1 From Fibonacci to Foxtrot:

2 Investigating Recursion Relations with Geometric Sequences

3

4

5 The Algebra Standard from the Principles and Standards for School Mathematics

6 (NCTM, 2000) states that all students should “Understand patterns, relations, and

7 functions.” In particular, to meet the grades 9 – 12 expectations, students should

8 “generalize patterns using explicitly defined and recursively defined functions.” The

9 Foxtrot comic strip shown below (Amend) provides a wonderful starting point to create

10 lessons that address the algebra standard.

11

12 In the first three panels of the cartoon, we see that Marcus has scored a

13 touchdown by identifying Jason’s sequence 0, 1, 1, 2, 3, 5, 8, 13… as the “Fibonacci

14 series” (though it would be preferable to replace “series” with “sequence,” since in

15 mathematics terminology a series designates the sum of the terms in a sequence). The

16 Fibonacci sequence is one of the most widely known in all of mathematics, recursively

=+ 17 defined by the recurrence relation FFFnnn++21. Thus, many readers of Foxtrot could 2

18 have emulated Marcus’ scoring success. However, there are at least two lingering

19 questions:

20 • Why did Jason begin his count with zero? It is more common to start the

== == 21 Fibonacci sequence with FF121, 1 instead of FF010, 1.

22 • How can Jason score a touchdown? In the last panel of the comic strip,

23 Marcus has challenged Jason with the sequence 3, 0, 2, 3, 2, 5, … . What

24 is this sequence?

25 The key to our investigation of these questions is the geometric sequence

234 = n 26 aarar,, , arar , ,K, which is defined explicitly by xn ar , n = 0, 1, 2, … .

= = 27 Alternatively, the sequence can be defined recursively by xnn+1 rx where x0 a . The

x 28 constant r = n+1 is the common ratio of the sequence. xn

29 We begin by first investigating a connection between the Fibonacci sequence and

30 geometric sequences. The techniques we develop provide us with a method to investigate

31 the mysterious sequence 3, 0, 2, 3, 2, 5, … .

32

33 IS THE FIBONACCI SEQUENCE A GEOMETRIC SEQUENCE?

34 The short answer to the question just raised is no. After all, FF10/ is not defined, and

== == 35 FF21/1/11 and FF32/2/12 are different ratios. Rather than being discouraged,

36 let’s examine several more ratios of successive Fibonacci numbers, as shown in the

37 following table.

n 0123 45678 F F/nF 0112 3 5 8 1321 38 n+1 n - 1.0 2.0 1.5 1.667 1.600 1.625 1.615 1.619 3

39 As n becomes larger, the Fibonacci sequence increasingly resembles a geometric

40 sequence with a common ratio of about 1.6. Since the early integer values of the

41 Fibonacci sequence seem to cause difficulty, suppose we not worry yet about the initial

= n 42 values of the sequence. Instead, we simply seek a geometric sequence xn ar that

=+ 43 satisfies the Fibonacci recurrence xnnn++21xx. That is, we want to determine values of

44 r for which arnnn++21=+ ar ar , since it is clear that an arbitrary value of the constant a is

45 allowed. Thus, we want values of r ≠ 0 for which rrrnnn++21=+. The terms in this

46 equation can be divided by the common factor r n , giving us the quadratic equation

47 rr2 =+1. Using the quadratic formula, we discover there are two roots r of the quadratic

15+ 15− 48 equation: p = and q = . As a decimal, p is about 1.618, and we suspect we 2 2

49 have identified the value we encountered in the numerical table above. For arbitrary

50 choices of the constants a and b, both of the geometric sequences apn and bqn solve the

=+ nnn++21=+ 51 Fibonacci recursion xnnn++21xx. Moreover, since ap ap ap and

52 bqnnn++21=+ bq bq , we can add these two equations to see that

53 ()()()apnn++22+=+++ bq ap nn ++ 11 bq ap nn bq .

=+nn 54 That is, for any choices of the constants a and b, xn ap bq solves the Fibonacci

=+ 55 recurrence relation xnnn++21xx.

