# The Arithmetic Derivative (Part 3)

Updated: Oct 14, 2021

__Let us count__

The nice thing about the arithmetic derivative is that it takes a natural number to another natural number. This leads to tons of questions about the growth of ** D(n)**, such as

When is

?*D(n) = n*How often is

?*D(n) < n*What is the significance of the differential equation

for different*D(n) = m*?*m*

I'll go ahead and answer the first question with no explanation, and we'll talk about it later.

### D(n) = n if and only if n = p^p for prime p

Let's focus on the second question for now. Here are all ** n < 100** for which

**.**

*D(n) < n*`1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 29, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 49, 50, 51, 53, 55, 57, 58, 59, 61, 62, 63, 65, 66, 67, 69, 70, 71, 73, 74, 75, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 98, 99`

For the first 10 numbers, 80% have

.*D(n) < n*For the first 1000 numbers, 68% have

.*D(n) < n*For the first 100000 numbers, 68.234% have

.*D(n) < n*

So it seems that MOST numbers get smaller when ** D** operates on them. Testing up to a million, 68.2183% numbers get smaller.

This graph shows the area where ** D(n) < n** in blue, which helps explain the 68% number - the points are very densely packed into that area, whereas the red points are more spread out. Another way to view this is to graph

**, as below.**

*D(n)/n*__Starter Differential Equations__

The general form of a *first-order* differential equation would be * D(n) = m*. If we fix the

*, then what can we say about the solutions*

**m***?*

**n**If

, then we know that*m = 0*.*n = 0,1*If

, then we know that*m = 1*, a prime number.*n = p*The first actual case:

. If*m = 2*, then*D(n) = 2*is not prime, so we can write*n*, and neither is 1. Then*n = ab*. If this is equal to 2, then both*D(n) = aD(b) + D(a)b*and*aD(b)*must be equal to 1. But we already know*D(a)b*and*a*aren't 1, so this is impossible!*b*

Therefore, there are NO solutions to ** D(n) = 2**. This brings a natural question:

**Question:**** **For which ** m **does

**have a solution?**

*D(n) = m*Here are the first few ** m** so that

**has no solution:**

*D(n) = m*`2, 3, 11, 17, 23, 29, 35, 37, 47, 53, 57, 65, 67, 79, 83, 89, 93, 97, 107, 117, 125, 127, 137, 145, 149, 157, 163, 173, 177, 179`

Or a more general question:

**Question:**** **Let ** S(m)** be the set of solutions

**such that**

*n***. What is the size of**

*D(n) = m***?**

*S(m)*Let's take our reasoning on ** m = 2** and expand it a bit. If

**, then**

*a,b > 1***, and**

*D(a), D(b) >= 1*### D(ab) = *aD(b) + D(a)b >= a + b*

*aD(b) + D(a)b >= a + b*

with equality only when ** a** and

**are both prime. If we want to maximize (in the usual sense)**

*b***with the constraint that**

*a + b***then we substitute**

*n = ab,***into**

*b = n/a***take the derivative with respect to**

*a + b,***, set it equal to 0, and solve.**

*a*### a + b = a + n/a

### 1 - n/a^2 = 0

### a = sqrt(n)

Therefore, we find that if ** n > 1** is not prime, then

### D(n) >= 2sqrt(n)

Often the derivative is MUCH higher. And we know that equality holds only if *n = p^2.*

This is very helpful as it gives us a bound when solving our differential equations. If we want to solve the equation ** D(n) = m** for a fixed

**, then we need**

*m***, meaning we only need to check**

*m >= 2sqrt(n)***. What this guarantees is that**

*n <= (m/2)^2***is finite for all**

*S(m)***. Specifically,**

*m > 1*### |S(m)| <= (m/2)^2

**Thank you for reading!**

**Jonathan Gerhard**

**J****Math****G**

**J**

**Math**

**G**