Re: IBM Ponder This August 2007
No, I don't think {f(n)/f(n-1)} as stated in the question
is the floor, for then the result isn't as interesting!
Hint: <font color="white"> Look at the sequence of fractions
j/k starting with f(1)/f(0), f(2)/f(1), etc. </font>
I haven't yet found an "elegant" proof.
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