Since a+b=6 and ab=1, the recursion Rn+1=6Rn−Rn−1 follows. Use this, together with R0=1 and R1=3, to calculate the units digits of R2,R3,R4,R5,R6,R7,… which are 7,9,7,3,1,3,…, respectively. An induction argument shows that Rn and Rn+6 have the same units digit for all nonnegative n. In particular, R3,R9,⋯,R12345 all have the same units digit, 9, since 12345=3+6⋅2057.
Note. Since (x−a)(x−b)=(x−3−22)(x−3+22)=x2−6x+1 it follows that a and b satisfy x2=6x−1, so an+1=an−1a2=an−1(6a−1)=6an−an−1 and similarly, bn+1=6bn−bn−1. This yields an alternate derivation of the recursion: