Timeline for Are the q-Catalan numbers q-holonomic?
Current License: CC BY-SA 2.5
5 events
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| May 10, 2010 at 13:27 | comment | added | Roland Bacher | Indeed. It can be modified in order to cope with the correct definition. The resulting proof is somewhat similar to Zudilin's solution except that the final step arises from the fact the $\prod_{n=1}^\infty (1-1/x^n)^{-1}$ is transcendental. | |
| May 7, 2010 at 6:21 | vote | accept | Johann Cigler | ||
| May 7, 2010 at 6:21 | |||||
| May 6, 2010 at 17:23 | comment | added | Johann Cigler | I think you have misunderstood the definition of $q$-holonomic. $F(z)$ is $q$-holonomic if there exist polynomials such that $p_0 (z)F(z) + p_1 (z)D_q F(z) + \cdots p_r (z)D_q^r F(z) = 0$ where $D_q $ denotes the q-differentiation operator defined by $D_q f(z) = \frac{{f(z) - f(qz)}} {{z - qz}}.$ | |
| May 6, 2010 at 16:30 | history | edited | Roland Bacher | CC BY-SA 2.5 |
added 53 characters in body; deleted 1 characters in body
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| May 6, 2010 at 16:21 | history | answered | Roland Bacher | CC BY-SA 2.5 |