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# Ruth-Aaron numbers

• Last Updated : 16 Jul, 2021

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 ``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

 ``
Output:
`Yes`

Time Complexity: O(N)

Auxiliary Space: O(N)

Reference: https://oeis.org/A006145

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