I'm trying to compute the sequence (and then trying to figure out the limit) of the maximum eigenvalues of the adjacency matrices of the infinite family of graphs build by starting with a triangle and adding a single vertex at each step
Maybe you know software that can help me ? maybe the result is known ?
any clue is appreciated
The context motivating this question is a variation of the Kostant game on graphs where the "mutation" of a vertex value is computed as follows: Let $x_v$ the value currently assigned to vertex $v$, the value after "mutation" is
$ m_v(x_v) = -\alpha \cdot x_v + \sum_{w \equiv v} x_w $
where $\alpha$ is a real or complex parameter, instead of the classical Kostant game mutation
$ m_v(x_v) = -x_v + \sum_{w \equiv v} x_w $
It is well known that the classical Kostant game, ends (see details on this game for more explanations) iff the graph is a Smith graph also known as the ADE graphs (2 infinite families of graphs $A_n$ and $D_n$ and 3 exceptional finte graphs $E_6, E_7$ and $E_8$. I'm trying to figure out what is the situation with this modified game and in particular for $\alpha \ne 1$ if there are other graphs with finite game.
Hope this gives more context to my question
I found no clear pattern for the largest eigenvalue, but the limit of the largest eigenvalue is easy to find. In general, the largest eigenvalue is between the average degree and the maximum degree of a graph. Here, the average degree in an $n$-vertex version of this graph is $4 - \frac6n$ (the degree sequence is $2,3,4,4,\dots,4,4,3,2$ whose sum is $4n-6$, and then we divide by $n$) and the maximum degree is $4$. Therefore the largest eigenvalue is between $4 - \frac6n$ and $4$, and converges to $4$ as $n\to \infty$.
Proof that the largest eigenvalue of $G$ is at most the maximum degree $\Delta(G)$.
This one is fairly standard. Suppose that $\mathbf x$ is an eigenvector of the adjacency matrix $A$ with eigenvalue $\lambda$, and let $x_i$ be its largest entry. (If there are ties, choose one.) We want to make sure $x_i$ is positive, so if $\mathbf x$ has all negative entries, replace it with $-\mathbf x$ first.
From the identity $A\mathbf x = \lambda\mathbf x$, take the $i^{\text{th}}$ row: $(A\mathbf x)_i$ is the sum $\sum_{j\in N(i)} x_j$, so $$\sum_{j \in N(i)} x_j = \lambda x_i.$$ The left-hand side has at most $\Delta(G)$ terms, each of which is at most $x_i$, so it is at most $\Delta(G)x_i$. We conclude that $\Delta(G)x_i \ge \lambda x_i$, and dividing by $x_i$, we get $\Delta(G) \ge \lambda$.
Proof that the largest eigenvalue of $G$ is at least the average degree $\frac{2|E(G)|}{|V(G)|}$.
Here, we characterize the largest eigenvalue of $A$ by the Rayleigh quotient: it is the maximum value of the quotient $\frac{\mathbf x^{\mathsf T}\!A \mathbf x}{\mathbf x^{\mathsf T}\mathbf x}$, where $\mathbf x$ can be any nonzero vector in $\mathbb R^{|V(G)|}$.
In particular, the largest eigenvalue is at least the value of the quotient when $\mathbf x$ is the all-ones vector.
When we set $\mathbf x = \mathbf 1$, the numerator $\mathbf 1^{\mathsf T}\!A \mathbf 1$ becomes the sum of all the entries of $A$; equivalently, the number of entries that are $1$. There are two $1$'s for every edge of $A$, so the numerator is $2|E(G)|$. The denominator $\mathbf 1^{\mathsf T}\mathbf 1$ is a sum $1 + 1 + \dots + 1$, adding up to $|V(G)|$: the number of entries of the vector.
So when $\mathbf x = \mathbf 1$, the Rayleigh quotient is $\frac{2|E(G)|}{|V(G)|}$: the average degree of $G$. The maximum value of the Rayleigh quotient (which gives us the largest eigenvalue) must be at least as large.