TITLE: Why does this sequence converges to $\pi$?
QUESTION [35 upvotes]: One of my daughters was having a small programming exercise. 
Let's consider following algorithm:

Take a list of length $n$: $\ (1\,\ 2\,\ \ldots\,\ n)$.
Remove every $2$nd number.
From the resulting list, remove every $3$rd number.
From the resulting list, remove every $4$th number.
... Follow on until the list remains unchanged and let $u_n$ be the number of remaining elements.

Example with $n=11$

$(\ 1\,\ 2\,\ 3\,\ 4\,\ 5\,\ 6\,\ 7\,\ 8\,\ 9\,\ 10\,\ 11\ )\quad \Rightarrow\quad (\ 1\ *\ 3\ *\ 5\ *\ 7\ *\ 9 \ *\ 11\ )$
$(\ 1\,\ 3\,\ 5\,\ 7\,\ 9\,\ 11\ )\quad \Rightarrow\quad (\ 1\,\ 3\ *\ 7\,\ 9\ *\ )$
$(\ 1\, 3\,\ 7\,\ 9\ )\quad \Rightarrow\quad (\ 1\,\ 3\,\ 7\ *\ )$
$(\ 1\,\ 3\,\ 7\ )\ $ -- will not be modified anymore, and therefore $u_n=3$.

QUESTION:   why do we have $\lim\limits_{n \to +\infty} \frac{n}{u_n^2}=\frac{\pi}{4}$ ?
Thanks!

REPLY [4 votes]: This is an extended comment: Interestingly enough, displaying the differences of consecutive terms of A000960 shows an amazing degree of fluctuation.