# Aliquot Divisors

When the idea came to write about this post, I had never came across the term “aliquot” to my recollection.  Let’s quickly deconstruct that sentence and assign some probabilities.

• I’ve never came across the term “aliquot”: probability = 0.01
• I don’t recall having come across the term “aliquot”: probability = 0.99

From this, I hope you’ve come to understand a few important things.

• Probabilities must sum to 1
• All probabilities must be non-negative and less than (or equal to) 1
• I most likely don’t have a great short term memory.
• I am most likely a pretty funny dude.

### Sum of Divisors Function

After searching the term “Sum of Divisors function” a page with a wiki on the Online Encyclopedia of Integer Sequences came up.  On this, they define a function for the sum of the divisors for a number.

Let’s do a quick example before introducing notation and look at the number 40 (that’s how old I am for another month and a half).  What numbers divide 40 evenly?  That is, if you push 40, and then $\div$, what numbers can you push next so that the result is an integer (not a decimal).

Here are the divisors of 40: 1, 2, 4, 5, 8, 10, 20, 40.  A function that computes the sum of the divisors of a number should give an output of $1+2+4+5+8+10+20+40 = 90$ whenever you input the value 40.

Now for the notation.  The sum of the divisors function is given as $\sigma(n)$.  To apply the above example, this means that $\sigma(40)=90$. In other words, you input the value 40 into the sum of divisors function $\sigma(\cdot)$, and the output is 90.

Let’s try a few others. What if we input 10?  That is, what is the value of $\sigma(10)$?

First, let’s find the divisors of 10: 1, 2, 5, 10. Next (and finally), sum them together: $1+2+5+10 = 18$.  Thus, $\sigma(10) = 18$.

How about $\sigma(11)$?  The only divisors of 11 are 1 and 11 (which, as a side note, means that 11 is a prime number).  This means $\sigma(11)=12$.  If you inspect this example thoroughly, you should be able to deduce that the sum of the divisors of any prime number will be one more than the number.

Since 5 is a prime number, $\sigma(5)= 6$ (the sum of 1 and 5).  Likewise, since 7 is a prime number, $\sigma(7)=8$.

Now, try a few on your own.  Find $\sigma(15)$ and $\sigma(24)$.  The answers are below, but you can also check them by typing “sigma(15)” or “sigma(24)” into WolframAlpha. Technically, the correct code is DivisorSigma[1,15] and DivisorSigma[1,24], but you need not concern yourself with that.

### So What are Aliquot Divisors?

The aliquot divisors of a number are what I have always known to be proper divisors of a number, which are all the divisors except the number itself.

So, the aliquot divisors of 40 are 1, 2, 4, 5, 8, 10, and 20 (we leave off 40).  And the sum of the aliquot divisors function is denoted by $s(\cdot)$.  For our example, then, $s(40)=1+2+4+5+8+10+20 =50$.

With this sum of aliquot divisor function, $s(n)$, we can now explore some fascinating things about different numbers.

### Prime Numbers

For every prime number (a number only divisible by 1 and itself), the only aliquot divisor is 1.  So, by plugging any prime number into the function $s(\cdot)$, the result will be 1.

### Deficient Numbers

The numbers 4 and 10 are deficient numbers. Why? When you plug them into the sum of aliquot divisors function, $s(\cdot)$, you will get a value smaller than the number.

The aliquot divisors of 4 are 1 and 2.  So, $s(4)=3$.  The aliquot divisors of 10 are 1, 2, and 5. So, $s(10) = 8$. Notice that each of the output values are smaller than the input values.  That means the numbers are deficient.

Let’s recall what you learned about prime numbers in the previous section.  If p is a prime number, than $s(p)=1$.  From this we can deduce that all prime numbers are deficient.

Furthermore, since we have the above stated examples of 4 and 10 as deficient numbers, and we know they are not prime since they are divisible by something other than themselves and 1, then we can also deduce that not all deficient numbers are prime.

### Abundant Numbers

The numbers 12 and 40 are abundant numbers? Why? When you plug them into the sum of aliquot divisors function, $s(\cdot)$, you will get a value larger than the number.

We have already seen that $s(40)=50$. The aliquot divisors of 12 are 1, 2, 3, 4, and 6.  Thus, $s(12) = 1+2+3+4+6=16$.  Each output value is larger than the input value. That means the numbers are abundant.

### Perfect Numbers

The numbers 6 and 28 are perfect numbers? Why? When you plug them into the sum of aliquot divisors function $s(\cdot)$, you will get the same value!

The aliquot divisors of 6 are 1, 2 and 3.  Thus $s(6)=1+2+3=6$.  The aliquot divisors of 28 are 1, 2, 4, 7, and 14.  Thus, $s(28) = 1+2+4+7+14 = 28$.  Each output value is the same as the input value!  That means the number is perfect.

### Amicable Chains

Recently, I was introduced to the concept of amicable chains.  What if we continue to apply the sum of aliquot divisor function $s(\cdot)$ over and over again?  For example, let’s try it with 12.  As we saw in the Abundant Numbers section, $s(12)=16$.

Now, what if we plug in 16?  Aliquot divisors of 16: 1, 2, 4, and 8.  Thus, $s(16)=15$.

Again? What if we plug in 15? Aliquot divisors of 15?  1, 3, and 5.  Thus, $s(15)=9$.

Again? Aliquot divisors of 9? 1 and 3.  Thus $s(9)=4$. The aliquot divisors of 4 are 1 and 2, so $s(4)=3$.  Since 3 is prime $s(3)=1$ and now we’re trapped at 1 forever.

This created the sequence $\displaystyle 12\rightarrow 16\rightarrow 15\rightarrow 9\rightarrow 4\rightarrow 3\rightarrow 1$

If you do a lot of exploration, a lot of numbers will produce a sequence that ends at 1 and becomes trapped. Some don’t. Take the perfect numbers 6 and 28 for example.

Since $s(6)=6$, then 6 has a chain of length 1 since it goes directly back to itself.  So does 28.

There are other numbers that do weirder things.  Take 220 and 284 for example (you will definitely need to use WolframAlpha on these if you want to check. However, you’ll need to type in “sigma(220)-220” and “sigma(284)-284”.)  The sum of the aliquot divisors of 220 is 284, and the sum of the aliquot divisors of 284 is 220.  This produces a chain of length 2. There are several chains like this one.

With the aid of a computer program, I’ve found one of length 5 and another of length 28.  Using a similar program, I also stumbled upon interesting numbers like 1230 and 1248.  Although they do not produce chains, they take a while to get back to 1.  Using a very cool trick of emailing a former colleague with a higher level of number theory understanding, he was able to find out that it takes 185 terms to get back to 1 starting at 1230 and 1076 terms to get back to 1 if you start at 1248.

The interesting thing about these sequences are how really, really, big they get.

#### Answers to Above Posed Questions $\sigma(15) = 24$, and $\sigma(24) = 60$.