Concrete mathematics : a foundation for computer science

544 ANSWERS TO EXERCISES
7.24 ntk,+...+k,rn kl . . . k,/m = Fz,+l + FznPl - 2. (Consider the
coefficient [znmll &ln(l/(l -G(z))), where G(z) =z/(l -z)~.)
7.25 The generating function is P(z)/(l - z'~), where P(z) = z + 2z2 +
. . + (m - 1 )z+’ = ((m - 1 )z”‘+’ - mz”’ + z)/( 1 - 2)‘. The denominator
is Q(z) = 1 - zm = (1 - cu’z)(l - w'z)...(l - cumm’z). By the rational
expansion theorem for distinct roots, we obtain
nmodm =zp + mg s .
k=l
7.26 (1 - z - z2)5(z) = F(z) leads to 5, = (2(n + l)F, + nF,+1)/5 as in
equation (7.60).
7.27 Each oriented cycle pattern begins with 8 or z or a 2 x k cycle (for
some k 3 2) oriented in one of two ways. Hence
Qn = Qn-I + Qn-2 +2Qnm 2 +2Qnp3 +...+ZQo
for n 3 2; Qo = QI = 1. The generating function is therefore
Q(z) = zQ(z) +z'Q(z) +2z'Q(z)/(l -z)+l
= l/(1-z-z2-222/(1-z))
(1 -z)
= (l-2z-2z2+23)
a2/5
=jqq+ c2/5 + 2/5
l-+2z l+z'
and Q,, = (+2n+2 + +~~2nP2 + 2(-1)“)/5 = (($‘l+l - $n+1)/&)2 = Fi,,.
7.28 In general if A(z) = (1 + z + ... + zmp' )B(z), we have A, + A,,, +
A rf2m + “’ = B(1) for 0 < r < m. In this case m = 10 and B(z) =
(1+z+~~~+z9)(1+z2+z4+z6+z8)(1+z5).
7.29 F(z)+ F(z)'+ F(z)~ +. . . =z/(l-z-z2-z)=(l/(l-(l+&)z)-
(l/(1 - (1 - vmlJ8, so the answer is ((1 + fi)" - (1 - fi)n)/J8.
7.30 XL=, (2”nm’,Pk) (anbnPk/(l -olz)k+anPkbn/(l -@z)~), by exercise 5.39.
7.31 The dgf is