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.

firstnumber, thepth and so on. Do you mean the zeroth number, thepth, the2pth, and so on? That would make more sense to me. That is, you want to think about an arithmetic progression with differencep, but regarded modn, and look at its complement. $\endgroup$ – Charles Matthews Mar 19 '11 at 10:05remaining numbersto find the next one that you remove? A small concrete example would be helpful. $\endgroup$ – Anthony Quas Mar 19 '11 at 23:35