Prime Numbers

3.7 Counting primes 153
Label the consecutive primes p1,p2,p3,...,wherep1 =2,p2 =3,p3 =5,
etc. Let
φ(x, y) =#{1 ≤ n ≤ x : each prime dividing n is greater than y}.
Thus φ(x, pa) is the number of integers left unmarked in the sieve of
Eratosthenes, applied to the interval [1,x], after sieving with p1,p2,...,pa.
Since sieving up to √ x leaves only the number 1 and the primes in ( √ x, x],
we have
π(x) − π √ x +1=φ x, √ x .
One could easily use this idea to compute π(x), the time taking O(x ln ln x)
operations and, if the sieve is segmented, taking O x 1/2 ln x space. (We shall
begin suppressing ln x and ln ln x factors for simplicity, sweeping them under a
rather large rug of O(x ɛ ). It will be clear that each x ɛ could be replaced, with
a little more work, with a small power of logarithm and/or double logarithm.)
A key thought is that the sieve not only allows us to count the primes, it
also identifies them. If it is only the count we are after, then perhaps we can
be speedier.
We shall partition the numbers counted by φ(x, y) by the number of prime
factors they have, counted with multiplicity. Let
φk(x, y) =#{n ≤ x : n has exactly k prime factors, each exceeding y}.
Thus, if x ≥ 1, φ0(x, y) is1,φ1(x, y) is the number of primes in (y, x], φ2(x, y)
is the number of numbers pq ≤ x where p, q areprimeswithy