re = /^(?!\d*(\d)\1{2,})(\d{1,5})/m
str = '1234567890
11234567890
11134567890
12345666890
81234567890
811234567890
811134567890
12345666890'
# Print the match result
str.scan(re) do |match|
puts match.to_s
end
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Ruby, please visit: http://ruby-doc.org/core-2.2.0/Regexp.html