# Divisor Summatory Function

# Problem

Given an integer $N$, find the Sum of Number of Divisors from $1 \to N$. That is, find $\sum_{i=1}^{N}NOD(i)$.

For example, find the Sum of Number of Divisors, $SNOD()$, when $N=5$. $SNOD(5)= NOD(1) + NOD(2) + NOD(3) + NOD(4) + NOD(5) = 1 + 2 + 2 + 3 + 2 = 10$.

# Divisor Summatory Function

“In number theory, the divisor summatory function is a function that is a sum over the divisor function.” – Wiki

Divisor Summatory Function is defined as $D()$ in Wiki, but we will use $SNOD()$ in this article. Divisor Summatory Function finds Sum of Number of Divisors from $1 \to N$.

# Brute Force Solution – $O(N^2)$ Approach

A simple solution is to run two nested loop which counts all divisors from $1 \to N$. It is simple but inefficient.

int SNOD( int n ) { int res = 0; for ( int i = 1; i <= n; i++ ) { // For each number from 1 - N for ( int j = 1; j <= i; j++ ) { // Find possible divisors of "i" if ( i % j == 0 ) res++; } } return res; }

# Using $NOD()$ - $O( N \times \frac{\sqrt{N}}{ln(\sqrt{N})})$ Approach

We can use the code for $NOD()$ to get the number of divisors for each integer from $1$ to $N$ instead of using a $O(N)$ loop. This will make it faster than the above solution.

int SNOD( int n ) { int res = 0; for ( int i = 1; i <= n; i++ ) { res += NOD(i); } return res; }

Since $NOD()$ has same complexity as `factorize()`

, the complexity for this code is $O( N \times \frac{\sqrt{N}}{ln(\sqrt{N})} )$.

# Using Divisor List - $O(N)$ Approach

Suppose we are trying to find $SNOD(10)$. Let us write down the divisors for each integer between $1$ and $10$.

$NOD(1) = 1, : {1}$

$NOD(2) = 2, : {1 , 2}$

$NOD(3) = 2, : {1 , 3}$

$NOD(4) = 3, : {1, 2, 4}$

$NOD(5) = 2, : {1, 5}$

$NOD(6) = 4, : {1, 2, 3, 6}$

$NOD(7) = 2, : {1, 7}$

$NOD(8) = 4, : {1, 2, 4, 8}$

$NOD(9) = 3, : {1, 3, 9}$

$NOD(10) = 4, : {1, 2, 5, 10}$

So $SNOD(10) = 1 + 2 + 2 + 3 + 2 + 4 + 2 + 4 + 3 + 4 = 27$.

Now, look at the divisor list for each of integer between $1$ and $N$. Each divisor in the divisor lists contributes $1$ to the result. There are $27$ divisors in total.

How many times do we find $1$ in the divisor lists? It will appear once for every integer it can divide from $1$ to $N$. $1$ can divide $\frac{N}{1}$ numbers. So it appears $\frac{10}{1} = 10$ times.

Repeat for $2 \to N$. How many times do we find $2$? $2$ can divide $\frac{10}{2} = 5$ numbers, namely ${2,4,6,8,10}$. So it will appear $5$ times.

By repeating this from $1$ to $N$ we can find $SNOD(N)$.

int SNOD( int n ) { int res = 0; for ( int i = 1; i <= n; i++ ) { res += n / i; } return res; }

We removed $NOD()$ from the code in line $4$ and replaced it with $\frac{n}{i}$.

# Using Divisor Pairs - $O(\sqrt{N})$ Approach

Let us create another list. Instead of making divisor list, we will make "Ordered Divisor Pair" list for each value between $1$ to $N$. A divisor pair of value $X$ is a pair $(A, B)$ such that $A \times B = X$. Ordered pair means if $A \neq B$, then $(A, B)$ is a different pair than $(B, A)$.

