SAMC

We have $a_2=3$, $a_3=5$, $a_4=8$, $\dots$ and $b_2=3$, $b_3=4$, $b_4=7$, $\dots$ It follows $a_0=b_0$, $a_1=b_1$, $a_2=b_2$, and $b_3<a_3<b_4<a_4<b_5$. We prove by induction of step 2 that for $m\ge 3$ we have $b_m<a_m<b_{m+1}$. The basis cases $m=3$, $m=4$ are verified above. Assume that $$b_k<a_k<b_{k+1}\text{ and }{b}_{k+1}<a_{k+1}<b_{k+2}$$ Adding these inequalities we get $b_k+b_{k+1}<a_k+a_{k+1}<b_{k+1}+b_{k+2}$, that is $b_{k+2}<a_{k+2}<b_{k+3}$. The sequences have in common only three integers: $1,2,3$.