the sequence, where each member is used at most once in any given representation. We state the following theorem without proof.

THEOREM I

The sequence defined by On = 2n (n > 0) is complete.

For example, since

ao = 2o = J,

a 1 = 2 1 = 2,

a 2 = 2 2 = 4,

a 3 = 2~ = 8,

we can write, for example:

I = I 5=4+1

2 = 2 6=4+2

3 = 2 + 1 7=4+2+1 4 = 4 8 = 8

a 4 = 24 = 16, a 5 = 2 5 = 32,

9=8+1 10 = 8 + 2 11 = 8+2+1 12 = 8 + 4

. . . '

It can be proved that each representation is unique, and you should recognize this as the basis for the binary system of numeration. For example :

l : 5: 101 9: 1001

2: 10 6: 110 10: 1010

3: II 7: 111 11: 1011

4: 100 8: 1000 12: 1100

69


The ancient Egyptians used the principle described on the preceding page in their method of multiplication. For example, to find

243 X 25,

they would write two columns- powers of2 and the corresponding products:

1 v 243v

2 486

4 972

8v I944v

I6v 3888v

Since 25 = 16 + 8 + I,

243 X 25 = 3888 + 1944 + 243 = 6075,

as you can see by applying the distributive property. We shall now prove an interesting theorem about the Fibonacci sequence.

THEOREM II

The Fibonacci sequence of numbers, where a 11 = fu (n > 1 ), is complete.

To discover a method of proof, let us look at the following table:

We observe that we can write:

I = F 1 = F2

2 = Fa= F2 + F1

3 = F4 = Fa+ F2

Fs 5 6 7

6 = F5 + F2 = F-1 + F3 + F2

7 = F.5 + F 3 = F -4 + F;{ + F '2 + F 1

8 = F6 = Fs + F4 = F5 + F'J + F2

Thus, each positive integer from I through 8 can be represented in at least two ways as a sum of Fibonacci numbers where each is used at most once in any representation.

Representation Theorems · 71

Proof of Theorem II. We know from (1 1) (page 52) that

Fn - I = F 1 + F 2 + F 3 + · · · + F n _ 2,

and we observe in the list on page 70 that for 3 < n < 6, each integer

m = 1, 2, 3, ... , Fn - I

can be represented as a sum of some or all of the Fibonacci numbers F 1. F 2 ,

F:>., ... , Fn-2· That is, for n = 3, Fn = F3 = 2, Fn - I = 2 - I = 1, Fn-'2 = F 1 = I, and we have

for n = 4, Fn - F4 - 3, Fn - I - 3 - I = 2, Fn-2 = F 2 - I, and we have

1 = Ft, 2 = F2 + Ft;

and so on. We shall use this as the basis of induction (recaiJ page 54). For the second part of the proof, we assume that every integer m = I,

2, 3, .. . , Fk - I, k > 3, is representable using the Fibonacci numbers F1, F 2 •... , Fk_ 2 • We must prove that every integer m = 1, 2, 3, ... , Fk+ 1 - I is representable using F h F 2 , ... , Fk- h k > 3. If we add FJ.:_ 1 to each of the given representations, we shall have representations for

... '

where Fk + F~.:_ 1 - I = Fk+ 1 - l. We now have representations for the two sequences of consecutive positive integers

I, 2, 3, .. . , F~.: - 1

and

... '

Are there any omissions between Fk - 1 and I + Fk-I? No, since fork = 3,

F~.: - I = F 3 - 1 = l

fork = 4,

Fk - J = F 4 - I = 2

and 1 + Fk-t = I + F2 = 2;

and I+ Fk-t- 1 + F3- 3;

and fork > 5, since F~.: - Fk-I > 2, we have

and so there is an overlap. In any case, every integer m = l, 2, 3, ... , Fk+ 1 - I, k > 3, is representable using F 1, F 2 , ••• , Fk- 17 which is what we set out to prove.

Thus, the proof is complete by mathematical induction.


THEOREM Ill

The fibonacci number sequence, where On - Fn (n > 1 ), with an arbitrary Fn missing is complete.

Observe, for example, that ifF 4 is omitted from the display on page 70, every integer from I through 8 can still be represented by a sum of other Fibonacci numbers.

Proof From the proof of the previous theorem we note that we can properly represent any number m = 1, 2, 3, ... , Fn+ I - I by using only the Fibonacci numbers Fh F 2 , F 3 , •.• , Fn-I, that is, without using Fn. Then Fn+ 1 can represent itself, and when we add F11 + 1 to the representations for m = I, 2, 3, ... , Fn+.1 - I, we have representations form = I, 2, 3, ... , 2Fn+ 1 - I. Since 2Fn+I - I > F11 +2 - I, we can easily proceed over the trouble spot.

THEOREM IV

The Fibonacci sequence of numbers, where a,t = F11 (n > 1 ), with any two arbitrary fibonacci numbers Fp and F,t missing is incomplete.

Proof Since

if Fp < Fk is missing, then

But with Fn > Fp also missing, we can reach

F1 + F2 + · · · + Fp - 1 + F11+t + · · · + Fu-i

= Fn+I- F11 - I< F"+l- I.

Since FP > I, there is a number Fn+ I - l without a proper representation. This happened because even if we used all of those ovailable munbers less than F11 at once, we cannot reach Fu+ 1 - l and any other F ibonacci numbers which are available are too large, since the Fibonacci numbers are a strictly increasing sequence for n > 2.

For example, suppose that we try to find a representation of 20 without using F 5 = 5 and F 7 = 13. Those available are then F 1 , F 2 , F 3 , Ft. F 6 .

The maximum attainable is 1 + l + 2 + 3 + 8 = 15. The next available Fibonacci number is 21 so that we cannot find representations for 16, 17,

1 18, 19, and 20.


Vv'e shall now consider the corresponding properties for the Lucas seq uence of numbers. First , let us look at the following table:

- Lt

2- L 0

L 0 Lt L 2 L 3 L 4

2 1 3 4 5 6 7 8

3 = Lz = L1 + Lo 4 = L 3 = L 2 + L 1

5 = La + L t = Lz + Lo

6 = La + Lo = Lz + L 1 + Lo

7 = L 4 = L :J + L 2 = La + L 1 + L o

8 = L4 + Lt = La + Lz + Lt

\Ve observe that each positive integer from 3 through 8 can be represented in at least two ways as a sum of Lucas numbers where each is used at most

. once in any representa tion .

