It all started with [Chris' answer](http://math.stackexchange.com/a/177865/19341) saying returning paths on cubic graphs without backtracking can be expressed by the following recursion relation:

> $$p_{r+1}(a) = ap_r(a)-2p_{r-1}(a)$$

$a$ is an eigenvalue of the adjacency matrix $A$. Chris mentions Chebyshev polynomials there.
It was [Will](http://math.stackexchange.com/a/1289646/19341) who found the generating function for the given recursion to be:

>$$G(x,a)=\frac{1-x^2}{1-ax+2x^2} $$

and just recently [Hamed](http://math.stackexchange.com/a/1532612/19341) put Chebychev back on the table:

> $$
\frac{1-x^2}{1-ax +2x^2} \xrightarrow{t=\sqrt{2}x}\frac{1-\frac{t^2}2}{1-2\frac{a}{\sqrt 8} t+t^2}=\left[1-\frac{t^2}{2}\right]\sum_{r=0}^\infty U_r\left(\frac{a}{\sqrt{8}}\right)t^r\\
=\sum_{r=0}^\infty \left(U_r\left(\frac{a}{\sqrt{8}}\right)-\frac12 U_{r-2}\left(\frac{a}{\sqrt{8}}\right)\right)t^r\\
$$
$$
\Rightarrow p_r(t)=\begin{cases}1 & \text{if $r=0$,}\\ 2^{r/2}U_r(t/2^{3/2})-2^{(r-2)/2}U_{r-2}(t/2^{3/2}) & \text{if $r\ge1$.}\end{cases}
$$
The final line as taken from [Will's community answer](http://math.stackexchange.com/a/1602432/19341)...

My question how to relate [Ihara's $\zeta$ function](https://en.wikipedia.org/wiki/Ihara_zeta_function) and Chebyshev seems therefore [mostly](http://math.stackexchange.com/questions/1580353/is-each-edge-interpreted-like-a-2-cycle) settled, but...:

Is it just a funny coincidence that the scaling factor of $\sqrt 8$ coincides with $\lambda_1\leq 2\sqrt 2$, which is the [definition of cubic Ramanujan graphs](https://en.wikipedia.org/wiki/Ramanujan_graph#Definition).

And, there is another interesting thing:

> As observed by Sunada, a regular graph is a Ramanujan graph if and only if its Ihara zeta function satisfies an analogue of the Riemann hypothesis.

What does this connection between Chebyshev, Ramanujan, Ihara and Riemann mean?