Find the sum of all Betrothed numbers up to N

Given a number N, the task is to find the sum of the all Betrothed numbers up to N.

Let, A and B are a Betrothed number pair, then the sum of the proper divisors of A is equal to B+1 and the sum of the proper divisors of B is equal to A+1.

Examples:

Input: 78 
Output: 123 
Explanation: 48 and 75 is the only pair of Betrothed Numbers upto 78.

Input: 5 
Output: 0 
Explanation: There is no pair of Betrothed Numbers up to 50. 

Approach:

• Initialise the array to store all Betrothed numbers upto to N
• Find the Betrothed numbers using sum of all its proper divisors and store the pair in the array.
• Then find the sum all the numbers less than and equal to N.
• The resultant sum is the required value.

Below code is the implementation of the above approach: 

C++

// C++ program to find the sum of the
// all betrothed numbers up to N
#include<bits/stdc++.h>
using namespace std;

// Function to find the sum of
// the all betrothed numbers
int Betrothed_Sum(int n)
{
    // To store the betrothed
    // numbers
    vector<int > Set;

    for(int number_1 = 1; number_1 < n;
                            number_1++)
    {

       // Calculate sum of
       // number_1's divisors
       // 1 is always a divisor
       int sum_divisor_1 = 1;

       // i = 2 because we don't
       // want to include
       // 1 as a divisor.
       int i = 2;

       while (i * i <= number_1)
       {
           if (number_1 % i == 0)
           {
               sum_divisor_1 = sum_divisor_1 + i;

               if (i * i != number_1)
                   sum_divisor_1 += number_1 / i;
           }
           i ++;
       }
       if (sum_divisor_1 > number_1)
       {
           int number_2 = sum_divisor_1 - 1;
           int sum_divisor_2 = 1;
           int j = 2;

           while (j * j <= number_2)
           {
               if (number_2 % j == 0)
               {
                   sum_divisor_2 += j;
                   if (j * j != number_2)
                       sum_divisor_2 += number_2 / j;
               }
               j = j + 1;
           }
           if (sum_divisor_2 == number_1 + 1 and
               number_1 <= n && number_2 <= n)
           {
               Set.push_back(number_1);
               Set.push_back(number_2);
           }
       }
    }

    // Sum all betrothed
    // numbers up to N
    int Summ = 0;
    for(auto i : Set)
    {
       if(i <= n)
          Summ += i;
    }
    return Summ;
}

// Driver code
int main()
{
    int n = 78;

    cout << Betrothed_Sum(n);
    return 0;
}

// This code is contributed by ishayadav181

Java

// Java program to find the sum
// of the all betrothed numbers
// up to N
import java.util.*;

class GFG{

// Function to find the sum of
// the all betrothed numbers
public static int Betrothed_Sum(int n)
{

    // To store the betrothed
    // numbers
    Vector<Integer> Set = new Vector<Integer>();

    for(int number_1 = 1;
            number_1 < n;
            number_1++)
    {

       // Calculate sum of
       // number_1's divisors
       // 1 is always a divisor
       int sum_divisor_1 = 1;

       // i = 2 because we don't
       // want to include
       // 1 as a divisor.
       int i = 2;

       while (i * i <= number_1)
       {
           if (number_1 % i == 0)
           {
               sum_divisor_1 = sum_divisor_1 + i;

               if (i * i != number_1)
                   sum_divisor_1 += number_1 / i;
           }
           i ++;
       }

       if (sum_divisor_1 > number_1)
       {
           int number_2 = sum_divisor_1 - 1;
           int sum_divisor_2 = 1;
           int j = 2;

           while (j * j <= number_2)
           {
               if (number_2 % j == 0)
               {
                   sum_divisor_2 += j;

                   if (j * j != number_2)
                       sum_divisor_2 += number_2 / j;
               }
               j = j + 1;
           }
           if (sum_divisor_2 == number_1 + 1 &&
               number_1 <= n && number_2 <= n)
           {
               Set.add(number_1);
               Set.add(number_2);
           }
       }
    }

    // Sum all betrothed
    // numbers up to N
    int Summ = 0;
    for(int i = 0; i < Set.size(); i++)
    {
       if (Set.get(i) <= n)
          Summ += Set.get(i);
    }
    return Summ;
}

// Driver code
public static void main(String[] args)
{
    int n = 78;

    System.out.println(Betrothed_Sum(n));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find the
# Sum of the all Betrothed
# numbers up to N

import math

# Function to find the Sum of
# the all Betrothed numbers
def Betrothed_Sum(n):

    # To store the Betrothed
    # numbers
    Set = []

    for number_1 in range(1, n):

