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Feb 21, 2018 at 2:45 review Suggested edits
Feb 21, 2018 at 4:26
Feb 21, 2018 at 2:40 comment added Richie Well changing it to (2..(1.0/0)).lazy.select{|i|!(2...i).any?{|j|i%j<1}}.first(n) will result in 61 chars.
Sep 2, 2015 at 5:55 comment added Qqwy Or even: (This makes the solution less efficient, but it does not change the upper O(n²) bound) (2..(1.0/0)).lazy.select{|i|(2..i).one?{|j|i%j<1}}.take(n).to_a. This shaves off two more characters.
Sep 2, 2015 at 5:50 comment added Qqwy You can shave one char off by changing it to (2..(1.0/0)).lazy.select{|i|!(2...i).any?{|j|i%j<1}}.take(n).to_a
S Feb 4, 2014 at 14:37 review Late answers
Feb 4, 2014 at 16:41
S Feb 4, 2014 at 14:37 review First posts
Feb 4, 2014 at 16:22
Feb 4, 2014 at 14:20 history answered Uri Agassi CC BY-SA 3.0