## August 2007

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Ponder This Solution:

The sequence {f(n)/f(n-1)} for positive integers n-s covers all the positive rational numbers in reduced form.

If n is odd (=2m+1), from the definition of f(n) (modulo 2) we get that 1 (=2^0) appears exactly once and hence

(1a) f(2m+1) = f(m).

If n is even (=2m), 1 either does not appear at all or appears twice. In the first case, we get f(m) possibilities and in the second we get f(m-1). Therefore,

(1b) f(2m)=f(m)+f(m-1).

Clearly f(n)>0 and therefore

(2) f(2n-1) = f(n-1) < f(n-1)+f(n) = f(2n).

Claim 1: gcd(f(n),f(n-1))=1

Proof: By induction on n. The base (n=1) is trivial: gcd(f(1),f(0))=gcd(1,1)=1

Suppose the induction hypothesis holds for k<n, then from (1a) & (1b):

gcd(f(n),f(n-1)) = gcd(f(2m+1),f(2m)) = gcd(f(m), f(m)+f(m-1)) = gcd(f(m),f(m-1)) = 1

or

gcd(f(n),f(n-1)) = gcd(f(2m),f(2m-1)) = gcd(f(m)+f(m-1), f(m-1)) = gcd(f(m),f(m-1)) = 1

qed.

Claim 2: f(n+1)/f(n) = f(n’+1)/f(n’) implies that n=n’

Proof: Again, by induction on max(n,n’). The base (max(n,n’)=0) is trivial since n=n’=0.

Assume the induction hypothesis holds for k<n, then from (2) we get that

f(n)/f(n-1) = f(n’)/f(n’-1) can only hold when n=n’(mod 2) and from (1a)&(1b) we get

f(m)/f(m-1) = f(m’)/f(m’-1) and by induction m=m’ yields n=n’.

qed.

Claim 3: For every positive rational number r, there exists a positive integer n such that r = f(n)/f(n-1).

Proof: Write r as a quotient of two relative prime natural numbers p and q.

We prove by induction on max(p,q).

The base (p=q=1) is trivial: n=1.

Assume the induction hypothesis holds for all k<n.

If p<q (q<p), then define r’ as p’/q’ where p’=p & q’=q-p (p’=q & q’=p-q). r’ satisfies the induction hypothesis and therefore r’ = f(m)/f(m-1). Hence r = f(2m+1)/f(2m) (f(2m)/f(2m-1))

qed.

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