Covering the natural numbers with sequences?

Question

I have got the following question:

Can you cover $\mathbb{N}$ with finite amount of arithmetic, disjunct sequences(their difference can't be the same, and $d>1$)?

The answer for countable amount was yes, since we have an example for that: $$\{2n, 4n+1,8n-1,\cdots \}$$ which are of the form $2^kn+u_k$ with $u_k$ is the residue closest to $0$ which has not been previously covered (thanks for the answer), but how about if we can only use finite amount?

My guess is no, but I just can't prove why not. :)

Any ideas? Thanks! :)

Perhaps what you're looking for are "covering systems" (http://en.wikipedia.org/wiki/Covering_system). The idea is that you have sets of congruence classes so that every natural number is in one of the congruence classes. — TravisJ, Mar 31 '15 at 16:53

user26486 · Accepted Answer · 2015-03-31T17:40:51.630

2

You're precisely searching for a disjoint, distinct covering system.

Mirsky-Newman theorem is a theorem that says there is no such system.

edited Mar 31 '15 at 17:40

answered Mar 31 '15 at 16:58

user26486

11,519

Where can I find a proof? :) – Atvin Mar 31 '15 at 17:01
http://math.stackexchange.com/questions/237909/understanding-a-proof-of-the-davenport-rado-mirsky-newman-theorem Is this connected? – Atvin Mar 31 '15 at 17:04
@Atvin Yes. It proves that if a covering system is disjoint, then there must exist a pair of equal moduli, so the covering system can't be distinct. – user26486 Mar 31 '15 at 17:12
My answer considers $\mathbb Z$ instead of $\mathbb N$. – user26486 Mar 31 '15 at 18:00

Covering the natural numbers with sequences?

1 Answers1