Given a closed Riemannian manifold and a generalized Laplace $\Delta$ operator, it is well known that $\Delta$ has discrete spectrum $(\lambda_n)_n$ (arranged in a increasing way, not counting multiplicities). By a (consequence of a) result of Colin de Verdière, given any finite strictly increasing sequence $a_1,\cdots,a_k$ of strictly positive numbers, there exists a Riemannian manifold and a Laplace operator on it such that the first k+1 eigenvalues are exactly, $0,a_1,\cdots,a_k$.
My question is about what's happening at infinity. More precisely, since usually the spectrum of a Laplace operator is a quadratic polynomial (in the sense that {$\lambda_n : {n\geq 0}$} is of the form {$P(n) : n\geq 0$} where $P$ is a quadratic polynomial), is there a Laplace operator (on a closed manifold) such that there is no $n_0$ such that {$(\lambda)_{n\geq n_0}$$\lambda_n : {n\geq n_0}$} is of the form {$P(n) : n\geq 0$} where $P$ is a quadratic polynomial ?
My question could be reformulated : do you know example where the explicit (exact) eigenvalues (not asymptotics of them) are (after a certain rank) not given by a formula of the form $a n^2 + bn +c$ where $a,b,c$ are constants (for example some kind of fraction P(n)/Q(n) where the degree of P is 3 and the degree of Q is one...)
[Edit : precisions]
