### The Fibonacci and Lucas Sequences

The Fibonacci numbers were first introduced by Fibonacci in his book Liber Abaci. As he was considering the growth of an idealized rabbit population, he noticed that there was one pair of rabbits at the end of the first month, two pairs of rabbits at the end of the second month, three rabbits at the end of the third month and five rabbits at the end of the fourth month. The first ten Fibonacci numbers are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 and 55.

The rule of the Fibonacci sequence can be viewed in the following way:

F0 = 0, F1 = 1, Fn = Fn−2 + Fn−1.

The Lucas numbers are very similar to the Fibonacci numbers – the only difference is the starting term. The Lucas numbers were studied by the mathematician Francois Edouard Anatole Lucas. Whereas the Fibonacci numbers start with 0, the Lucas numbers start with 2. However, both the Lucas and Fibonacci numbers have the property that the ratio between two consecutive terms converges to the golden ratio. The first ten Lucas numbers are: 2, 1, 3, 4, 7, 11, 18, 29, 47 and 76.

The rule of the Lucas sequence can be viewed in the following way:

L0 = 2, L1 = 1, Ln = Ln−2 + Ln−1.

The “period” of Fibonacci when reduced modulo n, or Ψ(n), for all integers greater than 1, is the number of terms in the Fibonacci sequence before the sequence repeats when reduced modulo n. One of the first patterns regarding the period is that the period of some composite numbers is equal to the product of the periods of its prime factors.

There are also many common patterns between the Lucas and Fibonacci sequences. Each Lucas number can be expressed as the sum of two Fibonacci numbers:

Theorem 5.1. Ln = Fn−1 + Fn+1  for all n>1.

However, the equation can be generalized further.

Conjecture 5.2. For any odd number k ∈ N, Ln · Fk+1 = Fn−k + Fn+k

Numerical Examples:

n = 6 and k = 5: L6 · F5+1 = F6−5 + F6+5 L6 · F6 = F1 + F11 ⇒ 11 · 5 = 0 + 55

n = 8 and k = 3: L8 · F3+1 = F8−3 + F8+3 L8 · F4 = F5 + F11 ⇒ 29 · 2 = 3 + 55

n = 9 and k = 7: L9·F7+1 = F9−7+F9+7 L9·F8 = F2+F16 ⇒ 47·13 = 1+610 ⇒ 611 = 611

What about for even differences?

Conjecture 5.3. For any even number j N, Lj+1 · Fn = Fnj + Fn+j

Numerical Examples:

n = 6 and k = 4: L4+1 · F6 = F6−4 + F6+4 L5 · F6 = F2 + F10 ⇒ 7 · 5 = 1 + 34 ⇒ 35 = 35

n = 10 and k = 2: L2+1 ·F10 = F10−2 +F10+2L3·F10 = F8+F12 ⇒ 3·34 = 13+89 ⇒ 102 = 102

n = 7 and k = 6: L6+1·F7 = F7−6+F7+6 L7·F7 = F1+F13 ⇒ 8·18 = 0+144 ⇒ 144 = 144

For some even natural number k − 2 (where k > 2), we assume that:

Fn−(k−2) + Fn+(k−2)= Lk−1· Fn

We also know that k − 1 will be odd, and so:

Fn−(k−1) + Fn+(k−1) = Ln · Fk

Is it possible to prove that

Fn−k + Fn+k = Lk+1 · F(k is even, as k − 2 is even)

and that

Fn−(k+1) + Fn+(k+1) = Ln · Fk+2 as k + 1 will be odd?

The Fibonacci and Lucas Sequences are two of the most powerful sequences in mathematics. Using these sequences, mathematicians and scientists have discovered patterns in bracts of a pine cone, number of petals in a flower and scales of a pineapple. In addition, these sequences have enhanced our understanding of the Golden Ratio and its significance in our lives.

| October 15th, 2017 | Posted in Uncategorized |