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Concrete mathematics : a foundation for computer science

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6.7 CONTINUANTS 289<br />

We also know that the total number of terms in a continuant is a Fibonacci<br />

number; hence we have the identity<br />

F,,+I = 2 (“; “)<br />

k=O<br />

(6.130)<br />

(A closed <strong>for</strong>m <strong>for</strong> (6.12g), generalizing the Euler-Binet <strong>for</strong>mula (6.123) <strong>for</strong><br />

Fibonacci numbers, appears in (5.74).)<br />

The relation between continuant polynomials and Morse code sequences<br />

shows that continuants have a mirror symmetry:<br />

K(x,, . . . , x2,x1) = K(x1,xr,...,xn). (6.131)<br />

There<strong>for</strong>e they obey a recurrence that adjusts parameters at the left, in addition<br />

to the right-adjusting recurrence in definition (6.127):<br />

K,(xI,... ,%I) = XI& 1(X2,...,&1) +Kn 2(x3,...,&). (6.132)<br />

Both of these recurrences are special cases of a more general law:<br />

K m+*(X1,...,X,,X,+1,~..,x~+~)<br />

= K,(xl,...,x,)K,(x,+~,...,x,+,)<br />

+kn I(xI,...,x, l)K, 1(~,+2,...,~rn+n). (6.133)<br />

This law is easily understood from the Morse code analogy: The first product<br />

K,K, yields the terms of K,+, in which there is no dash in the [m, m + 11<br />

position, while the second product yields the terms in which there is a dash<br />

there. If we set all the x’s equal to 1, this identity tells us that Fm+n+l =<br />

Fm+lF,+l + F,F,; thus, (6.108) is a special case of (6.133).<br />

Euler [90] discovered that continuants obey an even more remarkable law,<br />

which generalizes Cassini’s identity:<br />

K m+n(Xlr.~~ t Xm+n) Kk(Xm+l, . . . , %n+k)<br />

= kn+k(Xl, . . . rX,+k)K,(x,+l,...,x,+,)<br />

+ (-l)kKm I(XI,...,X, l)Kn k 1(%n+k+2,...,Xm+,). (6.134)<br />

This law (proved in exercise 29) holds whenever the subscripts on the K’s are<br />

all nonnegative. For example, when k = 2, m = 1, and n = 3, we have<br />

K(xl,x2,x3,x4)K(x2,x3) = K(Xl,X2,X?,)K(XL,X3,X4) +1<br />

Continuant polynomials are intimately connected with Euclid’s algorithm.<br />

Suppose, <strong>for</strong> example, that the computation of gcd(m, n) finishes

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