Numbers $1,2,3, ...., 2016$ arranged in a circle A student wrote numbers $1,2,3, ...., n$ arranged in a circle, then began with erasing number $1$ then he Leaves $2$ then he erased $3$ and leaves $4$ ......
exemple: if $n=10$ :
in the first round he erased $1,3,5,7,9$ , and leaves $2,4,6,8,10$
in the second round he erased $2,6,10$ and leaves $4,8$
in the third round he leaves $4$ because he erased the last number $(10)$ in the second round, then he erased $8$
The last remaining number in the circle is $4$
if $n=2016$ , what is the last remaining number in the circle?
My progress:
In the first round all the odd numbers will be erased $ n=1(\mod2) $
In the second round $ n=2(\mod4) $ will be erased, then $ n=4(\mod8) $....
but in the seventh round we have $ n=0(\mod128) $

Let it be that $a_{n}$ is the remaining number if we are dealing
with the numbers $1,2,3,4,5\dots,n$.
If $n$ is sufficiently large then after erasing $1$ and leaving
$2$ we have the numbers $3,4\dots,n,2$ ahead of us.
Comparing this with the situation in which we have the numbers $1,2,3,\dots,n-2,n-1$
ahead of us we conclude that $a_{n}=a_{n-1}+2$ if $a_{n-1}<n-1$
and $a_{n}=2$ otherwise.
So next to the evident $a_1=1$ we have the equality: $$a_{n+1}=\begin{cases}
a_{n}+2 & \text{ if }a_{n}<n\\
2 & \text{otherwise}\end{cases}$$
Looking at sequence $\left(a_{n}\right)$ the following conjecture
arises:
$$a_{n}=2n-2^{\lceil\log_{2}n\rceil}$$
This conjecture can be proved with induction and we conclude: $$a_{2016}=4032-2^{11}=1984$$
