The sequence of polynomials
$$P\_n=\sum\_{k=0}^{\lfloor(2n-1)/3\rfloor}
\frac{(2n-2k-1)!(2n-2k-2)!}{k!(n-k)!(n-k-1)!(2n-3k-1)!}x^k$$
satisfies apparently the identities
$$0=\sum\_{j=0}^nP\_{n-j}(P\_j-(-x)^j)$$
for all $n\geq 2$. (The previous condition
$n\geq 1$ was incorrect, as pointed out in the answer of Ira Gessel.) (This has probably a WZ proof since it involves hypergeometric stuff, see the answers below.)


It is easy to see that $P\_1,P\_2,\ldots$ evaluates to the sequence $1,1,2,5,14,\ldots$ of Catalan numbers at $x=0$. Leading coefficients are also closely related to Catalan numbers and central binomial coefficients.


All roots of the polynomials $P\_n$ are apparently in the real interval
$(-\infty,-16/27)$ and the
largest root of $P\_n$ converges rather quickly (for $n\rightarrow \infty$) to the rational number $-16/27$.


*Has this sequence of polynomials appeared elsewhere? Other interesting properties?*



A sequence of polynomials related to Catalan numbers
I find a slightly different initial condition for the recurrence:
$$0=\sum\_{j=0}^nP\_{n-j}(P\_j-(-x)^j)$$
for $n\ne 1$; for $n=1$ the sum is $-1$. It's easy to derive a formula for the generating function
$\sum\_{n=0}^\infty P\_n(x) z^n$ from this recurrence. We find that, as noted by Tewodros,
$$\sum\_{n=0}^\infty P\_n(x) z^n = \frac{1-\sqrt{4z(1+xz)^2}}{2(1+xz)}.$$


In terms of the Catalan number generating function $c(u) = (1-\sqrt{1-4u})/(2u)$, this is
$z(1+xz) c\bigl(z(1+xz)\bigr)$.
From this we can easily derive the OP's formula for $P\_n(x)$ in the form
$$P\_n(x) = \sum\_i C\_i \binom{2i+1}{n-i-1}x^{n-i-1},$$
where $C\_i$ is the Catalan number $\frac{1}{i+1}\binom{2i}{i}$.


