A counting problem of traversing a circular arrangement Let there be $n$ chairs, $C_1,C_2,\ldots,C_n$ around a circular table. A cat starts jumping from $C_1$ and after first jump it reaches to $C_3$, then after second jump it reaches to $C_6$ and so on. That is at the $k$th jump the cat skips $k$ chairs. Now my question:

Is there a bound on the number of jumps (depending on $n$) or an exact number of jumps such that the cat visits all the chairs at least once?

N.B I have proved that if $n$ is odd then there will at least be a chair that will not be visited ever. Thus $n$ must be even. Also there are instances of even $n$'s such that some chairs are not going to be visited ever. Thus I'm looking for a bound on the minimal number of jumps, whence the scenario is feasible for some even $n$.
 A: The cat's motions repeat every $2n$ jumps. Modulo $n$ the cat moves forward by one more seat per jump, so it visits all seats iff the triangular numbers are a complete residue system modulo $n$, which by this question happens iff $n$ is a power of two.
Then the cat always takes $2n-2$ jumps to visit every seat. Renumber the seats so the cat starts on seat 0 and write the numbers of the seats the cat visits in a loop ($n=8$ below):
 >2516433v
(0)      4
 ^7702516<

After any $n$ consecutive jumps covering $\frac{n(n-1)}2\equiv n/2\bmod n$ spaces the cat is on the opposite side of the table. Since all seats are visited, each residue modulo $n$ appears exactly twice in the loop. However, extending the jumps backwards we find that $n-1$ appears only in the two places before the initial 0, so moving forward the cat


*

*takes $2n-2$ jumps before landing on seat $n-1$,

*then jumps in place and

*finally jumps forward to seat 0.


All other seats appear twice in the first $2n-2$ jumps, so $2n-2$ is the answer.
