A script to generate and print perfect numbers (Python recipe) by Aditya Gopalakrishnan

ActiveState Code (http://code.activestate.com/recipes/437072/)

Here's the aforementioned url: http://code.box.sk/

Hi!

Here's a slightly shorter version ;)

n = 1
while True:
    factors = [1]
    [factors.append(i) for i in range(2,n+1) if n%i == 0]
    if sum(factors) == 2*n: print n
    n += 1

Regards, Markus

Even shorter :)

n = 1
while True:
    if sum([i for i in range(1,n+1) if n%i == 0]) == 2*n: print n
    n += 1

Thanks Markus! I'm new to python (and programming), and your solutions have taught me a lot about economy and style in writing code. Any further suggestions would be equally welcome. :) Regards, Aditya

Thanks Markus! I'm new to Python (and programming) and your solutions have taught me a lot about economy and style in writing code. Any further suggestions about style will be equally welcome. :) Regards, Aditya

For-loops simplify the code.

for c in xrange(1, 100000):
    if sum(x for x in xrange(1,c) if not c%x) == c:
        print c

Optimization. The following runs a lot faster, at the cost of some typing. (Finding perfect numbers with python is going to be slow no matter what, so I suppose it is a silly problem to optimize).

Mainly--you only need to search factors up to the square root of n.

from math import sqrt
import sys

try:
    max_n=int(sys.argv[-1])
except ValueError:
    max_n=50000

def factor(n):
    sr=int(sqrt(n)+1)
    factors=[i for i in xrange(2,sr) if n%i==0]
    factors.extend([n/i for i in factors][::-1])
    return factors

def perfect():
    print [n for n in xrange(1,max_n) if sum(factor(n))==n-1]

precomputed values. As with just about anything dealing with primes, the fastest way to generate values is to start with precomputed values. In this case the perfect numbers can be generated from Mersenne primes, and there are only 42 or so of those which are known. Testing Mersenne candidates is faster than the general prime factor generation algorithms shown here so if you don't want to stop with a ValueError you might want to implement one of them instead, or use a service which connects to the Great Mersenne Prime Search server.

mersenne_prime_powers = [2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521,
                   607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937,
                   21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433,
                   1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011,
                   24036583, 25964951]

def perfect_numbers():
    for p in mersenne_prime_powers:
        yield (2**p-1)*(2**(p-1))
    yield ValueError, "that's a large number"

for p in perfect_numbers():
    print p

More details at http://mathworld.wolfram.com/PerfectNumber.html .

yield -> raise. Err, change that "yield ValueError," into a "raise ValueError,". As you might guess, I didn't make it that far in my testing. :)