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 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? $$\sum_{k=0}^{2n}\mathcal{L}\binom{2n}k_{-1}=C_n \qquad \text{and} \qquad \sum_{k=0}^{2n+1}\mathcal{L}\binom{2n+1}k_{-1}=2C_n.$$

  • 1
    $\begingroup$ ${m\choose j}_{-1}$ is either a certain binomial coefficient or 0. Thus your questions reduce to binomial coefficient identities, which should be routine to prove. $\endgroup$ Aug 11, 2022 at 14:31
  • $\begingroup$ 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 $m>2.$ $\endgroup$ Aug 11, 2022 at 14:43
  • $\begingroup$ 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. $\endgroup$ Aug 14, 2022 at 22:25

1 Answer 1


This is just an expanded note based on Stanley's comment in the box above.

The calculation is based on the following easily verifiable facts:

$\frac{1-q^{2j}}{1-q^{2i}}\rightarrow \frac{j}{i}, \frac{1-q^{2j}}{1-q^{2i+1}}\rightarrow 0, \frac{1-q^{2j+1}}{1-q^{2i+1}}\rightarrow 0$ 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}

One may proceed in a similar manner to extract the claim made by Johann Cigler.


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