From Fibonacci to Foxtrot:
Investigating Recursion Relations with Geometric Sequences
The Algebra Standard from the Principles and Standards for School Mathematics
(NCTM, 2000) states that all students should “Understand patterns, relations, and
functions.” In particular, to meet the grades 9 – 12 expectations, students should
“generalize patterns using explicitly defined and recursively defined functions.” The
Foxtrot comic strip shown below (Amend) provides a wonderful starting point to create
lessons that address the algebra standard.
In the first three panels of the cartoon, we see that Marcus has scored a
touchdown by identifying Jason’s sequence 0, 1, 1, 2, 3, 5, 8, 13… as the “Fibonacci
series” (though it would be preferable to replace “series” with “sequence,” since in
mathematics terminology a series designates the sum of the terms in a sequence). The
Fibonacci sequence is one of the most widely known in all of mathematics, recursively
defined by the recurrence relation Fn 2  Fn1  Fn . Thus, many readers of Foxtrot could
2
have emulated Marcus’ scoring success. However, there are at least two lingering
questions:

Why did Jason begin his count with zero? It is more common to start the
Fibonacci sequence with F1  1, F2  1 instead of F0  0, F1  1 .

How can Jason score a touchdown? In the last panel of the comic strip,
Marcus has challenged Jason with the sequence 3, 0, 2, 3, 2, 5, … . What
is this sequence?
The key to our investigation of these questions is the geometric sequence
a, ar , ar 2 , ar 3 , ar 4 ,
, which is defined explicitly by xn  ar n , n = 0, 1, 2, … .
Alternatively, the sequence can be defined recursively by xn 1  rxn where x0  a . The
constant r 
xn 1
is the common ratio of the sequence.
xn
We begin by first investigating a connection between the Fibonacci sequence and
geometric sequences. The techniques we develop provide us with a method to investigate
the mysterious sequence 3, 0, 2, 3, 2, 5, … .
IS THE FIBONACCI SEQUENCE A GEOMETRIC SEQUENCE?
The short answer to the question just raised is no. After all, F1 / F0 is not defined, and
F2 / F1  1/1  1 and F3 / F2  2 /1  2 are different ratios. Rather than being discouraged,
let’s examine several more ratios of successive Fibonacci numbers, as shown in the
following table.
n
Fn
Fn +1/Fn
0
0
-
1
1
1.0
2
1
2.0
3
4
5
6
7
8
2
3
5
8
13
21
1.5 1.667 1.600 1.625 1.615 1.619
3
As n becomes larger, the Fibonacci sequence increasingly resembles a geometric
sequence with a common ratio of about 1.6. Since the early integer values of the
Fibonacci sequence seem to cause difficulty, suppose we not worry yet about the initial
values of the sequence. Instead, we simply seek a geometric sequence xn  ar n that
satisfies the Fibonacci recurrence xn 2  xn1  xn . That is, we want to determine values of
r for which ar n  2  ar n 1  ar n , since it is clear that an arbitrary value of the constant a is
allowed. Thus, we want values of r  0 for which r n2  r n1  r n . The terms in this
equation can be divided by the common factor r n , giving us the quadratic equation
r 2  r  1. Using the quadratic formula, we discover there are two roots r of the quadratic
equation: p 
1 5
1 5
and q 
. As a decimal, p is about 1.618, and we suspect we
2
2
have identified the value we encountered in the numerical table above. For arbitrary
choices of the constants a and b, both of the geometric sequences ap n and bq n solve the
Fibonacci recursion xn 2  xn1  xn . Moreover, since ap n 2  ap n1  ap n and
bq n 2  bq n1  bq n , we can add these two equations to see that
 ap
n2
 bq n  2    ap n 1  bq n 1    ap n  bq n  .
That is, for any choices of the constants a and b, xn  ap n  bq n solves the Fibonacci
recurrence relation xn 2  xn1  xn .
To summarize our progress, we have proved the following result:
Theorem. The Fibonacci recursion formula xn 2  xn1  xn , is solved by xn  ap n  bq n
where p 
1 5
1 5
, q
, and a and b are arbitrary constants.
2
2
4
The constant p 
1 5
 1.6180
2
is the famous Golden Ratio. It was known
(though not by that name*) in ancient Greek mathematics, since it solved this question:
determine the point C on line segment AB so that AB/AC = AC/BC.
The second solution q 
1 5
 0.6180
2
of the quadratic equation
r 2  r  1  0 can be investigated by factoring the quadratic polynomial. That is, since
r 2  r  1  (r  p)(r  q)  r 2   p  q  r  pq ,
then equating coefficients shows us that p  q  1 and pq = –1. It’s interesting to see how
these relationships can be found without requiring the explicit formulas for p and q that
involve 5.
THE LUCAS AND FIBONACCI SEQUENCES
A simple choice of constants for xn  ap n  bq n in the theorem above is a = 1 and
n
n
 1 5   1 5 
b = 1. This gives us the solution Ln  p  q  
  
