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Jul 7, 2010 at 12:25 comment added Pietro Majer why don't you? I posted an answer to stimulate other contributions... ;-)
Jul 6, 2010 at 23:16 comment added Wadim Zudilin At the end I find your identity surprising enough. +1. And 2 answers already, so I don't need to think of it. :-)
Jul 6, 2010 at 23:08 answer added David E Speyer timeline score: 5
Jul 6, 2010 at 22:33 answer added Pietro Majer timeline score: 3
Jul 4, 2010 at 11:25 comment added Brad Rodgers Ah, maybe that was Wadim's confusion as well. Probably the clarifying remark under the sum actually added confusion in retrospect. I made one more edit changing 'ordered set' to 'sequence.'
Jul 4, 2010 at 11:22 history edited Brad Rodgers CC BY-SA 2.5
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Jul 4, 2010 at 9:22 comment added darij grinberg I'm just realizing how messed up combinatorics must be when the word "ordered" may mean two completely different things... (I first thought it meant $p_0\geq p_1\geq ...$)
Jul 4, 2010 at 0:55 comment added Brad Rodgers Or more concretely: for r=2, the situation is to sum over the following sets for $(p_0,p_1,...)$: (2,0,0,...), (0,2,0,0,...), (0,0,2,0,...),... ;(1,1,0,0,...), (1,0,1,0,0...), ... ;(0,1,1,0,0,...), (0,1,0,1,0,...),...; ... Using partial fractions, it is not difficult to see that sum over all these entries is 1/2.
Jul 4, 2010 at 0:26 history edited Brad Rodgers CC BY-SA 2.5
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Jul 4, 2010 at 0:11 comment added Brad Rodgers I've written it slightly differently in an edit at the bottom; let me know if this is more transparent.
Jul 4, 2010 at 0:08 history edited Brad Rodgers CC BY-SA 2.5
added 207 characters in body
Jul 3, 2010 at 23:28 comment added Wadim Zudilin It's probbably to early for me, but I can't follow the pattern of your summand. I would guess that writing it clearly is a half of solution...
Jul 3, 2010 at 22:51 history asked Brad Rodgers CC BY-SA 2.5