Concrete mathematics : a foundation for computer science

4.9 PHI AND MU 137
m is large? How can we solve the recurrence (4.55)? Well, the function
g(m) = [m = 11 is obviously multiplicative-after all, it’s zero except when
m = 1. So the Mobius function defined by (4.55) must be multiplicative, by
Depending on bow what we proved a minute or two ago. Therefore we can figure out what k(m)
fast you read. is if we compute p(pk).
When m = pk, (4.55) says that
cl(l)+CL(P)+CL(P2)+...+CL(Pk) = 0
for all k 3 1, since the divisors of pk are 1, . . . , pk. It follows that
cl(P) = -1; p(pk) = 0 for k > 1.
Therefore by (4.52), we have the general formula
ifm=pjpz...p,;
if m is divisible by some p2.
(4.57)
That’s F.
If we regard (4.54) as a recurrence for the function q(m), we can solve
that recurrence by applying Mobius’s rule (4.56). The resulting solution is
v(m) = t Ad):.
d\m
For example,
(~(14 = ~(1)~12+~~(2)~6+~(3)~4+~(4)~3+~(6)~2+~(12)~1
=12-6-4+0+2+0=4.
(4.58)
If m is divisible by r different primes, say {p, , . . . , p,}, the sum (4.58) has only
2’ nonzero terms, because the CL function is often zero. Thus we can see that
(4.58) checks with formula (4.53), which reads
cp(m) = m(l - J-) . . . (I- J-) ;
if we multiply out the r factors (1 - 1 /pi), we get precisely the 2’ nonzero
terms of (4.58). The advantage of the Mobius function is that it applies in
many situations besides this one.
For example, let’s try to figure out how many fractions are in the Farey
series 3n. This is the number of reduced fractions in [O, l] whose denominators
do not exceed n, so it is 1 greater than O(n) where we define
Q(x) = x v(k).
l