(In reply to
Solution by Praneeth)
I think I have figured out Praneeth's original intent:
F(n) = F(n2) + F(n1)
We want to express F(n) with that number (i.e., F(n)) of F(n1)'s. We already have one F(n1) as the second term on the RHS, so we need F(n)1 more F(n1)'s.
Well, F(n2) = 1 + 1 + 1 + ... + 1 + 1, using a total of F(n2) 1's. For each of these 1's substitute F(n1)/F(n1). Now you have used an additional 2*F(n2) F(n1)'s.
So we still need to use an additional F(n)  1  2*F(n2) of the F(n1)'s.
In the 2/3 of cases, where F(n) is odd (as the Fib's sequence is odd, odd, even, odd, odd, even, etc.), this does work out as Praneeth has pointed out, as then the F(n)  1  2*F(n2) is even and these remaining F(n1)'s can be paired off in subtractions to contribute no further to the already achieved value of F(n).
However, when F(n) is even, such as 8 or 34, etc. this does not work, as the remaining F(n1)'s are odd in number and cannot be paired off in subtraction to produce zero. Praneeth had already pointed out F(n2) = 3 as an exception, but that is merely the case where F(n) = 8, an even number. But ALL even F(n) must be explained, not just F(n)=8, where F(n2)=3.

Posted by Charlie
on 20100710 03:26:55 