I had been fiddling around with a particular expression over the summer. Since I haven't had much luck with it, I thought it was about time I took an opinion over the question.
So here's where it all began. Consider the hexagonal lattice (also called the equilateral triangular lattice). Fix some point $P$ on the lattice to be the origin. Let $W_{n}$ be the number of walks of length $n$ which start and end at $P$. I was considering the simplest possible model, so there are no constraints (meaning no self-avoidance condition or anything of that sort). In trying to enumerate these walks, I came up with this 'cute' little identity
$$\displaystyle\sum_{j=0}^{n}\binom{n}{j}^{3} = \sum_{j=0}^{n}\binom{n}{j}2^{j}\times W_{n-j}$$ We have the initial constraints being $W_{0}=1$, $W_{1}=0$.
Next I tried to tweak out a q-deformation of this particular identity. Combinatory-wise I haven't got anything particularly exciting. But here is the q-version I am considering. The binomial coefficients are replaced with the q-binomial coefficients. Fix an integer $k$. Define $$2_{k}^{j} = \prod_{i=1}^{j} (1+q^{k+i})$$ with the norm that $2^{0}=1$. Now this allows one to obtain polynomials $W_{n,k}(q)$ recursively from the identity above with the initial conditions $W_{0}=1$, $W_{1}=0$. They don't necessarily have positive coefficients but as $n$ gets large, they do. Though one does obtain interesting polynomials for $k=-1,0$, here's the one I'll mention. As $k,n$ become large, one notices (formally) $$W_{n,k}(q)\rightarrow\prod_{i=1}^{\infty} \frac{1}{(1-q^i)^2}$$
Is there a good reason? I have a feeling my setup is slightly artificial, but I'd love to know what's happening!
Edit dated 6th Jan'2011:
$$W_{n}=\displaystyle\sum_{i=0}^{n} (-2)^{n-i}\binom{n}{i}(\displaystyle\sum_{j=0}^{i} \binom{i}{j}^3)$$
