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Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Following my earlier post on MO, one fine colleague asked me if there is a $q$-analogue of the identity formed by the so-called Shapiro's Catalan triangles $$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^2=C_{2n-1}$$ Unfortunately, it did not map out for me. The closest I have come up with is a $q$-analogue of $\sum_{k=1}^nk\left[\frac{k}n\binom{2n}{n-k}\right]^2=\binom{2n-1}{n-1}\binom{2n-2}{n-1}$ for which I like to ask:

QUESTION. Is there a combinatorial or conceptual proof of this identity? $$\sum_{k=1}^n q^{(k-1)^2}\frac{[2k]}{1+q^n} \left[\frac{[k]}{[n]} \binom{2n}{n-k}_q\right]^2 = \binom{2n-1}{n-1}_q \binom{2n-2}{n-1}_q,$$ where $[k]=1+q+q^2+\cdots+q^{k-1}$ and $\binom{n}k_q=\frac{[n]}{[k]\,[n-k]}$.

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  • $\begingroup$ Should the terms in the first sum be squared? $\endgroup$
    – Sam Hopkins
    Commented Feb 10, 2021 at 23:44
  • $\begingroup$ In your previous question, you write the same equality but with the terms in the sum squared. This seems strange to me. $\endgroup$
    – Sam Hopkins
    Commented Feb 11, 2021 at 14:00
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    $\begingroup$ @SamHopkins: I now understand what you meant. Yes, you are right there. Corrected. $\endgroup$
    – T. Amdeberhan
    Commented Feb 11, 2021 at 14:53
  • $\begingroup$ Not sure if this is of any help towards a combinatorial proof, but $1+q^n=[2n]/[n]$, so the identity may be rewritten as $$\sum_{k=1}^n q^{(k-1)^2}[2k] \left[\frac{[k]}{[n]} \binom{2n}{n-k}_q\right]^2 = \binom{2n}{n}_q \binom{2n-2}{n-1}_q.$$ $\endgroup$
    – Alexander Burstein
    Commented Feb 16, 2021 at 7:08
  • $\begingroup$ @AlexanderBurstein: that looks fine and perhaps would help. $\endgroup$
    – T. Amdeberhan
    Commented Feb 16, 2021 at 14:56

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