Aliquot sequence: An unsolved problem

Aliquot sequence: An unsolved problem

At school students get used to the idea that we know all the answers in mathematics – but the aliquot sequence is a simple example of an unsolved problem in mathematics.  The code above (if run for long enough on a super-computer!) might be enough to disprove a conjecture about this sequence.  Let’s look at this sequence and the associated conjecture.

Aliquot sequence: a defintion

We first choose a number k.  Let’s say 10.  We then add up the proper divisors of this number (the factors not including 10 itself).  This gives 1+2+5 = 8. 

We now add up the proper divisors of 8.  This gives 1+2+4 = 7.

We now add the proper divisors of 7.  This gives 1.

We can then see that the sequence will terminate here – as 1 has no proper divisors.

Aliquot sequence: what results could be possible?

1) Sequences might reach 1

2) Sequences might reach a number greater than 1 and then remain at that value

3) Sequences might oscillate between a set of 2 or more values.

4) Sequences might remain unbounded – an infinite sequence which does not repeat.

The Catalan-Dickson conjecture is that for all aliquot sequences either (1), (2) or (3) is true – i.e there are no numbers which generate infinite sequences which do not repeat.

Looking at the different cases

Let’s look at some numbers that give rise to the first 3 cases.  We’ve already seen that 10 leads to 1, let’s look at some other examples of this.  You can use the Python code at the top of the page to investigate for yourself – simply by changing the value of k in the first line.

A value of k which ends in a result of 1

I have written the code so it can simply be copied and pasted in Desmos to get some graphs of what is happening.

k = 30

We can see that when k =30 we do get over 200, before falling back down to 1.

A value of k which ends in a result greater than 1

Any perfect number (where the sum of the proper divisors equals the original number) will keep the same value for all iterations.  For example with 6: 1+2+3 = 6.

Any number that reaches a perfect number will also then remain at that number.

Some values of k which provide an oscillation

k = 220

The number 220 provides the oscillatory pattern where it oscillates between 220 and 284.  We call these numbers with period 2 amicable numbers.

k = 2122

Here we see that the sequence settles into an oscillation between 1184 and 1210.

k = 1,264,460

For this value of k we have an oscillatory period of 5.

Values for which the result is unknown

276 is the smallest number for which it is as yet unknown what happens to the sequence.  Running this on my laptop for about 30 minutes managed to find the first 44 terms of the sequence.  These numbers get very large!

k = 276

We can see that after 43 iterations the sequence is about to hit 5 billion – and the 44th iteration (not shown) will be over 9 billion. No-one has ever shown what happens to this sequence as the iterations increase.  With a computer large enough and enough time, we may be able to say one day if it satisfies (1), (2) or (3).  But it may satisfy (4) – in which case no amount of brute force checking will ever prove it.

Putting it all together:

k for all numbers less than 55

I then modified the code so that it would work out the sequences for all values less than 55 – and then plotted them on Desmos.  You can see the blue horizontal line for k = 28 (a perfect number).  Interestingly the 3 numbers that climb the highest all peak at the same number (259) before falling.  The majority of the numbers reach 1 within 5 iterations.

You can watch an excellent Numberphile video on this topic: