13. The integer sequence $a_{1}, a_{2}, a_{3}, \cdots$ is defined as follows: $a_{1}=1$, for $n \geqslant 1, a_{n+1}$ is the smallest integer greater than $a_{n}$, and for all $i, j, k \in\{1,2, \cdots, n+1\}$, it satisfies $a_{i} + a_{j} \neq 3 a_{k}$. Find the value of $a_{22006}$.
13. First, find the initial few $u_{n}$.
$a_{1}=1$.
Since $1+2=3 \times 1$, then $a_{2} \neq 2, a_{7}=3, a_{3}=4$.
Since $4+5=3+6=3 \times 3$, then $a_{1} \neq 5, a_{1} \neq 6, a_{1}=7$.
Since $1+8=3 \times 3,3+9=3 \times 4$, then $a: \neq 8, a_{5} \neq 9, a_{3}=10$.
Since $1+11=3 \times 4$, then $a_{5} \neq 11 \cdot a_{4}=12 \cdot a_{7}=13$.
Since $7+14=3 \times 7,15+15=3 \times 10$, then $a_{x} \neq 14, a_{x} \neq 15, a_{8}=16$.
Since $13+17=3 \times 10,18+3=3 \times 7$, then $a_{4} \neq 17, a_{0} \neq 18, a_{7}=19$.
Since $1+20=3 \times 7$, then $a_{1 . .} \neq 20, a_{10}=21, a_{11}=22$.
Since $7+23=3 \times 10.24+24=3 \times 16$, then $a_{12} \neq 23, a_{12} \neq 24, a_{12}=25$.
Up to this point, we have the first 12 values of $a_{n}$, listed as follows:
$$
\begin{array}{l}
n: 123456789101112 \\
a_{n}: 13471012131619212225
\end{array}
$$

If we consider every four terms as a group, then the terms in each group can be obtained by adding 9 to each term in the previous group. Below, we verify that $a_{n}$ satisfies the following formula:
$$
\begin{array}{l}
a_{1,-1}=9 r+1, a_{1} \ldots=9 r+3, \\
a_{1}-3=9 r+4 \cdot a_{1-\cdots}=9 r+7 . \\
\text { where } r=0,1,2, \cdots
\end{array}
$$

For $n=0,1,2$, the conclusion holds.
Assume that for $r=0,1, \cdots, m(m \geqslant 2)$, the conclusion holds, then in $a_{1}, a_{2}, \cdots, a_{i m-4}$, there is no term that is 2 modulo 3, and in the interval $[1,9 m+7]$, all numbers that are 1 modulo 3 are present in the above sequence.

Since $(9 m+8)+4=3(3 m+4)$, and $3 m+4=1(\bmod 3)$, which is in the above sequence, so $a_{1, m+5} \neq 9 m+8$. Similarly, since $(9 m+9)+3=3(3 m+4)$, so $a_{1 m+s} \neq 9 m+9$. In fact, $a_{1 m+s}=9 m+10$. (Because if $(9 m+10)+y=3 z$, then $y=2(\bmod 3)$, and $y$ is not in the above sequence, leading to a contradiction.)

Since $(9+m 11)+1=3(3 m+4)$, so $a_{1 m-6} \neq 9 m+11$. In fact, $a_{1 m+6}=9 m+12$. (Because if $(9 m+12)+y=3 \approx$, then $y \equiv 0(\bmod 3)$, hence $y=9 r+3$. And $(9 m+12)+(9 r+3)=3(3 m+3 r+5)$, but $3 m+3 r+5=2(\bmod 3)$ is not in the above sequence, leading to a contradiction.)
Since there is no $y \equiv 2(\bmod 3)$ in the above sequence, so $a_{1 m-7}=9 m+13$.
Since $(9 m+14)+7=3(3 m+7)$, and $3 m+7 \equiv 1(\bmod 3)$ is in the above sequence, so $a_{4 m-8} \neq 9 m+14$. Similarly, since $(9 m+15)+(9 m+15)=3(6 m+10)$, and $6 m+10=1(\bmod 3)$ is in the above sequence, so $a_{1 m+n} \neq 9 m+15$.
Finally, since there is no $y=2(\bmod 3)$ in the above sequence, so $a_{4 m+8}=9 m+16$.
By mathematical induction, the formula for $a_{n}$ holds.
Since $2006=4 \times 501+2$, then $a_{2100}=9 \times 501+3=4512$.