 . For example,
 2   2 
n
n
L0  p 0  q 0  1  1  2 and L1  p  q  1 , and we are delighted that no
5 term appears.
Moreover, since Ln 2  Ln1  Ln and L0  2 and L1  1 , the terms of the sequence are all
positive integers. We have rediscovered the Fibonacci-like sequence 2, 1, 3, 4, 7, 11, 18,
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
than the Fibonacci sequence, the Lucas sequence also has many properties of interest (see
the concluding section of this article.)
To obtain the Fibonacci numbers, we need to see if it possible to choose the
constants a and b so that Fn  ap n  bq n . Since F0 = 0 and ap 0  bq 0  a  b , we will
 1 5 1 5 

want b = – a. Also, since F1  1 and ap1  bq1  a  p  q   a 
a 5,
2 
 2
we must choose a 
1
. Thus, the nth Fibonacci number is given by the explicit formula
5
n
n
pn  qn
1  1  5   1  5  

Fn 

 
 .
5
5  2   2  


This formula is known as Binet’s formula for the Fibonacci numbers, named after
Jacques Binet (1786-1865) who discovered it in 1843. However, the formula was
discovered first in 1718 by Abraham DeMoivre (1667-1754).
Since Binet’s formula gives F0 = 0, we now know why Jason started his list of
Fibonacci numbers with 0. The formula also explains why the Fibonacci sequence,


though not geometric, is increasingly close to one. Since q  1  5 / 2  0.618 we see
that |qn| is increasingly small as n becomes large. Therefore, we have the approximate
equality Fn 
1 n
p . This explains why the Fibonacci sequence is nearly a geometric
5
series, as we noticed in the table of values computed earlier. If we let {x} denote the
 pn 
“round to the integer nearest to x” function, it is easy to check that Fn    for all n 
 5
0. Similarly, the Lucas numbers can be written as Ln   p n  when n  1.
6
INVESTIGATING MARCUS’ SEQUENCE 3, 0, 2, 3, 2, 5, …
It seems reasonable to guess that the new sequence is related to the Fibonacci sequence.
Therefore, suppose that we add consecutive pairs of the sequence 3, 0, 2, 3, 2, 5. The first
three sums are 3 + 0 = 3, 0 + 2 = 2, and 2 + 3 = 5 We seem to be on the right track, since
we get the next three terms 3, 2, and 5 of the sequence. This suggests that the term xn+ 3 is
the sum of the two consecutive terms xn and xn + 1. That is, Marcus’ sequence is defined
by the recursion formula
xn3  xn1  xn .
As before, we can search for a geometric sequence of the form xn  ar n that
solves the recurrence relation. As before, the constant a is arbitrary, but now we want r to
satisfy the equation r n3  r n1  r n . Dividing each term by the common factor r n , we get
the cubic equation r 3  r  1 . We could turn to a computer algebra system or a graphing
calculator to solve this cubic and discover there is one real root u =1.32472 … and a pair
of complex conjugate roots v = –0.662359+0.56228 i and w = – 0.662359 – 0.56228 i.
However, let us just suppose we factor the cubic polynomial to get
r 3  r  1  (r  u)  r  v  r  w .
By expanding the product of the three binomials on the right side, we get the equation
r 3  r  1  r 3  u  v  w r 2  uv  vw  uw r  uvw .
Since these two polynomials in the variable r are equal if and only if their coefficients are
equal, we obtain the equations
u + v + w = 0, uv + vw + uw = –1, and uvw = 1.
(*)
7
As with our earlier investigation of the Fibonacci recursion, we know that
xn3  xn1  xn is solved by xn  au n  bv n  cwn for any choice we make for the three
constants a, b, and c. Inspired by the choice that led us to the Lucas sequence, suppose
that we let a = b = c = 1, which gives us the sequence Pn  u n  v n  wn . The first term is
then P0  u 0  v 0  w0  1  1  1  3 . This looks promising, since 3 is indeed the first term
of Marcus’ sequence. The next term is P1  u  v  w  0 , where again we have used
equation (*). Happily enough, 0 is the next term of Marcus’ sequence!
If we can show that P2  2 , we will have unraveled Marcus’ sequence. We have
P2  u 2  v 2  w2 . Is this equal to 2? To find out, we again turn to the equations in (*),
where we see that
0   u  v  w   u 2  v 2  w2  2uv  2vw  2uw
2
Using (*) once again, we see that
P2  u 2  v 2  w2  2(uv  vw  uw)  2(1)  2 .
Therefore, the terms 3, 0, 2, 3, 2, 5, 5, 7, 10, … of Marcus’ sequence can either be
defined recursively by
P0  3, P1  0, P2  2,
, Pn3  Pn1  Pn ,
or can be given explicitly by Pn  u n  v n  wn , where u  1.32, v  –0.66+0.56 i, and w 
– 0.66 – 0.56 i are the roots of the cubic equation r 3  r  1 .
But what is the name of this sequence? Here Jason may want to use his wellknown Internet skills and access The On-Line Encyclopedia of Integer Sequences
(http://www.research.att.com/~njas/sequences/index.html) to find that the sequence is the
Perrin sequence, named for the French mathematician R. Perrin, who discussed the
8
sequence in a mathematical paper published in 1899 (although the sequence had already
been mentioned in 1878 by Lucas). Thus, Jason should yell out “Is it the Perrin
sequence?” to score a touchdown.
Of course, Jason might refer to the Perrin “series” instead of “sequence.” In the
next section, we will see that geometric sequences can also be examined by considering
their associated geometric series.
GEOMETRIC SERIES AND GENERATING FUNCTIONS
For t  1 , the terms of the infinite geometric sequence 1, t , t 2 ,
to give the formula 1  t  t 2 
 tn 

