How many trees have $n$ nodes with fewer than three neighbors? We want to know how many trees have $n$ nodes with fewer than three neighbors.  For $n=1$, the only possibility is a single node.  For $n=2$, the only possibility is two connected nodes.  For $n=3$, the only possibilities are a node adjacent to three nodes or a node adjacent to two nodes.  A set of drawings produces the following (possibly incorrect) sequence: $1$, $1$, $2$, $4$, $9$, $25$, $70$.  My question is whether this enumeration is a new problem or one that has been dealt with.  See OEIS A335342.
 A: Here are some observations which may inform a partial enumeration.
Such a tree has $t+n$ vertices, where the $l$ leaves are part of the $n$ nodes and the rest of the $n$ nodes look like they are interior vertices that are part of a path.  The $t$ vertices each have three or more branches, so we must have $3t \lt n+t+t-1$, where the $t-1$ is a correction for over counting edges between two branch vertices. This means $t \leq n-2$.
Now if $t$ is $0$, the only tree is a path, while if $t=1$ there is a bijection between these trees and partitions of $n$ into at least three parts.  For larger $t$, we need to decide on how to arrange and color the edges between the $t$ nodes. For $t=2$ we need to decide how many of the n nodes go between, and then make sure there are at least four leftover to distribute among the remaining branches.  For $t=3$, there are two internal edges to color with a total of $0$ up to $n-5$ nodes, and then distribute the remainder on the leafy portions, being sure to save at least one for the middle $t$ node.
We now see that this becomes a problem of counting branchy trees on $t$ nodes, and then partitioning $n$ nodes on the edges, and then worrying about overcounting.  At this point I leave the problem to the professionals.
Gerhard "Neither Professional Accountant Nor Gardener" Paseman, 2020.06.04.
