Joao Silva wrote in post #1067849:
> a = [1 2 2 3 5 4 6 2 6 5 4]
>
> I calculate the number that is repeated over and less repeats and if
> possible to calculate something like this for example:
>
> => With 0 repeats found the number 1.3, etc.
> => With 1 repetition was found the number: 4, etc.
> => With 2 repetitions was found the number 5, etc.

Do it in steps. First count the number of times each value is seen:

a = [1, 2, 2, 3, 5, 4, 6, 2, 6, 5, 4]
counts = Hash.new(0)
a.each { |val| counts[val] += 1 }
# optional: counts = counts.sort_by { |val,count| count }
counts.each do |val,count|
  puts "With #{count-1} repeats found the number #{val}"
end

This is a starting point. Then you can look at grouping together values 
with the same count.

a = [1, 2, 2, 3, 5, 4, 6, 2, 6, 5, 4]
counts = Hash.new(0)
a.each { |val| counts[val] += 1 }
groups = {}
counts.each do |val,count|
  groups[count] ||= []
  groups[count] << val
end
groups.each do |count,vals|
  puts "With #{count-1} repeats found #{vals.sort.join(", ")}"
end

When you're clear what's happening here then you can look at the more 
compact versions using group_by.

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