Because of a problem I ran into I am trying to get a quick start in covering with arithmetic progressions.

First I want to say I am aware of this previously asked question: Covering $\mathbb{N}$ with prime arithmetic progressions

Similarly to what is asked there I am interested in covering with arithmetic progressions of the type $A_{i}=k_{i}+np_{i}$ where $p_{i}$ is prime, $k_{i} \in \mathbb{N}$ and $n \in \mathbb{N_{0}}$.

Differently, my interests are however in covering of finite sets of the type $\{1, 2, \ldots, N\} \subset \mathbb{N}$ where $k_{i}<p_{i}$ and $p_{1} \leq p_{i} \leq p_{m}$.

By checking out some numbers it looks like that if all $k_{i}=p_{i}-1$ the covered set is $\{1, 2, \ldots, p_{m+1}-2 \}$.

So my questions are:

Let $p_{1} \leq p_{i} \leq p_{m}$ be the first $m$ consecutive primes and $\forall p_{i}$ let $A_{i}=k_{i}+np_{i}$ be $m$ arithmetic progressions, where $n \in \mathbb{N_{0}}$ and $k_{i} \in \mathbb{N}$ such that $1 \leq k_{i}<p_{i}$. Obviously is the arithmetic progression $A_{1}=k_{1}+np_{1}=1+2n$ fixed and covers all the odd numbers in all considerations.

- If $\forall p_{i}$ with $k_{i}=p_{i}-1$ is the covered set always $\{1, 2, \ldots, p_{m+1}-2 \}$, that is are all numbers up to $N=p_{m+1}-2$ covered?
- Will all sets of $k_{i}$ values where one or more $k_{i}<p_{i}-1$ give less coverage?
- (If "no" to 2.) Is there any set of $k_{i}$ values where one or more $k_{i}<p_{i}-1$ gives more coverage, that is covers numbers up to $N>p_{m+1}-2$?

Additional questions/help. I would appreciate all relevant information possible, like: Has this been proved?, Then by whom?; Any textbooks discussing this or very similar kind of problems; Websites; etc.

OM Question, let your new question explain each term independently, to avoid any confusion). $\endgroup$I am interesting. It is great that you areinterestingbut you can ask way moreSPECIFICquestions than just claiming than you are interesting (of course I believe that you are). $\endgroup$9more comments