The document discusses Fibonacci numbers modulo n and notes the following: 1) The Fibonacci sequence exhibits a repeating cycle when taken modulo any integer n, known as the Pisano period. 2) The Pisano period for modulo 2 is 3, with a cycle of 011, while for modulo 3 the period is 8 with a cycle of 01120221. 3) Fibonacci numbers modulo n will always be in a finite range and eventually sum to a previous pair, proving the sequence must repeat.

1. FIBONACCI NUMBERS IN
MODULO N

2. NOTICING A CYCLE
• First we look at the last digit of each number in the Fibonacci numbers (Modulo 10) :
n a(n)
0 0
1 1
2 1
3 2
4 3
5 5
6 8
7 3
8 1
9 4
10 5
n a(n)
11 9
12 4
13 3
14 7
15 0
16 7
17 7
18 4
19 1
20 5
21 6
n a(n)
49 9
50 5
51 4
52 9
53 3
54 2
55 5
56 7
57 2
58 9
59 1
n a(n)
60 0
61 1
62 1
63 2
64 3
65 5
66 8
67 3
68 1
69 4
70 5

3. PISANO PERIOD
• Fibonacci numbers:
• F(n)= F(n-1) + F(n-2)
• with seed values F(0)=0, F(1)=1
• Fibonacci numbers modulo n will always give a repeating sequence
• The period of this sequence is called a Pisano Period
• The nth Pisano period = π(n)
• Fibonacci numbers modulo 2:
• π(n) = 3
• Cycle = 011
• Sequence = 011011011011011011011011011…
• Fibonacci numbers modulo 3:
• π(n) = 8
• Cycle = 01120221
• Sequence =011202210112022101120221…

4. PROOF
• Let k and n be non-negative numbers, and let r = k mod n
• What is the range of r? R = { 0, 1, 2, … n-1} , |R| = n
• Each Fibonacci number is the sum of a pair numbers ( F(n-1), F(n-2) )
• In modulo n, each Fibonacci number is the sum of a pair of numbers ( F(n-1) mod n, F(n-2)
mod n )
• Is the number of pairs finite? How many possible pairs are there?
• If a + b = c and a and b each have a finite range it follows that c will also have a finite range
• Therefore Fibonacci numbers in modulo n will eventually be the sum of a pair of
numbers that have already been seen earlier in the series, this will conclude the cycle
and the sequence will repeat from that point on

5. main ()
{
int n, m, temp, f0= 0, f1= 1;
for (int k = 0; k < 10; k++);
{
cout << "Enter how many terms of the series to display:" << endl;
cin >> n;
cout << "Enter a value for m for the series F mod m:" << endl;
cin >> m;
cout << "The series F mod " << m << " is : " << endl;
for ( int i =0; i < n; i++)
{
if (i <= 1)
temp = i;
else
{
// f0 = f0 % m;
// f1 = f1 % m;
temp = f0 + f1;
temp = temp % m;
f0=f1;
f1=temp;
}
cout << i << " " << temp << endl;
}
}
return 0;
}