Is it true that for any integer $k\geq 3$ there are $\aleph_0$ many connected countably infinite, pairwise nonisomorphic $k$regular graphs?
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1$\begingroup$ For each $k\ge 3$ there are $2^{\aleph_0}$ nonisomorphic graphs of valency $k$ (necessarily countable), obtained as Cayley graphs of the same number of pairwise nonquasiisometric finitely generated groups. But of course without transitivity assumption it's much easier. $\endgroup$ – YCor Mar 6 at 17:38
Take an $n$cycle, add an infinite tree of the right degree at each vertex of the cycle (the vertex on the cycle having degree 2 less in the tree than the other vertices of the tree). This has only one cycle and it is of length $n$. So the graphs you get for two different values of $n$ are nonisomorphic.

4$\begingroup$ Start with a "chain of cycles", i.e. cycles $(C_n)_{n \in \mathbb N}$, with an edge from $C_n$ to $C_{n+1}$, and adapt the attached trees accordingly. Then altering the cycle lengths even gives you $2^{\aleph_0}$ many nonisomorphic such graphs. $\endgroup$ – Florian Lehner Jul 7 '18 at 18:11
