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 n

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}