There is a series of numbers "1, 1, 1, 3, 5, 9, 17, 31, 57, 105." what is the remainder of the number 2011 divided by 3

There is a series of numbers "1, 1, 1, 3, 5, 9, 17, 31, 57, 105." what is the remainder of the number 2011 divided by 3

This series of numbers starts from the fourth, and each number is the sum of the first three numbers. Therefore, the remainder of a number divided by three in the series is also equal to the remainder of the sum of the remainder of the first three numbers divided by three divided by three
Write the remainder of the first few terms divided by three
1、1、1、0、2、0、2、1、0、0、1、1、2,1,1、1、0、……
The remainder sequence is obviously a cycle of every 13 numbers
2011 ÷13 = 154 …… Yu 9
Therefore, the remainder of the number 2011 divided by 3 is equivalent to the remainder of the number 9 divided by 3, which is 0