Fibonacci Numbers

The Fibonacci sequence is an infinite mathematical sequence in which each term is the sum of the two previous terms, usually starting with 0 and 1.

The sequence begins like this: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on.

Numbers from the Fibonacci sequence, noted $ F_n $ ou $ F(n) $, are equal to the addition of the 2 previous terms, they follow the recurrence formula: $$ F(n) = F(n-1) + F(n-2) $$ that can be also written $$ F(n+2) = F(n) + F(n+1) $$

To initiate the sequence, by default, the two first terms are $ F(0) = 0 $ and $ F(1) = 1 $

Binet's formula allows you to calculate the nth term without having to use recursion or having to calculate the previous terms. The formula is $$ F(n) = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n $$ and involves the golden ratio.

Fibonacci sequence first numbers are:

For the next Fibonacci terms, use the calculator above.

Each term in the sequence is equal to the previous multiplied by approximately $ \varphi = 1.618 $ (golden number).

The rabbits' growth problem is a problem proposed by Leonardo Fibonacci in 1200.

There is a rabbit couple (male + female) and every month a couple breeds and gives birth to a new pair of rabbits which in turn can reproduce itself after 2 months. How many rabbits will be born after X months?

In the beginning there is 1 couple then

Each month, the total number of rabbits is equal to the sum of the numbers of the previous two months because it is the number of existing rabbits (the previous month) plus the number of babies born from rabbits couples who have at least two months (hence the number of rabbits 2 months ago). The numbers found are the numbers of the Fibonacci sequence.

The Lucas sequence is similar to the Fibonacci sequence, but it starts with 2 and 1 (instead of 0 and 1). The recurrence formulas are the same.

A source code to program the calculation of Fibonacci numbers by recurrence://Pseudo-code
function fibonacci(n) {
if (n == 0) return 0
if (n == 1) return 1
return fibonacci(n - 1) + fibonacci(n - 2)
}

And an iterative version of the algorithm //Pseudo-code
function fibonacci(n) {
if (n == 0) return 0
if (n == 1) return 1
prev = 0
curr = 1
for i from 2 to n {
next = prev + curr
prev = curr
curr = next
}
return curr
}