Denote by $ {n\brack {k}}$ a $q-$binomial coefficient.

Let ${D_{n,k}}(t,q) = \sum\limits_{j = 0}^{n - k} {{q^{{j^2} + kj}}}{n\brack {j}}{n\brack {k+j}}t^j $

${R_n}(x,t,q) = \sum\limits_{k = 0}^n {{{( - 1)}^k}{q^{\binom{k}{2}}{n\brack {k}}c(n,k,t)x^ {n-2k}}}$


$c(n,k,t)= \sum\limits_{j = 0}^k{k\brack {j}}q^{(n+1-k)j}{{n+j-k-1}\brack{j}}/{{n-1}\brack{j}} t^j$

for $k<n$ and

$c(n,n,t)=1+q^n t^n $.

Computations suggest that $$ \sum\limits_{k = 0}^n {{D_{n,k}}(t,q){R_k}(x,t,q)} = {x^n}.$$
This implies that the linear functional $L$ defined by $L(R_n)=0$ for $n>0$ has the moments

$L(x^n)= \sum\limits_{j = 0}^{n } {{q^{j^2} }}{n\brack {j}}^2t^j $

which are $q-$Narayana polynomials of type B.

Note that $D_{n,k}(1,1)=\binom{2n}{n+k}$ and $R_n(x,1,1)=L_{2n}(\sqrt{x})$ where $L_n(x)$ is a Lucas polynomial. There are proofs for $q=1$ and for $t=1.$ But these proofs use properties which are lacking in the general case.

I tried to prove this by comparing coefficients or by matrix inversion, but did only succeed in special cases. Are there $q-$ hypergeometric identities for double sums or other methods which can be applied?

Edit In the mean-time I could prove this identity by reducing it to a corresponding identity for $q-$Narayana polynomials of type A.

But I think there must be also a direct computational proof by comparing coefficients. Could anyone provide such a proof?

  • $\begingroup$ Is this related to diagonal harmonics and qt-Catalan stuff? $\endgroup$ – Per Alexandersson Nov 25 '16 at 14:21
  • $\begingroup$ As far as I know: not $\endgroup$ – Johann Cigler Nov 25 '16 at 15:21

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