Here is a short Python script to implement method 2 for generating the symbolic code for the nth iteration of the terdragon curve.
n = input("n = ") #order of iteration b = [''] for k in range(1,3**(n-1)+1): #mod 1,2 positions b += ['R','L',''] for k in range(1,3**(n-1)): #multiples of 3 positions b[3*k] = b[k] b = "".join(b[1:-1]) #convert list to string print(b)