Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 501
4.8 We want 1 Ox + 6y = 1 Ox + y (mod 15); hence 5y = 0 (mod 15); hence
y s 0 (mod 3). We must have y = 0 or 3, and x = 0 or 1.
4.9 32k+’ mod 4 � = 3, so (3 2k+’ -1)/2 is odd. The stated number is divisible
by (3’ - 1)(2 and (3” - 1)/2 (and by other numbers).
4.10 999(1 - ;)(l -- A) = 648.
4.11 o(O) = 1; o(1) = -1; o(n) = 0 for n > 1. (Generalized Mobius
functions defined on. arbitrary partially ordered structures have interesting
and important properties, first explored by Weisner [299] and developed by
many other people, notably Gian-Carlo Rota [254].)
4.12 xdim tkid P(d/k) g(k) = tk\,,, td\(m/k) CL(d) g(k) = &,,, g(k) X
[m/k= 11 = s(m), by (4.7) and (4.9).
4.13 (a) nP 6 1 for all p; (b) p(n) # 0.
4.14 True when k :> 0. Use (4.12), (4.14), and (4.15).
4.15 No. For example, e, mod 5 = [2or 31; e, mod 11 = [2,3,7, or lo].
4.16 l/e, +l/e~+~~~+l/e,=l-l/(e,(e,-l))=l-l/(e,+I -1).
4.17 We have f, mod f, = 2; hence gcd(f,, f,) = gcd(2,f,) = 1. (Incidentally,
the relation f, = fof, . ,. f,-l + 2 is very similar to the recurrence
that defines the Eucl.id numbers e,.)
4.18 Ifn= qmand q isodd, 2”+1 = (2m+1)(2n~m-2n~2m+~~~-2m+1).
4.19 Let p1 = 2 and let pn be the smallest prime greater than 2Pnm1. Then
2Pvl < pn < 2Pn-I t1 , and it follows that we can take b = lim,,, Igin) p,,
where Igin) is the function lg iterated n times. The stated numerical value
comes from p2 = 5, p3 = 37. It turns out that p4 = 237 + 9, and this gives
the more precise value
b FZ 1.2516475977905
(but no clue about ps).
4.20 By Bertrand’s, postulate, P, < 10". Let
K = x 10PkZPk = .200300005,. , .
k>l
Then 10nLK = P, + fraction (mod 10Znm ').
4.21 The first sum is n(n), since the summand is (k + 1 is prime). The
inner sum in the second is t,Gk