Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 531
6.49 Set z = i in (6.146); the partial quotients are 0, 2F3, 2F1, 2Fl, . .
(Knuth [172] noted that this number is transcendental.)
6.50 (a) f(n) is even tl 3\n. (b) If the binary representation of n is
(la'o"z... lam ‘o”m )2, where m is even, we have f(n) = K(a’,az,. ,a, ‘).
6.51 (a) Combinatorial proof: The arrangements of {l ,2,. . , p} into k subsets
or cycles are divided into “orbits” of 1 or p arrangements each, if we
add 1 to each element modulo p. For example,
{1,2,4I'J{3,51 + {2,3,5IuI4,11 + 13,4,lIu{5,21
--f 14,5,2P~u,31 + 15,1,31w2,41 + 11,2,41~{3,51.
We get an orbit of size 1 only when this transformation takes an arrangement
into itself; but then k = 1 or k = p. Alternatively, there’s an algebraic proof:
WehavexP-xfl-+xLandxE=xP-x (mod p), since Fermat’s theorem tells
us that x” -x is divisible by (x-0)(x - 1). . . (x - (p-l)).
(b) This result follows from (a) and Wilson’s theorem; or we can use
xp-’ E xF/(x-1) - l(XP -x)/(x- 1) =xp ' +xp-2 + . ..+x.
(c) We have {“l’} E [“:‘I z 0 for 3 6 k 6 p, then {“12} E [pl’] G 0
for 4 < k < p, etc. (Similarly, we have [‘PpP’] s -{2ppm’} E 1.)
(d) p! =pP= tk(-l)pPkpk[;] =pP[~]-pP~'[pp,]+...+p3[~]~
P'[;] +P[$ But P[:] = P!, so
[I] = P[I] --P$] +...+ppyP]
is a multiple of p2. (This is called Wolstenholme’s theorem.)
6.52 (a) Observe that H, = Hk + HlnlP~/p, where Hz = xc=,(k I p)/k.
(b) Working mod 5 we have H, = (0, 1,4,1,0) for 0 < r < 4. Thus the first
solution is n = 4. By part (a) we know that 5\a, =$ 5\ajn,sl; so the next
possible range is n == 20 + r, 0 < r 6 4, when we have H, = Ht + &H4 =
H;e+~Hq+H,+~~=, 20/k(20+k). The numerator of H;,, like the numerator
of HJ, is divisible by 25. Hence the only solutions in this range are n = 20
and n = 24. The next possible range is n = 100 + r; now H, = H; + &Hzo,
which is ;Hzo + H, plus a fraction whose numerator is a multiple of 5. If
$Hzo % m (mod 5), where m is an integer, the harmonic number H’~o+~ will
have a numerator divisible by 5 if and only if m + H, e 0 (mod 5); hence
m must be E 0, 1, or 4. Working modulo 5 we find $Hls = &Hz0 + &H4 3
-&Hh = & E 3; hence there are no solutions for 100 6 n < 104. Similarly
there are none for 120 < n 6 124; we have found all three solutions.
(By exercise 6.51(d), we always have p2\apP,, p\apzPp, and p\a,r-,,
if p is any prime 3 5. The argument just given shows that these are the only