UNDERSTANDING
FIB0NACCI NUMBERS
by Beryl Adamson, Wade Deacon High School, Widnes
Introduction
will be concentrated here on the rabbit problem alone in the
Suppose that rabbits breed according to the following model:
Every female produces a mixed pair of offspring, ie one male
and one female on the first day of every month, starting when
she is 2 months old.
A breeder starts with a new born mixed pair. If none die, how
many rabbit pairs are there after n months?
This famous problem, which is over 700 years old, often arises
hope that this will suffice to reveal some of the fascinating
properties of the Fibonacci numbers.
A first solution
To fix ideas and notation it may be helpful to examine the
solution, which many students might pursue. The following
diagrams and the table show the pattern of breeding, and the
number of rabbit pairs after n months.
as an illustration of the "Fibonacci numbers". It is known that
Key: 0 New born pair
A one month old pair
0 breeding pair
the solution Un forms the sequence
n=1 2 3 4 5 6 7 8.
Un=1 1 2 3 5 8 13 21 ...
1st month 2nd month 3rd month
*
The characteristic property of the sequence is that each term
for n > 3 is equal to the sum of the two preceding terms, ie
1
A
1
I
2
4th month 5th month 6th month
(1)
u, Un- - :Un n-
This recurrence relation (1) can be interpreted as follows. On
the first day of the nth month all the pairs of 2 months ago
are old enough to produce offspring, so there will be Un-_2
new pairs to add to the previous months un-1 pairs. However
in my experience this explanation is not self-evident to children.
Even after a hard look at the problem they will remain puzzled
New born pairs 1 0 1 1 2 3 5 8
regarding the formula for Un as an interesting coincidence.
Month old pairs 0 1 0 1 a 2 3 5
However, there is a revealing connection between the
Fibonacci numbers and the elements of Pascal's triangle which
appears to resolve this difficulty rather well. This takes the form
of an identity for Un as a sum of binomial coefficients, which
does not seem to be very well-known in schools. This identity
provides an explanation of the solution to the problem which
would satisfy even the most inquiring mind, and also helps to
3
Month
58
1
2
3
4
5
6
7
8
Breeding pairs 0 0 1 1 2 3 5 8
Total
1
1
2
3
5
8
13
21
Although the pattern seems clear here the sudden emergence
deepen one's understanding of the Fibonacci numbers them-
of the Fibonacci sequence satisfying (1) is nevertheless still
rather puzzling. To help to explain the way in which the
Fibonacci sequence arises it is now necessary to introduce
selves. Writing
Pascal's triangle.
S(r= 0, 1, 2,... n) (2)
r! (n - r)!
it turns out that the Fibonacci relation (1) is essentially a
consequence of the familiar and fundamental identity
1,
Fibonacci representation by binomial
coefficients
Pascal's triangle is familiar to many secondary school-children.
In its conventional form the binomial coefficients defined by
(2) are represented as the rows of the following triangle.
1
2
a=n-1 - (r 2 r n-) (3)
upon which Pascal's triangle is based. Using the result (3) and
this result only, it is possible to derive the relation (1) and to
identify each of the rabbit generations with a corresponding
binomial coefficient (2). Indeed the relation (3) can explain why
the total number of rabbit pairs must assume the Fibonacci
1
1
1 2 1
1 3 3 1
1
1 4 6 4 1
5 10 10 5 1
In general the n + 1 row of Pascal's triangle can be expressed
in the form
form.
It is also clear that the same kind of argument will help to
explain other well known Fibonacci problems, such as tiling a
2 x n path, and counting the leaves on a tree. However, attention
n) n (n n)
29
Now it is interesting to point out to students that the Fibonacci
numbers can actually be derived from Pascal's triangle by
adding all terms across the diagonals thus
ul
1
U2u3
1---
1
2
3
4
5
6
7
8
Zero generation 1 1 1 1 1 1 1 1
I st generation 0 0 1 2 3 4 5 6
1
2
Month
2nd generation 0 0 0 0 1 3 6 10
1
2
1
3rd generation 0 0 0 0 0 0 1 4
u4
U535
14
aa63
Total
S 13
811515
lo20
10
5611
15
u
us 21 1 7 21 35 35 21 7 1
This simple heuristic approach suggests we have the following
representation of a Fibonacci number as a natural sum of
binomial coefficients
+ + +
0( =2(n-2)
(n-1)
n-3
Un
1
1
2
3
5
8
13
Schematically the above table can be represented as follows:
Month
1
2
3
4
5
6
7
8
Zero generation 0 A 5 n n
1 st generation A I n I n
0
...
* A
(4)
The series ends when, by convention
2nd generation A I
0
("')=O
r-l
which occurs when
21
*
5
3 g a
n-r<r-1
ie when
r> 1/2(n+ 1)
3rd
generation
*
*
It requires a simple proof by induction to demonstrate that
the above series correctly describes a Fibonacci number in
general. The proof follows:
Assume (4) is true for n= k and n = k + 1, so
uk+1=+k-1) k-2) k-3+... (5)
k1 k-2 k-3
uk - 0 1 1 2 +... (6)
Using the recurrence formula
Explaining the Fibonacci solution
It is possible to use the representation for the rabbit problem
by generations to explain why the total number of rabbit pairs
un=Un_-1+Un_2. The relation actually follows as a consequence of relation (3) for binomial coefficients. I will use the
previous table to illustrate the argument. Looking at the case
n = 6 in particular.
