For the recurrence bn=an−an−1b_n = a_n - a_{n-1}bn=an−an−1 derived from an=3an−1−2an−2a_n = 3a_{n-1} - 2a_{n-2}an=3an−1−2an−2 with a0=1,a1=2a_0 = 1, a_1 = 2a0=1,a1=2, which recurrence does bnb_nbn satisfy?
bn=2bn−1b_n = 2b_{n-1}bn=2bn−1
bn=3bn−1−2b_n = 3b_{n-1} - 2bn=3bn−1−2
bn=2bn−1−bn−2b_n = 2b_{n-1} - b_{n-2}bn=2bn−1−bn−2
bn=bn−1+2bn−2b_n = b_{n-1} + 2b_{n-2}bn=bn−1+2bn−2