I noticed something interesting studying this Sturm-Liouville Problem:


$$ \frac{d}{dx}\left(\sqrt{(1-x^{2})}\frac{df}{dx} \right)+\frac{\left(n \alpha x+\alpha^2 x^{2} + \lambda\right)f}{\sqrt{(1-x^{2})}}= 0,$$
where $\alpha \in \mathbb{R}$ with periodic boundary conditions on $[-1,1]$. 
(Lambda is the eigenvalue)
From basic St.-Liouville theory we know that the spectrum will be discrete cause the Sturm-Liouville operator is self-adjoint and the problem is regular. 


Furthermore, we will have that $-\infty <\lambda\_0 < \lambda\_1 \le \lambda\_2 < \lambda\_3 \le .. \ .$


Now I noticed by extensive numerical calculations that for odd $n$ we will have that $\infty < \lambda\_0 < \lambda\_1 < \lambda\_2 <...< \lambda\_n,$ but after $\lambda\_n$ all eigenvalues occur pairwise $\lambda\_{n+1}=\lambda\_{n+2}$. Is there a way to explicitely show this?


Edit: Due to two good questions, I want to add some information to the question:
The eigenfunctions seem to a double eigenvalue seem to be even/odd functions respectively.
I assume $\alpha >0$, but maybe it is worth studying $\alpha=0$ first (is similar to Legendre's ODE).


Edit2: I think I could reduce my problem to a Linear Algebra problem. Due to the very different kind of question this observation includes I asked a new question about this: <https://mathoverflow.net/questions/177814/explain-eigenvalue-structure-of-these-sequences-of-matrices> .



Spectrum of this ODE
Let's rewrite your equation as a Schrödinger equation, as follows: Introduce the new variable $t\in (-\pi/2, \pi/2)$ by $x=\sin t$. Then if $f$ solves your boundary value problem (with periodic boundary conditions), then $y(t)=f(x(t))=f(\sin t)$ satisfies
$$
-y'' +V(t)y = \lambda y , \quad\quad y(-\pi/2)=y(\pi/2),\quad y'(-\pi/2)=y'(\pi/2) ,
$$
with the potential
$$
V(t) = -n\alpha\sin t -\alpha^2\sin^2 t = \frac{\alpha^2}{2}\cos 2t - n\alpha\sin t-\frac{\alpha^2}{2}.
$$
(We can do such a transformation for any SL equation; see for example [here](http://www.encyclopediaofmath.org/index.php/Liouville_normal_form).)


The additional change of variable $t=2s-\pi/2$ (so $0<s<\pi/2$) identifies $V$ as a *Whittaker-Hill potential* (up to an irrelevant shift):
$$
V(s) = -\frac{\alpha^2}{2}\cos 4s -n\alpha\cos 2s - \frac{\alpha^2}{2}
$$


The difference of two consecutive periodic or anti-periodic eigenvalues is a gap of the associated problem on $\mathbb R$ with $\pi/2$-periodic potential. The Whittakker-Hill potentials are indeed known to have the property you are interested in for suitable parameters, see [here](http://arxiv.org/abs/0906.1697).


PS: This last part came as quite a surprise to me; you can see from the edit history that I first conjectured the opposite. It's perhaps interesting in this context to mention that the Mathieu potential $W = g\cos 4s$ (corresponding to $n=0$ above) has all its gaps open.


