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Jul 16, 2018 at 21:17 comment added Marty I'm not sure -- I'd guess that people like Opdam and Dyson are aware of the fact that the modern MacDonald-conjecture style formulas specialize to Selberg's integrals, which are equivalent in a special case to Dixon's sum. Maybe ask Opdam directly?
Jul 16, 2018 at 16:01 comment added Noam D. Elkies Combining your and Dan Piponi's answers gives a proof of the formula for $\sum_{j=0}^{2n} (-1)^j {2n \choose j}^3$; is this a known proof?
Jul 14, 2018 at 17:56 history edited Marty CC BY-SA 4.0
Corrected typo.
Jul 14, 2018 at 17:36 history edited Marty CC BY-SA 4.0
Added some more history.
Jul 14, 2018 at 7:41 history answered Marty CC BY-SA 4.0