Peter wrote: A sequence of integers $ a_{1}, a_{2}, a_{3}, \cdots$ is defined as follows: $ a_{1} = 1$, and for $ n \ge 1$, $ a_{n + 1}$ is the smallest integer greater than $ a_{n}$ such that $ a_{i} + a_{j} \neq 3a_{k}$ for any $ i, j,$ and $ k$ in $ \{1, 2, 3, \cdots, n + 1 \}$, not necessarily distinct. Determine $ a_{1998}$. http://www.kalva.demon.co.uk/short/soln/sh98n4.html

chien than wrote: http://www.kalva.demon.co.uk/short/soln/sh98n4.html The link doesn't work in my computer because the link is wrong or maybe my computer is an ancient one. So I will post the solution. Initially, we can determine the first new values for $a_n$ are 1, 3, 4, 7, 10, 12, 13, 16, 19, 21, 22, 25. Since these are exactly the numbers of the forms $3k+1$ and $9k+3$, we conjecture that this is the general pattern. In fact, it is easy to see that the equation $x+y=3z$ has no solution in the set $K = \{ 3k+1, 9k+3 | k \in \mathbb{N} \} $ . We shall prove that the sequence $\{a_n\}$ is actually this set ordered increasingly. Suppose $a_n>25$ is the first member not belonging to $K$ . We have several cases : (i) $a_n=3r+2, r \in \mathbb{N}$. By the assumption, one of $r+1,r+2,r+3$ is of the form $3k+1$ (and smaller than $a_n$), and therefore is a member $a_i$ of the sequence. Then $3a_i$ equals $a_n+1$, $a_n+4$, or $a_n+7$, which is a contradiction because $1,4,7$ are in the sequence. (ii) $a_n=9r$, $r \in \mathbb{N}$. Then $a_n+a_2=3(3r+1)$, although 3r+1 is in the sequence, a contradiction. (iii) $a_n=9r+6$, $r \in \mathbb{N}$. Then one of the numbers $3r+3, 3r+6, 3r+9$ is a member $a_j$ of the sequence an thus $3a_j$ is equal to $a_n+3, a_n+12, a_n+21$, where $3,12,21$ are members of the sequence, again a contradiction. Once we have revealed the scructure of the sequence, it is easy to compute $a_{1998}$. We have $1998=4 \times 499 +2$, which implies $a_{1998}=9\times 499+a_2=4494$.