Consider an $n \times 1 - $rectangle where the $n$ squares are numbered $1$ to $n$. Cover this rectangle with white squares $a$, black squares $b$ and dominoes $dd$. To each covering of the rectangle associate the following weight: Each white square has weight 1, each black square at position $i$ has weight $q^ir$ and each domino at position $\left\{ {i,i + 1} \right\}$ has weight $q^i.$ The weight of a covering is the product of its components and the weight of a set of coverings is the sum of their weights.

Then it is easy to verify that the weight $u(n,k,r)$ of all coverings with precisely $k$ dominoes is the product $u(n,k,r)=u(n,k,0)v(n,k,r)$ with $u(n,k,0)= {q^{k^2}}{n-k\brack k} $ and $v(n,k,r) = (1 + {q^{k + 1}}r)(1 + {q^{k + 2}}r) \cdots (1 + {q^{n - k}}r).$

My question is: Does this product representation exist by lucky chance or is there a simple combinatorial reason for it?

  • $\begingroup$ +1 interesting, but what do you mean by ${n-k\brack k}$? $\endgroup$
    – draks ...
    May 7, 2012 at 10:56
  • 1
    $\begingroup$ @draks: Let $[n]=(1-q^n)/(1-q).$ Then ${n\brack k}=[1][2]\dots[n]/([1]\dots [k]\cdot[1]\dots[n-k]).$ $\endgroup$ May 7, 2012 at 11:06
  • 2
    $\begingroup$ @draks: It is the q-analog of the binomial coefficient. $\endgroup$
    – anon
    May 7, 2012 at 12:00


You must log in to answer this question.

Browse other questions tagged .