How to find if a number is perfect number with a recursive code

Question

So I want to write a func which get a number and return True if it a "perfect number". I need to do it without using for or while in my function. Here is what I tried to do -

def question_1(num):
i = 1

def checkDivide(n, num):
    if num % n == 0:
        return n
    else:
        return 0

def SumThemAll(num, i):

    if i == num:
        return 0
    else:
        return i + SumThemAll(num, checkDivide(i + 1, num))

if num == SumThemAll(num, i):
    return True
else:
    return False

The problem is when I get to an integer which does not divide with the number i want to check but I dont know how to fix it. any idea how to make it more fast will help too

Answer 1

You could do it like this:

def perfect(num: int):
    def recursive(num: int, i=1):
        if num == i:
            return 0
        elif num % i == 0:
            return i+recursive(num, i+1)
        else:
            return recursive(num, i+1)
    return False if num < 2 else recursive(num)==num

And the non-recursive solution just to check:

def non_recursive(num: int):
    return sum(i for i in range(1, num) if num%i==0) == num