A couple of months ago, we were studying graphs in Further Maths (not an \({xy}\) plane, rather a series of nodes and edges), when we decided to do a little investigation with \({tree}\) \({diagrams}\). The question goes like this: what are number of \({distinct}\) \({tree}\) \({diagrams}\) with \({n}\) nodes? Well we found that if a graph has (2,3,4,5,6,7) nodes then it would have (1,1,2,3,5,8) distinct tree diagrams respectively, which follows the Fibonacci sequence. We were not able to prove why, so could anybody be able to provide a proof? (I have a feeling it might be to do with the number of distinct ways you can write 2[\({n}\)-1] as a sum of \({n}\) numbers, if \({n}\) = number of nodes)

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TopNewestI count 6 trees with 6 nodes up to graph isomorphism. – Lee Gao · 2 years ago

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– Curtis Clement · 2 years ago

Check the order of your nodes - we didn't find any counter-examples as a class.Log in to reply

– Lee Gao · 2 years ago

OEIS A000055 agrees as well, the sequence proceed as 1, 1, 2, 3, 6, 11, 23, 47,...Log in to reply

– Curtis Clement · 2 years ago

What is OEIS A000055 ?Log in to reply

https://oeis.org/A000055 – Lee Gao · 2 years ago

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A series of nodes and edges such that the graph is connected and there are no cycles. Also, the nodes have to be connected using the least number of nodes. Examples: (order of each node) 2 nodes = {1,1} 3nodes = {2,1,1} 4 nodes = {1,2,2,1} and {3,1,1,1} 5 nodes = (4,1,1,1,1}, {3,2,1,1,1} and {2,2,2,1,1} etc – Curtis Clement · 2 years ago

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Can you define more precisely what you mean by "tree diagrams" ? It is known that (Cayley's formula) number of distinct labelled trees on \(n\) vertices is \(n^{n-2}\). – Abhishek Sinha · 2 years ago

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