Check if a number is Flavius Number

Given a series of integers from 1 to infinity and a number N.

The task is to remove every (i + 1)-th element from the remaining series at every i-th iterations and find that the given number N exists in the series or not.

Flavius Number:



Numbers in the Flavius Sieve are called Flavius Numbers.

Flavius sieve starts with the natural numbers and keep repeating the below step:
At the k-th sieving step, remove every (k+1)-st term of the sequence remaining of N natural numbers after the (k-1)-st sieving step.

For Example: 1, 3, 7, 13, 19, 27, 39, 49,

Examples:

Input: N = 17
Output: N0

Series after i-th iterations
1). 1, 3, 5, 7, 9, 11, 13, 15, 17, …
2). 1, 3, 7, 9, 13, 15, 19, 21, 25, …
3). 1, 3, 7, 13, 15, 19, 25, …
4). 1, 3, 7, 13, 19, 27, ….

Input: N = 3
Output: Yes

Approach:

  • If the given number is even so the answer is simply “No”. Because in the first iteration all the even number has been eliminated from the series.
  • Repeat this process.
    • Else remove the number of elements removed at 1st iteration i.e. (1/2)th the number and then check
      if it is divisible by 3 the answered should be “No”, else subtract the numbers before it that were
      removed i.e. (1/3)rd the number and so on.
    • Repeat above step for all iterations until we get answer.

Below is the implementation of the approach:

C++

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// C++ implementation
#include <bits/stdc++.h>
using namespace std;
  
// Return the number is
// Flavious Number or not
bool Survives(int n)
{
    int i;
  
    // index i starts from 2 because
    // at 1st iteration every 2nd
    // element was remove and keep
    // going for k-th iteration
    for (int i = 2;; i++) {
        if (i > n)
            return true;
        if (n % i == 0)
            return false;
  
        // removing the elements which are
        // already removed at kth iteration
        n -= n / i;
    }
}
  
// Driver Code
int main()
{
    int n = 17;
    if (Survives(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
  
    return 0;
}

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Java














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// Java implementation of the above approach 


class GFG  


{ 


  


// Return the number is 


// Flavious Number or not 


static boolean Survives(int n) 


{ 


  


    // index i starts from 2 because 


    // at 1st iteration every 2nd 


    // element was remove and keep 


    // going for k-th iteration 


    for (int i = 2;; i++) 


    { 


        if (i > n) 


            return true; 


        if (n % i == 0) 


            return false; 


  


        // removing the elements which are 


        // already removed at kth iteration 


        n -= n / i; 


    } 


} 


  


// Driver Code 


public static void main(String[] args) 


{ 


    int n = 17; 


    if (Survives(n)) 


        System.out.println("Yes"); 


    else


        System.out.println("No"); 


} 


}  


  


// This code is contributed by 29AjayKumar 



















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Python3














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# Python3 implementation of 


# the above approach  


  


# Return the number is  


# Flavious Number or not  


def Survives(n) : 


  


    # index i starts from 2 because  


    # at 1st iteration every 2nd  


    # element was remove and keep  


    # going for k-th iteration  


    i = 2


    while(True) : 


          


        if (i > n) : 


            return True;  


              


        if (n % i == 0) : 


            return False;  


  


        # removing the elements which are  


        # already removed at kth iteration  


        n -= n // i; 


        i += 1


  


# Driver Code  


if __name__ == "__main__" :  


  


    n = 17;  


      


    if (Survives(n)) : 


        print("Yes"); 


          


    else : 


        print("No"); 


          


# This code is contributed by AnkitRai01 



















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C#














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// C# implementation of the above approach 


using System; 


  


class GFG  


{ 


  


// Return the number is 


// Flavious Number or not 


static bool Survives(int n) 


{ 


  


    // index i starts from 2 because 


    // at 1st iteration every 2nd 


    // element was remove and keep 


    // going for k-th iteration 


    for (int i = 2;; i++) 


    { 


        if (i > n) 


            return true; 


        if (n % i == 0) 


            return false; 


  


        // removing the elements which are 


        // already removed at kth iteration 


        n -= n / i; 


    } 


} 


  


// Driver Code 


public static void Main(String[] args) 


{ 


    int n = 17; 


    if (Survives(n)) 


        Console.WriteLine("Yes"); 


    else


        Console.WriteLine("No"); 


} 


}  


  


// This code is contributed by PrinciRaj1992 



















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Output:


No








    

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