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I am looking for help:

Beginning with a single node ($\circ$), at each discrete time step I can add a node/link pair to any node currently in the tree. Nodes are unlabelled and the tree is undirected. This will produce a "tree trajectory", like this: $$ \circ \quad \rightarrow\quad \circ - \circ \quad \rightarrow\quad \begin{array}{c} \ \ \ \ \circ - \circ \\ | \\ \circ \end{array} \quad \rightarrow\quad \begin{array}{c} \ \ \ \ \circ - \circ \\ | \\\ \ \ \ \circ - \circ \end{array} $$ Another tree trajectory might look like this: $$ \circ \quad \rightarrow\quad \circ - \circ \quad \rightarrow\quad \begin{array}{c} \ \ \ \ \circ - \circ \\ | \\ \circ \end{array} \quad \rightarrow\quad \begin{array}{c} \circ - \circ \\ | \ \ \backslash \\ \circ \ \ \ \circ \end{array} $$

My questions are as follows:

1) How many possible distinct (non-isomporhic) graphs are there with N nodes and N-1 edges?

It's easy to find this by brute force up to 5: N=1: 1; N=2:1; N=3:1; N=4:2; N=5:3

2) For N nodes, how many different tree trajectories are there connecting the graph with N=1 node to the any graph of N nodes? (i.e., how many distinct tree trajectories are there; a tree trajectory is distinct from all others if for any other tree trajectory, there is at least a single step where the two are not isomorphic.) It's clear not just multiplying the number of non-ismorphic graphs at each step, since past graph topology will constrain future graph topologies.

While a solution would obviously be great, ANY help at all would be appreciated.

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    $\begingroup$ The answer to your first question is in OEIS. See here: oeis.org/A000055 $\endgroup$ Commented Feb 10, 2016 at 19:00
  • $\begingroup$ This was helpful - I should have thought to search unlabeled. Still thinking about question two... $\endgroup$ Commented Feb 11, 2016 at 0:07
  • $\begingroup$ A note: This is similar to counting number of Young diagrams of some size, and a trajectory of Young diagrams is just a standard Young tableau.... You can probably encode a trajectory as a linear extension of some poset. $\endgroup$ Commented Feb 16, 2016 at 16:26

1 Answer 1

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The total number of trajectories $T(N)$ (as a function of $N$) appears to be

N T(N) 1 1 2 1 3 1 4 2 5 4 6 13 7 48 8 235 9 1297 10 8628 11 63902 12 538454 13 4973090 14 50621738 15 557399709 16 6636723151 17 84584674076 18 1151419603932 19 16640050320703 20 254656634876886 21 4110614617328251 

I've added these counts to the OEIS as sequence A268951.

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  • $\begingroup$ The question is rather ambiguous (isomorphism must be defined more precisely), and your answer does not fit the first few terms given. $\endgroup$ Commented Feb 19, 2016 at 20:25
  • $\begingroup$ @F.C.: I defined it precisely in the OEIS. What given terms do you refer to? I see only terms given for question 1), but nothing for question 2). My counts are about question 2). $\endgroup$ Commented Feb 19, 2016 at 20:33

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