Concrete mathematics : a foundation for computer science

290 SPECIAL NUMBERS
in four steps:
@Cm, n) = gcd(no, nl 1
= gcd(nl , n2 1
Then we have
n4
= gcd(nr,n3’l
= gcd(n3, na‘i
= gcd(ns,O) = n4
== n4 = K()n4 ;
n3 =I q4n4 = K(q4h;
w =I qm +n4 = K(q3,q4h;
nl =T q2n2 +n3 = K(qZlq3,q4)n4;
no =T qlnl +w = K(ql,q2,q3,q4h
no = m, nl =n;
n2 = nomodn, = no-qlnl;
n3 = nl mod n2 = nl - q2n2 ;
n4 = nzmodn3 = nz-q3n3;
0 = n3 modn4 = n3 - q4n4.
In general, if Euclid’s algorithm finds the greatest common divisor d in k steps,
after computing the sequence of quotients ql, . . . , qk, then the starting numbers
were K(ql,qz,.. . ,qk)d and K(q2,. . . , qk)d. (This fact was noticed early
in the eighteenth century by Thomas Fantet de Lagny [190], who seems to
have been the first person to consider continuants explicitly. Lagny pointed
out that consecutive Fibonacci numbers, which occur as continuants when the
q’s take their minimum values, are therefore the smallest inputs that cause
Euclid’s algorithm to take a given number of steps.)
Continuants are also intimately connected with continued fractions, from
which they get their name. We have, for example,
a0 +
1 = K(ao,al,az,a3)
1 -K(al,az,a3) '
a1 + ~
1
a2 + G
(6.135)
The same pattern holds for continued fractions of any depth. It is easily
proved by induction; we have, for example,
K(ao,al,az,a3+l/a4) := K(ao,al,a2,a3,a4)
K(al, az, a3 + l/a41 K(al,az,as,ad) ’
because of the identity
K,(xl,. . . ,xn-lrxn+Y)
= K,(x,,...,xn~l,x,)+Kn-l(xl,...,xn~l)~
(This identity is proved and generalized in exercise 30.)
(6.136)