Concrete mathematics : a foundation for computer science

7.1 DOMINO THEORY AND CHANGE 315
How many pennies Obviously P, = 1 for all n 3 0. And a little thought proves that we have
are there, really? N, = Ln/5J + 1: To make n cents out of pennies and nickels, we must choose
If n is greater
than, say,
either 0 or 1 or . . . or Ln/5] nickels, after which there’s only one way to supply
10”)
I bet that P, = 0 the requisite number of pennies. Thus P, and N, are simple; but the values
in the “real world.” of Dn, Qn, and C, are increasingly more complicated.
One way to deal with these formulas is to realize that 1 + zm + 2’“’ +. . .
is just l/(1 - 2”‘). Thus we can write
P = l/(1 -2’1,
N = P/(1 -i’),
D = N/(1 - 2”) ,
Q = D/(1 - zz5) ,
C = Q/(1 -2”).
Multiplying by the denominators, we have
(l-z)P = 1,
(1 -z5)N = P,
(l-z”)D = N,
(~-z~~)Q = D,
(1-z5’)C = Q.
Now we can equate coefficients of 2” in these equations, getting recurrence
relations from which the desired coefficients can quickly be computed:
P, = P,-I + [n=O] ,
N, = N-5 + P,,
D, = Dn-IO -tN,,
Qn = Qn-25 -t D,,
Cn = G-50 + Qn.
For example, the coefficient of Z” in D = (1 - z~~)Q is equal to Q,, - Qnp25;
so we must have Qll - Qnp25 = D,, as claimed.
We could unfold these recurrences and find, for example, that Qn =
D,+D,-zs+Dn~5o+Dn~75+..., stopping when the subscripts get negative.
But the non-iterated form is convenient because each coefficient is computed
with just one addition, as in Pascal’s triangle.
Let’s use the recurrences to find Csc. First, Cso = CO + Q50; so we want
to know Qso. Then Q50 = Q25 + D50, and Q25 = QO + D25; so we also want
to know D50 and 1125. These D, depend in turn on DUO, DUO, DUO, D15,
DIO, D5, and on NSO, NC,, . . . , Ns. A simple calculation therefore suffices to