Let's denote the largest odd divisor of a positive integer n by lod(n).
Thus lod(1)=1; lod(72)=9; lod(2k+1)=2k+1; lod(2^k)=1.
Prove the following statement:
Sum of all values of lod(k), (k>1) from k=n+1 to k=2n, inclusive, equals n2.
Example: take n=7: lod(8,9,10,11,12,13,14)= (1,9,5,11,3,13,7), sum of the values within the last pair of brackets is 49, indeed.
Source: Shown to me as a trick i.e. "the wizard" guesses the result a priori.
(In reply to
Alternate approach by Paul)
I agree with chun. This is a very cool proof, and one which explains why this works. Thanks, very much Paul.
Now that I have seen the proof, it occurs to me that it can be simplified. There are easier ways to say that lod(x) <> lod(y) if x and y are both between (n+1) and 2n.
None of which is meant to take away from this being a great proof.
re: Alternate approach
