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 One type of proof by Jer)

Excellent proof!  Excellent problem!

Posted by Steve Herman on 2017-09-10 17:12:35