most recent 30 from http://mathoverflow.net 2013-05-25T08:47:40Z http://mathoverflow.net/feeds/question/102229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102229/are-the-following-q-genocchi-numbers-known Johann Cigler 2012-07-14T14:33:00Z 2013-03-30T05:22:00Z

<p>The sequence of Genocchi numbers<br> ${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$ </p> <p>can be defined by the generating function $z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{{{z^{2n}}}}{{(2n)!}}} .$</p> <p>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.</p> <p>Let $(a;q)_n=(1-a)(1-qa) \cdots (1-q^{n-1}a)$, $[n]=1+q+\cdots+q^{n-1}$ and $[n]!=[1][2] \cdots[n].$</p> <p>The q-analog can defined by the generating function</p> <p>$\sum\limits_{n \ge 1} {\frac{{{{( - 1)}^{n - 1}}{G_{2n}}(q){{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}} = $</p> <p>$\sum\limits_{n \ge 1} {\frac{{{{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}} $ divided by</p> <p>$\sum\limits_{n \ge 0} {\frac{{{{( - q;q)}_{2n}}}}{{[2n + 1]!}}} {z^{2n}}.$</p>

http://mathoverflow.net/questions/102229/are-the-following-q-genocchi-numbers-known/108365#108365 Serkan 2012-09-28T17:25:58Z 2012-09-28T17:25:58Z

<p>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: <a href="http://www.hindawi.com/journals/jfsa/2012/214961/" rel="nofollow">http://www.hindawi.com/journals/jfsa/2012/214961/</a></p>