THEOREM V

The lucas number sequence, where On - l 11 _ 1 (n > 1 ), ts complete.

We know from (J 2 ) (page 54) that

Lt + Lz + L3 + · · · + Ln = Ln+2- 3,

and so Lo + L1 + Lz + L3 + · · · + Ln = Ln+2 - l.

Since a 1 - L 0 = 2, a2 = Lt = 1, aa = L 2 = 3, ... , an+ 2 = an+ I + G11 ,

the sequence an is a generalized Fibonacci sequence whose sum is

Ot + Oz + 03 + ... + Gn = Gn+2 - 1.

Thus. this theorem can be proved by mathematical induction as Theorem II was proved.

Is the Lucas num ber sequence complete if any single term is missing? Clearly. withou t a 1 = L 0 we have no representation for 2, and without a 2 = L 1 we have no represen tation for L It is as yet an unanswered question as to which Lucas numbers can be left out without destroying completeness. For example, we can see that L3 - 4 can be left out:

I = l

2 = 2

3 = 3

4=3+1

5 = 3 + 2

6 = 3+2+1

7 = 7

8 = 7+ 1

9 = 7 +2

10 = 7 + 3

11 = 11

12 = 11+1

1t would seem that if any Lucas number Lk, k > I, is omitted, the resulting sequence is sti ll complete. One would expect that if any two Lucas numbers were missing, the resulting sequence would be incomplete.


We now ask: I( each integer m is to be represented by the least number of Fibonacci numbers what are the condit~ons for this minimal representation?

Certainly, we would never need both F 1 and F 2 in any minimal representation, and so we choose to use F 2 as the I.

If, for a given integer m, the Fibonacci numbers in any representation are arranged in order of size, let us direct our attention to the largest, say, Fn. Clearly, if the representation has more than one Fibonacci number, then it cannot contain both Fn and Fn_ 1 and be a minimal representation because we could replace Fn + Fn_ 1 by Fn+ 1 and thereby reduce the number used.

Now, if we have Fn but not Fn-h then we could have Fn_ 2 but not both Fn_ 2 and Fn-3 because then we could replace Fn-2 + Fn-3 by Fn-1 and next F11 _ 1 + Fn by Fn+ ~, thereby getting a double reduction. We can see, therefore, that any minimal representation cannot have two adjacent Fibonacci nun1bers.

These restdctions are precisely the conditions imposed by the following interesting theorem, which we state without proof.

THEOREM VI (E. Zeckendorf)*

Each positive integer m can be represented as the sum of distinct numbers in the sequence defined by a 11 = f 11 +1 (n > 1) using no two consecutive fibonacci numbers, and such a representation is unique.

From this we see that each positive integer has a minimal representation as described above, and such a representation is unique.

If in a representation of a positive integer musing the terms of the sequence

1, 2, 3, 5, 8, ... , F11 + 1, ...

we desire to use the maximum number of these Fibonacci numbers, thus obtaining a maximal representation, we should replace Fk by F~.:_ 1 + F~.:_ 2 whenever possible (avoiding repetitions, of course). This process results in the conditions described in the following theorem, which we state without proof.

THEOREM Vllf

Each positive integer m can be represented as the sum of distinct numbers in the sequence defined by a, = f 11 + 1 (n > 1) with the condition that whenever Fk (k > 4) is used, at least one of each pair F,1 , F,1 _ 1 (3 < q < k) must be used, and such a representation is unique.

*John L. Brown, Jr., "Zeckendorf's Theorem and Some Applications," Tlu' Fibunacci Quarterly, Vol. 2, No. 3 (October, 1964), pages 163-168.

t John L. Brown, Jr., "A New Characteristic of the Fibonacci Numbers," The Fibunacci Quarterly, Vol. 3, No. I (February, 1965), pages 1-8.


From Theorem VII we see that each positive integer has a maximal representation as described on page 74, and such a representation is unique.

It turns out that all positive integers have both a unique minimum and a unique maximum representation (while some have only one representation, which satisfies both conditions) from the sequence:

F2 F3 F4 F5 F6 F7 F8

I 2 3 5 8 13 21

We have the following representations (using each term in the sequence at most once) for 1-21 :

I =

2 =

Minimal

1 = F4

4= 5 = F5

6 = Fs + Fz 7= 8 = F6

9 = F6 + Fz

10 = F6 + F3

II = F6 + F4 I2 =

13 = F1

14 = F1 + Fz

I5 = F1 + F3

I6 = F1 + F4

17 = F7 + F4 + Fz

18 = F1 + Fs

19 = F1 + F5 + Fz 20 =

21 = Fp,

Same Maximal

Fs + F3 + Fz

Fs + F4 + Fz

Fs + F4 + F3

Fs + F4 + F3 + F2

F6 + F4 + F3

F 6 + F 4 + F 3 + F_2

F6 + Fs + F3

F6 + Fs + F3 + Fz

F6 + Fs + F4 + Fz

F6 + Fs + F4 + F3

F6 + Fs + F4 + F3 + Fz

Notice that each integer Fk - I (k > 3) has a single representation satisfying both minimal and maximal conditions, and these are the only positice integers for lrhich this is true.*

* David A. Klarner, "Representations of Nasa Sum of Distinct Elements from Special Sequences," Tire Fibonacci Quarterly, Vol. 4, No. 4 (December, 1966), pages 289- 306.

76 · Fibonacci and Lucas Numbers_

A standard puzzle problem is to determine the num~er of separate weights needed to weigh any given integral number of pounds, supposing that the thing to be weighed is placed on one side and the weights are to be placed on the other side of the balance system. You can see that the solution is related to the idea of completeness of a sequence of positive integers as defined at the beginning of this section.

For example, any integral number of pounds from l to 31 can be weighed with one l-pound weight, one 2-pound weight, one 4-pound weight, one 8-pound weight, and one 16-pound weight, because of Theorem I.

