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Let $a_k$ a an eigenvalue of the adjacency matrix $A$ of a planar cubic graph with $n$ vertices. For the returning paths without backtracking we get the generating function of

$$G(x,a_k)=\frac{1-x^2}{1-a_kx+2x^2} \tag{1} $$

Now multiply each of the $n$ expressions $G(x,a_k)$ and compare it with Ihara's $\zeta$ function for cubic graphs. Using $\chi-1=F-2$$\;=E-n$, there seems to be a mismatch $$ \zeta _{G}(x)={\frac {(1-x^{2})^{n-E}}{\det(I-Ax+2x^{2}I)}} \neq \frac{(1-x^2)^n}{\prod_{k=1}^n (1-a_kx+2x^2)}, $$ where you have to think of the determinant on the LHS written in a diagonal base.

$\hskip1.3in$Why do I have to divide by $(1-x^2)^{E}$ to get Ihara's formula?

$\hskip2.0in$Is each edge interpreted like a $2$-cycle?

Another thought: The Ihara $\zeta$ function written in another form (rightmost) $$ \zeta _{G}(x) = {\frac {(1-x^{2})^{n-E}}{\det(I-Ax+2x^{2}I)}} = \exp \left( \sum \limits_{k=1}^{\infty} N_k \frac{x^k}{k} \right). $$ The non-negative numbers $N_k$ count non-backtracking, tailless closed paths. Since we divide by $(1-x^2)^{E}$ we kind of remove the edges.

Is this edge-removement what makes the Ihara $\zeta$ function count tailless walks, since an edge is the beginning of a tail?

