Concrete mathematics : a foundation for computer science

332 GENERATING FUNCTIONS
Finally, since l/(1 -z")~ = xkao (k14)~'0k, we can determine the coefficient
of C, = [z”] C(z) as follows, when n = 1 Oq + r and 0 6 r < 10:
c lOq+r = ~Aj(k:4)[10q+r=10k+jl
= A:(‘:“) + A,+Io(‘;~) + A,+zo(~;‘) + A,+~o(‘;‘) . (7.40)
This gives ten cases, one for each value of r; but it’s a pretty good closed
form, compared with alterrratives that involve powers of complex numbers.
For example, we can u,se this expression to deduce the value of C50q =
Clog. Then r = 0 and we have
c50q = ("k") +45(q;3)+52(4;2) +2(“3
The number of ways to change 50# is (i) +45(t) = 50; the number of ways
to change $1 is ($) +45(i) -t 52(i) = 292; and the number of ways to change
$l,OOO,OOO is
= 66666793333412666685000001.
Example 5: A divergent series.
Now let’s try to get a closed form for the numbers gn defined by
40 = 1;
9 n = ngv1, for 11 > 0.
After staring at this for a Sew nanoseconds we realize that g,, is just n!; in Nowadayspeofact,
the method of summation factors described in Chapter 2 suggests this ~~~~‘e~c~~~
answer immediately. But let’s try to solve the recurrence with generating ~
functions, just to see what happens. (A powerful technique should be able to
handle easy recurrences like this, as well as others that have answers we can’t
guess so easily.)
The equation
9 n= ngn-1 + [n=Ol
holds for all n, and it leads to
G(z) = xgnz” = ~ng,-rz”+~z’.
n n n=O
To complete Step 2, we want to express t, ng, 1 2” in terms of G(z), and the
basic maneuvers in Table 320 suggest that the derivative G’(z) = t, ngnzn ’