56 To summarize our progress, we have proved the following result:

=+ =+nn 57 Theorem. The Fibonacci recursion formula xnnn++21xx, is solved by xn ap bq

15+ 15− 58 where p = , q = , and a and b are arbitrary constants. 2 2 4

15+ 59 The constant p ==1.6180 is the famous Golden Ratio. It was known 2 K

60 (though not by that name*) in ancient Greek mathematics, since it solved this question:

61 determine the point C on line segment AB so that AB/AC = AC/BC.

15− 62 The second solution q ==−0.6180 of the quadratic equation 2 K

63 rr2 −−=10 can be investigated by factoring the quadratic polynomial. That is, since

64 rr22−−=1( rprqr − )() − = −() pqrpq + + ,

65 then equating coefficients shows us that p +=q 1 and pq = –1. It’s interesting to see how

66 these relationships can be found without requiring the explicit formulas for p and q that

67 involve √5.

68

69 THE LUCAS AND FIBONACCI SEQUENCES

=+nn 70 A simple choice of constants for xn ap bq in the theorem above is a = 1 and

nn ⎛⎞⎛⎞15+− 15 71 b = 1. This gives us the solution Lpq=+=nn + . For example, n ⎜⎟⎜⎟ ⎝⎠⎝⎠22

=+=+=00 =+= 72 Lpq0 11 2 and Lpq1 1, and we are delighted that no 5 term appears.

=+ == 73 Moreover, since LLLnnn++21and LL012 and 1, the terms of the sequence are all

74 positive integers. We have rediscovered the Fibonacci-like sequence 2, 1, 3, 4, 7, 11, 18,

75 29, … named for Edouard Lucas (1842-1891). Though this sequence is less well known

* The entry for “Golden Ratio” in the MathWorld encylopedia (Weisstein) informs us that “The term "golden section" (in German, goldener Schnitt or der goldene Schnitt) seems to first have been used by Martin Ohm in the 1835 2nd edition of his textbook Die Reine Elementar-Mathematik (Livio 2002, p. 6). The first known use of this term in English is in James Sulley's 1875 article on aesthetics in the 9th edition of the Encyclopedia Britannica.” 5

76 than the Fibonacci sequence, the Lucas sequence also has many properties of interest (see

77 the concluding section of this article.)

78 To obtain the Fibonacci numbers, we need to see if it possible to choose the

=+nn 00+=+ 79 constants a and b so that Fapbqn . Since F0 = 0 and ap bq a b , we will

⎛⎞1515+− 80 want b = – a. Also, since F =1 and ap11+= bq a() p −= q a − = a 5 , 1 ⎜⎟ ⎝⎠22

1 81 we must choose a = . Thus, the nth Fibonacci number is given by the explicit formula 5

nn nn−+−⎡⎛⎞⎛⎞⎤ ==pq 115 − 15 82 Fn ⎢⎜⎟⎜⎟⎥ . 55⎢⎜⎟⎜⎟22⎥ ⎣⎝⎠⎝⎠⎦

83 This formula is known as Binet’s formula for the Fibonacci numbers, named after

84 Jacques Binet (1786-1865) who discovered it in 1843. However, the formula was

85 discovered first in 1718 by Abraham DeMoivre (1667-1754).

86 Since Binet’s formula gives F0 = 0, we now know why Jason started his list of

87 Fibonacci numbers with 0. The formula also explains why the Fibonacci sequence,

88 though not geometric, is increasingly close to one. Since q =−()1 5 / 2 ≈− 0.618 we see

89 that |qn| is increasingly small as n becomes large. Therefore, we have the approximate

1 90 equality Fp≈ n . This explains why the Fibonacci sequence is nearly a geometric n 5

91 series, as we noticed in the table of values computed earlier. If we let {x} denote the

n = ⎧⎫p ≥ 92 “round to the integer nearest to x” function, it is easy to check that Fn ⎨⎬ for all n ⎩⎭5

= n ≥ 93 0. Similarly, the Lucas numbers can be written as Lpn { } when n 1. 6

94 INVESTIGATING MARCUS’ SEQUENCE 3, 0, 2, 3, 2, 5, …

95 It seems reasonable to guess that the new sequence is related to the Fibonacci sequence.

96 Therefore, suppose that we add consecutive pairs of the sequence 3, 0, 2, 3, 2, 5. The first

97 three sums are 3 + 0 = 3, 0 + 2 = 2, and 2 + 3 = 5 We seem to be on the right track, since

98 we get the next three terms 3, 2, and 5 of the sequence. This suggests that the term xn+ 3 is

99 the sum of the two consecutive terms xn and xn + 1. That is, Marcus’ sequence is defined

100 by the recursion formula

=+ 101 xnnn++31xx.

