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S Feb 27, 2018 at 0:35 history suggested J.J. Green CC BY-SA 3.0
improved (I hope) displayed mathematics
Feb 26, 2018 at 20:09 review Suggested edits
S Feb 27, 2018 at 0:35
Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 26, 2010 at 6:53 comment added Pietro Majer Just a remark to recall that your p<sub>n</sub>(j,k) is the coefficient of x<sup>j</sup>&nbsp;y<sup>k</sup> in the expansion of (x + 1/x + 1 + y + 1/y )<sup>n</sup>.
May 26, 2010 at 1:03 vote accept Yakov Shlapentokh-Rothman
May 26, 2010 at 1:02 history edited Yakov Shlapentokh-Rothman CC BY-SA 2.5
added 335 characters in body
May 26, 2010 at 0:56 answer added George Lowther timeline score: 8
May 26, 2010 at 0:49 history edited Yakov Shlapentokh-Rothman CC BY-SA 2.5
added 147 characters in body
May 26, 2010 at 0:30 comment added Yakov Shlapentokh-Rothman That is a very good point! I will try to modify the question.
May 25, 2010 at 23:25 comment added George Lowther What you are suggesting implies that the probability of being at (3,4) is the same as being at (5,0) for all large n. That seems unlikely, and would guess that $C_n=5$ for n large.
May 25, 2010 at 22:41 answer added Tom LaGatta timeline score: 8
May 25, 2010 at 22:25 answer added Robby McKilliam timeline score: 7
May 25, 2010 at 18:06 history asked Yakov Shlapentokh-Rothman CC BY-SA 2.5