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Let $[x]_{q}=\frac{1-q^x}{1-q}$ and $\binom{x}{n}_{q}$ denote a $q$-binomial coefficient.

Let $A_n(x,q)$ be the $n\times n $ matrix with entries $$q^\binom{i-j}{2}\binom{i+j+x}{i-j+1}_{q},$$ $0 \le i,j \le n-1,$ and $n+1\le m \le 2n-1.$

Computer experiments suggest that $ A_n(-m,q) v_{n,m}=0$ if $v_{n,m}$ is the vector with entries $$q^{(\lfloor{m/2}\rfloor-j) (\lfloor{m/2}\rfloor+5-3j)/2}\frac{[m]_q}{[m-j]_q}\binom{m-j}{j}_{q}$$ for even $m$ and $$q^{(\lfloor{m/2}\rfloor-j) (\lfloor{m/2}\rfloor+7-3j)/2}\frac{[m]_q}{[m-j]_q}\binom{m-j}{j}_{q}$$ for odd $m.$

Any idea how to prove this?

For $q=1$ one can use Rothe’s formula, but I could not find a $q$-analogue which applies to the general case.

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  • $\begingroup$ Warren P. Johnson proved a q-version of Rothe, does not it help you? core.ac.uk/download/pdf/82453222.pdf $\endgroup$
    – Fedor Petrov
    Commented Mar 16, 2019 at 15:51
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    $\begingroup$ Yes, if $0 \le i \le n-1,$ and $n+1\le 2L \le 2n-1.$ $\endgroup$
    – Johann Cigler
    Commented Mar 16, 2019 at 19:31
  • $\begingroup$ Dear Johann, Did Christian K. see it? $\endgroup$
    – Wadim Zudilin
    Commented Mar 17, 2019 at 21:19
  • $\begingroup$ Dear Wadim, thank you for this advice. He has shown me a proof. $\endgroup$
    – Johann Cigler
    Commented Mar 19, 2019 at 16:09

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