Let $[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$ with $[0]_q!:=1$ and the Gaussian polynomials $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\,\cdot\,[n-k]_q!}$. Adopt the convention that $\binom{n}k_q=0$ whenever $k<0$ or $k>n$.

One of many facts: the Gaussian polynomials are log-concave, in the sense that the coefficients of $$\mathcal{L}\binom{n}k_q:=\binom{n}k_q^2-\binom{n}{k-1}_q\binom{n}{k+1}_q$$ are non-negative. At present, this is the source of motivation. First, observe if $q=1$ then $\binom{n}k_q=\binom{n}k$ and $$\sum_{k=0}^n\mathcal{L}\binom{n}k=C_{n+1}$$ where $C_n=\frac1{n+1}\binom{2n}n$ are the Catalan numbers. I would like to ask:

QUESTION. We take $q=-1$. Are these identities valid? $$\sum_{k=0}^{2n}\mathcal{L}\binom{2n}k_{-1}=C_n \qquad \text{and} \qquad \sum_{k=0}^{2n+1}\mathcal{L}\binom{2n+1}k_{-1}=2C_n.$$