# Sum of all divisors from 1 to n

• Difficulty Level : Medium
• Last Updated : 08 Sep, 2021
Given a positive integer n. Find the value of where function F(i) for number i be defined as the sum of all divisors of ‘i‘.

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Examples :

Input: 4
Output: 15
Explanation
F(1) = 1
F(2) = 1 + 2 = 3
F(3) = 1 + 3 = 4
F(4) = 1 + 2 + 4 = 7
ans = F(1) + F(2) + F(3) + F(4)
= 1 + 3 + 4 + 7
= 15
Input: 5
Output: 21

Naive approach is to traverse for every number(1 to n), find all divisors and keep updating the sum with that divisor. See this to understand more.

## C++

 // C++ program to find sum of all// divisor of number up to 'n'#includeusing namespace std; // Utility function to find sum of// all divisor of number up to 'n'int divisorSum(int n){    int sum = 0;     for(int i = 1; i <= n; ++i)    {                 // Find all divisors of i and add them        for(int j = 1; j * j <= i; ++j)        {            if (i % j == 0)            {                if (i / j == j)                    sum += j;                else                    sum += j + i / j;            }        }    }    return sum;} // Driver codeint main(){    int n = 4;    cout << " " << divisorSum(n) << endl;         n = 5;    cout << " " << divisorSum(n);         return 0;}

## Java

 // JAVA program to find sum of all// divisor of number up to 'n'import java.io.*; class GFG {     // Utility function to find sum of    // all divisor of number up to 'n'    static int divisorSum(int n)    {        int sum = 0;         for (int i = 1; i <= n; ++i) {             // Find all divisors of i            // and add them            for (int j = 1; j * j <= i; ++j) {                if (i % j == 0) {                    if (i / j == j)                        sum += j;                    else                        sum += j + i / j;                }            }        }        return sum;    }     // Driver code    public static void main(String args[])    {        int n = 4;        System.out.println(divisorSum(n));        n = 5;        System.out.println(divisorSum(n));    }} /*This code is contributed by Nikita tiwari.*/

## Python3

 # Python3 code to find sum of all# divisor of number up to 'n' # Utility function to find sum of# all divisor of number up to 'n'def divisorSum( n ):    sum = 0         for i in range(1, n + 1):                 # Find all divisors of i        # and add them        j = 1        while j * j <= i:            if i % j == 0:                if i / j == j:                    sum += j                else:                    sum += j + i / j            j = j + 1    return int(sum) # Driver coden = 4print( divisorSum(n))n = 5print( divisorSum(n)) # This code is contributed by "Sharad_Bhardwaj".

## C#

 // C# program to find sum of all// divisor of number up to 'n'using System; class GFG {     // Utility function to find sum of    // all divisor of number up to 'n'    static int divisorSum(int n)    {        int sum = 0;         for (int i = 1; i <= n; ++i) {             // Find all divisors of i            // and add them            for (int j = 1; j * j <= i; ++j) {                if (i % j == 0) {                    if (i / j == j)                        sum += j;                    else                        sum += j + i / j;                }            }        }        return sum;    }     // Driver code    public static void Main()    {        int n = 4;        Console.WriteLine(divisorSum(n));        n = 5;        Console.WriteLine(divisorSum(n));    }} /*This code is contributed by vt_m.*/

## PHP

 

## Javascript

 

Output :

15
21

Time complexity: O(n√(n)))
Auxiliary space: O(1)

Efficient approach is to observe the function and co-relate the pattern. For a given number n, every number from 1 to n contributes its presence up to the highest multiple less than n. For instance,

Let n = 6,
=> F(1) + F(2) + F(3) + F(4) + F(5) + F(6)
=> 1 will occurs 6 times in F(1), F(2),
F(3), F(4), F(5) and F(6)
=> 2 will occurs 3 times in F(2), F(4) and
F(6)
=> 3 will occur 2 times in F(3) and F(6)
=> 4 will occur 1 times in F(4)
=> 5 will occur 1 times in F(5)
=> 6 will occur 1 times in F(6)

From the above observation, it can easily be observed that number i is occurring only in their multiples less than or equal to n. Thus, we just need to find the count of multiples and then multiply it with i for full contribution in the final sum. It can easily be done in O(1) time by taking the floor of (n / i) and then multiply it with i for the sum.

