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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:

$$W_{n}=\displaystyle\sum_{i=0}^{n} (-2)^{n-i}\binom{n}{i}(\displaystyle\sum_{j=0}^{i} \binom{i}{j}^3)$$

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The usual way to find a reasonable $q$-version of a binomial identity is to write its detailed proof and replace the involved transformations by their $q$-counterparts. BTW, why don't you give an explicit formula for the Domb numbers ($W_n$ in your notation)? –  Wadim Zudilin Jan 6 '11 at 10:21
    
Dear Wadim, Just added the explicit formula that I obtained. Thanks for the comments. I'll try reworking the proof so as to obtain reasonable q-analogues. –  Vasu vineet Jan 6 '11 at 11:19
    
@Vasu, it seems that the only explicit formula is due to you (oeis.org/A002898). There is no need in proving it as it is equivalent to the earlier defining equation for $W_n$ by the inverse binomial transform. So, I can suggest searching the $q$-binomial transform and its inverse. –  Wadim Zudilin Jan 6 '11 at 13:57
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It seems to me that Gould's article from 1960 (projecteuclid.org/euclid.bams/1183523696) is an appropriate source. –  Wadim Zudilin Jan 6 '11 at 14:32
    
@Wadim, Thanks a lot for the pointers. I searched a bit just a little while back and actually managed to find a different (?) explicit formula in Anthony Guttmann's paper titled "Lattice green's functions in all dimensions (link: iopscience.iop.org/1751-8121/43/30/305205/pdf/…) –  Vasu vineet Jan 6 '11 at 14:47
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