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Determine whether a given number is a Hyperperfect Number

  • Last Updated : 29 Apr, 2021

Given a number, determine whether it is a valid Hyperperfect Number
A number n is called k-hyperperfect if: n = 1 + k ∑idi where all di are the proper divisors of n. 
Taking k = 1 will give us perfect numbers
The first few k-hyperperfect numbers are 6, 21, 28, 301, 325, 496, 697, … with the corresponding values of k being 1, 2, 1, 6, 3, 1, 12, …

Examples: 

Input :  N = 36, K = 1
Output :  34 is not 1-HyperPerfect
Explanation: 
The Divisors of 36 are 2, 3, 4, 6, 9, 12, 18
the sum of the divisors is 54. 
For N = 36 to be 1-Hyperperfect, it would 
require 36 = 1 + 1(54), which we see, is
invalid
 
Input :  N = 325, K = 3
Output :  325 is 3-HyperPerfect
Explanation: 
We can use the first condition to evaluate this 
as K is odd and > 1 so here p = (3*k+1)/2 = 5, 
q = (3*k+4) = 13 p and q are both prime, so we 
compute p^2 * q = 5 ^ 2 * 13 = 325
Hence N is a valid HyperPerfect number

C++




// C++ 4.3.2 program to check whether a
// given number is  k-hyperperfect
#include <bits/stdc++.h>
using namespace std;
 
// function to find the sum of all
// proper divisors (excluding 1 and N)
int divisorSum(int N, int K)
{
    int sum = 0;
 
    // Iterate only until sqrt N as we are
    // going to generate pairs to produce
    // divisors
    for (int i = 2 ; i <= ceil(sqrt(N)) ; i++)
 
        // As divisors occur in pairs, we can
        // take the values i and N/i as long
        // as i divides N
        if (N % i == 0)
            sum += ( i + N/i );
 
    return sum;
}
 
// Function to check whether the given number
// is prime
bool isPrime(int n)
{
    //base and corner cases
    if (n == 1 || n == 0)
        return false;
 
    if (n <= 3)
        return true;
 
    // Since integers can be represented as
    // some 6*k + y where y >= 0, we can eliminate
    // all integers that can be expressed in this
    // form
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    // start from 5 as this is the next prime number
    for (int i=5; i*i<=n; i=i+6)
        if (n % i == 0 || n % ( i + 2 ) == 0)
            return false;
 
    return true;
}
 
// Returns true if N is a K-Hyperperfect number
// Else returns false.
bool isHyperPerfect(int N, int K)
{
    int sum = divisorSum(N, K);
 
    // Condition from the definition of hyperperfect
    if ((1 + K * (sum)) == N)
        return true;
    else
        return false;
}
 
// Driver function to test for hyperperfect numbers
int main()
{
    int N1 = 1570153, K1 = 12;
    int N2 = 321, K2 = 3;
 
    // First two statements test against the condition
    // N = 1 + K*(sum(proper divisors))
    if (isHyperPerfect(N1, K1))
        cout << N1 << " is " << K1
             <<"-HyperPerfect" << "\n";
    else
        cout << N1 << " is not " << K1
             <<"-HyperPerfect" << "\n";
 
    if (isHyperPerfect(N2, K2))
        cout << N2 << " is " << K2
             <<"-HyperPerfect" << "\n";
    else
        cout << N2 << " is not " << K2
             <<"-HyperPerfect" << "\n";
 
    return 0;
}

Java




// Java program to check
// whether a given number
// is k-hyperperfect
import java.io.*;
 
class GFG
{
     
// function to find the
// sum of all proper
// divisors (excluding
// 1 and N)
static int divisorSum(int N,
                      int K)
{
    int sum = 0;
 
    // Iterate only until
    // sqrt N as we are
    // going to generate
    // pairs to produce
    // divisors
    for (int i = 2 ;
             i <= Math.ceil(Math.sqrt(N));
             i++)
 
        // As divisors occur in
        // pairs, we can take
        // the values i and N/i
        // as long as i divides N
        if (N % i == 0)
            sum += (i + N / i);
 
    return sum;
}
 
// Function to check
// whether the given
// number is prime
static boolean isPrime(int n)
{
    // base and corner cases
    if (n == 1 || n == 0)
        return false;
 
    if (n <= 3)
        return true;
 
    // Since integers can be
    // represented as some
    // 6*k + y where y >= 0,
    // we can eliminate all
    // integers that can be
    // expressed in this form
    if (n % 2 == 0 ||
        n % 3 == 0)
        return false;
 
    // start from 5 as this
    // is the next prime number
    for (int i = 5;
             i * i <= n; i = i + 6)
        if (n % i == 0 ||
            n % ( i + 2 ) == 0)
            return false;
 
    return true;
}
 
// Returns true if N is
// a K-Hyperperfect number
// Else returns false.
static boolean isHyperPerfect(int N,
                              int K)
{
    int sum = divisorSum(N, K);
 
    // Condition from the
    // definition of hyperperfect
    if ((1 + K * (sum)) == N)
        return true;
    else
        return false;
}
 
