(In reply to
Some Observations (partial solution) by Dustin)
I won't bother showing the proof of Dustin's assertion
When n is odd, F(n)=(n+1)/2
When n is even, F(n)=n/2
Other that to mention that each case can easily be proven by induction.
In other words f(2a)= a
and f(2a1) = a
Now lets consider each of the 3 possibilities for x and y:
 1) both are even x=2a, y=2b so x+y = 2a + 2b = 2(a+b)
 2) both are odd x=2a1, y=2b1 so x+y = 2(a+b1)
 3) one of each x=2a, y=2b1 (by symmetry it doesn't matter which) so x+y = 2(a+b)  1
Case 1) F(x) + F(y) + F(x+y) = a + b + (a+b) = 2(a+b)
In other words it is always negative if x and y are positive.
Case 2) F(x) + F(y) + F(x+y) = a + b + (a+b1) = 1
So you always get 1.
Case 3) F(x) + F(y) + F(x+y) = a + b + (a+b) = 2b
In other words the even one doesn't matter but the odd one increases by one (from 2b1 to 2b).
Since this 2b must be even, the sum of 2011 is not possible and the sum of 2012 requires 2b1 to be 2011.

Posted by Jer
on 20120107 20:29:52 