Fibonacci Sequence
The Fibonacci Sequence is the series of numbers:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
The next number is found by adding up the two numbers before it.
 The 2 is found by adding the two numbers before it (1+1)
 Similarly, the 3 is just (1+2),
 And the 5 is just (2+3),
 and so on!
Example: the next number in the sequence above would be (21+34) = 55
It is that simple!
Here is a longer list:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, ...
Can you figure out the next few numbers?
The Rule
The Fibonacci Sequence can be written as a "Rule" (see Sequences and Series):
The Rule is x_{n} = x_{n1} + x_{n2}
where:
 x_{n} is term number "n"
 x_{n1} is the previous term (n1)
 x_{n2} is the term before that (n2)
The terms are numbered form 0 onwards like this:
n = 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
... 
x_{n} = 
0 
1 
1 
2 
3 
5 
8 
13 
21 
34 
55 
89 
144 
233 
377 
... 
Example: term 6 would be calculated like this:
x_{6} = x_{61} + x_{62} = x_{5} + x_{4} = 5 + 3 = 8
Golden Ratio

And here is a surprise. If you take any two successive (one after the other) Fibonacci Numbers, their ratio is very close to the Golden Ratio "φ" which is approximately 1.618034...
In fact, the bigger the pair of Fibonacci Numbers, the closer the approximation. Let us try a few: 
A 
B 

B / A 
2 
3 

1.5 
3 
5 

1.666666666... 
5 
8 

1.6 
8 
13 

1.625 
... 
... 

... 
144 
233 

1.618055556... 
233 
377 

1.618025751... 
... 
... 

... 
Using The Golden Ratio to Calculate Fibonacci Numbers
And even more surprising is this formula for calculating any Fibonacci Number using the Golden Ratio:
Amazingly the answer always comes out as a whole number, exactly equal to the addition of the previous two terms.
Example:
When I used a calculator on this (only entering the Golden Ratio to 6 decimal places) I got the answer 8.00000033. A more accurate calculation would be closer to 8.
Try it for yourself!
Terms Below Zero
The sequence can be extended backwards!
Like this:
n = 
... 
6 
5 
4 
3 
2 
1 
0 
1 
2 
3 
4 
5 
6 
... 
x_{n} = 
... 
8 
5 
3 
2 
1 
1 
0 
1 
1 
2 
3 
5 
8 
... 
(Prove to yourself that adding the previous two terms together still works!)
In fact the sequence below zero has the same numbers as the sequence above zero, except they follow a ++ ... pattern. It can be written like this:
x_{−n} = (−1)^{n+1} x_{n}
Which says that term "n" is equal to (−1)^{n+1}^{ } times term "n", and the value (−1)^{n+1} neatly makes the correct 1,1,1,1,... pattern.