Now suppose, as a fairy tale, that Fibonacci was a professional weigher with weights F 1, F2 , F3 , F4 , ••• , F;u ... pounds, traveling from place to place weighing things for people. He had to be able to weigh any positive integral number of pounds.

Theorem II tells us that Fibonacci could always do his job if he had all of his weights with him.

Theorem III tells us that Fibonacci could still do his job if he somehow lost one of his Fibonacci weights.

Theorem IV tells us that Fibonacci could not do his job if he somehow lost any two of his Fibonacci weights.

Theorem VI (Zeckendorf) states that if Fibonacci did not use his F 1 weight and lined up the rest of his weights in order of size, then any job he might be given would have just one solution if he did not choose any two weights which were adjacent in the lineup.

Theorem VII states that if Fibonacci did. not use his F 1 weight and lined up the rest of his weights in order of size, then any job he might be given would have just one solution if whenever he used his Fk weight (k > 4), he also used at least one of each pair of weights, Fq and Fq-l (3 < q < k ).

Theorems for Lucas numbers corresponding to Theorems VI and VII on page 74 are the following.

THEOREM VIII

Each positive integer m can be represented as the sum of distinct numbers in the sequence On = Ln _ 1 (n > 1) with the conditions that no two consecutive lucas numbers are used in the same representation and that L0

and L2 are not both used in the same representation, and such a repre sentation is unique.


THEOREM IX

Each positive integer m car. be represented as the sum of distinct numbe rs in the sequence 0 11 = L11 _ 1 (n > 1) with the conditions that whenever

Lk (k > 2) is used, at least one of each pair Lq, Lq-l {1 < q < k) must be used and L 1 and L3 are not both to be used unless L0 or Lz is also in the same representation, and each such representation is unique.

From the sequence

Lr

I

we have the following representations (using each term in the sequence at most once) for 1- 18:

Theorem VIII

I

2 =

3 - Lz

4 = La

5 - La+ Lt 6 =

7- L4

8 = L4 + Lt

9 = L4 + Lo

10 = L 4 + Lz

11 = L 5

12 = L 5 +Lt

13 = L 5 + Lo

14 = L 5 + Lz

15 = L 5 + L3

16 = L 5 + L3 + Lr

17 - L 5 + L3 + Lo

18 = L6

Same Theorem IX

L3 + L1 + Lo

L3 + Lz + Lt

L3 + Lz + Lo

La + Lz + L1 + Lo

L4 + Lz + Lt

L4 + L2 + Lo

L4 +La+ Lo

L4 + La + Lt + Lo

L4 + La + Lz + L 1

L4 + L3 + Lz + Lo

L4 + La + L2 + Lt + Lo

Ls + La + Lt + Lo

Notice that the representation according to Theorem VIII is the minimal representation and the representation according to Theorem IX is the maximal representation.


EX ER C I S ES

1. Find a representation of each integer 1, 2, 3, .. . , 20 using Fibonacci numbers with distinct subscripts, including F1 = 1 but omitting F5 = 5.

2. Find the first integer not representable in terms of Fibonacci numbers with distinct subscripts if you are denied use of Fibonacci numbers F 4 = 3 and FG = 8.

3. Express 27 as the sum of three distinct Fibonacci numbers. In how many ways can you do this?

4. Express 1966 in Zeckendorf form. The available Fibonacci numbers are: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, and 1597.

5. Find the minimal and maximal representations of 32 using distinct terms of the sequence F2, F3, ....

6. Find the minimal and maximal representations of 32 using distinct terms of the sequence L1, Lo, L2, L3, .. ..

}3 • Fibonacci Numbers in Nature

Fibonacci numbers appear in nature in a number of unexpected ways. \Vhat do you notice about the number of petals (or petal-like parts) in

the nowers pictured in Figure 20?

E:"CIIJ\NTER"S NIGHTSHADE

HEPATICA

TRILLIUM

BLOODROOT

Figure 20

RLUET WILD ROSE

COSMOS

The enchanter's nightshade has 2 petals (even though they are deeply cleft and appear to be four). Thus, the flowers above have 2, 3, 5, or 8 (Fibonacci numbers), or 4 (a Lucas number), or 6 petals or petal-like parts. Others with 5 petals or parts are buttercups and columbine, for example. A day lily blossom has 2 sets of 3 petal-like parts, alternating. An iris blossom has 3 standards (vertical parts) alternating with 3 falls. It is interesting to observe that so many such numbers occurring in nature are Fibonacci numbers.

79


The numbers of petai-like parts of a flower, such as an aster, cosmos, daisy, or gaillardia, in the composite family is consistently a Fibonacci number or very close to one. A flower of a single variety of gaillardia may have 13 of these parts. Frank Land* reports finding 21, 34, 55, and 89 "petals" on daisies and other members of the composite family.

You can also find Fibonacci numbers in the arrangement of leaves (or twigs) on a stem. Select one leaf as a starting point and then count up the stem until you reach a leaf that is directly above your starting point. The number of leaves is usually a Fibonacci number. Then suppose that a string were wound around the stem following the leaves. The number of turns taken around the stem between the starting point and the leaf directly above it is also usually a Fibonacci number. (The turns may be clockwise or counterclockwfse.) The result is often stated as a ratio:

number of turns number of leaves

Some simplified examples are pictured in Figure 2 I:

ELM

---~ / leaf 2 \\\1

kaf 0

I Turn - Ratio +

4

2)r(3 0 I 5

2 Turns - Ratio ~

Figure 21

PEAR · .... ~.

~ ....... -~ ,.. . · · leaf 8

7~t, 6

5~ I g

kaf

3 Turns - Ratiof

* Frank Land, The Language of Mathematics (London: John 1\1 urray, 1960), Chapkr 13.

Fibonacci Numbers in Nature · 81

The beech tree has a ratio of!. and the ratio for the pussy willow is -fh·! Such arrangement of leaves on a stem is called phyllotaxis (from Greek phyl!on meaning leaf and taxis meaning arrangement).

