The real solution is writing a "checker" for this problem, and then BFS over possible states from 2^n-1 to reach 0 (on bitmask of where it can be located). That produces a nice pattern which is always 2*(n-2), and ofc. shortest possible since it's fair BFS.
2 3 ... n-2 n-1 2 3 .... n-2 n-1 for odd N
2 3 ... n-2 n-1 n-1 n-2 ... 3 2 for even N

for N this is a right solution
count is 2*(N - 2) and
2 2 3 3 4 4 .... (N-1) (N-1)

isn't it

Here's a simple program testing your answer. It outputs all the possible paths when the figure is still not found. You can modify constants in the header to test another one.

program test1789;
const nmoves = 12;
      nwidth = 5;
      mv: array[1..nmoves] of longint = (2, 2, 3, 1, 2, 4, 2, 3, 5, 3, 4, 5);
var i: longint;
    path: array[0..nmoves] of longint;
procedure Solve(v, z: longint);
var q: longint;
begin
    if (v < 1) or (v > nwidth) then exit;
    if z > nmoves then begin
        for q:=0 to nmoves do write(path[q], ' '); writeln;
    end else begin
        //if (v < 1) or (v > nwidth) then exit;
        if v = mv[z] then exit;
        path[z]:=v - 1;
        Solve(v - 1, z + 1);
        path[z]:=v + 1;
        Solve(v + 1, z + 1);
    end;
end;
begin
    for i:=1 to nwidth do begin
        path[0]:=i;
        Solve(i, 1);
    end;
end.

UPD: A small fix, moving one line into a proper place.

Edited by author 14.02.2016 16:28

It's work, but why? example
4
1 2 3 4 4 3 2
original 4->3->2->1->2->3->4
They don't meet

You say m is always 2 * (n-1) or is it (2*n) - 1
whichever it is, your initial logic doesn't even satisfy the given output for n = 3.
By your above method m should either be 2 * (3-1) = 4 or (2*3) - 1 = 7 whichever formula you intended to say - none of them satisfy the given output i.e. for n = 3 , m = 2

Да, уточните! Ведь если такая последовательность найдена, добавляя произвольное количество рандомных шагов сути задачи это не изменит. Но на АС это влияет.

Your solution can make unnecessary steps

give me some hints please..

the algo is very easy. i spent 2 minutes solving this problem.
try to find steps for n=3,4,5 without computer, using pen and sheet of paper. you will see the way)
PS poor english, i know

Just solve the problem for the case when the dodec. is on even place initially.

For example,

n = 4

2 2 3 3 won't work when the stone at 4


The final stone will be 4->3->2->1->2 for example

No

Видимо, по своему желанию может ходить за исключением 1-го и n-ого постамента

нет, это видно из примера