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# Fibonacci Investigation

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)

Note by Curtis Clement
2 years ago

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I count 6 trees with 6 nodes up to graph isomorphism. · 2 years ago

Check the order of your nodes - we didn't find any counter-examples as a class. · 2 years ago

OEIS A000055 agrees as well, the sequence proceed as 1, 1, 2, 3, 6, 11, 23, 47,... · 2 years ago

What is OEIS A000055 ? · 2 years ago

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 · 2 years ago

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 $$n^{n-2}$$. · 2 years ago