{x: x= 2^n + 1 }
I will write up the proof tomorrow.
outline:
basically after 1 round (last person gets his 2 coins) the configuration would be something like:
3,2,2,2,2,2,2,2,2,....2 in a circle and 2^(n-1) persons altogether.
and the guy with 3 coins is to pass 1 to the next guy...