Concrete mathematics : a foundation for computer science

508 ANSWERS TO EXERCISES
and (k- 1)/2 must be odd, etc. The necessary and sufficient condition is that
m be a product of distinct Nersenne primes.
4.59 Proof of the hint: If TL = 1 we have x1 = a = 2, so there’s no problem.
If n > 1 we can assume that x1 6 . . < x,. Case 1: xi’ + . . . + xi!, +
(x, - 1))’ 3 1 and x, > x+1. Then we can find p 3 x, - 1 3 x,-l such
that xl’ + ... +x;l, +P-' = 1; hencex, 6 p-t1 6 e, andxl...x, <
x1 . . . ~~~1 (p + 1) 6 el . . . e,, by induction. There is a positive integer m
such that a = x1 . . . x,/m; hence a 6 el . . . e, = e,+l - 1, and we have
x1 . . . ~~(~~+l) a( a + 1 ), because this identity is equivalent to
aa2-a’a+aa-a2+a+a > 0,
which is a consequence of aa( a - a) + (1 + a)a 3 ( 1 + a)a > a2 - a. Hence
we can replace x, and a by a - 1 and (3, repeating this transformation until
cases 1 or 2 apply.
Another consequence of the hint is that l/x, + . . . + l/x, < 1 implies
l/xl + ... +1/x, 6 l/e1 + ... +1/e,; see exercise 16.
4.60 The main point is that 8 < 5. Then we can take p1 sufficiently large
(to meet the conditions below) and pn to be the least prime greater than
p;-,. With this definition let a,, = 33”lnp, and b, = 3-nln(p, + 1). If we
can show that a,-1 < a,, < b, 6 b,-1, we can take P = lim,,, can as in
exercise 37. But this hypothesis is equivalent to pi-, < p,, < (p,-l + l)3. If
there’s no prime p,, in this range, there must be a prime p < p;-, such that
p + cpe > (p,-1 + 1 )3. But this implies that cpe > 3p213, which is impossible
when p is sufficiently large.
We can almost certainly take p1 = 2, since all available evidence indicates
that the known bounds on gaps between primes are much weaker than
the truth (see exercise 69). Then p2 = 11, p3 = 1361, p4 = 2521008887, and
1.306377883863 < P < 1.306377883869.
4.61 Let T?L and fi be the right-hand sides; observe that fin’ - m’fi = 1,
hence ??I. I T?. Also m/c >. m//n’ and N = ((n + N )/n’)n’ - n 3 li >
((n+N)/n’-l)n’- n - - -N-n’ 3 0. So we have T?-L/?L 3 m/‘/n”. If equality
doesn’t hold, we have n” = (&n’ - m’fi)n” = n’( tin” - m”fi) + fi(m”n’ -
m’n”) 3 n’ + fi > N, a contradiction.