= Solution
A nonzero <polynomial> solution of <degree> $n$ has $a_n\neq0$ but $a_{n+2}=0$. The <recurrence> therefore forces $2n+1-\epsilon=0$. Since $\epsilon=2E/(\hbar\omega)$,
$$
\boxed{E_n=\left(n+\frac12\right)\hbar\omega,\qquad n=0,1,2,\ldots.}
$$
Conversely, for this value of $\epsilon$, choose only the <parity> <sequence> containing <degree> $n$ and set the other initial <coefficient> to zero. The <recurrence> terminates at $n$, producing a nonzero <polynomial> times the decaying <Gaussian>. Thus every displayed <energy> occurs under the supplied normalizability criterion.
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