Answer: n=2^k-1 I solve this problem by using complex induction principle. I don't know how to use LATEX, so I can't write my solution.

And n=2.Also holds.

I don't understand... how can this permutation which has $ n+1 $ terms become a permutation with $ n $ terms?

anonymouslonely wrote: I don't understand... how can this permutation which has $ n+1 $ terms become a permutation with $ n $ terms? Good point. I fixed the problem.

does somebody have a solution? it looks so strange! frejer,how do you solve it more exactly?

This is (probably) the official solution. Let $P=(1,n,n-1,\ldots,3,2,0)$ denote the starting permutation, and let $R$ denote the desired finishing permutation $(1,2,\ldots,n,0)$. We observe that if $n=1$ or $2$, then $P$ is trivially regular. Now we suppose that $n>2$ and consider the following three cases: Case 1: $n>2$ is even. If $n$ is even, say $n=2k$ and $k>1$, then we obtain from $P$ after $k-1$ legal transportations the following permutation: $$(1,n,0,n-2,n-1,n-4,n-3,\ldots,4,5,2,3).$$ No further legal transportations are possible, and we conclude that $P$ is not regular in this case. Case 2: $n$ is of the form $2^j-1$ for some $j\ge 1$. For $i=1,2,\ldots, j$ define the permutation $P_i$ as follows: $$P_i=(1,2,\ldots,2^i-1,0; n-2^i+1,n-2^i+2,\ldots,n-1,n;n-2\cdot2^i+1,\ldots,n-2^i;\ldots; 2^i,\ldots,2\cdot 2^i-1).$$ The permutation $P_i$ consists of $(n+1)/2^i$ "blocks " of $2^i$ numbers, each one indicated by a semicolon. Note that the members within each block (except the first) are in increasing numerical order. We now note that $P_i$ is obtained from $P$ by precisely $\frac{n-1}{2}$ legal transportations, and that $P_{i+1}$ is obtained from $P_i$ by precisely $\frac{n+1}{2}$ legal transportations for $i=1,2,\ldots, j-1$. But $P_j=R$ and thus $R$ is obtained from $P$ by precisely $\frac{j(n-1)}{2}$ legal transportations. Hence, $P$ is regular whenever $n$ is of the form $2^j-1$. Case 3: $n$ is odd, but not of the form $2^j-1$. Then $n$ can be written uniquely as $n=(2k+1)2^j-1$, when $j,k\ge 1$. In this case we can define $P_1, P_2,\ldots, P_j$ as in Case 2; also as in case 2, $P_j$ is obtained by $P$ by precisely $\frac{j(n-1)}{2}-1$ legal transpositions. However, in this case $P_j\ne R$; but after exactly $k$ further legal transportations on $P_j$ we obtain the following permutation: $$(1,2,\ldots,2^j;n-2^j+1,\ldots,n;0,n-2\cdot 2^j+2,\ldots,n-2^j;\ldots; 2\cdot2^j,\ldots,3\cdot 2^j-1; 3\cdot 2^j,\ldots,2\cdot 2^j-1).$$ No further legal transportations are possible, and we conclude that $P$ is not regular. All numbers are now accounted for and we infer that $P$ is regular if $n=2$ or $n=2^j-1$ for $j=1,2,\ldots$.

shobber wrote: Given a permutation ($a_0, a_1, \ldots, a_n$) of the sequence $0, 1,\ldots, n$. A transportation of $a_i$ with $a_j$ is called legal if $a_i=0$ for $i>0$, and $a_{i-1}+1=a_j$. The permutation ($a_0, a_1, \ldots, a_n$) is called regular if after a number of legal transportations it becomes ($1,2, \ldots, n,0$). For which numbers $n$ is the permutation ($1, n, n-1, \ldots, 3, 2, 0$) regular? Does a_i=0 for i>0 mean only that a_0#0?