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May 31, 2020 at 21:13 vote accept Gerhard Paseman
May 31, 2020 at 21:12 answer added Gerhard Paseman timeline score: 1
May 31, 2020 at 15:00 answer added Iosif Pinelis timeline score: 7
May 31, 2020 at 8:07 history edited Gerhard Paseman CC BY-SA 4.0
Restated to fit my styling
May 31, 2020 at 7:51 history edited Alex B. CC BY-SA 4.0
Fixed grammar in the title
May 31, 2020 at 2:43 comment added Gerhard Paseman And the rough calculations suggest something greater than $e$, while the numerical computations suggest something less than $e$. If I can tweak the argument to get a related quantity less than 2, I am within spitting distance of combinatorial proof of. a theorem of Sylvester. Gerhard "More Combinatorial Than Sylvester's Proof" Paseman, 2020.05.30.
May 31, 2020 at 2:38 comment added Gerhard Paseman I often do strange things @Will. I guess I am just looking for confirmation. Gerhard "Anything For These Uncertain Times" Paseman, 2020.05.30.
May 31, 2020 at 2:35 answer added Gerhard Paseman timeline score: 0
May 31, 2020 at 2:35 comment added Will Sawin Your rough back-of-the-envelope asymptotic calculations suggest the limit is $e$, right? I think these calculations, unless you did them in a very strange way, are already a proof.
May 31, 2020 at 2:31 comment added Will Sawin Don't you just want to plug an effective form of the prime number theorem into Stirling's formula here?
May 31, 2020 at 2:16 comment added Gerhard Paseman I think it is upper bounded by 4 using a rough approximation to the prime counting function. I would appreciate verification. Gerhard "Worth An Acknowledgement To Me" Paseman, 2020.05.30.
May 31, 2020 at 2:06 comment added zeraoulia rafik nice question ,upvote
May 31, 2020 at 1:58 history asked Gerhard Paseman CC BY-SA 4.0