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.

good solution by mr. Particle

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.

Hello, Is it right to say that for any $n$, then Sum of the function is $$n^2 - 1$$ Thank you

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 $2^n-1$.