## C++

 // C++ program to find sum of all// divisor of number up to 'n'#includeusing namespace std; // Utility function to find sum of// all divisor of number up to 'n'int divisorSum(int n){    int sum = 0;    for (int i = 1; i <= n; ++i)        sum += (n / i) * i;    return sum;} // Driver codeint main(){    int n = 4;    cout <<" "<< divisorSum(n)<

## C

 // C program to find sum of all// divisor of number up to 'n'#include  // Utility function to find sum of// all divisor of number up to 'n'int divisorSum(int n){    int sum = 0;    for (int i = 1; i <= n; ++i)        sum += (n / i) * i;    return sum;} // Driver codeint main(){    int n = 4;    printf("%d\n", divisorSum(n));    n = 5;    printf("%d", divisorSum(n));    return 0;}

## Java

 // Java program to find sum of all// divisor of number up to 'n'import java.io.*; class GFG {     // Utility function to find sum of    // all divisor of number up to 'n'    static int divisorSum(int n)    {        int sum = 0;        for (int i = 1; i <= n; ++i)            sum += (n / i) * i;        return sum;    }     // Driver code    public static void main(String args[])    {        int n = 4;        System.out.println(divisorSum(n));        n = 5;        System.out.println(divisorSum(n));    }} /*This code is contributed by Nikita Tiwari.*/

## Python3

 # Python3 code to find sum of all# divisor of number up to 'n' # Utility function to find sum of# all divisor of number up to 'n'def divisorSum( n ):    sum = 0    for i in range(1, n + 1):        sum += int(n / i) * i    return int(sum)     # Driver coden = 4print( divisorSum(n))n = 5print( divisorSum(n)) # This code is contributed by "Sharad_Bhardwaj".

## C#

 // C# program to find sum of all// divisor of number up to 'n'using System; class GFG {     // Utility function to find sum of    // all divisor of number up to 'n'    static int divisorSum(int n)    {        int sum = 0;        for (int i = 1; i <= n; ++i)            sum += (n / i) * i;        return sum;    }     // Driver code    public static void Main()    {        int n = 4;        Console.WriteLine(divisorSum(n));        n = 5;        Console.WriteLine(divisorSum(n));    }} /*This code is contributed by vt_m.*/

## PHP

 

## Javascript

 // Javascript program to find sum of all// divisor of number up to 'n' // Utility function to find sum of// all divisor of number up to 'n'function divisorSum(n){    let sum = 0;    for (let i = 1; i <= n; ++i)        sum += Math.floor(n / i) * i;    return sum;} // Driver codelet n = 4;document.write(divisorSum(n) + "
");n = 5;document.write(divisorSum(n) + "
"); // This code is contributed by _saurabh_jaiswal.

Output :

15
21

Time complexity: O(n)
Auxiliary space: O(1)

More efficient solution:

We need to calculate To evaluate the above expression in O(sqrt(N)) we make use of The Harmonic Lemma.

Consider the harmonic sequence on integer division: {N/1, N/2, N/3, ….. ,N/N}

The lemma states that the above sequence is non-increasing, and there are at most 2*sqrt(N) different elements.

Consider floor(N/i) = k. Thus, k <= N/i < k+1. From this we get largest = floor(N/k). Therefore, we can find a range of values of i for which floor(N/i) is constant. And using The Harmonic Lemma we know that will be at most 2*sqrt(N) terms, thus we can calculate it programmatically in O(sqrt(N)) complexity. Consider the following example for better clarification.

