Fast doubling Fibonacci Python generator sequence

Question

I'm having troubles try to create a Fast doubling Fibonacci python generator, using the following.

Given F(k) and F(k+1), we can calculate these:

F(2k) = F(k)[2F(k + 1) − F(k)]
F(2k+1) = F(k+1)^2 + F(k)^2

I've got the following for the simplest (slow) Fibonacci generator:

def fib_generator():
    n = 1
    n0 = 1
    while True:
        yield n
        n, n0 = n + n0, n

Answer 1

An implementation could be:

from itertools import count

def fast_fib_generator():
    F = [1, 1]
    yield 1
    yield 1
    for k in count(1):
        F.append(F[k] ** 2 + F[k - 1] ** 2)
        yield F[-1]

        F.append(F[k] * (2 * F[k + 1] - F[k]))
        yield F[-1]


for x in fast_fib_generator():
    print x

First results:

1
1
2
3
5
8
13
21
34  

Answer 2

def fibonacci(number):
    numbers = [0, 1]
    while len(numbers) < number:
        numbers[len(numbers):len(numbers)] = [numbers[len(numbers)-2] + numbers[len(numbers)-1]]
    return numbers

First get a list of the fibonacci sequence... Then loop through.

def fib_gen(number):
    for iter in fibonacci(number):
        print iter

Answer 3

Try this

num=int(input("Enter length of series required"))
fiblist=[]
for i in range(num):
    if i==0 or i==1:
        fiblist.append(i)
    else:
        temp=fiblist[i-2]+fiblist[i-1]
        fiblist.append(temp)
print(*fiblist,sep=' ')