Write in ascending order the multiples of 3 that, when added to 1, are perfect squares, that is, $3,15,24,48, \ldots$ What is the multiple of 3 in the $2006^{\mathrm{th}}$ position?
Writing $a$ for any number in the list, we know that $a$ is a multiple of 3 and that $a+1$ is a perfect square, that is, $a+1=k^{2}$, for some positive integer $k$. Thus, $a=k^{2}-1=(k-1)(k+1)$.

Since $a$ is divisible by 3, then either $k+1$ or $k-1$ is divisible by 3. Therefore, $k$ cannot be divisible by 3, and thus, out of every three consecutive integers $k$ (starting from 2), two of them provide a number in the list: indeed, $k=2=1 \times 3-1$ gives the first number $a=2^{2}-1=3$ in the list; $k=4=1 \times 3+1$ gives $a=4^{2}-1=15$, which is the second; $k=5=2 \times 3-1$ gives $a=5^{2}-1=24$, which is the third; $k=7=2 \times 3+1$ gives the fourth $a=7^{2}-1=48$, and so on. In general, an even position $2n$ in the list is given by $k=n \times 3+1$, so the $2006^{\text{th}}$ even position is given by $k=1003 \times 3+1=3010$. Thus, the multiple of 3 at the $2006^{\text{th}}$ position in the list is $a=3010^{2}-1=9060099$.