I generated this sequence 2102012210101111222011212000200220212200120102112020222211102212100010011012, find next 4 digits and explain the reason.
Note that this is a ternary sequence. All digits are generated by few rules, no digit is random or arbitrary.
Hint:
The sequence generated is periodic and the period is equal to 80 digits, but 80 (as any number greater than 3) does not appear in the rules
After starting 2, 1, 0, 2,
each subsequent digit is obtained by summing modulo 3 the third- and fourth-to-last digits. Therefore, the next 4 digits are 1 + 0 = 1, 0 + 1 = 1, 1 + 2 = 0, 2 + 1 = 0.
This corresponds to a
ternary linear-feedback shift register based on the primitive polynomial x4 + x3 + 2 over the finite field GF(3), so it generates a ternary m-sequence. This means that the sequence of 4-digit strings produced (2102, 1020, 0201, 2012, ...) cycles through all 34 - 1 = 80 possibilities besides 0000 before repeating periodically.
The next four digits are
1100
Because
The first fourty digits are the same as the next thirty-six digits but with the 1s and 2s switched. So I'm assuming the next four continue that pattern.
Beyond that, I'm haven't figured out how the sequence was generated.
