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Find the number of integers x in range (1,N) for which x and x+1 have same number of divisors

  • Last Updated : 17 Jan, 2022

Given an integer N. The task is to find the number of integers 1 < x < N, for which x and x + 1 have the same number of positive divisors.

Examples:  

Input: N = 3 
Output: 1 
Divisors(1) = 1 
Divisors(2) = 1 and 2 
Divisors(3) = 1 and 3 
Only valid x is 2.

Input: N = 15 
Output: 2 
  

Approach: Find the number of divisors of all numbers below N and store them in an array. And count the number of integers x such that x and x + 1 have the same number of positive divisors by running a loop.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100005
 
// To store number of divisors and
// Prefix sum of such numbers
int d[N], pre[N];
 
// Function to find the number of integers
// 1 < x < N for which x and x + 1 have
// the same number of positive divisors
void Positive_Divisors()
{
    // Count the number of divisors
    for (int i = 1; i < N; i++) {
 
        // Run a loop upto sqrt(i)
        for (int j = 1; j * j <= i; j++) {
 
            // If j is divisor of i
            if (i % j == 0) {
 
                // If it is perfect square
                if (j * j == i)
                    d[i]++;
                else
                    d[i] += 2;
            }
        }
    }
 
    int ans = 0;
 
    // x and x+1 have same number of
    // positive divisors
    for (int i = 2; i < N; i++) {
        if (d[i] == d[i - 1])
            ans++;
        pre[i] = ans;
    }
}
 
// Driver code
int main()
{
    // Function call
    Positive_Divisors();
 
    int n = 15;
 
    // Required answer
    cout << pre[n] << endl;
 
    return 0;
}
Java



// Java implementation of the approach
class GFG
{
     
static int N =100005;
 
// To store number of divisors and
// Prefix sum of such numbers
static int d[] = new int[N], pre[] = new int[N];
 
// Function to find the number of integers
// 1 < x < N for which x and x + 1 have
// the same number of positive divisors
static void Positive_Divisors()
{
    // Count the number of divisors
    for (int i = 1; i < N; i++)
    {
 
        // Run a loop upto sqrt(i)
        for (int j = 1; j * j <= i; j++)
        {
 
            // If j is divisor of i
            if (i % j == 0)
            {
 
                // If it is perfect square
                if (j * j == i)
                    d[i]++;
                else
                    d[i] += 2;
            }
        }
    }
 
    int ans = 0;
 
    // x and x+1 have same number of
    // positive divisors
    for (int i = 2; i < N; i++)
    {
        if (d[i] == d[i - 1])
            ans++;
        pre[i] = ans;
    }
}
 
// Driver code
public static void main(String[] args)
{
    // Function call
    Positive_Divisors();
 
    int n = 15;
 
    // Required answer
    System.out.println(pre[n]);
}
}
 
/* This code contributed by PrinciRaj1992 */
Python3



# Python3 implementation of the above approach
from math import sqrt;
 
N = 100005
 
# To store number of divisors and
# Prefix sum of such numbers
d = [0] * N
pre = [0] * N
 
# Function to find the number of integers
# 1 < x < N for which x and x + 1 have
# the same number of positive divisors
def Positive_Divisors() :
     
    # Count the number of divisors
    for i in range(N) :
 
        # Run a loop upto sqrt(i)
        for j in range(1, int(sqrt(i)) + 1) :
 
            # If j is divisor of i
            if (i % j == 0) :
 
                # If it is perfect square
                if (j * j == i) :
                    d[i] += 1
                else :
                    d[i] += 2
 
    ans = 0
 
    # x and x+1 have same number of
    # positive divisors
    for i in range(2, N) :
        if (d[i] == d[i - 1]) :
            ans += 1
        pre[i] = ans
     