, tn,
can be summed
1
. If we then set t = px, where p is the
1 t
Golden Ratio, we find that
1  px  p 2 x 2 
 pn xn 

1
1
for x  .
1  px
p
Similarly, setting t = qx, we also have
1  qx  q 2 x 2 
Since both series converge for x 
 qn xn 

1
1
for x   p .
q
1  qx
1
, subtracting the series gives us
p
( p  q) x  ( p 2  q 2 ) x 2 
  pn  qn  xn 
But if we recall from Binet’s formula that Fn 
1
1
.

1  px 1  qx
pn  qn
is the nth Fibonacci number, we
5
see that the coefficient of the xn term of the series is
side of the equation above as follows:

5Fn . We can also simplify the right
9
1
1
( p  q) x
5x
5x
,




2
1  px 1  qx 1  px 1  qx  1   p  q  x  pqx 1  x  x 2
where we recall that p 
1 5
1 5
and q 
satisfy p  q  5, p  q  1, .
2
2
and pq  1 Altogether, we see that
x
 F0  F1 x  F2 x 2 
2
1 x  x
 Fn x n 
.
Since the coefficients of the series expansion of the function f  x  
x
are the
1  x  x2
Fibonacci numbers, f is called the generating function of the Fibonacci sequence.
If we had added rather than subtracted the two geometric series above, a similar
calculation shows that
g  x 
2 x
 2  x  3x 2  4 x3 
2
1 x  x
 Ln x n 
is the generating function for the Lucas numbers Ln  p n  q n .
We can also obtain the generating function for the Perrin numbers by letting t be
successively replaced with ux, vx, and wx. When the three geometric series are added we
find that
3  0 x  2 x 2  3x3  2 x 4  5 x5 
 Pn x n 

1
1
1


.
1  ux 1  vx 1  wx
The right side of this formula can be simplified by using the equations (*) found earlier.
We find that
3  2  u  v  w x   uv  ww  uw x 2
1
1
1
3  x2
.




1  ux 1  vx 1  wx 1   u  v  w x   uv  vw  uw  x 2  uvwx 3 1  x 2  x 3
That is,
3  x2
is the generating function for the Perrin sequence.
1  x 2  x3
10
NOTES ON THE FIBONACCI, LUCAS, AND PERRIN SEQUENCES
Each of the three sequences we have discussed—Fibonacci, Lucas, Perrin—is of
considerable mathematical interest. We’ll mention just a few items here, with the hope of
encouraging the reader to consult more extensive references. A particularly convenient
source of information is the online mathematical encyclopedia MathWorld (Weisstein).
For example, the entry for theFibonacci numbers informs us that
“A scrambled version 13, 3, 2, 21, 1, 1, 8, 5 (Sloane's A117540) of the
first eight Fibonacci numbers appears as one of the clues left by murdered
museum curator Jacque Saunière in D. Brown's novel The Da Vinci Code
(Brown 2003, pp. 43, 60-61, and 189-192). In the Season 1 episode
"Sabotage" (2005) of the television crime drama NUMB3RS, math genius
Charlie Eppes mentions that the Fibonacci numbers are found in the
structure of crystals and the spiral of galaxies and a nautilus shell.”
A search of MathWorld on the Lucas numbers would show that they are very
closely related to the Fibonacci numbers. For example,
Ln  Fn1  Fn1 and Fn  5( Ln1  Ln1 ) . The Lucas number Ln also answers this counting
problem:
Suppose n people are seated at a circular table. Including the empty set,
how many subsets of the people can be chosen which do not include any
two people seated side by side?
The most spectacular property of the Perrin sequence is its effectiveness as a test
for primality. In particular, if n is a prime, then it has been shown that n divides the Perrin
11
number Pn. For example, n = 11 divides the Perrin number P11 = 22, and n = 29 divides
the Perrin number P29 = 3480 = 29 120. Only rarely will a nonprime n divide Pn. Indeed,
the smallest Perrin pseudoprime is n = 277441 = 5212, which is a factor of P277441. This
was discovered quite recently, in 1982.
REFERENCES
Amend, B. "FoxTrot.com." Cartoon from Oct. 11, 2005. http://www.foxtrot.com/
The On-Line Encyclopedia of Integer Sequences.
www.research.att.com/~njas/sequences/index.html
Weisstein, Eric. MathWorld—A Wolfram Web Resource http://mathworld.wolfram.com/