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Sep 28, 2012 at 17:25 answer added Serkan timeline score: 1
Jul 20, 2012 at 8:57 comment added Johann Cigler 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 ${(- q;q)_ {2n - 1}} {G_{2n}}(q) = [2n] {T_{2n - 1}}(q).$
Jul 19, 2012 at 20:25 history edited John Wiltshire-Gordon
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Jul 15, 2012 at 6:45 comment added Johann Cigler The corresponding Seidel identity is $$\sum{(-1)^k}q^{\binom{2k}{2}} {n\brack{2k}}{{(-q^{n-2k+1};q)_{2k}}}/ {{(-q^{2n-2k};q)_{2k}}}{G_{2n-2k}}(q) =[n=1].$$
Jul 14, 2012 at 22:41 comment added Zack Wolske 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?
Jul 14, 2012 at 14:33 history asked Johann Cigler CC BY-SA 3.0