Assuming we are using the set of digits {1,2,3} no matter how many digits there are, and assuming a 1 cannot follow a 3 and a 3 cannot follow a 1, the number of ways an n-digit number can be constructed is:  (for 4,5,6 in bold)

numbers =  [0, 3, 7, 17, 41, 99, 239, 577, 1393, 3363]

(for n in  0,1,2, ... 9   n-digit numbers)

Algorithm:  I built a series of lists, each list having one more digit than the last, using the final digit of the element of the prior list to determine what choices of extra digit could be concatenated.

4-digit solution set:

[1111, 1112, 1121, 1122, 1123, 1211, 1212, 1221, 1222, 1223, 1232, 1233, 2111, 2112, 2121, 2122, 2123, 2211, 2212, 2221, 2222, 2223, 2232, 2233, 2321, 2322, 2323, 2332, 2333, 3211, 3212, 3221, 3222, 3223, 3232, 3233, 3321, 3322, 3323, 3332, 3333]

---------  code  ---------

ch = [['1','2'], ['1','2','3'], ['2','3']]

s1 = ['1','2','3']

s2 = []

s3 = []

s4 = []

s5 = []

s6 = []

s7 = []

s8 = []

s9 = []

numbers = [0] # zero ways if there are zero digits

for a in s1:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s2.append(a+nexdigit)

for a in s2:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s3.append(a+nexdigit)

for a in s3:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s4.append(a+nexdigit)

for a in s4:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s5.append(a+nexdigit)

for a in s5:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s6.append(a+nexdigit)

for a in s6:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s7.append(a+nexdigit)

for a in s7:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s8.append(a+nexdigit)

for a in s8:

    for nexdigit in ch[  int(a[-1]) - 1   ]:

        s9.append(a+nexdigit)

numbers.append(len(s1))

numbers.append(len(s2))

numbers.append(len(s3))

numbers.append(len(s4))

numbers.append(len(s5))

numbers.append(len(s6))

numbers.append(len(s7))

numbers.append(len(s8))

numbers.append(len(s9))

print(numbers)

# first method, above, from 0 to 9 digits

# second method, (verification)  below, more brute force 

#              way but this was much slower; up to 6 digits.

count = 0

valids = ['1','2','3']

valid = '123'

answers = [[] for i in range (7)]

for digits in range(1,7):

    for i in range(10**(digits-1), 10**digits):

        stri = str(i)

        countit = True

        for s in stri:

            if  valid.find(s) < 0:

                countit = False

        if '13' in stri or '31' in stri:

            countit = False

        if countit:

            count += 1

            answers[digits].append(i)

nnumbers = []

for a in answers:

    nnumbers.append(len(a))

print(nnumbers)

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