Non-backtracking closed walks and the Ihara zeta function (Updated with partial attempt) For a connected $d$-regular graph $G=(V,E)$ with adjacency matrix $A$, we defined a sequence of matrices
$$A_0,A_1,A_2,A_3,\dots$$
defined using powers of $A$ inductively as follows:
$$A_0=I$$
$$A_1=A$$
$$A_2=A^2-dI$$
For $k \geq 3$,
$$A_{k}=A_{k-1}A-(d-1)A_{k-2}$$
Just like $(A^k)_{v,w}$ counts the number of walks on $G$ from $v$ to $w$, the value $(A_k)_{v,w}$ counts the number of walks on $G$ from $v$ to $w$ without backtracking.
The recurrence relation above can be used to easily show that the ordinary (matrix) generating function for the above sequence is
$$\sum \limits_{k=0}^{\infty} t^k A_k = (1- t^2)I. \left( I-tA + (d-1)t^2 I \right)^{-1}$$
With some abuse of notation, we can rewrite this generating function as
$$\frac{1-t^2}{I-At+(d-1)t^2}$$

On the other hand, the Ihara zeta function of the graph $G$ is given by
$$\zeta_G(t) = exp \left( \sum \limits_{k=1}^{\infty} N_k \frac{t^k}{k} \right)$$
where $N_k$ is the number of closed non-backtracking walks on $G$ of length $k$. It is known that $\zeta_G(t)$ has an alternate expression using determinants as
$$\zeta_G(t) = \frac{(1-t^2)^{|V|-|E|}}{det(I-At+(d-1)t^2)}$$

My question is: can the determinant formula for the Ihara zeta function be derived from the generating function for the matrices $A_k$? What exactly is the relationship between $A_k$ and $N_k$? 
A similar question has been asked here How to get from Chebyshev to Ihara?
and I have also been trying out ideas from here Proof of 2 Matrix identities (Traces, Logs, Determinants)
But I am not interested in the Chebychev polynomial connection here: just whether the generating function can be manipulated using logarithms and traces to obtain the expression for the zeta function. Thanks.

