Concrete mathematics : a foundation for computer science

548 ANSWERS TO EXERCISES
1 + a. Hence p1 = (-b + dm)/Zc = 1 + 4; and this implies that
o = -c, b = -2c, p2 = 1 - fi. The generating function now takes the form
z(m - (r + m)z)
G(z) = (1 -22-z2)(1 -2)
-r + (m + 2r)z r
= 2(1-22-z') +2(1-z!
= mz+ (2m-r)z2 +... ,
where r = d/c. Since g2 is an integer, r is an integer. We also have
9 n= a(1 +Jz)n +a(1 -Jz)n+ tr = [cx(l+Jz)"],
and this can hold only if r = -1, because (1 - a)” alternates in sign as
it approaches zero. Hence (a, b, c, d) = *( 1,2, -1,l). Now we find o( =
i (1 + fi m), which is between 0 and 1 only if 0 < m 6 2. Each of
these values actually gives a solution; the sequences (g,,) are (O,O, 1,3,8,. . .),
(0,1,3,8,20 ,... ), and (0,2,5,13,32 ,... ).
7.49 (a) The denominator of (l/(1 - (1 + fl)z) + l/(1 - (1 - &!)z)) is
1 - 22 - z2; hence a,, = 2a,-l + a,-2 for n 3 2. (b) True because a,, is even
and -1 < 1 - d < 0. (c) Let
b, = (!?+!t?)n+(v)n
We would like b, to be odd for all n > 0, and -1 < (p - Jsi)/Z < 0. Working
as in part (a), we find bo = 2, bl = p, and b, = pb,~-l + i(q - p2)bn. 2 for
n 3 2. One satisfactory solution has p = 3 and q = 17.
7.50 Extending the multiplication idea of exercise 22, we have
Q = -+ QAQ + QfjQ + QQdQ +..
Replace each n-gon by znm ‘. This substitution behaves properly under multiplication,
because the pasting operation takes an m-gon and an n-gon into
an (m + n - 2)-gon. Thus the generating function is
Q = 1+zQ2+z2Q3+z3Q4+... = I.+*
1 -zQ
and the quadratic formula gives Q = (1 +z-dl - 6z + z2 ) /2z. The coefficient
of z” -’ in this power series is the number of ways to put nonoverlapping
diagonals into a convex n-gon. These coefficients apparently have no closed Give me Legenform
in terms of other quantities that we have discussed in this book, but ~~~~$~j~~~~ a
their asymptotic behavior is known [173, exercise 2.2.1-121.
closed form.