A Four Term Recursive Sequence

Note that no term in the sequence is equal to 0. By the recursive definition,

$\begin{aligned} t_n &= \dfrac{t_{n - 1} t_{n - 3}}{t_{n - 2} t_{n - 4}} \\ t_{n - 1} &= \dfrac{t_{n - 2} t_{n - 4}}{t_{n - 3} t_{n - 5}}. \end{aligned}$

Multiplying these two gives

$t_n t_{n - 1} = \dfrac{t_{n - 1} t_{n - 3}}{t_{n - 2} t_{n - 4}} \cdot \dfrac{t_{n - 2} t_{n - 4}}{t_{n - 3} t_{n - 5}} = \dfrac{t_{n - 1}}{t_{n - 5}}.$

This simplifies down to $t_n t_{n - 5} = 1.$ Replacing $n$ with $n - 5$ yields $t_{n - 5} t_{n - 10} = 1,$ giving us $t_n = t_{n - 10}.$ This tells us that the sequence repeats itself every ten terms.

Finally,

$t_{2^{2018}} = t_{2^{2018} \! \pmod{10}} = t_4 = \boxed{4}.$

The first several terms of the sequence are $1, 2, 3, 4, \frac{8}{3}, 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{3}{8}, 1, 2, 3, 4,$ and so on, which means the sequence repeats itself every $10$ numbers. Since we use a base $10$ numbering system, the value of $t_n$ is determined by the last digit of $n$ . In general, if $d$ is the last digit of $n$ , then $t_n = t_d$ .

The first several powers of $2$ are $2, 4, 8, 16, 32, 64,$ and so on, and the last digits of the powers of $2$ cycle through the $4$ numbers $2, 4, 8,$ and $6$ . Since $2018 \equiv 2 \mod 4$ , the last digit of $2^{2018}$ is $4$ .

Therefore, when $m = 2^{2018}$ , its last digit is $4$ , so $t_m = t_4 = \boxed{4}$ .