**Hint**
There is a well known solution: $$\\{2n, 4n+1,8n-1,\cdots \\}$$ which are of the form $2^kn+u_k$ with $u_k$ is the residue closest to $0$ which has not been previously covered
All integers are of the form $2n$ or $2n+1$ so we choose the **first sequence** $2n$ we cover all even numbers. We will divide the set of odd numbers into two sequences those of the form $4n+1$ or $4n-1$ so the **second sequence** is $4n+1$ and we have to divide $4n-1$ into two parts those of the form $8n-1$ and those of the form $8n+3$, so the **third sequence** is $8n-1$ and we will divide $8n+3$ into two parts $16n+3$ and $16n-7$ and we take the **fourth sequence** to be $16n+3$ and we divide the rest into two parts $\cdots\cdots$