Show that where $f(n)$ denotes the greatest odd divisor of a positive integer $n$, $f(n+1)+f(n+2)+...+f(2n)=n^2$.

edit: This post has been solved:

Merely induct on n, and you will find that as f(2n+1)=2n+1, we get (n+1)^2=n^2+2n+1 which is clearly true for all n. See solution below.


closed as off-topic by mfl, I am Back, Carl Schildkraut, Namaste, Brevan Ellefsen Oct 11 '17 at 2:34

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  • $\begingroup$ This is a pretty nice property. Absolutely not trivial... $\endgroup$ – Raffaele Oct 10 '17 at 20:56
  • $\begingroup$ I first attempted to do casework based on whether n was even or odd. This didn't get me very far though. $\endgroup$ – mathcounter Oct 10 '17 at 21:14

Induction: $n = 1$ is easy; both sides are equal to $1$.

Suppose true for some $n > 0$. Now we prove for $n + 1$: $$f(n+2) + \dots + f(2n) + f(2n + 1) + f(2n + 2) = f(n+1) + f(n+2) + \dots + f(2n) + (2n+1).$$ Here I've used $f(2n+2) = f(2(n+1)) = f(n+1)$ and $f(2n+1) = 2n + 1$.

Now the RHS of the indented equation is, by inductive hypothesis, $n^2 + (2n + 1) = (n+1)^2$, as required.

  • 1
    $\begingroup$ Wow, very smart approach. Thank you very much! $\endgroup$ – mathcounter Oct 10 '17 at 21:12

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