Perhaps the best known examples of the appearance of Fibonacci numbers (and occasionally Lucas numbers) in nature are the numbers of spirals in the seed patterns of sunflowers, the scale patterns of pine cones, and so on. A pine cone is pictured in Figure 22, and the two sets of spirals are pictured in the diagram beside it.

The seed-bearing scales of a pine cone grow outward in a spiral pattern from the point where it is attached to the branch.

Figzire 22

13 strips spiral to the left. 8 strips spiral to the right.

The author has raised a sunflower that had 89 spirals to the right and 55 spirals to the left and also one, eighteen inches across, that had 144 spirals to the right, 89 spirals to the left, and 55 shallow spirals to the right. _An excellent photograph of the head of a daisy is shown ori page 93 of the volume on Mathematics in the Life Science Library.* The diagram accompanying it shows clearly its 34 spirals to the left and 21 to the right.

A particularly interesting specimen of sunflower was given by the author to Brother Alfred Brousseau of St. Mary's College, California. After he had patiently counted the spirals, he found 123 spirals to the right, 76 spirals to the left, and a shallow set of 47 spirals to the right- a Lucas number sunflower! Brother Alfred Brousseau has also collected and classified by their Fibonacci spiral patterns cones from all but one of the twenty species of California pines. He reported his results in an article, "On the Trail of the California Pine,', in The Fibonacci Quarterly, Vol. 6, No. l (February, 1968), pp. 69- 76.

* David Bergamini· and the Editors of Life, Mathematics, Life Science Library (New York : Time Incorporated, 1963).


Figure 23

The spiral curves described on page 81 are called parastichies (from Greek para meaning beside and sticlws meaning row). Parastichies on pineapples are especially interesting. Since pineapple scales a re roughly hexagonal in shape, three sets of spirals can be fou nd, as pictured in Figure 23. Parastichies on pineapples are described on pages 21-22 of Mmhematical Diversions by J. A. H. Hunter and JosephS. Madachy (Princeton, N.J.: D. Van Nostrand Co. , Inc., 1963).

Other references for discussions of Fibonacci numbers in nature are:

E. J. Karchmar, "Phyllotaxis," The Fibonacci Quarterly, Vol. 3, No. 1 (February, 1965), pp. 64-66.

Sr. Mary de Sales McNabb, "Phyllotaxis," The Fibo11acci Quarterly, Vol. I , No. 4 (December, 1963), pp. 57-60.

A List of the First 100 Fibonacci Nun1bers

and the First 100 Lucas Numbers

Ft I Lt l

F"! I L:!. 3 F:~ 2 La 4 F, 3 L~ 7 F., 5 Ls 11 F,; 8 L6 · 18 F, 13 L1 29 Fs 21 Ls 47 Fn 34 L9 76 Fto 55 Lto 123

Ftt 89 Ltt 199 Ft"! 144 Ll2 322 F,a 233 L1a 521 FH 377 Lt., 843 Ft!i 610 L,s 1364

Ftr. 987 Lt6 2207 Ft7 1597 Ll7 3571 Ftf! 2584 Lts 5778 Ftn 4181 Lt9 9349 F2o 6765 Lzo 15127

Fzt 10946 L21 24476 F2z 17711 Lzz 39603 F'!.a 28657 L2a 64079 F~, 46368 LH 103682 F~s 75025 L?..s 167761

-F?.r. 121393 Lz6 271443 F:!., 196418 L21 439204 Fts 317811 . L2s 710647 F2!l 514229 L29 1149851 Fao 832040 L3o 1860498

Fat 1346269 Lat 3010349 Fa"! 2178309 La2 4870847 Faa 3524578 La a 7881196 Fa1 5702887 Lat 12752043 Fas 9227465 Las 20633239 Far. 14930352 LaG 33385282 Fa; 24157817 La1 54018521 Fas 39088169 Las 87403803 FaP 63245986 Lan 141422324 F.,o 102334155 L~o 228826127

FH 165580141 LH 370248451 F.,z 267914296 L.,2 5990.74578 F., a 433494437 L.,a 969323029 FH 701408733 LH 1568397607 F.,s 1134903170 L~s 2537720636

83


F .. 6 1836311903 L~6 4106118243 F41 2971215073 Ln 6643838879

F.s 4807526976 L~s I 0749957122 F.,g 7778742049 LHI I 7393796001 Fso 1 2586269025 Lso 2 8143753123

Fs1 2 0365011074 Ls1 4 5537549124 Fs2 3 2951280099 Ls2 7 3681302247 Fsa 5 3316291173 Lsa II 9218851371 Fs-t 8 6267571272 Ls~ 19 2900153618

. Fss 13 9583862445 Lss 31 2119004989

Fs6 22 5851433717 Ls6 50 5019158607 Fs1 36 5435296162 Ls7 81 7138163596

F ss 59 1286729879 Lss 132 2157322203

Fs9 95 6722026041 Lsv 213 9295485799

F6o 154 8008755920 L6o 346 .1452808002

F61 250 4730781961 L6t 560 0748293801 F62 405 2739537881 L62 906 2201101803 F6a 655 7470319842 L6a 1466 2949395604

F64 1061 0209857723 L6-t 2372 5150497407 F6s 1716 7680177565 L6s 3838 8099893011 F66 2777 7890035288 L66 6211 3250390418 F61 4494 5570212853 L6, 10050 1350283429 F6s 7272 3460248141 L(j';l, 16261 4600673847 F69 11766 9030460994 L(j'J 26311 5950957276 F1o 19039 2490709135 L1o 42573 0551631123

Fn 30806 1521170129 L;1 68884 6502588399 F12 49845 4011879264 L1'.! 111457 7054219522 F1a 80651 5533049393 L73 180342 3556807921 F1-1 130496 9544928657 LH 291800 061 1027443 F1s 211148 5077978050 L1s 472142 4167835364 F16 341645 4622906707 L16 763942 4778862807 F11 552793 9700884757 Ln 1236084 8946698171 F78 894439 4323791464 L,~> 2000027 3725560978 F19 1447233 4024676221 L1v 3236112 2672259149 Fso 2341672 8348467685 Lso 5236139 6397820127