// Driver Code
public static void main (String[] args)
{
    int N1 = 1570153, K1 = 12;
    int N2 = 321, K2 = 3;
     
    // First two statements test
    // against the condition
    // N = 1 + K*(sum(proper divisors))
    if (isHyperPerfect(N1, K1))
        System.out.println (N1 + " is " + K1 +
                            "-HyperPerfect" );
    else
        System.out.println(N1 + " is not " + K1 +
                               "-HyperPerfect" );
     
    if (isHyperPerfect(N2, K2))
        System.out.println( N2 + " is " + K2 +
                            "-HyperPerfect" );
    else
        System.out.println(N2 + " is not " + K2 +
                                "-HyperPerfect");
}
}
 
// This code is contributed by ajit

Python3




# Python3 program to check whether a
# given number is  k-hyperperfect
import math
 
# Function to find the sum of all
# proper divisors (excluding 1 and N)
def divisorSum(N, K):
     
    Sum = 0
     
    # Iterate only until sqrt N as we are
    # going to generate pairs to produce
    # divisors
    for i in range(2, math.ceil(math.sqrt(N))):
         
        # As divisors occur in pairs, we can
        # take the values i and N/i as long
        # as i divides N
        if (N % i == 0):
            Sum += (i + int(N / i))
 
    return Sum
 
# Function to check whether the given
# number is prime
def isPrime(n):
     
    # Base and corner cases
    if (n == 1 or n == 0):
        return False
         
    if (n <= 3):
        return True
 
    # Since integers can be represented as
    # some 6*k + y where y >= 0, we can eliminate
    # all integers that can be expressed in this
    # form
    if (n % 2 == 0 or n % 3 == 0):
        return False
         
    # Start from 5 as this is the next
    # prime number
    i = 5
    while (i * i <= n):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
             
        i += 6
         
    return True
 
# Returns true if N is a K-Hyperperfect
# number. Else returns false.
def isHyperPerfect(N, K):
     
    Sum = divisorSum(N, K)
 
    # Condition from the definition
    # of hyperperfect
    if ((1 + K * (Sum)) == N):
        return True
    else:
        return False
 
# Driver code
N1 = 1570153
K1 = 12
N2 = 321
K2 = 3
 
# First two statements test against the condition
# N = 1 + K*(sum(proper divisors))
if (isHyperPerfect(N1, K1)):
    print(N1, " is ", K1,
          "-HyperPerfect", sep = "")
else:
    print(N1, " is not ", K1,
          "-HyperPerfect", sep = "")
 
if (isHyperPerfect(N2, K2)):
    print(N2, " is ", K2,
          "-HyperPerfect", sep = "")
else:
    print(N2, " is not ", K2,
          "-HyperPerfect", sep = "")
 
# This code is contributed by avanitrachhadiya2155

C#




// C# program to check
// whether a given number
// is k-hyperperfect
using System;
 
class GFG
{
     
// function to find the
// sum of all proper
// divisors (excluding
// 1 and N)
static int divisorSum(int N,
                      int K)
{
    int sum = 0;
 
    // Iterate only until
    // sqrt N as we are
    // going to generate
    // pairs to produce
    // divisors
    for (int i = 2 ;
             i <= Math.Ceiling(Math.Sqrt(N));
             i++)
 
        // As divisors occur in
        // pairs, we can take
        // the values i and N/i
        // as long as i divides N
        if (N % i == 0)
            sum += (i + N / i);
 
    return sum;
}
 
// Function to check
// whether the given
// number is prime
static bool isPrime(int n)
{
    // base and corner cases
    if (n == 1 || n == 0)
        return false;
 
    if (n <= 3)
        return true;
 
    // Since integers can be
    // represented as some
    // 6*k + y where y >= 0,
    // we can eliminate all
    // integers that can be
    // expressed in this form
    if (n % 2 == 0 ||
        n % 3 == 0)
        return false;
 
    // start from 5 as this
    // is the next prime number
    for (int i = 5;
            i * i <= n; i = i + 6)
        if (n % i == 0 ||
            n % ( i + 2 ) == 0)
            return false;
 
    return true;
}
 
// Returns true if N is
// a K-Hyperperfect number
// Else returns false.
static bool isHyperPerfect(int N,
                           int K)
{
    int sum = divisorSum(N, K);
 
    // Condition from the
    // definition of hyperperfect
    if ((1 + K * (sum)) == N)
        return true;
    else
        return false;
}
 
// Driver Code
static public void Main ()
{
 
int N1 = 1570153, K1 = 12;
int N2 = 321, K2 = 3;
 
// First two statements
// test against the 
// condition N = 1 + K*
// (sum(proper divisors))
if (isHyperPerfect(N1, K1))
    Console.WriteLine(N1 + " is " + K1 +
                      "-HyperPerfect" );
else
    Console.WriteLine(N1 + " is not " + K1 +
                          "-HyperPerfect" );
 
if (isHyperPerfect(N2, K2))
    Console.WriteLine( N2 + " is " + K2 +
                       "-HyperPerfect" );
else
    Console.WriteLine(N2 + " is not " + K2 +
                           "-HyperPerfect");
}
}
 
