Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

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

share|cite|improve this question
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? –  Zack Wolske Jul 14 '12 at 22:41
The corresponding Seidel identity is $$\sum{(-1)^k}q^{\binom{2k}{2}} {n\brack{2k}}{{(-q^{n-2k+1};q)_{2k}}}/ {{(-q^{2n-2k};q)_{2k}}}{G_{2n-2k}}(q) =[n=1].$$ –  Johann Cigler Jul 15 '12 at 6:45
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 ${(- q;q)_ {2n - 1}} {G_{2n}}(q) = [2n] {T_{2n - 1}}(q).$ –  Johann Cigler Jul 20 '12 at 8:57

1 Answer 1

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:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.