Vol. 10 No 7 - Pi Mu Epsilon

542 PI MU EPSILON JOURNAL
We shall call the integer m the multiplier of the Fibonacci sequence in Z 0
•
Notice that m must be a unit in Z 0
, and if r is the order of m in Zn ·, and if a(n) is
the length of any row in the array above, then the period of A(n) satisfies the
equation A(n) = r · a(n).
The concept of the multiplier seems to be due to Carmichael [[2], pp. 354-
355], who also proved the fonnula which we have just stated.
We shall use this equation to compute the period. First consider two
representative examples.
The Fibonacci sequence in Z 8
:
0 1 I 2 3 5
0 5 5 2 7
0
The Fibonacci sequence in Z 5
:
0 1 1 2 3
0 3 3 l 4
0 4 4 3 2
0 2 2 4
0
Notice that each of these arrays has the general form which we have
described, and also that the multiplier in Z 8 has order 2 and that the multiplier
in Z 5 has order 4. The period in Z 8 is 12 and the period in Z., is 20.
Using the isomorphism
Z 10
-. Z 2
x Z 5
and the facts above we see that the period of the Fibonacci sequence in Z 10
is
l.c.m.(3,20) = 60
In general, we can find the period of the Fibonacci sequence in Zn for any
n if we can solve two problems:
The Divisibility Problem:
9iven a positive integer n. what is the smallest positive integer a(n) such
that n divides F a
The Multiplier Problem·
If m is the multiplier for the Fibonacci sequence in Zn, what is the order r of
min zn•
Before \Ve solve these two problems in the case where n is a power of 2 or
a power of5, and hence solve the problem of the digits, let us find the period of
the first two digits of the Fibonacci numbers. We will use the isomorphism.
n BONACCI NUMBERS, WASHBURN
Z 100 ... z~ x Z 2 'i
and since we know that the period in Z 4 is 6 \Ve need the period in Z::'i·
\\C have
0 1 2 3 5 8 13 21 9
5 14 19 8 2 10 12 22 9 6
15 21 11 7 18 0 18 18 11 4
15 19 9 3 12 15 2 17 19 19
5 16 21 12 8 20 3 23 24
0 24 24 23 22 20 17 12 4 16
20 11 6 17 23 15 13 3 16 19
10 4 14 18 7 () 7 7 14 21
10 6 16 22 l3 10 23 8 6 14
20 9 4 13 17 5 22 2 24
0 I
We see that the period in Z 25
is 100. Therefore the period in Z 100 is
l.c.m.(6JOO) = 300
Solutions to the Divisibility Problem and the Multiplier Problem can be
found in the papers of J.H. Halton (3] and J. Vinson [ 5]. We shall give our O\\on
solutions to these problems, first in the case of a power of 2. We have given
solutions for n = 2 and n = 4 above, so asswne that
n = 2d where d ' 3
The solution is given by the following Theorem.
Theorem S. If d 3 then
(i). 2d divides FHd-2,2d+I does not divide this Fibonacci number, and 2d
does not divide any smaller Fibonacci number.
(ii). The multiplier for the Fibonacci sequence in Z d 2
is congruent to 2d-t +
l(mod 2d).
In order to prove this Theorem we need to recall a few facts about Fibonacci
numbers and several identities.
First recall that the Lucas numbers are a companion sequence to the
Fibonacci numbers. The Lucas numbers are defined by the initial values
L 0
= 2 and L 1 = 1 ·..
and the recursion
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