How to find non-isomorphic graphs with specific orders? I work on a problem in my research. I have a graph, $G$, with $2n$ vertices. It has one connected component of order $2n-1$ and an isolated vertex. $\lambda_1\geq \lambda_2\geq \ldots \geq \lambda_{2n}$ are the eigenvalues of $G$. I have some bounds for them.
$$2n-3\leq \lambda_1<2n-2,\\
 0\leq \lambda_2\leq 1,\\
 -1\leq \lambda_i\leq \frac{1}{2}, \ \ \ 3\leq i \leq n+1,\\
 -3\leq \lambda_{n+2}\leq -1,\\
 -3\leq \lambda_i\leq \frac{-3}{2},\ \ \ n+ 3\leq i \leq 2n.$$
Also the maximum degree of this graph is $2n-2$.
How to find these graphs for $6\leq n \leq 10$ ?
Thanks for your help.
 A: So here is a family of graphs that satisfies your requirements.... is this the only family?

*

*Let $X_1$ be the graph consisting of $n-1$ disjoint copies of $K_2$. Then the spectrum of $X_1$ is $$\underbrace{1,1,\ldots,1}_{n-1},\underbrace{-1,-1,\ldots,-1}_{n-1}$$

*Next let $X_2$ be the complement of $X_1.$ As $X_1$ is regular, the eigenvectors of $A(X_2)$ are the same as those of $A(X_1)$, and there is a well-known transformation giving the eigenvalues of $X_2$ in terms of the eigenvalues of $X_1$. In particular, the spectral radius of $X_2$ is $2n-4$, the $n-2$ other eigenvalues equal to $+1$ all become $-1-1=-2$, while all the eigenvalues equal to $-1$ become $-(-1)-1 = 0$. So the spectrum of $X_2$ is $$2n-4, \underbrace{0,0,\ldots,0}_{n-1},\underbrace{-2,-2,\ldots,-2}_{n-2}$$ and so its characteristic polynomial is $$\varphi(X_2) = (x-(2n-4))x^{n-1}(x+2)^{n-2}$$

*Now let $X_3$ be the cone of $X_2$, which is the graph obtained by adding a new vertex to all existing vertices. Conveniently, Brouwer and Haemers (Graph Spectra, Chapter 1, Exercise 11) tell us that if $X$ is a $k$-regular graph on $v$ vertices and $Y$ is its cone, then $$\varphi(Y) = (x^2-kx-v)\varphi(X)/(x-k).$$ In our case we have $v=2n-2$ and $k=2n-4$ and so the result is that $$\varphi(X_3) = \left(x^2 - (2n-4)x - (2n-2)\right)\,x^{n-1} (x+2)^{n-2}.$$

*To find the eigenvalues of $X_3$ we need to know the roots of the quadratic factor $$p(x) =x^2 - (2n-4)x - (2n-2)$$ of $\varphi(X_3)$. Now it is straightforward to check that $p(2n-3)= -1$ and $p(2n-2) = 2(n-1)> 0$ and so there is one root $\lambda$ such that $$2n-3 < \lambda < 2n-2.$$ Next we see that $p(-1) = -1$ and $p(-2) = 2(n-1) > 0$ and so there is a second root $\mu$ between $-1$ and $-2$.

*Finally, form $X_4$ by adding an isolated vertex to the graph, which simply adds an additional $0$ to the spectrum. Putting everything together we get the following spectrum for $X_4$: $$\lambda, \underbrace{0,0,\ldots,0}_{n}, \mu, \underbrace{-2,-2,\ldots,-2}_{n-2}.$$

*Now let's check the conditions of the OPs problem:

*

*$2n-3 \leqslant \lambda_1 < 2n-2$ is true because $\lambda_1$ is the larger root of $p$.

*$0 \leqslant \lambda_2 \leqslant 1$ is true because $\lambda_2 = 0$.

*$-1 \leqslant \lambda_i \leqslant 1/2$ for $3 \leqslant i \leqslant n+1$ is true because all of these eigenvalues are equal to $0$.

*$-3 \leqslant \lambda_{n+2} \leqslant -1$ is true because $\lambda_{n+2} = \mu$ and we know that $\mu$ is between $-1$ and $-2$.

*$-3 \leqslant \lambda_i \leqslant -3/2$ for $n+3 \leqslant i \leqslant 2n$ is true because these eigenvalues are all exactly $-2$.

*The maximum degree of the graph is realised by the universal vertex added when the cone is formed, and that has degree $2n-2$.



