Prime Numbers

3.7 Counting primes 155
for b ≥ 2. We leave the simple proof for Exercise 3.33. Since φ(y, 2) =
⌊(y +1)/2⌋, we can continue to use (3.22) to eventually come down to
expressions φ(y, 2) for various choices of y. For example,
φ(1000, 7) = φ(1000, 5) − φ(142, 5)
= φ(1000, 3) − φ(200, 3) − φ(142, 3) + φ(28, 3)
= φ(1000, 2) − φ(333, 2) − φ(200, 2) + φ(66, 2)
− φ(142, 2) + φ(47, 2) + φ(28, 2) − φ(9, 2)
= 500 − 167 − 100 + 33 − 71 + 24 + 14 − 5
= 228.
Using this scheme, we may express any φ(x, pa) asasumof2 a−1 terms. In
fact, this bottom-line expression is merely the inclusion–exclusion principle
applied to the divisors of p2p3 ···pa, the product of the first a − 1 odd primes.
We have
φ(x, pa) =
n|p2p3···pa
µ(n)φ(x/n, 2) =
n|p2p3···pa
x/n +1
µ(n)
,
2
where µ is the Möbius function see Section 1.4.1.
For a = π(x 1/3 ), clearly 2 a−1 termsistoomany,andwewouldhavebeen
better off just sieving to x. However, we do not have to consider any n in the
sum with n>x, since then φ(x/n, 2) = 0. This “truncation rule” reduces
the number of terms to O(x), which is starting to be competitive with merely
sieving. By fiddling with this idea, we can reduce the O-constant to a fairly
small number. Since 2 · 3 · 5 · 7 · 11 = 2310, by computing a table of values
of φ(x, 11) for x =0, 1,...,2309, one can quickly compute any φ(x, 11): It is
ϕ(2310) ⌊x/2310⌋ + φ(x mod 2310, 11), where ϕ is the Euler totient function.
By halting the recurrence (3.22) whenever a b value drops to 11 or a y/pb
value drops below 1, we get
φ(x, pa) =
n|p6p7···pa
n≤x
µ(n)φ(x/n, 11).
If a = π x1/3 , the number of terms in this sum is asymptotic to cx with
c = ρ(3)ζ(2) −1 5 i=1 pi/(pi + 1), where ρ is the Dickman function (see Section
1.4.5), and ζ is the Riemann zeta function (so that ζ(2) = 6/π2 ). This
expression for c captures the facts that n has no prime factors exceeding x1/3 ,
n is squarefree, and n has no prime factor below 12. Using ρ(3) ≈ 0.0486, we
get that c ≈ 0.00987. By reducing a to π x1/4 (and agreeing to compute
1/4
φ3 x, x
1/4 in addition to φ2 x, x ), we reduce the constant c to an
expression where ρ(4) ≈ 0.00491 replaces ρ(3), so that c ≈ 0.000998. These
machinations amount, in essence, to the method of Meissel, as improved by
Lehmer, see [Lagarias et al. 1985].