If $p$ is a positive prime number, put $p! = \textrm{lcm}(2, 3, 5, \dots, p) = p\cdots3\cdot2$. Then every non-negative integer can be written uniquely as:

$x\_1 + 2\cdot x\_2 + 3\cdot 2 x\_3 + \dots + p! \cdot x\_p$, where each $x\_p$ is indexed over the primes $p$ such that $p! \leq x$ and $0\leq x\_i \lt \sigma(i)!, \ \forall i$ where $\sigma(i)$ is the next prime after $i$.

Divide up the numbers into divisions of $2$ integers, then divisions of $2\cdot 3$ integers, and so on, $x\_p$ addresses what $p!$ block to be on and within that block, $x\_{p-1}$ addresses what sub-$x\_{p-1}$ block to be in and so on... to add up to the number $x$.

It is unique since each $x$ is in one and only one box on each level.