Fs1 3788906 2373 143906 L'dt 8472251 9070079276 Fsz 6130579 0721611591 L'd'!. 13708391 5467899403 Fsa 9919485 3094755497 Ls:J 22180643 4537978679 Fs-t 16050064 3816367088 Ls4 35889035 0005878082 Fss 25969549 6911122585 L'ds 58069678 4543856761 Fs6 42019614 0727489673 L'du 93958713 45497348~3 Fs7 67989163 7638612258 La, 152028391 9093591604 Fss 110008777 8366101931 LBs 245987105 3643326~7 Fsu 177997941 6004714189 L8'J 398015497 2736918051 F9o 288006719 4370816120 Luo 644002602 6380244498

F9t 466004661 0375530309 L!tt 1042018099 9117162549 Fllz 754011380 4746346429 Lu~ 1686020702 54974070~7 Foa 1220016041 5121876738 LJ3 2728038802 4614569596 F94 1974027421 9868223167 L!H 4414059505 0111976643 Fgs 3194043463 4990099905 Lus 7142098307 4726546239 F96 5168070885 4858323072 Luu 1556157812 4838522882 F91 8362114348 9848422977 L<J7 I 8698256119 9565069121 Fgs 1 3530185234 4706746049 Lu'r!. 3 0254413932 4403592003 Fog 2 1892299583 4555169026 Luu 4 8952670052 3968661124 F1oo 3 5422484817 9261915075 L too 7 9207083984 8372253127

Partial Solutions Section 2

1-2. See list on page 83. 3. Several will be developed later in the text. 4. I, 4, 5, 9, 14, ... , 7375, 11,933, 19,308

Section 3

1. Using the quadratic formula with a = 1, b = -1, and c = -1, you have

1 ± '\It - 4(1 )(- I ) I ± VS 1 + Vs 1 - VS x= = ·a = •{3= · 2 2 ' 2 2

2. a..:_ 1 + ;·236 · 1_.618; {3 ..:_ 1 - } ·236 ..:_ - .61 8

1 + \IS I - "Is 3. a. a + {3 = --2- + --2- = 1

1 +\IS t -v's .r b. a - {3 = --2- - 2 = v 5

'· af3 ~ c +2 Vs) ('- 2 V~ ~ (1)2 -4 (V5)2 - I 4 5 ~ -I

4 . I + ! = l + 1 _ = l + 2(1 _- vS) _ _ I + Vs = a

a 1 + "15 0 + v5)(l - V5) 2 2

5. a. a 3 = a(a 2) = a(a + l) = a 2 +a = (a + 1) +a = 2a + 1 b. a 4 = a(a 3) = a(2a + 1) = 2a 2 + a = 2(a + 1) + a = 3a + 2 c. a 5 = a(a 4 ) = a(3a + 2) = 3a 2 + 2a = 3(a + l) + 2a = 5a + 3 Note: It can be proved by mathematical induction (reviewed on page 54) that,

in general, a" = F,a + Fn-I, n > l.

.l 1 t - 1 (2 - vs) 6. a· - 3 = 2a + 1 - 2 1 = 2 + V 5 - _ a · a+ 2+VS 2-V5

= 2 + vs + 2 - v5 = 4 = L:~ 4 -4

S. {3 -1 a -a 7. mce a = -1, a = -{3· - ---' vs

Section 4

1. In any triangle, we must have a + b > c, b + c > a, c + a > b. For any three consecutive Fibonacci numbers, Fn + Fn+1 = Fn+2, and so there can be no triangle with sides having measures Fn, F"+ 1, Fn+2· In general, consider Fibonacci numbers Fp, Fq, Fr, where Fp < Fq-1 and Fq+ 1 < Fr. Since F,_I + F4 = Fq+l and Fp < Fq-1, we have FP + Fq < Fq+1, and since F,+1 s Fr, we have Fp + Fq < Fr. Therefore, there can be no triangle with sides having measures F1,, Fq, and Fr.

2. The measures of the sides of the triangles must be (page 17) a, ra, r2a and ra, r 2a, r=1a with r -=;£ l (page 21). Therefore, it is impossible to have a = ra or a = r 2a or ra = r 2a, and so on, and so neither triangle can be isosceles.

85


3 . a . y = (x - a)(x - b), a < b If a < x < b, then x - a > 0 and x - b < 0, and so y < 0. If x < a, then x - a ~ 0 and x - b < 0, and so y > 0. If x > b, then x- a> 0 and x- b > 0, and soy > 0.

b. If a= b, theny = (x- a)2 ; (x- a) 2 > 0 for all x. 4. a. r2 - r - 1 = 0 has roots a and {:3, {:3 < a. Thus, r2 - r - 1

(r - a)(r - {:3), and from Ex. 3a, r2 - r - I < 0 for {:3 < r < a. b. r 2 + r - 1 = 0 has roots -a and -{:3, -a < -{:3. Thus, r 2 + r - I

[r - ( -a)][r - ( -{:3)], and from Ex. 3a, r 2 + r - I > 0 when r < -a (both factors negative) or r > -{:3 (both factors positive). (Of course, in the application in Prob. 4, r > 0.)

S. R v'5 - 1 d 1 + v'5 R s· R R c. mce -fJ = 2 an a = 2 , -fJ < a. mce fJ < -fJ < a,

{r: {:3 < r < a} n {r: r < -a or r > -{J} = {r: -{:3 < r < a}. . . . . 1 1

Smce-afj = -1, we have -{:3 = -, and- < r < a. - a a

S. From Ex. 3b, (r - 1)2 > 0 for all r. Thus, r"L - 2r + I > 0, or r 2 + I > 2r, for all r. If r > 0, 2r > r, and so r 2 + I > r for r > 0.

<> a a 6 . If r > I, a < ar < ar- and -:J < - <a. The sides can be paired by measures:

r- r a a

. a a r2 r a a~- ar~- ar2 ~a - = - = -- = - • the ratio of similarity.

r 2 ' r ' · a ar ar2 r2

. CG s + 1 s 7 Smce CG = DB = s + 1 and CD = GB = s - = -- = - = a and

. ' GB s 1 '

so DCG B is a Golden Rectangle. . .. AF FE s

8 . Smce ~AFB"' ~FEB, FB = EB = t = a .

