(In reply to solution
by Dej Mar)
The pattern has a three important features:
First on an arbitrarily long string of lamps at every phase 2^n - 1 there is a single string of on lamps. This only happens at phase numbered by powers of 2. [this needs proving]
Second, the highest number of lit lamp is one more than the phase number. ie at phase 0 lamp L1 is lit, at phase 1 lamp L2 is lit etc...
[proof: L(a) will light the phase after L(a-1) since it is at a different state]
Third, only way the lamps can all turn off is if they all turn on.
[proof: this is the only way they call all be the same state.]
So if n is a power of two then on phase n-1 the lights will all be on and on phase n they will all turn off.
If n is not a power of two, then on phase n-1 when the last light is lit they are not all lit. So they will not all turn off on phase n.
[What I thought I had shown was phase n would be a repeat of a previous phase or a repeat in reversed order. This is not the case. Proving a repeat eventually occurs this will be necessary.]
Posted by Jer
on 2010-09-03 15:34:40