We have a set of consecutive numbers from ```1``` to ```n```.
We want to count all the subsets that do not contain consecutive numbers.
E.g.

If our set ```S1``` is equal to ```[1,2,3,4,5]```
The subsets that fulfill these property are:
```
[1],[2],[3],[4],[5],[1,3],[1,4],[1,5],[2,4],[2,5],[3,5],[1,3,5]
```
A total of      ```12   ``` subsets.

From the set ```S2``` equals to```[1,2,3]```, it is obvious that we have only ```4``` subsets and are:
```
[1],[2],[3],[1,3]
```
Make a code that may give the amount of all these subsets for any integer       ```n >= 2       ```.

Features of the random tests:
```
number of tests = 100 
10 <= n <= 120
```

Difficulty: introductory

Starter code:

def f(n):
	

Solutions:
def f(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b + 1
    return a

def f(n):
    a, b = 1, 2
    for _ in range(n):
        a, b = b, a + b
    return a - 1

f=lambda n,a=0,b=1:f(n-1,b,a+b) if n!=-2 else a-1 

def f(n):
    xs = [0, 1, 2]
    for i in range(n - len(xs) + 1):
        xs.append(2 * xs[-1] - xs[-3])
    return xs[n]

fib = [1,2]
for _ in range(1000): fib.append(fib[-2]+fib[-1])
def f(n): return fib[n]-1

FIBS = [1, 1]

def f(n):
    # I dedicate this method to oeis.org :)
    while len(FIBS) <= n + 1:
        FIBS.append(FIBS[-2] + FIBS[-1])
    return FIBS[n + 1] - 1


def f(n):
    if n==1:
        return 1
    elif n == 0:
        return 0
    
    a,b = 1,2
    for i in range(0,n-2):
        a,b = b,a+b+1
    return b

def f(n):
    a,b = 1,1
    for _ in range(n+1):
        a,b = b,a+b
    return a-1

def f(n):
    n=n+2
    a, b = 0, 1
    while n > 0:
        a, b = b, a + b
        n -= 1
    return a-1