Numbers and numbers

Let $a_1=x, a_2=y, a_3=z$ . First note that if any absolute value equals 0, then $a_n=0$ . Also note that if at any position, $a_n=a_{n-1}$ , then $a_{n+2}=0$ . Then, if any absolute value equals 1, then $a_n=0$ . Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria. Assume that to be the only way the criteria is met. To prove, let $|y-x|>1$ , and $|z-y|>1$ . Then, $a_4 \ge 2z$ , $a_5 \ge 4z$ , and $a_6 \ge 4z$ . However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$ , $|y-x|=2$ . Again assume that any other scenario will not meet criteria. To prove, divide the other scenarios into two cases: $z>1$ , $|y-x|>1$ , and $|z-y|>1$ ; and $z=1$ , $|y-x|>2$ , and $|z-y|>1$ . For the first one, $a_4 \ge 2z$ , $a_5 \ge 4z$ , $a_6 \ge 8z$ , and $a_7 \ge 16z$ , by which point we see that this function diverges. For the second one, $a_4 \ge 3$ , $a_5 \ge 6$ , $a_6 \ge 18$ , and $a_7 \ge 54$ , by which point we see that this function diverges. Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $|y-x|<2$ (280 options) $|z-y|<2$ (280 options, 80 of which coincide with option 1) $z=1$ , $|y-x|=2$ . (16 options, 2 of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}.$