I have competing answers on my question about "Returning Paths on Cubic Graphs Without Backtracking". Assuming Chris is right the following should work. Up to one thing:
The number of returning paths on 3-regular graphs of length $r$ without backtracking may be written as $2^{-r/2}p_r(x/\sqrt{2})$ which is a Chebyshev Polynomial of the Second Kind $U_r(x)$.
The linked MathWorld page also says that
the defining generating function of the Chebyshev polynomials of the second kind is
$$ g(t,x)=\frac1{1-2xt+t^2}=\sum_{r=0}^\infty U_r(x) t^r \tag{8,9} $$
So, if we substitute $U_r(x)$ with $2^{-r/2}p_r(x)$, next $x$ with $x/\sqrt{2}$ and sum up over $r$ we get $$ \sum_r \left(\frac t{\sqrt 2}\right)^r \cdot p_r\left(\frac x{\sqrt{2}}\right) \tag{10} $$
Now we plug in the eigenvalues $a_k$ of $G$'s adjacency matrix $A$ for $x$. Then the result should equal $$ \frac1{\operatorname{det}(I-\text{Diag}(a_k)t+\frac {t^2}2I)}\tag{11}. $$ where the denominator can be transformed, so that it overall represents a polynomial $$P_{\text{Diag}(a_k)}(t)=P_{A}(t).$$
To match
Ihara $\zeta$ function: If $G$ is $3$-regular with adjacency matrix $A$ then $$ \zeta_G(t) = \frac{(1-t^2)^{1-\chi(G)}}{\det(I - At + 2t^2I)} \tag{12} $$ where $\chi(G)$ is the circuit rank, which is $r=m-n+c$ where m is the number of edges in G, n is the number of vertices, and c is the number of connected components.
$G$ is cubic and shall have one component, so $2m=3n$ and $c=1$ and $\chi(G)-1=r-1=\frac32n-n+1-1=\frac12n$. The numerator is $(1-t^2)^{n/2}$ then and let's say call $Q_A(t)=\det((1+2t^2)I - At)={t^{-n}}\det((t^{-1}+2t)I - A)={t^{-n}}Q^\prime_A(t)$.
Question:
Did I go wrong somewhere or is it possible to prove:
that $$ \begin{eqnarray*} (1-t^2)^{n/2}P_A(t)&=&Q_A(t)\\ (1-t^2)^{n/2}P_A(t)&=&{(t^2)^{-n/2}}Q^\prime_A(t)\\ (t^2)^{n/2}(1-t^2)^{n/2}P_A(t)&=&Q^\prime_A(t)\\ (t^2-t^4)^{n/2}P_A(t)&=&Q^\prime_A(t) \end{eqnarray*} $$ is independant of $A$?
or
where is difference between $({11})$ and $(12)$?