[xenophanes]

Thanks! I worked out a different way, too.

SPOILER

For just positive numbers, have the first number count up to the second number. Then increment the second number and reset the first to 0. Then adjusting for negatives is no big deal. Pretty simple in concept. Here's non-simple ruby code for calculating the pair for any step directly:

  def inefficient(n)
    count = 0
    while n > count
      count += 1
      n -= count
    end
    [n, count]
  end
  
  def with_negatives(n)
    # wastes some steps trying to get the negative version of 0. whatever.
    tmp = inefficient(n/4)
    if n%4 == 1
      tmp[0] *= -1
    elsif n%4 == 2
      tmp[1] *= -1
    elsif n%4 == 3
      tmp[0] *= -1
      tmp[1] *= -1
    end
    tmp
  end
  
  50.times {|n| puts "#{n}: " + with_negatives(n).inspect}

[xenophanes]

Worked out the math to get rid of the loop.

  def efficient(x)
    num = (Math.sqrt(2 * x + 0.25) - 0.5).floor
    num2 = (num**2 + num)/2
    [x-num2, num]
  end

Edit: Or:

  def efficient2(x)
    num = (Math.sqrt(2 * x + 0.25) - 0.5)
    num2 = ((num - num.floor) * num).round
    [num2, num.floor]
  end