2 . s s + 2sr

9. a. Smce - = a, 2 + 2 t s I

') 2 ') ~

s.2 + 2 (~) r2 1

s2 - + 1 t 2 -

')

a-+ 2a

a2 +I

t- - 2st s + 1- s- + 2s1 b. = - = 1 - a = {:3

52 + 12 s2 + 12 52 + 12

a(a + 2)

a+2

2 2 2 ') c. s + 4st - I = s + 1st _ C_~ 2st = a _ {:3 = VS

52 + 12 52 + 12 52 + ( 2

Section 5

2n R2n a -fJ

1. F2n = --- a-{:3

a -fJ " n (

11 R") ~~ (a + {:3 ) = F,.L,., n > 1

(a"- 1 - {:3" - I) + (a " + ! - J}" + ') 2. F n - 1 + Fn+l = R ----

a-fJ

= a.

a'1+l +an-I _ {J" + l _ J]" -1 a"+I _ (a{J)an- l + (af})J}" - 1 - (J"+l -

a-f) a -f)

= a"(a - {:3) + (J"(a - {J) = a " + {:3" = L,, since a f} = - 1. a - f)

Partial Solutions · 87

a-{3

-{3n+lan + an+l/3" _ {3n+lan + an+l{3n 2(an+l/3n _ a"/3n+l) = -

a-{3 a-{3

= 2(a 11/3")(a - /3) = 2( -l )n a-/3

7 F,±2._Ln+t=Fn+tLn-Ln+tFn=2(-l)"b E 6 dl . F L F L F y x. an . n n n n 2n

8. If n < 0, then -n > 0. For -n > 0, F -<-n> = ( -l)-n+lp _,,

Fn ( - 1 )n+l Fn n+l or F_, = (-l)-"+1 = (-1)2 =(-I) Fn.

9. If n < 0, then -n > 0. For -n > 0, L-<-n> = ( -1)-"L-n,

or L_n = ( -~}-n = ( - ·1 )"L,..

10. From Ex. 4, SF~ = Lz,. - 2( -1) 11 • From Ex. S, L2n = L~ - 2( -1)". Thus, SF~ = L~ - 4( -1)".

11. L,. = F,.+ 1 + Fn-l = (Fn+2 - F,.) + (F, - Fn-2) = Fn+2 - Fn-2

Section 6

12 12

1. F 12 • a _. log a _ · 12(0.20898) - 0.3494 · 2.S078 - 0.3494 .:_ 2.l S84. vs vs

Thus, F12 = 144. 2. L 12 · a 12• loga 12 .:_ 2.S078. Thus, L12 = 322.

:H

3. F:l4 • a _ . S70 X 104

vs IfF = IS 7 h F = [1S97 + 1 + V127S204S] = [1S98 + 3S71]

5. n 9 ' t en n+l 2 2

~ [ 51269] ~ [2584.5] ~ 2584.

If L _ 2207 h L _ [2207 + 1 + \.1243S424S] _ [2208 + 4935] 6. n - 't en n+l - 2 - 2

~ [ 71; 3] ~ [3571.5] ~ 3571.


Section 7

1. F1 = 13; Fu = 13(29); F2 1 = 13(842); F2s = 13(24,447) 2. F10 = 55; F2o = 55(123); F3o ~ 55(15,128); F.w = 55(1,860,621) 3. F24 = 46,368 = 2(23,184) = 3(15,456) = 8(5796) = 2I(2208) = 144(322) 4. Fao = 832,040 = 2(416,020) = 5(166,408) = 8(104,005)

= 55(15,128) = 610(1364) 5. L4 = 7; Fs = 7(3); Ftu = 7(141) 6. L1 = 29; Fu = 29(13); F28 = 29(10,959) 7. L4 = 7; L12 = 7(46); L2o = 7(2161) 8. Ls = 11; Lts = 11(124); L2s = 11(15,251) 9. F12 = 144 = 2(76 - 4) = 2(4)(18) = 3(47 + 1) = 8(18)

10. Fts = 2584 = 2(1364 - 76 + 4) = 8(322 + I) = 34(76) 11. (Ftu, F24) = Fo6. 24> = Fs = 21 12. (F24, Fau) = F<24, 36> = F12 = 144

Section 8

1. Since 13 is F1, the entry point is 7. The remainders on dividing by 13 are Ro = 0, Rt = 1, ... , R1 = 0, Rs = 8, ... , Rt4 = 0, Rts = 12, ... , R21 = O,R22 = 5, ... ,R28 = O,R2u = 1, ... , andsoKta = 28.

2. The remainders on dividing by 13 are R o = 2, R 1 = 1, ... , R u1 = 2, R2o = 8, ... , R2a = 2, R24 = 7, ... , R2s = 2, R29 = 1, ... , and so K1a = 28. No remainder is 0, and so 13 does not divide any Lucas number.

3. The remainders on dividing by 10 are Ro = 0, R 1 = 1, . .. , R 15 = 0, R 1 G = 7, ... , Rao = 0, Rat= 9, . .. , R4s = 0, R4u = 3, ... , Roo= 0, Rut= 1, ... , and so Kto = 60. The entry point is 15.

4. The remainders on dividing by 6 are Ro = 2, R 1 = 1, ... , Ru = 0, ... , R 15 = 2, R 1 u = 5, ... , RIB = 0, . .. , R21 = 2, R22 = 3, ... , R2-1 = 2, R2s = 1, ... , and so ku = 24. The entry point is 6.

5. The remainders on dividing by 10 are Ro = 2, R 1 = 1, .. . , R 12 = 2, R13 = 1, ... , and so Kto = 12. No remainder is 0, and so 10 has no entry point in the Lucas numbers. (Also, since no Lucas number is divisible by 5 (page 45), no Lucas number is divisible by 10.)

