= Solution
For an <integer> $n\geq0$, choose the <series> whose <integer parity> agrees with $n$ and set the other initial coefficient to zero. At $m=n$, the numerator in the recurrence vanishes, so $a_{n+2}=0$ and every later coefficient in that <integer parity> vanishes. With a nonzero coefficient of $x^n$, this gives a <polynomial> of degree exactly $n$. Normalize it by $P_n(1)=1$.
For $n=0$, the even branch is constant; for $n=1$, the odd branch is proportional to $x$. For $n=2$, the recurrence gives $a_2=-3a_0$, so normalization fixes $a_0=-1/2$. Hence
$$
\boxed{P_0(x)=1,\qquad P_1(x)=x,\qquad P_2(x)=\frac12(3x^2-1).}
$$
The opposite-parity branch generally does not terminate; <polynomial> termination does not mean that both independent solutions become <polynomials>.
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