I know in this post "Recurrence relation for number of ternary strings that contain two consecutive zeros" Pat wrote "$a_n = a_{n-1} + a_{n-2} + 2^{n-2}$".
But I cannot understand how the $2^{n-2}$ appear. If the rightmost characters are 00, removing the "00" will get the string of n-2. The remaining string can be "01" "10" or "11"'s combination. Why the cardinality of this group is $2^{n-2}$?