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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Wed, 15 Apr 2015 09:59:51 -0600
Subject: Re: find all multiplicands and multipliers for a number
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Xref: csiph.com comp.lang.python:88981

On Tue, Apr 14, 2015 at 8:37 PM, Paul Rubin <no.email@nospam.invalid> wrote:
> Steven D'Aprano <steve+comp.lang.python@pearwood.info> writes:
>> def turner():
>>     nums = itertools.count(2)
>>     while True:
>>         prime = next(nums)
>>         yield prime
>>         nums = filter(lambda v, p=prime: (v % p) != 0, nums)
>
> This is nice, though it will still hit the nesting limit about equally
> soon, because of the nested filters.  I like the faster versions in the
> O'Neill paper.

Nope. You do end up with a lot of nested filter objects, but there's
no recursion in the Python code, which means that you're not piling up
frame objects, and you'll never hit the interpreter's recursion limit.
You might eventually get a seg fault when the C stack space runs out,
however.