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Check if a number is Flavius Number

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  • Last Updated : 05 May, 2021

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 answer should be “No”, else subtract the numbers before it that were 
      removed i.e. (1/3)rd the number and so on. 
    • Repeat the above step for all iterations until we get an answer. 
       

Below is the implementation of the approach:
 

C++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
 
// Javascript implementation
 
// Return the number is
// Flavious Number or not
function Survives(n)
{
    let i;
 
    // index i starts from 2 because
    // at 1st iteration every 2nd
    // element was remove and keep
    // going for k-th iteration
    for (let 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 -= parseInt(n / i);
    }
}
 
// Driver Code
    let n = 17;
    if (Survives(n))
        document.write("Yes<br>");
    else
        document.write("No<br>");
 
</script>
Output: 
No

 


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