How to write the Fibonacci Sequence?

Question

I had originally coded the program wrongly. Instead of returning the Fibonacci numbers between a range (ie. startNumber 1, endNumber 20 should = only those numbers between 1

Answer 1

How about this one? I guess it's not as fancy as the other suggestions because it demands the initial specification of the previous result to produce the expected output, but I feel is a very readable option, i.e., all it does is to provide the result and the previous result to the recursion.

#count the number of recursions
num_rec = 0

def fibonacci(num, prev, num_rec, cycles):

    num_rec = num_rec + 1

    if num == 0 and prev == 0:
        result  = 0;
        num = 1;
    else:
        result = num + prev

    print(result)

    if num_rec == cycles:
        print("done")
    else:
        fibonacci(result, num, num_rec, cycles)

#Run the fibonacci function 10 times
fibonacci(0, 0, num_rec, 10)

Here's the output:

0
1
1
2
3
5
8
13
21
34
done

Answer 2

These all look a bit more complicated than they need to be. My code is very simple and fast:

def fibonacci(x):

    List = []
    f = 1
    List.append(f)
    List.append(f) #because the fibonacci sequence has two 1's at first
    while f<=x:
        f = List[-1] + List[-2]   #says that f = the sum of the last two f's in the series
        List.append(f)
    else:
        List.remove(List[-1])  #because the code lists the fibonacci number one past x. Not necessary, but defines the code better
        for i in range(0, len(List)):
        print List[i]  #prints it in series form instead of list form. Also not necessary

Answer 3

Using for loop and print just the result

def fib(n:'upto n number')->int:
    if n==0:
        return 0
    elif n==1:
        return 1
    a=0
    b=1
    for i in range(0,n-1):
        b=a+b
        a=b-a
    return b

Result

>>>fib(50)
12586269025
>>>>
>>> fib(100)
354224848179261915075
>>> 

Print the list containing all the numbers

def fib(n:'upto n number')->int:
    l=[0,1]
    if n==0:
        return l[0]
    elif n==1:
        return l
    a=0
    b=1
    for i in range(0,n-1):
        b=a+b
        a=b-a
        l.append(b)
    return l

Result

>>> fib(10)
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]

Answer 4

Fibonacci sequence is: 1, 1, 2, 3, 5, 8, ....

That is f(1) = 1, f(2) = 1, f(3) = 2, ..., f(n) = f(n-1) + f(n-2).

My favorite implementation (simplest and yet achieves a light speed in compare to other implementations) is this:

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

Test

>>> [fibonacci(i) for i in range(1, 10)]
[1, 1, 2, 3, 5, 8, 13, 21, 34]

Timing

>>> %%time
>>> fibonacci(100**3)
CPU times: user 9.65 s, sys: 9.44 ms, total: 9.66 s
Wall time: 9.66 s

Edit: an example visualization for this implementations.

Answer 5

use recursion:

def fib(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fib(n-1) + fib(n-2)
x=input('which fibonnaci do you want?')
print fib(x)

Answer 6

Basically translated from Ruby:

def fib(n):
    a = 0
    b = 1
    for i in range(1,n+1):
            c = a + b
            print c
            a = b
            b = c

...