Fredrick wrote approximately:

> \[ J : \textbf{Graph} \rightarrow \mathbb{N}^\mathbb{N} \]

> Does \\(J\\) have an inverse? Does \\(J\\) have a right adjoint?

(Here \\(\mathbb{N}^\mathbb{N} \\) is my preferred notation for the set of functions from \\(\mathbb{N}\\) to itself - that is, sequences of natural numbers

This is a great thing to think about. The questions as stated don't make sense, for two reasons. Both these problems can be fixed:

A. It is not a graph that gives a sequence of numbers, but a **pointed graph**: a graph with a chosen node. This lets us count the number of paths of length \\(n\\) starting and ending at that node, and get a sequence of natural numbers. There is a category \\(\textbf{Graph}_\ast\\) of pointed graphs, where a morphism is a map of graphs that sends the chosen node of one to the chosen node of the other.

B. \\(\textbf{Graph}\_\ast \\) is a _category_ while \\(\mathbb{N}^\mathbb{N} \\) is a mere _set_. Thus, it's not clear what \\(J\\) should be: a functor (a map between categories) or a function (a map between sets). We can cure this problem in two equivalent ways:

1. Turn the category \\(\textbf{Graph}_\ast \\) into a set, namely the set of isomorphism classes of pointed graphs.

2. Turn the set \\(\mathbb{N}^\mathbb{N} \\) into a category with only identity morphisms.

Since they're equivalent we can use the first, less stressful approach.

There is indeed a well-defined function \\(J\\) from the set of isomorphism classes of pointed graphs to the set of sequences of natural numbers, where we count the number of paths of length \\(n\\) from the chosen node to itself.

This function \\(J\\) is neither one-to-one nor onto. The interesting question are thus:

1. Which sequences of natural numbers arise by counting the number of paths of length \\(n\\) from the chosen node in a pointed graph to itself?

1. When do two pointed graphs give the same sequence?

I know a nice answer to the first, which I'll explain if nobody can guess it, but I'm pretty clueless about the second.

I think Matthew Doty at least can guess the answer to the first question here, since he rapidly translated the Pell sequence into a pointed graph that gives it.

I think the easiest way to tackle the first question is to figure out a nice general formula for the number of paths of length \\(n\\) from any node in a graph to any other node.