For every positive integer $k$ let $a(k)$ be the largest integer such that $2^{a(k)}$ divides $k$. For every positive integer $n$ determine $a(1)+a(2)+\cdots+a(2^n)$.
Problem
Source: Pan African MO 2006 Q4
Tags: floor function, number theory unsolved, number theory
Particle
01.05.2013 18:39
The sum is equal to $v_2(n!)=\sum _{k\ge1}\left \lfloor \dfrac {n}{2^k}\right \rfloor$ @djb86, this problem has nothing to do with order, so it would be better if you change the topic name.
mathuz
01.05.2013 19:26
good solution by mr. Particle
djb86
05.05.2013 21:29
Particle wrote: @djb86, this problem has nothing to do with order, so it would be better if you change the topic name. Yes, you are right. In my mind I was mixing terminology between order and valuation. Hope this is better.
JoelBinu
15.06.2017 20:28
Hello, Is it right to say that for any $n$, then Sum of the function is $$n^2 - 1$$ Thank you
Rukevwe
09.05.2023 14:04
JoelBinu wrote: Hello, Is it right to say that for any $n$, then Sum of the function is $$n^2 - 1$$ Thank you I believe it should be $2^n-1$.