$NOD(1) = 1, : (1,1)$

$NOD(2) = 2, : (1,2) (2,1)$

$NOD(3) = 2, : (1,3)(3,1)$

$NOD(4) = 3, : (1,4)(2,2)(4,1)$

$NOD(5) = 2, : (1,5)(5,1)$

$NOD(6) = 4, : (1,6)(2,3)(3,2)(6,1)$

$NOD(7) = 2, : (1,7)(7,1)$

$NOD(8) = 4, : (1,8)(2,4)(4,2)(8,1)$

$NOD(9) = 3, : (1,9)(3,3)(9,1)$

$NOD(10) = 4, : (1,10)(2,5)(5,2)(10,1)$

The "Ordered Divisor Pair" list is almost same as "Divisor List". The only difference is that each divisor $A$ is now paired with $B = \frac{X}{A}$.

Now, $NOD(X)$ represents number of ordered divisor pairs $(A,B)$ such that $A \times B = X$. So, $SNOD(N)$ is same as number of ordered divisor pairs such that $A \times B \leq N$.

## How to find Number of Ordered Divisor Pairs $\leq N$?

We can find this by following three steps.

### 1. Find number of divisor pairs $(A, B)$ where $A < B$

What are the possible values for $A$? Can it ever be $> \sqrt{N}$? No. If $A > \sqrt{N}$, then with $B>A$, $B$ will be $> \sqrt{N}$ and $A \times B$ will be $ > N$. So $A$ can only be between $1$ and $\lfloor \sqrt{N} \rfloor$.

Let $u = \lfloor \sqrt{N} \rfloor$. Then for each value of $A$ between $1$ and $u$, what are the possible values of $B$?

For any value of $A$, there are $\frac{N}{A}$ possible values of $B$ for which $A \times B$ does not exceed $N$. For $N=10$ and $A=2$, there are $\frac{10}{2}=5$ possible values for $B$, and they are ${1,2,3,4,5}$. Multiplying $A=2$ with any of these values produces value $\leq N$. But we need $B > A$. So we omit ${1,2}$ from the list. We can only use ${3,4,5}$ as $B$.

So, for any value of $A$, there will be $\lfloor \frac{N}{A} \rfloor$ possible values of $B$. The values will be from $1$ to $\lfloor \frac{N}{A} \rfloor$. Out of those, we have to omit values that are $\leq A$. So we can use $\lfloor \frac{N}{A} \rfloor - A$ values for $B$.

For example, $N=10$. Then $u=\lfloor \sqrt{10} \rfloor = 3$. When $A=1$, number of legal values we can use is $\lfloor \frac{10}{1} \rfloor - 1 = 9$. Namely, $B$ can be one of ${2,3,4,5,6,7,8,9,10}$.

When $A=2$, legal values are $\lfloor \frac{10}{2} \rfloor - 2 = 3$. $B$ can be ${3,4,5}$.

When $A=3$, legal values are $\lfloor \frac{10}{3} \rfloor - 3 = 0$.

$A$ cannot be bigger than $u$. So number of $(A, B)$ pairs such that $A \times B \leq 10$ and $A < B$ is 12.

### 2. Multiply the result with $2$

Once we find the number of pairs $(A, B)$ such that $A < B$ and $A \times B \leq N$, we multiply that number with $2$. That is because we want to find the number of ordered divisor pairs. When $A \neq B$, each pair $(A, B)$ will occur in "Ordered Divisor Pair" list as $(B, A)$ again.

Once we double the number, our result contains the number of ordered pair $(A, B)$ such that $A \neq B$ and $A \times B \leq N$.

### 3. Add the number of pairs $(A, B)$ where $A = B$

We still did not count pairs where $A = B$ and $A \times B \leq N$. How many pairs can there be? Since $A$ must be between $1$ and $u$, there cannot be more than $u$ such pairs. The pairs will be $(1,1),(2,2)...(u,u)$.

## Code for $O(\sqrt{N})$ Approach

int SNOD( int n ) { int res = 0; int u = sqrt(n); for ( int i = 1; i <= u; i++ ) { res += ( n / i ) - i; //Step 1 } res *= 2; //Step 2 res += u; //Step 3 return res; }

# Reference

- forthright48 - Number of Divisors of an Integer - https://forthright48.com/2015/07/number-of-divisors-of-integer.html
- Wiki - Divisor Summatory Function - https://en.wikipedia.org/wiki/Divisor_summatory_function