UPDATE:
Here's a partial attempt of mine. Using Chebychev polynomials of the second kind defined as 
$$U_0(x)=1$$
$$U_1(x) = 2x$$
and for $k \geq 2$,
$$U_k(x) = U_{k-1}(x)U_1(x) - U_{k-2}(x)$$
and with generating function
$$\sum \limits_{k=0}^{\infty} U_k(x)t^k = \frac{1}{1-2xt+t^2}$$
we can express the matrix $A_k$ as
$$A_k = (d-1)^{k/2} U_k \left( \frac{A}{2 \sqrt{d-1}} \right) - (d-1)^{k/2-1} U_{k-2} \left( \frac{A}{2 \sqrt{d-1}} \right)$$
and so
$$Tr(A_k) = (d-1)^{k/2} \sum \limits_{i=0}^{n-1} U_k \left( \frac{\mu_i}{2 \sqrt{d-1}} \right) - (d-1)^{k/2-1} \sum \limits_{i=0}^{n-1} U_{k-2} \left( \frac{\mu_i}{2 \sqrt{d-1}} \right) $$
where 
$$d=\mu_0 \geq \mu_1 \geq \dots \geq \mu_{n-1} \geq d$$
are the $n$ eigenvalues of the adjacency matrix $A$. Thus we have an expression for the trace of $A_k$ as a polynomial in the eigenvalues of $A$. 
In a similar way, working backwards fro the definition of the Ihara zeta function, we can define another matrix sequence $B_k$ as follows: 
$$B_1=A$$
$$B_2 = A^2-dI$$
and for $k \geq 3$,
$$B_k = \begin{cases}
B_{k-1}A - (d-1)B_{k-2} - (d-2)A & \text{ if k is odd}\\
B_{k-1}A - (d-1)B_{k-2} + d(d-2)I & \text{ if k is even}
\end{cases}$$
Just like $A_k$ could be expressed in terms of Chebychev polynomials $U_k$ of the second kind, the matrices $B_k$ can be expressed using Chebychev polynomials $T_k$ of the first kind defined by the recurrence
$$T_0(x)=1$$
$$T_1(x)=x$$
and for $k \geq 2$,
$$T_k(x)=2xT_{k-1}(x)-T_{k-2}(x)$$ 
The expression for $B_k$ is
$$B_k=\begin{cases}
2(d-1)^{k/2}T_k \left( \frac{A}{2\sqrt{d-1}} \right) & \text{ if k is odd}\\
2(d-1)^{k/2}T_k \left( \frac{A}{2\sqrt{d-1}} \right)+(d-2)I & \text{ if k is even}
\end{cases}$$
All this is simply by reverse engineering the expression for the Ihara zeta function to obtain $N_k$ as the trace of some matrix. My question now, modulo correctness of my calculations, is whether the matrices $B_k$ as defined above have any natural interpretation in terms of walks on the graph.
 A: A matrix generating function is a matrix over a ring (of power series). Trace and determinant are defined for matrices of arbitrary commutative rings, and so they can be used freely in this context.
A: This is not an answer, but I thought it too long for the chat room.
The formula
$$\exp \left( \sum \limits_{k=1}^{\infty} \text{Tr}\,A_k \frac{t^k}{k} \right) = \frac{1}{\det(I-At+(d-1)t^2)}
$$
cannot be correct, as it does not work when you try it on a particular example.  Consider the $4$-regular graph consisting of a single vertex with two loops.  We have $d=4$, $A=[4]$, and $A_k=\left[4\cdot3^{k-1}\right]$.  So the left side reduces to
$$
\exp\left(\frac{4}{3}\sum_{k=1}^\infty\frac{(3t)^k}{k}\right)=\exp\left(-\frac{4}{3}\log(1-3t)\right)=(1-3t)^{-4/3},
$$
which is not equal to the right hand side, $(1-4t+3t^2)^{-1}$.  In fact, the left hand side is not even a rational function of $t$, although it is algebraic.
In this example, it is not too hard to eliminate the walks with tails using inclusion-exclusion.  By the recurrence for $A_k$, there are $4\cdot3^{k-1}$ walks of length $k$ that have no backtracking except for a possible tail.  This can be easily understood by noting that each of the two loops may be traversed in either of two directions, for a total of four possible steps, but that no step may be the reverse of the one just taken.  Let $s_1s_2\ldots s_k$ be the sequence of steps taken.  The walk has a tail if $s_k=s_1^{-1}$.  We want to subtract such walks  A set containing all such walks with tails can be obtained by appending $s_1^{-1}$ to each of the $4\cdot3^{k-2}$ nonbacktracking walks of length $k-1$.  Some of the walks obtained in this way contain backtracking, however, namely those in which $s_{k-1}=s_1$, and therefore should not have been included in the subtraction.  So we add these back in.  Such walks may be obtained by appending $s_1$ to one of the $4\cdot3^{k-3}$ nonbacktracking walks of length $k-2$.  Some of these walks now contain backtracking, namely those in which $s_{k-2}=s_1^{-1}$.  So we need to subtract these, and so on.  The end result of these subtractions and additions is
$$
\begin{aligned}
N_k&=4\cdot3^{k-1}-\sum_{j=1}^{\lfloor(k-1)/2\rfloor}(4\cdot3^{k-2j}-4\cdot3^{k-2j-1})\\
&=4\cdot3^{k-1}-3^{k-1}\left[1-\left(\frac{1}{3}\right)^{2\lfloor(k-1)/2\rfloor}\right]\\
&=4\cdot3^{k-1}-(3^{k-1}-3^{\epsilon_k})\\
&=3^k+3^{\epsilon_k},
\end{aligned}
$$
where $\epsilon_k=0$ for $k$ odd and $1$ for $k$ even.
Inserting this into the expression for the zeta function gives
$$
\begin{aligned}
\exp \left( \sum \limits_{k=1}^{\infty} N_k \frac{t^k}{k} \right) &=\exp \left( \sum \limits_{k=1}^{\infty} \frac{(3t)^k}{k} +\sum \limits_{k=1}^{\infty} \frac{t^k}{k}+\sum \limits_{k=1}^{\infty} \frac{2t^{2k}}{2k}\right)\\
&=\exp\left(-\log(1-3t)-\log(1-t)-\log(1-t^2)\right)\\
&=\frac{(1-t^2)^{-1}}{1-4t+3t^2}\\
&= \frac{(1-t^2)^{|V|-|E|}}{\det(I-At+(d-1)t^2)},
\end{aligned}
$$
confirming the determinant formula for the Ihara zeta function in this example.
I have not tracked down what went wrong in your derivation of the expression for
$$\exp \left( \sum \limits_{k=1}^{\infty} \text{Tr}\,A_k \frac{t^k}{k} \right),
$$
but I'm not sure I follow even the first step, where an integration is performed.  Can you elaborate on that?
