Prime Numbers

3.2 Sieving 127
}
S = S \ (pS ∩ [1,N]); // Now S = Sl.
l = l +1;
6. [Return the set of primes in [1,N]]
return (S \{1}) ∪{p1,p2,...,pm};
Each set Sl consists of the numbers in [1,N] that are coprime to p1,p2,...,pl.
Thus the first member after 1 is the (l + 1)-th prime, pl+1. Let us count
the number of operations in Algorithm 3.2.2. The number of operations for
Step [Setup] is O((N/ ln N)
i≤k 1/pi) = O(N ln ln N/ ln N). (In fact, the
expression ln ln N may be replaced with ln ln ln N, but it is not necessary for
the argument.) For Step [Roll wheel], the number of operations is #Sk ≤
⌈N/Mk⌉ϕ(Mk) =O(Nϕ(Mk)/Mk), where ϕ is Euler’s function. The fraction
ϕ(Mk)/Mk is exactly equal to the product of the numbers 1 − 1/pi for
i =1uptok. By the Mertens Theorem 1.4.2, this product is asymptotically
e−γ / ln pk. Further, from the prime number theorem, we have that pk is
asymptotically equal to ln N. Thus the number of operations for Step [Roll
wheel] is O(N/ ln ln N).
It remains to count the number of operations for steps [Find gaps] and
[Find special set]. Suppose S = Sl−1, andletGl = G. The number of members
of S ∩ [1,N/pl] isO(N/(pl ln pl−1)), by Mertens. Thus, the total number of
steps to find all sets Gl is bounded by a constant times
m
l=k+1
N/(pl ln pl−1) =O(N/ ln pk) =O(N/ ln ln N).
The number of additions required to compute gpl for g in Gl by the repeated
doubling method is O(ln g) =O(ln N). The sum of all of the values of g in Gl
N/pl
is at most N/pl, so that the number of members of Gl is O .Thus
the total number of additions in Step [Find gaps] is bounded by a constant
times
m
l=k+1
N
pl
ln N ≤
⌊ √ N⌋
i=2
N
i
ln N ≤
√ N
1
N
t ln Ndt=2N 3/4 ln N.
We cannot be so crude in our estimation of the number of operations in Step
[Find special set]. Each of these operations is the simple bookkeeping step
of deleting a member of a set. Since no entry is deleted more than once, it
suffices to count the total number of deletions in all iterations of Step [Find
special set]. But this total number of deletions is just the size of the set Sk
√N,N lessthenumberofprimesin . This is bounded by #Sk, whichwe
have already estimated to be O(N/ ln ln N).