From: Faith Tarcha <faith@...>
Date: 2012-02-07T20:03:47+09:00
Subject: Re: Easy loop question

Yes, this is the easiest way to do it:

def f n # this function calculates next number in your sequence
  if n % 2 == 0 # n is even
    return n / 2
  else
    return n * 3+1
  end
end

num = gets.to_i # read in a number
while num < 1 # infinitely...
  num = f(num) # find the next number
  puts num # print it
  sleep 1 # wait one second
end


Does anyone know how you can make it count all the numbers in the loop
(until it reaches number 1)?

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