Improving pure Python prime sieve by recurrence formula

Question

I am trying to optimize further the champion solution in prime number thread by taking out the complex formula for sub-list length. len() of the same subsequence is too slow as len is expensive and generating the subsequence is expensive. This looks to slightly speed up the function but I could not yet take away the division, though I do the division only inside the condition statement. Of course I could try to simplify the length calculation by taking out the optimization of starting marking for n instead of n*n...

I took out xrange as it is being discontinued, but then my function lost for earlier version. How I replaced division / by integer division // to be compatible with Python 3 or

from __future__ import division

Also I would be interesting if this recurrence formula could help to speed up the numpy solution, but I have not experience of using numpy much.

If you enable psyco for the code, the story becomes completely different, however and Atkins sieve code becomes faster than this special slicing technique.

def rwh_primes1(n):
    # http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Returns  a list of primes < n """
    sieve = [True] * (n//2)
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i//2]:
            sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
    return [2] + [2*i+1 for i in xrange(1,n/2) if sieve[i]]

def primes(n):
    # http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    # recurrence formula for length by amount1 and amount2 Tony Veijalainen 2010
    """ Returns  a list of primes < n """
    sieve = [True] * (n//2)
    amount1 = n-10
    amount2 = 6

    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i//2]:
             ## can you make recurrence formula for whole reciprocal?
            sieve[i*i//2::i] = [False] * (amount1//amount2+1)
        amount1-=4*i+4
        amount2+=4

    return [2] + [2*i+1 for i in xrange(1,n//2) if sieve[i]]