Concrete mathematics : a foundation for computer science

6.6 FIBONACCI NUMBERS 281
form kn if F, mod F, = 0.) First we have F, mod Fi = F,; that’s not zero.
Next we have
F2n = FnFn+l + F,-lF, = 2F,F,+l (mod Fi) ,
by (6.108), since F,+I E F,-l (mod F,). Similarly
F2,+1 = Fz+l + Fi E Fi+l (mod F,f).
This congruence allows us to compute
F3n = F2,+1 Fn + FznFn-I
= Fz+lF, + (ZF,F,+I)F,+I = 3Fz+,F, (mod Fi) ;
F3n+1 = F2n+1 Fn+l + F2nFn
=- F;t+l + VFnF,+l IF, = F:+l
In general, we find by induction on k that
Fkn E kF,F,k+; and Fk,,+l E F,k+, (mod F:).
Now Fn+l is relatively prime to F,, so
Fkn = 0 (mod Fz) tl kF, E 0 (mod F:)
We have proved Matijasevich’s lemma.
W k E 0 (mod F,).
(mod F,f) .
One of the most important properties of the Fibonacci numbers is the
special way in which they can be used to represent integers. Let’s write
j>>k j 3 k+2. (6.112)
Then every positive integer has a unique representation of the form
n = h, + Fkz + . . . + Fk, , kl > kz >> . . . > k, >> 0. (6.113)
(This is “Zeckendorf’s theorem” [201], [312].) For example, the representation
of one million turns out to be
1~~0000 = 832040 + 121393 + 46368 + 144 + 55
= F30 + F26 + F24 + FIZ +Flo.
We can always find such a representation by using a “greedy” approach,
choosing Fk, to be the largest Fibonacci number 6 n, then choosing Fk2
to be the largest that is < n - Fk,, and so on. (More precisely, suppose that