= n 102 As before, we can search for a geometric sequence of the form xn ar that

103 solves the recurrence relation. As before, the constant a is arbitrary, but now we want r to

104 satisfy the equation rrrnnn++31=+. Dividing each term by the common factor r n , we get

105 the cubic equation rr3 =+1. We could turn to a computer algebra system or a graphing

106 calculator to solve this cubic and discover there is one real root u =1.32472 … and a pair

107 of complex conjugate roots v = –0.662359+0.56228 i and w = – 0.662359 – 0.56228 i.

108 However, let us just suppose we factor the cubic polynomial to get

109 rr3 −−=1( rurvrw − )()() − − .

110 By expanding the product of the three binomials on the right side, we get the equation

111 rr33−−=1 r −()() uvwruvvwuwruvw ++ 2 + + + − .

112 Since these two polynomials in the variable r are equal if and only if their coefficients are

113 equal, we obtain the equations

114 u + v + w = 0, uv + vw + uw = –1, and uvw = 1. (*) 7

115 As with our earlier investigation of the Fibonacci recursion, we know that

=+ =++nn n 116 xnnn++31xx is solved by xn au bv cw for any choice we make for the three

117 constants a, b, and c. Inspired by the choice that led us to the Lucas sequence, suppose

=++nn n 118 that we let a = b = c = 1, which gives us the sequence Pun v w. The first term is

=++00 0 =++= 119 then Pu0 v w111 3. This looks promising, since 3 is indeed the first term

=++ = 120 of Marcus’ sequence. The next term is Puvw1 0 , where again we have used

121 equation (*). Happily enough, 0 is the next term of Marcus’ sequence!

= 122 If we can show that P2 2 , we will have unraveled Marcus’ sequence. We have

=++22 2 123 Pu2 v w. Is this equal to 2? To find out, we again turn to the equations in (*),

124 where we see that

125 0222=++()u v w2 =++ u22 v w 2 + uv + vw + uw

126 Using (*) once again, we see that

=++=−++22 2 =−−= 127 P2 u v w2( uv vw uw ) 2( 1) 2.

128 Therefore, the terms 3, 0, 2, 3, 2, 5, 5, 7, 10, … of Marcus’ sequence can either be

129 defined recursively by

== = =+ 130 PPP0123, 0, 2,K , PPPnnn++ 31,

=++nn n ≈ ≈ ≈ 131 or can be given explicitly by Pun v w, where u 1.32, v –0.66+0.56 i, and w

132 – 0.66 – 0.56 i are the roots of the cubic equation rr3 =+1.

133 But what is the name of this sequence? Here Jason may want to use his well-

134 known Internet skills and access The On-Line Encyclopedia of Integer Sequences

135 (http://www.research.att.com/~njas/sequences/index.html) to find that the sequence is the

136 Perrin sequence, named for the French mathematician R. Perrin, who discussed the 8

137 sequence in a mathematical paper published in 1899 (although the sequence had already

138 been mentioned in 1878 by Lucas). Thus, Jason should yell out “Is it the Perrin

139 sequence?” to score a touchdown.

140 Of course, Jason might refer to the Perrin “series” instead of “sequence.” In the

141 next section, we will see that geometric sequences can also be examined by considering

142 their associated geometric series.

