Concrete mathematics : a foundation for computer science

Lemmanow,
dilemma later.
3.5 FLOOR/CEILING SUMS 93
When we experimented with small values of m, these three columns led respectively
to a[x/aJ, bn, and c.
In particular, we can see how b arises. The second column is an arithmetic
progression, whose sum we know-it’s the average of the first and last terms,
times the number of terms:
(m- 1)n ;o+ ( m 1
.m = (m-lb
2
So our guess that b = (m - 1)/2 has been verified.
The first and third columns seem tougher; to determine a and c we must
take a closer look at the sequence ofnumbers
Omodm, nmodm, 2nmodm, . . . . (m-1)nmodm.
Suppose, for example, that m = 12 and n = 5. If we think of the
sequence as times on a clock, the numbers are 0 o’clock (we take 12 o’clock
to be 0 o’clock), then 5 o’clock, 10 o’clock, 3 o’clock (= 15 o’clock), 8 o’clock,
and so on. It turns out that we hit every hour exactly once.
Now suppose m = 12 and n = 8. The numbers are 0 o’clock, 8 o’clock,
4 o’clock (= 16 o’clock), but then 0, 8, and 4 repeat. Since both 8 and 12 are
multiples of 4, and since the numbers start at 0 (also a multiple of 4), there’s
no way to break out of this pattern-they must all be multiples of 4.
In these two cases we have gcd( 12,5) = 1 and gcd( 12,8) = 4. The general
rule, which we will prove next chapter, states that if d = gcd(m,n) then we
get the numbers 0, d, 2d, . . . , m - d in some order, followed by d - 1 more
copies of the same sequence. For example, with m = 12 and n = 8 the pattern
0, 8, 4 occurs four times.
The first column of our sum now makes complete sense. It contains
d copies of the terms [x/m], 1(x + d)/mJ, . . . , 1(x + m - d)/m], in some
order, so its sum is
This last step is yet another application of (3.26). Our guess for a has been
verified:
a = d = gcd(m, n)