For $n\in \mathbb{N}$ numbers $I\_{n}=(1,2,3..n)$ and prime $p$, we define operation $(1,2,3..n)$ to $A=(a\_{1},a\_{2}...a\_{p-1})$ as follows:


We arrange the $n$ numbers in a circle, then we eliminate the first number, the $p$th number, the 2$p$th number, etc, until there is only $p-1$ numbers left and the process terminated. We identify this subset as $A$. 


My question is, for given $p$, does $A$ being equidistributed in $I\_{n}$ with $n\rightarrow \infty$? I feel that "equidistributed" in arbitrarily set seems to be not well defined. In this one I want at least for a subset of $I\_{n}$ of the form $S=(s,s+1...s+t-1)$. $|S\cap A|\rightarrow \frac{t}{n}\*(p-1)$ with $n\rightarrow \infty$. I do not know whether this is possible. A few simple cases (like $p$=3, $n$=2011) is already in need of programming and the result seemed to be very random, I feel "intuitively" this should be true, but I do not know how to prove it.


There is some confusion which is obvious from the comment. I mean a circular process that eliminate a certain number, jump $p-1$ numbers in between, and then terminate the next number. This process will stop at the place there are $p-1$ numbers left. 


An example: $n=20$, $p=5$, we have $(1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19)$ in the first elimination process. Then we have $(1,2,3,4,7,8,9,11,13,14,16,17,19)$ in the second round elimination process, and $(1,2,3,7,11,14,16,17,19)$ in the third round, and $(2,3,7,11,16,17,19)$ in the fourth round, finally yielding $(2,7,11,17)$ in the end. 



Is this set equidistributed?
Not quite an answer to the question but see <http://en.wikipedia.org/wiki/Josephus_problem>


There is a huge literature on the Josephus problem -- the wiki article is a good start. See also:


<http://doc.utwente.nl/67513/1/pospp.pdf>


and the very cool:


<ftp://ftp.cis.upenn.edu/pub/wilf/josephus.ps>


