Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 191
There is one solution for each way to write 2100 = x ‘y where x is even and
y is odd; we let a == :1x-yI + i and b = i(x+y) + i. So the number of
solutions is the number of divisors of 3. 52 .7, namely 12. In general, there
are n,,2(n, + 1) ways to represent n,, pn”, where the products range over
primes.
2.31 tj,k>2j k = tja21/j2(l -l/j) :=tj,21/j(j-l). Thesecondsumis,
similarly, 3/4.
2.32 If2n~x~2~n+l,thesumsareO+~~~+n+(x-n-l)+~~~+(x-2n) =
n(x-n) = (x-l) + (x-3) + ... + (x-2n+l). If 2n - 1 < x < 2n they are,
similarly, both equa:i to n(x - n). (Looking ahead to Chapter 3, the formula
Li(x + l)] (x - [i(x + l)]) covers both cases.)
2.33 If K is empty, AkEK ak = M. The basic analogies are:
kEK
H /j (C -t ak) = C + A ak
kCK kEK
~!ak+bk! = t%+tbk H A min(ak,bk)
ktK kcK kEK kEK
t
kcK
jEJ
kEK
t
kEK
ak =
>, apikl
PikICK
jEJ kEK
ak = ta,[kEK]
k
&K
= min(r\ ak, A bk)
kCK
H A ak = // aplki
PIUEK
H A ai.k = A // aj.k
iEl ieJ kEK
kEK
kEK
H A ak = Aak.aslkeKi
A permutation that 2.34 Let K+ = {k 1ok 3 0} and K = {k 1 ak < 0). Then if, for example, n
consumes terms of
one sign faster than
is odd, we choose F, to be F, -I U E,, where E, g KP is sufficiently large that
those of the other 1 kc(F, ,nK* ) ak - t-k@,, tpak) < A
can steer the sum
toward any value
that it l&s.
2.35 Goldbach’s sum can be shown to equal
as follows: By unsumming a geometric series, it equals xkEP,La, k ‘; therefore
the proof will be complete if we can find a one-to-one correspondence
between ordered pairs (m, n) with m,n 3 2 and ordered pairs (k, 1) with
k E P and 1 3 1, where m” = k’ when the pairs correspond. If m 4 P we let
(m,n) H (m”, 1); but if m = ah E P, we let (m,n) H (an,b).
kEK
k