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Mathematical mysteries: Hailstone sequences


Here’s a little game to play. Starting with any positive whole number $n$ form a sequence in the following way:

  • If $n$ is even, divide it by $2$ to give $n^\prime = n/2$.
  • If $n$ is odd, multiply it by $3$ and add $1$ to give $n^\prime = 3n + 1.$

Then take $n^\prime $ as the new starting number and repeat the process. For example, $n = 5$ gives the sequence

  \[ 5, 16, 8, 4, 2, 1, 4, 2, 1,... \]    

and $n = 11 $ gives the sequence

  \[ 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1,... . \]    

Sequences formed in this way are sometimes called hailstone sequences because they go up and down just like a hailstone in a cloud before crashing to Earth. However, it seems that all hailstone sequences eventually end in the endless cycle

  \[ 4, 2, 1, 4, 2, 1. \]    

The ones for $n=5$ and $n=11$ both do, though other values for may $n$ generate a very long sequence before the repeating cycle begins. For example, try starting with $n = 27.$ You can do this by hand, or use our hailstone calculator: enter any positive whole number and the hailstone sequence will be returned.

The question is whether every hailstone sequence eventually settles on the 4, 2, 1 cycle, no matter what starting value you use. Experiments certainly suggest that they all do. Computers have checked all starting values up to 5 x 260, a number that is 19 digits long, and found that the 4, 2, 1 cycle eventually appears. The trouble is that nobody has been able to prove that this is the case for all sequences. This open question is known as the Collatz conjecture after the mathematician Lothar Collatz, who first proposed it in 1937. It's amazing that such an easy recipe for forming sequences leads to a question even the best mathematicians haven't been able to answer yet.

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