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I am trying to find all the natural numbers satisfying $2^(n-1)\mid n!$. I managed to prove that all naturals of the form $2^m$ satisfy this property:

$$ \nu_2((2^m)!) = \sum\limits_{k=1}^{m}\left\lfloor\frac{2^m}{2^k}\right\rfloor = \sum\limits_{k=1}^m 2^{m-k} = 2^m\left(\frac{1-\left(\frac{1}{2}\right)^{m+1}}{1-\frac{1}{2}}-1\right) = 2^m\left(1-\frac{1}{2^m}\right) = 2^m-1 $$ and therefore $2^{2^m-1}$ divides $\left(2^m\right)!$

My goal is to prove that this is the only case. I have managed to bound $\nu_2(n!)$ by $n$, but I want to prove that if $n\neq 2^m$, then $\nu_2(n!)$ is strictly bounded by $n-1$.

Joshhh
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1 Answers1

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Let $n$ be so that $2^{l-1} \leq n < 2^{l}$.

Then, as in the question to which this is a followup, we have $$n-1=\nu_2(n!) = \sum_{k=1}^{\lfloor\log_2(n)\rfloor} \Big\lfloor \frac{n}{2^k} \Big\rfloor \leq \sum_{k=1}^{l-1} \frac{n}{2^k} = n (1-\frac{1}{2^{l-1}})=n-\frac{n}{2^{l-1}}$$

This gives $n \leq 2^{l-1}$.

N. S.
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  • Why is the inequality true? I would assume the sum needs to be until $k=l$, or am I wrong? – Joshhh Mar 14 '16 at 09:03
  • @Joshhh We have $2^{l-1} \leq n < 2^l$. Then $\frac{n}{2^l}<1$ so the term with $k=l$ is zero. – N. S. Mar 14 '16 at 22:22