## C++

 // C++ program to calculate sum of divisors// of numbers from 1 to N in O(sqrt(N)) complexity#include using namespace std; #define ll long long#define mod 1000000007 /*Function to calculate x^y usingModular exponentiationRefer to https://www.geeksforgeeks.org/modular-exponentiation-power-in-modular-arithmetic/*/ll power(ll x, ll y, ll p){         // re x^y if p not specified    // else (x^y)%p    ll res = 1;    x = x % p;    while (y > 0)    {        if (y & 1)            res = (res * x) % p;        y = y >> 1;        x = (x * x) % p;    }    return (res + p) % p;} // Function to find modular// inverse of a under modulo m// Assumption: m is primell modinv(ll x){    return power(x, mod - 2, mod);} // Function to calculate sum from 1 to nll sum(ll n){    // sum 1 to n = (n*(n+1))/2    ll retval = ((((n % mod) * ((n + 1) %        mod)) % mod) * modinv(2)) % mod;    return retval;} ll divisorSum(ll n){    ll l = 1;    ll ans = 0;     while (l <= n)    {        ll k = n / l;        ll r = n / k;        k %= mod;                 // For i=l to i=r, floor(n/i) will be k        ans += ((sum(r) - sum(l - 1) %                        mod) * k) % mod;                 // Since values can be very large        // we need to take mod at every step        ans %= mod;        l = r + 1;    }    ans = ans % mod;      // ans can be negative      // for example n = 831367 ans would be -534577982    if (ans < 0){        return ans+mod;    }else{        return ans;    }} /* Driver program to test above function */int main(){    int n = 5;    cout << "The sum of divisors of all \                numbers from 1 to " << n << " is: " \                            << divisorSum(n) << '\n';     n = 14;    cout << "The sum of divisors of all \                numbers from 1 to " << n << " is: " \                            << divisorSum(n) << '\n';}

## Java

 // Java program to calculate// sum of divisors of numbers// from 1 to N in O(sqrt(N))// complexityimport java.util.*;class Main{     static int mod = 1000000007;     /*Function to calculate x^y using Modular exponentiationRefer to https://www.geeksforgeeks.org/modular-exponentiation-power-in-modular-arithmetic/*/public static long power(long x,                         long y,                         long p){  // re x^y if p not specified   // else (x^y)%p  long res = 1;  x = x % p;     while (y > 0)  {    if ((y & 1) != 0)      res = (res * x) % p;    y = y >> 1;    x = (x * x) % p;  }  return (res + p) % p;} // Function to find modular // inverse of a under modulo m// Assumption: m is primepublic static long modinv(long x){  return power(x, mod - 2, mod);} // Function to calculate sum// from 1 to npublic static long sum(long n){  // sum 1 to n = (n*(n+1))/2  long retval = ((((n % mod) * ((n + 1) %                    mod)) % mod) * modinv(2)) %                    mod;  return retval;}       public static long divisorSum(long n){  long l = 1;  long ans = 0;   while (l <= n)  {    long k = n / l;    long r = n / k;    k %= mod;     // For i=l to i=r,    // floor(n/i) will be k    ans += ((sum(r) - sum(l - 1) %             mod) * k) % mod;     // Since values can be very    // large we need to take mod    // at every step    ans %= mod;    l = r + 1;  }  ans = ans % mod;  return ans;} // Driver code   public static void main(String[] args){  int n = 5;  System.out.println("The sum of divisors of" +                     " all numbers from 1 to " +                     n + " is: " + divisorSum(n));   n = 14;  System.out.println("The sum of divisors of all" +                     " numbers from 1 to " + n +                     " is: " + divisorSum(n));}} // This code is contributed by divyeshrabadiya07

## Python3

 # Python program to calculate# sum of divisors of numbers# from 1 to N in O(sqrt(N))# complexitymod = 1000000007; # Function to calculate x^y using Modular exponentiation Refer to# https:#www.geeksforgeeks.org/ modular-exponentiation-power-in-# modular-arithmetic/def power(x, y, p):         # re x^y if p not specified    # else (x^y)%p    res = 1;    x = x % p;     while (y > 0):        if ((y & 1) != 0):            res = (res * x) % p;        y = y >> 1;        x = (x * x) % p;         return (res + p) % p; # Function to find modular# inverse of a under modulo m# Assumption: m is primedef modinv(x):    return power(x, mod - 2, mod); # Function to calculate sum# from 1 to ndef sum(n):       # sum 1 to n = (n*(n+1))/2    retval = ((((n % mod) * ((n + 1) % mod)) % mod) * modinv(2)) % mod;    return retval; def divisorSum(n):    l = 1;    ans = 0;     while (l <= n):        k = n // l;        r = n // k;        k %= mod;         # For i=l to i=r,        # floor(n/i) will be k        ans += ((sum(r) - sum(l - 1) % mod) * k) % mod;         # Since values can be very        # large we need to take mod        # at every step        ans %= mod;        l = r + 1;         ans = ans % mod;    return ans; # Driver codeif __name__ == '__main__':    n = 5;    print("The sum of divisors of all numbers from 1 to " , n , " is: " ,int( divisorSum(n)));     n = 14;    print("The sum of divisors of all numbers from 1 to ", n ," is: " , int(divisorSum(n))); # This code contributed by aashish1995 Write

