Let $n\ge 2r$ be positive integers.
Is there a closed form for following finite summation involving in q-binomial coefficients
$$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-s}_q\,\,\,\,\,\, ?$$
I found this while studying q-Fibonacci/ Lucas polynomials.
What is the general approach for evaluating this type of series?
Any suggestion would be appreciated.
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$\begingroup$ What is your definition of $\big[..\big]_q$ ? $\endgroup$– Konstantinos KanakoglouJan 23, 2017 at 1:10
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$\begingroup$ I use the definition $[n]_q=1+q+q^2+\cdots+q^{n-1}$ for q-integers . $\endgroup$– BumblebeeJan 23, 2017 at 1:15
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1$\begingroup$ Out of curiosity: where did you find this sum? I mean did you find it mentioned in some paper or is it a byproduct of some calculations of yours? It would be helpful to add some more context. $\endgroup$– Konstantinos KanakoglouJan 23, 2017 at 2:04
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$\begingroup$ I an also very curious about the sum - it looks like the major index statistic on some family of lattice paths with some restriction (using inclusion-exclusion)... $\endgroup$– Per AlexanderssonJan 26, 2019 at 22:33
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1 Answer
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Doron Zeilberger has written a Maple code for checking and proving ordinary binomial identities and their $q$-analogues. What you need in the present case is the package called qEKHAD.
I just tested your sum and it leads to a quadratic recurrence in $r$, so you may not expect a closed form as an answer. The lucky cases are linear recurrences in the "external" parameter (here, it is $r$). The latter situation fits the so-called Wilf-Zeilberger pair.
I also sought for other factors $q^{\mu(s)}$ instead of $q^{\binom{s+1}2}$, none of the usual suspects lead to a closed form. Sorry.
answered Jan 23, 2017 at 2:31
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$\begingroup$ Thank you for your help. I am tring to simplify this summation bit further. Numerical values suggest that above sum equal to $\sum_{s=0}^{r-1}(-1)^{s+1}q^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-s-1}_q.$ And also I fund that this equality holds for ordinary binomial coefficients (in which $q=1$). Do you think that they are actually equals for q-binomials? $\endgroup$ Jan 23, 2017 at 3:10
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2$\begingroup$ I just checked: the two sums satisfy the same quadratic recurrence (in $n$) and matching initial conditions. They are indeed equal. $\endgroup$ Jan 23, 2017 at 3:56
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$\begingroup$ I found this paper by Wilf and Zeilberger. Does the same algorithm works for q-analogies? $\endgroup$ Jan 24, 2017 at 1:32
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