Construct the $n$-tuple Cartesian product of the ternary set $X_n=\{0,1,2\}\times\cdots\times\{0,1,2\}=\{0,1,2\}^n$. Define its subset $W_n$ according to the rule (here $y=(y_1,\dots,y_n)$ is made use of)
$$W_n=\{y\in X_n: y_1\leq1, y_1+y_2\leq2,\dots,y_1+\cdots+y_{n-1}\leq n-1, y_1+\cdots+y_n=n\}.$$
Introduce the one-variable polynomial
$$P_n(x)=\sum_{y\in W_n}x^{\prod_{j=1}^n\binom{2}{y_j}}.$$
QUESTION. If $C_n$ stands for the Catalan numbers, is this true?
$$\frac{dP_n}{dx}(1)=C_{n+1}.$$
Examples. $P_1(x)=x^2, P_2(x)=x^4+x, P_3(x)=x^8+3x^2, P_4(x)=x^{16}+6x^4+2x$.
Best Answer
Yes, this is true.
Write each $y_j = 1 - x_j$ with $x_j \in \{-1, 0, 1\}$, so the condition is that $\sum_{j=1}^n x_n = 0$ and no partial sum is negative. This can be viewed as an n-move king path from $(0,0)$ to $(n,0)$ that never goes below the horizontal axis. We want the sum over such $(x_1,\ldots,x_n)$ of $2^h$ where $h$ is the number of indices $j$ for which $y_j = 1$, i.e. $x_j = 0$ (the number of horizontal king steps).
If we drop the nonnegativity condition, then for each $k$ the sum of $2^h$ over paths from $(0,0)$ to $(n,k)$ is the $t^k$ cofficient in the generating function $(t^{-1} + 2 + t)^n$; since that's just $(t+1)^{2n} / t^n$, the sum is $2n \choose n+k$.
By a reflection argument familiar from the Catalan enumeration of Dyck paths, it follows that for $k=0$ restricted to nonnegative paths is the difference between the unrestricted $k=0$ and $k=-1$ sums, which is ${2n \choose n} - {2n \choose n-2} = C_{n+1}$.