## C#

 // C# program to calculate// sum of divisors of numbers// from 1 to N in O(sqrt(N))// complexityusing System; class GFG{     static int mod = 1000000007; /*Function to calculate x^y using Modular exponentiationRefer to https://www.geeksforgeeks.org/modular-exponentiation-power-in-modular-arithmetic/*/static long power(long x, long y, long p){         // re x^y if p not specified     // else (x^y)%p    long res = 1;    x = x % p;         while (y > 0)    {        if ((y & 1) != 0)            res = (res * x) % p;                     y = y >> 1;        x = (x * x) % p;    }    return (res + p) % p;} // Function to find modular // inverse of a under modulo m// Assumption: m is primestatic long modinv(long x){    return power(x, mod - 2, mod);} // Function to calculate sum// from 1 to nstatic long sum(long n){         // sum 1 to n = (n*(n+1))/2    long retval = ((((n % mod) * ((n + 1) %                  mod)) % mod) * modinv(2)) %                  mod;    return retval;}    static long divisorSum(long n){    long l = 1;    long ans = 0;         while (l <= n)    {        long k = n / l;        long r = n / k;        k %= mod;                 // For i=l to i=r,        // floor(n/i) will be k        ans += ((sum(r) - sum(l - 1) %                   mod) * k) % mod;                 // Since values can be very        // large we need to take mod        // at every step        ans %= mod;        l = r + 1;    }    ans = ans % mod;    return ans;} // Driver codestatic void Main(){    int n = 5;    Console.WriteLine("The sum of divisors of" +                      " all numbers from 1 to " +                      n + " is: " + divisorSum(n));         n = 14;    Console.WriteLine("The sum of divisors of all" +                      " numbers from 1 to " + n +                      " is: " + divisorSum(n));}} // This code is contributed by divyesh072019

## Javascript

 

Output:

The sum of divisors of all numbers from 1 to 5 is: 21
The sum of divisors of all numbers from 1 to 14 is: 165

Time complexity: O(sqrt(N))

Auxiliary space: O(1)

Another sqrt(n) approach:

Anywhere division is used in the below article, it means integer division.

Let’s start with an example assume that n = 20, now let’s see how each number  from 1 to 20 appears as the factor of some other number.

 1 : 1 * 1, 1 * 2, 1 * 3, 1 * 4..., 1 * (20 / 1)
2 : 2 * 1, 2 * 2, 2 * 3, 2 * 4,..., 2 * (20 / 2)
3 : 3 * 1, 3 * 2, 3 * 3, 3 * 4...., 3 * (20 / 3)

our goal is to add every number each time it appears as the factor of some other number. For example 3 appears as the factor of (3 * 1), (3 * 2), (3 * 3)…, (3 * (20 / 3)). Now let’s start from 1 and add 1 to our sum each time it appears and also we will add all the numbers that appeared with 1, we’ll do the same thing with 2 and when we reach 3 we have already added 3 to our sum when it appeared with 1 and 2 so now we will only add 3 when it appears with numbers greater than 2 i.e. 3, 4, 5, 6 also we will add the numbers that appeared with 3 so we’ll add 4, 5 and 6 as well (notice here we will not add 3 twice because of 3 * 3). Similarly when we reach 4 we have already added 4 when it appeared with 1, 2 and 3 so we’ll add it only when it appears with numbers >= itself and add the numbers that appear with 4.

Finally we can say that when we are at a number i, we have already processed numbers from1 to  i – 1 and hence we have added i every time it appears with numbers 1 to i – 1 so this time we only need to add i every time it appears with numbers >= i also we have to add all the numbers that appear together with i and they are  > i.

