Consider two different integers $d\neq c>2$ and the two Fibonacci-like recursions (notice the negative signs): $$ a(i)=ca(i-1)-a(i-2)\text{ and } b(i)=db(i-1)-b(i-2),$$ where $a(0)=b(0)=0$ and $a(1)=b(1)=1$.
I want to know if its possible to have $a(i)=b(j)$ for some $i,j>2$. Moreover, what is the smallest number $N$ for which every number bigger than $N$ is in the orbit of some sequence as these with the restrictions above?
Numerical evidence suggests both claims above are true, but I don't know how to show it. This seems like something that should be well known, references would be appreciated.
About the restrictions:
I consider $c>2$ becase the sequence for $c=2$ is simply $0,1,2,3\dots$ and both questions above become trivial.
Sequences are strictly increasing so for disjointness we need $i>1$. If $i=2$ then $a(2)=c-1$ and the claims above become both trivial as we vary $c$. I think the minimal interesting example is $i=3$.
Perhaps I should make some more assumptions. Will report back as soon as I know.