smc

$(a+b)R_n=\frac{1}{2}(a^{n+1}+b^{n+1})+ab\cdot \frac{1}{2}(a^{n-1}+b^{n-1})=R_{n+1}+ab{R}_{n-1}$ but $a+b=6$ and $ab=1$, so this means that $R_{n+1}=6R_n-R_{n-1}$. Since $R_0=1$ and $R_1=3$, all terms are integers and we can continue the recurrence $\mathrm{mod}10$ to get the repeating sequence $1,3,7,9,7,3$. The number $12345$ is divisible by three but not by six, so the required element in the sequence is $9$ which is answer $\boxed{9}$