**Puzzle 105**: I can't see Matthew's answer in [comment #5](https://forum.azimuthproject.org/discussion/comment/18796/#Comment_18796), so I will write my own. Following what John wrote in [comment #4](https://forum.azimuthproject.org/discussion/comment/18795/#Comment_18795), it seems like it should be sufficient to add an arrow \\(k\\) from \\(x\\) to itself. That way, every path of length \\(n+2\\) will come from composing either \\(f\\) or \\(k\\) with a path of length \\(n+1\\), or composing \\(h\circ g\\) with a path of length \\(n\\). Since this gives the recurrence relation $$P_{n+2}=2P_{n+1}+P_n$$ that is satisfied by the Pell numbers (with \\(P_0=1\\) and \\(P_1=2\\) due to \\(\\{\text{id}_x\\}\\) being the only path starting and ending at \\(x\\) of length zero and \\(\\{f\\},\\{k\\}\\) being the only paths of length one).

Generalizing slightly, one can get any recursion relation of the form $$a_{n+2}=\alpha a_{n+1}+\beta a_n\qquad\text{for }\alpha,\beta\in\mathbb{N}$$ by starting with the node \\(x\\) and its identity arrow, and adding \\(\alpha\\) arrows from \\(x\\) to itself and \\(\beta\\) nodes, each with a pair of arrows connecting each node to the node \\(x\\) (restricting to the case with just two nodes, \\(x\\) and \\(y\\), one can get any \\(\alpha\in\mathbb{N}\\) and square betas \\(\beta=m^2, \forall m\in\mathbb{N}\\) with the appropriate arrows).

To every recursion relation with specified base case(s) there is associated a generating function. For example, for $$F_{n+2}=F_{n+1}+F_n\qquad\text{with }P_0=1,\,P_1=1$$ there is associated the generating function $$F(z)=\frac{1}{1-z-z^2},$$ which has the Fibbonacci sequence as the sequence of coefficients in its formal power-series expansion. Does the generating function associated with a graph tell us anything interesting about the graph?