We know that the number $2^n+2^m$ is unique for $n,m\in\mathbb{N}$. Is there any explicit way of writing a function $\sigma:\mathbb{N}\to\mathbb{N}^2$ such that $$ \sigma(2^n+2^m)=(n,m), $$ for all $n,m\in\mathbb{N}$?
Remark (edit): Here, $(n,m)$ is to be interpreted as the unordered pair $\{n,m\}$, since I'm only interested in the pair. More specifically, I want to find functions $\sigma,f$ and $g$ such that $$ \sigma(k)=(g(k),f(k)), $$ where $k=2^{g(k)}+2^{f(k)}$, $\forall k \in\mathbb{N}$.
What do you think of $$\forall k \in \mathbb{N}, \quad \sigma(k)=\left( \lfloor \log_2(k) \rfloor, \lfloor \log_2 \left( k-2^{\lfloor \log_2(k) \rfloor} \right)\rfloor \right)$$
This is well defined except when $k$ is a power of $2$.
In the other case where $k=2^p$, choose $\sigma(k)=(p-1,p-1)$.