http://artofproblemsolving.com/community/c6h155692p874978 similar to this

Apparently $n=2^k+1$ works and we get a Sierpienski's triangle

Wasn't this too easy for a P3??! Or maybe I am missing smth here . Here's my solution: We claim that $n=4k+3$ works (for $k \in \mathbb{N}$). There are two crucial observations:- If at any moment, all the lamps are on, then in the next second, all lamps will become off. $\text{ }$ A lamp $L$ will remain off, till the lamp just one step closer to $L_1$ than $L$ is switched on. So it suffices to show that all the lamps will become on at some instant. We proceed by induction on $k$. The base case (i.e. $n=7$) can be checked by hand easily. Assume that the result is true for $n=4k-1$. We show that it is true for $n=4k+3$ also. Consider the lamps $L_{-(2k-1)},L_{-(2k-2)}, \dots ,L_0,L_1,L_2, \dots ,L_{2k-1}$ (where indices are taken modulo $n$). Then they form a system of $4k-1$ lamps, and so by the induction hypothesis, all of them will be switched on after a finite number of moves. Until then, the $5$ lamps $L_{2k},L_{2k+1},L_{2k+2},L_{2k+3},L_{2k+4}$ will remain off. Now, in the next moment, all lamps, except for $L_{2k-1},L_{2k},L_{-(2k-1)},L_{-2k}$ will get closed. Then, one can easily see that, in the next second, all lamps will become on, as desired. Hence, done. $\blacksquare$

Can someone post a correct solution because the one above is wrong ( you can’t apply the induction hypothesis because those lamps aren’t in a circle )

when n=5,11 it is wrong 00000100000 00001110000 00011011000 00111111100 01100000110 11110001111 00011011000 00111111100 01100000110 11110001111 ...... 00100 01110 11011 01110 11011 ...... so all the answer above are wrong......

we can prove that n=2^k-1,2^k-2 is right,while n=2^k (k>=2) is wrong