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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 $n$.

Parcly Taxel
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mudok
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  • After $2n$ jumps, things must start repeating. It will have returned to the starting chair (total distance travelled is a multiple of n) and its next jump will have length $2n+1$ which is 1 modulo n. So if it hasn't visited all chairs by then, it won't ever do so. It may start repeating earlier for some n. – Jaap Scherphuis May 04 '17 at 13:07
  • Consider whether $n$ is a power of $2$ or not – Henry May 04 '17 at 13:22
  • It's worth noting that the cat is at the same chair after $k$ jumps as after $k'$ jumps if and only if $n$ divides $\frac{1}{2}(k-k')(k+k'+1)$. – Servaes May 04 '17 at 13:34
  • You know does not work for odd n. It is easy to see that it then also does not work if some odd prime divides n. So @Henry is right. It can only work if n is a power of 2. It does always work in n-1 jumps when n is a power of 2, which can no doubt be proved by induction on the exponent. – Jaap Scherphuis May 04 '17 at 13:53
  • @JaapScherphuis No it takes $2n-2$ jumps. A whole cycle minus two. See my answer. – Parcly Taxel May 04 '17 at 14:19
  • @ParclyTaxel You're right. I thought the first jump was distance 1 instead of distance 2. – Jaap Scherphuis May 04 '17 at 14:52

1 Answers1

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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.

Parcly Taxel
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