Concrete mathematics : a foundation for computer science

294 SPECIAL NUMBERS
numbers are 1, 4, 6, 9, 12, 14, 17, and 19. If k,(n) is even, then n - 1 is
F-even, by (6.114); similarly, if k,(n) is odd, then n - 1 is F-odd. Therefore
k,(n) is even M n - 1 is F-even.
Furthermore, if k,(n) is even, (6.144) implies that kT( [n+]) = 2; if k,(n) is
odd, (6.144) says that kr( [rt@]) = k,(n) + 1. Therefore k,.( [n+J) is always
even, and we have proved that
In@] - 1 is always F-even.
Conversely, if m is any F-even number, we can reverse this computation and
find an n such that m + 1 == Ln@J. (First add 1 in F-notation as explained
earlier. If no carries occur, n is (m + 2) shifted right; otherwise n is (m + 1)
shifted right.) The right-hand sum of (6.143) can therefore be written
x z LQJ = zt zm [m is F-even] ,
TL>l ll@O
(6.145)
How about the fraction on the left? Let’s rewrite (6.143) so that the
continued fraction looks like (6.141), with all numerators 1:
zcFfi +
1
-=-
1-Z
,lMJ .
1 z z
z-h + ' 1
z-F2 + '-
lI>l
(6.146)
(This transformation is a bit tricky! The numerator and denominator of the
original fraction having zFn as numerator should be divided by zFnmI .) If
we stop this new continued fraction at l/zPFn, its value will be a ratio of
continuants,
K,,.z(O, 2~~0, zPFI,. . . ,zPFn) K,(z/ , . . . , z-~,)
-=
K,+, (z-~o,z~~I,. . . ,zpFn) K,+, (z-~o, z-~I,. . , z-~,) ’
as in (6.135). Let’s look at the denominator first, in hopes that it will be
tractable. Setting Qn = K,+l z Fo,. . ,zPFn), we find Q. = 1, Q, = 1 + z-l,
Q 2 = 1 -tz--’ + -2 Q = 1 ‘-I
z, 3 $ z + z-2 + zP3 + zP4, and in general everything
fits beautifully and gives a geometric series
Q,, = 1 + z-’ + z-2 + . . . + z-(Fn+2-l 1 .