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, thep*th and so on. Do you mean the zeroth number, the *p*th, the *2p*th, and so on? That would make more sense to me. That is, you want to think about an arithmetic progression with difference *p, but regarded modn, and look at its complement. – Charles Matthews Mar 19 '11 at 10:05remaining numbersto find the next one that you remove? A small concrete example would be helpful. – Anthony Quas Mar 19 '11 at 23:35