        # Calculate sum of
        # number_1's divisors

        # 1 is always a divisor
        sum_divisor_1 = 1

        # i = 2 because we don't
        # want to include
        # 1 as a divisor.
        i = 2

        while i * i <= number_1:
            if (number_1 % i == 0):
                sum_divisor_1 = sum_divisor_1 + i

                if (i * i != number_1):
                    sum_divisor_1 += number_1 // i
            i = i + 1

        if (sum_divisor_1 > number_1):

            number_2 = sum_divisor_1 - 1
            sum_divisor_2 = 1
            j = 2
            while j * j <= number_2:
                if (number_2 % j == 0):
                    sum_divisor_2 += j
                    if (j * j != number_2):
                        sum_divisor_2 += number_2 // j
                j = j + 1

            if (sum_divisor_2 == number_1 + 1
                and number_1<= n and number_2<= n):
                Set.append(number_1)
                Set.append(number_2)

    # Sum all Betrothed
    # numbers up to N
    Summ = 0
    for i in Set:
        if i <= n:
            Summ += i
    return Summ


# Driver Code
n = 78
print(Betrothed_Sum(n))

C#

// C# program to find the sum
// of the all betrothed numbers
// up to N
using System;
using System.Collections;

class GFG{

// Function to find the sum of
// the all betrothed numbers
public static int Betrothed_Sum(int n)
{

    // To store the betrothed
    // numbers
    ArrayList set = new ArrayList();

    for(int number_1 = 1;
            number_1 < n;
            number_1++)
    {

        // Calculate sum of
        // number_1's divisors
        // 1 is always a divisor
        int sum_divisor_1 = 1;

        // i = 2 because we don't
        // want to include
        // 1 as a divisor.
        int i = 2;

        while (i * i <= number_1)
        {
            if (number_1 % i == 0)
            {
                sum_divisor_1 = sum_divisor_1 + i;

                if (i * i != number_1)
                    sum_divisor_1 += number_1 / i;
            }
            i ++;
        }

        if (sum_divisor_1 > number_1)
        {
            int number_2 = sum_divisor_1 - 1;
            int sum_divisor_2 = 1;
            int j = 2;

            while (j * j <= number_2)
            {
                if (number_2 % j == 0)
                {
                    sum_divisor_2 += j;

                    if (j * j != number_2)
                        sum_divisor_2 += number_2 / j;
                }
                j = j + 1;
            }

            if (sum_divisor_2 == number_1 + 1 &&
                number_1 <= n && number_2 <= n)
            {
                set.Add(number_1);
                set.Add(number_2);
            }
        }
    }

    // Sum all betrothed
    // numbers up to N
    int Summ = 0;
    for(int i = 0; i < set.Count; i++)
    {
        if ((int)set[i] <= n)
            Summ += (int)set[i];
    }
    return Summ;
}

// Driver code
static public void Main()
{
    int n = 78;

    Console.WriteLine(Betrothed_Sum(n));
}
}

// This code is contributed by offbeat

Javascript

<script>
// Javascript program to find the sum of the
// all betrothed numbers up to N

// Function to find the sum of
// the all betrothed numbers
function Betrothed_Sum(n)
{
    // To store the betrothed
    // numbers
    let Set = [];

    for(let number_1 = 1; number_1 < n;
                            number_1++)
    {

       // Calculate sum of
       // number_1's divisors
       // 1 is always a divisor
       let sum_divisor_1 = 1;

       // i = 2 because we don't
       // want to include
       // 1 as a divisor.
       let i = 2;

       while (i * i <= number_1)
       {
           if (number_1 % i == 0)
           {
               sum_divisor_1 = sum_divisor_1 + i;

               if (i * i != number_1)
                   sum_divisor_1 += parseInt(number_1 / i);
           }
           i++;
       }
       if (sum_divisor_1 > number_1)
       {
           let number_2 = sum_divisor_1 - 1;
           let sum_divisor_2 = 1;
           let j = 2;

           while (j * j <= number_2)
           {
               if (number_2 % j == 0)
               {
                   sum_divisor_2 += j;
                   if (j * j != number_2)
                       sum_divisor_2 += parseInt(number_2 / j);
               }
               j = j + 1;
           }
           if ((sum_divisor_2 == number_1 + 1) &&
               number_1 <= n && number_2 <= n)
           {
               Set.push(number_1);
               Set.push(number_2);
           }
       }
    }

    // Sum all betrothed
    // numbers up to N
    let Summ = 0;
    for(let i = 0; i < Set.length; i++)
    {
       if(Set[i] <= n)
          Summ += Set[i];
    }
    return Summ;
}

// Driver code
    let n = 78;
    document.write(Betrothed_Sum(n));

// This code is contributed by rishavmahato348.
</script>

Output:

123

Time Complexity: O(N * sqrt(N))

Auxiliary Space: O(n)