A4. In the sequence of positive integers, starting with $2018,121,16, \ldots$ each term is the square of the sum of digits of the previous term. What is the $2018^{\text {th }}$ term of the sequence?
Solution: The answer is 256 .
We will notice that from the $5^{\text {th }}$ term on, the series oscillates between 169 for odd terms and 256 for even terms: $2018,121,16,49,169,256,169, \ldots$ The $2018^{\text {th }}$ term is thus 256 .