Concrete mathematics : a foundation for computer science

6.7 CONTINUANTS 291
Moreover, continuants are closely connected with the Stern-Brocot tree
discussed in Chapter 4. Each node in that tree can be represented as a
sequence of L’s and R'S, say
RQO La’ R”Z L”’ . . . Ran-’ LO-“-’ , (6.137)
where a0 3 0, al 3 1, a2 3 1, a3 3 1, . . . , a,-2 3 1, an 1 3 0, and n is
even. Using the 2 x 2 matrices L and R of (4.33), it is not hard to prove by
induction that the matrix equivalent of (6.137) is
K,-2(al,. . . ) an-21 Kn-l(al,...,an-2,an I)
K,-l(ao,al,...,an-2) Kn(ao,al,...,an~~2,an~l)
(The proof is part of exercise 80.) For example,
R”LbRcLd = bc + 1 bcd+b+d
abc + a + c abcd+ab+ad+cd+l
Finally, therefore, we can use (4.34) to write a closed form for the fraction in
the Stern-Brocot tree whose L-and-R representation is (6.137):
f(R"" ., .L"-') := Kn+l(ao,al,...~an~l,l)
K,(al,. . . , an-l, 1 I
(6.139)
(This is “Halphen’s theorem” [143].) For example, to find the fraction for
LRRL we have a0 = 0, a1 = 1, a2 = 2, a3 = 1, and n = 4; equation (6.13~)
gives
K(O, 1,&l, 1) KC4 1,l) U&2) 5
K(l,Ll,l) = K(1,2,1,1) =-=-
K(3,2) 7 ’
(We have used the rule K,(xl,. . . ,x,-l, x, + 1) = K,+, (XI,. . . ,x,-r ,x,,, 1) to
absorb leading and trailing l’s in the parameter lists; this rule is obtained by
setting y = 1 in (6.136).)
A comparison of (6.135) and (6.13~) shows that the fraction corresponding
to a general node (6.137) in the Stern-Brocot tree has the continued
fraction representation
f(Rao.. . Lo-+’ ) = a0 +
al +
a2 +
1
. . . +
1
1
1
1
an I+- 1
(6.140)