// This code is contributed
// by akt_mit

PHP




<?php
// PHP 4.3.2 program to check
// whether a given number is
// k-hyperperfect
 
// function to find the sum
// of all proper divisors
// (excluding 1 and N)
function divisorSum($N, $K)
{
    $sum = 0;
 
    // Iterate only until
    // sqrt N as we are
    // going to generate
    // pairs to produce
    // divisors
    for ($i = 2 ;
         $i <= ceil(sqrt($N)) ; $i++)
 
        // As divisors occur in
        // pairs, we can take the
        // values i and N/i as long
        // as i divides N
        if ($N % $i == 0)
            $sum += ( $i + $N / $i);
 
    return $sum;
}
 
// Function to check whether
// the given number is prime
function isPrime($n)
{
    // base and corner cases
    if ($n == 1 || $n == 0)
        return false;
 
    if ($n <= 3)
        return true;
 
    // Since integers can be
    // represented as some 6*k + y
    // where y >= 0, we can
    // eliminate all integers that
    // can be expressed in this form
    if ($n % 2 == 0 || $n % 3 == 0)
        return false;
 
    // start from 5 as this
    // is the next prime number
    for ($i = 5;
         $i * $i <= $n; $i = $i + 6)
        if ($n % $i == 0 ||
            $n % ($i + 2) == 0)
            return false;
 
    return true;
}
 
// Returns true if N is a
// K-Hyperperfect number
// Else returns false.
function isHyperPerfect($N, $K)
{
    $sum = divisorSum($N, $K);
 
    // Condition from the
    // definition of hyperperfect
    if ((1 + $K * ($sum)) == $N)
        return true;
    else
        return false;
}
 
// Driver Code
$N1 = 1570153;
$K1 = 12;
$N2 = 321;
$K2 = 3;
 
// First two statements test
// against the condition
// N = 1 + K*(sum(proper divisors))
if (isHyperPerfect($N1, $K1))
    echo $N1 , " is " , $K1,
           "-HyperPerfect" , "\n";
else
    echo $N1 , " is not " , $K1,
         "-HyperPerfect" , "\n";
 
if (isHyperPerfect($N2, $K2))
    echo $N2 , " is " , K2,
          "-HyperPerfect" , "\n";
else
    echo $N2 , " is not " , $K2 ,
          "-HyperPerfect" , "\n";
 
// This code is contributed
// by akt_mit
?>

Javascript




<script>
 
// Javascript program to check
// whether a given number
// is k-hyperperfect
 
// Function to find the
// sum of all proper
// divisors (excluding
// 1 and N)
function divisorSum(N, K)
{
    let sum = 0;
 
    // Iterate only until sqrt N as
    // we are going to generate
    // pairs to produce divisors
    for(let i = 2;
            i <= Math.ceil(Math.sqrt(N));
            i++)
 
        // As divisors occur in
        // pairs, we can take
        // the values i and N/i
        // as long as i divides N
        if (N % i == 0)
            sum += (i + parseInt(N / i, 10));
 
    return sum;
}
 
// Function to check
// whether the given
// number is prime
function isPrime(n)
{
     
    // base and corner cases
    if (n == 1 || n == 0)
        return false;
 
    if (n <= 3)
        return true;
 
    // Since integers can be
    // represented as some
    // 6*k + y where y >= 0,
    // we can eliminate all
    // integers that can be
    // expressed in this form
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    // Start from 5 as this
    // is the next prime number
    for(let i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Returns true if N is
// a K-Hyperperfect number
// Else returns false.
function isHyperPerfect(N, K)
{
    let sum = divisorSum(N, K);
 
    // Condition from the
    // definition of hyperperfect
    if ((1 + K * (sum)) == N)
        return true;
    else
        return false;
}
 
// Driver code
let N1 = 1570153, K1 = 12;
let N2 = 321, K2 = 3;
 
// First two statements
// test against the
// condition N = 1 + K*
// (sum(proper divisors))
if (isHyperPerfect(N1, K1))
    document.write(N1 + " is " + K1 +
               "-HyperPerfect" + "</br>");
else
    document.write(N1 + " is not " + K1 +
                   "-HyperPerfect" + "</br>");
 
if (isHyperPerfect(N2, K2))
    document.write(N2 + " is " + K2 +
               "-HyperPerfect" + "</br>");
else
    document.write(N2 + " is not " + K2 +
                   "-HyperPerfect" + "</br>");
                    
// This code is contributed by decode2207
 
</script>

Output: 

1570153 is 12-HyperPerfect
321 is not 3-HyperPerfect

Given k, we can perform a few checks in special cases to determine whether the number is hyperperfect: 

  1. If K > 1 and K is odd , then let p = (3*k+1)/2 and q = 3*k+4 . If p and q are prime, then p2q is k-hyperperfect
  2. If p and q are distinct odd primes such that K(p + q ) = pq – 1 for some positive integral value of K, then pq is k-hyperperfect

Reference : 
https://en.wikipedia.org/wiki/Hyperperfect_number

This article is contributed by Deepak Srivatsav. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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