3 small correction

The resulting graph is the linegraph of the subdivision graph of $G$. This survey paper of Bojan Mohar tells how to obtain the Laplacian spectrum of the linegraph of a semiregular graph and the subdivision graph of a regular graph.

Let's generalize. Let $G$ be a regular graph of $n$ vertices, degree $d$, and therefore $m=nd/2$ edges. Let $\mu(G,x)$ denote the characteristic polynomial of the Laplacian matrix, and let $\kappa(G)$ be the number of spanning trees.

The blowup $B(G)$ of $G$, formed by replacing each vertex by a $d$-clique, is the linegraph of the subdivision graph of $G$. Using Theorems 3.8 and 3.9 in the survey paper of Mohar, we find $$\mu(B(G),x) = (-1)^n (x-d)^{m-n} (x-d-2)^{m-n} \mu(G,x(d+2-x)).$$

We know that $\kappa(G) = n^{-1} (-1)^{n-1} \mu'(G,0)$. Differentiating and using $\mu(G,0)=0$, we find $$\kappa(B(G)) = d^{m-n-1} (d+2)^{m-n+1} \kappa(G).$$

For the $k$-fold blowup, we have $$\kappa(B^k(G)) = d^{d_k(m-n)-1d^{d_k(m-n)-k} (d+2)^{d_k(m-n)+1} d+2)^{d_k(m-n)+k} \kappa(G),$$ where $d_k=1+d+\cdots+d^{k-1}$.

For $d=3$, I believe this gives $$\kappa(B^k(G)) = (5/3)^k 15^{(3^k-1)n/4} \kappa(G).$$

2 fix and expand

The resulting graph is the linegraph of the subdivision graph of $G$. This survey paper of Bojan Mohar tells how to obtain the Laplacian spectrum of the linegraph of a semiregular graph and the subdivision graph of a regular graph.I'm just about to board

Let's generalize. Let $G$ be a planeregular graph of $n$ vertices, but I think what it gives is this:

Write degree $d$, and therefore $m=nd/2$ edges. Let $\mu(G,x)$ for denote the characteristic polynomial of the Laplacian matrix, and let $\kappa(G)$ be the number of spanning trees.

The blowup $B(G)$ of $G$, formed by replacing each vertex by a cubic $d$-clique, is the linegraph of the subdivision graph of $G$. Then for Using Theorems 3.8 and 3.9 in the blowup $L(S(G))$ survey paper of Mohar, we have find $\mu(L(S(G)),x)$ \mu(B(G),x) = (-1)^{n/2} -1)^n (2-x)^n x-d)^{m-n} (x-d-2)^{m-n} \mu(G,x(5-x))$. From this mu(G,x(d+2-x)). $$We know that \kappa(G) = n^{-1} (-1)^{n-1} \mu'(G,0). Differentiating and using \mu(G,0)=0, we find$$ \kappa(B(G)) = d^{m-n-1} (d+2)^{m-n+1} \kappa(G). $$For the number of spanning trees should follow easilyk-fold blowup, but I we have to catch my plane.$$ \kappa(B^k(G)) = d^{d_k(m-n)-1} (d+2)^{d_k(m-n)+1} \kappa(G), $$where d_k=1+d+\cdots+d^{k-1}. For d=3, I believe this gives$$ \kappa(B^k(G)) = (5/3)^k 15^{(3^k-1)n/4} \kappa(G).$$1 The resulting graph is the linegraph of the subdivision graph of$G$. This survey paper of Bojan Mohar tells how to obtain the Laplacian spectrum of the linegraph of a semiregular graph and the subdivision graph of a regular graph. I'm just about to board a plane, but I think what it gives is this: Write$\mu(G,x)$for the characteristic polynomial of the Laplacian matrix of a cubic graph$G$. Then for the blowup$L(S(G))$we have$\mu(L(S(G)),x) = (-1)^{n/2} (2-x)^n \mu(G,x(5-x))\$.

From this the number of spanning trees should follow easily, but I have to catch my plane.