A referee on a paper of mine showed me the following recurrence for polynomials $P_{n,k}\in\mathbb Q[q,q^{-1}]$ for $n\geq 0$ and $0\leq k\leq n/2$. \begin{align} P_{0,0}&=1\\ \text{for $n\geq 1$}\quad P_{n,k}&=\begin{cases} qP_{n-1,k-1}+q^{-1}P_{n-1,k}&2k+2\leq n\cr qP_{n-1,k-1}+\frac 12(1+q^{-1})P_{n-1,k}&2k+1=n\cr 2P_{n-1,k-1}&2k=n \end{cases} \end{align} where we interpret $P_{n,k}$ as zero if $k<0$. Note that we have $P_{n,k}(1)=\binom nk$, so these are some sort of deformed binomial coefficients.

Do these polynomials have a name? Has anyone come across them before?

My context for the polynomials is the following. I calculated the $s$-filtration on the Lee homology of the torus links $T_{n,n}$ and formed their Poincar\'e polynomials. The result is: \begin{align}\sum_{i,j}d_{T_{1,1}}(i,j)\cdot t^iq^j&=q+q^{-1}\cr\sum_{i,j}d_{T_{2,2}}(i,j)\cdot t^iq^j&=[1+q^2]+(tq^3)^2[q^{-2}+1]\cr\sum_{i,j}d_{T_{3,3}}(i,j)\cdot t^iq^j&=[q^3+q^5]+(tq^3)^4[q^{-3}+3q^{-1}+2q]\cr\sum_{i,j}d_{T_{4,4}}(i,j)\cdot t^iq^j&=[q^8+q^{10}]+(tq^3)^6[q^{-2}+4+3q^2]+(tq^3)^8[q^{-4}+3q^{-2}+2]\cr\sum_{i,j}d_{T_{5,5}}(i,j)\cdot t^iq^j&=[q^{15}+q^{17}]+(tq^3)^8[q+5q^3+4q^5]+(tq^3)^{12}[q^{-5}+5q^{-3}+9q^{-1}+5q]\cr\sum_{i,j}d_{T_{6.6}}(i,j)\cdot t^iq^j&=[q^{24}+q^{26}]+(tq^3)^{10}[q^6+6q^8+5q^{10}]\cr&+(tq^3)^{16}[q^{-4}+6q^{-2}+14+9q^2]+(tq^3)^{18}[q^{-6}+5q^{-4}+9q^{-2}+5]\end{align} The above looked like a mess to me, and I figured there was no pattern, but the referee on my paper pointed out the that: $$\sum_{i,j}d_{T_{n,n}}(i,j)\cdot t^iq^j=\sum_{k=0}^n(tq^3)^{2k(n-k)}q^{(n-2k)^2}P_{n,\min(k,n-k)}(q^2)$$ where $P_{n,k}$ is defined by the recurrence above!

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    $\begingroup$ uh what a coincidence, I've seen that very sequence in a paper I've recently refereed. $\endgroup$ – Pietro Majer Dec 20 '11 at 17:51
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    $\begingroup$ just kidding :) $\endgroup$ – Pietro Majer Dec 20 '11 at 17:52
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    $\begingroup$ A paper by Anonymous? $\endgroup$ – Igor Rivin Dec 20 '11 at 17:55
  • $\begingroup$ There is an inconsistency between the value at 1 (claimed to be a binomial coefficient) and the last formula, as illustrated in the examples, which seems to imply that the value at 1 is twice a binomial coefficient. Maybe there is a factor $1-q^2$ missing in the last formula ? $\endgroup$ – F. C. Dec 20 '11 at 20:50
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    $\begingroup$ In case anyone cares: arxiv.org/pdf/1107.4702.pdf $\endgroup$ – Benjamin Dickman Nov 8 '12 at 6:01

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