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 Lots of lods (Posted on 2017-09-10)
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.

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 Alternate approach | Comment 5 of 7 |

Personally, I think Jer's inductive proof is much more elegant, but here's one of those alternate proofs he hoped for.

If lod(x) = z then x = z * 2^p
similarly if lod(y) = lod(x) = z then y = z * 2^q
(Also, lod(x) is necessarily <= x)

Assume x > y (which implies p > q), and consider x - y

x - y = z*2^p - z*2^q = z * 2^q * (2^(p-q) - 1)

2^(p-q) - 1 >= 1 (since p > q) and so
x - y >= z * 2^q
x - y >= y
x >= 2y

Suppose y is in the range [n+1, 2n]. Then x cannot be in the same range since it's at least 2y.

As a result, no two numbers in the range [n+1, 2n] can have the same lod.

There are n numbers in the range [n+1, 2n] whose lod's must come from [1,3,5...2n-1] and each of these must be unique. Since both sets contain n members, and repeats are forbidden, they must be permutations of each other, and therefore they have the same sum. That sum is just the sum of the first n odds: n^2

 Posted by Paul on 2017-09-11 11:34:05

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