Python Perfect Numbers

Question

So I am supposed to write a Python program that will identify and print all the perfect numbers in some closed interval [ 2, n ], one per line. We only have to use nested while loops/ if-else statements. I did it somehow using a for loop, but can't figure out the same using a while loop. I'd appreciate your help if you could show me how to translate my code into a while loop. Thanks guys. Here's what I have:

limit = int(input("enter upper limit for perfect number search: "))

for n in range(2, limit + 1):
    sum = 0
    for divisor in range(1, n):
        if not n % divisor:
            sum += divisor
        if sum == n:
            print(n, "is a perfect number")

Answer 1

Here is a (somewhat more efficient) sieve version:

# search all numbers in [2..limit] for perfect numbers
# (ones whose proper divisors sum to the number)
limit = int(input("enter upper limit for perfect number search: "))

# initialize - all entries are multiples of 1
#   (ignore sieve[0] and sieve[1])
sieve = [1] * (limit + 1)

n = 2
while n <= limit:
    # check n
    if sieve[n] == n:
        print(n, "is a perfect number")
    # add n to all k * n where k > 1
    kn = 2 * n
    while kn <= limit:
        sieve[kn] += n
        kn += n
    n += 1

---

Running it to 10000 finds

6 is a perfect number
28 is a perfect number
496 is a perfect number
8128 is a perfect number

and factorizing those shows an interesting pattern:

6          3 * 2                         (  4 - 1) * (  4 / 2)
28         7 * 2 * 2                     (  8 - 1) * (  8 / 2)
496       31 * 2 * 2 * 2 * 2             ( 32 - 1) * ( 32 / 2)
8128     127 * 2 * 2 * 2 * 2 * 2 * 2     (128 - 1) * (128 / 2)

where the first factor (3, 7, 31, 127) is a prime which is one less than a power of two, and it is multiplied by half that same power of two. Also, the powers involved are prime (2**2, 2**3, 2**5, 2**7).

In fact Euclid proved that (2**p - 1) * 2**(p - 1) is a perfect number if 2**p - 1 is prime, which is only possible (though not assured) if p is prime. Euler went further, proving that all even perfect numbers must be of this form.

This suggests an incredibly more efficient version - I am going to go ahead and use for loops, feel free to rewrite it without. First, we need a source of primes and an is_prime test:

def primes(known_primes=[7, 11, 13, 17, 19, 23, 29]):
    """
    Generate every prime number in ascending order
    """
    # 2, 3, 5 wheel
    yield from (2, 3, 5)
    yield from known_primes
    # The first time the generator runs, known_primes
    #   contains all primes such that  5 < p < 2 * 3 * 5
    # After each wheel cycle the list of known primes
    #   will be added to.
    # We need to figure out where to continue from,
    #   which is the next multiple of 30 higher than
    #   the last known_prime:
    base = 30 * (known_primes[-1] // 30 + 1)
    new_primes = []
    while True:
        # offs is chosen so  30*i + offs cannot be a multiple of 2, 3, or 5
        for offs in (1, 7, 11, 13, 17, 19, 23, 29):
            k = base + offs    # next prime candidate
            for p in known_primes:
                if not k % p:
                    # found a factor - not prime
                    break
                elif p*p > k:
                    # no smaller prime factors - found a new prime
                    new_primes.append(k)
                    break
        if new_primes:
            yield from new_primes
            known_primes.extend(new_primes)
            new_primes = []
        base += 30

def is_prime(n):
    for p in primes():
        if not n % p:
            # found a factor - not prime
            return False
        elif p * p > n:
            # no factors found - is prime
            return True

then the search looks like

# search all numbers in [2..limit] for perfect numbers
# (ones whose proper divisors sum to the number)
limit = int(input("enter upper limit for perfect number search: "))

for p in primes():
    pp = 2**p
    perfect = (pp - 1) * (pp // 2)
    if perfect > limit:
        break
    elif is_prime(pp - 1):
        print(perfect, "is a perfect number")

which finds

enter upper limit for perfect number search: 2500000000000000000
6 is a perfect number
28 is a perfect number
496 is a perfect number
8128 is a perfect number
33550336 is a perfect number
8589869056 is a perfect number
137438691328 is a perfect number
2305843008139952128 is a perfect number

in under a second ;-)

Answer 2

You can replace your for loops with the following:

n = 2
while n < limit + 1:
   ...
   divisor = 1
   while divisor < n:
      ...
      divisor += 1
   ...
   n += 1

Hint: You can also use n/2 as the upper limit for the second loop as any divisor of n cannot be greater than n/2.