Month
Uk+2 -Uk,1+ Uk
and by adding (5) and (6) we have
6
Zero generation 1=(5)
1 st generation
Uk+2 k- k-2
1 2 3 +
4= 4
4=
*
Here we have used relation (3) repeatedly. Therefore if (4) is
true for n-= k+ 1, it is also true for n= k+ 2. If
k=l ul- )1
and if
2nd generation A3= 3
Total u6 = 8
The black squares and triangles represent repeats from the
2-
2
)=1
-
Now (4) is true for k = 1 and k = 2, and it follows by induction
that (4) is true for all positive integral values of k.
Having seen the Fibonacci numbers expressed in this way it
now becomes possible for students to comprehend the solution
to the rabbit problem in terms of binomial coefficients. This
arises since it is quickly noticed that the individual binomial
coefficients are readily identified with generations of the
breeding rabbits. Thus
( ')=I
O
denotes the original pair (parents)
n -21) denotes the first generation (children)
(n- 3 denotes the second generation (grandchildren)
2)
and so on. This can be illustrated as follows.
30
A
A1
case n= 5. The black circles represent new offspring. Consider
the three generations separately.
(i) Of course the zero generation must always be the original
parents and hence is one pair. This can be written 1= 1 + 0
which is a special case of relation (3)
5 ) 4 ) (taking n)= 0 for all n 0)
0) (0) (- 1) - 1)
(ii) The first generation consists of four pairs. These can be
considered as three pairs repeated from the previous month
plus one new-born pair. Furthermore this new-born pair is
offspring of the original parents from the zero generation. The
total of "repeated pairs" and "new-born" pairs is again a
special case of relation (3), i.e. 4 = 3 + 1 or
C)-( )(~)
(iii) The second generation is three pairs, one repeated from
the previous month plus two new-born pairs. Parents of the
new-born pairs are two pairs from the first generation capable
of giving birth, i.e. they themselves were already living or
new-born at n = 4. Again this is just relation (3)
Each generation consists of two kinds of rabbit - pair in
general. These are the black square or triangle pairs (merely
G)=( 2)G)
Schematically the three generations are derived from the case
n= 5 as follows:
generation. A glance at the table in section 3 illustrates this
Month 5 6
convincingly. By inspection, it is always seen that
Zero generation (4 )= (5)
0
repeats from the previous month), and the black circle pairs
(new-born pairs from all parents living or new-born two months
previously). It is relation (3) which derives the total for each
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1st generation 3 4 > (0)-+
Total number of Total number of Total number of
pairs for generation = pairs for generation + pairs for generation
r, month n ( r, month n-I1 ( r-l, month n-2
Then, adding the totals for each generation in any given month
and using the relation (4), the formula un, = u, n- 1 + u, n- 2 follows
straightaway.
So it can be seen more clearly from this analysis than the
intuitive explanation given in the introduction that the total
number of rabbit pairs after n months is the sum of the previous
2nd
generation
1
+
3
2
months pairs and the new offsprings from pairs of 2 months
ago which are now ready to breed. The solution to the rabbit
problem can only be the Fibonacci number in the binomial
representation derived above.
An SLUTON
to Pzzle, Pstim s an Prblem on . 1
1. Jigsaw puzzles with tetrominoes
(a) Shape A can also be made with 4 T (Fig. i), or 4 L, or
4 J (Fig. ii), but not with S or Z.
(b) Shape B can also be made with 4 J or 4 L (Fig. iii).
(c) Shape C: 4 R, or 4 S (Fig. iv), or 4 T (Fig. v).
Fig. i
Fig. ii
Fig. iii
Shape D: 4 Q, or 4 S (Fig. vi), or 4 L (Fig. vii and Fig. viii).
Shape E: 4 T (Fig. ix), or 4 L, or 4 J (Fig. x).
Shape F: 4 T only.
Shape G: 4 L only.
Shape H: 4 J only.
Shape K: 4 T (Fig. xi), or 4 Z (Fig. xii).
(d) Shape E only has asymmetrical solutions, as shown in
Figure x, made with 4 J, or the corresponding arrangement
Fig. iv
made with 4 L.
Fig. v
Fig. vi
2. Fun with fractions
(a) Each fraction is '/3. n2/3n'2 1/3.
(b) Each fraction is 1/4. n2/4n2 = 1/4.
(c) 1, 3/4, 2/3, %5/, 3/, 7/12. (n + 1)/2n.
(d) l/, 3/7 , %3, 3/8, 7/19. (n + 1)/(3n + 1).
The fraction following (n + 1)/2n is (n + 2)/(2n + 2) and the
nth term exceeds the (n+ 1)th term by 1/2n(n+ 1). Also
(n + 1)/2n = 1/ + 1/2n, and so each term is greater than 1/z.
Fig. vii
Fig. viii
Fig. ix
(n + 1)/(3n + 1) exceeds the next term (n + 2)/(3n + 4) by
2/(3n + 1)(3n + 4), and (n + 1)/(3n + 1) exceeds 1/3 by 1/3(3n + 1).
3. Powers of three
(i) All powers of 3 leave a remainder 3 when divided by 6.
(ii) The next powers of 3 are 81, 243, 729, 2 187, 6 561.
(iii) (3, 1), (3, 4, 2, 1), (3), (3, 2, 6, 4, 5, 1), (3, 1), (3, 9, 7, 1).
(iv) 4, 10, 3, 21, 4, 20. All are factors or multiples of the
divisor.
(v) 3n=3n-2+8x3n-2. So 3n when divided by 4 or 8
Fig. x
Fig. xi
Fig. xii
leaves the same remainder as 3n- 2 or 3n-4, etc.
(vi) All numbers ending in 5 have a factor 5, but 3n has 3 as
its only prime factor.
31