Covering a finite subset of $\mathbb{N}$ with prime arithmetic progressions

Question

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.

1. 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?
2. Will all sets of $k_{i}$ values where one or more $k_{i}<p_{i}-1$ give less coverage?
3. (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.

Answer 1

This is equivalent to studying large intervals of numbers that are not coprime to a product of some selected prime numbers. When one has such a large interval, one can translate it by subtraction to begin at 1, and can compute the requisite k_i. Taking m to be 5, n the product of the first m primes, in this case n=2310, we find the interval of nontotients of n from 114 to 126: translating the k's in increasing order of p_i are 1,1,2,6,8. This tiles 13 integers, and 13 is greater than 13-2.

Again searching MathOverflow for Jacobsthal's function should give you more examples