143

144 GEOMETRIC SERIES AND GENERATING FUNCTIONS

2 n 145 For t <1, the terms of the infinite geometric sequence 1,tt , ,KK , t , can be summed

1 146 to give the formula 1++tt2 + + tn + = . If we then set t = px, where p is the LL1− t

147 Golden Ratio, we find that

1 1 148 1++px p22 x ++ pnn x += for x < . LL1− px p

149 Similarly, setting t = qx, we also have

1 1 150 1++qx q22 x ++ qnn x += for x <=p . LL1− qx q

1 151 Since both series converge for x < , subtracting the series gives us p

11 152 ()(pqxp−+−222 qx ) ++−() pnnn qx += − . LL11−−pxqx

pnn− q 153 But if we recall from Binet’s formula that F = is the nth Fibonacci number, we n 5

n 154 see that the coefficient of the x term of the series is 5Fn . We can also simplify the right

155 side of the equation above as follows: 9

11 ()p − qx 5 x 5 x 156 −= = = , 11−−px qx()() 11 −− px qx 1 −++() p q x pqx22 1 −− x x

15+− 15 157 where we recall that pq== and satisfy pq−=5, pq += 1, . 22

158 and pq =−1Altogether, we see that

x 159 =+FFxFx +2 ++ Fxn +. 1−−xx2 01 2 LLn

x 160 Since the coefficients of the series expansion of the function fx()= are the 1−−x x2

161 Fibonacci numbers, f is called the generating function of the Fibonacci sequence.

162 If we had added rather than subtracted the two geometric series above, a similar

163 calculation shows that

− ()==++++++2 x 23 n 164 gx2 234 x x xLL Lxn 1−−x x

=+nn 165 is the generating function for the Lucas numbers Lpqn .

166 We can also obtain the generating function for the Perrin numbers by letting t be

167 successively replaced with ux, vx, and wx. When the three geometric series are added we

168 find that

111 169 302325++xx2345 + x + x + x ++ Pxn += + + . LLn 111−−−ux vx wx

170 The right side of this formula can be simplified by using the equations (*) found earlier.

171 We find that

11132−+++++()()u v w x uv ww uw x2 3− x2 172 ++ = = . 111−−−ux vx wx 1 −+++++−()() u v w x uv vw uw x2323 uvwx 1 −− x x

3− x2 173 That is, is the generating function for the Perrin sequence. 1−−x23x 10

174 NOTES ON THE FIBONACCI, LUCAS, AND PERRIN SEQUENCES

175 Each of the three sequences we have discussed—Fibonacci, Lucas, Perrin—is of

176 considerable mathematical interest. We’ll mention just a few items here, with the hope of

177 encouraging the reader to consult more extensive references. A particularly convenient

178 source of information is the online mathematical encyclopedia MathWorld (Weisstein).

179 For example, the entry for theFibonacci numbers informs us that

180 “A scrambled version 13, 3, 2, 21, 1, 1, 8, 5 (Sloane's A117540) of the

181 first eight Fibonacci numbers appears as one of the clues left by murdered

182 museum curator Jacque Saunière in D. Brown's novel The Da Vinci Code

183 (Brown 2003, pp. 43, 60-61, and 189-192). In the Season 1 episode

184 "Sabotage" (2005) of the television crime drama NUMB3RS, math genius

185 Charlie Eppes mentions that the Fibonacci numbers are found in the

186 structure of crystals and the spiral of galaxies and a nautilus shell.”

187 A search of MathWorld on the Lucas numbers would show that they are very

188 closely related to the Fibonacci numbers. For example,

=+ = + 189 LFnn+−11 F n and F n 5( L n +− 11 L n ) . The Lucas number Ln also answers this counting

190 problem:

191 Suppose n people are seated at a circular table. Including the empty set,

192 how many subsets of the people can be chosen which do not include any

193 two people seated side by side?

194 The most spectacular property of the Perrin sequence is its effectiveness as a test

195 for primality. In particular, if n is a prime, then it has been shown that n divides the Perrin 11

196 number Pn. For example, n = 11 divides the Perrin number P11 = 22, and n = 29 divides

197 the Perrin number P29 = 3480 = 29 ⋅120. Only rarely will a nonprime n divide Pn. Indeed,

2 198 the smallest Perrin pseudoprime is n = 277441 = 521 , which is a factor of P277441. This

199 was discovered quite recently, in 1982.

200

201

202 REFERENCES

203 Amend, B. "FoxTrot.com." Cartoon from Oct. 11, 2005. http://www.foxtrot.com/

204 The On-Line Encyclopedia of Integer Sequences.

205 www.research.att.com/~njas/sequences/index.html

206 Weisstein, Eric. MathWorld—A Wolfram Web Resource http://mathworld.wolfram.com/

207