Find the nth lucky number generated by a sieve in Python

Question

I'm trying to make a program in Python which will generate the nth lucky number according to the lucky number sieve. I'm fairly new to Python so I don't know how to do all that much yet. So far I've figured out how to make a function which determines all lucky numbers below a specified number:

def lucky(number):
    l = range(1, number + 1, 2)
    i = 1
    while i < len(l):
        del l[l[i] - 1::l[i]]
        i += 1
    return l

Is there a way to modify this so that I can instead find the nth lucky number? I thought about increasing the specified number gradually until a list of the appropriate length to find the required lucky number was created, but that seems like a really inefficient way of doing it.

Edit: I came up with this, but is there a better way?

def lucky(number):
    f = 2
    n = number * f
    while True:
        l = range(1, n + 1, 2)
        i = 1
        while i < len(l):
            del l[l[i] - 1::l[i]]
            i += 1
        if len(l) >= number:
            return l[number - 1]
        f += 1
        n = number * f

Answer 1

I came up with this, but is there a better way?

Truth is, there will always be a better way, the remaining question being: is it good enough for your need?

One possible improvement would be to turn all this into a generator function. That way, you would only compute new values as they are consumed. I came up with this version, which I only validated up to about 60 terms:

import itertools


def _idx_after_removal(removed_indices, value):
    for removed in removed_indices:
        value -= value / removed
    return value


def _should_be_excluded(removed_indices, value):
    for j in range(len(removed_indices) - 1):
        value_idx = _idx_after_removal(removed_indices[:j + 1], value)
        if value_idx % removed_indices[j + 1] == 0:
            return True
    return False


def lucky():
    yield 1
    removed_indices = [2]
    for i in itertools.count(3, 2):
        if not _should_be_excluded(removed_indices, i):
            yield i
            removed_indices.append(i)
            removed_indices = list(set(removed_indices))
            removed_indices.sort()

If you want to extract for example the 100th term from this generator, you can use itertools nth recipe:

def nth(iterable, n, default=None):
    "Returns the nth item or a default value"
    return next(itertools.islice(iterable, n, None), default)

print nth(lucky(), 100)

I hope this works, and there's without any doubt more room for code improvement (but as stated previously, there's always room for improvement!).