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: