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Ruth-Aaron numbers

  • Last Updated : 16 Jul, 2021
Geek Week

A number N is said to be Ruth-Aaron numbers if sum of prime divisors of N is equal to the sum of prime divisors of N+1.
The first few Ruth-Aaron numbers are:

5, 24, 49, 77, 104, 153, 369, 492, 714……..

Check if N is a Ruth-Aaron number

Given a number N, the task is to find if this number is Ruth-Aaron number or not. 
Examples: 
 

Input: N = 714 
Output: YES
Input: N = 25 
Output: No

Approach: The idea is to find the sum of all proper divisors of N and N + 1 and check if the sums of proper divisors of N and N+1 are equal or not. If the sum of proper divisors of N and N+1 are equal then the number is Ruth-Aaron number.
For Example: 
 



For N = 714 
Sum of Proper Divisors of N (714) = 2 + 3 + 7 + 17 = 29 
Sum of Proper Divisors of N+1 (715) = 5 + 11 + 13 = 29
Therefore, N is a Ruth-Aaron number.

Below is the implementation of the above approach:
 

C++




// C++ implementation of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find prime divisors of
// all numbers from 1 to N
int Sum(int N)
{
    int SumOfPrimeDivisors[N + 1] = { 0 };
 
    for (int i = 2; i <= N; ++i) {
 
        // if the number is prime
        if (!SumOfPrimeDivisors[i]) {
 
            // add this prime to all
            // it's multiples
            for (int j = i; j <= N; j += i) {
 
                SumOfPrimeDivisors[j] += i;
            }
        }
    }
    return SumOfPrimeDivisors[N];
}
 
// Function to check Ruth-Aaron number
bool RuthAaronNumber(int n)
{
    if (Sum(n) == Sum(n + 1))
        return true;
    else
        return false;
}
 
// Driver code
int main()
{
 
    int N = 714;
    if (RuthAaronNumber(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java implementation of the above approach
class GFG{
     
// Function to find prime divisors of
// all numbers from 1 to N
static int Sum(int N)
{
    int SumOfPrimeDivisors[] = new int[N + 1];
 
    for (int i = 2; i <= N; ++i)
    {
 
        // if the number is prime
        if (SumOfPrimeDivisors[i] == 1)
        {
 
            // add this prime to all
            // it's multiples
            for (int j = i; j <= N; j += i)
            {
                SumOfPrimeDivisors[j] += i;
            }
        }
    }
    return SumOfPrimeDivisors[N];
}
 
// Function to check Ruth-Aaron number
static boolean RuthAaronNumber(int n)
{
    if (Sum(n) == Sum(n + 1))
        return true;
    else
        return false;
}
 
// Driver code
public static void main (String[] args)
{
 
    int N = 714;
    if (RuthAaronNumber(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Ritik Bansal

Python3




# Python3 implementation of the above approach
 
# Function to find prime divisors of
# all numbers from 1 to N
def Sum(N):
 
    SumOfPrimeDivisors = [0] * (N + 1)
 
    for i in range(2, N + 1):
 
        # If the number is prime
        if (SumOfPrimeDivisors[i] == 0):
 
            # Add this prime to all
            # it's multiples
            for j in range(i, N + 1, i):
                SumOfPrimeDivisors[j] += i
 
    return SumOfPrimeDivisors[N]
 
# Function to check Ruth-Aaron number
def RuthAaronNumber(n):
 
    if (Sum(n) == Sum(n + 1)):
        return True
    else:
        return False
 
# Driver code
N = 714
 
if (RuthAaronNumber(N)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by vishu2908

C#




// C# implementation of the above approach
using System;
class GFG{
     
// Function to find prime divisors of
// all numbers from 1 to N
static int Sum(int N)
{
    int []SumOfPrimeDivisors = new int[N + 1];
 
    for (int i = 2; i <= N; ++i)
    {
 
        // if the number is prime
        if (SumOfPrimeDivisors[i] == 1)
        {
 
            // add this prime to all
            // it's multiples
            for (int j = i; j <= N; j += i)
            {
                SumOfPrimeDivisors[j] += i;
            }
        }
    }
    return SumOfPrimeDivisors[N];
}
 
// Function to check Ruth-Aaron number
static bool RuthAaronNumber(int n)
{
    if (Sum(n) == Sum(n + 1))
        return true;
    else
        return false;
}
 
// Driver code
public static void Main()
{
    int N = 714;
    if (RuthAaronNumber(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
// Javascript implementation of the above approach
 
    // Function to find prime divisors of
    // all numbers from 1 to N
    function Sum( N) {
        let SumOfPrimeDivisors = Array(N + 1).fill(0);
 
        for ( let i = 2; i <= N; ++i) {
 
            // if the number is prime
            if (SumOfPrimeDivisors[i] == 1) {
 
                // add this prime to all
                // it's multiples
                for (let  j = i; j <= N; j += i) {
                    SumOfPrimeDivisors[j] += i;
                }
            }
        }
        return SumOfPrimeDivisors[N];
    }
 
    // Function to check Ruth-Aaron number
    function RuthAaronNumber( n) {
        if (Sum(n) == Sum(n + 1))
            return true;
        else
            return false;
    }
 
    // Driver code
    let N = 714;
    if (RuthAaronNumber(N)) {
        document.write("Yes");
    } else {
        document.write("No");
    }
 
// This code is contributed by Rajput-Ji
</script>
Output: 
Yes

 

Time Complexity: O(N)

Auxiliary Space: O(N)

Reference: https://oeis.org/A006145
 

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