Therefore for every number i we want to add the following terms to our sum

t1 : (add i each time it appears with numbers >= itself) -> i * (num / i - (i - 1))

(recall i will appear with numbers 1 to num / i
and we have already added i each time it appeared with a numbers less than itself)

t2 : (add numbers that appear with i) -> (i + 1) + (i + 2) ... + (num / i)
(numbers 1 to num / i will appear with i but
we have already processed numbers 1 to i - 1 and added them
when they appeared with i so now we only have to add the numbers
that appear with i and are greater than i,
here we will not add i itself because when i appears with itself
it should be added only once and we have added it once in t1)

we need to calculate t2 in O(1) time, here's how to do that
t2 = (i + 1) + (i + 2) + (i + 3) + ... + (num / i)
add and subtract 1 + 2 + 3 ... + i
=> t2 = 1 + 2 + 3 + ... + i + (i + 1) + (i + 2) + ... + (num / i) - (1 + 2 + 3 + ... + i)
=> t2 = (1 + 2 + 3 + .. + (num / i)) - (1 + 2 + 3 .. + i)
=> t2 = ((num / i) * (num / i + 1)) / 2 - (i * (i + 1)) / 2 

Finally, let’s look at the numbers that are greater than sqrt(num). These numbers will only appear with numbers that are lesser than sqrt(num). Let’s say x is a number greater than sqrt(num)

we have,
x > sqrt(num)
multiply sqrt(num) on both sides
=> x * sqrt(num) > sqrt(num) * sqrt(num)
=> x * sqrt(num) > num

we want to add x each time it appears, from above proof we see that x multiplied by root of num itself is greater than num hence x will only appear with numbers less than root of num so if we process all the numbers from 1 to sqrt(num) we will add every time x appears. For example take n = 100 now consider 11, 11 * 10 > 100 so 11 appears only with 1 to 9 i.e. as a factor of 11, 22, 33,…, 99 same is true for rest of the numbers that are greater than 10 they will only appear with numbers lesser than 10 and hence we only need to process numbers from 1 to 10 to add the numbers greater than 10 for n = 100.

Finally, our solution is this

for each i in 1 to sqrt(num) //no need to visit numbers greater than the root
add t1 and t2 to the sum

below is the c++ code

## C++

 #include using namespace std;long long sum_all_divisors(long long num){    long long sum = 0;    for (long long i = 1; i <= sqrt(num); i++) {        long long t1 = i * (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself        long long t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2); // adding numbers that appear with i and are greater than i        sum += t1 + t2;    }    return sum;}int main(){    int n;    long long sum = sum_all_divisors(n);    cout << sum << '\n';    return 0;}

## Java

 import java.io.*; class GFG { public static int sum_all_divisors(int num){    int sum = 0;    for (int i = 1; i <= Math.sqrt(num); i++) {        int t1 = i * (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself        int t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2); // adding numbers that appear with i and are greater than i        sum += t1 + t2;    }    return sum;}   // Driver code public static void main (String[] args){    int n = 1;    int sum = sum_all_divisors(n);    System.out.println(sum);}} // This code is contributed by shivanisinghss2110

## Python3

 import mathdef sum_all_divisors(num):     sum = 0;    for i in range(1,math.floor(math.sqrt(num))+1):        t1 = i * (num / i - i + 1) # adding i every time it appears with numbers greater than or equal to itself        t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2) # adding numbers that appear with i and are greater than i        sum += t1 + t2;         return sum; n = 1sum = sum_all_divisors(n)print(sum) # This code is contributed by shivanisinghss2110

## C#

 using System; class GFG { public static int sum_all_divisors(int num){    int sum = 0;    for (int i = 1; i <= Math.Sqrt(num); i++) {        int t1 = i * (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself        int t2 = (((num / i) * (num / i + 1)) / 2) - ((i * (i + 1)) / 2); // adding numbers that appear with i and are greater than i        sum += t1 + t2;    }    return sum;}   // Driver code public static void Main (String[] args){    int n = 1;    int sum = sum_all_divisors(n);    Console.Write(sum);}} // This code is contributed by shivanisinghss2110

## Javascript

 `

Time complexity: O(sqrt(N))

Auxiliary space: O(1)

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