Project Euler Forum

I can't understand how $C(3) = 3$ unless "different" doesn't mean "distinct".

Clearly a clock sequence must start $1$, so the only clock sequence of length $1$ is $1, 1, 1, \ldots$.

An irreducible clock sequence of period $2$ would have to start $1, 2$ but that fails because the seventh segment has to sum to $6$ or $8$.

An irreducible clock sequence of period $3$ could have two elements in the second segment, but then they're both $1$. So it must start $1, 2, 3$, and that breaks down because the fourth segment must sum to $3$ or $6$.

I conclude that either $C(3) = 1$ or that the three "different clock sequences" which contribute to $C(3)$ are all the same sequence: $1, 1, 1, 1, 1, 1, \ldots$. Am I missing something?

Thanks for you post pjt33; I think I can clarify.

The sequence 121212... is actually valid as a clock sequence. There should be no difficulty in making the seventh segment sum to 7.

The sequence 123123... is invalid as you correctly observe, but there is another period-3 sequence that has been overlooked.

Thanks. I think I must have skipped an element from (12)* when doing it by hand just after waking up.

I had an issue understanding what the period of six meant in the example given. I now understand that it is the six numbers "123432" that keeps on repeating itself. I hope this helps others the struggle to get to grips with this puzzle.