(In reply to
Solution by Praneeth)
I don't think that is the same thing.
Praneeth's solution was to produce an F(n2) using F(n) F(n1)'s.
My question asked for production of an F(n) using F(n) F(n1)'s.
So instead of making a 5 using 5 3's, Praneeth's solution (if I understand it correctly) would be to make a 2 using 5 3's. Even then, the 1st 4 3's can make the 2: 3/3 + 3/3, but then there's a pesky 3 leftover, not enough to make a zero, ie 33.

Posted by Larry
on 20100709 21:19:06 