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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?

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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 $\frac{1}{i+1}\binom{2i}{i}$.

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    $\begingroup$ This solves the mystery, thanks. You are of course right concerning the value of the sum. $\endgroup$ Commented Nov 4, 2022 at 21:03
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Just a remark regarding a recurrence for $P_n$ (found by the Wilf-Zeilberger methodology):

$$(n + 4)P_{n+4} + (nx - 4n + 4x - 10)P_{n+3} - 6(2n + 3)xP_{n+2} - 6(2n + 1)x^2P_{n+1} - 2(2n - 1)x^3P_n=0.$$

Another remark is this: we may write $$P_n=\sum_{i=0}^{n-1}\binom{2i+1}i\binom{2i+1}{n-1-i}\frac{x^{n-1-i}}{2i+1} \qquad \text{or}$$ $$P_n=\sum_{j=0}^{n-1}\sum_{k=0}^j\binom{2k+2}{j-k}\binom{2k}k \frac{(-1)^{n-1-j}x^{n-1-k}}{k+1}.$$ One more: a bi-variate generating function $$\sum_{n\geq0}P_n(x)t^n =\frac{1-\sqrt{1-4t(tx + 1)^2}}{2t(tx + 1)}.$$

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    $\begingroup$ Very nice recursive identity! $\endgroup$ Commented Nov 4, 2022 at 18:13
  • $\begingroup$ @RolandBacher then you may need to accept that nice answer . $\endgroup$ Commented Nov 4, 2022 at 18:16
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    $\begingroup$ Not necessarily: I believe there is something deeper behind this. Waiting what else pops out. $\endgroup$ Commented Nov 4, 2022 at 18:18

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