Find a sequence $(s_n)$ such that for any $a,b \in \mathbb{N}$, $\exists n \in \mathbb{N}$ where $s_n = a + b(n-1)$

Question

I’m fairly sure such a sequence should exist. I’ve tried devising a sequence using primes or fibannoci numbers but haven’t been able to make progress.

Answer 1

To summarize the discussion in the comments: Since the possible pairs $(a,b)$ form a countable set, we can identify them with natural numbers, or with a subset of the natural numbers. One simple way to do that is via the function $(a,b)\mapsto 2^a\times 3^b$. That suffices to allow us to construct a sequence $s_n$ that works.

For instance: $$s_n=v_2(n)+v_3(n)\times (n-1)$$

Where, as usual, $v_p(n)$ denotes the order to which a prime $p$ divides $n$. Thus $v_2(2)=1$, $v_3(18)=2$, $v_7(11)=0$ and so on.

This sequence suffices. In particular, given any pair $(a,b)$ of natural numbers, we have $s_{2^a3^b}=a+b\times (n-1)$ as desired.

Of course, this $n$ is not defined uniquely by $(a,b)$. If $n$ works, then so does $5n$, for instance. But uniqueness wasn't specified in the post.