Let us say that the sequence $a_n$ is partitioned by the subsequences $a_{i_1},a_{i_2},...a_{i_m}$ if for every $n_0 \in \mathbb N$ there is $i_j \in \{i_1,i_2,...,i_m\}$ such that $a(n_0)=a_{i_j}(n_0)$ and for every pair $(i_j,i_k)$ such that $i_j,i_k \in \{i_1,i_2,...,i_m\}$ and $i_j \neq i_k$ we have $a_{i_j}(n) \neq a_{i_k}(n)$, for every $n \in \mathbb N$.
The question is:
Is it true that for every $n$-tuple $(k_1,k_2,...,k_n)$ such that $k_i \in \mathbb N$ ; $i=1,2,...,n$ there exist coprime natural numbers $a$ and $b$ such that the sequence $s(m)=a^m+b$ can be partitioned into $n$ subsequences $s_1(m),s_2(m),...s_n(m)$ such that $k_i \mid s_i(m)$; $i=1,2,...,n$?
I somehow believe that this is not the case although I might be wrong, if there are some results in this direction or if this is some known fact I would appreciate your response about this problem I created.
The problem is to me interesting because if this is not true then we would have that sequences of the form $a^n+b$ (with $a$ and $b$ coprime) can satisfy only certain divisibility criteria (and not all), and that would be something that is interesting in itself and of itself.