(In reply to Some Observations (partial solution)
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(2a-1) = 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=2a-1, y=2b-1 so x+y = 2(a+b-1)
- 3) one of each x=2a, y=2b-1 (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+b-1) = 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 2b-1 to 2b).
Since this 2b must be even, the sum of 2011 is not possible and the sum of 2012 requires 2b-1 to be 2011.
Posted by Jer
on 2012-01-07 20:29:52