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

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++

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// 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;
}

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Java














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// 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 */



















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Python3














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# 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 



















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C#














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// 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 



















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PHP














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<?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 


?> 



















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Output:


2







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