Let's explain the following facts. The sequence of digits "1221" is first considered as a number in the base-13, then in the base-12, followed by the base-11, and finally in the base-10 number system. In each case, we convert our number to the number system with the base one less. In the first three cases, the result of the conversion is the same sequence of digits (the same digits in the same order). Why can the final conversion not result in the same sequence as the first three cases? - Also provide another such sequence of digits.
I. The number 1221 written in base $b$ is converted to base $b-1$ by rearranging the corresponding expression $K=b^{3}+2 b^{2}+2 b+1$ according to the powers of $b-1$. Based on the identities

$$
(b-1)^{3}=b^{3}-3 b^{2}+3 b-1, \quad(b-1)^{2}=b^{2}-2 b+1
$$

we have

$$
b^{3}=(b-1)^{3}+3 b^{2}-3 b+1, \quad b^{2}=(b-1)^{2}+2 b-1
$$

Applying these successively,

$$
\begin{aligned}
K & =(b-1)^{3}+5 b^{2}-b+2=(b-1)^{3}+5(b-1)^{2}+9 b-3= \\
& =(b-1)^{3}+5(b-1)^{2}+9(b-1)+6
\end{aligned}
$$

Thus, the short form of the number $K$ in base $(b-1)$ is 1596, provided the coefficients 5, 9, and 6 are digits in the base $(b-1)$ system, i.e., smaller than the base. This is true for $b-1=12$, 11, and 10, but not for the conversion from base 10 to base 9. In this case,

$$
1221_{10}=9^{3}+5 \cdot 9^{2}+9 \cdot 9+6=9^{3}+6 \cdot 9^{2}+6=1606_{9}
$$

Furthermore, we see that the equality $1221_{b}=1596_{b-1}$ holds for any base greater than 10.

II. Let a sequence of digits be $m n p q$, where $m, n, p, q$ are integers, $1 \leq m < b$ and $0 \leq n, p, q < b$. By a similar transformation,

$$
\begin{aligned}
m b^{3} & +n b^{2}+p b+q=m(b-1)^{3}+(n+3 m) b^{2}+(p-3 m) b+q+m= \\
& =m(b-1)^{3}+(n+3 m)(b-1)^{2}+(p+2 n+3 m) b+(q-n-2 m)= \\
& =m(b-1)^{3}+(n+3 m)(b-1)^{2}+(p+2 n+3 m)(b-1)+ \\
& +(q+p+n+m)
\end{aligned}
$$

Therefore, according to our requirement, the coefficients

$$
m, \quad n+3 m, \quad p+2 n+3 m, \quad q+p+n+m
$$

none of them should be greater than 9:

$$
m \leq 9, \quad n+3 m \leq 9, \quad p+2 n+3 m \leq 9, \quad p+q+n+m \leq 9
$$

and at least one of them should be equal to 9.

The largest valid case is $3006_{b}=3999_{b-1}$, and it is easy to derive from the given sequence of digits $1222, 1223, 1224$, and 1220, and a case with all different digits $1035_{b}=1369_{b-1}$.

It is not difficult to generate all valid (four-digit) forms, as well as those that allow further reduction of the base of the number system.