Program to check whether a number is Proth number or not

Given a positive integer N, the task is to check if it is a Proth number. If the given number is a Proth number then print ‘YES’ otherwise print ‘NO’.

Proth Number: In mathematics, a Proth number is a positive integer of the form 

n = k * 2ⁿ + 1

where k is an odd positive integer and n is a positive integer such that 2ⁿ > k .

The first few Proth numbers are – 

3, 5, 9, 13, 17, 25, 33, 41, 49, ……

Examples:

Input: 25
Output: YES
    Taking k= 3 and n= 3, 
    25 can be expressed in the form of 
    (k.2n + 1) as (3.23 + 1) 

Input: 73
Output: NO
    Taking k=9 and n=3
    73 can be expressed in the form of
    (k.2n + 1 ) as  (9.23 + 1)
    But 23 is less than 9 
    (it should be greater than k to be Proth Number) 

Approach 1:

1. Iterate through values of k from 1 to n.
2. For each k, iterate through exponent from 0 to n.
3. Calculate prothNumber using the formula k * (1ULL << exponent) + 1.
4. If prothNumber is equal to the given number n, return true.
5. If prothNumber is greater than n, break the inner loop and continue with the next k.
6. If no Proth number is found, return false.

Note: This brute-force approach has high time complexity and is not efficient for large values of n.

C++

#include <iostream>
#include <cmath>

// Utility function to check if a number is a power of two
bool isPowerOfTwo(unsigned long long int n) {
    return n && !(n & (n - 1));
}

// Function to check if a number is a Proth number using brute-force
bool isProthNumber(unsigned long long int n) {
    if (n < 3)
        return false;

    for (unsigned long long int k = 1; k <= n; ++k) {
        for (unsigned long long int exponent = 0; exponent <= n; ++exponent) {
            unsigned long long int prothNumber = k * (1ULL << exponent) + 1;
            if (prothNumber == n)
                return true;
            if (prothNumber > n)
                break;
        }
    }

    return false;
}
// Nikunj Sonigara
// Driver code
int main() {
    // Get n
    unsigned long long int n = 25;

    // Check n for Proth Number
    if (isProthNumber(n - 1))
        std::cout << "YES" << std::endl;
    else
        std::cout << "NO" << std::endl;

    return 0;
}

Java

public class GFG {

    // Utility function to check if a number is a power of two
    public static boolean isPowerOfTwo(long n) {
        return n != 0 && (n & (n - 1)) == 0;
    }

    // Function to check if a number is a Proth number using brute-force
    public static boolean isProthNumber(long n) {
        if (n < 3)
            return false;

        for (long k = 1; k <= n; ++k) {
            for (long exponent = 0; exponent <= n; ++exponent) {
                long prothNumber = k * (1L << exponent) + 1;
                if (prothNumber == n)
                    return true;
                if (prothNumber > n)
                    break;
            }
        }

        return false;
    }

    // Nikunj Sonigara
    public static void main(String[] args) {
        // Get n
        long n = 25;

        // Check n for Proth Number
        if (isProthNumber(n - 1))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

Python3

# Utility function to check if a number is a power of two
def isPowerOfTwo(n):
    return n != 0 and (n & (n - 1)) == 0

# Function to check if a number is a Proth number using brute-force
def isProthNumber(n):
    if n < 3:
        return False

    for k in range(1, n+1):
        for exponent in range(0, n+1):
            prothNumber = k * (1 << exponent) + 1
            if prothNumber == n:
                return True
            if prothNumber > n:
                break

    return False

# Nikunj Sonigara
# Driver code
n = 25

# Check n for Proth Number
if isProthNumber(n - 1):
    print("YES")
else:
    print("NO")

Output

YES

Time Complexity: O(N²)
Auxiliary Space: O(1)

Approach 2:

1. Deduct 1 from the number. This would give a number in the form k*2ⁿ, if the given number is a proth number.
2. Now, loop through all odd numbers starting form k=1 to n/k and check if k can divide n in such a way that ( n/k ) is a power of 2 or not.
3. If found, print ‘YES’
4. If no such value of k is found then Print ‘NO’

Below is the implementation of the above idea 

C++

// CPP program to check Proth number

#include <bits/stdc++.h>
using namespace std;

// Utility function to check power of two
bool isPowerOfTwo(int n)
{
    return (n && !(n & (n - 1)));
}

// Function to check if the
// Given number is Proth number or not
bool isProthNumber(int n)
{

    int k = 1;
    while (k < (n / k)) {

        // check if k divides n or not
        if (n % k == 0) {

            // Check if n/k is power of 2 or not
            if (isPowerOfTwo(n / k))
                return true;
        }

