Let $G$ be a countably infinite discrete group. Let us call $G$ sequentially amenable, if there is a bijection $\mathbb{N}\to G,\; j\mapsto g_j$ such that the sets $F_k = \{g_1,\dots,g_k\}$ form a Folner sequence for $G$.
How big is the class of sequentially amenable groups?
The only obvious examples I can think of are the groups $\mathbb{Z}^m$ for $m\in\mathbb{N}$ or (a little bit more general) finitely generated abelian groups. (For $\mathbb{Z}^m$, you can see this by counting all $v\in\mathbb{Z}^m$ with $\|v\|_\infty=n$ in increasing order $n=0,1,2,\dots$.
Can you think of interesting non-abelian examples? Any answer would be very helpful.
Edit:
One fairly big class of examples are finitely generated groups with subexponential growth. Let's give a quick proof:
Let $G$ be such a group. Pick some finite generating set $S=S^{-1}\subset G$ and define the corresponding length metric $$\ell: G\to\mathbb{N},\quad \ell(g)=\min\{k \;|\; g=s_1\cdot\dots\cdot s_k\;\text{for some}\; s_i\in S \}.$$ Then it is well-known that the unit balls $$B_n = \ell^{-1}(\{0,\dots,n \})$$ form a Folner sequence for $G$. This is because of subexponential growth, which means that $\lim_{n\to\infty} |B_n|/|B_{n+1}|=1$. Then define the bijection $j\mapsto g_j$ of $G$ by counting all group elements $g$ with $\ell(g)=n$ in increasing order of $n$. Define $F_k=\{g_1,\dots,g_k \}$ for all $k$. Then it is clear that $$\forall\; k: \exists n: B_n\subset F_k\subset B_{n+1}.$$ It follows for all $g\in G$ that $$\lim_{k\to\infty} \frac{|F_k\cap gF_k|}{|F_k|} \geq \lim_{n\to\infty} \frac{|B_n\cap gB_n|}{|B_{n+1}|} = \lim_{n\to\infty} \frac{|B_n|}{|B_{n+1}|}\cdot \frac{|B_n\cap gB_n|}{|B_n|} = 1,$$ which confirms that the $F_k$ form a Folner sequence.
