Isn't that the same as the last one, with the additional constraint (looked at as a binary expression) that the first and last bits can't both be 1 (because they wrap round)? In which case, it must be S(n-2)-1, because that situation can only happen once.
So whereas in the straight-line case 3 was the minimal solution, here 3 gives
1 - 1 = 0
which is no good, as you would expect for 3 seats at a round table. 4 gives:
1 + 2 - 1 = 2
which seems to make sense:
10 0 1and 01 1 0
so 8 would be
(1 + 2 + 3 + 4+ 5 + 6) - 1 = 20
I don't think that can be right though.
If there were 5 seats and Galahad sits down first he has 5 choices. Mordred can then not sit at the one Galahad has chosen, or the two on either side, but there are 2 seats remaining and he can choose either. Hence there are 5 x 2 = 10 possible arrangements, not 1 + 2 + 3 - 1 = 5.
Ah yes, I missed that we have two distinct individuals in this case rather than two indistinguishable cats. Need to think about that a bit...