# Gaussian at $q=\pm1$, log-concave polynomials, Catalan numbers

Let $$[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$$ with $$_q!:=1$$ 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 my motivation. First, observe that 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? $$\sum_{k=0}^{2n}\mathcal{L}\binom{2n}k_{-1}=C_n \qquad \text{and} \qquad \sum_{k=0}^{2n+1}\mathcal{L}\binom{2n+1}k_{-1}=2C_n.$$

• ${m\choose j}_{-1}$ is either a certain binomial coefficient or 0. Thus your questions reduce to binomial coefficient identities, which should be routine to prove. Aug 11, 2022 at 14:31
• This is very interesting. I have tried to compute $H(n,m,q)= \sum_{k=0}^{mn}\mathcal{L}\binom{mn}{k}_{q}$ if $q=z_m$ is a primitive m-th root of unity for some small $m,n$. It seems that $H(n,m,z_m)=\binom{2n}{n}$ for $m>2.$ Aug 11, 2022 at 14:43
• The two identities asked by OP and another one mentioned by J. Cigler are indeed easy to prove. On my request, my graduate student Wei Xia has proved them easily. Aug 14, 2022 at 22:25

$$\frac{1-q^{2j}}{1-q^{2i}}\rightarrow \frac{j}{i}, \frac{1-q^{2j}}{1-q^{2i+1}}\rightarrow 0, \frac{1-q^{2j+1}}{1-q^{2i+1}}\rightarrow 0$$ as $$q\rightarrow -1$$. So, if one pairs up such factors (carefully) in the Gaussian polynomials then it becomes clearer that
$$\binom{2n}{2k+1}_{-1}=0, \binom{2n}{2k}_{-1}=\binom{n}k, \binom{2n+1}{2k+1}_{-1}=\binom{2n+1}{2k}_{-1}=\binom{n}k$$. Consequently, we have \begin{align} \sum_{k=0}^{2n}\mathcal{L}\binom{2n}k_{-1} &=\sum_{j=0}^n\binom{n}j^2-\sum_{j=0}^n\binom{n}{j-1}\binom{n}{j+1} \\ &=\binom{2n}n-\binom{2n}{n-1}=C_n, \\ \sum_{k=0}^{2n+1}\mathcal{L}\binom{2n+1}k_{-1} &=\sum_{j=0}^n\mathcal{L}\binom{2n+1}{2j+1}_{-1} +\sum_{j=0}^n\mathcal{L}\binom{2n+1}{2j}_{-1} \\ &=2\sum_{j=0}^n\binom{n}j^2-2\sum_{j=0}^n\binom{n}{j-1}\binom{n}{j+1} \\ &=2\binom{2n}n-2\binom{2n}{n-1}=2C_n. \end{align}