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    <p>The question is, why would it even be faster? In both examples you are filtering multiples of two, the hard way. It doesn't matter whether you hardcode <code>candidates[4::2] = [None] * (n // 2 - 1)</code> or that it gets executed in the first loop of <code>for i in range(2, fin + 1):</code>.</p> <p>If you are interested in an optimized sieve of Eratosthenes, here you go:</p> <pre><code>def primesbelow(N): # https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188 #""" Input N&gt;=6, Returns a list of primes, 2 &lt;= p &lt; N """ correction = N % 6 &gt; 1 N = (N, N-1, N+4, N+3, N+2, N+1)[N%6] sieve = [True] * (N // 3) sieve[0] = False for i in range(int(N ** .5) // 3 + 1): if sieve[i]: k = (3 * i + 1) | 1 sieve[k*k // 3::2*k] = [False] * ((N//6 - (k*k)//6 - 1)//k + 1) sieve[(k*k + 4*k - 2*k*(i%2)) // 3::2*k] = [False] * ((N // 6 - (k*k + 4*k - 2*k*(i%2))//6 - 1) // k + 1) return [2, 3] + [(3 * i + 1) | 1 for i in range(1, N//3 - correction) if sieve[i]] </code></pre> <p>Explanation here: <a href="https://stackoverflow.com/questions/5293238/porting-optimized-sieve-of-eratosthenes-from-python-to-c">Porting optimized Sieve of Eratosthenes from Python to C++</a></p> <p>The original source is <a href="https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188">here</a>, but there was no explanation. In short this primesieve skips multiples of 2 and 3 and uses a few hacks to make use of fast Python assignment.</p>
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    1. POWhy is this algorithm worse?
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    1. POWhy is this algorithm worse?
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    1. COThanks for your input and sparing your time. I bypass all the even numbers. Isn't that significant?
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    2. CO@blackened: The problem is you don't bypass the even numbers. You allocate storage for them, and you even filter them out, wasting precious CPU time! Why give stinky even numbers this special treatment they don't deserve? If you would bypass the even numbers you would create an array with `^[N/2]` (^[] is rounding up) elements. The element at the key `i` would mark the number `2*i + 1` prime or non-prime.
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    3. COWow, this algorithm is fast! Now I need to decode it.
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