Concrete mathematics : a foundation for computer science

520 ANSWERS TO EXERCISES
The infinite series converges because the terms for fixed j are dominated by
a polynomial in j divided by 2j. Now sum over k, getting
Absorb the j + 1 and apply (5.57) to get the answer, 4(m+ 1).
5.67 3(2nntt52) by (5.26), because
(‘i’) = 3(y).
5.68 Using the fact that
we get “(2” ’ - (,zij,)).
[n is even] ,
5.69 Since (k:‘) + (‘y’) < (:) + (i) W k < 1, the minimum occurs
when the k’s are as equal as possible. Hence, by the equipartition formula of
Chapter 3, the minimum is
(n mod m) -t (n - (n mod m))
b/ml
-4 2 >
$- (n mod m) : .
L i
A similar result holds for any lower index in place of 2.
5.70 This is F(-n, i; 1;2); but it’s also (-2)Pn(F)F(-n, -n; i -n; i) if we
replacekbyn-k. NowF(-n,-n;i-n;:) =F(-f,-l;&n;l)byGauss’s
identity (5.111). (Alternatively, F(-n,-n; i-n; i) = 2-“F(-n, i; i-n; -1)
by the reflection law (5.101), and Kummer’s formula (5.94) relates this to
(5.55).) The answer is 0 when n is odd, 2-“(,,y2) when n is even. (See [134,
$1.21 for another derivation. This sum arises in the study of a simple search
algorithm [ 1641.)
5.71 (a) S(z) = EkZO okzm-+k/(l -Z)m+Zk+’ = Zm(l -2) -“-‘A@/(1 -z)‘).
(b) Here A(z) = x k20 (2,“)(-z)k/(k + 1) = (dm - 1)/2z, so we have
A(z/(l -z)‘) = 1 -z. Thus S, = [z”] (z/(1 - 2))“’ = (;I;).
5.72 The stated quantity is m(m - n) . . . (m - (k - l)n)nkPYik’/k!. Any
prime divisor p of n divides the numerator at least k - y(k) times and divides
the denominator at most k - v(k) times, since this is the number of