        // update k to next odd number
        k = k + 2;
    }

    // If we reach here means
    // there exists no value of K
    // Such that k is odd number
    // and n/k is a power of 2 greater than k
    return false;
}

// Driver code
int main()
{

    // Get n
    int n = 25;

    // Check n for Proth Number
    if (isProthNumber(n - 1))
        cout << "YES";
    else
        cout << "NO";

    return 0;
}

Java

// Java program to check for Proth number

class GFG {

    // Utility function to check power of two
    static boolean isPowerOfTwo(int n)
    {
        return n != 0 && ((n & (n - 1)) == 0);
    }

    // Function to check if the
    // Given number is Proth number or not
    static boolean isProthNumber(int n)
    {

        int k = 1;
        while (k < (n / k)) {

            // check if k divides n or not
            if (n % k == 0) {

                // Check if n/k is power of 2 or not
                if (isPowerOfTwo(n / k))
                    return true;
            }

            // update k to next odd number
            k = k + 2;
        }

        // If we reach here means
        // there exists no value of K
        // Such that k is odd number
        // and n/k is a power of 2 greater than k
        return false;
    }

    // Driver code
    public static void main(String[] args)
    {

        // Get n
        int n = 25;

        // Check n for Proth Number
        if (isProthNumber(n - 1))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

Python3

# Python3 program to check for Proth number

# Utility function to Check
# power of two
def isPowerOfTwo(n):

    return (n and (not(n & (n - 1))))


# Function to check if the
# Given number is Proth number or not
def isProthNumber( n):


    k = 1

    while(k < (n//k)):

        # check if k divides n or not
        if(n % k == 0):

            # Check if n / k is power of 2 or not
            if(isPowerOfTwo(n//k)):
                    return True


        # update k to next odd number
        k = k + 2


    # If we reach here means
    # there exists no value of K
    # Such that k is odd number
    # and n / k is a power of 2 greater than k
    return False



# Driver code

# Get n
    int n = 25;

# Check n for Proth Number
if(isProthNumber(n-1)):
    print("YES");
else:
    print("NO");

C#

// C# program to check Proth number

using System;
class GFG {

    // Utility function to check power of two
    static bool isPowerOfTwo(int n)
    {
        return n != 0 && ((n & (n - 1)) == 0);
    }

    // Function to check if the
    // Given number is Proth number or not
    static bool isProthNumber(int n)
    {

        int k = 1;
        while (k < (n / k)) {

            // check if k divides n or not
            if (n % k == 0) {

                // Check if n/k is power of 2 or not
                if (isPowerOfTwo(n / k))
                    return true;
            }

            // update k to next odd number
            k = k + 2;
        }

        // If we reach here means
        // there exists no value of K
        // Such that k is odd number
        // and n/k is a power of 2 greater than k
        return false;
    }

    // Driver code
    public static void Main()
    {

        // Get n
        int n = 25;

        // Check n for Proth Number
        if (isProthNumber(n - 1))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}

PHP

<?php
// PHP program to check Proth number

// Utility function to check
// power of two
function isPowerOfTwo($n)
{
    return ($n && !($n & ($n - 1)));
}

// Function to check if the
// Given number is Proth
// number or not
function isProthNumber($n)
{
    $k = 1;
    while ($k < ($n / $k))
    {

        // check if k divides n or not
        if ($n % $k == 0)
        {

            // Check if n/k is power
            // of 2 or not
            if (isPowerOfTwo($n / $k))
                return true;
        }

        // update k to next odd number
        $k = $k + 2;
    }

    // If we reach here means
    // there exists no value of K
    // Such that k is odd number
    // and n/k is a power of 2
    // greater than k
    return false;
}

// Driver code

// Get n
$n = 25;

// Check n for Proth Number
if (isProthNumber($n - 1))
    echo "YES";
else
    echo "NO";

// This code is contributed
// by inder_verma
?>

Javascript

<script>

// Javascript program to check Proth number

// Utility function to check power of two
function isPowerOfTwo(n)
{
    return (n && !(n & (n - 1)));
}

// Function to check if the
// Given number is Proth number or not
function isProthNumber(n)
{
    let k = 1;
    while (k < parseInt(n / k))
    {

        // Check if k divides n or not
        if (n % k == 0)
        {

            // Check if n/k is power of 2 or not
            if (isPowerOfTwo(parseInt(n / k)))
                return true;
        }

        // Update k to next odd number
        k = k + 2;
    }

    // If we reach here means
    // there exists no value of K
    // Such that k is odd number
    // and n/k is a power of 2 greater than k
    return false;
}

// Driver code

// Get n
let n = 25;

// Check n for Proth Number
if (isProthNumber(n - 1))
    document.write("YES");
else
    document.write("NO");

// This code is contributed by souravmahato348

</script>

Output

YES

Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)