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The sequence of Genocchi numbers
${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$

can be defined by the generating function $z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{{{z^{2n}}}}{{(2n)!}}} .$

Many different q-analogs of these numbers have been studied. Does anyone know if the following q-analog ${({G_{2n}(q)})_{n \ge 0}}$ is known? It is intimately related with q-Chebyshev polynomials.

Let $(a;q)_n=(1-a)(1-qa) \cdots (1-q^{n-1}a)$, $[n]=1+q+\cdots+q^{n-1}$ and $[n]!=[1][2] \cdots[n].$

The q-analog can defined by the generating function

$\sum\limits_{n \ge 1} {\frac{{{{( - 1)}^{n - 1}}{G_{2n}}(q){{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}} = $

$\sum\limits_{n \ge 1} {\frac{{{{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}} $ divided by

$\sum\limits_{n \ge 0} {\frac{{{{( - q;q)}_{2n}}}}{{[2n + 1]!}}} {z^{2n}}.$

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  • $\begingroup$ Is this the same as the q-analog you get by rewriting $\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k {n \choose 2k}G_{2n-k} = 0$ to a relation involving q-binomial coefficients? $\endgroup$
    – Zack Wolske
    Commented Jul 14, 2012 at 22:41
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    $\begingroup$ The corresponding Seidel identity is $$\sum{(-1)^k}q^{\binom{2k}{2}} {n\brack{2k}}{{(-q^{n-2k+1};q)_{2k}}}/ {{(-q^{2n-2k};q)_{2k}}}{G_{2n-2k}}(q) =[n=1].$$ $\endgroup$
    – Johann Cigler
    Commented Jul 15, 2012 at 6:45
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    $\begingroup$ In the mean time I have seen that these q-Genocchi numbers are related to the usual $q-$tangent numbers ${T_{2n - 1}}(q)$ by ${(- q;q)_ {2n - 1}} {G_{2n}}(q) = [2n] {T_{2n - 1}}(q).$ $\endgroup$
    – Johann Cigler
    Commented Jul 20, 2012 at 8:57

1 Answer 1

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Different definitions of the q-Genocchi numbers and polynomials have been studied by many mathematicians for a long time, for instance: T. Kim, L. C. Jang, C. S. Ryoo, Y. Simsek, S. Araci, H. Jolany,...etc. So that, the readers can refer to the link: http://www.hindawi.com/journals/jfsa/2012/214961/

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