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 ∑_id_i where all d_i 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:

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
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

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