Section 9

From (A) on page 49:

1. With x = l and y = x, 2. With X = ] ' y = I' 3. With x= J,y = - I,

±(~)xi=(l+xf. i= O 1

± ('~) = 2n. i = O l

t ('~) (-I )1 = 0. i = O I

4" @ ~ O!(n 1~ 0)! ~ I; (:) ~ n!(n n~ n)! ~ I. (;;,) ~ m!(;~ m)!'

( n - I) ( n - I ) (n - 1)! (n - 1)! m + m - 1 = m!(n - I - m)! + (m - I )!(n - m)!

= (n - m)(n - 1 )! + m(n - I)! = n! = (n) . m!(n - m)! m!(n - m)! m

s. tot) F; ~ (~) Fo + G) F, + G) F2 + e) F;~ + (:) F,

= Fo + 4Fr + 6Fz + 4FJ + F4 = 21 = F'b

-'


6. f ('~) F.-+i = _ _!__[f ('~) ai+i- ± ('~) rl+i] i = 0 1 a {3 i =0 I i = 0 I

= 1 [c/(1 +a)" _ {3j(1 + {3)" ] = __ 1 _ [a2n+i _ {32n+i] a-{3 a-{3

= Fz,.+i• since 1 +a = a 2 and I + {3 = (32.

7. f ('~) L ;+i = t ('~) ai+i + f ('~) {3i+i i=O I idl I _ i=O 1

Section /0

n n

1. L F; = Fn+2 - 1,11 > l 3. L F7 = FnFn+t,n > 1 i =l i = l

2 2 2 4. Fn+l = F,. + Fn-I; Fn+ l = Fn + 2FnFn-1 + Fn-1·

2 2 From (b), F,.Fr.-l = Ft + · · · + Fn - 1·

2 2 2 2 2 Thus, Fn+I = Fn + 3Fn-l + 2(Fn-2 + · · · + FI). The number of squares is 1 + 3 + 2(n - 2) = 2n.

n

7. L F2i-l = F2n, 11 > 1 i=l

10. Lt = L2- Lo La = L4 - L2 L5 = LG - L4

L2n-:~ = Lz,._z - L zn- t

Lzn-1 = Lzn - Lz11-Z

" L Lzi-1 = Lzn - 2 i=l

n

9. L F2.- = Fzn +l - 1, n > 1

11. Lz = L3 - Lt L4 = L5- L3 Lc, = L, - L5

Lz,._z = Lzn-1 - Lzn-:3 Lzn = Lzn+l - Lzn-1

n

L Lz.- = Lzn+l - 1 i=l


12. Ln = Fn-1 + Fn+I ([g); Ln- 1 = Fn-2 + F"; Ln+1 = Fn + Fu+2· 1;(Ln-l + Ln+l) = 1;(Fn-2 + 2Fn + Fn+2)

= 1;[(Fn - Fn-1) + 2Fn + (F,. + F,.+J)] = 1;(4Fn + Fn) = Fn

13. F2n = FnLn (h); Ln = F,._1 + Fn+I (Is). F2n = Fu(Fn-1 + Fn+1) = (F,.+l - Fn-I)(Fu+t + F,._J)

2 2 = Fn+l - Fn-l

2 2 14. F2n = Fn+1 - Fn-l (Iw).

? ? ? ?

F2n+I = F2<n+1> - F2n = F;;+2 - F,; - F;;+l + F;;_l 2 2 ? 2 2 2

= (Fn+1 + Fn) - Fn - F;;+t + (F,.+t - F,.) = Fn+t + Fn 1 1 1 . ? ? •)

15. h = <-'or -8' smce g- + r > s-. vs2 + 32 V82 ·

17. To prove: Hn+2 = qFn+1 + pFn H1 = p; H2 = q; Hn+2 = H,.+I + H,, n 2:: 1 n = 1: H a = H 2 + H 1 = q + p = qF 2 + pF 1

n = 2: H4 = Ha + H2 = qF2 + pF1 + q = qF:~ + pF2 Thus, we have a basis for induction. We assume (inductive hypothesis): P(k - 1): Hk+I = qFk + pFk-l

and P(k): H"+2 = qF~.:+1 + pFk Adding, we have: P(k + 1): H~.-+a = qFk+2 + pF~.-+1 The proof is complete by mathematical induction.

Note: Observe the variation from the usual pattern of mathematical induction. The inductive basis has two (consecutive) validations, and the inductive hypothesis has two (consecutive) assumptions.

18. F<k+1>n = FknFn+l + Fkn-tFn To prove: Fn divides Fnk. k > 0. k = I: Fn divides Fn. Thus, we have a basis for induction. We assume (inductive hypothesis): P(p): F,. divides F,.P. But F<p+l>n = (Fpn)Fn+l + Fpu-I(Fu). Since Fn divides Fn and is assumed to divide F1,, F,. must divide F<p+ll"' and the proof is complete by mathematical induction.

Section 11

1. A + Z = (~ ~) + (g g) Z + A = (g g) + (~ ~) =

2 . A + B = (a b) + (e f) c d g lz

(a+ 0 b + 0) c+O d+O

( 0 +a 0 +b) O+c O+d

(a+ e c+g

b + l) d+h

_ (e + a I + b) _ (e f) (a - g+c lz+d - g h + c ~)

3. (A + B) + C = [ (~ ~) + (; -~;)] + (~ ~)

=A

= A

B+A

( (a + e) + i (e + /) + j) _ (a + (e + i) e + (f + j)) (c + g) + k (d + h) + I - c + (g + k) d + (h + /)

(~ !) + [ (; {) + (~ ~)] = A + (B + C)


4_ A+ (-A) = (a b) + (-a -b) = (a - a b- b) = (0 0) = z c d -c -d c - c d - d 0 0

5_ (sr)A = ((sr)a (sr)b) = (..\{ra) s(rb)) = ·s(rA) (sr)c (sr)d s(rc) s(rd)

6 _ -A= (-a -b)= ((-l)a (-l)b) = (-l)(A) - c - d (- I )c ( -l)d

7 • A! = (a b) (1 0) (a(l) + b(O) a(O) + b(l)) c d 0 • 1 c(l) + d(O) c(O) + d(l)

!A = (1 0) (a h) (I(a) + O(c) l(b) + O(d)) = 0 L c d O(a) + l(c) O(b) + l(d)