# Driver code
if __name__ == "__main__" :
 
    # Function call
    Positive_Divisors()
 
    n = 15
 
    # Required answer
    print(pre[n])
 
# This code is contributed by Ryuga
C#



// C# implementation of the approach
using System;
 
class GFG
{
     
static int N =100005;
 
// To store number of divisors and
// Prefix sum of such numbers
static int []d = new int[N];
static int []pre = new int[N];
 
// Function to find the number of integers
// 1 < x < N for which x and x + 1 have
// the same number of positive divisors
static void Positive_Divisors()
{
    // Count the number of divisors
    for (int i = 1; i < N; i++)
    {
 
        // Run a loop upto sqrt(i)
        for (int j = 1; j * j <= i; j++)
        {
 
            // If j is divisor of i
            if (i % j == 0)
            {
 
                // If it is perfect square
                if (j * j == i)
                    d[i]++;
                else
                    d[i] += 2;
            }
        }
    }
 
    int ans = 0;
 
    // x and x+1 have same number of
    // positive divisors
    for (int i = 2; i < N; i++)
    {
        if (d[i] == d[i - 1])
            ans++;
        pre[i] = ans;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    // Function call
    Positive_Divisors();
 
    int n = 15;
 
    // Required answer
    Console.WriteLine(pre[n]);
}
}
 
// This code has been contributed by 29AjayKumar
PHP



<?php
 
// PHP implementation of the approach
 
$N = 100005;
 
// To store number of divisors and
// Prefix sum of such numbers
$d = array_fill(0,$N,NULL);
$pre = array_fill(0,$N,NULL);
 
// Function to find the number of integers
// 1 < x < N for which x and x + 1 have
// the same number of positive divisors
function Positive_Divisors()
{
    global $N,$d,$pre;
    // Count the number of divisors
    for ($i = 1; $i < $N; $i++) {
 
        // Run a loop upto sqrt(i)
        for ($j = 1; $j * $j <= $i; $j++) {
 
            // If j is divisor of i
            if ($i % $j == 0) {
 
                // If it is perfect square
                if ($j * $j == $i)
                    $d[$i]++;
                else
                    $d[$i] += 2;
            }
        }
    }
 
    $ans = 0;
 
    // x and x+1 have same number of
    // positive divisors
    for ($i = 2; $i < $N; $i++) {
        if ($d[$i] == $d[$i - 1])
            $ans++;
        $pre[$i] = $ans;
    }
}
 
// Driver code
 
    // Function call
    Positive_Divisors();
 
    $n = 15;
 
    // Required answer
    echo $pre[$n] ;
 
    return 0;
     
// This code is contributed by ChitraNayal
?>
Javascript



<script>
 
// Javascript implementation of the approach
const N = 100005;
 
// To store number of divisors and
// Prefix sum of such numbers
let d = new Array(N).fill(0);
let pre = new Array(N).fill(0);
 
// Function to find the number of integers
// 1 < x < N for which x and x + 1 have
// the same number of positive divisors
function Positive_Divisors()
{
     
    // Count the number of divisors
    for(let i = 1; i < N; i++)
    {
 
        // Run a loop upto sqrt(i)
        for(let j = 1; j * j <= i; j++)
        {
             
            // If j is divisor of i
            if (i % j == 0)
            {
                 
                // If it is perfect square
                if (j * j == i)
                    d[i]++;
                else
                    d[i] += 2;
            }
        }
    }
 
    let ans = 0;
 
    // x and x+1 have same number of
    // positive divisors
    for(let i = 2; i < N; i++)
    {
        if (d[i] == d[i - 1])
            ans++;
             
        pre[i] = ans;
    }
}
 
// Driver code
 
// Function call
Positive_Divisors();
let n = 15;
 
// Required answer
document.write(pre[n]);
 
// This code is contributed by souravmahato348
 
</script>
Output: 
2
 
Time complexity: O(nlogn)
Auxiliary Space: O(n)

My Personal Notes
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pawan_asipu
@pawan_asipu
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