Finding sets such that the cartesian product gives the natural numbers I just noticed that for these two sequences:
$$
A={1,2,4,8,16,32\dots2^n\dots}\\
B={1,3,5,7,9,11\dots2n+1\dots}
$$
The following holds:
$$
\forall x \in \mathbb{N}, x=a_i·b_j \text{ for unique values of i and j}\\ 
a_{n+1}-a_{n}\geq a_n-a_{n-1}\\
b_{n+1}-b_{n}\geq b_n-b_{n-1}\\
$$
Are there any other increasing sequences of integers A and B that satisfy this?
Edit:
I'm looking for sequences in which the first condition holds for a unique value of $i$ and $j$, but, even if we leave that aside, I can't find any sequence that doesn't contain the original $A$ and $B$.
 A: The two sequences which you have found are the only sequences which satisfy all three of your constraints.
First we notice that since we must be able to write $1=a_i b_j$ for some $i$ and $j$, that $1$ must be an element of both sequences. Thus $a_1=b_1=1$. (Otherwise $1$ would not appear in one of the sequences since the sequences are increasing)
Then the following is true:
Each natural number $n>1$ can be in at most one of the sequences.
This is because if $a_i=b_j=n$, then we would have $a_1b_j = a_i b_1 = n$, contradicting the uniqueness requirement for writing $n$ as a product of elements from each sequence.
We will also make repeated use of the fact that if $n$ is a natural number such that all of the proper factors of $n$ other than possibly $n$ itself appear in one of the sequences (i.e. either all in $A$ or all in $B$) then $n$ must appear in one of the sequences. 
To prove this, suppose without loss of generality that all of the proper factors of $n$ are in $A$. We must be able to write $n=a_ib_j$ for some $i$ and $j$. Suppose that both $a_i\neq n$ and $b_j \neq n$. Then $b_j$ is a proper factor of $n$ that is not equal to $1$ or $n$ and so $b_j$ appears in $A$. But then $b_j$ appears in both sequences, which is a contradiction.
We note that since $2=a_i b_j$ for some $i$ and $j$, that $2$ must appear in one of the sequences, and since each sequence is strictly increasing, either $a_2=2$ or $b_2 = 2$. Without loss of generality, suppose that $a_2=2$.
Since we have that $3=a_i b_j$ for some $i$ and $j$, we see that $3$ must also be in one of the sequences. (Indeed, any prime number must be in one of the sequences) Since the sequences are increasing, either $a_3=3$ or $b_2=3$.
Suppose that $a_3=3$. Then I claim that $a_n=n$ for all $n$. I will prove this by induction.
The base cases are true since $a_n=n$ for $n=1,2,3$. Suppose that $a_k=k$ for all $k \leq n$ where $n \geq 3$. We will show that $a_{n+1}=n+1$.
Suppose that $a_{n+1} \neq n+1$. (And hence that $n+1$ does not appear in $a$) For any $k\leq n$ we see that $k$ is not in the second sequence since each natural number can appear in at most one of the sequences. Thus all of the factors of $n+1$ occur only in the first sequence. We see that in order for us to write $n+1$ as a product of elements from each sequence, that $n+1$ itself must be in one of the sequences, and so $b_2=n+1$. The constraints then require that $b_3 \geq 2b_2-b_1 = 2n+1$, and so $n+2$ does not appear in the second sequence. Since $n+1$ is not a factor of $n+2$, we see that $n+2$ must be in the first sequence, and so $a_{n+1}=n+2$. Again, since $n+1$ is not a factor of $n+3$, we see that $n+3$ must be in one of the sequences. But $a_{n+2}\geq 2a_{n+1}-a_n=n+4$ and $b_3 \geq 2n+1 > n+3$ since $n\geq 3$, and so $n+3$ can not be in either sequence, a contradiction.
Thus $a_{n+1}=n+1$, and so $a_n=n$ for all $n$ by induction. But this is clearly also a contradiction since then the second sequence can only contain the number $1$!.
Thus we must have that $b_2=3$. We can then also see that $a_3=4$ since $4$ must appear in one of the sequences (indeed, any square of a prime number must appear in one of the sequences), and $b_3 \geq 5$.
We will now prove the following claim by induction:
For some natural number $n$, if $k \leq 2^n$ then $k$ appears in $A$ iff $k$ is a power of $2$, $k$ appears in $B$ iff $k$ is odd, and $k$ appears in neither sequence otherwise.
We see that this would imply that the sequences must coincide with the two which you have already found.
Our claim is true for $n=1$ and $n=2$ as we showed above. Suppose that it is true for some $n$.
We first note that any even numbers between $2^n$ and $2^{n+1}$ can already be written as a product of elements from each list, since any such number is of the           form $m2^k$ where $k\leq n$ and $m$ is an odd number less than $2^n$. Thus these even numbers do not appear in either sequence.
Now note that that the inductive hypothesis implies that $a_{n}=2^{n-1}$ and $a_{n+1}=2^n$, and so $a_{n+2}\geq 2a_{n+1}-a_n = 2^n + 2^{n-1}$.
Suppose that $m$ is an odd number such that $0<m<2^{n-1}$. I claim that $2^n+m$ lies in $B$. Any proper factor of $2^n+m$ must be odd, and at most $\frac{2^n+m}{2}<2^n$, and hence appears in $B$ by the inductive hypothesis. Thus $2^n+m$ must lie in one of the sequences. But $a_{n+1}<2^n+m<a_{n+2}$, and so $2^n+m$ must appear in $B$.
Now we will show that if $m$ is an odd number with $2^{n-1}<m<2^n$ that $2^n+m$ also appears in the second sequence. We proceed inductively. Suppose that all of the odd numbers less than $2^n+m$ have already been shown to lie in $B$. Then all of the proper factors of $2^n+m$ are odd and hence lie in $B$, and so $2^n+m$ appears in one of the sequences. Suppose that it is the first one. i.e. that $a_{n+2}=2^n+m$ We see then that $a_{n+3}\geq 2^n+2m \geq 2^{n}+2(2^{n-1}+1)=2^{n+1}+2$ and so $2^{n+1}$ does not appear in the first sequence. But all of its proper factors appear in the first sequence, and so $B$ contains $2^{n+1}$. But now consider the odd number $s=2^{n+1}+1$. All of its proper factors are at most $\frac{s}{2}<2^n+1$ and so lie in $B$. Thus $s$ appears in one of the sequences, and can not appear in $A$. Thus $s$ appears in $B$, and so $2^{n+1}$ and $2^{n+1}+1$ are consecutive terms in $B$. But their difference is $1<b_2-b_1$, which is a contradiction.
Thus $2^n+m$ must be in the second list, and so we see that all odd numbers less than $2^{n+1}$ are in the second list. Finally, $2^{n+1}$ has all of its factors in the first sequence, and so must appear in one of the sequences. It can not appear in the second sequence, since the element following $2^{n+1}-1$ must be at least $2^{n+1}+1$, and so $2^{n+1}$ appears in the first sequence, and so the results is true for $n+1$ as well.
This proves our claim.