8. A(B + C) = (a b) (e + i J + j) c d g+k h+l

( a(e + i) + b(g + k) a(f + j) + b(h + /)) c(e + i) + d(g + k) c(f + j) + d(lz + I)

=A

=A

( (ae + bg) + (ai + bk) (of+ bh) + (aj + bl)) = AB + AC (ce + dg) + (ci + dk) (cf + dh) + (cj + dl)

Similarly, (B + C)A = BA + CA. 9. To prove: det (Q") = (-l)", n 2: I.

det Q = -1 ; det Q:? = det ((Q)(Q)) = det Q det Q = (-1)2

Thus, we have a basis for induction. We assume (inductive hypothesis): P(k): det Qk = ( -l)k We wish to prove P(k + 1): det Qk+I = (-l)k + I. det Qk+I = det ((Q)(Qk)) = det Q det Qk = (-1)(-l)k = (-l)k+I The proof is complete by mathematical induction.

10 Q"2 = (l I) (J 1) = (2 1) . Q +I= (2 l) = Q 2 . 1010 II' 11

Section 12

1. 1 = Fz 2 = F:1 3 = F4 4 = F.1 + F2 5 = F-1 + F:{

11 = Fr. + F4 12 = Fn + F4 + Fz 13 = F7 14 = F7 + Fz

6 = F-1 + F:l + Fz 15 = F1 + F3 16 = F, + F4

7 = F-t + F:l + F2 -t- Ft 8 = Fr.

17 = F7 + F.t + Fz 18 = F7 + F-t + F:1

9 = Fr.+ Fz 10 = Fu + F:1

2. 1 = F2 2 = F:1 3 = F:1 + Fz

19 = F1 + F4 + F3 + F2 20 = F 7 + F.t + F 3 + F 2 + F 1

4 = F:1 + Fz + Ft 7 = Fs + F3 5 = F;; 8 = Fs + F3 + F1 6 = Ft. + Fz 9 = Fs + F3 + Fz + Ft

There is no possible representation for 10 with F4 and FG missing. The next available one is F; = 13, which is too large, and 9 is the sum of all the smaller available Fibonacci numbers.

3. There is only one representation of 27 using distinct Fibonacci numbers: 27 = 21 + 5 + I. This may be expressed as 27 = Fs + Fs + Fz.

4. 1966 = 1597 + 233 + 89 + 34 + l3 = F17 + F1:1 + F11 + Fg + F1 5. 32 = 21 + 8 + 3 = Fp, + Fu + F4 (minimal, or Zeckendorf)

= lJ + 8 + 5 + 3 + 2 + 1 = F7 + Fu + Fs + F4 + F3 + F2 (max.) 6. 32 = 29 + 3 = L, + L 2 (minimal)

= 18 + 7 + 4 + 2 + 1 = L 6 + L4 +La + L 0 + Lt (maximal)

Index

Alpha, 9, 1 I See also Golden Section.

Arithmetic progression, 4, 49

Beta, 9 relation to alpha, 1 I

Binet form for Fibonacci numbers, II for Lucas numbers, 26

Binomial coefficients, 49, 50

Carroll, Lewis, 57 Completeness of a sequence, 69

Fibonacci, 10: 72 Lucas, 73

Determinant of a matrix, 64 Divisibility of Fibonacci numbers

by Fibonacci numbers, 37 by Lucas numbers, 40

Divisibility of Lucas numbers by ·Lucas numbers, 40

Entry point, 45

Factorial, 50 Fibonacci numbers, 5

Binet form for, I I how to find large, 3 I with negative subscripts, 28 in Pascal's triangle, 50 relation to Lucas numbers, 27 remainder properties of, 46 sum of, 6

Fibonacci quadratic equation, 11 characteristic equation of Q-matrix, 66

Fibonacci sequence, 5 generalized, 7

Flowers petal count, 79-80 spirals in, 8 1

Geometric progression, 49 Geometrical paradox, 57 Golden Rectangle, 12, 16 Golden Section, II

in polygons, 23 ratios which approximate, 28- 2tJ

Golden Triangle, 22 Greatest common divisor, 39 Greatest integer not greater than a

number, 30

Identities for Fibonacci numbers, 7, 52- 57, 59-60

for Lucas numbers, 54- 56, 59- 60

Incompleteness of a sequence, 72-73 Inequalities, propositions for, 30

Leaf arrangement, 80-81 Lucas numbers, 7

Binet form for, 26 how to find large, 32 with negative subscripts, 28 relation to Fibonacci numbers, 27 remainder properties, 46

Mathematical induction, 54-55 Matrix, two-by-two, 62 Maximal representation, 74-77 Minimal representation, 74-77

Parastichics, 82 Pascal's triangle, 48

rising diagonals in, and Fibonacci numbers, 50

Period of a number in a sequence, 45 Phyllotaxis. 81

Q-matrix, 65

Rabbit problem, 2-6 Rank of apparition, 45 Ratios of successive Fibonacci and Lucas

numbers, 28-29 Recurrence (recursive) formula, 4 Rectangle

in geometrical paradox, 57 Golden. 12. 16 removing triangles from, 14-16

Remainder properties, 46 Representation of integers as sums

of Fibonacci numbers, 70, 74, 75 of Lucas numbers, 73, 76-77

Residues, 46

Scaiar, 63 Semicircle, square inscribed in, 24 Summation notation . 41) Spiral count, 81 Square

in geometrical paradox, 57 inscribed in semicircle, 24

Triandc(s) witi~ just five parts congruent, 17- 19 Golden, 22 inscribed in a rectangle, 14-16

Zeckendorf form o f representation, 74 92

Documents

ffil~©©~ ~ffi)~ ~oo©~~ OOoornru~~~~ - Alyoops! and Lucas Numbers.pdf1 Columbia University Library, D. E. Smith Collection 81 American Museum of Natural History, photo by Franklyn

ffil~©©~ ~ffi)~ ~oo© OOoornru~~ - Alyoops! and Lucas Numbers.pdf1 Columbia University Library, D. E. Smith Collection 81 American Museum of Natural History, photo by Franklyn