TITLE: Exotic spectrum of Laplace operator
QUESTION [7 upvotes]: 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 {$\lambda_n : {n\geq n_0}$} 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]

REPLY [5 votes]: For a generic metric on an $m$-dimensional the manifold the eigenvalues of the Laplacian are all simple.  Fix such a metric and denote the coresponding eigenvalues by
$$ \lambda_1, \lambda_2,\cdots $$
Using Weyl's asymptotic expansion we conclude that
$$\lambda_n\sim \text{const}\cdot n^{m/2}. $$
Thus for any polynomial $P$ of degree $d>m/2$ we have
$$ \lim_{n\to\infty} \lambda_n/P(n) = 0. $$