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draks ...
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    The generating function in the linked post and the Ihara zeta function count related, but different quantities, so they aren't expected to agree. More precisely, I don't see a combinatorial reason for expecting the determinant of one to equal the other. The only way to sort this out is to do a detailed combinatorial analysis. – Will Orrick Dec 20 '15 at 17:24
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    To be clearer, I'm not actually sure how to formulate what it is the the Ihara zeta function counts. Just looking at the definition, it's enumerating products of equivalence classes of primitive nonbacktracking circuits according to their total length, but I'm not clear on how to understand "product" in terms of graphs. The generating function in the linked post, on the other hand, is easier to understand: it is matrix valued, and the coefficient of $x^n$ in the $(i,j)$ element of this matrix is the number of nonbacktracking paths of length $n$ starting at vertex $i$ and ending at vertex $j$. – Will Orrick Dec 20 '15 at 21:55
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    A detailed study of the proof of the determinant formula in this paper, or of some of the papers it cites, would probably clear this up. – Will Orrick Dec 20 '15 at 21:56
  • Hey @WillOrrick why did you delete this answer, I was about to accept it. Also your comments in MO are deleted. Did I offend you? Sorry if so... – draks ... Dec 30 '15 at 22:28
  • Sorry about that. I wasn't offended and I've restored the deleted answer, in case it is useful. I'm feeling a certain personal futility regarding this problem. You ask good questions, which, unfortunately, are beyond my expertise to answer. I don't want my minor observations and corrections to clutter the comment sections of your posts, possibly discouraging better answers from being posted. In the MO post, it would be good if you added the missing powers of $2$ in front of the Chebyshev polynomials. The other comment was just pointing out that the scale factor coincides with the... – Will Orrick Jan 01 '16 at 13:15
  • ...factor in the definition of Ramanujan graph for all $k$, not just $k=3$. Feel free to add the to your original post. – Will Orrick Jan 01 '16 at 13:15
  • @WillOrrick, thanks. you mean the powers of 2 that also Chris once mentioned...? – draks ... Jan 01 '16 at 21:39
  • Yes, but it's better to refer to Hamed's answer, since, insightful as Chris's answer is, it is incorrect in a number of details. The coefficient of $x^r$ in the original generating function in Hamed's answer is $p_r(a)$. But after the change of variable $t=\sqrt{2}x$, you should say that the coefficient of $t^r$ is $2^{-r/2}p_r(a)$. So Hamed's final expression for the generating function in $t$ implies that $$p_r(a)=2^{r/2}U_r(a/\sqrt{8})-2^{(r-2)/2}U_{r-2}(a/\sqrt{8}).$$ You can also get the correct expression from one of my... – Will Orrick Jan 02 '16 at 18:44
  • ...comments on Chris's answer. It may also be worth mentioning that that expression for $p_r(a)$ is only correct for $r\ge1$; it does not hold for $r=0$: the expression gives $p_0(a)=3/2$, whereas the correct value is $p_0(a)=1$. – Will Orrick Jan 02 '16 at 18:44
  • The generalization to $k$-regular graphs is the following: the recurrence for the number of nonbacktracking paths is $p_{r+1}(a) = ap_r(a)-(k-1) p_{r-1}(a)$, the generating function is$$ G(x,a)=\frac{1-x^2}{1-ax+(k-1)x^2}, $$ and an explicit form is $$ p_r(a)=(k-1)^{r/2}U_r\left(\frac{a}{2\sqrt{k-1}}\right)-(k-1)^{(r-2)/2}U_{r-2}‌​\left(\frac{a‌​}{2\sqrt{k-1}}\right).$$ The condition for a $k$-regular graph $G$ to be Ramanujan is that $\lambda(G)\le 2\sqrt{k-1}$. – Will Orrick Jan 02 '16 at 18:46
  • @draks... Hello. So as I was saying, we have a generating function $$\sum \limits_{k=0}^{\infty} t^k A_k= \frac{1-t^2}{I-At+(d-1)t^2}$$ and my goal is to derive the relation $$exp \left( \sum \limits_{k=1}^{\infty} Tr(A_k) \frac{t^k}{k} \right)=\frac{(1-t^2)^{|E|-|V|}}{det(I-At+(d-1)t^2)}$$ from that, using elementary reasoning if possible. Simply taking a trace doesn't seem to help as far as I see it. Also, did have a look at Copeland's question too, but will go through it in more detail. Thanks. – BharatRam Apr 05 '17 at 16:12
  • @BharatRam start from your last expression. Diagonalize A then the determinant is a simple product of quadratic polynomials in t that depend on the eigenvalues of A. Simply take the log of this polynomial and prove it by using Möbius inversion for Ihara's zeta function. What do you think? It's definitely more tricky but sounds doable... – draks ... Apr 05 '17 at 18:24
  • @BharatRam if you still worry about the circuit rank, this is the right place to discuss about it. Concerning your other question Tom or Chris are definitely the ones the should be able to help here... – draks ... Apr 06 '17 at 06:32
  • @draks... Tom Copeland seems to be using a different definition which makes the result almost trivial. He defines the zeta function as $$\zeta(t) = exp \left( \sum N_k \frac{t^k}{k} \right)$$ where $N_k = Tr(A^k)$, which counts the number of closed walks of length $k$ including backtracking. In this case, the generating function for $A^k$ is trivially $$\sum t^k A^k= \frac{1}{1-At}$$ so simply taking a determinant will work. Isn't the whole definition wrong here? – BharatRam Apr 06 '17 at 08:55
  • @BharatRam right, I misread that. But he is definitely able to answer your question. As Chris Godsil is. If you like we can also try to work it out on our own (in a chat?)... – draks ... Apr 06 '17 at 11:44
  • @draks... Sure, I'd be more than happy to work it out together. Although for the moment, I have some kind of calculation which seems convincing for the most part, but doesn't quite get the circuit rank involved. I'll write it out as an update to my question, and maybe we could start from there? I suspect much would be cleared up if we can settle on the definitions at least. Link for your convenience again: http://math.stackexchange.com/questions/2215888/non-backtracking-paths-and-the-ihara-zeta-function – BharatRam Apr 06 '17 at 17:02

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