In our blog we’ve already covered Kaprekar’s constant, as well as the (not-so) magic number 10. Here’s another cool little trick to sink your teeth into…

First, pick a positive integer (whole number).

Now, we are going to generate a sequence of numbers, starting with our starting number. Each term in the sequence will depend on the previous term.

If the previous term is even, we divide the previous term by 2 to get the next term.

If the previous term is odd, we multiply the previous term by 3 and add 1 to get the next term.

This can be expressed this way:

For the previous term n, the next term of the sequence will be

n ÷ 2 if n is even

3n + 1 if n is odd

Then we repeat.

So let’s do an example.__Example 1__

Say I chose my starting number to be 5…

5 is odd, so our next term will be 3 × 5 + 1 = **16**.

16 is even, so our next term will be 16 ÷ 2 = **8**.

8 is even, so our next term will be 8 ÷ 2 = **4**.

4 is even, so our next term will be 4 ÷ 2 = **2**.

2 is even, so our next term will be 2 ÷ 2 = **1**.

1 is odd, so our next term will be 3 × 1 + 1 = **4**.

But we’ve already been here before. We know what’s going to happen next..

4 is even, so our next term will be 4 ÷ 2 = **2**.

2 is even, so our next term will be 2 ÷ 2 = **1**.

1 is odd, so our next term will be 3 × 1 + 1 = **4**.

And the process will just go around in a circle forever..

So our final sequence we have generated is 5, 16, 8, 4, 2, 1, 4, 2, 1, … (and it will continue 4, 2, 1 forever!)

What we have discovered here is that if the sequence reaches 1, the sequence starts repeating on a 4, 2, 1 loop.

Let’s do another example.__Example 2__

Say I chose the starting number to be 6…

6 is even, so our next term will be 6 ÷ 2 = **3**.

3 is odd, so our next term will be 3 × 3 + 1 = **10**.

10 is even, so our next term is 10 ÷ 2 = **5**.

We don’t actually need to do any more calculations, because from Example 1, we know exactly what happens after 5…

So our sequence generated in Example 2 will be 6, 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, … (and it will continue 4, 2, 1 forever!). It’s the same as Example 1, but with a few extra numbers added at the start.

What this means is that we now know how the sequence will progress if any of the numbers in the above sequence appears.

This allows us to draw a diagram (I have created the example below) showing the paths that each of the numbers we have encountered so far will follow.

This group of numbers generated are known as hailstone numbers. They are called hailstone numbers because the numbers go up and down like hailstones in a cloud.

We have already found that a number of integers will end up at 1. But the question is – do all positive integers end up at 1? This is called the Collatz conjecture, and is one of the longest-running currently unsolved mathematical problems. The conjecture is named after German mathematician Lothar Collatz after he introduced the problem in 1937.

In mathematics, a conjecture is a principle which is widely expected to be true, but as of yet has not been proven or disproven. In the case of the Collatz conjecture, no positive integer has yet been found that disproves it but there is also no proof that this is true for all positive integers. You might think why can’t we prove it by showing that every positive integer ends up at 1, however that is actually impossible because there are infinitely many positive integers! So the conjecture is in a state of limbo until either someone proves it for all positive integers or somebody finds an example that doesn’t end up at 1. For years, mathematicians have been checking positive integers to try to find the elusive number that makes the Collatz conjecture fall down. As of 2017, all numbers up to 87 × 2^{60} have been checked (that’s a very large number – try not to think about how large it is, it’ll hurt your brain). Checking numbers this large takes immense computer processing power, so is no mean feat.

Another interesting property of hailstone numbers is how some of them end up at 1 very quickly, whereas others take much longer. Take our two examples from earlier – the number 5 fittingly took 5 moves to get to 1, while the number 6 took 8 moves. Is it the case that the larger the number, the more moves it takes to get to 1? Well, no, it isn’t actually. We already know this from the diagram from earlier – the number 16 comes after the number 5 in the diagram, meaning it will take 1 move fewer. Likewise, the number 10 comes after the number 3, meaning that will take 1 move fewer.

In the table below, I have calculated how many moves each number from 1 to 30 takes to get to 1.

You probably notice the anomaly in this table… the number 27 takes 111 moves to reach 1. That’s 88 more moves than the next highest in the table (the number 25 takes 23 moves). Throughout the marathon 111-move journey it reaches a highest number of 9,232 (incidentally proving that 9,232 takes fewer moves than 27, which is bizarre in itself) before eventually making its way down to 1. In fact, 27 takes the longest of any number until 54, which takes only one more move. This should come as no surprise, as if you start at 54, the first move will be to halve it which takes you to 27. After this, you will follow the same path as 27.

This seemingly contrived and meaningless mathematical problem has puzzled and fascinated mathematicians for decades. And more importantly, it’s interesting and a good way of practicing your numerical skills!

As a bonus, find the hailstone sequence for each of the following numbers and how many moves it takes to reach 1 (don’t worry, I’m not giving you any that will take anywhere as many as 27!).

1. 64

2. 42

3. 80

4. 96

5. 92

Hint: you will be able to use shortcuts for some of them – look out for any numbers that you have seen before!

Have a go at them and put your answers in the comments or email them to sam@metatutor.co.uk and I’ll let you know if you got them right.

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