Analyzing the Pisano Period
Autor: Sara17 • June 17, 2018 • 3,247 Words (13 Pages) • 662 Views
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memorize the 60 number long period. While information is limited, we have learned little things from the Pisano Period that were not very obvious with just the sequence alone. Not to mention, we could expand to modulo 100 and 1000 but at that point, it becomes so absurd that it wouldn’t really help us that much.
The Pisano period is still being talked about today so we don’t have many theorems about them currently. However, according to Renault, we have proved some things about the Pisano Period that are actually very helpful. The main point saying that a Pisano period of any length no matter what, will have zeroes equally spaced from one another. Not only that, but there will be exactly one, two or four zeros in each Pisano period (Renault, 2013, p372). The reason why this is so helpful is that if there is a long, challenging period that you have figure out, once the cycle reaches a third zero, you can just stop, and multiply the length of the cycle before that third zero, by 2 and get the length of the whole cycle.
There is also a minor theorem about Fibonacci that is good to know. It is a variation on Lamé’s Theorem which states, “The number of steps (i.e., divisions) in an application of the Euclidean algorithm never exceeds 5 times the number of digits in the lesser.” The variation however, made by Knuth, said, ”For n≥1, let integers u and v, u>v>0, be such that processing u and v by the Euclidean algorithm takes exactly n division steps. Moreover, assume that u is the least possible number satisfying that requirement. Then u=Fn+2 and v=Fn+1, where {Fk} is the Fibonacci sequence,” (Bogomolny). To sum it up, this is just explaining how fibonacci sequence works, as one step leads to another, which is the whole premise of the sequence.
That was the base data and information I had to start my researching and experimentation. I decided to first do random things and see if I can notice any properties immediately. Something I found, was that in a way, a Pisano period cycle follows the rule that makes the Fibonacci sequence, take the two previous terms and add them to get your next term. Take modulo 4 as an example.
x
1
1
2
3
5
8
13
21
34
55
89
144
π(x)
1
1
2
3
1
0
1
1
2
3
1
0
Just inspect π(x). Keep in mind that we can’t have a term above 3 and that all multiples of 4 are 0 as 4x mod 4 = 0. The start is exactly the same, they both start with 1,1. Add those together and you get 2, the third term,. Add 2 and 1 and you get 3, the fourth term. Add 3 and 2, you get 5. Once you do 5 mod 4, you get an answer of 1. That’s when the loop is evident. Not to mention, that afterwards, you add 3 and 1, which is 4. Do 4 mod 4and you get 0. Basically, what you can do is, as long as the previous two terms are under modulo modulo value, which they should always be when working with a Pisano period, add those two terms together. If your result is a number equal to or greater than the modulo value, subtract the result by the modulo value.
After random testings with the Pisano period, one thing is true of all Pisano period except for 2. While taking modulo 2 will give you a Pisano period of 3, an odd number, using any other modulo value, that’s an integer, will result in an even number. This and Renaults writings about zeroes makes it easier to identify mistakes when trying to compute the Pisano period of some modulo value.
After, figuring out how the Fibonacci sequence is directly related to the Pisano period, there were still many questions about actually getting the Pisano period of any modulo value. I started to first work with primes as their speciality is found in division and factoring. I wanted to see if the same applied here. I started with 2 and went onward and tried out the first few prime numbers.
x
2
3
5
7
11
13
17
19
23
31
37
41
43
47
53
π(x)
3
8
20
16
10
28
36
18
48
30
76
40
88
32
108
Remember, modulo 2 and 5 are classified as exceptions as they have their own distinct Pisano period. As we look at these relationships, the term in their one’s value place determines a special property of some prime numbers. A one’s place of 3 sticks out the most as the rule in this set is always 2x + 2. An expansion of this set is necessary to validate this theory.
x
3
13
23
43
53
73
83
103
113
233
263
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