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Let $k$ be a positive integer and let $$h(n,k,q)=\frac{1-(1+q^{k})q^{2k(n-1)+1}+q^{2}}{1-q^{2n-1}}h(n-1,k,q)-\frac{(1-q^{k(2n-3)})(1-q^{2k(n-1)})q^2}{(1-q^{2n-1})(1-q^{2n-3})}h(n-2,k,q)$$ with initial values $h(0,k,q)=1$ and $h(1,k,q)=1+q^2{\frac{1-q^{k-1}}{1-q}}.$

Remark: This question is related to question 212926 since $h(n,k,q)$ can also be written as $\frac{\sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{j} \binom{2n}{j}_{q^k}}{(q;q^2)_{n}}.$

The rational functions $h(n,k,q)$ converge for $q\to1$ to $h(n,k,1)=k^n$ and for $q\to-1$ to $h(n,2k,-1)=2^n$ and $h(n,2k+1,-1)=1.$

It seems that if $k=2^m$ is a power of $2$ then $h(n,2^m,q)$ is a polynomial with (non- negative) integer coefficients.

My question is: Which methods can be used to prove such facts?

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  • $\begingroup$ This kind of question can sometimes be very difficult. A general strategy is to find some kind of interpretation of the coefficients of the polynomial. The alternating-sum expression for $h(n,k,q)$ suggests that there may be an interpretation in terms of Betti numbers, although I do not have a concrete suggestion. $\endgroup$ Aug 27, 2015 at 23:35

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